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Theory and Practice

Gilles Brassard and Paul Bratley
Departement d'informatique et de recherche operationnelle
Universitd de Montreal

PRENTICE HALL, Englewood Cliffs, New Jersey


Library of Congress Cataloging-in-Publication Data
Brassard, Gilles
Algorithmics : theory and practice.

1. Recursion theory. 2. Algorithms. I. Bratley,
Paul. II. Title.

QA9.6.B73 1987
ISBN 0-13-023243-2

51 I'.3


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© 1988 by Prentice-Hall, Inc.
A division of Simon & Schuster
Englewood Cliffs, New Jersey 07632

All rights reserved. No part of this book may be
reproduced, in any form or by any means,
without permission in writing from the publisher.

Printed in the United States of America
10 9 8 7 6 5 4 3 2 1

ISBN 0-13-023243-2



for Isabelle and Pat

1. Data Structures Xiii Preface 1 Preliminaries 1 1.4. What Is an Algorithm? 1.7. When Is an Algorithm Specified? 1. 1. 15 19 20 20 VII .1.7. Average and Worst-Case Analysis 7 1. 1. Calculating the Fibonacci Sequence. 13 Multiplication of Large Integers. Some Practical Examples 1. 16 Fourier Transforms.2. 1 5 12 Sorting.3.2. Why Do We Need Efficient Algorithms? 1. 1.3. 1.9. What Is an Elementary Operation? 9 1.6. 14 Calculating the Greatest Common Divisor.1.6. 13 Evaluating Determinants.4.5.9. 19 1. Problems and Instances 4 1. The Efficiency of Algorithms 1. 11 Lists.

2.viii Contents 1. 1. 2.2. 21 Rooted Trees. 47 Constructive Induction. Minimal Spanning Trees. 2. 3. 52 Homogeneous Recurrences.3.1. 65 100 Colouring a Graph. 2. 3.2.9. Graphs. 1.5.5. 43 Conditional Asymptotic Notation.4. 37 Other Asymptotic Notation.1. 3.7.of " .1. Introduction 3. 3.1.9. Heaps.1. 2.4. 45 Asymptotic Recurrences. 101 The Travelling Salesperson Problem.2. 25 1.3. Greedy Algorithms and Graphs 3. 72 Range Transformations. 75 Supplementary Problems.5.9. 43 51 2. 2. 2. Analysis of Algorithms 2. 76 References and Further Reading 78 3 Greedy Algorithms 79 3.5.4. 48 For Further Reading.8. Operations on Asymptotic Notation. 2. 87 81 81 92 Minimizing Time in the System. References and Further Reading 104 102 .3. 2.2.3. 95 Greedy Heuristics 3. Disjoint Set Structures. 65 Inhomogeneous Recurrences. 2.3. 2.4. 68 Change of Variable. 3. 92 Scheduling with Deadlines. 37 37 A Notation for "the order. 79 Greedy Algorithms for Scheduling 3.6. Solving Recurrences Using the Characteristic Equation 2. 2. 23 30 References and Further Reading 35 2 Analysing the Efficiency of Algorithms 2.3. 41 Asymptotic Notation with Several Parameters. Asymptotic Notation 2.3. Shortest Paths.

Supplementary Problems 5.2.5. Exponentiation : An Introduction to Cryptology 4. Supplementary Problems 4.8.8. Sorting by Merging 4. Traversing Trees 6.7. The World Series 5.10.5.! 128 169 170 Articulation Points.3. Exchanging Two Sections of an Array 4. The Travelling Salesperson Problem 5. Shortest Paths 5. Matrix Multiplication References and Further Reading 142 144 146 150 154 159 162 164 167 6 Exploring Graphs 169 6.3. Quicksort 4.3.6. References and Further Reading 105 107 109 115 116 119 124 132 134 136 140 5 Dynamic Programming 142 5. Introduction 6. Chained Matrix Multiplication 5.1.2. 174 171 .11. Memory Functions 5.4. Binary Searching 4. Determining the Threshold 4.3. Introduction 5. Arithmetic with Large Integers 4.1.12. Introduction 4.ix Contents 4 Divide and Conquer 105 4. Optimal Search Trees 5.6. Depth-First Search : Undirected Graphs 6. Selection and the Median 4.2.4.

2. Classification of Probabilistic Algorithms 8. Supplementary Problems 8.5. 223 238 Selection and Sorting. 8.4. 7.4. 7.3.1. 230 Probabilistic Counting.4. 232 More Probabilistic Counting. Numerical Probabilistic Algorithms 228 8.2. Breadth-First Search 6. 228 Numerical Integration. 7.3. 216 References and Further Reading 222 8 Probabilistic Algorithms 223 8. 178 Strongly Connected Components. 176 182 184 Backtracking. 213 The Boyer-Moore Algorithm.1.7. 8.3. 245 Las Vegas Algorithms 247 226 237 .4. 185 Graphs and Games: An Introduction.4.2. 240 Searching an Ordered List. 199 Buffon's Needle. 235 Numerical Problems in Linear Algebra. 6. Implicit Graphs and Trees 6.2. 8. 179 6. 242 Universal Hashing.5. Acyclic Graphs: Topological Sorting.1.6.1. 8. 8.4.4. 238 Stochastic Preconditioning.3.x Contents 6. Sherwood Algorithms 8. Introduction Depth-First Search : Directed Graphs 6. 7. 205 Ancestry in a rooted tree. 205 Signatures.3. References and Further Reading 204 189 202 7 Preconditioning and Precomputation 7. 6. Branch-and-Bound.6. 209 Precomputation for String-Searching Problems 211 7.2.8. 8. Preconditioning 205 7.4. 211 The Knuth-Morris-Pratt Algorithm. 207 Repeated Evaluation of a Polynomial.2. 8.2. 6.3.2. Introduction.1.

6. 10.5. 9 10 248 262 Majority Element in an Array. 304 10.3.3. Reductions Among Graph Problems. 8.6.3. Cook's Theorem.3. 256 Choosing a Leader.5. 252 8. Introduction 9.3. 328 10.5. The Eight Queens Problem Revisited. 8. 8.1. Multiplication of Large Integers 9. NP-Complete Problems.4.5. Factorizing Integers.2.1.3. References and Further Reading 290 277 280 284 286 Introduction to Complexity 10. Decision Trees 10. The Classes P and NP. 269 A Probabilistic Test for Set Equality. Reductions Among Matrix Problems.4. 1. 8. Square Roots Modulo p.4.6. 274 References and Further Reading 274 Transformations of the Domain 277 9. 332 10.2. 316 10.1. 325 10.3.7. 268 Probabilistic Primality Testing.2. Reductions Among Arithmetic and Polynomial Problems.6. Nondeterminism.2. 324 References and Further Reading 336 Table of Notation 338 Bibliography 341 Index 353 .4.2. Reduction 292 292 300 10. Symbolic Operations on Polynomials 9. Some Reductions. 260 Monte Carlo Algorithms 8. 302 10. 10. 271 Matrix Multiplication Revisited.2.2. 8.2.xi Contents 8.3.1. 8. Introduction to NP-Completeness 308 315 10.3. The Discrete Fourier Transform 279 9.1. The Inverse Transform 9.3.6.


we are aware of no dictionary of the English language that has an entry for algorithmics.) In a nutshell. however.000 items in 30 seconds using a good algorithm.) Although the Concise Oxford Dictionary offers a more upto-date definition for the word algorithm. (This situation is not corrected in the OED Supplement. calculations once infeasible become routine. has had an even more important effect in extending the frontiers of feasible computation: the use of efficient algorithms. There are other examples of tasks that can be completed in a small fraction of a second. The same word was coined independently by several people. We chose the word algorithmics to translate the more common French term algorithmique.Preface The explosion in computing we are witnessing is arousing extraordinary interest at every level of society. (Although this word appears in some French dictionaries. the definition does not correspond to modern usage.7. For instance. even on a machine a thousand times faster. Another factor.3 for more detail). For instance. quoted in the opening sentence of Chapter 1. sometimes with slightly different meanings. The Oxford English Dictionary defines algorithm as an "erroneous refashioning of algorism" and says about algorism that it "passed through many pseudo- etymological perversions. Harel (1987) calls algorithmics "the spirit of xiii . the subject matter of this book. using a more naive algorithm. but that would require millions of years with less efficient algorithms (read Section 1. today's typical medium-sized computers can easily sort 100. including a recent algorithm". algorithmics is the systematic study of the fundamental techniques used to design and analyse efficient algorithms. As the power of computing machinery grows. whereas such speed would be impossible.

you should have some previous programming experience. This and the general. we suggest that the material be supplemented by attacking some subjects in greater depth. Still less is it a "cookbook" containing a long catalogue of programs ready to be used directly on a machine to solve certain specific problems. the aim of our book is to give the reader some basic tools needed to develop his or her own algorithms. operations research. Each technique is first presented in full generality. we do not neglect the needs of practitioners: besides illustrating the design techniques employed. perhaps using the excellent texts by Garey and Johnson (1979) or Tarjan (1983). Although our approach is rigorous and theoretical. A certain mathematical maturity is more important still. knowledge and expertise that concerns algorithms". artificial intel- ligence. you should not expect to be able to use the algorithms we give directly: you will always be obliged to make the necessary effort to transcribe them into some appropriate programming language. If used as the basis for a course at the graduate level.xiv computing". Our book is neither a programming manual nor an account of the proper use of data structures. nor are the examples for any particular machine. numerical analysis. It is unrealistic to hope to cover all the material in this book in an undergraduate course with 45 hours or so of classes. symbolic computation. Berkeley. Thus we concentrate on the techniques used to design and analyse efficient algorithms. From time to time a passage requires more advanced mathematical knowledge. most of the algorithms presented also have real-life applications. Some of the chapters. an introductory undergraduate course in algebra and another in calculus should provide sufficient background. Our book can also be used for independent study: anyone who needs to write better. In making a choice of subjects. more efficient algorithms can benefit from it. linear algebra. Thereafter it is illustrated by concrete examples of algorithms taken from such different applications as optimization. To profit fully from this book. and the concept of a graph. computing in the humanities. cryptography. we use no particular programming language. adopting the wider perspective that it Preface is "the area of human study. contain original material. On the other hand. in whatever field of application they may be required. Generally speaking. On the contrary. in particular the one concerned with probabilistic algorithms. but such passages can be skipped on the first reading with no loss of continuity. However. The use of Pascal or similarly structured language will help reduce this effort to the minimum necessary. Some basic mathematical knowledge is required to understand this book. fundamental treatment of the material ensure that the ideas presented here will not lose their relevance. We take it for granted that the reader is familiar with such notions as mathematical induction. We have used preliminary versions at both the University de Montreal and the University of California. set notation. Our book is intended as a textbook for an upper-level undergraduate or a lowerlevel graduate course in algorithmics. and so on. the teacher should bear in mind that the first two chapters are essential to understanding the rest of . but giving at best a vague idea of the principles involved in their design.

We originally wrote our book in French. perhaps to lay the ground before going into detail in a subsequent graduate class.Preface xv the book. It is crucial to read the problems: their statements form an integral part of the text. nor do we think it advisable to provide a solutions manual. Although we give the origin of a number of algorithms and ideas. Our warmest thanks. and Luis Miguel and Dan Philip in Berkeley. without necessarily going over each and every example given there of how the techniques can be applied. must go to those who carefully read and reread several chapters of the book and who suggested many improvements and corrections: Pierre . We are also grateful to those people who used the preliminary versions of our book. It would be a pity to study this material without carrying out at least one such experiment. our primary aim is not historical. however. Their level of difficulty is indicated as usual either by the absence of an asterisk (immediate to easy). Although less than a year separates the first French and English printings. however. Several problems call for an algorithm to be implemented on a computer so that its efficiency may be measured experimentally and compared to the efficiency of alternative solutions. In this form it was published by Masson. An elementary course should certainly cover the first five chapters. or by the presence of one asterisk (takes a little thought) or two asterisks (difficult. The first printing of this book by Prentice Hall is already in a sense a second edition. was crucial in improving the presentation of some topics. Particular thanks are due to those who kindly allowed us to copy their course notes: Denis Fortin. is not always consistent between the French and English versions. The solutions to many of the difficult problems can be found in the references. The numbering of problems and sections. No solutions are provided for the other problems. The references from each chapter are combined at the end of the book in an extensive bibliography including well over 200 items. the teacher may find it interesting to discuss these briefly in an undergraduate class. Paris. Laurent Langlois. and Sophie Monet in Montreal. whether they were our own students. both at the undergraduate and graduate levels. deal with more advanced topics. The choice of the remaining material to be studied depends on the teacher's preferences and inclinations. The comments and suggestions we received were most valuable. Almost 500 exercises are dispersed throughout the text. maybe even a research project). We hope the serious teacher will be pleased to have available this extensive collection of unsolved problems from which homework assignments can be chosen. in particular at an international summer school in Bayonne. and in spotting occasional errors. although most of Chapter 1 can probably be assigned as independent reading. Our thanks go first to the students who have followed our courses in algorithmics over the years since 1979. however. The other chapters are to a great extent independent of one another. Each chapter ends with suggestions for further reading. You should therefore not be surprised if information of this kind is sometimes omitted. Our goal is to suggest supplementary reading that can help you deepen your understanding of the ideas we introduce.The last three chapters. the experience gained in using the French version. Writing this book would have been impossible without the help of many people. or colleagues and students at other universities.

understanding. and Jean-Jacques Quisquater and Philips Research Laboratory. Andre Chartier. We also thank Eugene L. Dan Joraanstad. Isabelle and Pat. we particularly appreciate the help we received from James Fegen. for putting up with us -while we were working on the French and English versions of this book. and Production. Paul Bratley thanks Georges Stamon and the Universite de Franche-Comte. Claude Goutier. provided unstinting support. It was her misfortune to help us struggle with the text editing system through one translation and countless revisions. We are also grateful to those who made it possible for us to work intensively on our book during long periods spent away from Montreal. Bruxelles. Claude Crepeau.xvi Preface Beauchemin.-Michel deserves our special thanks. We thank the entire team at Prentice Hall for their exemplary efficiency and friendliness. of Editing. Lawler for mentioning our French manuscript to Prentice Hall's representative in northern California. was no less misfortuned to help us struggle with the last stages of production. and exemplary patience-in short. Amsterdam. Jean-Marc Robert. and Lise DuPlessis who so many times made her country house available. Santiago Miro. Inc. Design. Gilles Brassard Paul Bratley . who taught him so much of the material included in this book. and Alan Sherman. Gilles Brassard thanks Manuel Blum and the University of California. for their encouragement. Berkeley. Annette Hall. Pierre L'Ecuyer. Michel Maksud and Robert Gerin-Lajoie. Bennett Fox. we owe a considerable debt of gratitude to our wives. its sylvan serenity provided the setting and the inspiration for writing a number of chapters. The heads of the laboratories at the Universite de Montreal's Departement d'informatique et de recherche operationnelle. Last but not least. The Natural Sciences and Engineering Research Council of Canada provided generous support. Pierre McKenzie. David Chaum and the CWI. even before we plucked up the courage to work on an English version. Denise St.. He also thanks John Hopcroft.



the chances are that you will multiply the multiplicand successively by each figure of the multiplier. it is important to decide which algorithm for its solution should be used. Take elementary arithmetic as an example. and so on. other systematic methods for solving problems could be included. and that finally you will add all these rows to obtain your answer. provided they do not include instructions like "Add salt to taste". The most famous algorithm in history dates from the time of the Greeks : this is Euclid's algorithm for calculating the greatest common divisor of two integers. The answer can depend on many factors : the size of the instance to be solved. the speed and memory size of the available computing equipment. the way in which the problem is presented. machine) calculation". When we talk about algorithms. the methods we learn at school for multiplying and dividing integers are also algorithms. Suppose you have to multiply two positive integers using only pencil and paper. taken from right to left. It is even possible to consider certain cooking recipes as algorithms. For example. When we set out to solve a problem. If you were raised in North America.1 WHAT IS AN ALGORITHM? The Concise Oxford Dictionary defines an algorithm as a "process or rules for (esp. This is the "classic" multiplication algorithm. that you will write these intermediate results one beneath the other shifting each line one place left.I Preliminaries 1. The execution of an algorithm must not include any subjective decisions. nor must it require the use of intuition or creativity (although we shall see an important exception to this rule in Chapter 8). Nonetheless. we shall mostly be thinking in terms of computers. 1 .

For example. one under each operand. we shall not confine ourselves to the use of one particular programming language : in this way. the essential points of an algorithm will not be obscured by the relatively unimportant programming details. ignoring any fractions. We did not so much as try to describe the classic multiplication algorithm in any detail. or Boolean) are usually omitted. In this example we get 19+76+152+608 = 855. To use it. Make two columns. it is essentially the method used in the hardware of many computers. and then add up the numbers that remain in the column under the multiplicand. We shall see in Section 4. real.1. However.7 that there exist more efficient algorithms when the integers to be multiplied are very large.I.2 Preliminaries Chap. Sometimes we write. Even our description of an algorithm as simple as multiplication a la russe is not completely clear. Scalar parameters of functions and procedures are passed by value unless a different specification is given explicitly. Although this algorithm may seem funny at first. we rapidly discover that natural languages are not at all suited to this kind of thing. Multiplication a la russe. sometimes called "multiplication a la russe ".I. here is quite a different algorithm for doing the same thing. multiplying 19 by 45 proceeds as in Figure I. Finally. To avoid confusion. these more sophisticated algorithms are in fact slower than the simple ones when the operands are not sufficiently large. If we try to describe them in English. for instance procedure proc1(T : array) . Declarations of scalar quantities (integer. 1 However. and how to double a number or divide it by 2. The notation used to specify that a function or a procedure has an array parameter varies from case to case. and arrays are passed by reference. Write the multiplier and the multiplicand side by side. and double the number under the multiplicand by adding it to itself. These phrases should not be confused with comments on the program. we shall in future specify our algorithms by giving a corresponding program. 45 19 19 22 --- 11 38 76 5 152 152 2 304 ----- 1 608 608 76 855 Figure 1. cross out each row in which the number under the multiplier is even.1. At this point it is important to decide how we are going to represent our algorithms. which will always be enclosed within braces. there is no need to memorize any multiplication tables : all we need to know is how to add up. by repeating the following rule until the number under the multiplier is 1 : divide the number under the multiplier by 2. However. We shall use phrases in English in our programs whenever this seems to make for simplicity and clarity.

b ] : integers) In such cases n. the range of a statement such as if. and their values are determined by the bounds of the actual parameter corresponding to T when the procedure is called.. These bounds can be specified explicitly. The operators div and mod represent integer division (discarding any fractional result) and the remainder of a division. for example. m ]) . or record. In such a case #T denotes the number of elements in the array T. The statement return marks the dynamic end of a procedure or a function. B ) arrays X. respectively. or changed. we write procedure proc3(T [ 1 . is shown by indenting the statements affected. If the bounds or the type of T are important. as well as that of a declaration such as procedure. We assume that the reader is familiar with the concepts of recursion and of pointers.. here is a formal description of multiplication a la russe. n ]) or more generally procedure proc4(T [a . and b should be considered as formal parameters. Y[1] -B i (-- I make the two columns } while X [i] > 1 do X[i+1]E-X[i]div2 Y[i+l] <-. function russe (A.1 3 or even procedure proc2(T) if the type and the dimensions of the array T are unimportant or if they are evident from the context. The latter are denoted by the symbol " T ". while. A reader who has some familiarity with Pascal. and in the latter case it also supplies the value of the function. will have no difficulty understanding the notation used to describe our algorithms. For instance. by a procedure call of the form proc3(T [ l .Y[i] +Y[i] i -i +1 { add the appropriate entries } prod F. function.0 while i > 0 do if X [i ] is odd then prod F prod + Y [i ] <-- return prod .. 1. To avoid proliferation of begin and end statements.What Is an Algorithm? Sec. Y { initialization } X[1] -A. or for. a.

An algorithm must work correctly on every instance of the problem it claims to solve. Y[1] . It gives a general solution to the problem of multiplying positive integers. To show that an algorithm is incorrect. B ) arrays X. describes quite a different algorithm. function not -russe (A. Nonetheless. Y { initialization } X[1] . It is important not to lose sight of the fact that in this book we are interested in algorithms. exponentiation. and that this program could easily be simplified. despite a superficial resemblance to the one given previously. we preferred to follow blindly the preceding description of the algorithm. we shall occasionally consider finite problems such as that of playing a perfect game of chess. the following program. 1. T [1] R<-+/(2I LA=T)/BXT-1. and multiplication by powers of 2 to describe an algorithm for multiplying two integers . V R-A RUSAPL B.2*t[2*A V On the other hand..2 PROBLEMS AND INSTANCES Multiplication a la russe is not just a way to multiply 45 by 19. The following APL program describes exactly the same algorithm (although you might reasonably object to a program using logarithms. 19) is an instance of this problem. and that different programs can be used to describe the same algorithm.0 while i > 0 do if X [i ] > 0 then prod .prod + Y [i ] i -i-1 return prod We see that different algorithms can be used to solve the same problem.4 Preliminaries Chap. On the .B iF1 { make the two columns) while X [i ] > 1 do X[i+1]<-X[i]-1 Y[i+1]-B i-i+1 { add the appropriate entries } prod F-. not in the programs used to describe them..) . even if this is more suited to a calculation using pencil and paper than to computation on a machine.A . you will probably have noticed that the arrays X and Y are not really necessary. Most interesting problems ihclude an infinite collection of instances. We say that (45. 1 If you are an experienced programmer. However. we need only find one instance of the problem for which it is unable to find a correct answer.

memory space. This is often not the case with the empirical approach. 1. The advantage of the theoretical approach is that it depends on neither the computer being used. It saves both the time that would have been spent needlessly programming an inefficient algorithm and the machine time that would have been wasted testing it. To make our analyses clearer. 19) is not an instance of the problem being considered. It is also possible to analyse algorithms using a hybrid approach. When we talk about numerical problems.1). It also allows us to study the efficiency of an algorithm when used on instances of any size. For example. this limit cannot be attributed to the algorithm we choose to use. however. so we can choose the best. an instance involving n items is generally considered to be of size n. The theoretical (or a priori) approach. this does not invalidate the algorithm since (-45. that is. which we favour in this book. etc. when we talk about sorting (see Section 1. This raises the question of how to decide which of several algorithms is preferable. using some precisely defined and reasonably compact encoding. denoted by I x 1. This last point is particularly important since often a newly discovered algorithm may only begin to perform better than its predecessor when both of them are used on large instances. the set of instances to be considered. nor even the skill of the programmer. where the form of the function describing the algorithm's efficiency is determined theoretically. it is important to define its domain of definition. even though each item would take more than one bit when represented on a computer. 1. where practical considerations often force us to test our algorithms only on instances of moderate size. corresponds formally to the number of bits needed to represent the instance on a computer.7. However. Any real computing device has a limit on the size of the instances it can handle. Once again we see that there is an essential difference between programs and algorithms. Although multiplication a la russe will not work if the first operand is negative. When we specify a problem. it is obviously of interest to find several algorithms that might be used. it is usually more difficult to prove the correctness of an algorithm.3 THE EFFICIENCY OF ALGORITHMS When we have a problem to solve. The empirical (or a posteriori) approach consists of programming the competing algorithms and trying them on different instances with the help of a computer. we sometimes give the efficiency of our algorithms in terms of the value of the instance being considered. we often use the word "size" to mean any integer that in some way measures the number of components in an instance.Sec.) needed by each algorithm as a function of the size of the instances considered. rather than its size (which is the number of bits needed to represent this value in binary). The size of an instance x. nor the programming language.3 The Efficiency of Algorithms 5 other hand. consists of determining mathematically the quantity of resources (execution time. and then any required numerical parameters are determined empirically for a particular .

an algorithm is quadratic. If such an extrapolation is made solely on the basis of empirical tests. since we do not have a standard computer to which all measurements might refer. 1 program and machine. a change of machine may allow us to solve a problem 10 or 100 times faster. then there always exists a positive constant c such that tl(n) <_ ct2(n) whenever n is sufficiently large. For example. where k and c are appropriate constants. The use of seconds in this definition is obviously quite arbitrary. cubic. Sections 1. polynomial. We say that an algorithm takes a time in the order of t (n). Certain orders occur so frequently that it is worth giving them a name. Consider.6 and 1. although the multiplicative constant may change from one implementation to another.6 Preliminaries Chap. for a given function t. it is likely to be less precise. usually by some form of regression. according to which two different implementations of the same algorithm will not differ in efficiency by more than some multiplicative constant. for example. ignoring all theoretical considerations. This approach allows pred- ictions to be made about the time an actual implementation will take to solve an instance much larger than those used in the tests. It will be clear from the formal definition why we say "in the order of" rather than the more usual "of the order of ". to solve an instance of size n . two algorithms whose implementations on a given machine take respectively n 2 days and n 3 seconds to solve an . we say that it takes linear time. In this case we also talk about a linear algorithm. n k . respectively. This principle remains true whatever the computer used (provided it is of a conventional design). It is natural to ask at this point what unit should be used to express the theoretical efficiency of an algorithm. regardless of the programming language employed and regardless of the skill of the programmer (provided that he or she does not actually modify the algorithm!). or c. Similarly. since we only need change the constant to bound the time by at (n) years or bt (n) microseconds. respectively. Coming back to the question of the unit to be used to express the theoretical efficiency of an algorithm. An answer to this problem is given by the principle of invariance. but only a change of algorithm will give us an improvement that gets more and more marked as the size of the instances being solved increases. if not plain wrong. where n is the size (or occasionally the value.7 illustrate the important differences between these orders of magnitude. say. More precisely. for numerical problems) of the instance considered. By the principle of invariance any other implementation of the algorithm will have the same property. where n is the size of the instance to be solved. or exponential if it takes a time in the order of n 2. if two implementations take t i(n) and t2(n) seconds. if there exist a positive constant c and an implementation of the algorithm capable of solving every instance of the problem in a time bounded above by ct (n) seconds. if an algorithm takes a time in the order of n. there will be no such unit : we shall only express this efficiency to within a multiplicative constant. n 3. The hidden multiplicative constant used in these definitions gives rise to a certain danger of misinterpretation. In the next chapter we give a more rigorous treatment of this important concept known as the asymptotic notation. There can be no question of expressing this efficiency in seconds. Thus.

however. the former is asymptotically better than the latter. 6. To illustrate this. respectively. 3. V represents the worst case . Problem 1. whereas V is in descending order. It may also be interesting to study the possibility of a trade-off between time and memory space : using more space sometimes allows us to reduce the computing time. 1.4.2 to n do x. and V = [6.4 Average and Worst-Case Analysis 7 instance of size n.T[i].Sec. we concentrate on execution time. 1. 4..1 while j > 0andx <T[j] doT[j+l] <. 4. 5. 2. 1.1 shows that both these algorithms take more time on V than on U.i .. 5.i. 3.4. In this book. The other resources needed to execute an algorithm. 2. n ]) for i F-.-T[j] j F-j-1 T[j+1]-x and procedure select (T [1 . 3] . 1]. As is more usual. can be estimated theoretically in a similar way.4 AVERAGE AND WORST-CASE ANALYSIS The time taken by an algorithm can vary considerably between two different instances of the same size. j. 9. 4. 6] . U = [1. Finally. n ]) for i E-. procedure insert (T [1 . "In" and "log" denote natural logarithms and logarithms to the base 10. 0 Let U and V be two arrays of n elements.i + I to n do if T [ j ] < minx then minj i. its performance is better on all sufficiently large instances. 1. 5. from a theoretical point of view. memory space in particular.T [i] for j E-. 5. and conversely. note that logarithms to the base 2 are so frequently used in the analysis of algorithms that we give them their own special notation : thus "lg n " is an abbreviation for 1og2n.1 to n -1 do minj .1.j minx -T[j] T[minjI -T[i] T [i] f. 2. It is only on instances requiring more than 20 million years to solve that the quadratic algorithm outperforms the cubic algorithm ! Nevertheless.minx Problem 1. minx F. In fact. that is to say. such that U is already sorted in ascending order. Simulate these two algorithms on the arrays T = [3. Make sure you understand how they work. consider two elementary sorting algorithms inser-: tion and selection.

it is among the fastest algorithms known on the average. although in certain cases it can be much faster. The algorithm therefore performs in linear time. we shall have an idea of the likely time taken to sort an array initially in random order. 1 for these two algorithms : no array of n elements requires more work.2. Nonetheless. because the while loop is executed i -1 times for each value of i (see Example 2.3 The insertion sorting algorithm thus that this average time is also in the order of n2. insert (U) is very fast. takes quadratic time both on the average and in the worst case. if it is a question of controlling a nuclear power plant. such an analysis of average behaviour . On the one hand. On the other hand. we programmed this algorithm in Pascal on a DEC VAx 780. The variation in execution time is only due to the number of times the assignments in the then part of this test are executed. As Example 2. An implementation in Pascal on the DEC VAx 780 shows that insert (U) takes less than one-fifth of a second if U is an array of 5.1 will show. We shall see in Example 2. Thus we say that insertion sorting takes quadratic time in the worst case. it is crucial to know an upper limit on the system's response time. it increases with the number of elements to be sorted. It is usually harder to analyse the average behaviour of an algorithm than to analyse its behaviour in the worst case. but that requires only a time in the order of n log n on the average. how can we talk about the time taken by an algorithm solely in terms of the size of the instance to be solved? We usually consider the worst case of the algorithm. regardless of the initial order of the elements. whereas insert (V) takes three and a half minutes when V is an array of 5. for each size we only consider those instances of that size on which the algorithm requires the most time. The variation in time is therefore considerable.2.000 elements already in ascending order. in a situation where an algorithm is to be used many times on many different instances.5 we shall see another sorting algorithm that also takes quadratic time in the worst case. In Section 4. because the condition controlling the while loop is always false at the outset. We saw that the time taken by the insertion sort algorithm varies between the order of n and the order of n 2. Also. it may be more important to know the average execution time on instances of size n. Worst-case analysis is appropriate for an algorithm whose response time is critical.2. For example. and moreover.000 elements in descending order. regardless of the particular instance to be solved. If such large variations can occur.3). insert (V) takes quadratic time. the time required by the selection sorting algorithm is not very sensitive to the original order of the array to be sorted : the test "if T [ j ] < minx " is executed exactly the same number of times in every case. On the other hand. The situation is quite different if we compare the times taken by the insertion sort algorithm on the arrays U and V. Even though this algorithm has a bad worst case.Preliminaries 8 Chap. that is. If we can calculate the average time taken by the algorithm on the n! different ways of initially ordering n elements (assuming they are all distinct). the time required by select (T) is quadratic. To verify this. We found that the time required to sort a given number of elements using selection sort does not vary by more than 15% whatever the initial order of the elements to be sorted.

initially ordering n elements is equally likely fails. and their behaviour made independent of the specific instances to be solved.5 WHAT IS AN ELEMENTARY OPERATION? An elementary operation is an operation whose execution time can be bounded above by a constant depending only on the particular implementation used (machine. In what follows we shall only be concerned with worst-case analyses unless stated otherwise. 1. function Wilson (n) { returns true if and only if n is prime } if n divides ((n -1)! + 1) exactly then return true else return false Can we consider addition and multiplication to be unit cost operations ? In theory these operations are not elementary since the time needed to execute them . it is only the number of elementary operations executed that matters in the analysis.5 What Is an Elementary Operation? 9 can be misleading if in fact the instances to be solved are not chosen randomly when the algorithm is used in practice. A useful analysis of the average behaviour of an algorithm therefore requires some a priori knowledge of the distribution of the instances to be solved. In Chapter 8 we shall see how this difficulty can be circumvented for certain algorithms. In the description of an algorithm it may happen that a line of program corresponds to a variable number of elementary operations.Sec. not the exact time required by each of them. and that for some reason it might mostly be asked to sort arrays whose elements are already nearly ordered. if T is an array of n elements. Since we are only concerned with execution times of algorithms defined to within a multiplicative constant. the hypothesis that each of the n! ways of. 1. Wilson's theorem would let us test an integer for primality with astonishing efficiency. For example. For example.2 to n do if T [i ] < x then x . we say that elementary operations can be executed at unit cost. since it is an abbreviation for x . If we allowed ourselves to count the evaluation of a factorial and a test for divisibility at unit cost. In this case. and this is normally an unrealistic requirement.T [i ] Similarly.T[l] for i F. the time required to compute x -min {T[i]ll <-i <-n } increases with n. some mathematical operations are too complex to be considered elementary. and so on). programming language. Equivalently. regardless of the operand's size. it could happen that a sorting algorithm might be used as an internal procedure in some more complex algorithm.

function Not-Gauss (n) { calculates the sum of the integers from 1 to n } sum E. given an integer n.l ton do sum E-sum + i return sum and function Fibonacci (n) { calculates the n th term of the Fibonacci sequence (see section 1. the algorithm should work for all possible values of n. we must attribute to them a cost proportional to the length of the operands concerned. however.5. as a practical matter. rather. In theory. In Example 2.j .2.i + j " cause arithmetic overflow on a 32-bit machine. In your analysis you . to consider that these additions can be carried out at unit cost . so that no real machine can in fact execute these additions at unit cost if n is chosen sufficiently large.1)! + 1 ).1.496 bits are needed to hold the result corresponding to n = 65. however. The following problem illustrates this danger. In the case of multiplication. If we are using a 32-bit machine. it is even more important to ensure that arithmetic operations do not overflow : it is easier to produce large operands by repeated multiplication than by addition. j E.535. The situation is quite different in the case of Fibonacci.7.0 for i <.5)) i E. Two examples will illustrate what we mean. Use Wilson's theorem (n is prime if and only if it is a divisor of (n . It is therefore not realistic. this algorithm (there called fib2) is shown to take quadratic time. ** Problem 1. and the divide-and-conquer technique discussed in Chapter 4 to design an algorithm capable of deciding in a time in the order of log n .i+ j i E. As many as 45.Preliminaries 10 Chap. although it may still be reasonable to consider this an elementary operation for sufficiently small operands. even though at first glance its execution time appears to be linear.i return j In the algorithm called Not-Gauss the value of sum stays quite reasonable for all the instances that the algorithm can realistically be expected to meet in practice.0 fork E. Newton's binomial theorem. all the additions can be executed directly provided that n is no greater than 65. In practice.1.8.l to n do j E. 1 increases with the length of the operands. The analysis of the algorithm must therefore depend on its intended domain of application. whether or not n is prime. it may be sensible to consider them as elementary operations so long as the operands concerned are of a rea- sonable size in the instances we expect to encounter. It suffices to take n = 47 to have the last addition "j E.535.

6 WHY DO WE NEED EFFICIENT ALGORITHMS? As computing equipment gets faster and faster. when you figure out that now. modulo operations. Solving an instance of size 20 will take nearly two minutes. however. Suppose. You find a cubic algorithm that can solve your problem. You may feel you have wasted your money. With the same algorithm you can now solve an instance of size n in only 10-6 x 2" seconds. In most practical situations. To sum up. you buy a new computer one hundred times faster than the first.1. Your program can thus solve an instance of size 10 in one-tenth of a second. Since you need to solve bigger instances than this.6. it is reasonable to count such arithmetic operations at unit cost. A similar problem can arise when we analyse algorithms involving real numbers if the required precision increases with the size of the instances to be solved. subtractions.7. when you run your new machine for a whole year. you cannot even solve an example of size 45. the use of single precision floating point arithmetic proves satisfactory despite the inev- itable loss of precision. multiplications. Supposing you were able to run your computer without interruption for a year. . 1. Suppose you decide instead to invest in algorithmics. Imagine. however.Sec.500. that using the original machine this new algorithm can solve an instance of size n in 10-2 x n 3 seconds. One typical example of this phenomenon is the use of De Moivre's formula to calculate values in the Fibonacci sequence (see Section 1. it may seem hardly worthwhile to spend our time trying to design more efficient algorithms.5). To solve an instance of size 30. Would it not be easier simply to wait for the next generation of computers? The remarks made in the preceding sections show that this is not true. even deciding whether an instruction as apparently innocent as "j F i + j " can be considered as elementary or not calls for the use of judgement. and assignments at unit cost unless explicitly stated otherwise. In general. Boolean operations. comparisons. to illustrate the argument. In what follows we count additions. When this is so. if you were previously able to solve an instance of size n in some given time. regardless of the size of the operands involved. In one day you can now solve instances whose size is greater than 200. 1. divisions. This is illustrated by Figure 1. your new machine will solve instances of size at best n + 7 in the same time. multiplications. with one year's computation you can almost reach size 1. and tests of divisibility by an integer (but not calculations of factorials or exponentials) can be carried out in unit time.6 Why Do We Need Efficient Algorithms ? 11 may assume that additions. for example. that to solve a particular problem you have available an exponential algorithm and a computer capable of running this algorithm on instances of size n in 10-4 x 2' seconds. you would only just be able to solve an instance of size 38. even a whole day's computing will not be sufficient.

6. Such problems can always be solved by using multiple-precision arithmetic (see Sections 1. there have been cases where even more spectacular improvements have been made. 1.1. .7. make such a purchase much more profitable. Additions and multiplications are therefore generally taken to be elementary operations in the following paragraphs (except. On the original machine the new algorithm takes 10 seconds to solve an instance of size 10. in particular on the rather small ones. the new algorithm should not be used uncritically on all instances of the problem. of course. which is one hundred times slower than the old algorithm. a machine one hundred times faster than the old one will allow you to solve instances four or five times bigger in the same length of time. we shall simplify our presentation by ignoring the problems that may arise because of arithmetic overflow or loss of precision on a particular machine. Chap. In fact. Nevertheless.Preliminaries 12 Figure 1.7. even for well-established algorithms. it will also. The new algorithm is faster only for instances of size 20 or greater. In fact.2). Not only does the new algorithm offer a much greater improvement than the purchase of new machinery. 1 Algorithmics versus hardware. Some of the following examples use large integers or real arithmetic. thanks to your new algorithm. supposing you are able to afford both. Unless we explicitly state the contrary. it is possible to com- bine the two algorithms into a third one that looks at the size of the instance to be solved before deciding which method to use.7 SOME PRACTICAL EXAMPLES Maybe you are wondering whether it is really possible in practice to accelerate an algorithm to the extent suggested in the previous section.2 and 4. for Section 1.7). Naturally.

The time required is therefore in the order of mn. we programmed insertion sort and quicksort in Pascal on a DEC VAx 780. provided we choose the smaller operand as the multiplier . Both these algorithms. Among others. To sort 1. and in particular in algorithmics. When we have 5. and that it executes approximately one elementary addition for each of these multiplications.7 Some Practical Examples 13 1. The classic algorithm of Section 1. Sorting problems are often found inside more complex algorithms. and three times as fast when sorting 100 elements. take quadratic time both in the worst case and on the average. our estimate is that it would take nine and a half hours to carry out the same task using insertion sorting. All these algorithms take a time in the order of n log n on the average . We have already seen two classic sorting algorithms in Section 1. we can use a representation such as FORTRAN's "double precision".000 elements to sort.2 Multiplication of Large Integers When a calculation requires very large integers to be manipulated. mergesort (see Section 4. it can happen that the operands become too long to be held in a single word of the computer in use. insertion takes more than three seconds. In this case. more generally. respectively. or Hoare's quicksort algorithm (see Section 4.4). 1.4: insertion sorting and selection sorting. other sorting algorithms are more efficient when n is large. this choice does not alter our analysis of the order of efficiency of the algorithms in question. the first two take this same amount of time even in the worst case. To have a clearer idea of the practical difference between a time in the order of n 2 and a time in the order of n log n . as we saw.1 Sorting The sorting problem is of major importance in computer science. compared to little more than one second for quicksort. Such operations thereupon cease to be elementary. we must ask ourselves how the time necessary to multiply two large integers increases with the size of the operands.2. Although these algorithms are excellent when n is small. or. multipleprecision arithmetic.000 elements .4 and Problem 2. Multiplication a la russe also takes a time in the order of mn.Sec.1 can easily be transposed to this context. 1.7. the inefficiency of insertion sorting becomes still more pronounced : one and a half minutes are needed on average.5). are to be multiplied. We are required to arrange in ascending order a collection of n objects on which a total ordering is defined. whereas quicksort requires less than one-fifth of a second.3). quicksort can handle 100. of sizes m and n.000 elements. In 30 seconds. (This last remark would be false should we be considering exponential time algorithms-can you see why?) Suppose two large integers. we might use Williams's heapsort algorithm (see Example 2.7.2. When this occurs. Since these measures differ only by a multiplicative constant. We see that it multiplies each word of one of the operands by each word of the other. The difference in efficiency between the two algorithms is marginal when the number of elements to be sorted is small. We can measure this size by either the number of computer words needed to represent the operands on a machine or the length of their representation in decimal or binary. Quicksort is already almost twice as fast as insertion when sorting 50 elements.

To take account of the architecture of the machine. The determinant of the matrix M.. is often defined recursively : if M [i. The theoretically better algorithm of Section 4. If both operands are of size n.7.1. even more sophisticated algorithms exist for much larger operands. which we shall study in Section 4.Preliminaries 14 Chap. more efficient algorithms exist to solve this problem.2 .. the algorithm thus takes a time in the order of n 1.7.n j be an n x n matrix. 1 and the larger as the multiplicand.. then . an . To verify this. the size of an operand is measured in terms of the number of 20-bit segments in its binary representation. a1. Accordingly.3 Evaluating Determinants Let a1. The gain in efficiency continues to increase as the size of the operands goes up.59 is less spectacular than that between the order of n 2 and the order of n log n . however. there is no reason for preferring it to the classic algorithm. we carried out the calculations in base 220 rather than in base 10.1 a2..59. . Thus.7 in Pascal on a CDc CYBER 835 and tested them on operands of different sizes. the fast algorithm is some three times more efficient than the classic algorithm : they take about 15 seconds and 40 seconds. denoted by det(M). Integers of 20 bits are thus multiplied directly by the hardware of the machine.599 which is preferable to the quadratic time taken by both the classic algorithm and multiplication a la russe.n M = I an l an 2 . which we saw in the case of sorting algorithms. For operands ten times this length. takes a time in the order nm0. How much time does multiplication a la russe take if the multiplier is longer than the multiplicand ? As we mentioned in Section 1. The difference between the order of n 2 and the order of n 1. The simplest.n a2.1 a1. we programmed the classic algorithm and the algorithm of Section 4. 1. As we shall see in Chapter 9. yet at the same time space is used quite efficiently (the machine has 60-bit words). particularly as the hidden constant is likely to be larger. respectively. where n is the size of the larger operand and m is of nmlg(3/2).. or approximately the size of the smaller.1) submatrix obtained from M by deleting the i th row and the j th column.2 a2. j ] denotes the (n ..7 gives little real improvement on operands of size 100 (equivalent to about 602 decimal digits) : it takes about 300 milliseconds. Problem 1.1) x (n . whereas the classic algorithm takes about 400 milliseconds.7.1.

it therefore takes a time in the order of n. On the contrary.min(m . On the other hand. n) is obtained directly from the definition. function ged (m. we estimate that it would take more than 10 million years to calculate the determinant of a 20 x 20 matrix. If we use the recursive definition directly.1 det(M[1. A classic algorithm for calculating gcd(m. n). .Sec. In particular. Gauss-Jordan elimination.1 . denoted by gcd(m. and their lower powers are 2 and 1. The Gauss-Jordan algorithm finds the determinant of a 10 x 10 matrix in one-hundredth of a second . 700) we first factorize 120=2 3 x 3 x 5 and 700=2 2 x5 2 x 7. another classic algorithm. On the other hand. is the largest integer that divides both m and n exactly. respectively. Strassen discovered in 1969 a recursive algorithm that can calculate the determinant of an n x n matrix in a time in the order of n 1g7 or about n 2.2. n) consists of first factorizing m and n. The greatest common divisor of m and n. each prime factor being raised to the lower of its powers in the two arguments. (-1)j+'a1.81. 1.i .1 until i divides both m and n exactly return i The time taken by this algorithm is in the order of the difference between the smaller of the two arguments and their greatest common divisor. j ]) . and then taking the product of the prime factors common to m and n. For example. and we need to know how to calculate them efficiently. the recursive algorithm takes more than 20 seconds on a 5 x 5 matrix and 10 minutes on a 10 x 10 matrix . This is even worse than exponential. 1. n) + 1 repeat i F. a task accomplished by the Gauss-Jordan algorithm in about onetwentieth of a second ! You should not conclude from this example that recursive algorithms are necessarily bad. does the computation in cubic time. For example.7 Some Practical Examples = det(M) 15 n Y. The greatest common divisor of 120 and 700 is therefore 22 x 51 = 20. we obtain an algorithm that takes a time in the order of n! to calculate the determinant of an n x n matrix (see Example 2. the determinant is defined by det(M) = a 1. The common factors of 120 and 700 are therefore 2 and 5. j=1 If n = 1. Chapter 4 describes a technique where recursion plays a fundamental role in the design of efficient algorithms. Determinants are important in linear algebra.7. 21) = 1. to calculate gcd(120. We programmed the two algorithms in Pascal on a Coc CYBER 835.4 Calculating the Greatest Common Divisor Let m and n be two positive integers. gcd(6. we say that m and n are coprime. it takes about five and a half seconds on a 100 x 100 matrix.15) = 3 and gcd(10.5). n) i . The obvious algorithm for calculating gcd(m . When m and n are of similar size and coprime. n) =1. thus proving that Gauss-Jordan elimination is not optimal. When gcd(m.

fI = 1 fn fn -I +fn-2 and for n ? 2.5 Calculating the Fibonacci Sequence The Fibonacci sequence is defined by the following recurrence : fo0. The first ten terms of the sequence are therefore 0. 13. n) while m > 0 do tF. Since 4 < 1. 2. 1. function fib 1(n) if n < 2 then return n else return fibl(n . which is much faster than the preceding algorithms.2) : L = _1 [on . 3. This sequence has numerous applications in computer science. Euclid's original algorithm works using successive subtractions rather than by calculating a modulo. Nevertheless. 1. 8. On the CDC CYBER 835. an operation we do not know how to do efficiently. De Moivre's formula is of little immediate help in calculating fn exactly. there exists a much more efficient algorithm for calculating greatest common divisors.2. which means that the value of fn is in the order of 0n . the term (-4)-n can be neglected when n is large. The algorithm obtained directly from the definition of the Fibonacci sequence is the following. the greater is the degree of precision required in the values of 5 and 0. since the larger n becomes. 1. in mathematics.3. 34. This is Euclid's famous algorithm. To be historically exact. 21. 1 Even though this algorithm is better than the one given previously. a single-precision computation programmed in Pascal produces an error for the first time when calculating f66 . De Moivre proved the following formula (see Example 2. and in the theory of games. this algorithm takes a time in the order of the logarithm of its arguments. 5.(-4)-n where 4 _ (I +x(5)/2 is the golden V rJatio.nmodm n F-m mt return n Considering the arithmetic operations to have unit cost. It is when applied to two consecutive terms of the Fibonacci sequence that Euclid's algorithm takes the longest time among all instances of comparable size.6).1) + fib 1(n-2) . it requires us to factorize m and n.7. even in the worst case (see Example 2. However. function Euclid (m.Preliminaries 16 Chap.

assuming we count each addition as an elementary operation (see Example 2. to calculate fib l (5) we need the values of fib 1(4) and fib l (3) . In fact. (All these times are approximate.2. which at first sight appears quite mysterious.Sec. fibl(2) three times.7. itself. However. For instance. with the exception of the case n = 100 on which fib 1 would take well over 109 years ! Note that fib2 is more efficient than fib3 on small instances. j . takes a time in the order of the logarithm of n (see Example 2.h2 h F. in the order of 0" (see Example 2. This is much better than the first algorithm.5. Times greater than two minutes were estimated using the hybrid approach.7 Some Practical Examples 17 This algorithm is very inefficient because it recalculates the same values many times. To avoid problems caused by arithmetic overflow (the Fibonacci sequence grows very rapidly : fl00 is a number with 21 decimal digits). we carried out all the computations modulo 107. j E-0.7). but fib 1(4) also calls for the calculation of fib 1(3). This third algorithm. k -0. function fib3(n) i .1. the time required to calculate f.2kh + t k E--k2+t n E-n div 2 return j Once again. Table 1. that is to say.9). using this algorithm is in the order of the value of f. To avoid wastefully recalculating the same values over and over.1.jh j . we programmed the three algorithms in Pascal on a CDC CYBER 835 in order to compare their execution times empirically. 1. and fibl(0) three times. fibl(1) five times.2. function fib2(n) i E. It will be explained in Chapter 4. .8).ik+t t f.0 fork -ltondoj -i+j return j i 4j-i This second algorithm takes a time in the order of n.1 eloquently illustrates the difference that the choice of an algorithm can make.2. there exists a third algorithm that gives as great an improvement over the second algorithm as the second does over the first. which is to say that we only obtained the seven least significant figures of the answer. it is natural to proceed as in Section 1.) The time required by fib l for n > 50 is so long that we did not bother to estimate it. h E-1 while n > 0 do if n is odd then t F-.ih+jk+t i . We see that fib 1(3) will be calculated twice.

000 100.581 0.1 PERFORMANCE COMPARISON BETWEEN MODULO 107 FIBONACCI ALGORITHMS n 10 fibl 8 msec I msec I msec fib2 fib3 6 3 20 30 50 I sec 2 min I msec 21 days ! msec 2 msec I msec i msec 1. and t3(n) = 1logn milliseconds.019 10.011 0.7.005 0.000 times larger to make fib3 take one extra millisecond of computing time.7.087 0. 25 . TABLE 1.000 fib2 fib3 0.9.2 compares the times taken by these three algorithms when they are used in conjunction with an efficient implementation of the classic algorithm for multiplying large integers (see Problem 2. We could also have calculated all the figures in the answer using multipleprecision arithmetic. PERFORMANCE COMPARISON BETWEEN EXACT FIBONACCI ALGORITHMS* n 5 10 15 20 fib l fib2 fib3 0. Writing t.11).941 0.020 118.017 0. and Problem 2.177 3.Preliminaries 18 Chap. but their joint advantage over fibi remains just as striking.000. It takes a value of n 10.000 5.-20 seconds. we find t1(n) ^ 0.000 10.000 fib2 fib3 1 i msec 150 msec msec I msec 2 1.000 3 5 n 100 10. Table 1.021 0. t2(n) = 15n microseconds. we can estimate approximately the time taken by our implementations of these three algorithms.892 29.041 0.664 298.000.2. (n) for the time taken by fibi on the instance n.348 76.007 0.2.017 0.132 0.109 1.2.msec 4 2 2 25 min 15 sec msec 1 2 msec Z Using the hybrid approach. 1 TABLE 1. In this case the advantage that fib3 enjoys over fib2 is less marked.7.457 0.013 0.021 n 100 500 1.107 0.766 0. Example 2.553 * All times are in seconds.

How are the recursive calls to be set up? The obvious approach requires a time in the order of n 2 to be used before each recursive call.3 is another example of an incompletely presented algorithm. systems theory. nor must it require the use of intuition or creativity".7.5 describes an algorithm ? The problem arises because it is not realistic to consider that the multiplications in fib3 are elementary operations. And if that were not sufficient.6. 1. but all the n recursive calls. 1.2. In one early test of the "new" algorithm the Fourier transform was used to analyse data from an earthquake that had taken place in Alaska in 1964.11 and 4. And what should we say about De Moivre's formula used as an algorithm ? Calculation of a determinant by the recursive method of Section 1. Thus the development of numerous applications had been hindered for no good reason for almost a quarter of a century. the choice may be considered a subjective decision.7. Any practical implementation must take this into account. That this distinction is not merely academic is illustrated by Problems 2. However. The "discovery" by Cooley and Tukey in 1965 of a fast algorithm revolutionized the situation : problems previously considered to be infeasible could now at last be tackled. can we reasonably maintain that fib3 of Section 1.7.Sec. probably by using a program package allowing arithmetic operations on very large integers. all the necessary theoretical groundwork for Danielson and Lanczos's algorithm had already been published by Runge and Konig in 1924! . We shall come back to this subject in Chapter 9. Since the exact way in which these multiplications are to be carried out is not specified in fib3.6 Fourier Transforms The Fast Fourier Transform algorithm is perhaps the one algorithmic discovery that had the greatest practical impact in history. Ironically it turned out that an efficient algorithm had already been published in 1942 by Danielson and Lanczos. For years progress in these areas was limited by the fact that the known algorithms for calculating Fourier transforms all took far too long.8 When Is an Algorithm Specified ? 19 1. and hence fib3 is not formally speaking an algorithm. and signal processing including speech recognition.8 WHEN IS AN ALGORITHM SPECIFIED? At the beginning of this book we said that "the execution of an algorithm must not include any subjective decisions. which show that indeed the order of time taken by fib3 depends on the multiplication algorithm used. acoustics. In this case. . We shall see in Problem 2. For the moment let us only mention that Fourier transforms are of fundamental importance in such disparate applications as optics. the "new" algorithm was able to perform the same task in less than two and a half seconds.5 that it is possible to get by with a time in the order of n to set up not just one. Although the classic algorithm took more than 26 minutes of computation. telecommunications.7. this added subtlety does not alter the fact that the algorithm takes a time in the order of n! to calculate the determinant of an n x n matrix.2. quantum physics.

1 Lists A list is a collection of nodes or elements of information arranged in a certain order.9 DATA STRUCTURES The use of well-chosen data structures is often a crucial factor in the design of efficient algorithms.10). The different computer implementations that are commonly used differ in the quantity of memory required. 1 To make life simple. . maxlength ] : information the elements of a list occupy the slots value [ 1 ] to value [counter ].4.. Chosen because they will be used in subsequent chapters. and Example 2. which is the last. this section concentrates on the less elementary notions of heaps and disjoint sets. we shall continue to use the word algorithm for certain incomplete descriptions of this kind. these two structures also offer interesting examples of the analysis of algorithms (see Example 2. this book is not intended to be a manual on data structures. pointers. and lists. We also suppose that he or she has already come across the mathematical concepts of directed and undirected graphs. and so on. and which are the predecessor and the successor (if they exist) of any given node. We suppose that the reader already has a good working knowledge of such basic notions as arrays. and in the greater or less ease of carrying out certain operations. Here we content ourselves with mentioning the best-known techniques. which is the first node in the structure.2.. Such lists are subject to a number of operations : we might want to insert an additional node. After a brief review of some important points. Such a structure is frequently represented graphically by boxes and arrows. Implemented as an array by the declaration type tablist = record counter : 0 .9. A list. 1. The information attached to a node is shown inside the corresponding box and the arrows show transitions from a node to its successor. as in Figure 1. Using this implementation. The details will be filled in later should our analyses require them. structures.Preliminaries 20 Chap. alpha l-il beta as gamma 31 delta Figure 1.1. and knows how to represent these objects efficiently on a computer. 1. to delete a node. to copy a list. for example.3. to count the number of elements it contains. maxlength value[ I. Problem 2. Nevertheless. and the order of the elements is given by the order of their indices in the array. The corresponding data structure must allow us to determine efficiently.

2 for instance. as we can the predecessor and the successor of a given node. without having to follow k pointers and thus to take a time in the order of k. b } . beta and delta. In the case of a directed graph the nodes are joined by arrows called edges. inserting new nodes or deleting an existing node can be done rapidly. In every case.Sec. In our example. however. the space needed to represent the list can be allocated and recovered dynamically as the program proceeds. Even if additional pointers are used to ensure rapid access to the first and last elements of the list. a graph is a set of nodes joined by a set of lines or arrows. b). Formally speaking. the nodes are usually represented by some such structure as type node = record value : information next : T node . it is difficult when this representation is used to examine the k th element. then they must go in opposite directions). 1. for arbitrary k. .2 Graphs Intuitively speaking. it suffices to add a second pointer to each node to allow the list to be traversed rapidly in either direction. In the example of Figure 1.9 Data Structures 21 we can find the first and the last elements of the list rapidly. once an element has been found. it presents the major disadvantage of requiring that all the memory space potentially required be reserved from the outset of a program. Consider Figure 1. a graph is therefore a pair G = < N. a single pointer is used in each node to designate its successor : it is therefore easy to traverse the list in one direction. if pointers are used to implement a list structure. the nodes are joined by lines with no direction indicated.2 there exists an edge from alpha to gamma and another from gamma to alpha. In this case. However. inserting a new element or deleting one of the existing elements requires a worst-case number of operations in the order of the current size of the list. However. also called edges. We distinguish directed and undirected graphs. On the other hand. If a higher memory overhead is acceptable. In the case of an undirected graph. An edge from node a to node b of a directed graph is denoted by the ordered pair (a. This implementation is particularly efficient for the important structure known as the stack. where each node includes an explicit pointer to its successor. whereas an edge joining nodes a and b in an undirected graph is denoted by the set { a. but not in the other. On the other hand. which we obtain by restricting the permitted operations on a list : addition and deletion of elements are allowed only at one particular end of the list.9. There are never more than two arrows joining any two given nodes of a directed graph (and if there are two arrows.9. provided a suitably powerful programming language is used. 1. the edges may form paths and cycles. and there is never more than one line joining any two given nodes of an undirected graph. are joined only in the direction indicated. A > where N is a set of nodes and A c N x N is a set of edges.9.

independently of the number of edges that exist involving this particular node. j ] = true . This takes a time in the order of nbnodes. A directed graph. the number of nodes in the graph.9. the matrix is necessarily symmetric.. If the number of edges in the graph is small. we have to scan a complete row in the matrix. 1 alpha gamma delta Figure 1. that is to say of those nodes j such that an edge from i to j (in the case of a directed graph) or between i and j (in the case of an undirected graph) exists. this representation is preferable from the point of view of the memory space used. The first is illustrated by type adjgraph = record value [ 1 . Equivalently. which is less efficient than looking up a Boolean value in an array. In the case of an undirected graph. On the other hand. connected. There are at least two obvious ways to represent a graph on a computer. On the other hand. to determine whether or not two given nodes i and j are connected directly. undirected graph.. Here we attach to each node i a list of its neighbours. too).1 . we have to scan the list of neighbours of node i (and possibly of node j. j ] = false. It may also be possible in this case to examine all the neighbours of a given node in less than nbnodes operations on the average. The memory space required is quadratic in the number of nodes. A tree is an acyclic. . A second possible representation is as follows : type lisgraph = array[ L. With this representation it is easy to see whether or not two nodes are connected. should we wish to examine all the nodes connected to some given node. nbnodes ] : information adjacent [ 1 . then adjacent [i . a tree may be defined as an undirected graph in which there exists exactly one path between any given pair of nodes. nbnodes ] of record value : information neighbours : list . otherwise adjacent [i.. nbnodes ] : Booleans If there exists an edge from node i of the graph to node j.Preliminaries 22 Chap.2. nbnodes. The same representations used to implement graphs can be used to implement trees.

then G is a rooted tree and r is its root.9. that alpha is an ancestor of epsilon. It is usual to represent a rooted tree with the root at the top.5. A rooted tree.3 would be represented as in Figure 1.) Extending the analogy with a family tree. If there exists in G a vertex r such that every other vertex can be reached from r by a unique path. lambda lambda Figure 1. On a computer. A leaf of a rooted tree is a node with no children.4 may therefore be considered as different.3. As in the case of lists. we say that beta is the parent of delta and the child of alpha. 1. Any rooted tree with n nodes contains exactly n -1 edges.9. to the parent or the eldest sibling of a given node) may speed up certain operations at the price of an increase in the memory space needed. Although nothing in the definition indicates this. but rather the pointers used in the computer representation. In this example alpha is at the root of the tree. Figure 1. any rooted tree may be represented using nodes of the following type : type treenode = record value : information eldest-child.9. the branches of a rooted tree are often considered to be ordered : in the previous example beta is situated to the left of gamma. like a family tree. Two distinct rooted trees. The two trees in Figure 1. the other nodes are called internal nodes.4.9. where now the arrows no longer represent the edges of the rooted tree. the use of additional pointers (for example. we shall use the simple term "tree" instead of the more correct "rooted tree". that epsilon and zeta are the siblings of delta.3.9.9 23 Data Structures 1. (When there is no danger of confusion. and so on.9. .9. next-sibling :T treenode The rooted tree shown in Figure 1. and (by analogy with a family tree once again) delta is the eldest sibling of epsilon and zeta. as in Figure 1.Sec.3 Rooted Trees Let G be a directed graph.

3. Two distinct binary trees. whereas in the second case b is the younger child of a and the elder child is missing. we naturally tend to talk about the left-hand child and the right-hand child. and thus also the depth of its deepest leaf. In this case. height 0. the positions occupied by the children are significant. and level 1 in the tree of Figure 1.9.9. the binary trees of Figure 1. Finally. gamma has depth 1.. n ] : T n-ary-node Figure 1. The height of a rooted tree is the height of its root.5. There are several ways of representing an n-ary tree on a computer. . although the metaphor becomes somewhat strained. One obvious representation uses nodes of the type type n-ary-node = record value : information child [ 1 . zeta VV M Possible computer representation of a rooted tree. If each node of a rooted tree can have up to n children.6. For instance. the level of a node is equal to the height of the tree minus the depth of the node concerned.9. The height of a node is the number of edges in the longest path from the node in question to a leaf.6 are not the same : in the first case b is the elder child of a and the younger child is missing. we say it is an n-ary tree. The depth of a node in a rooted tree is the number of edges that need to be traversed to arrive at the node starting from the root. For example.9. In the important case of a binary tree.Preliminaries 24 Chap. 1 alpha beta delta gamma epsilon 1A Figure 1.

This kind of tree can be represented using an array T by putting the nodes of depth k. This structure is interesting because it allows efficient searches for values in the tree. T [2k+1]. and less than or equal to the values contained in its right-hand descendants. A binary tree is a search tree if the value contained in every internal node is larger than or equal to the values contained in its left-hand descendants. that is. without destroying the search tree property. one on the left and one on the right. A binary tree is essentially complete if each of its internal nodes possesses exactly two children. right-child : T binary-node It is also sometimes possible. if it exists. with the possible exception of a unique special node situated on level 1. is to the right of all the other level 1 internal nodes. Problem 1. .Sec. all the leaves are either on level 0. .1.9. The parent of the node represented in T [i] is found in T [i div 2] for i > 1.1. to represent a rooted tree using an array without any explicit pointers. as well as the efficient implementation of certain dynamic priority lists. The unique special node. called heapsort (see Problem 2. For instance. This interesting structure lends itself to numerous applications. Moreover. if this is done in an unconsidered fashion. it can happen that the resulting tree becomes badly unbalanced. However. here we only mention their existence. which possesses only a left-hand child and no right-hand child. in the sense that the height of the tree is in the order of the number of nodes it contains. in the positions T [2k]. which may be incomplete).2. to delete nodes or to add new values. from left to right. Suppose the value sought is held in a node at depth p in a search tree. 1. T[2 k+1_11 (with the possible exception of level 0.4 Heaps A heap is a special kind of rooted tree that can be implemented efficiently in an array without any explicit pointers. Figure 1. Since these concepts are not used in the rest of this book.9. and .5. Design an algorithm capable of finding this node starting at the root in a time in the order of p. and no leaf is found on level 1 to the left of an internal node at the same level. It is possible to update a search tree. as we shall see in the following section. including a remarkable sorting technique..7 shows how to represent an essentially complete binary tree containing 10 nodes. such as the use of AVL trees or 2-3 trees.9 Data Structures 25 In the case of a binary tree we can also define type binary-node = record value : information left-child.. An example of a search tree is given in Figure 5. or else they are on levels 0 and 1. More sophisticated methods.9.3). allow such operations as searches and the addition or deletion of nodes in a time in the order of the logarithm of the number of nodes in the tree in the worst case. These structures also allow the efficient implementation of several additional operations. 1.

so that it becomes greater than the value of its parent. the value of a node is decreased so that it becomes less than the value of at least one of its children. A heap. The heap property is that the value of each internal node is greater than or equal to the values of its children.8. the children of the node represented in T[i] are found in T[2i] and T [2i + 1 ]. An essentially complete binary tree.7. it suffices to exchange these two values and then to continue the same process upwards in the tree until the heap property is restored.9. . We say that the modified value has been percolated up to its new position (one often encounters the rather strange term sift-up for this process). The subtree whose root is in T [i] is also easy to identify. If. 1 T[I0] Figure 1. whenever they exist. each of whose nodes includes an element of information called the value of the node.9.8 gives an example of a heap. it suffices to exchange the modified value with Figure 1. on the contrary. This same heap can be represented by the following array : 10 7 9 4 7 5 2 2 1 6 The fundamental characteristic of this data structure is that the heap property can be restored efficiently after modification of the value of a node. Figure 1. A heap is an essentially complete binary tree.9.Preliminaries 26 T[7] T(6] T[5] T[4] Chap. If the value of the node increases.

and then to continue this process downwards in the tree until the heap property is restored. i ) { this procedure percolates node i so as to re-establish the heap property in T [I . i ) {this procedure sifts node i down so as to re-establish the heap property in T [I . n ]. n] is a heap . adding a new node. If the reader wishes to make use of heaps for a "real" application. removing it.j div 2 exchange T [ j ] and T [k] until j = k The heap is an ideal data structure for finding the largest element of a set. we also suppose that 1 <_ i 5 n } k F-i repeat j *-k { find the larger child of node j } if 2j <_ n and T [2j ] > T [k] then k . n ] .. the value of T[i] is set to v and the heap property is re-established . v ) { T [I . they are written so as to reflect as closely as possible the preceding discussion. n I.T[i] T[i]<-v if v < x then sift-down (T.. the parameter n is not used here } k-i repeat j *-k if j > 1 and T [j div 2] < T [k] then k . we suppose that T would be a heap if T [i ] were sufficiently large . i. n ]. procedure alter-heap (T [ 1 .. For the purpose of clarity. The event with highest priority is always found at the root of . we also suppose that 1 <_ i <_ n .Sec.9 Data Structures 27 the larger of the values in the children. These are exactly the operations we need to implement dynamic priority lists efficiently : the value of a node gives the priority of the corresponding event. we encourage him or her to figure out how to avoid the inefficiency resulting from our use of the "exchange" instruction. i) procedure sift-down (T [1 . n ].. we suppose that 1<_ i<_ n } x . We say that the modified value has been sifted down to its new position. or modifying a node. we suppose that T would be a heap if T [i] were sufficiently small . 1. i ) else percolate (T. The following procedures describe more formally the basic heap manipulation process. then the node has arrived at its final position) until j = k procedure percolate (T [1 .2j if2j <n andT[2j+1]>T[k]thenk -2j+1 exchange T [ j ] and T [k] { if j = k...

v percolate (T [1 .1) procedure insert-node (T [ 1 ... The obvious solution is to start with an empty heap and to add elements one by one. for example. n ]) { this procedure makes the array T [I . n -1 ]. n ] into a heap. procedure slow-make-heap (T [ 1 . and the priority of an event can be changed dynamically at all times. but rather inefficiently) for i ... also by sifting down their roots.. 1 the heap.2 to n do percolate (T [1 ... There exists a cleverer algorithm for making a heap.9a.9. This is particularly useful in computer simulations....9b. Figure 1. function find-max (T [1 .9. n + 1]. n ]. This results in an essentially complete binary tree corresponding to the array : 1 10 9 7 7 5 2 2 4 6 It only remains to sift down its root in order to obtain the desired heap.2. The final process thus goes as follows : . Suppose. this approach is not particularly efficient (see Problem 2. i ]. v) { adds an element whose value is v to the heap T [ 1 .9. The subtrees at the next higher level are then transformed into heaps. n -1 ] } T[1]-T[n] sift-down (T [ 1 . We begin by making each of the subtrees whose roots are at level 1 into a heap. n ] and restores the heap property in T [I ..2). as illustrated in Figure 1. The other subtree at level 2 is already a heap. n ] and restores the heap property in T [I .Preliminaries 28 Chap.. n ] of elements in an undefined order. n return T[1] procedure delete-max (T [1 . n + 1 ] } T[n+1] .9c shows the process for the left-hand subtree. n ]) { removes the largest element of the heap T [I . by sifting down those roots... i ) However. n ]) { returns the largest element of the heap T [ 1 . n + 1) It remains to be seen how we can create a heap starting from an array T [I . that our starting point is the following array : 1 6 9 2 7 5 2 7 4 10 represented by the tree of Figure 1.

Figure 1. i) We shall see in Example 2. procedure make-heap (T [ 1 . . Making a heap.9. (c) One level 2 subtree is made into a heap (the other already is a heap)..Data Structures Sec.9.4 that this algorithm allows the creation of a heap in linear time.(n div 2) downto 1 do sift-down (T. n ] into a heap } for i .8. This algorithm can be described formally as follows. (a) The starting situation.. 1.9 10 1 9 10 7 10 7 29 7 7 5 9 1 7 5 9 4 7 5 2 2 2 2 4 6 2 4 6 2 1 6 whose tree representation is shown in the previous example as Figure 1.9. n ]) { this procedure makes the array T [ 1 .2. (b) The level t subtrees are made into heaps.

m) Draw the heap after each operation. The advantage of this procedure is that it requires only one comparison at each level in order to sift down the empty node. Finally. The Fibonacci heap allows also the merging of priority lists in constant time. 12] be an array such that T [i ] = i for each i <-12. this data structure allows the implementation of Dijkstra's algorithm in a time in the order of a+ n log n for finding the length of the shortest path from a designated source node to each of the other nodes of a graph with n nodes and a edges (Section 3. 12. the basic heap operations needed to implement a dynamic priority list can also be handled by data structures completely different from the heap we have considered so far.) 1. the preceding times for Fibonacci heaps are correct in the amortized sense-a concept not discussed here.. It is still possible to represent such heaps in an array without explicit pointers. 10) alter-heap (T. the Fibonacci heap (or lazy binomial queue) can process each insert-node. each object being in exactly one set at any given time. 5. Let T [I .5 Disjoint Set Structures Suppose we have N objects numbered from 1 to N.. and each delete-max in logarithmic time.9. 1. 6) alter-heap (T.2.2). but care is needed to do it correctly. 8) . For applications that have a tendency to sift down the (updated) root almost to the bottom level. find-max.2. At this point. such that the following sequence results in a different heap : m . For applications that need percolate more often than sift-down (Problems 3. (To be precise.. As an application. it pays to ignore temporarily the new value stored at the root. n ] containing distinct values. Problem 1. We wish to group these objects into disjoint sets. 1 Problem 1.2. Exhibit a heap T [I .12 and 3.3). an operation beyond the (efficient) reach of classic heaps. and to sift it all the way down to a leaf. In particular.2. The basic concept of a heap can be improved in several ways. In each set . put back the relevant value into the empty leaf and percolate it to its proper position.Preliminaries 30 Chap. treating this node as an empty location. n -1]. Experiments have shown this approach to yield an improvement in the classic heapsort algorithm (Problem 2.2. Exhibit the state of the array after each of the following procedure calls : make-heap (T) alter-heap (T.find-max (T) delete-max (T) insert-node (T [l . rather than two with the usual procedure (the children are compared to each other but not to their father).9.9.3. and percolate operation in constant time.13) it pays to have more than two children per internal node : this speeds up percolate (because the heap is shallower) at the cost of slowing down any operation that must consider every child at each level.

we execute a series of operations of two kinds : for a given object. N ].i We wish to know the time required to execute an arbitrary series of n operations of the type find and merge. we can represent each set as an "inverted" rooted tree. 16. To merge two sets. each containing exactly one object.10.a.9 Data Structures 31 we choose a canonical object. 51. we need now only change a single value in the array .6. starting from the initial situation.. b) {merges the sets labelled a and b } i f.9).7. it is harder to find the set to which an object belongs. j -b if i > j then exchange i and j fork E. A series of n operations therefore takes a time in the order of nN in the worst case. it suffices to place the label of the set corresponding to each object in the appropriate array element. Thereafter. which is necessarily the label for its set.Sec. The array 1 2 3 2 1 3 4 3 4 therefore represents the trees given in Figure 1. if set [i] =j# i. We adopt the following scheme : if set [i ] = i. which in turn represent the sets 11. then j is the parent of i in some tree.10) and (3. . (2.4.9. merge the two corresponding sets. find which set contains it and return the label of this set . on the other hand. 1. the N objects are in N different sets.8. function find 1(x) [ finds the label of the set containing object x } return set [x] procedure merge 1(a . and given two distinct labels. If we now declare an array set [ 1 .3. Suppose we decide to use the smallest member of each set as the label : thus the set (7. Still using a single array.91 will be called "set 3". then i is both the label of a set and the root of the corresponding tree . If consulting or modifying one element of an array counts as an elementary operation. We can do better than this. How can we represent this situation efficiently on a computer? One possible representation is obvious. it is clear that find l takes constant time and that mergel takes a time in the order of N. which will serve as a label for the set. Initially. The two operations can be implemented by two procedures.I to N do if set [k] = j then set [k] F.

Using this technique. b) [ merges the sets labelled a and b } if a < b then set [b] .b Problem 1. then after an arbitrary sequence of merge operations starting from the initial situation. 1 Figure 1. we may find ourselves confronted by a tree of height k. the trees do not grow as rapidly. Prove by mathematical induction that if this tactic is used. prove that the time needed to execute an arbitrary sequence of n operations find2 or merge2 starting from the initial situation is in the order of n 2 in the worst case. Tree representation for disjoint sets. it would be better to arrange matters so that it is always the root of the tree whose height is least that becomes a child of the other root. If each consultation or modification of an array element counts as an elementary operation.5. In this way the height of the resulting merged tree will be max(h 1. we have chosen arbitrarily to use the smallest member of a set as its label. (The figure shows the direction of the pointers in the array.4.Preliminaries 32 Chap. not of the edges in the tree. The problem arises because after k calls on merge2. Let us therefore try to limit the height of the trees produced. . or h 1+1 if h 1=h2.) function find2(x) { finds the label of the set containing object x } i -x while set [i ] 91. When we merge two trees whose heights are respectively h 1 and h2.set [i ] return i procedure merge2(a. we have not gained anything over the use of findl and mergel. In the case when n is comparable to N. a tree containing k nodes will have a height at most Llg kJ . so that each subsequent call on find2 takes a time in the order of k.9.10. Problem 1.i do i .9.a else set [a] .9. h 2) if h 1 #h2. So far.

9. we can make our operations faster still. This technique is called path compression. Once we know the root.11. we can now traverse the same edges again. procedure merge3(a.b Problem 1.9 33 The height of the trees can be maintained in an additional array height [1 . 1. modifying each node encountered on the way to set its pointer directly to the root. N] so that height [i] gives the height of node i in its current tree.9. Initially. the result is the tree of 4 11 4 9 10 1 8 11 9 1 G )a (b) after 20 12 21 16 (a) before Figure 1. Path compression.height [a] + 1 set [b] <. when we execute the operation find (20) on the tree of Figure 1.Data Structures Sec.1 la. Prove that the time needed to execute an arbitrary sequence of n operations find2 and merge3 starting from the initial situation is in the order of n log n in the worst case. height [a] therefore gives the height of the corresponding tree.6.9. When we are trying to determine the set that contains a certain object x. b) { merges the sets labelled a and b we suppose that a # b } if height [a] = height [b] then height [a] .a else if height [a] > height [b] then set [b] F. 20 16 . height [i] is set to zero for each i. For example.a else set [a] F. we first traverse the edges of the tree leading up from x to the root. By modifying find2. Whenever a is the label of a set.. The procedure find2 is still relevant but we must modify merge accordingly.

2. when we use this combination of an array and of the procedures find3 and merge3 to deal with disjoint sets of objects. we say we are using a disjoint set structure. On the other hand.x while i # r do j E.set [i] set [i ] . ** Problem 1. and 9. The pointers of the remaining nodes have not changed. the new find operation takes about twice as long as before.9. Analyse the combined efficiency of find3 together with your merge4 from the previous problem. Problem 1.10. A canonical object has no parent. Is path compression a good idea? The answer is given when we analyse it in Example 2.Preliminaries 34 Chap.) . 10. it is no longer true that the height of a tree whose root is a is given by height [a].9. now point directly to the root.9.r i F-j return r From now on.1lb: nodes 20. Path compression does not change the number of nodes in a tree.7. We call this value the rank of the tree. A second possible tactic for merging two sets is to ensure that the root of the tree containing the smaller number of nodes becomes the child of the other root. Our function becomes function find3(x ) { finds the label of the set containing object x } r Fx while set [r ] r do r . This technique obviously tends to diminish the height of a tree and thus to accelerate subsequent find operations. Problem 1.9. and change the name of the array accordingly. this remains an upper bound on the height. However. Write a procedure merge4 to implement this tactic.9. which lay on the path from node 20 to the root. Use this remark to implement a disjoint set structure that uses only one length N array rather than the two set and rank. and give a result corresponding to the one in Problem 1. so that it is easy to store this value exactly (whereas we could not efficiently keep track of the exact height of a tree after path compression). (Hint : use negative values for the ranks. Using path compression.9.set { r is the root of the tree } [r ] i <-. and we make no use of the rank of any object that is not canonical.8.5. 1 Figure 1.

Although we do not use any specific programming language in the present book.10 REFERENCES AND FURTHER READING We distinguish between three kinds of books on algorithm design. Hop- croft. however. which offers experimental proof that intelligent use of algorithmics may allow a TRs-80 to run rings round a CRAY-1. and Ullman (1974). we are aware of two more works on algorithmics : Horowitz and Sahni (1978) and Stinson (1985). and Ullman (1974). Gondran and Minoux (1979). a remarkable little French book of popular mathematics. To reinforce our remarks in Sections 1. 1973). For more information about data structures. searching. Fourier transforms. Even (1980). Brigham (1974). Graphs and trees are presented from a mathematical standpoint in Berge (1958. 1984b. Hopcroft. we suggest that a reader unfamiliar with Pascal would do well to look at one of the numerous books on this language. Nievergelt. 1984c). The most complete collection of algorithms ever proposed is no doubt found in Knuth (1968. Problem 1. Horowitz and Sahni (1976). Papadimitriou and Steiglitz (1982). Borodin and Munro (1975). For an introduction to Fourier transforms. The heap was introduced as a data structure for sorting in . they are described in detail in Knuth (1973) . and Tarjan (1983). and so on. Besides our own book and Harel (1987). 1969. and Gonnet (1984). Nilsson (1971). AVL trees come from Adel'son-Vel'skii and Landis (1962) . see Knuth (1977) and Lewis and Papadimitriou (1978). Stone (1972). Reingold. read Demars (1981). Specific books will be referred to in the following chapters whenever they are relevant to our discussion .1 comes from Shamir (1979). Finally. We give more references on this subject in Chapter 9.10 References and Further Reading 35 1. we may mention.Sec. Lawler (1976). 1. such as Jensen and Wirth (1985) or Lecarme and Nebut (1985). 1970). Multiplication a la russe is described in Warusfel (1961). and Melhorn (1984a.5. Baase (1978). Several other well-known general books are worth mentioning : in chronological order Aho. Specific books cover algorithms that are useful in a particular application area : sorting. Aho. we encourage the reader to look at Bentley (1984). For a more popular account of algorithms. The excellent book of Harel (1987) takes a broader view at algorithmics and considers it as no less than "the spirit of computing". books on algorithmics concentrate on the techniques of algorithm design : they illustrate each technique by examples of algorithms taken from various applications areas. Hopcroft. Sedgewick (1983). consult Knuth (1968. and Deo (1977). Knuth (1973) contains a large number of sorting algorithms. The fast algorithm for calculating the Fibonacci sequence is explained in Gries and Levin (1980) and Urbanek (1980). General books cover several application areas : they give algorithms useful in each area.7. originally intended to consist of seven volumes. graph theory. Standish (1980). 2-3 trees come from Aho. The algorithm capable of calculating the determinant of an n x n matrix in a time in the order of n 2'81 is given in Strassen (1969) and Bunch and Hopcroft (1974). 1973).6 and 1. and Ullman (1983). Dromey (1982). Christofides (1975).

1976). Hopcroft. Fredman and Tarjan (1984). For ideas on building heaps faster. for more applications see Hopcroft and Karp (1971) and Aho. 1987). The improvements suggested at the end of the sub-section on heaps are described in Johnson (1975). Gonnet and Munro (1986). we give only some of the possible uses of disjoint set structures . 1 Williams (1964). consult McDiarmid and Reed (1987).36 Preliminaries Chap. . In this book. Carlsson (1986) also describes a data structure. that allows finding efficiently the largest and the smallest elements of a set. or deap. and Carlsson (1986. and Ullman (1974. which he calls the double-ended heap.

we talk about the order of f (n) even 37 .1 A Notation for "the order of " Let IN and IR represent the set of natural numbers (positive or zero) and the set of real numbers. of programming language. To this end. We define O (f (n)) = { t : IN -4 IR* I (3c E IR+) (2 no E IN) (b'n ? no) [t(n) ! cf (n)] 1. we now introduce formally the asymptotic notation that will be used throughout the present book. we say that t(n) is in the order of f (n) even if t(n) is negative or undefined for some values n < n 0.2 Analysing the Efficiency of Algorithms 2. The set { true. we allow ourselves to misuse the notation from time to time. 0 (f (n)) (read as "the order of f (n)") is the set of all functions t(n) bounded above by a positive real multiple of f (n). For convenience. and the set of nonnegative real numbers by 1R* (the latter being a nonstandard notation). provided that n is sufficiently large (greater than some threshold no).1. the set of strictly positive real numbers by IR+. respectively. We denote the set of strictly positive natural numbers by IN+. theoretical analyses of the efficiency of algorithms are carried out to within a multiplicative constant so as to take account of possible variations introduced by a change of implementation. Let f : IN -* JR* be an arbitrary function.1 ASYMPTOTIC NOTATION As we mentioned in Chapter 1. Similarly. 2. For instance. In other words. or of computer. false 1 of Boolean constants is denoted by 13.

since t(n) E O (t(n)). Problem 2. it is allowable to talk about the order of n / log n . we try to express the order of the algorithm's running time using the simplest possible function f such that t(n) E O (f (n)). n3EO(n2) iii. even though this function is not defined when n =0 or n = 1. in this case we must choose no sufficiently large to be sure that such behaviour does not happen for n ? no. Which of the following statements are true? Prove your answers. The principle of invariance mentioned in the previous chapter assures us that if some implementation of a given algorithm never takes more than t(n) seconds to solve an instance of size n. (n + I)! E 0 (n!) v. n2E0(n3) ii.1.1.3. Problem 2.Analysing the Efficiency of Algorithms 38 Chap. 2 ' ' Q ( 2 ) iv. for any function f : IN -+ IR*. Some implementation of a certain algorithm takes a time bounded above by t(n) = 3 seconds . 2 when f (n) is negative or undefined for a finite number of values of n . and it is correct to write n3-3n2-n -8 E 0(n). In particular. for any function f : IN -* IR*. For example. (Such behaviour is unlikely since t(n) decreases as n increases for n <. however.) Find the simplest possible function f : IN -4 lR* such that the algorithm takes a time in the order of f (n). f (n) E O (n) = [f (n)]2 E O (n 2) vi. it takes a time in the order of t (n) itself. then any other implementation of the same algorithm takes a time in the order of t(n) seconds. Prove that t(n) E O (f (n)). .333.1.l8n milliseconds + 27n2 microseconds to solve any instance of size n. f (n) E O (n) 2f (n) E O (2n) Prove that the following definition of 0 (f (n)) is equivalent to Problem 2. provided that f (n) is strictly positive for all n E IN: O(f(n))={t:IN -4 ft*1(3cjR)(b'nEIN)[t(n)<_cf(n)]}.1. i.2. We say that such an algorithm takes a time in the order of f (n) for any function f : IN -4 IR* such that t(n) E O (f (n)). the preceding one. In general.

0 (f (n)) = 0 (g(n)) if and only if f (n) E 0 (g(n)) and g(n) E O (f (n)). prove that 0 (f (n) + g(n)) = 0 (max(f (n).1. 2.1. we do have to ensure that f (n) and g(n) only take nonnegative values (possibly with a finite number of exceptions) to avoid false arguments like the following : 0 (n2)=O(n3+(n2-n3)) = 0 (max(n 3.1. However. Find two functions f and g : IN -* IN+ such f (n) 0 0 (g(n)) and g(n) 0 0 (f (n)). Prove that the relation " E 0 " is transitive : if f (n) E 0 (g(n)) and g(n) E 0 (h (n)). n3+3n2+n +8 E O(n3+(3n2+n +8)) = 0 (max(n 3. For arbitrary functions f and g : IN -4 IR*. however.Sec. prove that i.n <. n 2. then f (n) E 0 (h (n)).6. For arbitrary functions f and g : IN -> IR*. g(n))). where the sum and the maximum are to be taken pointwise. as suggested by the following exercises.4. 3n 2 + n + 8)) = 0 (n 3). Problem 2. even though it is often useful in practice. to allow us to conclude that n3-3n2-n -8 E O(n3) because . that Problem 2. Problem 2. The result of the preceding problem is useful for simplifying asymptotic calculations.5. 3n2 + n +8)#n 3 when 0 <. because the asymptotic notation only applies when n is sufficiently large. A little manipulation is sufficient. 0 (f (n)) C 0 (g(n)) if and only if f (n) E 0 (g(n)) but g(n) e 0 (f (n)). For instance.7.1 Asymptotic Notation 39 In other words. Prove your answer. then 0 (g(n)) c 0 (h (n)) This asymptotic notation provides a way to define a partial order on functions and consequently on the relative efficiency of different algorithms to solve a given problem. and ii. Conclude that if g(n) e 0 (h (n)). Problem 2. the threshold no is not necessary in principle.n 3)) = O (n 3). The last equality holds despite the fact that max(n 3.3.1.

n . Given f and g : IN -* R+. .9 to but that e O (log n). De l'Hopital's rule is often useful in order to apply the preceding problem.8)) = O(In3)=0 (n3).2. and n->- ii.3n 2 . then provided na- n-+ - that the domains off and g can be extended to some real interval [no. Let a be an arbitrary real constant. prove that i.1.1. n3-3n2-n .1. n i = 1+2+ Find the error in the following argument : +n E 0(1+2+ +n) = 0(max(1. lim f (n)/g(n) = 0 = O (f (n)) c O (g(n)) = O (g(n) ± f (n)). is never zero for x E [no. X-->- provided that this last limit exists. The notion of a limit is a powerful and versatile tool for comparing functions.1. . Here again. lim f (n)/g(n) E R+ 0 (f (n)) = 0 (g(n)). False arguments ofzthe kind illustrated in the following problem are also to be avoided. +oo).. Problem 2.8 is negative when 0 <.11.1. +oo) in such a way that the corresponding new functions f and g are differentiable on this interval and also that g'(x).. Use the relaProblem 2. Recall that if lim f (n) = lim g(n) = 0.1. or if both these limits are infinite.9. and iv.n <. it can happen that 0 (f (n)) C 0 (g(n)) when the limit of f (n)lg(n) does not exist as n tends to infinity and when it is also not true that 0 (g(n)) = 0 (g(n) -f(n)).40 Analysing the Efficiency of Algorithms Chap. Problem 2.6 is of no concern. Problem 2. but n-i- iii. it can happen that 0 (f (n)) = 0 (g(n)) although the limit of f (n)/g(n) does not exist as n tends to infinity. then lira f (n)lg(n) = lim f '(x)lg'(x). 0 < e < 1. prove that log n e 0 Use de l'Hopital's rule and Problems 2.8. the derivative of g(x).5 and 2.10. n)) = O(n). 2 n3-3n2-n -8 E 0(n3-3n2-n -8) = O(Zn3+(Zn3-3n2-n -8)) = 0(max(fn3. tions " c " and "_ " to put the orders of the following functions into a sequence : . the fact that n 3 .

1. a fundamental asymmetry between the notation 0 and Q. n 41 1+e. 2. there exists a real positive constant d such that the algorithm takes a time longer than . since they cannot take more time than the worst case. f (2n) 2f (n) + cg(n). and let c be a strictly positive real constant such that for every integer n i. for each sufficiently large n. provided n is sufficiently large. Q(f (n)).2 Other Asymptotic Notation The notation we have just seen is useful for estimating an upper limit on the time that some algorithm will take on a given instance. 2. f (n) e O (g(n)).R+ be two increasing functions. Let f and g : IN -. 4g(n) g(2n) <_ 8g(n) and ii. If an algorithm takes a time in 0 (f (n)) in the worst case.3. and n 2/ log n.1. Prove that iii.1. In a worst-case analysis there is. The following notation is proposed to this end : Q(f(n))= {t:IN --> IR* I (2ce1R+)(3noEIN)(Vn ?no)[t(n)?cf(n)] }. (n 2+8n +Iog3n)4. For arbitrary functions f and g : IN -a IR*. there exists a real positive constant c such that a time of cf (n) is sufficient for the algorithm to solve the worst instance of size n. Prove your answer.Sec. In other words. if an algorithm takes a time in Q(f (n)) in the worst case. These instances can all be taken care of as suggested by Problem 2. Assuming only a finite number of instances of each given size exists. n s. The following exercise brings out the symmetry between the notation 0 and Q. however. (1+E)" . On the other hand. by using a bigger constant. It is also sometimes interesting to estimate a lower limit on this time. there can thus be only a finite number of instances.1 Asymptotic Notation n log n. on which the algorithm takes a time greater than cf (n). which we read unimaginatively as omega of f (n) is the set of all the functions t(n) bounded below by a real positive multiple of f(n). * Problem 2.1. This time is obviously also sufficient to solve all the other instances of size n. Problem 2.12. all of size less than the threshold.13. prove that f (n) E O (g(n)) if and only if g(n) e Q(f (n)).

prove that the following statements are equivalent : i. despite the fact that a time in the order of n is sufficient to solve arbitrarily large instances in which the items are already sorted. For arbitrary functions f and g : IN -4 R*. f (n) E O(g(n)). prove that if f and g : IN -* IR Problem 2.1. H. We shall be happiest if.4. lim f (n)/g(n) a 1R+ = f (n) E O(g(n)).14. are two arbitrary functions. Insertion sort. 2 df (n) to solve the worst instance of size n. For this reason we introduce a final notation O(f(n)) = O(f(n)) n f (f(n)). Thus there can exist an infinity of instances for which the algorithm takes a time less than df (n). its execution time is bounded simultaneously both above and below by positive real multiples (possibly different) of the same function. The following problem shows that the O notation is no more powerful than the 0 notation for comparing the respective orders of two functions. na00 . Continuing Problem 2. for each sufficiently large n. then I.1. lim f (n)/g(n) = 0 = f (n) E O (g(n)) but f (n) 0 O(g(n)). d e R+)(3n0EIN)(Vn ? n0) [cg(n) 5 f(n) 5 dg(n)].15. and n-4iii. ii. Prove that f (n) E O(g(n)) if and only if (3c.42 Analysing the Efficiency of Algorithms Chap.9. called the exact order of f (n). which we saw in Section 1.= f (n) E f (g(n)) but f (n) 0 O(g(n)). Problem 2.1. 0 (f (n)) = 0 (g(n)). This in no way rules out the possibility that a much shorter time might suffice to solve some other instances of size n.16. lim f (n)/g(n) = +. Problem 2. and iii. when we analyse the asymptotic behaviour of an algorithm. O(f (n)) = O(g(n)).1. provides a typical example of such behaviour : it takes a time in S2(n 2) in the worst case.

noE N) (Vn?n0)(Vm ?mo)[t(m.n)l } Other generalizations are defined similarly. 2.0 such that n >_ no is not true. log. but 210g°n 0 O(21°g'n) if a # b.3. bother to specify the base of a logarithm in an asymptotic expression)..18. n > such that m >. there are in general an infinite number of pairs < m . Give an example of a function f : INxIN 1R+ such that O(f(m.m o and n >. For instance. This is explained by the fact that while there are never more than a finite number of values of n >. 0 2. We define O(f(m.nEIN)[t(m. we can manipulate the asymptotic notation using arithmetic operators.0 and n >. b > 1 (so that we generally do not iii. iv. 43 Prove the following assertions : i.1. This situation is typical of certain algorithms for problems involving graphs. In such cases the notion of the "size of the instance" that we have used so far may lose much of its meaning.1.3 Asymptotic Notation with Several Parameters It may happen when we analyse an algorithm that its execution time depends simultaneously on more than one parameter of the instance in question.1 Asymptotic Notation Problem 2. n (=_O(logb n) whatever the values of a.n)<_cf(m. iii.n))= [t:INxJ-4JR' I(3cElR+)(3mo. Let f : INxIN -* 1R* be an arbitrary function. r=i the hidden constant in the 0 notation may depend on the value of k).1. For this reason the asymptotic notation is generalized in a natural way to functions of several variables.n)l }.no are not both true. and n i -' E O(log n).1. it can happen that the thresholds mo and no are indispensable.Sec.17.n)5cf(m.1. where the time depends on both the number of vertices and the number of edges. n ik E O(nk+') for any given integer k >_ 0 (this works even for real k > -1.n))# {t:INx1N-4 1R* 1(3cER+)(`dm. There is nevertheless an essential difference between an asymptotic notation with only one parameter and one with several : unlike the result obtained in Problem 2. log(n!) E O(n log n).4 Operations on Asymptotic Notation To simplify some calculations. * Problem 2. v.0 yet such that m >. 0 (f (n))+O (g(n)) represents the set of functions . 2.

To understand the first notation. . we use X op a to denote X op Ida.O (n 2) mean. where Ida : N -f IR* is the constant function Ida (n) = a for every integer n. if a E IR*. If g is a function from IN into IR*. reserving the symbol "\' to denote set difference : A \ B = { x E A I x e B 1. [0 (f (n))]2 does not denote the set of pairs of functions chosen from the set 0 (f (n)). a genuinely ambiguous situation would arise : what would O (n 3) . Prove the following identities : i. Still. For instance. On the other hand. n x O(f (n)) denotes { t :IN -4 JR I (3g(n)EO(f(n)))(3noEIN)(Vn ?no) [t(n) = n xg(n)] }. if N is the set of vertices of a directed graph.44 Analysing the Efficiency of Algorithms Chap. If the symbol "-" were used to denote the differ- ence of two sets. Furthermore. Similarly. we stretch the notation by writing X op g to denote X op { g } . and all this theory of operations on sets is extended in the obvious way to operators other than binary. The hidden constants that multiply f (n) and g(n) may well be different. but this is of no consequence (see Problem 2.1.19(i)). Intuitively this represents the order of the time taken by an algorithm composed of a first stage taking a time in the order of f (n) followed by a second stage taking a time in the order of g(n). More formally. think of it as O(f (n)) exp { Id 2 } . where " exp " denotes the binary exponentiation operator and " Id 2 " is the constant function 1d2(n)=2 for all n . O(f (n))+O(g(n)) = O(f (n) + g(n)) = O(max(f (n). for example ? To avoid this ambiguity. Although they both denote the same set as 0([f this requires a proof (Problem 2. To belong to [O(f (n))]2. a function g(n) must be the pointwise square of (n)]2). then N x N denotes as usual the set of possible edges between these vertices. +O(f (n)). it suffices for g(n) to be the pointwise product of two possibly different functions. g(n))) = max(O(f (n)). In every case but one the context removes any potential ambiguity. We also use the symmetrical notation g op X and a op X. there is one case that must be treated cautiously. 2 obtained by adding pointwise any function in 0 (f (n)) to any function in 0 (g(n)). which is not at all the same as yj 1 O(f (n)) = O(f (n))+O(f (n)) + Problem 2. This notation occasionally conflicts with the usual mathematical conventions. some function in O(f (n)) . however. each a member of O(f (n)).19. Notice the subtle difference between [O(f (n))]2 and O(f (n)) x O(f (n)). Similarly.1. 0 (f (n)) x 0 (g(n)) does not denote the Cartesian product of O (f (n)) and 0 (g(n)). then "X op Y " denotes { t :IN IR* I (3f EX )(3gEY)(3n0EIN)(b'n>no)[t(n) = f(n) op g(n)] }. we use "-" only to denote arithmetic subtraction. if X and Y are sets of functions from IN into IR* and if op is any binary operator. to belong to O(f (n)) x O(f (n)). O(g(n))) . Let f and g be arbitrary functions from IN into IR*. including pointwise subtraction of functions.1.19(ii)).

Although this expression can be simplified. 2.1.X)[te0(f(n))] }. such as being a power of 2. Example 2.1 Asymptotic Notation 45 H.5 Conditional Asymptotic Notation Many algorithms are easier to analyse if initially we only consider instances whose size satisfies a certain condition. but [0(1)]n # 2°(") . Conditional asymptotic notation handles this situation. [I +0(1)]n = 2°(n). A function f : IN -* R* is .5. The notation S2(f (n) I P(n)) and O(f (n) I P(n)) is defined similarly. the natural way to express the execution time required by Dixon's integer factorization algorithm (Section 8. 2. Let X be a set of functions from IN into R*.1.12. iii.1.1. and iv. In other words. The principal reason for using this conditional notation is that it can generally be eliminated once it has been used to make the analysis of an algorithm easier. We define O(f(n) I P(n)) t :lN -4 R* I (3cER+)(3noeIN)(Vn ? no) [P(n) = t(n) <_ cf (n)] }.Sec. as is the notation with several parameters.R* I (3f (=. O([f(n)]2) _ [0(f(n))]2=O(. Let f : IN -4 R* be any function and let P : IN -* 113 be a predicate. possibly defined by some asymptotic notation. Now 0 (X) denotes LJ0(f(n)) = { t :IN . is the set of all functions t(n) bounded above by a real positive multiple of f (n) whenever n is sufficiently large and provided the condition P(n) holds. f(n) E Hi=o0(1) Another kind of operation on the asymptotic notation allows it to be nested. 0 (f (n) I P(n)).3) is 0(eo(mi )) where n is the value of the integer to be factorized. You probably used this idea for solving Problem 2. which we read as the order of f (n) when P(n).f(n))xO(f(n)). The notation 0 (f (n)) defined previously is thus equivalent to 0 (f (n) I P(n)) where P(n) is the predicate whose value is always true. 12(X) and O(X) are defined similarly.

First. 2 eventually nondecreasing if (3 n 0 E IN) ('d n > n 0) [f (n) <_ f (n + 1)]. S2. One might be tempted to claim that t(n) is eventually nondecreasing because such is obviously the case with n log n . 0 We illustrate this principle using an example suggested by the algorithm for merge sorting given in Section 4. By the induction particular. This argumentation is irrelevant and fallacious because the relation between t(n) and n log n has only been demonstrated thus far when n is a power of 2. Therefore In 1. and let t : IN --> IR* be an eventually nondecreasing function such that t(n) EX (f (n) I n is a power of b). Example 2. Give two specific examples to illustrate that the conditions "t(n) is eventually nondecreasing" and "f (bn ) E O (f (n))" are both necessary to obtain these results.3. in particular Problem 2. where X stands for one of 0. or e. Let t (n) be defined by the following equation : t(n) _ Ja ifn = 1 t(Ln/2J)+t([n/21)+bn otherwise. which implies by mathematical induction that (3 n0 E l) (`d n >_ n0) (V m >_ n) [f (n) <_ f (m)]. Furthermore.4. The following problem assembles these ideas. it satisfies the condition f (bn ) E O (f (n)). The techniques discussed in Section 2. Prove that t(n) E X (f (n)). The proof that (Vn >_ 1) [t(n) <_t(n + 1)] is by mathematical induction. The proof that t(n) is nondecreasing must use its recursive definition. we shall therefore in future simply refer to such functions as being smooth. if we only consider the cases when n is a power of 2. where a and b are arbitrary real positive constants. Let b ? 2 be any integer.Analysing the Efficiency of Algorithms 46 Chap. The presence of floors and ceilings makes this equation hard to analyse exactly. .2. It turns out that any function that is b-smooth for some integer b >_ 2 is also c-smooth for every integer c ? 2 (prove it!). t (Ln / 2j) t(n)=t(Ln/2j)+t([n/2])+bn St(L(n+1)12j)+t([(n+l)/21)+b(n+l)=t(n+l). as well as being eventually nondecreasing. Let n be greater than hypothesis. In order to apply the result of the previous problem to conclude that t(n) E O(n log n). let f : IN -4 IR* be a smooth func- tion. <_ t (L(n + l )/2]) and t ([n / 21) <_ t (((n + 1)12 ).20. if 1(n) E O(f (n)). * Problem 2.1.6. we need only show that t(n) is an eventually nondecreasing function and that n log n is smooth.3. A word of caution is important here. note that t (1) = a <_ 2(a +b) = t (2). assume that (V m < n) [t(m) <_ t (m + 1)]. prove that t (n) is also smooth. However.1. the equation becomes t (n) Ja ifn=l 2t (n / 2) + bn if n > I is a power of 2. Such a function is b-smooth if. Let b ? 2 be any integer. allow us to infer immediately that t (n) E O(n log n { n is a power of 2).

Coming back now to the function t(n) satisfying the preceding inequalities. More importantly. and simultaneously t(n) for some constants c. This change from the original inequalities to a parametrized equation is useful from two points of view. This could not have been done directly with the original inequalities.1.20. .1.2 in the preceding section.min {t7(n)/f(n) I n Sn0}). n log n)) = O (n log n). We immediately conclude that t(n) E O(f (n)) = O(n log n). and for appropriate initial functions t 1. 2. We saw in the previous section that f (n) E O(n log n). n log n is smooth because it is clearly eventually nondecreasing and 2n log(2n) = 2n (log2 + log n) = (21og2)n + 2n log n E O (n + n log n) = O (max(n.1 Asymptotic Notation 47 Finally. t2: IN -f IR+.20. It is easy to prove by mathematical induction that v <_ t(n)/f(n) <_ u for every integer n. More often we have to deal with inequalities such as t(n) < tI(n) ifn :5 no t (Ln / 2]) + t (Fn / 21) + cvi otherwise t2(n) ifn <_n0 t (Ln / 2]) + t ([n / 21) + do otherwise. Our asymptotic notation allows these constraints to be expressed succinctly as t(n) E t(Ln/2])+t([n121)+0(n). n 0 E IN. to generalize our results automatically to the case where n is an arbitrary integer.Sec. using the conditional asymptotic notation and the technique explained in Problem 2. which in turn prevents us from applying Problem 2.1. d E 1R+. it is convenient to convert them first to equalities. since they do not allow us to conclude that t(n) is eventually nondecreasing. however. 2.1. To this end. let u = max(c. it allows us to confine our analysis in the initial stages to the easier case where n is a power of 2. we do not always find ourselves faced with equations as precise as those in Example 2. define f : IN -* IR by f(n)= 1 ifn = 1 f (Ln / 2]) + f ([n / 21) + n otherwise. To solve such inequalities. max I t I (n) I n Sn0}) and v =min(d. Obviously it saves having to prove independently both t (n) c O (n log n) and t (n) E Q (n log n).6 Asymptotic Recurrences When analysing algorithms. It is then possible.

3. We now have therefore a new. We therefore try the partially specified induction hypothesis HI (n) according to which f (n) = an 2 + bn + c.1) is true for some n ? 1. In the case of the preceding example it would have been simpler to determine the values of a. Knowing that f (O) = 0. more complete hypothesis. Example 2.1) otherwise.7 Constructive Induction Mathematical induction is used primarily as a proof technique. By applying this technique.2.f (n) = ifn =0 n + f (n . By equating the coefficients of each power of n.b + c = c.1. and c are not yet known. and 2. which we continue to call HI (n) : f (n) = n 2/ 2 + n / 2 + c. Pretend for a moment that you do not know that f (n) = n (n + 1)/ 2. The technique of constructive induction consists of trying to prove this incomplete hypothesis by mathematical induction.4 and 2. and c by constructing three linear equations using the values of HI (n) forn =0. We begin with a simple example. 1. b. As we shall see in Examples 2. Now HI (0) says precisely that f (O) = a 02 + b 0 + c = c. If we are to conclude HI (n). Obviously f (n) = YJ%o i <_ J70n = n 2. the value of c being as yet unconstrained. mathematical induction is a tool sufficiently powerful to allow us to discover not merely the proof of a theorem. Let the function f : IN -> IN be defined by the following recurrence : 0 . We have just shown that if HI (n -1) is true for some n ? 1.1. It remains to establish the truth of HI (0) in order to conclude by mathematical induction that HI (n) is true for every integer n. Along the way we hope to pick up sufficient information about the constants to determine their values. Too often it is employed to prove assertions that seem to have been produced from nowhere like a rabbit out of a hat.1. we know that f(n)=n +f(n-1)=n +a(n-1)2+b(n-1)+c =an2+(1+b-2a)n +(a-b+c). However. it must be the case that f (n) = an 2 + bn + c. but also its exact statement. this technique of constructive induction is especially useful for solving certain recurrences that occur in the context of the analysis of algorithms. we obtain two nontrivial equations for our three unknowns : 1 + b . It is clear that f (n) = Yi"_o i.5. While the truth of these assertions is thus established.2a = b and a . we conclude that c = 0 and that f (n) = n 2/ 2 + n / 2 is true for every integer n . . we can simultaneously prove the truth of a partially specified assertion and discover the missing specifications thanks to which the assertion is correct. their origin remains mysterious. This hypothesis is partial in the sense that a.Analysing the Efficiency of Algorithms 48 Chap. 2 2. then so is HI (n). b. and so f (n) E O (n 2 ). Supposing that HI (n . From these it follows that a = b = '-f. but that you are looking for some such formula. This suggests that we formulate a hypothesis that f (n) might be a quadratic polynomial.

we now try to show that t (n) E O (n!) by proving the existence of a real positive constant v such that t(n) 5 vn! for every positive integer n. Thus we see that t(n) ? un! is always true. and thus that t (n) E U(M). The technique of constructive induction is useful in both cases. As usual. regardless of the value of u. b =! given that t(n) :. since we are only interested in establishing an upper bound on the quantity of interest. Although this equation is not easy to solve exactly. Since t (l) = a. we shall prove independently that t (n) E O (n!) and that t(n) E S2(n!). However.1. there exists a real positive constant u such that t (n) >.W. To establish this.1 Asymptotic Notation 49 c=0 a+b+c=1 4a +2b +c=3 Solving this system gives us immediately a = 2. For simplicity.1) ? u (n . and c = 0. It seems then that constructive induction has nothing to offer in this context. Thus once the constants are determined we must in any case follow this with a proof by mathematical!.u.1)! for some n > 1. that is. provided that t(n .u (n -1)! for some n > 1. it is sufficiently similar to the recurrence that characterizes the factorial (n! = n x(n -1)!) that it is natural to conjecture that t (n) E O(n!). Suppose by the partially specified induction hypothesis that t(n -1) <.1) otherwise. By definition of t(n). using this approach does not prove that f (n) = n 2/ 2 + n / 2.1)!. since nothing allows us to assert a priori that f (n) is in fact given by a quadratic polynomial.a. this allows us to affirm that t(n) = bn2 + nt(n -1) <. or perhaps even that the hypothesis to the effect that t(n) EO (n!) is false.bn2 + vn!. In order to conclude that t(n) >.4. Suppose by the partially specified induction hypothesis that t(n -1) >. 2. Taking u = a.bn2 + vn!. However our aim is to show that t(n) :.un! for every positive integer n. this is the same as saying that u <. we begin by proving that t(n)ES2(n!).un! for every positive integer n. an exact solution of the recurrence equations is generally unnecessary.1. Even the techniques we shall see in Section 2. we have established that t(n) ? an! for every positive integer n .3. In this setting constructive induction can be exploited to the hilt. Some recurrences are more difficult to solve than the one given in Example 2. it suffices to show that this is true for n = 1. Example 2. . Let the function t : IN+ -> IR+ be given by the recurrence t(n) _ (a ifn=1 jl bn z + nt(n .Sec. However. where a and b are arbitrary real positive constants. we know that t(n) = bn2 + nt(n -1) >. in the context of asymptotic notation. Encouraged by this success.bn2 + nu (n -1)! = bn2 + un! >.3 will prove insufficient on occasion. t (l) >.v(n . that is. Unfortunately no positive value of v allows us to conclude that t(n) <.

Since n /(n -2):5 3 for every n >.vn! . 11 The following problem is not so easy . independently of the value of! . defining t(n).a + 3b. we can apply the recurrence definition of t(n) to find t(2) = 4b + 2t(1) = 4b + 2a . Verify that a <.3.-w.3b. then so too is the induction hypothesis it allows us to use. you may wish to prove this assertion. When n = 2.v .22. 2 In fact. This idea may seem odd.v (n .c <. it is necessary and sufficient that v >. All that remains is to adjust the constant v to take care of the cases n <.21. This inequality holds if and only if n >. g : IN+ Let k E IN and a. solve exactly the recurrence * Problem 2. We may hope for success.3.w (n -1)) = vn! + ((b -w )n + w )n. it is necessary and sufficient that v >. on the grounds that if the statement to be proved is stronger. since t(n) <.w (n -1).vn! .bn 2 + n (v (n -1)! . If we are to conclude that t(n) <. which is stronger than the previous condition. provided that n >. it is possible to obtain the result we hoped for.1)! . Determine in terms of a and b the real positive constant c such that t(n) lim n-*= n! = c. If you got lost in the preceding argument.2v .2.6b . which we were unable to prove. it illustrates well the advantage obtained by using constructive induction.1.w (n -1) for some n > 1.3 and w > bn /(n -2). In particular. Suppose then by the partially specified induction hypothesis that t(n -1) < v (n -1)! . Using the definition of t(n). Problem 2.wn is a consequence of the hypothesis t(n -1) <.wn is a stronger statement than t(n) <. The conclusion from all this is that t(n) E 9(n!) since an! S t(n) S (a + 5b)n! . we use constructive induction to determine real positive constants v and w such that t(n) <.1. we may in particular choose w = 3b to ensure that t(n) <.a + 5b. Rather than trying to prove directly that t(n) <. which is now completely! .Analysing the Efficiency of Algorithms 50 Chap. by straightforward mathematical induction.wn. it is necessary and sufficient that (b -w )n + w <.3bn for every positive integer!. To complete Example 2. Let IR+ be the function defined by the recurrence .4. b E IR+ be arbitrary constants. we know that t(l) = a. thanks to which we were able to prove that t(n) E O(n!) without ever finding an exact expression for t(n). If we are to conclude that t(n) <. When n = 1. we conclude that t (n) = bn 2 + nt (n -1) <.1.!.wn for any positive integer n. To conclude that t(n) <.a + 5b. we may choose v = a + 5b.

a statement such as O(f (n)) + O(g(n)) = O(max(f (n). the meaning of "such-and-such an algorithm takes a time in 0 (n 2)" does not change since algorithms cannot take negative time. the notation becomes difficult to handle.23. 2.1. The second difference is less striking but more important. This corresponds more closely to our intuitive idea of what a lower bound on the performance of an algorithm should look like. We often find 0(. The most striking is the widespread use of statements such as n 2 = O (n 3) where we would write n 2 E O (n 3). 1 2.8 For Further Reading The notation used in this chapter is not universally accepted.n 2 E n 3 + O (n 2).13 is neatly avoided. Prove that g(n) e O(n!). Some authors define Q(f(n)) = { t:IN -> IR* I (3ce1R+)(VnoeIN)(3n ?no) [t(n)?cf(n)] I.1 Asymptotic Notation g(n) = 51 a ifn=1 bnk + ng (n . With this definition it suffices that there exist an infinite number of instances x that force some algorithm to take at least cf ( I x I) steps in order to conclude that this algorithm takes a time in Q (f (n)). since it can lead an incautious reader astray. The third difference concerns the definition of 0. Unfortunately.1) otherwise.Sec. one would write n 3 . 11 . Furthermore.1. Why ? When we want the equivalent of this definition of 0 (f (n)) we write ±O (f (n)). and because it makes O asymmetric. with this definition.IR 1 (3cclR+)(2noEIN)(Vn ? n0)[It(n)I <_ cf(n)] 1 where It(n)I denotes (here only) the absolute value of t(n). With this definition we say that the execution time of some algorithm is of the order of f (n) (or is 0 (f (n))) rather than saying it is in the order of f (n). Problem 2.f(n)) _ { t :IN . Of course. You may encounter three major differences in other books.1. Notice the quantifier reversal. Use of such "one-way equalities" (for one would not write O (n3) = n2) is hard to defend except on historical grounds. the asymmetry between 0 and Q noted after Problem 2. Using this definition. g(n))) is no longer true. On the other hand. in particular because Q thus defined is not transitive.

The next step is to simplify this function using the asymptotic notation as well as the techniques explained in Section 2. A similar analysis for the lower bound shows that in fact it takes a time in O(n 2). Consider the following algorithm (which is reminiscent of the countsort algorithm discussed in Section 10. one possible barometer is the test in the inner loop. Example 2. The time taken by each trip round the inner loop can be bounded above by a constant a. Choosing a barometer.Analysing the Efficiency of Algorithms 52 Chap.1)/ 2 times when n items are sorted. intuition.2. Details like the initialization of the loops are rarely considered explicitly.0 for i . from which we conclude that the algorithm takes a time in 0(n2). However. that such simplifications should not be made incautiously. The complete execution of the inner loop for a given value of i therefore takes at most a time b + a (n -i ). How much time does the algorithm take? .T[i] <. including the implicit control statements for this loop.2. Let s be the sum of the elements of T. and finally. It is largely a question of judgement.1): k E. where T is an array of n integers such that 0 <. and experience. The following example shows.i for every i <.n. the complete algorithm takes a time not greater than d + 7-11 [c + b + a (n -i )]. provided that the time taken to execute the chosen instruction can itself be bounded above by a constant. It is often sufficient to choose some instruction in the algorithm as a barometer and to count how many times this instruction is executed.1. where c is a third constant. 0 In this first example we gave all the details of our argument. One trip round the outer loop is therefore bounded above by c +b + a (n-i ). In the selection sort example. however. which is executed exactly n (n . When an algorithm includes several nested loops. there are cases where it is necessary to take account of the implicit control of the loops.4. for a fourth constant d. Selection sort. 2 2. We can simplify this expression to 2 n 2 + (b +c -a /2)n + (d -c -b ). involving summations or recurrences. Here are some examples. as is the case with selection sort. Often a first analysis gives rise to a complicated-looking function. any instruction of the inner loop can usually be used as a barometer.1 to T [i ] do k +-k +T[j] . This figure gives us the exact order of the execution time of the complete algorithm. Consider the selection sorting algorithm given in Section 1.3.2 ANALYSIS OF ALGORITHMS There is no magic formula for analysing the efficiency of an algorithm. where b is a second constant introduced to take account of the time spent initializing the loop. Most of the execution time is spent carrying out the instructions in the inner loop.2.1 to n do for j f. Example 2.

We use the comparison "x < T [ j ] " as our barometer. In this case s = [I J. Coming back to our algorithm. including the time spent on loop control. Thus the algorithm makes i -1 comparisons. With a little experience the same algorithm can be analysed more succinctly. as in the algorithm. Next. To this we must add n to take account of the control of the outer loop and of the fact that the inner loop is initialized n times. where the execution time often depends on both the number of vertices and the number of edges. where the constant b represents the initialization time for the loop. To execute the inner loop completely for a given value of i therefore takes a time b + aT [i]. T [i -2] .1. the algorithm clearly takes a time in 11(n) since each element of T is considered at least once. s )). Suppose that T [i] = 1 whenever i is a perfect square and that T [i] = 0 otherwise. As Section 1. since in this case we have to compare x to T [i . Finally. The number of comparisons carried out between elements is a good measure of the complexity of most sorting algorithms. (Here we do not count the implicit comparisons involved in control of the for loop.2. Problem 2. The worst case arises when x is less than T [ j ] for every j between 1 and i . we would conclude that the algorithm takes a time in the exact order of s. The total number of times it is executed is therefore E° T [i] = s times.2 Analysis of Algorithms 53 For each value of i the instruction " k . nor the comparison j>0".4. . The total time taken by the algorithm is therefore in O(n+s). as we shall see in Chapter 10. The time therefore depends on two independent parameters n and s and cannot be expressed as a function of just one of these variables. . where c is a new constant. the complete algorithm takes a time d + J:i" 1(c + b + aT [i]). However.1. Insertion sort.. Let x = T[i]. 2. The time that this algorithm takes to sort n elements depends on their original order. This time is not zero when T[i]=O. this expression yields (c +b )n + as + d . The detailed analysis of this algorithm is as follows. analyses of average behaviour are usually harder than analyses of the worst case. This can happen for .1 ] . given in Section 1. for yet another constant d. the time taken to execute one trip round the outer loop is c + b + aT [i ]. The problem arises because we can only neglect the time spent initializing and controlling the loops provided we include something each time the loop is executed. We now give an example of analysis of the average behaviour of an algorithm. T [ 1 ] before we leave the while loop because j = 0. and they presuppose that we know a priori the probability distribution of the instances to be solved. The instruction in the inner loop is executed exactly s times.4 points out.3.Sec. When simplified. A simple example is sufficient to convince us that this is not the case.7 says that we can express its execution time in asymptotic notation in two ways : O(n +s) or O(max(n . This situa- tion is typical of algorithms for handling graphs. If indeed this instruction could serve as a barometer.. Consider the insertion sorting algorithm Example 2.k + T [ j ] " is executed T [i ] times.) Suppose for a moment that i is fixed. Let a be the time taken by one trip round the inner loop.

we often have to evaluate the sum of arithmetic. and the first comparison x < T [j ] gets us out of the while loop. On the other hand. T [i -1].k <. T [2]. With probability 1/i .1) = n (n . Although the algorithm takes a time in c2(n2) both on the average and in the worst case. the n th term of the harmonic series.1. . geometric. and other series.1 comparisons will be carried out. the probability is 2/i that i . since this happens both when x < T [ 1 ] and when T [ 1 ] S x < T [2]. T[i] < T [i -1 ] is false at the outset. . The same probability applies to any given number of comparisons up to i-2 included.17. The total number of comparisons is therefore It 2(i . .. Notice that selection sorting systematically makes the same number of comparisons between elements that insertion sorting makes in the worst case.. To determine the time needed by the insertion sorting algorithm on the average.. a time in 0 (n) is sufficient for an infinite number of instances.1 . Here H" = E i "_ I i.i. the probability that T[i] is the kth largest element among T [1]. k k=I _ (i -1)(i+2) _ i + I _ I 2i 2 i These events are independent for different values of i. T [i] is 1/i because this happens for [n] (i -1)! (n -i )! = n!/i of the n! possible permutations of n elements. but this number is still in O(n2). If i and k are such that 1 <. T [2]. . T[i] can therefore be situated with equal probability in any position relative to the items T [I]. The average number of comparisons made for a given value of i is therefore i-2 ci = . suppose that the n items to be sorted are all distinct and that each permutation of these items has the same probability of occurrence. as shown in Problem 2. . In the worst case insertion sorting thus takes a time in O(n2). The average number of comparisons made by the algorithm when sorting n items is therefore " " Eci=z i=2 i+1 2 i=2 - -1 i n2+3n -H" 4 E 0(n2). For a given value of i. The insertion sorting algorithm makes on the average about half the number of comparisons that it makes in the worst case. is negligible compared to the dominant term n 2/4 because H" E O(log n).2(i-1)+ Y. 2 every value of i from 2 to n when the array is initially sorted in descending order..1)/ 2 E O(n 2).Analysing the Efficiency of Algorithms 54 Chap. When we analyse an algorithm.

I > lg (n / 8). Denote by j. so that this can be done in a time in 0 (n). therefore k >. if 1 < t <. = i. But it is impossible for k (and thus j ) to exceed n. Consider the "make-heap " algorithm given at the end of Section 1. notice first that for any k ? 0 2--. As a barometer we use the instructions in the repeat loop of the algorithm used to sift down a node.. But d = Llg (n / 2)] implies that d + 1 <.m. ? 2jr_1 for 1 < t <. the value of j after execution of the assignment "j . n. try to see how you might have discovered it for yourself. Consequently n >_Jm >2Jm-I?4Jm-2? .1). 55 Prove that for any positive integers n and d d k=o Rather than simply proving this formula by mathematical induction. This shows that j.lg n and d .2. 2.4: this algorithm constructs a heap starting from an array T of n items.E2 k lg(n/2k) < 2dllg(n/2d i=1 1) k=O (by Problem 2. Hence Ln/2J lg(n/i)<-3n From (*) we thus conclude that Ln / 2j + 3n trips round the repeat loop are enough to construct a heap. Moreover. X lg(n/i)<-2klg(n/2k). Since any algorithm for constructing a heap must look at each element of the array at least once. then at the end of the (t -1)st trip round the loop we had j # k . Making a heap.2 Analysis of Algorithms Problem 2.Sec. Let d = Llg (n / 2)]. Obviously j. which implies that m <.1.1 + lg (n / i ). The total number of trips round the repeat loop when constructing a heap can now be bounded above by Ln/2] (I +lg(n/i))..2. (*) To simplify this expression. we obtain our . i). Example 2. >2m-li Thus 2m n / i.2j.4.k " on the t th trip round the loop. Let m be the largest number of trips round the loop that can be caused by calling sift-down (T. Ln/2i d E lg(n/i) <. % =GA The interesting part of the sum (*) can therefore be decomposed into sections corresponding to powers of 2.m.2.9.

The techniques of Section 2. additions and multiplications are considered to be elementary operations. 1) { T is sorted) What is the order of the execution time required by this algorithm in the worst case ? * Problem 2.9.4 we saw another algorithm for making a heap (slow-make-heap ). Recursive calculation of determinants. Analyse the worst case for this algorithm and compare it to the algorithm analysed in Example 2. can be used to conclude that t (k) E O (2k). procedure heapsort (T [ 1 . in particular Example 2.1 (the right hand subtree could be of height k . Williams invented the heap to serve as the underlying data structure for the following sorting algorithm.4. For a given number of elements.2.2. 2 final result that the construction of a heap of size n can be carried out in a time in O(n). .3. most of the work done by the algorithm consists of calling itself recursively n times to work on (n -1) x (n -1) matrices.n. But a heap containing n elements is of height Llg nj. Assume k >.. Problem 2. We therefore ignore for the time being the problems posed by the fact that the size of the operands can become very large during the execution of the algorithm. The algorithm then sifts the root down a path whose length is at most k. When n is greater than 1. Find the exact order of the execution time for Williams's heapsort.2.3.. the algorithm first transforms each of the two subtrees attached to the root into heaps of height at most k .7.5.2. We thus obtain the asymptotic recurrence t (k) E 2t (k -1) +O (k). In our analysis. We now analyse the algorithm derived from the recursive definition of a determinant (Section 1. which is in O(n) since 2UgnJ <.56 Analysing the Efficiency of Algorithms Chap.2).3). which takes a time in the order of k in the worst case. i -1].3.4. In Section 1.2.5. Let t (k) stand for the time needed to build a heap of height at most k in the worst case. Let t(n) be the time taken by some implementation of this algorithm working on an n x n matrix. Problem 2.2. A different approach yields the same result. both in the worst case and on the average. Analysis of heapsort. hence it can be built in at most t (Llg n J) steps. In order to construct the heap. n ]) IT is an array to be sorted) make-heap (T) for i F n step -1 to 2 do exchange T [1] and T [i ] sift-down (T [1. what are the best and the worst ways to arrange the elements initially insofar as the execution time of the algorithm is concerned? Example 2.2.

then (n mod m)<m <-n/2. Analyse the algorithm again. let ni and mi be the values of n and m at the end of the i th trip round the loop.5. 2. The values of mi and ni are defined by the following equations for 1 <. it is always true that n modm <n12.22.2. and so Ln / m] = 1. Ifm<_n/2.5 supposes that the time needed to compute a determinant. If m > n / 2. Mk = 0 causes the algorithm to ter- minate and mi > 1 for every i < k.i <.2. This gives us the following asymptotic recurrence : t (n) E nt(n .6. . By Problem 2. Example 2.Sec.4).k.1) + O(n 3 ).2.m m -t return n We first show that for any two integers m and n such that n >_ in. For each integer i <.22 the algorithm therefore takes a time in O(n!) to calculate the determinant of an n x n matrix. which takes a time in O(n2) for each of the n recursive calls if we do this without thinking too much about it (but see Problem 2.5). By Problem 2. * Problem 2. In particular. taking account this time of the fact that the operands may become very large during execution of the algorithm. this does not affect the fact that the complete algorithm takes a time in O(n!). Let k be the number of trips round the loop made by the algorithm working on the instance < m. Assume that you know how to add two integers of size n in a time in O(n) and that you can multiply an integer of size m by an integer of size n in a time in O(mn). Analysis of Euclid's algorithm. is in O(n3). which means that n modm =n -m < n -n/2=n12. Show that this time can be reduced to O(n).2 Analysis of Algorithms 57 Besides this. n) while m > 0 do t . n >.6.1.1. where mo and no are the initial values of m and n : ni =mi_i mi =ni_i modmi_1.2. however.n mod m n <. Example 2.k. excluding the time taken by the recursive calls. Problem 2. function Euclid (m.7. then 1 <_ n / m < 2. Recall that Euclid's algorithm calculates the greatest common divisor of two integers (Section 1.2. the matrices for the recursive calls have to be set up and some other housekeeping done.

_1 <n1_1/2=m1_2/2 for every i >. 0(4") to calculate the Using constructive induction. .1. In conclusion. b.t(n) <. Prove that the algorithm fibl takes a time in even if we Problem 2.2d.2. Analysis of the algorithm fib2. 2 1.8.b. the recurrence looks so like the one used to define the Fibonacci sequence that it is tempting to suppose that t(n) must be in 0(f ). its size is in 0(n lg o) = E) (n).7. k = 2d + 1 <. for each i Chap.7. Suppose for the moment that k is odd. still not taking account of the large size of the operands involved. On the other hand. .8. > m. prove that af <. Analysis of the algorithm fibl. the number of trips round the loop made by Euclid's algorithm working on the integers in and n. . provided the instructions in the loop can be considered as elementary. and c such that a f. where 0 = (1+' )/ 2. take into account that we need a time in O(n) to add two integers of size n. Let t(n) be the time taken by some implementation of this algorithm working on the integer n. Define d by k = 2d + 1. The algorithm fib 1 therefore takes a time in n th term of the Fibonacci sequence. b. and therefore the time required by the algorithm. It is clear that the algorithm fib2 takes a time equal to a + bn on any instance n . we have mi =n.. . Then mk_1 < mk_3/2 < Mk-5/4 < .2. and so m o >. for appropriate constants a and b. We give without explanation the corresponding asymptotic recurrence: t(n) E t(n-1)+t(n-2)+O(1). Finally. and c.7.2.) Example 2.2.1 + 21g m o . and give values for these constants. * Problem 2.t (n) <. Prove that the worst case for Euclid's algorithm arises when we calculate the greatest common divisor of two consecutive numbers from the Fibonaccisequence.d. Once again. . Using the preceding observation. for appropriate constants a. However..9. (Since the value of f is in O(4"). are in 0 (log m).Analysing the Efficiency of Algorithms 58 Clearly n. < mo/2d.2. <. it is easy to use this technique to find three real positive constants a.c for any positive integer n.5._1 mod . remembering that m 1 = no mod m o < m 0.c Problem 2.2. The case when k is even is handled similarly. But recall that Mk-1 ? 1. We now analyse the algoExample 2. rithm fib l of Section 1. This time is therefore in 0(n). as in Example 2.4. constructive induction cannot be used directly to find a constant d such that t(n) <.

ab (2k -1) = abn 2 + (d + 2c +4a .. If you find the result disappointing. take the instructions in the while loop as our barometer. To evaluate the number of trips round the loop. which implies that the algorithm fib3 takes a time in 0 (log n). if we take account of the size of the operands involved? Let a be a constant such that the time to add two numbers of size n is bounded above by an. of course. Consequently nt 5 nr-1/2 5 nr-2/4 <_ .4ab) .1) for k >. It is obvious that nr = Lnr_1/ 2]<.3. Assume that addition of two integers of size n takes a time in O(n) and that multiplication of an integer of size n by an integer of size m takes a time in O(mn). The k th trip round the loop therefore consists of calculating fk_2 + fk _ 1 and fk . We conclude that the loop is executed at most m times.11. and so nn = 0. the time is bounded above by c +2a. Compare your result to that obtained in Example 2. Analysis of the algorithm fib3. 2.. It is easy to see by symmetry that the algorithm takes a time in 0(n 2). For each of the first two trips round the loop. Let d be an appropriate constant to account for necessary initializations. and let b be a constant such that the size of fn is bounded above by bn for every integer n ? 2.t <_ m.10. Notice first that the values of i and j at the beginning of the k th trip round the for loop are respectively fk_2 and fk. The preceding equation shows that nn 5 n/2m < 1. To see this. which is the condition for ending the loop. Prove that the execution time of the algorithm fib3 on an integer n is in @(log n) if no account is taken of the size of the operands. look back at the table at the end of Section 1. plus some constant time c to carry out the assignments and the loop control. Then the time taken by ffb2 on an integer n > 2 is bounded above by n d + 2(c + 2a) + Y.2.n12' .7.2.7 we shall see a multiplication algorithm that can be used to improve the performance of the algorithm fib3 (Problem 4. k=3 which is in 0 (n 2). ** Problem 2.2 Analysis of Algorithms 59 What happens. which takes a time bounded above by ab (2k . in particular n 1 = Ln / 2j.8. that of fib2 (why not?). But nn is a nonnegative integer. < n 1/ 2t-1 <.5 and remember that the hidden constants can have practical importance! In Section 4.1 (where we take f _1 = 1). Let m = I + LlgnJ. however. Example 2.7.5).n. Determine the exact order of the execution time of the algorithm fib3 used on an integer n. let n. Problem 2. be the value of n at the end of the t th trip . .-1/2 for every 2 <. but not. The analysis of fib3 is relatively easy if we do not take account of the size of the operands.Sec.

The array rank [ 1 . N ] keeps the meaning given to it in algorithms find3 and merge3: set [i ] gives the parent of node i in its tree. Finally.i while j# 0 do j E. We also introduce a strictly increasing function F : IN -* IN (specified later) and its "inverse" G : IN -* N defined by G (n) = min { m E IN I F (m) >_ n ). Example 2. time for the algorithm Answer the same question as in the preceding problem. Chap. N ] plays the role of height [ 1 . assignments. The algorithms become procedure init (initializes the trees) global <.2. Analysis of disjoint set structures.2. N ] in algorithm merge3 : rank [i] denotes the rank of node i (see Section 1. The analysis of these algorithms is the most complicated case we shall see in this book. We begin by introducing a counter called global and a new array cost [ 1 ...Analysing the Efficiency of Algorithms 60 Problem 2. except when set [i ] = i.9.12.2. Find the exact order of its execution time.. define the group of an element of rank r as G (r).i while j is odd do j F.j div 2 Supposing that integer division by 2. which indicates that i is the root of its tree. Their purpose will be explained later.i rank [i] 0 cost [i ] .5). it is clear that this algorithm takes a time in U(n) n 0 (n log n). The array set [1 .1 to N do set [i] F.5.. Prove your answer.13.j div 2 Show a relationship between this algorithm and the act of counting from 0 to n + 1 in binary. It can happen that the analysis of an algorithm is facilitated by the addition of extra instructions and counters that have nothing to do with the execution of the algorithm proper.0 .0 fori F.10. For instance. this is so when we look at the algorithms find3 and merge3 used to handle the disjoint set structures introduced in Section 1.9. 2 Consider the following algorithm : for i F-0ton do j F. this for i -0ton do j F. * Problem 2. N J. and loop control can all be carried out at unit cost.

is in N O (N + n + global + cost [i ]).2 61 function find (x) { finds the label of the set containing object x I r -x while set [r r do r F. The time required for a call on the procedure merge can be bounded above by a constant. b) { merges the sets labelled a and b we suppose that a b [ if rank [a] = rank [b] then rank [a] t. 2. the following remarks are relevant : 1. including initialization.Analysis of Algorithms Sec.cost [i ] + 1 j . . In order to obtain an upper bound on these values. Therefore the total time required to execute an arbitrary sequence of n calls on find and merge.a else if rank [a] > rank [b] then set [b] F a else set (a] b With these modifications the time taken by a call on the procedure find can be reckoned to be in the order of I plus the increase of global + EN cost [i] occasioned by the call.x while i * r do if G (rank [i]) < G (rank [set [i]]) or r = set [i] then global E. 2.r i-j return r procedure merge (a.set [i ] set [i] F. once an element ceases to be the root of a tree.rank [a] + 1 set[b] f. it never becomes a root thereafter and its rank no longer changes .global + 1 else cost [i] F.set [r] { r is the root of the tree) i E. the rank of a node that is not a root is always strictly less than the rank of its parent . i=1 where global and cost [i] refer to the final values of these variables after execution of the sequence.

Analysing the Efficiency of Algorithms


Chap. 2

3. the rank of an element never exceeds the logarithm (to the base 2) of the number
of elements in the corresponding tree ;

4. at every moment and for every value of k, there are not more than N i 2k elements of rank k ; and

5. at no time does the rank of an element exceed L1gNJ, nor does its group ever
exceed G ( Llg NJ).

Remarks (1) and (2) are obvious if one simply looks at the algorithms. Remark

(3) has a simple proof by mathematical induction, which we leave to the reader.
Remark (5) derives directly from remark (4). To prove the latter, define subk (i) for
each element i and rank k : if node i never attains rank k, subk (i) is the empty set ; oth-

erwise sub, (i) is the set of nodes that are in the tree whose root is i at that precise
moment when the rank of i becomes k. (Note that i is necessarily a root at that
moment, by remark (1).) By remark (3), subk (i) # 0
# subk (i) ? 2k . By remark
(2), i # j
subk (i) n subk (j) = 0. Hence, if there were more than N/ 2k elements i
such that subk (i) 0, there would have to be more than N elements in all, which
proves remark (4).

The fact that G is nondecreasing allows us to conclude, using remarks (2) and
(5), that the increase in the value of global caused by a call on the procedure find
cannot exceed 1 + G ( LlgNJ). Consequently, after the execution of a sequence of n
operations, the final value of this variable is in 0 (1 + nG (L1g NJ)). It only remains to
find an upper bound on the final value of cost [i] for each element i in terms of its final

Note first that cost [i ] remains at zero while i is a root. What is more, the value
of cost [i ] only increases when a path compression causes the parent of node i to be
changed. In this case the rank of the new parent is necessarily greater than the rank of
the old parent by remark (2). But the increase in cost [i ] stops as soon as i becomes
the child of a node whose group is greater than its own. Let r be the rank of i at the
instant when i stops being a root, should this occur. By remark (1) this rank does not
change subsequently. Using all the preceding observations, we see that cost [i ] cannot
increase more than F (G (r)) - F (G (r) -1) -1 times. We conclude from this that the
final value of cost [i] is less than F (G (r)) for every node i (=-final (r), where final (r)
denotes the set of elements that cease to be a root when they have rank r >- 1 (while,
on the other hand, cost [i] remains at zero for those elements that never cease to be a
root or that do so when they have rank zero). Let K = G (Llg NJ) - 1. The rest is
merely manipulation.



cost [i ] = Y

E cost [i ]

g=0 r=F(g)+1 iefina((r)




I F (G (r))

g=0 r=F(g)+I iefinal(r)

Analysis of Algorithms

Sec. 2.2




E (N12`)F(g+1)

<_ 1

g=O r=F(g)+l


It suffices therefore to put F (g + 1) = 2F(g) to balance global and 1"' cost [i] and so
to obtain E 7l cost [i] NG (Llg NJ). The time taken by the sequence of n calls on
find and merge with a universe of N elements, including the initialization time, is
therefore in

O(N +n +global +

cost[i]) c O(N +n +nG(L1gNJ)+NG(L1gNJ))

= O(max(N,n)(l +G(L1gN]))).
Now that we have decided that F (g + 1) = 2F(g), with the initial condition
F (0) = 0, what can we say about the function G ? This function, which is often
denoted by lg*, can be defined by

G(N)=lg*N =min{k I lglg ...lgN <0)
k times

The function lg* increases very slowly : lg*N <- 5 for every N 5 65,536 and
lg*N <- 6 for every N < 265,536 Notice also that lg*N - lg*(L1gN j) 2, so that
lg*(Llg NJ) E O(lg*N ). The algorithms that we have just analysed can therefore execute a sequence of n calls on find and merge with a universe of N elements in a time
in 0 (n lg*N ), provided n >_ N, which is to most intents and purposes linear.
This bound can be improved by refining the argument in a way too complex to
give here. We content ourselves with mentioning that the exact analysis involves the
use of Ackermann's function (Problem 5.8.7) and that the time taken by the algorithm
is not linear in the worst case.







Figure 2.2.1


The towers of Hanoi.



Analysing the Efficiency of Algorithms


Example 2.2.11. The towers of Hanoi.

Chap. 2

It is said that after creating the

world, God set on Earth three rods made of diamond and 64 rings of gold. These rings
are all different in size. At the creation they were threaded on one of the rods in order

of size, the largest at the bottom and the smallest at the top. God also created a
monastery close by the rods. The monks' task in life is to transfer all the rings onto
another rod. The only operation permitted consists of moving a single ring from one
rod to another, in such a way that no ring is ever placed on top of another smaller one.
When the monks have finished their task, according to the legend, the world will come
to an end. This is probably the most reassuring prophecy ever made concerning the
end of the world, for if the monks manage to move one ring per second, working night
and day without ever resting nor ever making a mistake, their work will still not be
finished 500,000 million years after they began!

The problem can obviously be generalized to an arbitrary number of rings. For
example, with n = 3, we obtain the solution given in Figure 2.2.1. To solve the general problem, we need only realize that to transfer the m smallest rings from rod i to
rod j (where 1 <- i <- 3, 1 <- j <- 3, i # j, and m ? 1), we can first transfer the smallest
m -1 rings from rod i to rod 6 - i -j, next transfer the m th ring from rod i to rod j,

and finally retransfer the m -1 smallest rings from rod 6 - i -j to rod j. Here is a
formal description of this algorithm ; to solve the original instance, all you have to do
(!) is to call it with the arguments (64, 1, 2).

procedure Hanoi(m, i, j)
(moves the m smallest rings from rod i to rod j }

ifm > 0 then Hanoi(m -1, i, 6 - i -j)
write i "-*" j
Hanoi(m - 1, 6 - i j, j)
To analyse the execution time of this algorithm, let us see how often the instruction write, which we use as a barometer, is executed. The answer is a function of m,
which we denote e (m). We obtain the following recurrence :


2e(m-1)+1 ifm > 1,

from which we find that e (m) = 2" - 1 (see Example 2.3.4). The algorithm therefore
takes a time in the exact order of 2" to solve the problem with n rings.
Problem 2.2.14. Prove that the algorithm of Example 2.2.11 is optimal in the
sense that it is impossible with the given constraints to move n rings from one rod to
another in less than 2" - 1 operations.
* Problem 2.2.15.
Give a nonrecursive algorithm to solve this problem. (It is
cheating simply to rewrite the above algorithm using an explicit stack to simulate the
recursive calls.)

Sec. 2.3

Solving Recurrences Using the Characteristic Equation



We have seen that the indispensable last step when analysing an algorithm is often to
solve a system of recurrences. With a little experience and intuition such recurrences
can often be solved by intelligent guesswork. This approach, which we do not illustrate here, generally proceeds in four stages : calculate the first few values of the
recurrence, look for regularity, guess a suitable general form, and finally, prove by
mathematical induction that this form is correct. Fortunately there exists a technique
that can be used to solve certain classes of recurrence almost automatically.

2.3.1 Homogeneous Recurrences
Our starting point is the resolution of homogeneous linear recurrences with constant
coefficients, that is, recurrences of the form

aptn +altn-1 + ... +aktn-k =0



i. the t, are the values we are looking for. The recurrence is linear because it does
not contain terms of the form t, ti+j , t, 2, and so on ;
ii. the coefficients a, are constants ; and
iii. the recurrence is homogeneous because the linear combination of the t, is equal
to zero.

After a while intuition may suggest we look for a solution of the form


where x is a constant as yet unknown. If we try this solution in (*), we obtain


+a1xn-I + ... +akx"-k = O.

This equation is satisfied if x = 0, a trivial solution of no interest, or else if


+alxk-I + ... +ak =0.

This equation of degree k in x is called the characteristic equation of the recurrence

rk of this characteristic
Suppose f o r the time being that the k roots r I , r2r 2 ,-.
equation are all distinct (they could be complex numbers). It is then easy to verify that
any linear combination

t" _



ci rin

Analysing the Efficiency of Algorithms


Chap. 2

of terms rrn is a solution of the recurrence (*), where the k constants c 1 , c 22 ,-.- Ck
are determined by the initial conditions. (We need exactly k initial conditions to determine the values of these k constants.) The remarkable fact, which we do not prove
here, is that (*) has only solutions of this form.
Example 2.3.1.

Consider the recurrence

n >2

to -3tn-1 -4tn_2=0
subject to to = 0, t i = 1.
The characteristic equation of the recurrence is

x2-3x -4=0
whose roots are -1 and 4. The general solution therefore has the form
=C1(-1)n +C24n.


The initial conditions give

c1+ c2=0
that is, c 1 =- 5. C2 =

n =0



We finally obtain

[4n - (-1)n j.

to =

Example 2.3.2. Fibonacci.

Consider the recurrence

to = to-1 + to-2

n >2

subject to to = 0, t 1 = 1.

(This is the definition of the Fibonacci sequence ; see Section 1.7.5.)

The recurrence can be rewritten in the form to - to-1 - to-2 = 0, so the characteristic equation is

x2-x - 1 =0
whose roots are

2 and


The general solution is therefore of the form

to =Cir +c2r2
The initial conditions give

Sec. 2.3

Solving Recurrences Using the Characteristic Equation

c1+ c2=0

n =0


n =1


from which it is easy to obtain




Thus t" = 5 (r i - r2" ). To show that this is the same as the result obtained by De
Moivre mentioned in Section 1.7.5, we need to note only that r I = 0 and r2 = - 0-I.

* Problem 2.3.1.

Consider the recurrence

t" =2t"_I -2tn-2

n >2

subject to to = 0, t I = 1.

Prove that t, = 2'12sin(n it/4), not by mathematical induction but by using the
characteristic equation.

Now suppose that the roots of the characteristic equation are not all distinct. Let



I + ...

+ ak

be the polynomial in the characteristic equation, and let r be a multiple root. For every
n >- k, consider the n th degree polynomial defined by
h (x) = x [x" -k p (x)]' = a o n x" + a 1(n - 1)x"- I +

+ ak (n -k


Let q (x) be the polynomial such that p (x) = (x -r )2q (x). We have that
h (x) = x [(x-r )2 x


q (x)]' = x [2(x-r

)x"-k q (x) + (x-r

)2[x"-k q (x)]7

In particular, h (r) = 0. This shows that

aonr" +ai(n-1)r"-I+

+ak(n-k)rn-k =0,

that is, to = nr" is also a solution of (*). More generally, if m is the multiplicity of the
root r, then t" = r" , t" = nr" , t" = n 2r"
t" = n m - I r n are all possible solutions
of (*). The general solution is a linear combination of these terms and of the terms
contributed by the other roots of the characteristic equation. Once again there are k
constants to be determined by the initial conditions.
Example 2.3.3.

Consider the recurrence

tn =5tn_I-Stn-2+4tn-3
subject to to = 0, t I = 1, t2 = 2.
The recurrence can be written

n >_3

Analysing the Efficiency of Algorithms


Chap. 2

to - Stn -I + 8tn-2 - 4tn-3 = 0
and so the characteristic equation is

x3-5x2+8x -4=0
or (x-1)(x-2)2 = 0.
The roots are 1 (of multiplicity 1) and 2 (of multiplicity 2). The general solution
is therefore
to =C11n +C22n +C3n2n.

The initial conditions give


n =0

c1+ C2

n =2

from which we find c 1 = -2, c 2 = 2, c 3 = - z . Therefore
to = 2n +1

- n2n- - 2.

2.3.2 Inhomogeneous Recurrences
We now consider recurrences of a slightly more general form.

aotn +a1tn-1+ ... +aktn-k = b n p(n)

( ** )

The left-hand side is the same as (*), but on the right-hand side we have bn p (n),

i. b is a constant ; and
H. p (n) is a polynomial in n of degree d.
For example, the recurrence might be

tn -2tn_1=3n.

In this case b = 3 and p (n) = 1, a polynomial of degree 0. A little manipulation
allows us to reduce this example to the form (*). To see this, we first multiply the
recurrence by 3, obtaining
3tn - 6tn - 1 =

3n + 1

If we replace n by n + 1 in the original recurrence, we get

Finally, subtracting these two equations, we have

The number of movements of a ring required in the Towers of Hanoi problem (see Example 2.4. Adding these three equations. Once again. replace n in the recurrence by n + 1 and then multiply by . which can be solved by the method of Section 2.6. +ak)(x-b)d+1=0. Once this equation is obtained. Example 2.1. (x-2)(x-3) = 0. Intuitively we can see that the factor (x-2) corresponds to the left-hand side of the original recurrence. Generalizing this approach. Here is a second example.-I = (n+5)3n The necessary manipulation is a little more complicated : we must a.2tn+l = -6th+I + 12th (n+5)3n+2 (n+7)3' 2 = -6(n+6)3n+1 . obtaining respectively 9th .Sec. we obtain th+2 . we can see that the factor (x-2) comes from the left-hand side of the original recurrence.3. (x-2)(x-3)2 = 0.11) is given by .18tn_I = to+2 .2. The characteristic equation is x2-5x +6=0 that is..3 Solving Recurrences Using the Characteristic Equation 69 to+1 -5th +6th_I =0. whereas the factor (x-3)2 is the result of our manipulation. replace n in the recurrence by n +2. we can show that to solve (**) it is sufficient to take the following characteristic equation : (aoxk+alxk-I+ .18tH-I = 0.3.2t. proceed as in the homogeneous case. multiply the recurrence by 9 b. 2.8th+1 + 21tn .. and c. whereas the factor (x-3) has appeared as a result of our manipulation to get rid of the right-hand side. The characteristic equation of this new recurrence is X3-8X2+21x-18=0 that is. th .

The characteristic equation is therefore (x-2)(x-1) = 0 where the factor (x-2) comes from the left-hand side and the factor (x-1) comes from the right-hand side. since clearly to >. to find a second initial condition we use the recurrence itself to calculate t1=2to+I=1. 2 n>?1 subject to to = 0. the number of movements of a ring required.Analysing the Efficiency of Algorithms 70 tn=2tn_1+I Chap.n. once we know that to = C I In + C22n we can already conclude that to E 8(2n). to We need two initial conditions. a polynomial of degree 0. we find 1 = to . We finally have c1+ c2=0 n =0 c1+2c2=1 n=1 from which we obtain the solution to=2n-1. which is of the form (**) with b = 1 and p (n) = 1. We know that to = 0. -I = 1. In the previous example. it is therefore always the case that c 1 must be equal to -1. For this it is sufficient to notice that to .2tn -I =c1+C22n -2(c1+C22n-') Whatever the initial condition. so the general solution of the recurrence is =Ciln +C22n. Therefore C2 > 0. Substituting the general solution back into the original recurrence. In fact we can obtain a little more. . The roots of this equation are 1 and 2. If all we want is the order of to .2t.. there is no need to calculate the constants in the general solution. The recurrence can be written to . is certainly neither negative nor a constant. and the conclusion follows.

A further generalization of the same type of argument allows us finally to solve recurrences of the form aotn +aitn .3.6. Example 2. Conclude that all the interesting solutions of the recurrence must have c i > 0.. . prove that in the preceding example C2 = -2 and C3 = -1 whatever the initial condition. and to solve the problem as before. +ak)(X-b1)d'+I(x-b2)°.1 subject to to = 0.2tn -I = n. The general solution is to =c12" +C2In +C3nIn. If this is so.Sec.0 for every n.. . 2. and hence that they are all in O(2"). The recurrence can be written t -2tn_I =n +2n. (***) where the b.. we can conclude immediately that to must be in 0 (2").3.5. (n) are polynomials in n respectively of degree d. .3. The characteristic equation is therefore (x-2)(x-1)2 = 0 with roots 2 (multiplicity 1) and 1 (multiplicity 2). +aktn-k = b?p1(n)+bnP2(n)+ .I+ .2. we are always looking for a solution where to >. Consider the recurrence to = 2tn -I + n. It suffices to write the characteristic equation (aoxk +aIxk . This can be written to . which is of the form (**) with b = 1 and p(n) = n. =0 which contains one factor corresponding to the left-hand side and one factor corresponding to each term on the right-hand side.I + .3 Solving Recurrences Using the Characteristic Equation 71 Problem 2. are distinct constants and the p. . . on the contrary! Why? Example 2. a polynomial of degree 1.3. In the problems that interest us.3. Problem 2. By substituting the general solution back into the recurrence.+I ... There is nothing surprising in the fact that we can determine one of the constants in the general solution without looking at the initial condition . Solve tn =2tn_I+n +2" n >.

7.3. Here is how we can find the order of T (n) if n is a power of 2 and if T(n)=4T(n12)+n n > 1. =-2-n +2n+1+n2". In the following examples we write T (n) for the term of a general recurrence. t2 = 12. and p2(n) is of degree 0. b2 = 2.4. The general solution of the recurrence is therefore of the form to =C1I" +C2nI" +C32" +Cgn2". Using the recurrence. p2(n) = 1. pi(n) = n . we can calculate t 1 = 3. 2 which is of the form (***) with b 1 = 1. The degree of p 1(n) is If the characteristic equation of the recurrence (***) is of degree m =k +(d1+1)+(d2+1)+ . Example 2.1) + 2k . The characteristic equation is (x-2)(x-1)2(x-2) = 0. This can be written . C3 and C4 from =0 n =0 c1+ c2+2c3+ 2c4= 3 n=1 c1 +2c2+4c3+ 8c4= 12 c1+3c2+8c3+24c4=35 n =2 n =3 + c3 Cl arriving finally at t. Problem 2. We could obviously have concluded that t o e O (n 2") without calculating the constants. C 2 . both of multiplicity 2. then the general solution contains m constants C 1 . regardless of the initial condition. How many constraints on these constants can be obtained without using the initial conditions ? (See Problems 2. and tk for the term of a new recurrence obtained by a change of variable. cm . Problem 2.Analysing the Efficiency of Algorithms 72 Chap. t3 = 35. We can now determine C1 .4.3 Change of Variable It is sometimes possible to solve more complicated recurrences by making a change of variable.3. Replace n by 2k (so that k = lg n) to obtain T (2k) = 4T(2'.3.3 and 2. which has roots 1 and 2.3.) 2. Prove that all the solutions of this recurrence are in fact in O(n 2"). C2.

The characteristic equation is (x-4)2 = 0. Here is how to find the order of T (n) if n is a power of 2 and if T(n)=4T(n/2)+n2 n > 1. Example 2. Here is how to find the order of T (n) if n is a power of 2 and if T (n) = 2T (n / 2) + n lg n n > 1. T (n) E 0 (n log2n I n is a power of 2). Hence. we obtain T(2k)=2T(2k-1)+k2k tk =2tk-I+k2k The characteristic equation is (x-2)3 = 0. We know how to solve this new recurrence : the characteristic equation is (x-4)(x-2) = 0 and hence tk = c 14k + c 22k . Example 2.9. 2. Thus T (n) e O (n 2log n I n is a power of 2). As before. and so tk =c14k +c2k4k T(n) =c1n2+C2n2lgn.3. .8. T (n) is therefore in 0 (n 2 1 n is a power of 2). we obtain successively T(2k)=4T(2k-1)+4k tk =4tk-I +4k.3 Solving Recurrences Using the Characteristic Equation 73 tk =4tk-1+2k if tk = T (2k) = T (n). and so tk =c12k +C2k2k +C3k22k T(n)=cin +c2nlgn +c3nlg2n. Putting n back instead of k. Proceeding in the same way. we find T(n)=C1n2+C2n.Sec.3.

3. 0 Finally.20 since the functions n 2.3.3. Remark.3. This follows from problem 2. n log2n and n I g 3 are smooth. It is therefore inevitable that the solution obtained should be in conditional asymptotic notation. T (n) = 3T(n12) + cn We obtain successively T(2k)=3T(2k-l)+c2k tk =3tk_i+c2k. In each of these four cases. 2 We want to find the order of T (n) if n is a power of 2 and if (c is constant. whereas a and c are positive real numbers. The characteristic equation is (x-3)(x-2) = 0. it is sufficient to add the condition that T (n) is eventually nondecreasing to be able to conclude that the asymptotic results obtained apply unconditionally for all values of n. Chap.0 are integers.10. T (n) E O (n 'g 3 I n is a power of 2).Analysing the Efficiency of Algorithms 74 Example 2.1.13. Let T : IN -) 1R+ be an eventually nondecreasing function such that T(n)=aT(n/b)+cnk n > no when n In 0 is a power of b. however. n = 2k > 1).3. * Problem 2. Show that the exact order of T (n) is given by T (n) E O(nk) ifa < hk O(n k log n) ifa = b k 'ogn a ifa > b k O(n ) Rather than proving this result by constructive induction. and so tk = c 13k + C22k T(n)=c13'g" +c2n and hence since algb = b1g° T(n)=cInIg3+c2n. In Examples 2. . b >.6. obtain it using the techniques of the characteristic equation and change of variable. The constants no ? 1.10 the recurrence given for T (n) only applies when n is a power of 2.2 and k >. n 2 log n .7 to 2. This result is generalized in Problem 2.

4 Range Transformations When we make a change of variable. Problem 2.3 Solving Recurrences Using the Characteristic Equation Problem 2. none of the techniques we have seen applies to this recurrence since it is not linear. 75 Solve the following recurrence exactly for n a power of 2: T(n)=2T(n/2)+lgn n >>-2 subject to T(1) = 1.3.Sec. and hence Vk = (3 + lg 3)2k . one of the coefficients is not constant. Finally. To transform the range. We give just one example of this approach.7. and c 3 =-J. = 3 + 21g 3. It is sometimes useful to transform the range instead in order to obtain something of the form (***). V. and so Vk =c12k +c2lk +c3klk. 2. At first glance. We want to solve n>1 T (n) = n T 2(n / 2) subject to T (l) = 6 for the case when n is a power of 2. and V2 = 8 + 41g 3 we obtain c i = 3 + lg 3. 2. The characteristic equation is (x-2)(x-1)2 = 0.8. which gives k>0 tk =2ktk2 I subject to to = 6. and furthermore. Express your solution as simply as possible using the 0 notation.3. which yields k>0 -I subject to V() = lg 6. The first step is a change of variable : put tk = T (2k ). Express your solution as simply as possible using the O notation. we create a new recurrence by putting Vk = lg tk . we transform the domain of the recurrence. using tk = 2t'" and T(n) = tlgn .k .2. we obtain 23n-23n T(n)= n . Solve the following recurrence exactly for n of the form )+lgn 22A n >_4 subject to T(2) = 1. From VO= I + lg 3. C2=-2.3.

make the change of variable T '(n) = T (n+2). Hint : observe that T (n) < 3 T (1 + [n / 21) + cn for n > n 0. 2 2.5 Supplementary Problems Problem 2.15n + 106 for 0!5 n :5 2.6. Define X={n ENIlogb(n/no)EIN}={nENNI(3iEIN)[n=nob`I). whereas a and d are real positive constants. Consider any constants c E R+ and no E N.3. Problem 2. Let f : X .3. and 4.12.7.Analysing the Efficiency of Algorithms 76 Chap.2 are integers.1. (Multiplication of large integers: see Sections 1.11.3. In some cases. Express your answer as simply as possible using the 0 notation.2. Problem 2. Let T AN 4 R* be an eventually nondecreasing function such that T(n) ST(Ln/2])+T([n/2])+T(1+[n/21)+cn n > no. Problem 2.10.3. a more general result is required (and the technique of the characteristic equation does not always apply).R* be an arbitrary function. use Example 2.13. The constants no >-1 and b >. Express your answer as simply as possible using the O notation. Solve the following recurrence exactly : tn =tn_I +tn_3-tn_4 n >>>4 subject to to = n for 0 5 n < 3. * Problem 2.20 to conclude for T (n). Solve the following recurrence exactly for n a power of 2: T (n) = 5T (n / 2) + (n lg n)2 n >.3. 4. Express your answer as simply as possible using the 0 notation. Recurrences arising from the analysis of divide-and-conquer algorithms (Chapter 4) can usually be handled by Problem 2. Prove that T (n) E O (n ig 3 ).2tn-3 n >3 subject to to = 9n 2 .1.3.n>no .3. Solve the following recurrence exactly : to = to -1 + 2tn-2 . Define the function T : X -* R* by the recurrence T(n)= d if n =no aT (nlb)+ f(n) if nEX. however.9.2 subject to T(1) = 1.7).10 to solve for T '(n) when n is a power of 2.3. and use problem 2.

14.) T (n) = E a' f (n l b') i=o ii. Solve the following recurrence exactly : t = 1/(4-tn_1) n>1 subject tot 1 = 1/4. the value of T (n) is given by a simple summation when n EX logn(n In. 2. T (n) e O(n P) whenever f (n) E O (n' ) for some real constant r < p . If we set f (n o) = d (which is of no consequence for the definition of T ). If you disprove it. Problem 2. The last alternative can be generalized to include cases such as f (n) E O(n P+q log n) or f (n) E O(n P+q / log n) .3. find the simplest but most general additional constraint on f (n) that q 1 <_ q 2 such that f (n) E O (n P (log n)q' 1) and f (n) E Q(n P (log n)' suffices to imply T (n) E O(f (n) log n). n >2 . Note that the third alternative includes f (n) E O(nP) by choosing q=1. all asymptotic notation is implicitly conditional on n E=X.3.3 Solving Recurrences Using the Characteristic Equation 77 Let p = loge a . It turns out that the simplest way to express T (n) in asymptotic notation depends on how f (n) compares to n P . Solve the following recurrence exactly as a function of the Problem 2. As a special case of the first alternative. then T (n) E < O(no) if f(n) E O(nP/(logn)1+q ) O(f (n) log n log log n) if f (n) E O(n P / log n) O(f(n)logn) if f(n) E O(nP(log n)q Off (n)) if f (n) E O(n p+q) 1) . W. Prove or disprove that the third alternative can be generalized as follows : T(n) E O(f (n) log n) whenever there exist two strictly positive real constants ). Let q be any strictly positive real constant .15. Prove that i. T (I) = b. In what follows. initial conditions a and b : T(n+2)=(1+T(n+l))/T(n) subject to T (O) = a.Sec. we also get T (n) E O(f (n)) if there exist a function g : X -* IR* and a real constant a> a such that f(n)EO(g(n)) and g(bn)>-ag(n) for all v. iv.

n?3 El 2.2. . consult any book on mathematical analysis.2. Rudin (1953).2. The analysis of disjoint set structures given in Example 2. Example 2. Chap. For a more rigorous mathematical treatment see Knuth (1968) or Purdom and Brown (1985). 1983).15. The paper by Bentley. The main mathematical aspects of the analysis of algorithms can also be found in Greene and Knuth (1981).3. Knuth (1976) gives an account of its history and proposes a standard form for it.1. and Saxe (1980) is particularly relevant for recurrences occurring from the analysis of divide-and-conquer algorithms (see Chapter 4).10 is adapted from Hopcroft and Ullman (1973).1. Several techniques for solving recurrences. 2 Solve the following recurrence exactly : T(n)='T(n/2)-zT(n/4)-1/n subject to T (1) = 1 and T (2) = 3/2. including the characteristic equation and change of variable.1 corresponds to the algorithm of Dixon (1981). The more precise analysis making use of Ackermann's function can be found in Tarjan (1975. Haken.16. For information on calculating limits and on de l'Hopital's rule. The book by Purdom and Brown (1985) presents a number of techniques for ana- lysing algorithms. Problem 2.3 comes from Williams (1964). are explained in Lueker (1980). Buneman and Levy (1980) and Dewdney (1984) give a solution to Problem 2.4 REFERENCES AND FURTHER READING The asymptotic notation has existed for some while in mathematics: see Bachmann (1894) and de Bruijn (1961). for instance.20 are introduced by Brassard (1985).Analysing the Efficiency of Algorithms 78 Problem 2. who also suggests that "one-way inequalities" should be abandoned in favour of a notation based on sets. Conditional asymptotic notation and its use in Problem 2.

find the shortest route in a graph. the length of the path we have found. this is the function we are trying to optimize. a function that checks whether a particular set of candidates provides a solution to our problem. as the case may be) the value of the 79 . a selection function that indicates at any time which is the most promising of the candidates not yet used . or whatever . a function that checks whether a set of candidates is feasible.1 INTRODUCTION Greedy algorithms are usually quite simple. the nodes of the graph. and so on. They are typically used to solve optimization problems : find the best order to execute a certain set of jobs on a computer. and an objective function that gives the value of a solution (the time needed to execute all the jobs in the given order. In the most common situation we have a set (or a list) of candidates : the jobs to be executed.3 Greedy Algorithms 3. ignoring questions of optimality for the time being . whether or not it is possible to complete the set in such a way as to obtain at least one solution (not necessarily optimal) to our problem (we usually expect that the problem has at least one solution making use of candidates from the set initially available). we look for a set of candidates constituting a solution that optimizes (minimizes or maximizes. and so on). that is. the set of candidates that have already been used . To solve our optimization problem.

The selection function is usually based on the objective function . without worrying about the future. we shall see in the following examples that at times there may be several plausible selection functions. 3 objective function. then the candidate we just added stays in the set of chosen candidates from now on. and containing at least one coin of each type .Greedy Algorithms 80 Chap. they may even be identical. we check whether the set now constitutes a solution to our problem. it is there for good . the procedure chooses the best morsel it can swallow. We want to give change to a customer using the smallest posExample 3.0 (S is a set in which we construct the solution) while not solution (S) and C # 0 do x .1. so that we have to choose the right one if we want our algorithm to work properly. Each time we enlarge the set of chosen candidates. representing for instance 1. if the enlarged set is still feasible. When a greedy algorithm works correctly. If the enlarged set of chosen candidates is no longer feasible. 10. the first solution found in this way is always optimal. 5. a solution : the total value of the chosen set of coins is exactly the amount we have to pay . Initially. and the objective function : the number of coins used in the solution. our choice being guided by the selection function. and 25 units. The elements of the problem are the candidates: a finite set of coins. It never changes its mind : once a candidate is included in the solution. we try to add to this set the best remaining candidate. once a candidate is excluded from the solution. A greedy algorithm proceeds step by step. the candidate we tried and removed is never considered again. 0 .an element of C maximizing select(x) C-C\{x} if feasible (S u {x }) then S .1. sible number of coins. we remove the candidate we just added . the set of chosen candidates is empty. However.S u {x } if solution (S) then return S else return "there are no solutions" It is easy to see why such algorithms are called "greedy": at every step. Then at each step. it is never reconsidered. function greedy (C : set) : set { C is the set of all the candidates) S f. a feasible set : the total value of the chosen set does not exceed the amount to be paid the selection function : choose the highest-valued coin remaining in the set of candidates. However.

if the nodes of G represent towns. T > is called a minimal spanning tree for the graph G. Prove that with the values suggested for the coins in the preceding example the greedy algorithm will always find an optimal solution provided one exists. Prove that the partial graph < N. Show that it can even happen that the greedy algorithm fails to find a solution at all despite the fact that one exists. Prove. b } is the cost of building a road from a to b. Finally. We give two greedy algorithms to solve this problem. Each edge has a given non-negative length. In particular. Lemma 3. Obviously. In the terminology we have used for greedy algorithms.1. and it is feasible if it does not include a cycle.1 Minimal Spanning Trees Let G = < N. A > be a connected undirected graph where the length of each edge is given.2 GREEDY ALGORITHMS AND GRAPHS 3.1. Let G = < N.2. and the cost of an edge (a.1. by giving specific counterexamples.) Problem 3. This problem has many applications. then a minimal spanning tree of G shows us how to construct at the lowest possible cost a road system linking all the towns in question.2 Greedy Algorithms and Graphs 81 * Problem 3. we can associate a cost to each edge. The graph < N.1. The following lemma is crucial for proving the correctness of the forthcoming algorithms. an edge touches a given set of nodes if exactly one end of the edge is in the set. (Instead of talking about length. It is obviously more efficient to reject all the remaining 25-unit coins (say) at once when the remaining amount to be represented falls below this value. The problem is to find a subset T of the edges of G such that all the nodes remain connected when only the edges in T are used. Using integer division is also more efficient than proceeding by successive subtractions. that the greedy algorithm no longer gives an optimal solution in every case if there also exist 12-unit coins. A > be a connected undirected graph where N is the set of nodes and A is the set of edges. and the sum of the lengths of the edges in T is as small as possible. For instance. Moreover. Let . Let B C N be a strict subset of the nodes of G.2.Sec. or if one type of coin is missing from the initial set. 3. a set of edges is a solution if it constitutes a spanning tree. the empty set is always promising since G is connected. T > formed by the nodes of G and the edges in T is a tree. 3. a feasible set of edges is promising if it can be completed so as to form an optimal solution. on the other hand.2. In this case the problem is to find a subset T whose total cost is as small as possible. this change of terminology does not affect the way we solve the problem.

that also touches B (otherwise the cycle could not close . Therefore U' is also a minimal spanning tree. The set T of edges is initially empty. Otherwise when we add the edge e to U. If e e U. cannot be in T. (Initially. Kruskal's algorithm. so that T is then a minimal spanning tree for all the nodes of G.Greedy Algorithms 82 Chap. Proof. edges are added to T. As the algorithm progresses.2. At the end of the algorithm only one connected component remains. Let e be the shortest edge that touches B (or any one of the shortest if ties exist). N\B B Figure 3.1). we note that T c U' because the edge e'.see Figure 3. Each edge is either included in the set that will eventually form the solution or eliminated from further consideration.) The elements of T that are included in a given connected component form a minimal spanning tree for the nodes in this component. At every instant the partial graph formed by the nodes of G and the edges in T consists of several connected components. Let U be a minimal spanning tree of G such that T E_ U (such a U must exist since T is promising by assumption). which touches B. there is nothing to prove. Then T v f e } is promising. we create exactly one cycle (this is one of the properties of a tree). e'. and it includes e. when T is empty. each node of G forms a distinct trivial connected component. the total length of the edges in U' does not exceed the total length in U. since e touches B. The main difference between the various greedy algorithms to solve this problem lies in the order in which the edges are selected.2. 3 T c A be a promising set of edges such that no edge in T touches B. A greedy algorithm selects the edges one by one in some given order. there necessarily exists at least one other edge. say. But since the length of e is by definition no greater than the length of e'. the cycle disappears and we obtain a new tree U' that spans G. . In this cycle. A cycle is created if we add edge e to U. To complete the proof. If we now remove e'. The initial set of candidates is the set of all the edges.1.

Prove that Kruskal's algorithm works correctly. (2.2. (1. Otherwise the edge is rejected : it joins two nodes in the same connected component and cannot therefore be added to T without forming a cycle since the edges in T form a minimal spanning tree for each component. (5. (3. .5). A graph and its minimal spanning tree.7) 6 t2.5).2.4. we examine the edges of G in order of increasing length.5.2.7) (1. 5) .4). The algorithm stops when only one connected component remains. 12. is by induction on the number of edges selected until now. The algorithm proceeds as follows.7).2.2.5} (6.4} (1.3} (41 (5} (6) 171 (1. If an edge joins two nodes in different connected components. 2) . El Figure 3. its total length is 1 2 T contains the chosen edges (1.2. 3. and (4. which uses lemma 3.3) (4.3} (4.5) rejected 7 14. consider the graph in figure 3. the two connected components now form only one component. Problem 3. Step Edge considered Initialization - Connected components (11 (2) (3) (4) 151 (6) f7) 11.51 (61 (7) 4 16. This minimal spanning tree is shown by the heavy lines in Figure 3.3. (5.2. 14. we add it to T.5) (1. (2.3} (4.5 } . 3 } .71.6}.4).2. 31. and consequently.7).71 5 (1.5} (6. The proof. (6. (4.2.2. (2. 13. (4.7) (1. (1.6.21 11.2).7).2.2 Greedy Algorithms and Graphs 83 To build bigger and bigger connected components.21 (3} (4} 151 (61 (71 3 (2.71.4}. In increasing order of length the edges are : (1. (6. To illustrate how this algorithm works.

Here is the algorithm. which tells us in which component the node x is to be found.2. function Kruskal (G = < N.1. B) to merge two disjoint sets.v}} until #T = n -1 return T Problem 3. A graph may have several different minimal spanning trees. by mistake. vcomp) T -T u {{u. and at worst. On a graph with n nodes and a edges the number of operations is in 0 (a log a) to sort the edges. where is this possibility reflected in the algo- 0 rithm ? To implement the algorithm.2.Greedy Algorithms 84 Chap.find (u ) vcomp f. each containing one distinct element of N { greedy loop } repeat { u . v) f. which is equivalent to 0 (a log n) since n-1 <-a Sn(n-1)/2.5). we run the algorithm on a graph We can estimate the execution time of the algorithm as follows. We conclude that the total time for the algorithm is in 0 (a log n) because 0 (lg*n) c 0 (log n). 0 (n) to initialize the n disjoint sets . A > : graph . 3 Problem 3. O (a) for the remaining operations. in the worst case 0 ((2a+n -1)lg*n) for all the find and merge operations.9. We have to carry out rapidly the two operations find (x). We therefore use disjoint set structures (Section 1. For a connected graph we know that a >.3. since there are at most 2a find operations and n -1 merge operations on a universe containing n elements .find (v) if ucomp # vcomp then merge (ucomp.n . we have to handle a certain number of sets : the nodes in each connected component.0 1 will contain the edges of the minimal spanning tree) initialize n sets.shortest edge not yet considered ucomp F-.4. For this algorithm it is preferable to represent the graph as a vector of edges with their associated lengths rather than as a matrix of distances.2. Is this the case in our example. length : A -i IR*) : set of edges 1 initialization) Sort A by increasing length n -#N T . and merge (A . that is not connected? What happens if.10. Although this does not . and if so. by the analysis given in example 2.

v J such that u E N \ B and v E B .2. leaving to the algorithm the job of working out which edges exist ? Prim's algorithm. although each search for a minimum in the repeat loop will now take a time in 0 (log a) = 0 (log n).here the heap property should be inverted so that the value of each internal node is less than or equal to the values of its children). is by induction on the number of nodes in B. Here is an informal statement of the algorithm. 3. it is preferable to keep the edges in a heap (Section 1.2. We continue thus as long as B # N. In such cases.4 .6.2. The proof.0 (will contain the edges of the minimal spanning tree } B (-. function Prim (G = < N. There results a forest of trees that grows somewhat haphazardly. In Prim's algorithm.2.(an arbitrary member of N } while B # N do find { u . consider once again the graph in Figure 3. v) to T . In this way the edges in T form at any instant a minimal spanning tree for the nodes in B. A > : graph .Sec. Prove that Prim's algorithm works correctly. length : A -> IR*) : set of edges ( initialization) T F. We arbitrarily choose node I as the starting node.9.5. In Kruskal's algorithm we choose promising edges without worrying too much about their connection to previously chosen edges. the user supplies a matrix of distances. and the set T of edges is empty. starting from an arbitrary root.2 Greedy Algorithms and Graphs 85 change the worst-case analysis. This is particularly advantageous in cases when the minimal spanning tree is found at a moment when a considerable number of edges remain to be tried. the set B of nodes contains a single arbitrary node. Initially. rithm if. At each step Prim's algorithm looks for the shortest possible edge { u . on the other hand.v}} B F-B U (u) return T Problem 3. It then adds u to B and (u. the minimal spanning tree grows "naturally". the original algorithm wastes time sorting all these useless edges. which again uses Lemma 3. What can you say about the time required by Kruskal's algoProblem 3. This allows the initialization to be carried out in a time in 0 (a).1. At each stage we add a new branch to the tree already constructed. . and the algorithm stops when all the nodes have been reached. v } of minimum length such that u E N \ B and v E B T 6-T u {(u. except that we are careful never to form a cycle. To illustrate how the algorithm works. instead of providing a list of edges.2.



3 gives the length of each directed edge : L [i.. The modifications to the algorithm are simple : Figure 3. j ] ? 0 if the edge (i..2.some element of C minimizing D [v ] C (-. D would not change if we did one more iteration to remove the last element of C.20. n.i] { greedy loop } repeat n -2 times v .min(D[w]. 10] Initialization Clearly.. where P [v ] contains the number of the node that precedes v in the shortest path.31 3 3 {2) [ times..30.30. .4. If we want not only to know the length of the shortest paths but also where they pass.w]) return D The algorithm proceeds as follows on the graph in Figure 3.20.S u { v } } for each w e C do D[w] f. Here is the algorithm. n ] (initialization) C F.{ 2.10] 1 5 (2. n ) { S = N \ C exists only by implication ) for i -2tondoD[i]*-L[l.100.3.Greedy Algorithms 88 Chap. which is why the main loop is only repeated n .3. function Dijkstra (L [1 . 10] (35.2. To find the complete path.. Step V - C D (2. n ]) : array[2.5) [50. j ] = 00 otherwise.30.3 A directed graph.10] 2 4 12. j) exists and L [i.4) [50. . .C \ { v) { and implicitly S . simply follow the pointers P backwards from a destination to the source. it suffices to add a second array P [2.. D [v] +L[v. 3. I . n ].3.

We therefore have to verify that the shortest path from the source to v does not pass through a node that does not belong to S. consider the inductive step.D[v] +L[v. part (i) of the induction remains true.2 89 initialize P [i] to 1 for i = 2. In the latter case it seems at first . the first node encountered that does not belong to S is some node x distinct from v (see Figure 3. Problem 3. or else it now passes through v.w] P[w]-v. it will now belong to S. if a node i is not in S. then D [i] gives the length of the shortest special path from the source to i. Proof of correctness.3. Suppose the contrary : when we follow the shortest path from the source to v. there are two possibilities for the shortest special path from the source to w : either it does not change.w] then D[w] <-. . the base for our induction is thus obtained.2. if a node i is in S. is a special path. We must therefore check that D [v ] gives the length of the shortest path from the source to v. replace the contents of the inner for loop by ifD[w]>D[v]+L[v. We have thus verified that when v is added to S. We prove by mathematical induction that i.10. Consequently. Consider now a node w 0 S different from v When v is added to S. 3. . then D [i] gives the length of the shortest path from the source to i H. as far as x. The initial section of the path. ii. This follows immediately from the induction hypothesis for each node i that was already in S before the addition of v. Next.2. I. By the induction hypothesis D [v ] certainly gives the length of the shortest special path. n .. Look at the initialization of D and S to convince yourself that these two conditions hold at the outset . Show how the modified algorithm works on the graph of Figure 3. .Greedy Algorithms and Graphs Sec. and suppose by the induction hypothesis that these two conditions hold just before we add a new node v to S. the total distance to v via x is distance to x (since edge lengths are non-negative) D [x] (by part (ii) of the induction) (because the algorithm chose v before x) D [v] and the path via x cannot be shorter than the special path leading to v. 3.4).2. As for node v..

we need only note that when its execution stops all the nodes but one are in S (even though the set S is not constructed explicitly). because D [x] <_ D [v].11. glance that there are again two possibilities : either v is the last node in S visited before arriving at w or it is not.5) that v is visited. then Dijkstra's algorithm does not always work correctly. but not just before arriving at w : a path of this type cannot be shorter than the path of length D [x] + L [x. However. At this point it is clear that the shortest path from the source to the remaining node is a special path. w] that we examined at a previous step when x was added to S. Is it still sensible to talk about shortest paths if negative distances are allowed? S Figure 3.2. can be negative. The shortest path from the source to v cannot go through node x. We have to compare explicitly the length of the old special path leading to w and the length of the special path that visits v just before arriving at w . we can ignore the possibility (see Figure 3. To complete the proof that the algorithm works.Greedy Algorithms 90 Chap. 3 The shortest path The shortest special path Figure 3. Thus the algorithm ensures that part (ii) of the induction also remains true when a new node v is added to S.5 The shortest path from the source to w cannot visit x between v and w. . the algorithm does this. Show by giving an explicit example that if the edge lengths Problem 3.

we have to remove the root of the heap exactly n . . n -2. a > n -1 and this time is in 0 (a log n). we must modify D [w] and percolate w up the heap. but how are we to avoid taking a time in c(n2) to determine in succession the n -2 values taken by v ? The answer is to use a heap containing one node for each element v of C. 2 values of D on successive iterations. This is interesting when we remember that eliminating the root has for effect to sift down the node that takes its place. i.1 . whereas less than n roots are eliminated. it now consists of looking. giving a total time in 0 ((a +n) log n). If we remember to invert the heap.2. for each element w of C adjacent to v. If a << n 2.9. If so. 1 iterations. to see whether D [v] + L [v. 3. Initialization takes a time in O (n).C \ { v } " consists of eliminating the root from the heap. 1 . we saw that up to a nodes can be percolated. The instruction " C . since we only have to consider those nodes w adjacent to v.2 Greedy Algorithms and Graphs 91 Analysis of the algorithm.. ordered by the value of D [v]. at the cost of slowing down sifting. . . the element v of C that minimizes D [v] will always be found at the root. In a straightforward implementation. Suppose Dijkstra's algorithm is applied to a graph having n nodes and a edges.1. We might therefore consider modifying the definition of a heap slightly to allow percolation to run faster still. Let k = max(2.. giving for each node its direct distance to adjacent nodes (like the type lisgraph of Section 1. whereas it is preferable to use a heap if the graph is sparse.. This does not happen more than once for each edge of the graph. we compare the value of a node to the value of its parent rather than making comparisons with both children).2. for a total also in 0 (n 2). If the graph is connected. Show how your modification allows you to calculate the shortest paths from a source to all the other nodes of a graph in a time in . giving a total time in 0 (n 2).. The time required by this version of the algorithm is therefore in O (n 2) the matrix L... This allows us to save time in the inner for loop. the choice of algorithm may depend on the specific implementation. To sum up. which takes a time in O (log n). which again takes a time in 0 (log n ). With this in mind. As for the inner for loop. w ] < D [w]. the instance is given in the form of a matrix L [1 . Initialization of the heap takes a time in 0 (n). and that percolating up is somewhat quicker than sifting down (at each level. The straightforward implementation is therefore preferable if the graph is dense. What modification to the definition of a heap do you suggest ? ii. choosing v in the repeat loop requires all the elements of C to be examined. it could be preferable to represent the graph by an array of n lists.Sec. La In j ). In the preceding analysis. If a E O(n 2/ log n). n . so that we look at n . . it seems we might be able to avoid looking at the many entries containing . The inner for loop does n . * Problem 3. n J.2 times and to percolate at most a nodes. Using the representation suggested up to now. n .2).

customer 1 is served immediately. i 2. (Still faster algorithms exist.. a petrol pump. .1. . Show that Prim's algorithm to find minimal spanning trees can also be implemented through the use of heaps. Suppose that after scheduling customers i I . a cashier in a bank. and so on) has n customers to serve. Order T 123 : 5 + (5+10) + (5+10+3)=38 5 + (5+3) + (5+3+10)=31 10 + (10+5) + (10+5+3)=43 132: 213: 312: 10 + (10+3) + (10+3+5) = 41 3 + (3+5) + (3 + 5 + 10) = 29 E. then six orders of service are possible. if we have three customers with t1=5.) Problem 3. Imagine an algorithm that builds the optimal schedule step by step. we add customer j. Problem 2. show that the modification suggested in the previous problem applies just as well to Prim's algorithm.3 GREEDY ALGORITHMS FOR SCHEDULING 3. 3.optimal 321: 3 + (3+10) + (3+10+5)=34 231 : In the first case.. The service time required by each customer is known in advance : customer i will take time t..1 Minimizing Time in the System A single server (a processor. minimizing the total time in the system is equivalent to minimizing the average time. i. For example. Show that it then takes a time in 0(a log n ). We want to minimize n T= (time in system for customer i ). t3=3. it therefore gives the best of both worlds. customer 2 waits while customer 1 is served and then gets his turn. . just as Kruskal's algorithm would. The increase in T at this stage is . and customer 3 waits while both 1 and 2 are served and then is served himself: the total time passed in the system by the three customers is 38. Since the number of customers is fixed.Greedy Algorithms 92 Chap. t2=10.13. Note that this gives 0 (n 2) if a = n 2 and 0 (a log n) if a = n . 1 <_ i <_ n. Finally. 3 O (a logk n)..2.3.17(i) does not apply here since k is not a constant.

.1 . k=1 Suppose now that I is such that we can find two integers a and b with a < b and t.. because n T(I')=(n-a+1)ti.3. > t. t. tin After exchange of is and !b I Service duration Served customer (from I') Figure 3. in ) be any permutation of the integers { 1 ..... .+ =nti +(n-1)ti. tea . the a th customer is served before the b th customer even though the former needs more service time than the latter (see Figure 3.. t' . In the preceding example this algorithm gives the correct answer 3. 2.1).1 l tab 1o Exchanging the positions of two customers.+ti....+(n-b+1)ti + E (n-k+l)t. the total time passed in the system by all the customers is T(I)=ti. in other words. 2. .3 Greedy Algorithms for Scheduling 93 +ti +tj . t" . b a n i. 1. we obtain a new order of service I' obtained from I by interchanging the items is and ib This new order is preferable : ..+(ti. If customers are served in the order 1. . n 1. We now prove that this algorithm is always optimal. +ti. 3.Sec.+ To minimize this increase. 11 (from I) Service duration 2 t..b Service order 1 Served customer . If we exchange the positions of these two customers..+(n-2)ti + n _ k(n-k+1)t. add to the end of the schedule the customer requiring the least service among those who remain. fin . we need only minimize tj .. k=1 kta.. t' .+ti)+(ti... Let I = (i I i 2 .3.. This suggests a simple greedy algorithm : at each step.

3. In this context.. . the average time required to load a program is n I T =cE Pr Eli - j=1 k=1 where the constant c depends on the recording density and the speed of the drive. i + 2s .. Prove by giving an explicit example that it is not necessarily optimal to hold the programs in order of decreasing values of pi iii. as can the algorithm.3.3. If the programs are held in the order i 1. A magnetic tape contains n programs of length l 1 . suppose the customers are numbered so that s . (This of course implies that Ji 1 pi = 1.. n ] as data and produces an optimal schedule ? The problem can be generalized to a system with s servers. i. We know how often each program is used : a fraction pi of requests to load a program concern program i. We want to minimize T. the tape is rewound to the beginIn ning. in that order. The only schedules that remain are those obtained by putting the customers in nondecreasing order of service time.) Information is recorded along the tape at constant density. All such schedules are clearly equivalent. l2 Problem 3. Problem 3. . and therefore they are all optimal. 1 <. Prove that T is minimized if the programs are held in order of decreasing pi /li .. We can therefore improve any schedule in which a customer is served before someone else who requires less service. .Greedy Algorithms 94 Chap. and the speed of the tape drive is also constant.t2 <S to . How much time (use the 0 notation) is required by a greedy algorithm that accepts n and t [I . i 2.n. ii. must serve customers number t 1 <.2.tin ) >0. server i. loss of generality. . Prove that this algorithm always yields an optimal schedule. Without Problem 3..1. 1 i9 i +s .i <. . Each time a program has been loaded. i.-. 3 and thus (b -a )(ti. 1 2 .3.. Prove by giving an explicit example that it is not necessarily optimal to hold the programs in order of increasing values of 1i .

after its deadline d2 = 1. so job 2 is also rejected. 1. 3. Happily this is not the case. 1 <_ i <_ n . provided that the chosen set of jobs remains feasible.Sec. Next. Finally we try (1.. 1. Job i.2 Scheduling with Deadlines We have a set of n jobs to execute. The sequence 3. which in fact can only be done in the order 4.1 65 4. 2. which turns out not to be feasible . earns us a profit g. 3. Let J be a set of k jobs. 2. 2. 3 the schedules to consider and the corresponding profits are Sequence: 1 2 Profit: 50 10 3 15 4 30 1. we should execute the schedule 4. 41 is feasible because it can be executed in the order 4. if and only if it is executed no later than time d. we try the set 11. For example. . adding at each step the job with the highest value of g. 41. we can execute exactly one job. At first glance it seems we might have to try all the k ! possible permutations of these jobs to see whether J is feasible. for instance.. among those not yet considered. which is also infeasible .3 Greedy Algorithms for Scheduling 95 3. is not considered because job 2 would be executed at time t = 2.1 65 60 2.3 25 3. 1. 41.41.3 2. It remains to prove that this algorithm always finds an optimal schedule and to find an efficient way of implementing it. .is therefore to execute the set of jobs 11. An obvious greedy algorithm consists of constructing the schedule step by step. In the preceding example we first choose job 1. d. we choose job 4: the set 11.optimum 4. To maximize our profit in this example.3 45. . Next.3. A set of jobs is feasible if there exists at least one sequence (also called feasible) that allows all the jobs in the set to be executed in time for their respective deadlines. Our solution . . with n = 4 and the following values : i 50 2 10 15 4 30 2 1 2 1 1 g.optimal in this case . each of which takes unit time. At any instant t = 1. 1 80 . job 3 is therefore rejected.

and S are distinct sequences since I # J.1. the set I u { b } is feasible. and S. each having at least one more position in agreement with a. The "if " is obvious. such that every job common to both I and J is scheduled for execution at the same time in the two sequences.. See Figure 3. it is clear that b > a . Continuing thus.3. 3 Lemma 3. We now prove that the greedy algorithm outlined earlier always finds an optimal schedule. Finally. . . from right to left. in order of increasing deadlines.. Also dr. Let J be a set of k jobs. which is therefore feasible. . and S. The only remaining possibility is that some task a is scheduled in S.. If some task b is scheduled in S. . Then the set J is feasible if and only if the sequence a is feasible. whereas a different task b is scheduled in S.. If gQ > gb . there exists at least one sequence of jobs p = (r 1r2 that dr.. and S. one could substitute a for b in J and improve it. after a maximum of k -1 steps of this type.) S. Consider two feasible sequences S. Suppose a # p. Suppose that the greedy algorithm chooses to execute a set of jobs I whereas in fact the set J # I is optimal. The result is a new feasible sequence. the set J u { a } is feasible and would be more profitable than J. at the time when rb is at present scheduled. and let a = (s 1s 2 . If some task a is scheduled in S. ? d. this implies that a does not appear in J and that b does not appear in I. This is also impossible since it did not do so. we can obtain two feasible sequences S. sk) be a permutation of these jobs such that ds <. we can interchange the items ra and rb in p. 2. and S.2 for an example. This is not possible since J is optimal by assumption. . 11 This shows that it suffices to check a single sequence... hence the greedy algorithm should have included b in I. Proof. Let us consider an arbitrary time when the task scheduled in Sj is different from that scheduled in S.Greedy Algorithms 96 Chap. . > i . we obtain a series of feasible sequences. for the two sets of jobs in question. (by the construction of a and the minimality of a) (by the definition of b). Proof of optimality. The job ra could therefore be executed later. . (and therefore task a does not belong to J ). which is the same as a at least in positions 1. The necessary interchanges are easily found if we scan the two sequences S. 1 <.3. . Again.i S k. Let a be the smallest index such that Sa # rQ and let b be defined by rb = sQ . For the "only if " : rk) such If J is feasible. to know whether a set of jobs is or is not feasible. Since rb can certainly be executed earlier than planned. whereas there is a gap in S. By making appropriate interchanges of jobs in S. . whereas there is a gap in S. a. = dr. (We may have to leave gaps in the schedule. we obtain a itself.ds2 ds. This goes against the optimality of J..

if this task is a x Y P r u S P r S.Sec. function sequence (d [0.g .n . k ] array j [0. This is not possible either since it did not include h in I. array[ 1 . 1 <_ i <.. q V q w that one will be b common tasks Figure 3..k] P Y q x r r s r P u S.. and S.g 2 ? >.. and thus I is optimal as well. S" . 3. . V q w Si after reorganization. the same task.The only remaining possibility is therefore that go = gn In conclusion. n ] d [0]. j[I . For our first implementation of the algorithm suppose without loss of generality that the jobs are numbered so that g I >. j [ l ] . sequences S.2 to n do { in decreasing order of g I r-k while d [ j Ir ]] > max(d [i]. j [0] f.2. This implies that the total worth of I is identical with that of the optimal set J. suppose further that n > 0 and that di > 0.0 (sentinels) k. or two distinct tasks yielding the same profit.3. for each time slot. either schedule no tasks. n ]) : k.. r) do r F. . < gl.1 { task 1 is always chosen } {greedy loop I for i <-.If g.3 Greedy Algorithms for Scheduling 97 . Rearranging schedules to bring identical tasks together.i k4-k+I return k. the greedy algorithm should have chosen h before even considering a since (I \ { a }) u { h } would be feasible. To allow us to use sentinels.r - ifd[j[r]]<-d[i]andd[i]>r then for/ F-kstep-] tor+lldoj[I+1]E-j[l] j[r+I] .

For any position t. Lemma 3. which is always free. 11 The lemma suggests that we should consider an algorithm that tries to fill one by one the positions in a sequence of length l = min(n.2. Sets of positions. Clearly. max({ di 1 <_ i <_ n 1)). the schedule can be compressed once all the jobs are included.3. and add it to the schedule being built as late as possible.5. = nj j Free position Occupied position Figure 3.Greedy Algorithms 98 Chap. This obviously does not affect the feasibility of a set of jobs. let F (K) be the smallest member of K.3. For a given set K of positions.3.2. Problem 3.3. these sets will merge to form larger sets . and show that it requires quadratic time in the worst case. then the set J is infeasible. Prove Lemma 3. If we so wish. more efficient algorithm is obtained if we use a different technique to verify whether a given set of jobs is feasible. .3. It may happen that there are gaps in the schedule thus constructed when no job is to be executed. 3 The exact values of the g. A second.4. where t is the largest integer such that 0 < t <_ min(n. We obtain an algorithm whose essential steps are the following : I Positions of the same set i n. A set of jobs J is feasible if and only if we can construct a feasible sequence including all the jobs in J as follows : for each job i e J. as we assign new jobs to vacant positions. define n. but no later than its deadline. are unnecessary provided the jobs are correctly numbered in order of decreasing profit.3.3). define a fictitious position 0. consider each job i e J :n turn. disjoint set structures are intended for just this purpose. If a job cannot be executed in time for its deadline. Verify that the algorithm works. di) and the job to be executed at time t has not yet been decided. Problem 3. = max{k <_ t I position k is free }. Finally. Also define certain sets of positions : two positions i and j are in the same set if ni = nj (see Figure 3. execute i at time t. In other words.3.

3 1 1 3 1 3.j[i] . function sequence 2(d [ i n ]) : k. j[l .3. k ] array j. 99 2. . 1 ] 1 -min(n.. .Sec.1 to n do { in decreasing order of g } k .merge K and L . Here is a more precise statement of the fast algorithm.3.max{d[i]I1<i <n1) { initialization) for i F.assign the job to position F (K) . 3. let this be set K . array[1 .k] j[k] E. ifF(K) 0.1. 1.0 F[i] .i 1 <--find (m -1) F[k] F-F[1] merge (k.3 Greedy Algorithms for Scheduling i.. 1 ) { it remains to compress the solution } k -0 for i F. addition of a job with deadline d : find the set that contains min(n.. d ). .1.5 illustrate the workings of the slow and fast algorithms.. ii.0<-i <<-l.4 and 3.find (min(n. initialization : each position 0. F [0. Example 3. Figures 3.i initialize set { i } {greedy loop } for i F.. . 1 is in a different set and F({i})=i.3. Consider a problem involving six jobs : i 1 2 3 4 5 6 gi 20 15 10 7 5 3 d. . respectively. reject the job . the value of F for this new set is the old value of F (L).0 to l do j [i] E.find the set that contains F (K) . let this be set L (it is necessarily different from K) . we have assumed that the label of the set produced by a merge operation is necessarily the label of one of the sets that were merged. To simplify the descrip- tion. d [i])) m .I to 1 do if j[i]>0then k-k+1 return k. if F (K) = 0.F [k] if m * 0 then j[m]F.

4 GREEDY HEURISTICS Because they are so simple. which is essentially linear.Greedy Algorithms 100 Chap. so that we have to begin by sorting them. most of the time will be spent manipulating disjoint sets. value = 42 Figure 3. If the instance is given to us with the jobs already ordered by decreasing profit.3.4. . We content ourselves with giving two examples of this technique. 4. and since n > 1. 3. If. on the other hand. so that an optimal sequence can be obtained merely by calling the preceding algorithm. Since there are at most n+1 find operations and I merge operations to execute. greedy algorithms are often used as heuristics in situations where we can (or must) accept an approximate solution instead of an exact optimal solution. These examples also serve to illustrate that the greedy approach does not always yield an optimal solution. Illustration of the slow algorithm. 1. 3 dU[il] 3 Initialization: Ad k 3 Try 2: 2 Try 3: unchanged 3 Try 4: 2 3 4 Try 5: unchanged Try 6: unchanged Optimal sequence: 2. the jobs are given to us in arbitrary order. the required time is in 0 (n lg*1). we need a time in 0 (n log n) to obtain the initial sequence.

assign task 1 to position 3 F= 0 1 2 Fry 2: d2 = 1. 4.Sec. no free position available Optimal sequence: 2.4. When no further nodes can be painted. then they must be of different colours. An obvious greedy algorithm consists of choosing a colour and an arbitrary starting node. value = 42 Figure 3. 1. Illustration of the fast algorithm. the graph in Figure 3.5. max(d)) = 3 F= 0 1 2 3 Try 1: d. and blue for nodes 2 and 5.4 Greedy Heuristics 101 Initialization: I = min(6.1 Colouring a Graph Let G = < N. no free position available since the Fvalue is 0 Try 4: d4 = 3. no free position available Try 6: d6 = 3. 3. For instance. A > be an undirected graph whose nodes are to be coloured. we choose a new colour and a new .4. and then considering each other node in turn. If two nodes are joined by an edge. painting it with this colour if possible.3. 3. 3 and 4. assign task 2 to position 1 F= 0 2 Try 3: d3 = 1. = 3.1 can be coloured using only two colours: red for nodes 1. Our aim is to use as few different colours as possible. assign task 4 to position 2 F= 0 Try 5: d5 = 1.

This is an example of the NP-complete problems that we shall study in Chapter 10. and we are forced to make do with an approximate method (of which this greedy heuristic is among the least effective).1. Prove the following assertions : i. For a graph G and an ordering a of the nodes of G.2 The Travelling Salesperson Problem We know the distances between a certain number of towns. 3 A graph to be coloured. be the number of colours used by the greedy algorithm. the greedy heuristic may find the optimal solution.Greedy Algorithms 102 Figure 3. The algorithm is therefore no more than a heuristic that may possibly. 4. in this case the result is not optimal. If we start again at node 2 using blue paint. but it may also give an arbitrarily bad answer. (t/(XEIR+)(3G)(3o)[ca(G)1c(G) > a].4. 5.4. this is an optimal solution. Find two or three practical problems that can be expressed in Problem 3:4. Chap. then node 2 is painted blue. we can colour nodes 2 and 5 and finish the job using only two colours . to visit each other town exactly once. but not certainly.2. we paint as many nodes as we can with this second colour. 3. starting node that has not yet been painted. let c(G) Problem 3. The travelling salesperson wants to leave one of these towns.4. 3. but now nodes 3 and 4 require us to use a third colour.1. find a "good" solution. we are not allowed to paint node 2 with the same colour. terms of the graph colouring problem. We . (VG)(3(. nodes 3 and 4 can be red. In our example if node 1 is painted red.)[c6(G)=c(G)]. and so on. In other words. having travelled the shortest total distance possible. 2. if we systematically consider the nodes in the order 1. and lastly node 5 may not be painted. However. Let C (G) be the optimal (smallest) number of colours. we get a different answer: nodes I and 5 are painted red. and to arrive back at the starting point. R. Why should we be interested by such algorithms? For the colouring problem and many others the answer is that all the exact algorithms known require exponential computation time. For a large-scale instance these algorithms cannot be used in practice.

6.4. and k. if chosen. it is true that distance (i .4 Greedy Heuristics 103 assume that the distance between two towns is never negative. In some instances it is possible to find a shorter optimal tour if the salesperson is allowed to pass through the same town several times.5).2).4. 2. 3. As for the previous problem. j. (1. In this instance the greedy algorithm does not find an optimal tour since the tour (1. was not kept when we looked at it because it would have completed a circuit (1. 3. Give an explicit example illustrating this. Problem 3. if it is not possible to travel directly between certain pairs of towns ? Problem 3. On the other hand. 2. a distance matrix is said to be Euclidean if the triangle inequality holds : for any towns i. For example. it will not be the third chosen edge incident on some node. 5 ). 6. 5. i.3). 3. 5. (2. Show that in this case it is never advantageous to pass through the same town several times.Sec. (4. 4. all the known exact algorithms for this problem require exponential time (it is also NP-complete). 5. that is. (3. 1) whose total length is 58. which completes the salesperson's tour) . j ). 2.5). Problem 3.) One obvious greedy algorithm consists of choosing at each step the shortest remaining edge provided that it does not form a cycle with the edges already chosen (except for the very last edge chosen. and also because it would have been the third edge incident on node 5. Your algorithm must find a solution whose length is not more than double the length of an optimal tour.4.3. (4. if our problem concerns six towns with the following distance matrix : From 1 2 3 To : 2 3 4 5 6 3 10 11 7 25 6 12 8 26 9 4 20 5 15 4 5 18 edges are chosen in the order (1. j) <_ distance (i . 4. The problem can be represented using a complete undirected graph with n nodes.5. Edge (1. 3. k) + distance (k .6. Give a heuristic greedy algorithm to solve the travelling salesperson problem in the case when the distance matrix is Euclidean.6) to make the circuit (1. for example.) . (The graph can also be directed if the distance matrix is not symmetric: see Section 5. 1) has a total length of only 56.6). What happens to this greedy algorithm if the graph is not complete. (Hint : start by constructing a minimal spanning tree and then use the answer to the previous problem. Hence they are impractical for large instances. ii. 1).4.

6.5 is given in Christofides (1976) . Problem 3. Other ideas concerning shortest paths can be found in Tarjan (1983).2. the same reference gives an efficient heuristic for finding a solution to the travelling salesperson problem with a Euclidean distance matrix that is not more than 50% longer than the optimal tour. see also Problem 5. Prove that if a directed graph is complete (that is. Other greedy algorithms for a variety of problems are described in Horowitz and Sahni (1978).5 of this book. There exists a greedy algorithm (which perhaps would better be called "abstinent") for solving the problem of the knight's tour on a chessboard : at each step move the knight to the square that threatens the least possible number of squares not yet visited.1 can be found in Wright (1975) and Chang and Korsh (1976) . The first algorithm proposed (which we have not described) is due to Boruvka (1926). Other more sophisticated algorithms are described in Yao (1975).2. The problem of minimal spanning trees has a long history. see Schwartz (1964).4.12 can be found in Johnson (1977). and Tarjan (1983).Greedy Algorithms 104 Chap.5 REFERENCES AND FURTHER READING A discussion of topics connected with Problem 3. The details of the improvement suggested in Problem 3. 3 Invent a heuristic greedy algorithm for the case when the disProblem 3. The algorithm to which Prim's name is attached was invented by Jarnik (1930) and rediscovered by Prim (1957) and Dijkstra (1959).4. which is discussed in Graham and Hell (1985). in particular.7. tance matrix is not symmetric. Faster algorithms for both these problems are given in Fredman and Tarjan (1984).4.8. .4. Cheriton and Tarjan (1976). In a directed graph a path is said to be Hamiltonian if it passes exactly once through each node of the graph. Try it ! Problem 3. and give an algorithm for finding such a path in this case. if each pair of nodes is joined in at least one direction) then it has a Hamiltonian path.8. use of the Fibonacci heap allows them to implement Dijkstra's algorithm in a time in 0 (a + n log n ). but without coming back to the starting node.1. The implementation of Dijkstra's algorithm that takes a time in 0 (n 2) is from Dijkstra (1959). Similar improvement for the minimal spanning tree problem (Problem 3.13) is from Johnson (1975). Kruskal's algorithm comes from Kruskal (1956). An important greedy algorithm that we have not discussed is used to derive optimal Huffman codes. The solution to Problem 3. 3.

you have not managed to change the order of the time required : algorithm B still takes quadratic time. 105 .1 INTRODUCTION Divide-and-conquer is a technique for designing algorithms that consists of decomposing the instance to be solved into a number of smaller subinstances of the same problem. You discover that it would be possible to solve such an instance by decomposing it into three subinstances each of size Fn/21. you obtain a new algorithm B whose implementation takes time tB(n)=3tA([n/21)+t(n)<_3c((n+l)/2)2+dn =4cn2+(4c+d)n+4c The term 4cn 2 dominates the others when n is sufficiently large. solving successively and independently each of these subinstances. solving these subinstances. which means that algorithm B is essentially 25% faster than algorithm A. The efficiency of the divide-and-conquer technique lies in the answer to this question. By using both your old algorithm and your new idea. The first question that springs to mind is naturally "How should we solve the subinstances ?".4 Divide-and-Conquer 4. and then combining the subsolutions thus obtained in such a way as to obtain the solution of the original instance. Let d be a constant such that the time needed to carry out the decomposition and the recombination is t(n) <_ dn. Let c be a constant such that your particular implementation requires a time tA (n) <_ cn2 to solve an instance of size n. and combining the results. Suppose you already have some algorithm A that requires quadratic time. Although this improvement is not to be sneezed at. nevertheless.

We shall see that it is sometimes possible to replace the recursivity inherent in divide-and-conquer by an iterative loop. this chapter shows how divide-and-conquer is used to solve a variety of important problems and how the resulting algorithms can be analysed. . and 4. However. When implemented in a conventional language such as Pascal on a conventional . 's to obtain a solution y for x return y . For this approach to be worthwhile a number of conditions are usually required : it must be possible to decompose an instance into subinstances and to recombine the subsolutions fairly efficiently. Here then is the general outline of the divide-and-conquer method: function DQ (x) { returns a solution to instance x ) if x is sufficiently small or simple then return ADHOC (x) decompose x into smaller subinstances x 1. but instead.59 The improvement compared to the order of n 2 is therefore quite substantial.) recombine the y. gives us a time in the order of ntg3 which is approximately n 1. the more this improvement is worth having. where ADHOC. We should also mention that some divide-and-conquer algorithms do not follow the preceding outline exactly. it is hard to justify calling the technique divide-and-conquer. 4.8.3. 4 To do better than this. is usually both small and also independent of the particular instance to be solved. When k = 1.106 Divide-and-Conquer Chap.6). xk fori . After looking at the question of how to choose the optimal threshold. the basic subalgorithm. The number of subinstances. it is possible that algorithm A may still be the best way to proceed. the decision when to use the basic subalgorithm rather than to make recursive calls must be taken judiciously. and in this case it goes by the name of simplification (see sections 4.10. they require that the first subinstance be solved even before the second subinstance is formulated (Section 4.10). we are also concerned to make the hidden constant that multiplies nlg3 as small as possible. k.3. and the subinstances should be as far as possible of about the same size. and the bigger n is. This equation. x 2. Although this choice does not affect the order of the execution time of our algorithm. We shall see in the following section how to choose no in practice. might it not be better to use our new algorithm recursively? The idea is analogous to profiting from a bank account that compounds interest payments ! We thus obtain a third algorithm C whose implementation runs in time tc (n) _ tA(n) ifn <no 3tc( [ n/2 1 )+t(n) otherwise where no is the threshold above which the algorithm is called recursively. is used to solve small instances of the problem in question. which is similar to the one in Example 2.l tokdoy1 E-DQ(x. we come back to the question posed in the opening paragraph : how should the subinstances be solved? If they are small. when the subinstances are sufficiently large.

This example shows that the choice of threshold can have a considerable influence on the efficiency of a divide-and-conquer algorithm. To determine the value of the threshold no that minimizes tc (n). For instance. In this case. consider once again algorithm C from the previous section. Problem 4. it is not sufficient to know that to (n) E O(n 2) and that t (n) E O(n). then for all 1 >_ k the implementation considered previously takes 2k 31-k (32+2k) -21 +5 milliseconds to solve an instance of size 21. if no = 1.2 DETERMINING THE THRESHOLD An algorithm derived by divide-and-conquer must avoid proceeding recursively when the size of the subinstances no longer justifies this. Suppose we have an instance of size 1024 to solve. the answer is no : in our example. Must we conclude that divide-and-conquer allows us to go from a quadratic algorithm to an algorithm whose execution time is in 0(n 1g 3 ). that is. whose execution time is given by tc (n) _ ifn <no f tA(n) 3tc ([n / 21) + t (n) otherwise. 4. it may be possible to save a substantial amount of memory space in this way : for an instance of size n. This is ridiculous. where to (n) is the time required by the basic subalgorithm.2 Determining the Threshold 107 machine. by setting no = oo. since the instance can be solved in little more than a quarter of an hour by using the basic subalgorithm directly. an iterative algorithm is likely to be somewhat faster than the recursive version.1. the recursive algorithm uses a stack whose depth is often in 1 (log n) and in bad cases even in S2(n). provided we choose the threshold no intelligently. Problem 4.2. Choosing the threshold . although only by a constant multiplicative factor.Sec. 4. consider an implementation for which the values of to (n) and t(n) are given respectively by n2 and 16n milliseconds. that is.2. but only at the cost of an increase in the hidden constant so enormous that the new algorithm is never economic on instances that can be solved in a reasonable time? Fortunately. it is better to apply the basic subalgorithm. To illustrate this. If the algorithm proceeds recursively until it obtains subinstances of size 1. the instance of size 1024 can be solved in less than 8 minutes. it takes more than half an hour to solve this instance. and t (n) is the time taken to do the decomposition and recombination. Find all the values of the threshold that allow an instance of 0 size 1024 to be solved in less than 8 minutes.2. On the other hand. Prove that if we set no= 2k for some given integer k >_ 0.

000 (1982) Canadian dollars worth of machine time ! On the other hand. as Problem 4. however. This approach may require considerable amounts of computer time. Given a particular implementation. Obviously. a purely theoretical calculation of the optimal threshold is rarely possible. . It is often possible to estimate an optimal threshold simply by tabulating the results of these tests or by drawing a few diagrams. a threshold larger than 66 is optimal for instances of size 67.6 describes an algorithm for which the ultimate threshold is less obvious. Several groups of students tried to estimate the optimal threshold empirically. since we saw in Problem 4.2. consists of determining theoretically the form of the recurrence equations.108 Divide-and-Conquer Chap. that it is not usually enough simply to vary the threshold for an instance whose size remains fixed. whether we apply the basic subalgorithm directly or whether we go on for one more level of recursion. which we recommend. Moreover. If we neglect this difficulty. On the other hand. Finally.2 makes it clear. This remark may appear trivial. the optimal threshold can be found by solving to (n) = 3tA ([n / 21) + t (n). So how shall we choose no? One easy condition is that we must have no > 1 to avoid the infinite recursion that results if the solution of an instance of size 1 requires us first to solve a few other instances of the same size. The optimal threshold can then be estimated by finding the value of n at which it makes no difference. The presence of a ceiling in this equation complicates things. we must avoid thresholds below the ultimate threshold. corresponding to the fact that the average value of Fn / 21 is (2n + 1)/4. if we systematically replace In / 21 by (n + 1)/2. but also on the particular implementation. Coming back to our example. is to choose n o = 67 for our threshold. the preceding problem shows that. given that it varies from one implementation to another. A reasonable compromise. each group using in the attempt more than 5.2 that in fact no uniformly optimal threshold exists. the optimal threshold can be determined empirically. There is nothing surprising in this.6. whereas it is best to use a threshold between 33 and 65 for instances of size 66. for an instance of size n. over a certain range.2. and then finding empirically the values of the constants used in these equations for the implementation at hand. because tc ([n / 21) = to ([n / 21) if [n / 21 <_ no. there is in general no uniformly best value of the threshold : in our example. We vary the value of the threshold and the size of the instances used for our tests and time the implementation on a number of cases.1.8 makes clear. we find n = 70. We shall in future abuse the term "optimal threshold" to mean nearly optimal. in order to compare it with the classic algorithm from Section 1. but Section 4. changes in the value of the threshold may have no effect on the efficiency of the algorithm when only instances of some specific size are considered. Problem 4.7. 4 is complicated by the fact that the best value does not generally depend only on the algorithm concerned. The hybrid approach. We once asked the students in an algorithmics course to implement the algorithm for multiplying large integers given in Section 4. we obtain n = 64.

we wish to find the index i such that 0 <_ i <_ n and T [i] S x < T [i + I].1R* defined by the recurrence x2 f(x)= if X <_ s 3fs (x / 2) + bx otherwise. It is probably the simplest application of divide-and-conquer. that is. that is.Sec. It is therefore usually insufficient merely to estimate the constant a. Prove by mathematical induction that if u = 4b /a and if v is an arbitrary positive real number. (x) for every real number x. with the logical convention that T [0] _ . Show that this choice of no = 67 has the merit of being suboptimal for only two values of n in the neighbourhood of the threshold. Supposing. precisely because infinite recursion is not a worry. then fu (x) 5 f.3. we mean that these values are not in fact present as sentinels in the array. Formally. and let x be some item. Furthermore. The problem consists of finding x in the array T if indeed it is there. for instance. 4.) In practice. Although bn + c becomes negligible compared to an 2 when n is large. it is the algorithm used to look up a word in a dictionary or a name in a telephone directory. consider the function f5 : IR* -. (For purists : even if the domain of f. measure to (n) a number of times for several different values of n . (Notice that this would not cause an infinite recursion because the threshold is strictly larger than zero. that to (n) is quadratic.) Problem 4. and f is not countable. 1 < j 5 n = T [i] <.2. then instead we want to find the position where it might be inserted. If the item we are looking for is not in the array. Notice that this u is chosen so that au 2 = 3a (u / 2)2 + bu ..3 BINARY SEARCHING Binary searching predates computers. b.) The obvious . n] be an array sorted into increasing order.T [ j ]. The following problem shows that a threshold of 64 would be optimal were it always possible to decompose an instance of size n into three subinstances exactly of size n / 2. Let T [I . and c depending on the implementation. In essence.2. (By logical convention. the problem can be solved without recourse to transfinite induction. and then estimate all the necessary constants. one more complication arises. it may happen that tA (n) = an 2 + bn + c for some constants a. if instances of fractional size were allowed. probably using a regression technique. Let a and b be real positive constants. 4. the basic subalgorithm is used in fact precisely on instances of moderate size.4. prove that there are no instances that take more than I % longer with threshold 67 than they would with any other threshold.3 Binary Searching 109 * Problem 4. For each positive real number s. Instead.oo and T [n + 1] _ + oo.

where r is the index returned : this is S2(n) in the worst case and O (1) in the best case... If we assume that all the elements of T are distinct. x) f binary search for x in array T [ if n = 0 or x < T [1] then return 0 return binrec (T.. . Because the recursive call is situated dynamically at the very end of the algorithm. function sequential (T [ 1 . j ]. function binsearch (T [ 1 . that x is indeed somewhere in the array. we compare x to an element in the middle of the array : if x < T [ 1+ Ln / 2j]. j ]... On the average.3. j ] Prove too that when binrec (T [i . then the average number of trips round the loop is (n 2+ 3n -2)/2n. with j < i . 4 approach to this problem is to look sequentially at each element of T until we either come to the end of the array or find an item bigger than x.i. otherwise it is sufficient to search T [1+Ln / 2j . To find out which of these searches is appropriate. as well as in the worst case.2. so that technically it is an example of simplification rather than of divide-and-conquer. x) else return binrec (T [k . x) function binrec (T [i .x) Prove that the function binrec is never called on T [i . k -1 ]. then the search for x can be confined to T [I .. x) makes a recursive call binrec (T [u . it is always true that v .. and T [n + 1] are never used (except in the comments ! )..3. [n12]]. x)..u < j . Show finally that the values T [0] Problem 4. n J. n ]. v ]. x) [ sequential search for x in array T [ for i F. Show that the algorithm takes a time in 0( log n) to find x in T [1 . j ]. To speed up the search.I to n do if T[i] >x then return i -1 return n This algorithm clearly takes a time in O(1 + r). and that it is to be found with equal probability at each possible position. x) [ binary search for x in subarray T [i . We obtain the following algorithm. it is easy to produce an iterative version..110 Divide-and-Conquer Chap. therefore. n ]. j this procedure is only called if T[i] S x < T [j + 1] and i j[ if i = j then return i k -(i+j+1)div2 if x < T [k] then return binrec (T [i ... Problem 4. Conclude from these two results that a call on binsearch always terminates.1. sequential search takes a time in O(n). n ] whatever the position of x in T. The algorithm in fact executes only one of the two recursive calls. divide-and-conquer suggests that we should look for x either in the first half of the array or in the second half.

The comparison between x and T [9] causes the assignment i . and k = 9. On the first trip round the loop. Two more trips round the loop are necessary before we leave with i = j = 13. j <-.Sec.(i+j+1)div2 if x < T [k] then j <-. "k F(i+j+1)div2" by " k <-. n].k "...3 Binary Searching 111 function iterbin (T [1 .k return i It is easy to go wrong when programming the concept of Problem 4. simple though this is. while it is possible that the variant will only make one or two trips round the loop if x is favourably .n while i < j do (T[i] <-x <T [j+1]} k <-(i+j)div2 case x< T [k] : k-1 x?T[k+1]:i -. 11 A first inspection of this algorithm shows what is apparently an inefficiency. j = 17. The following algorithm leaves the loop immediately after we find the element we are looking for. "j F. This comparison could allow us to end the search immediately. Show by examples that the preceding algorithm would be incorrect if we replaced i. Suppose T contains 17 distinct elements and that x = T [13]. or iii. function iterbin 2(T [1 . and k = 13.k-1 else i <. but no test is made for equality.l.k+1 otherwise : i . On the second trip round the loop i = 9.13 is carried out.(i+j)div2". 4. A comparison is made between x and T [13].j<-. ii.x) ( variant on iterative binary search } if n = 0 or x < T [1] then return 0 iE.j'. regardless of the position of x in T. i = 1. binary searching.3. "i k" by " i E--k+l". n ]. and so the assignment i <.k return i Which of these algorithms is better ? The first systematically makes a number of trips round the loop in O(log n).n while i < j do ( T [ i ] Sx <T [j+1] } k <-.9 to be executed.3.k -1" by " j <-.x) ( iterative binary search for x in array T } if n = 0 or x < T [1] then return 0 iE--1. j = 17.

respectively. which yields a (n) = n lg n using the techniques of Section 2. Let k = 1 + Ln / 2j.k. already shown to hold when n is a power of 2.3. occupying each possible position with equal probability. respectively.k .1. 4 situated. taking k = In / 21. guessing the likely form of the answer and determining the missing parameters in the course of a tentative proof by mathematical induction. Exact analysis for arbitrary n is harder. In a similar way. b(2)=2. What might we add to n Llg n j to arrive at a (n) ? Let n * denote the largest power of 2 that is less than or equal to n. is that n Llg n j <. (*) The first equation is easy in the case when n is a power of 2. after which the algorithm starts over on an instance reduced to n -k + 1 elements. To compare them. With probability (k -1)/n . Let A (n) and B (n) be the average number of trips round the loop made by the first and the second iterative versions. We proceed by constructive induction. Analysis of the Second Version. since it then reduces to a(n)=2a(n/2)+n. Llg n] = lg n *. Analysis of the First Version. n ?2 a(1)=0. n >-3 B(1)=0. which causes the assignment i E. which causes the assignment j . A likely hypothesis. n >-2 b(n)=n +b([n/21-1)+b(Ln/2j). after which the algorithm starts over on an instance reduced to k -1 elements. Suppose to make life simpler that T contains n distinct elements and that x is indeed somewhere in T. One trip round the loop is carried out before the algorithm starts over. n >-3 a(1)=b(1)=0.Divide-and-Conquer 112 Chap. It seems reasonable to hope that . a trip round the loop in the variant will take a little longer to execute on the average than a trip round the loop in the first algorithm. we shall analyse exactly the average number of trips round the loop that each version makes. x < T [k]. we obtain the recurrence B(n)= 1+ Fn/ 21 . On the other hand. B(2)=I. Define a (n) and b (n) as n A (n) and n B (n).n Fig n 1. The equations then become a(n)=n +a(Ln/2])+a([n/2l). n?2 A(1)=0. With probability I . x > T [k]. so the average number of trips round the loop is given by the recurrence A(n)=1+ Ln/2]A(Ln/2])+ rn/21A(Ln/21).1)/n.B(Ln/2j).a (n) <. In particular.IB(Fn/21-1)+ Ln/2.(k .

(Ln / 2] )* =(n+1)14 and Fn/21 = (rn / 21)* = n * = (n + 1)/2. d. c and d being still unconstrained.Sec. that is A (n) = Llg n] + 2(1. They allow us to conclude that d = . e. our initial guess holds : Llg nJ <. If our hypothesis is correct.3 113 Binary Searching a (n) = n lg n * + cn + do * + e lg n * +f for appropriate constants c.2.A (n) <.rig n I . The average number of trips round the loop executed by the first iterative algorithm for binary searching. To prove HI(n) in this case. it is thus necessary and sufficient that n lgn*+cn +dn*+e lgn*+f = n lgn*+cn +dn*+2e lgn*+(2f -2e) that is. we need e = 0 and f = 0. To prove HI(n) in this case. which gives c = 2 and implies that the general solution of the recurrence (*) for a (n) is a(n)=nlgn*+2(n-n*). then (Ln / 2j)* _ (rn / 21)* = n */ 2. is given by A (n) = a (n)l n .n */ n) In particular. we therefore know a (n) = n lg n * + cn + do * . then we shall have proved by mathematical induction that it is true for every positive integer n.2n * true for the base n = 1. there still being no constraints on c. when looking for an element that is in fact present with uniform probability distribution among the n different elements of an array sorted into increasing order. and f . Denote this hypothesis by HI (n). using the recurrence (*). At this point we know that if only we can make the hypothesis a (n) = n lg n * + cn . 4. . These two equations are not linearly independent. 2 4 2 2 2 2 4 2 that is 4c + 2d = 4c + 3d + 2 and 2d = 3d+2. When n > 1 is not of the form 2'-1. When n > 1 is of the form 2'-1. Our final constraint is therefore 0=a(1)=c -2. it is necessary and sufficient that n1gn+1 +(c+d)n+d = nlgn+l +(c+3d+!)n+(3d+ i). HI (Ln / 2]) and HI (rn / 2 1 ). then Ln / 2] _ (n -1)/2.

A simple modification of the argument allows us to conclude that nlgn*+n/2-2n*+lgn*+3/2 <. Unfortunately. and 4 for n in the hypothesis HI (n). is whereas b (2) = 2 which inconsistent and shows that the original c=3. and f .3.2. The problem arises because the two basic cases are incompatible in the sense that it is impossible to obtain them both from the recurrence starting from some artificial definition of b (0). Explain why this is insufficient. e. e . we conclude that the .5.3.3. and f .(n + 1)/2 for every positive integer n.2n* + lg n * + (1 +c).6. 2. Given that the first algorithm takes less time on the average than the variant to execute one trip round the loop.nlgn*+2n/3-2n*+lgn*+5/3. Nonetheless.3.6. hypothesis was wrong.3. It seems reasonable to formulate the same incompletely specified induction hypothesis : b (n) = n lg n * + cn + do * + e lg n * + f for some new con- stants c. The case when n ? 4 is a power of 2 obliges us to choose d = . Equivalently lt(n -1) = [n * + On / n *j -2)(2n -3n*)] / 2 We are finally in a position to answer the initial question : which of the two algorithms for binary searching is preferable? By combining the preceding analysis of the function a (n) with the solution to Problem 4.2n * + lg n * + 2 . Show that the function n(n) of the previous exercise is given by lt(n -1) = jl n */ 2 if 2n < 3n * n-n * otherwise for all n > 2. The general solution of the recurrence for b (n) is more difficult than the one we have just obtained. * Problem 4. solve equation (*) exactly for b (n) when n is of the form 2' -1. we obtain A(n)-B(n)=1+ lt(n)-LngnJ-2 3 Thus we see that the first algorithm makes on the average less than one and a half trips round the loop more than the second. Problem 4. The hypothesis therefore becomes b (n) = n lg n * + cn .4. which can easily be solved to give the same c.n(n). b (1) = 0 = c = 2 . our efforts were not entirely in vain.3. Using the techniques presented in Section 2. and such that the exact solution of the recurrence is b (n) = n lg n * + n . Constructive induction yields e = 1 and f = 1 + c to take account of the case when n >. d. d.b(n) <.Divide-and-Conquer 114 Chap.7. Stated more elegantly : * Problem 4.3 is not a power of 2. A seemingly simpler approach to determine the constants would be to substitute the values 1. 4 Problem 4. 3. thus obtaining four linear equations in four unknowns.n(n) <. 1N+ such that Show that there exists a function it : IN+ (n + 1)/3 <.

. V [ 1 . the threshold beyond which the first algorithm is preferable to the variant can be very high for some implementations. n ] of an array are sorted independently. The obvious divide-and-conquer approach to this problem consists of separating the array T into two parts whose sizes are as nearly equal as possible.. When the number of elements to be sorted is small.. and merge (T.4 SORTING BY MERGING Let T [I . n div 2]. We are interested in the problem of sorting these elements into ascending order. where insert (T) is the algorithm for sorting by insertion from Section 1. You may only use a fixed number of working variables to solve the problem.4 Sorting by Merging 115 first algorithm is more efficient than the second on the average whenever n is sufficiently large. 4. n ]) { sorts array T into increasing order } if n is small then insert (T) else arrays U [ 1 . n I.2. (n + 1) div 2] U F T[ 1 . and then merging the solutions for each part.3). Problem 4. V) merges into a single sorted array T two arrays U and V that are already sorted.. in a time in the exact order of the sum of the lengths of U and V.2. 4.. n div 2] V -T[1+(n div 2) . We have already seen that the problem can be solved by selection sorting and insertion sorting (Section 1. and you wish to sort the whole array T [I . but without using an auxiliary array : the sections T [I . U. sorting these parts by recursive calls...2. However. being careful to preserve the order.. ** Problem 4. and your algorithm must work in linear time. mergesort (V ) merge (T.4. when this is justified by the number of elements. .4. or by heapsort (Example 2. n ] be an array of n elements for which there exists a total ordering. Recall that an analysis both in the worst case and on the average shows that the latter method takes a time in O(n log n).U.4 and Problem 2. On the other hand. k ] and T [k + 1 . The situation is similar if the element we are looking for is not in fact in the array. n] mergesort (U) .4.V).Sec. whereas both the former methods take quadratic time.4). that is.1. We obtain the following algorithm : procedure mergesort (T [ 1 .. Repeat the previous problem. a relatively simple algorithm is used. This sorting algorithm is a good illustration of all the facets of divide-andconquer. Give an algorithm capable of merging two sorted arrays U and V in linear time.

and analyse its execution time. Consequently. Problem 4. To see why. Separating T into U and V takes linear time. and then combines the two sorted half-arrays to obtain the solution to the original instance. allows us to conclude that the time required by the algorithm for sorting by merging is in O(n log n).5.116 Divide-and-Conquer Chap. and why? 4. As a first step. We obtain t'(n) E t'(n . Develop this idea. usually known as "quicksort".1) + t'(1) + O(n). and L(n + 2)/3].5 QUICKSORT The sorting algorithm invented by Hoare. 4 mergesort separates the instance into two subinstances half the size. is also based on the idea of divide-and-conquer. The fact that the sum of the sizes of the subinstances is equal to the size of the original instance is not typical of algorithms derived using divide-and-conquer. we might choose to separate T into about LI arrays. ** Problem 4. the fact that the original instance is divided into subinstances whose sizes are as nearly as possible equal is crucial if we are to arrive at an efficient algorithm.6.1. The merge sorting algorithm we gave. Let t'(n) be the time required by this variant to sort n items.4. Give a more formal description of this algorithm. Rather than separate T into two half-size arrays. Problem 4. L(n + 1)/3j. this algorithm chooses one of the items in the array to be sorted as the pivot. This equation. have two points in common. we might choose to separate it into three arrays of size Ln/3].3. and analyse its performance. By the result of Problem 4.4. and then to merge the three sorted arrays. to sort each of these recursively.4.6. The array is then partitioned on either side of the pivot : elements . as we shall see in several subsequent examples. t(n) E t(Ln / 2j) + t (Fn / 21) + 0(n).4.4. and those suggested by the two previous problems. the nonrecursive part of the work to be done is spent constructing the subinstances rather than combining their solutions. Simply forgetting to balance the sizes of the subinstances can therefore be disastrous for the efficiency of an algorithm obtained using divide-and-conquer. Show that t'(n) E O(n 2). the final merge also takes linear time. Which one.4. look at what happens if instead we decide to separate T into an array U with n -1 elements and an array V containing only 1 element. respectively. solves each of these recursively. each containing approximately L' elements. Let t(n) be the time taken by this algorithm to sort an array of n elements. Unlike sorting by merging.1. On the other hand. Problem 4. Following up the previous problem. which we analysed in Section 2. This poor sorting algorithm is very like one we have already seen in this book.

For this reason we simply use the first element of the array as the pivot.5 117 Quicksort are moved in such a way that those greater than the pivot are placed on its right.. the elements of T [i .l <. If now the two sections of the array on either side of the pivot are sorted independently by recursive calls of the algorithm.1... no subsequent merge step being necessary..i is small then insert (T [i j ]) { Section 1. whereas all the others are moved to its left. procedure pivot (T [i . i <. Finally. ... i <. crucial in practice that the hidden constant be small..4 } else pivot (T [i .p.p while k < I do interchange T [k] and T [1] repeat k F. j ] are greater than p. j ] into increasing order } if j ..j repeat I E.1 until T [1] p interchange T [i] and T [1] Invent several examples representing different situations that Problem 4. but starting at both ends. j ] in such a way that. j ] . see Section 4. and pointer I is decremented until T [1] <.k + 1 until T [k] > p or k >.5. (For a definition of the median. at the end. It is. j ] just once. however. 4. j ]) { sorts array T [i . and the elements of T J1+1 . 1-1] are not greater than p. Pointers k and 1 are initialized to i and j + 1.1. j ].1<-j+1 repeat k. To balance the sizes of the two subinstances to be sorted.k < 1 1 <k 5j and quicksort(T [i . I . j ]) = T[k]>T[1]} Designing a linear time pivoting algorithm is no challenge. respectively. var 1) { permutes the elements in array T [i . Here is the algorithm. Pointer k is then incremented until T [k] >p.Sec.1] and T [1 + 1 ..1.1) T [k] 5 T [l] { after pivoting. where p is the initial value of T [i ] } p -T[i] kF-i. j ] to be sorted are severely unbalanced.. One good way of pivoting consists of scanning the array T [i . T[11 =p. we would like to use the median element as the pivot... the final result is a completely sorted array.1 until T [1] <. I -1]) quicksort (T [1 +1 . This process continues as long as k < 1. Let p = T [i ] be the pivot. T [i] and T [1] are interchanged to put the pivot in its correct position.j. might arise.) Unfortunately. procedure quicksort (T [i . Quicksort is inefficient if it happens systematically on most recursive calls that the subinstances T [i .k + 1 until T [k] > p repeat 1 f. Now T [k] and T [1] are interchanged.6. finding the median takes more time than it is worth. and simulate the pivoting algorithm on these examples.

Divide-and-Conquer 118 Chap. taking f (x) = x lg x. n ]. Let t (m) be the average time taken by a call on quicksort (T [a + 1 . we obtain n-1 E i lg i i=no+1 n <- f x lg x dx x=no+1 . a < b . reasonable to hope that t(n) will be in 0 (n log n) and to apply constructive induction to look for a constant c such that t (n) S cn lg n .b. it is likely that most of the time the subinstances to be sorted will be sufficiently well balanced.5. To use this approach we need an upper bound on i =no+ 1 i lg i.) Let a and b be real numbers. This is obtained with the help of a simple lemma. respectively. and let f : [a. it is. The value of I returned by the pivoting algorithm after the initial call pivot (T [I .do + n t (k) for n > n 0 k=0 An equation of this type is more difficult to analyse than the linear recurrences we saw in Section 2... t(n) E O(n) + n 1=1 A little manipulation yields t(n)eO(n)+ n n-1 t(k) k =O To make this more explicit. This pivoting operation takes a time in O(n). nevertheless. b ] . (t(1-1) + t(n -1)) . Let j and k be two integers such that a <. each value having probability 11n. 4 Problem 4. It remains to sort recursively two subarrays of size 1 -1 and n -1. The average time required to execute these recursive calls is t(I -1) + t(n -1).IR be a nondecreasing function. Then k-1 k f(i) < J f(x) dx .m <. On the other hand. if the array to be sorted is initially in random order.n -m. Consequently.n and 0:5 a <. Give an explicit example of an array to be sorted that causes such behaviour. By analogy with sorting by merging.2.1) can therefore be any integer between 1 and n. (We suggest you find a graphical interpretation of the lemma. we assume that all the elements of T are distinct and that each of the n! possible initial permutations of the elements has the same probability of occurring. j = no and k = n .j < k <.3. Show that in the worst case quicksort requires quadratic time. The pivot chosen by the algorithm is situated with equal probability in any position with respect to the other elements of T. let d and no be two constants such that n-1 t (n) <. a + m ]) for 0 <. In particular. To determine the average time required by quicksort to sort an array of n items.

n ] I T [ i ] < m } <n/2 and # { i E [ I . 4. or that the elements of T may not all be lg n for all n > n o >_ 1. Intuitively. Outline a modification to the algorithm to avoid this. The hidden constant is in practice smaller than those involved in heapsort or in merge sort. .Igex2 4 2 n . T[(i+j)div21 andT[j]. Show by a simple argument that. we can obtain a version of quicksort that takes a time in 0 (n log n) even in the worst case. By combining the modification hinted at in the previous problem with the linear algorithm from the following section.6 Selection and the Median 119 [-lx . We mention this possibility only to point out that it should be shunned : the hidden constant associated with the "improved" version of quicksort is so large that it results in an algorithm worse than heapsort in every case. Complete the Problem 4. whatever the choice of pivot. the median of T is that element m in T such that there are as many items in T smaller than m as there are items larger than m.Ige n2 provided no > 1. at the price of a small increase in the hidden constant. The probability of suffering an execution time in U(n 2) can be greatly diminished.6 SELECTION AND THE MEDIAN Let T [I . by choosing as pivot the median of T [i].Sec. The formal definition takes care of the possibility that n may be even. Problem 4. quicksort as described here always takes a time in S2(n2) in the worst case. 4. t (n) <.x=no+1 2 < 2 lgn . . What could be easier than to find the smallest element or to calculate the mean of all the elements ? However. .4. Thus we define m to be the median of T if and only if m is in T and # I i E [ I . If an occasional long execution time can be tolerated. k=o Quicksort can therefore sort an array of n distinct elements in an average time in O (n log n). this is the sorting algorithm to be preferred among all those presented in this book. it is not obvious that the median can be found so easily.. where c= 2d lge + proof by 4 (no+l)2lge mathematical induction that t (k) . n ] be an array of integers. n ] I T [ i ] < m f >_n/2.5.5.

n ] I T [i] < p } v #{i a [1.v then { got it ! } return p otherwise { k > v } array V[1 .some element of T [I . Using ideas from the iterative version of binary searching seen in Section 4. n ]. k -v ) Problem 4. and greater than p.. which is not yet completely specified. Let T be an array of n elements. If we use heapsort or merge sort. The kth # { i e [ l . the median of T is its In / 21th smallest element. n ] { to be specified later) u ... it is the k th item in T if the array is sorted into ascending order.6. Can we do better? To answer this question. n -v ] V .. 4 The naive algorithm for determining the median of T consists of sorting the array into ascending order and then extracting the In / 21th entry.. The following algorithm. The values i and j should be returned by the pivoting procedure. u ] U F the elements of T smaller than p { the k th smallest element of T is also the k th smallest element of U } return selection(U. T [I . Problem 4. give a nonrecursive version of the selection algorithm.. T [i . Your algorithm is allowed to alter the initial order of the elements of T.3 and the pivoting procedure of the previous problem. k) { finds the k th smallest element of T . this algorithm takes a time in 0 (n log n) to determine the median of n ele- ments. this algorithm assumes that 1 k <_ n } if n is small then sort T return T RI p . n] element m such that I T [i ] <. containing the elements of T that are smaller than p..# { i e [l .2. Generalize the notion of pivoting from Section 4.. n i smallest element of T is that l T[i] <m } <k.. j]. n ]. . and no auxiliary arrays should be used.m } >_ k ..Divide-and-Conquer 120 Chap.1. i -1 ]. not calculated beforehand. whereas # { i e [ I .6.the elements of T larger than p { the k th smallest element of T is also the (k-v )th smallest of V } return selection(V. equal to p. we consider a more general problem : selection.. and let k be an integer between 1 and n.5 to partition the array T into three sections. k) if k <. and T [j + 1 . In other words. For instance. Your algorithm should scan T once only. function selection(T [ 1 . respectively.n]IT[i]Sp } if k S u then array U[1 . Do not use any auxiliary arrays. solves the selection problem in a way suggested by quicksort.

Let tm (n) be the time required by this method in the worst case to find the k th smallest element of an array of at most n elements. We have tm(n)E0(n)+max{ tm(i) I i 5 Ln/2] }. Problem 4.. Show that tm (n) is in O (n). the arrays U and V therefore contain a maximum of Ln / 2] elements.) Note that the technique hinted at in this exercise only aprlies because the average time turns out to be linear : the average time taken on several instances is not otherwise equal to the time taken on an average instance. n ]. k) stand for the expected size of the subarray involved in the first recursive call produced by a call on selection (T [I .Ln / 2]. the algorithm works by simplification. In this problem.T [I] ".4. not by divide-and-conquer.selection (T. If there is a recursive call.3.5. by definition of the median. still supposing that the median can be obtained magically. . independently of the value of k. Thus we have an algorithm capable of finding the k th smallest element of an array in a time linear in the size of the array. * Problem 4. But what shall we do if there is no magic way of getting the median? If we are willing to sacrifice speed in the worst case in order to obtain an algorithm reasonably fast on tf. notice first that. When we do this. What happens to the selection algorithm if the choice of p is Problem 4. we can once again borrow an idea from quicksort and choose simply p F-T[1]. u < [n / 21 and v ? [n / 21. therefore. The remaining operations. taking the size to be zero if there is no recursive call. so that the sizes of the arrays U and V will be as similar as possible (even if at most one of these two arrays will actually be used in a recursive call). (The hidden constant must not depend on k.6. For the time being. and k > 1. prove that this selection algorithm takes linear time on the average. Let E (n . Assume the n elements of the array T are distinct and that each of the n! permutations of the elements has the same probability of occurring.Sec. we hope that on the average the sizes of the arrays U and V will not be too unbalanced.6. v = 1. To analyse the efficiency of the selection algorithm. whatever the value of k. (n + 1) div 2) "? Suppose first that the median can be obtained by magic at unit cost. take a time in O (n).e average. made using "p <. Consequently. which causes a recursive call on n -1 elements.6 Selection and the Median 121 Which element of T should we use as the pivot p ? The natural choice is surely the median of T.k) < n + k (n-k) < 3n 2 n 4 Assuming that the pivoting algorithm preserves random ordering in the subarrays it produces for the recursive calls. k). even if occasionally we meet an instance where u = 0.6. consider the selection algorithm obtained by choosing "p f. n . Prove that E(n.v <. 4.

i3 between Si .n div 5 array S [I . Similarly. To visualize how these factors arise. yet its rank is approximately between 3n / 10 and 7n / 10. Let m be this approximation. as is each of the Ln / 5] columns. the smallest elements going to the left and to the top. to the value of m returned by the algorithm. Chap. 4 Show. Note that the time taken by adhocmed5 is bounded above by a constant. there are three i 1. This quadratic worst case can be avoided without sacrificing linear behaviour on the average : the idea is to find quickly a good approximation to the median.6.. The conclusion is that although m is perhaps not the exact median of T. The middle row corresponds to the array S in the algorithm.1 to s do S [i] t. s ] for i E. Therefore # { i E [ 1 . that is. But each element of S is the median of five elements of T.6. 5i ]) return selection(S.. consider the following algorithm : function pseudomed (T [1 . we have #{ie[l...adhocmed5(T [5i-4 . with the possible exception of one to four elements left aside (Figure 4..6. n ] Problem 4. about 3n / 10 elements. each of the elements in the box is less than or equal to m. Notice that the box contains approximately three-fifths of one-half of the elements of T. .n]IT[i]<m }<(7n-3)/10.s]IS[i]<-m }> Is/21.m . n ]) { finds an approximation to the median of array T) s . that in the worst case this algorithm requires quadratic time. This can be done with a little cunning. (s + 1) div 2) . although nothing in the execution of the algorithm pseudomed really corresponds to this illustration.7.1). Assuming n ? 5.pseudomed(T) . for every i such that S [i ] <. Since m is the exact median of the array S. however. Consequently.4 and 5i such that T [i 1] <_ T [i 2] <_ T [i 3] = S [i ] m.Divide-and-Conquer 122 Problem 4. respectively. where adhocmed 5 is an algorithm specially designed to find the median of exactly five elements. Show similarly that #{iE[I. We now look at the efficiency of the selection algorithm given at the beginning of this section when we use p E.. that is. I T[i]m }>31s/21=31Ln/5]/21>(3n-12)/10. By the transitivity of ":5 ". Now suppose that the middle row is sorted by magic. i2. imagine that all the elements of T are arranged in five rows. We look first at the value of the approximation to the median found by the algo- rithm pseudomed. the element in the circle corresponds to the median of this array.6.

3)/ 10 and v >.6 Selection and the Median Figure 4. Hence. Calculating u and v also takes linear time.6. Let p and q be two positive real constants such that p + q < 1. In particular.8.9.6. * Problem 4. so n -v <. Let t (n) be the time required in the worst case to find the k th smallest element in an array of n elements using the selection algorithm discussed earlier.6. First let us solve a more general. and let b be some positive real constant. Let n be the number of elements in T. We have to ensure that n > 4 in the preceding equation.9 (with p =1/5 and q = 3/4) such that t(n) <_ f(n) for every integer n. yet simpler problem of the same type. Argue that t(n)ES2(n).(3n . Conclude that t(n)eO(n). let no be a positive integer. The version of the algorithm suggested by Problem 4.2 is preferable in prac- . Problem 4.6. Use constructive induction to prove that f (n) E O(n).6. Let f : IN -* IR* be any function such that f(n) =.f (LpnJ) +.12)/10. because the array S can be constructed in linear time. 123 Visualization of the pseudomedian.(7n .(7n + 12)/10. Give explicitly a nondecreasing function f(n) defined as in Problem 4. Problem 4. At the first step. still independently of the value of k.10.f (Lqn]) + bn for every n > n 0. and let t (n) be the time required in the worst case by this algorithm to find the k th smallest element of T. 4. The initial preparation of the arrays U and V takes linear time. it is possible to find the median of an array of n elements in linear time. The recursive call that may follow therefore takes a time bounded above by max { t (i) I i <.6.(7n + 12)/10 } . What is it in the algorithm that makes this restriction necessary? This equation looks quite complicated. calculating pseudomed (T) takes a time in O (n) + t( Ln / 5]).7 and the preceding discussion show that u <. there exists a constant c such that t(n)<-t(Ln/5j)+max{t(i)Ii <-(7n+12)/10}+cn for every sufficiently large n. and thus t(n)EO(n). l7 Problem 4.Sec.6.1.

the classic algorithm and multiplication a la russe both take quadratic time to multiply these same operands. as well as additions and subtractions.124 Divide-and-Conquer Chap. From a more practical point of view.7.1.2 on a computer using the representation you invented for Problem 4. If results have to be calculated exactly and all the figures count.7. Problem 4. To use still less auxiliary space.8). This is only reasonable if the size of the operands is such that they can be handled directly by the hardware. large integers are of crucial importance in cryptology (Section 4. Your solution to Problem 4.7. This was necessary.1. we can also construct the array S (needed to calculate the pseudomedian) by exchanging elements inside the array T itself. Your representation should use a number of bits in O (n) for an integer that can be expressed in n decimal digits. 4 tice. and it must be possible to carry out in linear time multiplications and integer divisions by positive powers of 10 (or another base if you prefer). For some applications we have to consider very large integers. Can we do better? Let u and v be two integers of n decimal digits to be multiplied. we are obliged to implement the arithmetic operations in software.7.7. a computer. when the Japanese calculated the first 134 mil- lion digits of It in early 1987. Representing these numbers in floating-point is not useful unless we are concerned solely with the order of magnitude and a few of the most significant figures of our results. Problem 4. alas. this feat constitutes an excellent aerobic exercice for the computer !) The algorithm developed in this section is not. for instance. Give an algorithm able to add an integer with m digits and an integer with n digits in a time in O(m + n). Divide-and-conquer suggests that we should separate . that is. Implement your solution to Problem 4. Although an elementary multiplication takes scarcely more time than an addition on most computers. sufficiently efficient to be used with such operands (see Chapter 9 for more on this).2 shows how to add two integers in linear time. Also implement the classic algorithm for multiplying large integers (see Section 1.7.3. It must also allow negative numbers to be represented. Design a good data structure for representing large integers on Problem 4.1). On the other hand. 4.7 ARITHMETIC WITH LARGE INTEGERS In most of the preceding analyses we have taken it for granted that addition and multiplication are elementary operations.2. this is no longer true when the operands involved are very large. even though this does not constitute an iterative algorithm : it avoids calculating u and v beforehand and using two auxiliary arrays U and V. (At the very least. the time required to execute these operations is bounded above by a constant that depends only on the speed of the circuits in the computer being used.

a multiplication.2.7.7. we have only managed to increase the hidden constant ! The trick that allows us to speed things up consists of calculating wy. return mult (w .1 and the algorithms of Problem 4. y )) x 101 +mult(x. each of which serves to multiply two integers whose size is about n12.) n u w x V y z fn/21 Ln/2 Figure 4. The integers w and y therefore both have Fn/21 digits.z) Let tb (n) be the time required by this algorithm in the worst case to multiply two n digit integers. Splitting the operands for large integer multiplication. In fact. where 0<_ x< 10'. x .7. the integer divisions and multiplications by 102s and 10'. so we have not made any improvement compared to the classic algorithm.7. Thus tb (n) E 3tb ([n / 21) + tb (Ln / 2j) + O(n). 4. The product that interests us is uv = 102swy + 10S(wz+xy)+xz We obtain the following algorithm. as well as the additions. wz +xy. and a subtraction. By Example 2. z) + mult (x .u div lOs . since these are equivalent to an integer division. and xz by executing less than four half-length multiplications. If we use the representation suggested in Problem 4.7 the time taken by the preceding algorithm is therefore quadratic.3.1. This equation becomes tb (n) E 4tb (n / 2) + O(n) when n is a power of 2. The last statement of the algorithm consists of four recursive calls. and s = Ln / 2j. y) x 102s + (mult (w . (For convenience. 0!5z< 10S . See Figure 4. The same is true of the modulo operations. even if it is not greater than or equal to 10i -1.7 Arithmetic with Large Integers 125 each of these operands into two parts of as near the same size as possible : u = 10S w + x and v =10'y +z. we say that an integer has j digits if it is smaller than 10j.smallest integer so that _u and v are of size n if n is small then multiply u by v using the classic algorithm return the product thus computed s f-n div 2 w <-. even if this means that .Sec. are executed in linear time. v : large-integers) : large-integer n .u mod IOs y E--v div 101z 4-v mod 10. function mult (u .1.

The initial Example 4.345 and v = 6.7. but rather.005)x102 + 4. w = 23.J)+t([n/21)+t(l+Fn121)+cn for every n ? no.t(Ln12. Finally.608. s = 2.mult(w.205 . and z = 89.Divide-and-Conquer 126 Chap.1. This gives an execution time in 0 (n'3). However.789.410. p = 23 x 67 = 1541. we conclude that t (n) E 0 (n 1g3).q -mult(x. Using Problem 2.000 + 506. x = 45.4.005 = 15. y = 67. respectively. Taking account of the fact that w + x and y + z may have up to 1 + Fn / 21 digits. 4 we have to do more additions. It is thus possible to multiply two n digit integers in a time in 0 (n 1g3).mult (w +x. The classic algorithm obtains the product of u by v in a time in 0 (mn). the required product uv is obtained by calculating 1. which we met at the beginning of the chapter.541-4.3. which is in 0 (n 1. q = 45 x 89 = 4. decomposition of the operands gives n = 4. y +z) p <. we find that there exist constants c E IIt+ and no E IN such that t(n) <. We obtain successively and r = (23 + 45)(67 + 89) = 68 x 156 = 10.608 -1.2. This suggests we should replace the last statement of the algorithm by r E. which is unacceptable if m is very much smaller .7.005. our example is so small that it would be quicker to use the classic multiplication algorithm in this case. This resembles the equation tc (n) E 3tc J n / 21) + 0 (n). Two other multiplications are needed to isolate these terms.11 (notice that t(n) is nondecreasing by definition).541x104 + (10.z) return 102sp + l0s (r -p -q) + q Let t(n) be the time required by the modified algorithm to multiply two integers of size at most n . the algorithm we have just seen can multiply u and v. the largest base for which the hardware allows two "digits" to be multiplied directly. since it will simply treat u as though it too were of size n.200 + 405 = 15.7.y). the hidden constants are such that this algorithm only becomes interesting in practice when n is quite large.59).n. We require to multiply u = 2. Of course. Even when m # n. Consider the product r =(w+x)(y+z)=wy +(wz+xy)+xz .920. Suppose without loss of generality that m <. After only one multiplication this includes the three terms we need in order to calculate uv. Let u and v be integers of exactly m and n digits. Problem 4. This is sensible because addition is much faster than multiplication when the operands are large. A good implementation will probably not use base 10. Recall that the performance of this algorithm and of the classic algorithm are compared empirically at the end of Section 1.

2. (Remark: Integer division of an n digit integer by an arbitrary constant k can be carried out in a time in O(n).6. Your algorithm must allow for the possibility that the operands may differ in size.7.9. consider the following algorithm for multiplying large integers.5).7. Generalize the algorithm suggested by Problem 4. and compare your answer once again to Example 2.7. Although the idea tried in Problem 4. Analyse the efficiency of the algorithm suggested by this idea. for every real number a > 1.10.7. Rework Problem 2. it is nevertheless possible to multiply two n digit integers in a time in 0 (n log n log log n) by separating each operand to be multiplied into about parts of about the same size and using Fast Fourier Transforms (Section 9.5.7 Arithmetic with Large Integers 127 than n.7.7. so as to obtain the required product using five multiplications of integers about one-third as long (not nine as would seem necessary at first sight). Show that it is possible in this case to multiply u and v in a time in O (nm lg(3i2)) ** Problem 4. that is.smallest integer so that u and v are of size n if n is small then multiply u by v using the classic algorithm elsea .7.8. v : large-integers) : large-integer n . * Problem 4. where a = 1 + (lg lg n) / lg n.7.7.11 (analysis of the algorithm fib3) in the context of our new multiplication algorithm.3. Find two fundamental errors in this analysis of supermul.) ** Problem 4.7.3. that can multiply two n digit integers in a time in the order of n a. function supermul (u. Compare empirically this implementation and the one you made for Problem 4. * Problem 4. although the value of the hidden constant may depend on k.7.1+(lglgn)/lgn multiply u and v using algorithm A a return the product thus computed At first glance this algorithm seems to multiply two n digit numbers in a time in the order of n a.2.7. Show by a simple argument that the preceding problem is impossible if we insist that the algorithm A a must take a time in the exact order of n a. implement on your machine the algorithm we have just discussed. an algorithm Aa. Following up Problem 4.Sec.8. Following up Problem 4.8.7. Show that it is possible to separate each of the operands to be multiplied into three parts rather than two. Problem 4. 4. by showing that there exists. in a time in 0 (n log n).10 does not work. Problem 4. .

although Eve can overhear conversations. As a first step. Alice and Bob agree openly on some integer p with a few hundred decimal digits.2.8. Clearly. Let p be an odd prime number. they are not treated here. (If you wish to think about the problem. Their problem is complicated by the fact that the only way they can communicate is by using a telephone.Divide-and-Conquer 128 Chap.3). Bob sends Alice the value b = g B mod p. Clearly. similarly. which they suspect is being tapped by Eve. may well be inherently harder to compute .g A mod p and transmits this result to Bob . ** Problem 4. the secret value exchanged can now be used as the key in a conventional cryptographic system. They do not want Eve to be privy to their newly exchanged secret. and the calculation of the integer part of a square root can all be carried out in a time whose order is the same as that required for multiplication (Section 10. 4 Multiplication is not the only interesting operation involving large integers. Nevertheless. Alice computes x = b A mod p and Bob calculates y = a B mod p. * Problem 4. Such a generator always exists. Some other important operations. Find a protocol by which Alice and Bob can attain their ends. Integer division. neither of them can control directly what this value will be. such as calculating the greatest common divisor. modulo operations. Several other protocols have been proposed since. This value is therefore a piece of information shared by Alice and Bob. The security of the secret they intend to establish is not compromised should Eve learn these two numbers.1. we assume that.1. For some reason they wish to establish such a secret.2. They cannot therefore use this protocol to exchange directly a message chosen beforehand by one or the other of them. The cyclic multiplicative group ZP is defined as { x E N I 1 <. At the second step Alice and Bob choose randomly and independently of each other two positive integers A and B less than p. delay reading the rest of this section!) A first solution to this problem was given in 1976 by Diffie and Hellman.x < p } under multiplication modulo p. a malevolent third party. An integer g in this group is called a generator if each member of the group can be obtained as some integral power of g. 4. she can neither add nor modify messages on the communications line. To simplify the problem.8. Finally. Next Alice computes a =. and on some other integer g between 2 and p . Now x = y since both are equal to g' mod p.8 EXPONENTIATION : AN INTRODUCTION TO CRYPTOLOGY Alice and Bob do not initially share any common secret information. the condition that g be a generator is necessary if Alice and Bob require that the secret .

If Eve needs to be able to calculate discrete logarithms efficiently to discover the secret shared by Alice and Bob. g a generator of ZP . it seems therefore that this method of providing Alice and Bob with a shared secret is sound. For the time being. Problem 4. For instance. and b that does not involve calculating a discrete logarithm. At the end of the exchange Eve has been able to obtain directly the values of p. and 1 <. (If the logarithm does not exist. There exists an obvious algorithm to solve it.2 hold. even if A and B are chosen randomly with uniform probability between 1 andp -1. this average time is more than the age of Earth even if p only has two dozen decimal digits. 4. Furthermore.Sec.) function dlog (g. g and a is called the problem of the discrete logarithm.3. p) A <. and b only.A' < p . there is no integer A such that 3=2Amod7. If p is an odd prime. still b A mod p =b A' mod p provided that g A mod p = g A' mod p and that there exists a B such that b = g B mod p. g. although no one has yet been able to prove this. g. the algorithm returns the value p. Calculating A' from p. One way for her to deduce x would be to find an integer A' such that a = g A'mod p . If each trip round the loop takes 1 microsecond.8. The obvious algorithm for this is no more subtle or efficient than the one for discrete logarithms. then A' is necessarily equal to A. none of them is able to solve a randomly chosen instance in a reasonable amount of time when p is a prime with several hundred decimal digits. there is no known way of recovering x from p. a. since it makes p 12 trips round the loop on the average when the conditions of Problem 4. a. An attentive reader may wonder whether we are pulling his (or her) leg. and sox =x and the secret is correctly computed by Eve in this case.O. a.8. however.8 Exponentiation : An Introduction to Cryptology 129 exchanged by the protocol could take on any value between 1 and p . The value x' calculated by Eve in this way is therefore always equal to the value x shared by Alice and Bob. that regardless of the choice of p and g some secrets are more likely to be chosen than others.x 4.1. it is equally true that Alice and Bob must be able to calculate efficiently exponentiations of the form a = g A mod p.l repeat A<-A+1 x 4. . Although there exist other more efficient algorithms for calculating discrete logarithms.xg until (a = x mod p) or (A = p) return A This algorithm takes an unacceptable amount of time. Prove. Show that even if A #A'. and then to proceed like Alice to calculate x' = b A' mod p .

x25 = (((x2x)2)2)2x Thus x25 can be obtained with just two multiplications and four squarings. assume that calculating a modulo takes a time in the exact order of that required for multiplication. p ). and p allows us to avoid accumulation of extremely large integers in the loop. (The same improvement can be made in dlog.1 for i F. By inspection of the algorithm we find . p ) a<-. A .p) return (ag mod p) else a f. Use the classic algorithm for multiplying large integers. Repeat the problem using the divide-and-conquer algorithm from Section 4.A/2. which is necessary if we hope to execute each trip round the loop in 1 microsecond. For simplicity. function dexpo (g . We leave the reader to work out the connection between 25 and the sequence of bits 11001 obtained from the expression (((x2x)2x 1)2x1)2x by replacing every x by a 1 and every 1bya0. which consequently takes a time in O (h (A) x M (p )).dexpo(g. x 24 = (x 12)2.1 for i . This idea can be generalized to obtain a divide-and-conquer algorithm. These operations dominate the execution time of the algorithm. 4 function dexpo l (g . z.4. there exists a more efficient algorithm for computing the exponentiation.Divide-and-Conquer 130 Chap. The preceding formula for x 25 arises because x 25 = x 24x .7 for the multiplications. An example will make the basic idea clear.p) return (a2 mod p) Let h (A) be the number of multiplications modulo p carried out when we calculate dexpo (g . and so on.1 to A do a F.A-l. In both cases.8.1 to A do a E. dexpo2 as a function of the value of A and of the size of mod p return a Analyse and compare the execution times of dexpol and * Problem 4. A . y. 0 Happily for Alice and Bob. p ) a 4.dexpo(g.) function dexpo2(g. where M (p) is an upper bound on the time required to multiply two positive integers less than p and to reduce the result modulo p. A . suppose that g is approximately equal to p/2. A . including the return a mod p The fact that x y z mod p = ((x y mod p) x z ) mod p for every x. p ) if A = 0 then return 1 if A is odd then a .

A . let us just say that h (A) is situated between once and twice the length of the binary representation of A. Find an explicit formula for h (A) and prove your answer by Problem 4. On the other hand. the computation of dexpo (g . As a function of the size of the base and of the value of the exponent. This means that Alice and Bob can use numbers p. p ) n F.000 multiplications of 200-digit numbers and 3. (Do not try to use characteristic equations. which corresponds intuitively to calculating x25 as x 16x8x 1.8. algorithms.) Without answering Problem 4. In both cases the number of multiplications can easily be reduced to six by avoiding pointless multiplications by the constant 1 and the last squaring carried out by dexpoiter. The algorithms dexpo and dexpoiter do not minimize the number of multiplications (including squarings) required. we obtain algorithms for handling large integers capable of calculating efficiently all the digits of g'. which is entirely reasonable.8.ay mod p y <__ y2modp n(-ndiv2 return a Problem 4. This recursive call is not at the dynamic end of the algorithm. More generally. dexpo calculates x 15 as (((lx)2x)2x)2x.g.7. that is. dexpoiter calculates x 15 as 1 x x 2x 4x 8.8 Exponentiation : An Introduction to Cryptology 0 h(A)= 131 ifA =0 l + h (A -1) if A is odd 1 + h (A /2) otherwise . For example. 4.a . which makes it harder to find an iterative version. A and B of 200 decimal digits each and still finish the protocol after less than 3.5. Show that in fact x 15 can be calculated with only five multiplications.6. provided A >_ 1. As was the case for binary searching.A. By suppressing all the reductions modulo p in the preceding * Problem 4.y F. for an arbitrary base g and an arbitrary exponent A. P) takes a time in 0 (M (p ) x log A ).5.1 while n > 0 do if n is odd then a .8.8. It is therefore an example of simplification rather than of divide-and-conquer. with seven multiplications. mathematical induction. A . how much time do the algorithms corresponding to dexpo2 and dexpo .000 computations of a 400-digit number modulo a 200-digit number. function dexpoiter (g . there exists a similar iterative algorithm. The efficiency of these algorithms depends in turn on the efficiency of the algorithm used to multiply large integers. Nonetheless.Sec. the algorithm dexpo only requires one recursive call on a smaller instance. which involves eight multiplications (the last being a useless computation of x 16).

We leave the reader to verify that the required product AB is given by the following matrix : m2+m3 m1+m2+m5+m6 m 1+m2+m4-m7 m 1+m2+m4+m5 J . Towards the end of the 1960s. The basic idea is similar to that used in the divide-and-conquer algo- rithm of Section 4. Since there are n2 entries to compute in order to obtain C. The preceding problem shows that it is sometimes not sufficient to be only halfclever ! 4. Strassen caused a considerable stir by improving this algorithm. Let A= all a12 a21 a22 and B= b11 b12 b21 b22 be two matrices to be multiplied. Consider the following operations : m1 =(a21 +a22-a11)(b22-b12+b11) m2=a11b11 M3 = a 12b21 m4=(a11-a21)(b22-b12) m5=(a21+a22)(b12-b11) m6=(a12-a21 +all -a22)b22 m7=a22(b11+b22-b12-b21).Divide-and-Conquer 132 Chap. The classic algorithm comes directly from the definition : Cij = E Aik Bkj k=1 Each entry in C is calculated in a time in O(n).9 MATRIX MULTIPLICATION Let A and B be two n x n matrices to be multiplied. First we show that two 2 x 2 matrices can be multiplied using less than the eight scalar multiplications apparently required by the definition. and let C be their product. the product of A and B can be calculated in a time in 8(n 3).7. assuming that scalar addition and multiplication are elementary operations.7 for multiplying large integers. 4 take when the classic multiplication algorithm is used? Rework the problem using the divide-and-conquer multiplication algorithm from Section 4.

9 Matrix Multiplication 133 It is therefore possible to multiply two 2 x 2 matrices using only seven scalar multiplications. this algorithm does not look very interesting : it uses a large number of additions and subtractions compared to the four additions that are sufficient for the classic algorithm. provided n is sufficiently large. once again based on divide-and-conquer: he found a way to multiply two 70 x 70 matrices that involves only 143. 4.2. Following publication of Strassen's algorithm.9. have been discovered subsequently. Almost a decade passed before Pan discovered a more efficient algorithm.9. The number of additions and subtractions needed to calculate the product of two 2 x 2 matrices using this method seems to be 24. it was discovered by Coppersmith and Winograd in September 1986. taking account of the idea from Problem 4. The algorithm discussed in this section is a variant discovered subsequently by Shmuel Winograd. however. Given that matrix additions can be executed much faster than matrix multiplications.000.1.1. * Problem 4.3. show that it is possible to multiply two n x n matrices in a time in 0 (n 2.376) . At first glance. the few additional additions compared to the classic algorithm are more than compensated by saving one multiplication. If we now replace each entry of A and B by an n x n matrix.9. This is possible because the basic algorithm does not rely on the commutativity of scalar multiplication. The asymptotically fastest matrix multiplication algorithm known at the time of this writing can multiply two n x n matrices in a time in 0 (n 2. Show that this can be reduced to 15 by using auxiliary variables to avoid recalculating terms such as m1+m2+m4.81). Because of the hidden constants. Problem 4. Bearing in mind what you learnt from Section 4. Building on the preceding discussion. none of the algorithms found after Strassen's is of much practical use. Strassen's original algorithm takes 18 additions and subtractions as well as 7 multiplications.) Numerous algorithms.640 scalar multiplications. find the exact number of scalar additions and multiplications executed by your algorithm for Problem 4.2. (Compare this to 702. asymptotically more and more efficient. What do you do about matrices whose size is not a power of 2 ? Problem 4. .2.000 and to which exceeds 150. Assuming that n is a power of 2. a number of researchers tried to improve the constant w such that it is possible to multiply two n x n matrices in a time in 0 (n °).9.81. 703 = 343. we obtain an algorithm that can multiply two 2n x 2n matrices by carrying out seven multiplications of n x n matrices. propose a threshold that will minimize the number of scalar operations.Sec. Your answer will depend on the threshold used to stop making recursive calls. as well as a number of additions and subtractions of n x n matrices.9.

j ) i 4i -j else j4-j-i exchange (k -i ..10 EXCHANGING TWO SECTIONS OF AN ARRAY For this additional example of an algorithm based on simplification. m) to interchange the elements and T [ j .j-i) ifi=j if i > j if i < j For instance.l do interchange T [i +p ] and T [ j +p ] procedure transpose (T [I .j <. For instance. Let T (i . the required result in T is d e f g i g 0 h k i a j k b c It is easy to invent an algorithm exchange (i . we ignore the recursive formulation and give only an iterative version. j.1). k) i . j . k .2. 4. m ) forp *.k.k 4-k+1 while i # j do if i > j then exchange (k -i .j) i +T(i. we can solve our problem as illustrated in Figure 4. After each exchange this part is smaller than before : thus we can affirm that each exchange simplifies the solution of the instance.j)= j + T(ij. 4 4. Figure 4.Otom .10. We wish to interchange the first k elements and the last n -k. n ]. without making use of an auxiliary array.Divide-and-Conquer 134 Chap. if T is the array d a e f and k = 3. k+j. i) exchange (k -i . We have .10. provided that m < i +m <. j +m -1] in a time in 0(m). j -n -k. j) be the number of elementary exchanges that have to be made to transpose a block of i elements and a block of j elements. i ) The analysis of this algorithm is interesting.. k .. Let T be an array of n elements. With its help.n -m + 1 (see Fig. i+m -1] procedure exchange (i.2 shows how a block of three elements and a block of eight elements are transposed.10. The general algorithm is as follows. T [i . Here the arrows indicate the part of the array where there are still some changes to make. Then i T(i.

1) d e g Figure 4. 1) = 10.2. where gcd(i. 1) h e exchange (3.2) = 8+T(l.7. 6.2)=9+T(1. 3).j). 2) d e I h I a c a c exchange (3.10 Exchanging Two Sections of an Array 135 T(3.1.Sec. b c . j) denotes the greatest common divisor of i and j (Section 1. h i Progress of transpose (T.10. The progression of the parameters of the function T recalls an application of Euclid's algorithm and leads to the following result : Problem 4. Effect of the exchange algorithm. 5. 4. 4.j)=i + j -gcd(i. Prove that T(i.10. in elements m elements T J exchange Figure 4.10.1. a c d e I g h i j k h a b c k a b c exchange (1.5) = 6 + T(3. 4.4). 3) k J d e I g exchange (1. 3) I h J exchange (1. 9. 8) = 3 + T(3.

The Fibonacci sequence. as well as another algorithm capable of multiplying two polynomials of degree i in a time in 0 (i log i ) Problem 4. d ] containing its coefficients.. space ? 4. You may ..11. j) and the matrix F? What happens if i and j are two consecutive numbers from the Fibonacci sequence? Use this idea to invent a divide-and-conquer algorithm to calculate this sequence. Let T [1 . Give an efficient algorithm based on divide-and-conquer to find the unique monic polynomial p (n) of degree d such that p (n 1) = p (n 2) = = p (nd) = 0. n ] be an array of n elements.2. (A polynomial is monic if its coefficient of highest degree ad = 1. ... Problem 4.5 ? Polynomial interpolation. nd be any integers.ind t-1 for i +. 2 . max FT[1].Divide-and-Conquer 136 Chap.11 SUPPLEMENTARY PROBLEMS Problem 4. Let n 1.T [i ].2.) Analyse the efficiency of your algorithm. Represent the polynomial p (n) = a o + a 1 n + a 2 n + + ad n d of degree d by an array P [0 .7. T[ind]*-T[1] min *.11. Consider the matrix F=(011.2 to n do if max < T [i ] then max 4.3. Does this help you to understand the algorithm fib3 of Section 1. .10. Suppose you already have an algorithm capable of multiplying a polynomial of degree i by a polynomial of degree 1 in a time in 0 (i ). we exclude implicit comparisons in the control of the for loop.11. Ill 1 Let i and j be any two integers.2 more comparisons. ind F i We only count comparisons between elements . 4 Can we do better if we are allowed unlimited auxiliary Problem 4. Smallest and largest elements. We could subsequently find the smallest element of T by making n .1.T [2] for i F-3tondo if min > T [i] then min T[i] Find an algorithm that can find both the largest and the smallest elements of an array of n elements by making less than 2n -3 comparisons between elements.see Chapter 9. What is the product of the vector (i . It is easy to find the largest element of T by making exactly n -1 comparisons between elements. n 2.

including the initial conditions.4.5. 4.11.see Figure 4. n ] be an array of n elements. provided such an index exists. j = 5. Tally circuit. and if so find it. one j bit input. The number of switches used must be in 0 (n log n ). Let T [1 .Sec.11. Let T [I . You may therefore not assume that an order relation exists between the elements. For example. or input A with output B and input B with output A. i.11. and two outputs. Exactly how many comparisons does your algorithm require? How would you handle the situation when n is not a power of 2 ? Problem 4. * Problem 4. For example. Problem 4. You may not suppose that n has any special form. show how to build an efficient n-tally. but allow your algorithm to take a time in 0 (n log n). Use these switches to construct a network with n inputs and n outputs able to implement any of the n! possible permutations of the inputs. It connects input A with output A and input B with output B. j) 3-tallies. Problem 4. a control.7. Using the O notation.5 with the supplementary constraint that the only comparisons allowed between elements are tests of equality. * Problem 4. for the number of 3-tallies needed to build your n-tally. It counts (in binary) the number of bits equal to 1 among the inputs. An (i .11 Supplementary Problems 137 assume that n is a power of 2.11. if n = 9 and the inputs are 011001011. some of which may be negative but all of which are different. j )-adders as primitive elements. Majority element. Telephone switching. and one [I+ max(i. j)-adder using exactly max(i. A switch is a circuit with two inputs. It adds its two inputs in binary. try again. depending on the position of the control . j )-adders you might have used. if i = 3. It is always possible to construct an (i . Rework Problem 4. Your algorithm must run in linear time. If you could not manage the previous problem.11.. Give an algorithm that can decide whether an array T [I . the output is 011100. and the inputs are 101 and 10111 respectively. Problem 4. An n-tally is a circuit that takes n bits as inputs and produces 1 + LIg n] bits as output.11. An element x is said to be a majority element in T if # [ i T [i ] = x } > n 12. Justify your answer.9. For this reason the 3-tally is often called a full adder. . Your algorithm should take a time in 0 (log n) in the worst case. give the simplest expression you can for the number of 3-tallies needed in the construction of your n-tally. Using full adders and (i .. j)]-bit output. iii.11. j)-adder is a circuit that has one i bit input.6..1.8. Give an algorithm that is able to find an index i such that I <_ i <_ n and T [i] = i. Give the recurrence. Do not forget to count the 3-tallies that are part of any (i . n ] be a sorted array of integers.11. the output is 0101. ii. n ] includes a majority element (it cannot have more than one).

the F4 merge circuit shown in Figure 4. and the depth is the largest number of comparators an input may have to pass before it reaches the corresponding output. which is of size 5 and depth 3.3 gives an example of S4. inputs are on the left and outputs are on the right.11. 4 Divide-and-Conquer 138 4 N A A A A B B B B Figure 4. Problem 4.2 shows an F4 circuit. Figure 4. Each rectangle represents a comparator. show how to construct a merge circuit F whose size and depth are exactly 1 + n lg n and 1 + lg n .Chap. Merge circuit.11.11.1. and the larger input appears on the lower output. For a given integer n. 7 sorted outputs . By convention. For n a power of 2. Batcher's sorting circuit. Figure 4. respectively.11. a sorting circuit S. it sorts the inputs presented to it. For instance. The depth is interesting because it determines the reaction time of the circuit. The smaller input appears on the upper output. then each input appears on one of the outputs. There are two ways to measure the complexity of such a circuit : the size of the circuit is the number of comparators it includes.11. illustrating how the inputs are transmitted to the outputs. A comparator is a circuit with two inputs and two outputs.11. Following up the previous problem. show how to construct an efficient sorting circuit S .11. and the outputs are also sorted. a merge circuit F has two groups of n inputs and a single group of 2n outputs. has n inputs and n outputs . For n a power of 2.2. By "efficient" we mean that the depth of your circuit must be significantly less than n 3 first sorted group 1 1 3 3 3 3 6 4 4 4 2 of inputs L37: 6 4 It 2 5 5 2 2 4 5 second sorted group of inputs 2 5 5 5 6 6 7 7 7 7 7 7 6 2 4 Figure 4. For example.11.10. * Problem 4. A merge circuit. Provided each of the two groups of inputs is already sorted.2 has size 9 and depth 3. Telephone switches.

11. Player n=5 1 2 3 4 1 2 1 - 5 2 3 5 1 - 2 3 4 3 2 1 - 4 5 - 4 3 1 - 4 5 2 3 2 3 4 5 6 Player 1 n=6 5 Day 4 Day 1 2 1 6 5 4 3 2 3 5 1 6 2 4 3 4 3 2 1 6 5 4 5 6 4 3 1 2 5 6 4 5 2 3 1 Figure 4.11. Solve these 0 equations exactly. 4. for the size T and the depth P of your circuit S. . O Problem 4.12. and express T and P in O notation as simply as possible.11. Each competitor must play exactly once against each of his opponents.Sec. show that it is pos- ** Problem 4. Tournaments.4. A sorting circuit. whenever n is sufficiently large.13. Timetables for five and six players.11.11 Supplementary Problems 2 3 139 2 1 3 1 2 3 2 2 1 2 2 4 2 4 1 /\3-3 4 1 3 3 3 4 4 4 1 Figure 4. You are to organize a tournament involving n competitors. sible to construct a sorting circuit for n elements whose size and depth are in 0(n log n) and 0(logn).3. including the initial conditions. Give recurrences. respectively. . but their depth and size must then be taken into account. Continuing the two previous problems. You may use merge circuits to your heart's content.

or in n days if n is odd.6 is examined in Knuth (1969).8 is discussed in Knuth (1969). The algorithm of Section 4. .8 is crucial. efficient exponentiation as described in Section 4. Mergesort and quicksort are discussed in detail in Knuth (1973). The importance for cryptology of the arithmetic of large integers and of the theory of numbers was pointed out by Rivest. If n is a power of 2. Closest pair of points. and Tarjan (1972). Problem 4.1 is due to Diffie and Hellman (1976). On the other hand.4. Kranakis (1986). and Adleman (1978). 1. For any integer n > 1 give an algorithm to construct a timetable allowing the tournament to be finished in n -1 days if n is even. however. Floyd. The survey article by Brassard.11. The original solution to Problem 4.59) is attributed to Karatsuba and Ofman (1962). and the algorithm that is asymptotically the most efficient known at present is by Coppersmith and Winograd (1987).7 to be worthwhile. as described in Hellman (1980).12 REFERENCES AND FURTHER READING Quicksort is from Hoare (1962). The algorithm for multiplying large integers in a time in 0 (n 1.2 was solved by Kronrod. Denning (1983). Pratt. The answer to Problems 4.7.7 and 4. The natural generalization of Problem 4. Give an of points in a time in O (n log n ). Monet. Rivest. consult the introductory papers by Gardner (1977) and Hellman (1980) and the books by Kahn (1967).8. with the possible exception of a single day when he does not play at all. has since been broken. Bear in mind. Subsequent efforts to do better than Strassen's algorithm began with the proof by Hopcroft and Kerr (1971) that seven multiplications are necessary to multiply two 2x2 matrices in a non-commutative structure . and Zuffellato (1986) covers computation with very large integers. and Brassard (1988). The algorithm that multiplies two n x n matrices in a time in O (n 2.7. the first positive success was obtained by Pan (1978). points in the plane. You are given the coordinates of n algorithm capable of finding the closest pair 4.8. see the solution to Exercise 18 of Section 5. give an algorithm to construct a timetable allowing the tournament to be finished in n -1 days.4 gives possible timetables for tournaments involving five and six players. Figure 4.14. 4 Moreover.2. For example.10 comes from Gries (1981).81) comes from Strassen (1969).4 of Knuth (1973). that the cryptosystem based on the knapsack problem. The algorithm linear in the worst case for selection and for finding the median is due to Blum. each competitor must play exactly one match every day. 2. ** Problem 4. Notice too that the integers involved in these applications are not sufficiently large for the algorithms of Section 4. For more information about cryptology. Shamir.11.140 Divide-and-Conquer Chap.

11. Problem 4. Komlos.Sec. but consult Section 8.11 are solved in Batcher (1968).1 can be found in Gries and Levin (1980) and Urbanek (1980).11.12 References and Further Reading 141 The solution to Problem 4. Problem 4.11.12 is solved.3 is discussed in Pohl (1972) and Stinson (1985).11. Problems 4.10 and 4.14 is solved in Bentley and Shamos (1976). and Szemeredi (1983).11.11. . in Ajtai. at least in principle. 4.7 for more on this problem. Problem 4.

142 . If we solve each of these independently. We usually start with the smallest. we immediately attack the complete instance. It sometimes happens that the natural way of dividing an instance suggested by the structure of the problem leads us to consider several overlapping subinstances. If. If we pay no attention to this duplication.5 Dynamic Programming 5. until finally we arrive at the solution of the original instance.1 INTRODUCTION In the last chapter we saw that it is often possible to divide an instance into subinstances. Divide-and-conquer. By combining their solutions. is a top-down method. they will in turn create a large number of identical subinstances. and then to combine the solutions of the subinstances so as to solve the original instance. usually by keeping a table of known results. we take advantage of the duplication and solve each subinstance only once. we obtain the answers to subinstances of increasing size. on the other hand. which we then divide into smaller and smaller subinstances as the algorithm progresses. The underlying idea of dynamic programming is thus quite simple : avoid calculating the same thing twice. which we fill up as subinstances are solved. to solve the subinstances (perhaps by further dividing them). saving the solution for later use. subinstances. Dynamic programming is a bottom-up technique. then a more efficient algorithm will result. on the other hand. and hence the simplest. it is likely that we will end up with an inefficient algorithm. When a problem is solved by divide-and-conquer.

we obtain a more efficient algorithm.7.Introduction Sec. Although this principle may appear obvious.2. Thus the algorithm takes a time in 0 (nk) and space in 0 (k).I k 0 1 C(n-1. In fact.5 uses dynamic programming? Dynamic programming is often used to solve optimization problems that satisfy the principle of optimality: in an optimal sequence of decisions or choices. If we calculate [ k] directly by function C (n . If.1). k . j < k are calculated over and over. Since the final result is obtained by adding up a certain number of Is.k) C(n. 0<k <n k otherwise . Prove that the total number of recursive calls made during the computation of C (n.1. Calculating the Fibonacci sequence affords another example of this kind of technique.5 and Example 2.2. see Figure 5. We have already met a similar phenomenon in algorithm fib] for calculating the Fibonacci sequence (see Section 1..1. it is not even necessary to store a matrix : it is sufficient to keep a vector of length k. The table should be filled line by line.7). the execution time of this algorithm is certainly in S2( [ k 1).1) + C (n . k) many of the values C (i. representing the current line.1. k .l . Problem 5. on the other hand. j). i < n . k) is exactly 2 [ n -2- Problem 5.1. we use a table of intermediate results (this is of course Pascal's triangle . k ) if k = 0 or k = n then return 1 else return C (n . .1.1. 143 Consider calculating the binomial coefficient + n Inkl j.1 Example 5.1.7.. 0 1 1 1 1 2 1 2 2 3 . if we assume that addition is an elementary operation.1. Which of the algorithms presented in Section 1. 5.1. which we update from left to right.k-1)C(n-1. k) Figure 5. each subsequence must also be optimal. it does not always apply. Pascal's triangle.

it is not immediately obvious that the subinstance consisting of finding the shortest route from Montreal to Ottawa is irrelevant to the shortest route from Montreal to Toronto. then that part of the journey from Montreal to Kingston must also follow the shortest route between these two cities : the principle of optimality applies.2. We assume that there are no tied games. Argue that this is due to the fact that one cannot in general splice two simple paths together and expect to obtain a simple path. whereas team B still needs j more victories if they are to win. dynamic programming efficiently solves every possible subinstance in order to figure out which are in fact relevant. j) be the probability that team A will win the series given that they still need i more victories to achieve this. For example. but rather concentrate on the control structure and the order of resolution of the subinstances. Imagine a competition in which two teams A and B play not more than 2n -1 games. Instead. and the principle of optimality does not apply. maybe we have to stop to fill up somewhere on the second half.) The principle of optimality can be restated as follows for those problems for which it applies : the optimal solution to any nontrivial instance is a combination of optimal solutions to some of its subinstances. For this reason the problem considered in this section is not one of optimization. Without this restriction the longest path might be an infinite loop. let us not worry about the principle of optimality. However. The difficulty in turning this principle into an algorithm is that it is not usually obvious which subinstances are relevant to the instance under consideration. if the fastest way to drive from Montreal to Toronto takes us first to Kingston. 5 Example 5. 5. Problem 5. it does not follow that we should drive from Montreal to Kingston as quickly as possible : if we use too much petrol on the first half of the trip.2 THE WORLD SERIES As our first example of dynamic programming. (A path is simple if it never passes through the same place twice. that the results of each match are independent.1. Let P (i. before the first game of the series the probability that .3.1. and only then are these combined into an optimal solution to the original instance. Coming back to Example 5. If the shortest route from Montreal to Toronto goes via Kingston.2. The subtrips Montreal-Kingston and Kingston-Toronto are not independent. and that for any given match there is a constant probability p that team A will be the winner and hence a constant probability q = 1 -p that team B will win. This difficulty prevents us from using a divide-and-conquer approach that would start from the original instance and recursively find optimal solutions precisely to those relevant subinstances.144 Dynamic Programming Chap. losing more time than we gained by driving hard. the winner being the first team to achieve n victories.1. Show that the principle of optimality does not apply to the problem of finding the longest simple path between two cities.

1. Problem 5. j) P(i. P (0. if we look at the way the recursive calls are generated.1) k . ll the required time is thus in S2( [ 2n n JJ P(i. n) is in 0(4 n ) and Q(4 n In).j-1) i ? l. j) if i = 0 then return 1 else if j = 0 then return 0 else return pP (i -1. 5. 1 < i <. The total number of recursive calls is therefore exactly 2(' ' I .1) + d .1 matches left that call P(i . j) k matches left calls P(i .2 (Problem 5. j) = C ((i -1) + j . j).i < n. T (k) is therefore in 0 (2k) = 0 (4") if i = j = n. n) : both teams still need n victories. j . If team A has already won all the matches it needs. this problem does have other applications ! ) .2.4" /(2n + 1).The World Series Sec. n) that team A will win given that the series has not yet started.1. 0) = 0.2 145 team A will be the overall winner is P (n . In fact. since team A wins any given match with probability p and loses it with probability q.j)+gP(i. j) P(i . j -1).j ? 1 Thus we can compute P (i. Prove that (nn. j). Combining these results. (Although sporting competitions with n > 4 are the exception. Recursive calls made by a call on function P (i . j . 1 <. Similarly P (i.2. k>1 where c and d are constants. j -1) Let T (k) be the time needed in the worst case to calculate P (i. then it is of course certain that they will win the series : P (0.1) P(i. >. To calculate the probability P (n . 0) is undefined.1.n.2) k .1). j . which is identical to that followed by the naive calculation of the binomial coefficient C (i + j .2.2 matches left etc.2. Finally. j) using function P (i. we find the pattern shown in Figure 5. This method is therefore not practical for large values of n. we see that the time required to calculate P (n . j) + qP (i. Figure 5.1. i) = 1. P(i. j) + C (i + (j -1).j)=pP(i-1. With this method.1. we see that T(1)=c T(k):5 2T(k .1. where k = i +j.

and since a constant time is required to calculate each entry.. 5 * Problem 5.) 5.1 to s-1 do P[k. To speed up the algorithm.0. Problem 5.n.6. instead of filling the array line by line. calculate the probability that team A will win the series if p = 0. (Hint : use a completely different approach ..45 and if four victories are needed to win.3.see Section 8. Here is the algorithm to calculate P (n . Using this algorithm.s -k . n) in a time in O(n). n) using the preceding algorithm is in 0(4"I[ ).3 CHAINED MATRIX MULTIPLICATION We wish to compute the matrix product M=M1M2. however. n ] Problem 5. This time..1. we proceed more or less as with Pascal's triangle : we declare an array of the appropriate size and then fill in the entries.M" Matrix multiplication is associative. Problem 5.s-k] + gP[k.1] for s 1 to n do fork .. Show that a memory space in O(n) is sufficient to implement this algorithm.2.n-k] .s-k] <.pP[s+k-1.pP[k-l. Since in essence the algorithm has to fill up an n x n array. Show how to compute P (n.I to n do P[0.5.2.n-k-1] return P [n.2. Prove that in fact the time needed to calculate P (n . n) .0]-0 fork . p ) array P[0. function series (n .2.n-k]+qP[s+k.0 to n -s do P[s+k.P[s. we work diagonal by diagonal. so we can compute the product in a number of ways : .4.2.s].n] qE-1-p for s F. its execution time is in 0(n 2).146 Dynamic Programming Chap.

856 4. here is the corresponding number of scalar multiplications. ) (Mi + 1 Mi +2 ..Mn) = (M1(M2(M3 . we obtain the following recurrence for T (n) : nI T(n)= E T(i)T(n-i)... ))) The choice of a method of computation can have a considerable influence on the time required.201 2. (In the second case that follows. There are five essentially different ways of calculating this product. M.1. Problem 5. We wish to calculate the product ABCD of four matrices : A is 13 x 5. Since i can take any value from 1 to n .326 multiplications for a total of 10. For example. we obtain successively (AB) (AB )C ((AB )C )D 5. Let T (n) be the number of essentially different ways to parenthesize a product of n matrices.. B is 5 x 89.582 54.3. we could simply parenthesize the expression in every possible fashion and count each time how many scalar multiplications will be required.3. we count the number of scalar multiplications that are involved.1.785 multiplications 3... Mn) .) In each case. we do not differentiate between the method that first calculates AB and the one that starts with CD. There are now T (i) ways to parenthesize the left-hand term and T (n . Example 5.418 The most efficient method is almost 19 times faster than the slowest.. 0 To find directly the best way to calculate the product.i) ways to parenthesize the right-hand term.1. i=1 ... 5. C is 89 x 3. and D is 3 x 34.055 26.Sec... To measure the efficiency of the different methods.3 Chained Matrix Multiplication 147 M=(.582 multiplications.((M1M2)M3). ((AB )C )D (AB) (CD) (A (BC ))D A ((BC )D) A (B (CD )) 10. Suppose we decide to make the first cut between the i th and the (i + 1)st matrices of the product : M = (M 1 M 2 . Show that calculating the product AB of a p x q matrix A and a q x r matrix B by the direct method requires pqr scalar multiplications.471 multiplications 1. if we calculate M = ((AB )C )D. (Mn-1Mn) ..

i = 1. Prove that T(n)= 1 (2n-2] n-1 n For each way that parentheses can be inserted it takes a time in U(n) to count the number of scalar multiplications required (at least if we do not try to be subtle). if the best way of multiplying all the matrices requires us to make the first cut between the ith and the (i + l)st matrices of the product.785. We build the table mid diagonal by diagonal : diagonal s contains the elements mid such that j .078. we can thus calculate all the values of T...2..3. . The values of T (n) are called Catalan numbers. Since T(n) is in S2(4"/n2) (from Problems 5.n. then both the subproducts M1M2 Mi and Mi + A12 . Fortunately.89.that is. Mn must also be calculated in an optimal way.Dynamic Programming 148 Chap.2. the principle of optimality applies to this problem.k < i + s .2. It is only for clarity that the second case is written out explicitly. 5 Adding the initial condition T (l) = 1. This suggests that we should consider using dynamic programming.3. . m34 = 9.i <. .34).674. finding the best way to calculate M using the direct approach requires a time in Q(4"/n). for s = 2 we obtain .862 2. Suppose the dimensions of the matrices Mi are given by a vector di . . Next..5. such that the matrix Mi is of dimension di -I by di .i <.j <.2). i =1. i = 1. We thus obtain in succession : s = 0: mii = 0. Example 5.i = s.for the part Mi Mi +I duct. We construct a table mid . *Problem 5. 2. i<k<i+s The third case represents the fact that to calculate Mi Mi+l .=(13.n-s. Mi +s) and choose the best for i <.. n s = 1 : mi.3. as it falls under the general case with s = 1. i + 1 = di -1di di + 1....1 and 5. the required MM of the required pronumber of scalar multiplications . Continuation of Example 5..i+s+di-ldkdi+s). We have d =(I 3.. The solution to the original problem is thus given by m 1. This method is therefore impracticable for large values of n : there are too many ways in which parentheses can be inserted for us to look at them all.440. 2. we find n 1 2 3 4 5 10 15 T (n) 1 1 2 5 14 4.n. 0 <.3. where mid gives the optimal solution . . we find m 12 = 5. For instance. Among other values..3. For s = 1. m23 = 1.1.. Mi+s we try all the possibilities (Mi Mi + 1 .3.2.1) 1 <S <n:mi.335. n -1 (see Problem 5. I <. Mk ) (Mk + 1 .i+s= min (mik+mk+l. .

3.1. 5. Write the algorithm to calculate m In . {k=3} m +m44+ 13x3x34) = min(4055.n (n . .1)/6 = (n 3-n )/6 The execution time is thus in 0(n 3).3.s elements to be computed in the diagonal s . for each.3. {k=2} m12 m34+ 13x89x34 . 2 856 s=3 2 s=2 3 S=1 4 s=0 Figure 5.530 m24=min(m22+m34+5x89x34. for s = 3 m14=min({k=1} mll+m24+13x5x34. Example of the chained matrix multiplication algorithm.845 Finally.1.856 .4. Problem 5. 9256) = 1. m12+m33+ 13x89x3) = min(1530. there are n .Sec. but also to know how to calculate the product M in the most efficient way ? For s > 0. j=1 2 4 3 0 1 \ 5 785 \ 1 530 i=1 .3.y s2 = n 2 (n . 2856) = 2.3 Chained Matrix Multiplication 149 m13 =min(mll +m23+ 13x5x3. How must the algorithm be modified if we want not only to Problem 5.3. calculate the value of m In . 1845) = 1.1)(2n .5 01. The execution time of the algorithm is therefore in the exact order of n-1 s=l n-1 n-I s=1 s=l (n-s)s =n2:s .1)/2 . The array m is thus given in figure 5. we must choose between s possibilities (the different possible values of k). m23+m44+5x3x34) = min(24208.

3.b. It is this duplication of effort that causes the inefficiency of minmat. 2. 5 * Problem 5. . 2. to all the others.2. i ] = 0. n }. n ] is global. a call on minmat (1. . At itera- tion k. 2. j] ? 0 if i # j. Let Dk be the matrix D after the k th iteration. n) of Problem 5. j)) return ans .min(ans.. k) + minmat (k + 1. d [i -1 ]d [k]d [ j ] + minmat (i. A > be a directed graph . j) does not exist. both of which recursively solve BCDEF from scratch. (Hint: for the "0 " part.5 is faster than naively trying all possible ways to parenthesize the desired product.. minmat recursively solves 12 subinstances. The necessary check can be written as . and L [i.3. k) as intermediate nodes. where we were looking for the length of the shortest paths from one particular node. We want to calculate the length of the shortest path between each pair of nodes.00 fork <--i toj-1 do ans F. The algorithm initializes D to L.2. The principle of optimality applies : if k is a node on the shortest path from i to j. (Compare this to Section 3.) As before. N is the set of nodes and A is the set of edges.Dynamic Programming 150 Chap. and that from k to j. Prove that with the following recursive algorithm function minmat (i. suppose that the nodes of G are numbered from 1 to n. n) takes a time in O(3" ). L [i. . including the overlapping ABCDEF and BCDEFG. . D gives the length of the shortest paths that only use nodes in 11. k -1 }. In order to decide on the best way to parenthesize the product ABCDEFG. Each edge has an associated nonnegative length... j) if i = j then return 0 ans .4 SHORTEST PATHS Let G = <N. must also be optimal. It then does n iterations. j) whether or not there exists a path passing through node k that is better than the present optimal path passing only through nodes in { 1. . N = { 1. This behaviour illustrates a point made in the first paragraph of this chapter. 5.) Although a call on the recursive minmat (1.. the source. use ccnstructive induction to find constants a and b such that the time taken by a call on minmat (1. . j] = oc if the edge (i. with L [i ..5. where the array d [0. the algorithm has to check for each pair of nodes (i. and that a matrix L gives the length of each edge. We construct a matrix D that gives the length of the shortest path between each pair of nodes. After iteration k . it is still much slower than the dynamic programming algorithm described previously.. n) is no greater than a 3" . then that part of the path from i to k. After n iterations we therefore obtain the result we want.

If the graph is not very dense (a << n 2).n.j]).k]+D[k. whereas at first sight a matrix n x n x 2 (or even n x n x n) seems necessary. each time choosing a different node as the source. j] then D[i. j] . I. since D [k . known as Floyd's algorithm.Dk-i[i.j]-k . where a is the number of edges in the graph.Sec. The innermost loop of the algorithm becomes ifD[i. where we make use of the principle of optimality to compute the length of the shortest path passing through k. In this case we have to apply the algorithm n times. if the graph is dense (a = n 2). it may be preferable to use Dijkstra's algorithm n times . n ]) : array[ l . if we use the version of Dijkstra's algorithm that works with a heap. If we use the version of Dijkstra's algorithm that works with a matrix of distances. and hence with lists of the distances to adjacent nodes. At the k th iteration the values in the k th row and the k th column of D do not change. j] <--D[i. In this case we use a second matrix P initialized to 0. j] P[i.1 to n do for j E. k ] is always zero.. in O ((an +n 2) log n). The algorithm.n] D -L fork <-Iton do for i .j]. the total time is in n x O ((a +n) log n). j]) return D Figure 5. procedure Floyd(L [1 . that is. but the simplicity of Floyd's algorithm means that it will probably be faster in practice.min(D [i..2...l. This allows us to get away with using only a twodimensional matrix D.. k ] +D [k. n. n. On the other hand. the total computation time is in n x O(n 2). not just its length.4. I. D [i.k]+D[k. The order is the same as for Floyd's algorithm. n ] array D[I. It is obvious that this algorithm takes a time in O(n3). We can also use Dijkstra's algorithm (Section 3. j] <D[i.4 Shortest Paths 151 Dk[i. j].1 to n do D [i. We have also implicitly made use of the fact that an optimal path through k does not visit k twice.1 gives an example of the way the algorithm works. it is better to use Floyd's algorithm.k]+Dk-1[k.. follows. in O(n 3 ).j] =min(Dk_i[i. that is. 5. It is therefore not necessary to protect these values when updating D.2) to solve the same problem. We usually want to know where the shortest path goes. .

If P [i. otherwise. . P [i. j) . j]. j ] contains the number of the last iteration that caused a change in D [i. if P [i. j]. 5 50 0 15 5 30 00 0 15 15 00 5 0 0 0 5 50 0 D2 = 5 30 35 0 15 15 20 5 0 5 20 0 D4 = 20 10 15 0 5 15 oo o 15 10 10 5 30 35 0 15 15 20 5 0 Floyd's algorithm at work. the shortest path from i to j passes through k.4. j] = 0. look at P [i. To recover the shortest path from i to j. j ] = k. the shortest path is directly along the edge (i. k ] and P [k.1. 5 15 0 Do=L = 0 D D 1 = 5 00 00 50 0 15 5 30 35 0 15 15 20 5 0 0 20 10 5 45 0 3 = 15 30 35 0 15 20 5 Figure 5. Look recursively at P [i.Dynamic Programming 152 Chap. j] to find any other intermediate nodes along the shortest path. When the algorithm stops.

case when the matrix L is symmetric (L [i.1. the shortest path from 1 to 3 passes through 4. Problem 5. Suppose we allow edges to have negative lengths. only their existence is important.2 in the * Problem 5. In this case. and L [i . 5. the length of the edges is of no interest . These two problems are NP-complete (see Section 10. 3. Looking now at P [1.1. j] = false otherwise. j) exists. . but that does not include a negative cycle ? Prove your answer.Sec. we discover that between 1 and 4 we have to go via 2. On a graph that has some edges whose lengths are negative.4.) Find a significantly better algorithm for Problem 5.4. 2. The shortest path from I to 3 is thus 1.4.4. Problem 5.4 153 Shortest Paths Example 5. No efficient algorithm is known for finding shortest simple paths in graphs that may have edges of negative length. For the graph of Figure 5. On a graph that includes a negative cycle ? ii.3. the notion of "shortest path" loses much of its meaning : the more often we go round the negative cycle. the notion of a shortest simple path still makes sense. 3] = 4.2. the shorter our path will be ! Does Floyd's algorithm work i.2. Initially. If G includes cycles whose total length is negative. Even if a graph has edges with negative length.4. but that from 4 to 3 we proceed directly. j] = L [ j. P becomes 0 0 4 2 4 0 4 0 P = 0 1 0 0 0 1 0 0 Since P [1.3). L [i.1. (We shall see an asymptotically more efficient algorithm for this problem in Section 10. We want to find a matrix D such that D [i.3. Finally we see that the trips from 1 to 2 and from 2 to 4 are also direct.4. j] = false otherwise. i ]). j] = true if there exists at least one path from i to j. 4] and P [4.1. 3]. 4. j] = true if the edge (i. This is the situation we encountered in Problem 5. Warshall's algorithm. (We are looking for the reflexive transitive closure of the graph G. and D [i.) Adapt Floyd's algorithm for this slightly different case.2.

5 5. A binary search tree.5. the tree in Figure 5. we first examine the key held in the root. search trees will be understood to be binary.1 shows an example of a binary search tree containing the keys A. and less than or equal to the values contained in its righthand descendants. we only need look at the left-hand subtree . . If X=R. For a given set of keys. and the search stops . and if X > R. Suppose this key is R. A binary tree each of whose nodes contains a key is a search tree if the value contained in every internal node is greater than or equal to (numerically or lexicographically) the values contained in its left-hand descendants.5.5 OPTIMAL SEARCH TREES We begin by recalling the definition of a binary search tree. Write a procedure that looks for a given key in a search tree and returns true if the key is present and false otherwise. and less than or equal to the key of its righthand child.) To determine whether a key X is present in the tree.5.2. A recursive implementation of this technique is obvious. . Problem 5.2 contains the same keys as those in Figure 5. tinct keys ? How many different search trees can be made with eight dis- .5. The nodes may also contain further information related to the keys : in this case a search procedure does not simply return true or false.154 Dynamic Programming Chap. we only need look at the right-hand subtree. the information attached to the key we are looking for. but rather.1. if X < R. H. (It provides an example of simplification : see chapter 4. Show by an example that the following definition will not do : "A binary tree is a search tree if the key contained in each internal node is greater than or equal to the key of its left-hand child.) Problem 5. we have found the key we want. B. several search trees may be possible : for instance. Figure 5.5." Figure 5.3. (For the rest of this section.1.5.1. Problem 5. C ..5.

Suppose that in an optimal tree containing all the n keys this sequence of j-i + 1 keys occupies the nodes of a subtree. give a tree that minimizes the average number of comparisons needed. and (4+3+2+3+1+3+2+3)/8=21/8 comparisons on the average in Figure 5.1 two comparisons are needed to find the key E . 2.. . i=1 This is the function we seek to minimize. the average number of comparisons carried out in this subtree when we look Figure 5. Consider the sequence of successive keys ci . Problem 5.5. Recall that the depth of the root of a tree is 0. is held in a node of depth dk in the sub- tree. we shall solve a more general problem still. i = 1. If T (n) is the number of different search trees we can make with n distinct keys. suppose that Ein=1 pi = 1.5.4. all the requests refer to keys that are indeed present in the search tree.) In Figure 5. it takes (2+3+1+3+2+4+3+4)/8 = 22/8 comparisons on the average to find a key in Figure 5. ci + i . In fact.5.5. .5. n .i. i < k < j. For the case when the keys are equiprobable. . For a given tree the average number of comparisons needed is set c 1 < c 2 < * C= pi (di + 1).2 Another binary search tree. Let the probability that a request refers to key c. a single comparison suffices.. cj . the depth of its children is 1. on the other hand. and so on. that is..5.2. If all the keys are sought with the same probability. .5.3.5. 5. .5 155 Optimal Search Trees *Problem 5. (Hint: reread Section 5.1.Sec.2.. be pi . in Figure 5. Suppose we have an ordered < cn of n distinct keys. Repeat the problem for the general case of n equiprobable keys. j >. If the key Ck . For the time being. If some key ci is held in a node at depth di . find either an explicit formula for T (n) or else an algorithm to calculate this value. then di + 1 comparisons are necessary to find it.

ci + I . . To obtain a dynamic programming scheme. L and R. must occupy the root of the subtree. .I and R is an optimal subtree containing Ck +I . cJ when a key is sought in the main tree. .Dynamic Programming 156 Chap. L is an optimal subtree con- taining the keys ci .3. cJ . J where the three terms are the contributions of the root. .) One of these keys.5.. Cii = pi mm n ISk<J (Ci.. . In this case one comparison is made with Ck .. The average number of comparisons carried out is therefore CiJ = mil + Ci. c j is mil . To find the optimal search tree if the probabilities associated Example 5. with five keys c1 to c5 are . A subtree of the optimal search tree is optimal. When we look for a key in the main tree.. In Figure 5. . . ci +I . and let CiJ be the average number of comparisons carried out in an optimal subtree containing the keys ci . Ck .5. 5 for a key in the main tree (the key in question is not necessarily one of those held in the subtree) is CY. Let mij = Yk=i Pk .J) (*) .. say. it remains to remark that the root k is chosen so as to minimize Cij : CiJ =mil + In particular.5.. Pk(dk+l).1. We thus arrive at the principle of optimality : in an optimal tree all the subtrees must also be optimal with respect to the keys they contain. the probability that it is in the sequence ci .1. .. ci + I . k=i We observe that this expression has the same form as that for C. Figure 5. and a change in the subtree does not affect the contribution made to C by other subtrees of the main tree disjoint from the one under consideration. k.. k - I+ Ck + I . and others may then be made in L or R. (It is convenient to define CiJ = 0 if j = i .k-I +Ck+I. respectively.3.

43 0.58 + min(0.00 + min(1.20.88 1. 1 <_ i <_ 5.69. 0.70 0.18.45 5 0. C12+C45.09. 0.05 0.4).50. 1. 0. 0.40 Similarly C23=0.73.53 0. C14+C65) = 1. C22+C45. C13+C55. we note that Cii = pi .73 comparisons on the average to find a key (see Figure 5.61) = 0.12 Now.85. 1.30.05 0.5 157 Optimal Search Trees i 1 2 3 Pi 0. C24+C65) = 0.08 0. 0. C13+C54) = 0. C11+C35.49) =1.70 + min(0. 1.35 0.05.88 + min(0. and next. but how do we find the form of this tree ? 0 .76 C35 = m35 + min(C32+C45.61 C24 = m24 + min(C21+C34.35 + min(0.40) = 0. 0.38.61. 0.58 0. C12 = m12 + min(CI0+C22. CII+C32) = 0.18) = = 0. 0.73. necessary in the optimal tree. 0.13 0.91.12 we first calculate the matrix m. C12+C43) = 0. m = 0.08 4 0.18.57 0.85 C14=m14+min(C1o+C24.5.00 0.43 + min(0.00 C15=m15+min(C1o+C25.30 0.5. 0. C11 +C34.74. 5. C11+C33. We know how to find the minimum number of comparisons Problem 5. 0. C23+C54) = 0.65 0. C22+C44.65 + min(0. 0. The optimal search tree for these keys requires 1.69. we use (*) to calculate the other values of Cii. C34=0. C12+C44. 0.30 0.45 0. C45=0.76) = 1.Sec.15. Then C13 = m13 + min(C10+C23. 0. C23+C55.61) = 1.6. C33+C55.76.49 C25 = m25 + min(C21 +C35. C34+C65) = 0.

. . We now have n n . each involving a choice among m + 1 possibilities.. i = 1. .5.. Problem 5. * Problem 5. and let qj . we do not need their exact values. if it is present in the tree.j <.n. . m=1 Problem 5.k <. i = 1 .j and C. An optimal binary search tree. Prove this last equality. 1. =Mid + Ci .I+ Ck + 1.4. When j -i = m ..(In-t (n-m)(m+l)) =8(n3) . 5 Write an algorithm that accepts the values n and pi .Dynamic Programming 158 Chap. 2. . and so on.. n . be the probability that a request concerns a key ci that is in the tree. Figure 5.9.8. i = 0. The optimal tree must minimize the average number of comparisons required to either find a key. let pi .i <. ] } . 2. there are n -m values of C.5. and that produces a description of the optimal search tree for these probabilities. to calculate.pi +Iqi=1. ** Problem 5. let rid = max { k I i <. n .. For 1 <. The required computation time is therefore in . Specifically. then for j -i = 2. n.10.5. k . Give an algorithm that can determine the optimal search tree in this context.7. Generalize the preceding argument to take account of the possibility that a request may involve a key that is not in fact in the tree..) In this algorithm we calculate the values of Cij first for j -i = 1. or to ascertain that it is missing. (Provided the keys are sorted. be the probability that it concerns a missing key situated between ci and ci+1 (with the obvious interpretation for q0 and qn ). .5..5. 2.

If the circuit is optimal (as short as possible).5. recursively in the same way. followed by a path from j to 1 that passes exactly once through each node in N \ 11.10 to show how to calculate an optimal search tree in a time in 0(n2). c. Let G = < N.5. . S # N \ { 1 ).12. i ] = 0. . Consider a set of nodes S c N \ 111 and a node i EN \ S. after having gone exactly once through each of the other nodes. say.j +g(j. c. we take N = { 1. Ck+2 .+1. (Problems 5. if i # 1.11.S)=rni (L.5. c 2.. we see that g(1.j})). +I . j # 1. with i =1 allowed only if S = N \ { 1 } . i. n j. j). .... assuming the keys are already sorted ? ii. More generally. . Problem 5.11 generalize to the case discussed in Problem 5. Prove that r.and right-hand subtrees for c i . Write also r. . 5. at the root. we are required to find the shortest possible circuit that begins and ends at the same node.j _i <_rj <r.j for every 1 <_i j Sn. then so is the path from j to 1 : the principle of optimality holds.4. explicit example that this greedy algorithm does not always find the optimal search tree.S\ { j})).. j] = oo if the edge (i. and the lengths of the edges are denoted by L. Given a graph with nonnegative lengths attached to the edges.. Give an optimal search tree for your example. and calculate the average number of comparisons needed to find a key for both the optimal tree and the tree found by the greedy algorithm.. with L [i . and i 0 S. j }.N\{l. Suppose without loss of generality that the circuit begins and ends at node 1. and construct the left. cj . By the principle of optimality. g (1. 5. S) as the length of the shortest path from node i to node 1 that passes exactly once through each node in S. . S # 0. There is an obvious greedy approach to the problem of constructing an optimal search tree : place the most probable key.10 and 5. Define g (i.Sec. N \ { 1 }) is the length of an optimal circuit. . Using this definition. . Ck. ck _ 1 and Ck + i . g(i..2. As usual.) Problem 5.9.N\{1})=2<j<n min(Ljj +g(j. A > be a directed graph.. 2. jES (*) . j) does not exist.j . It therefore consists of an edge (1.5.6 THE TRAVELLING SALESPERSON PROBLEM We have already met this problem in Section 3.5. j ] >_ 0 if i and L [i. L [i . How much time does this algorithm take in the worst case. Show with the help of a simple. Use the result of Problem 5.5.6 The Travelling Salesperson Problem 159 be the root of an optimal subtree containing c. -I = i .

3. N \ { 11) and solve the problem. and so on. 0) = 6. we can use (*) to calculate g (1.1: 0 10 15 20 0 9 10 6 13 0 12 8 9 0 5 L = 8 We initialize g (2..1. We can apply (**) to calculate the function g for all the sets S that contain exactly one node (other than 1).6. Example 5.1. i =2. 15 6 Figure 5. 5 Furthermore. A directed graph for the travelling salesperson problem. 0) = 8 . g (3. Let G be the complete graph on four nodes given in Figure 5. N \ { 1. g (4. then we can apply (**) again to calculate g for all the sets S that contain two nodes (other than 1).. n . The values of g (i.6.. S) are therefore known when S is empty.Dynamic Programming 160 Chap. g(i.0)=Li1. j }) is known for all the nodes j except node 1. . 0) = 5.6.. Once the value of g ( j.

3. L 14 + g (4.14))=20 g(4.(2. (Continuation of Example 5.41) = min(L 12 + g (2.4)) = min(L23 + g (3.25) = 25 g (3. 5.4})=4 -3J(4. S) is the value of j chosen to minimize g at the moment when we apply (*) or (**) to calculate g(i. (3. L43 + g (3.3 )) = min(L42 + g (2.0)= 18 and similarly g(3.2. L 13 + g (3.{2})=18. (4)).S).3.31) =2 J(3. g(3.Sec.4)) J(1.4))=2 J(2. To know where this circuit goes.{3})= 15. 43) = 35 The optimal circuit in Figure 5.1 has length 35. { 2 } )) = min(31.25) = 25 g (4. we have g (2.13 1) =L23+g(3. (2.6.4 1).12. { 3) )) = min(29. L 34 + g (4.{3. { 2.) J(2. {2.4)) = min(L 32 + g (2. Example 5.6 The Travelling Salesperson Problem 161 Using (**). ( 2.{41)=L24+g(4.4)) = 2 and the optimal circuit is 1 -+J(1. { 3) ). g(4. we obtain g(2.3) )) = min(35.12))) = min(23. { 3.0)= 15 g(2.3. { 4) ).6.(31)=3 -1 The required computation time can be calculated as follows : In this example we find . 40. 1 2. 27) = 23 Finally we apply (*) to obtain g (1.{2. L24 + g (4.4)) =4 =4 J(4. Next. (2. using (**) for sets of two nodes.6.1.{3. we need an extra function : J (i .4 } ).{2})= 13.

5. S) that has already been calculated. which is not very practical either. Time: Direct method n! Time: Dynamic programming n22" Space: Dynamic programming n 2" 5 120 800 160 10 3.Dynamic Programming 162 calculate g (1.400 491. which data structure do you suggest to hold the values of g ? With your suggested structure.628.n . . as would be the case if we simply tried all the possible circuits. TABLE 5.400 10. 13 in Q(n 2" ). 0) : n -1 consultations of a table.520 20..240 15 1. Verify that the space required to hold the values of g and J is Problem 5. S) such that 1 :5 #S = k <.1. but it is still far from offering a practical algorithm.7 MEMORY FUNCTIONS If we want to implement the method of Section 5. n SOLVING THE TRAVELLING SALESPERSON PROBLEM.6. For instance.6. Since S is a set.1. 5 .520 Problem 5. it is easy to write a function that calculates g recursively.971.800 102.1 illustrates the dramatic increase in the time and space necessary as n goes up. how much time is needed to access one of the values of g ? Table 5.800 419.430. The preceding analysis assumes that we can find in constant time a value of g (j .6. whereas 20! microseconds exceeds 77 thousand years. 202220 microseconds is less than 7 minutes.2: (n -1) (n k 21 k additions in all.372.. What is more .(n -1)k In k 21) = 19(n22") k=1 since k1 k (k1 = r 2r-1 This is considerably better than having a time in SZ(n !).2.31 x 1012 20 2.6. The computation time is thus in 0(2(n -1) + .6 on a computer. N \ (1 }) : n -1 additions. These operations can be used as a calculate g ( j.43x1018 7. consider . For example. .to calculate all the g (i .

To the recursive function we add a table of the necessary size. (In fact. S ) if S = 0 then return L [i .distviaj return ans . Formulated in the following way : function g (i. In this way it is never necessary to calculate the function twice for the same values of its parameters. S ] > 0 then return gtab [i.1] ans <. all the entries in this table hold a special value to indicate that they have not yet been calculated. whenever we call the function. we return the value held in the table. we go ahead and calculate the function.7 Memory Functions 163 function g (i.L [i. S \ { j } ) if distviaj < ans then ans . If so. j ] + g ( j. it is not immediately obvious how to set about it. we save it at the appropriate place in the table.L [i.oo for each j E S do distviaj F.ans return ans . Before returning the calculated value. if we calculate g in this top-down way. . then all the sets containing two elements from N \ { 1). we come up once more against the problem outlined at the beginning of this chapter : most values of g are recalculated many times and the program is very inefficient. Initially. and so on.1 ] if gtab [i.distviaj gtab [i.1)!). One easy way to take advantage of the simplicity of a recursive formulation without losing the efficiency offered by dynamic programming is to use a memory function. S ] ans F oo for each j E S do distviaj . S ) if S = 0 then return L [i .) So how can we calculate g in the bottom-up way that characterizes dynamic programming ? We need an auxiliary program that generates first the empty set. Although it is maybe not too hard to write such a generator.6 let gtab be a table all of whose entries are initialized to -1 (since a distance cannot be negative). S ] . Unfortunately. however. For the algorithm of Section 5. we first look in the table to see whether it has already been evaluated with the same set of parameters. If not. j ] + g ( j. Thereafter. it ends up back in S2((n .Sec. 5. S \ { j } ) if distviaj < ans then ans . the function g combines the clarity obtained from a recursive formulation and the efficiency of dynamic programming. then all the sets containing just one element from N \ { I).

8. if any . Show how to calculate (i) a binomial coefficient and (ii) the function series (n .1. If we are willing to use a little more space (the space needed is only multiplied by a constant factor.8 SUPPLEMENTARY PROBLEMS Let u and v be two strings of characters.Dynamic Programming 164 Chap.) Show how an array T [1 . transform u into v with the smallest possible number of operations of the following types : delete a character. We sometimes have to pay a price for using this technique.. We want to * Problem 5..2. Show that this transformation is not optimal. n ] and a few pointers. n ] } procedure store (i . but we do not know in advance which ones.. . 5 Problem 5. This is particularly desirable when in fact only a few values of the function are to be calculated.7. abbac -* abac ababc abcbc (delete b) (add b ) (change a into c ).. n ] and P [1 . Implemented using a memory function. v ) { sets T [i ] to the value v } function val (i) { returns the last value given to T [i ]. add a character. For instance. it is possible to avoid the initialization time needed to set all the entries of the table to some special value. that we can calculate a binomial coefficient (k] using a time in 0 (nk) and space in 0(k). returns a default value (such as -1) otherwise } A call on any of these procedures or functions (including a call on init !) should take constant time in the worst case. You should write three algorithms. * Problem 5. however).6.2. n ] can be virtually initialized with the help of two auxiliary arrays B [1 . procedure init { virtually initializes T [1 . (For an example.7. for instance. see Section 6.2 using a memory function. we can transform abbac into abcbc in three stages. 5. p) of Section 5.1. the calculation takes the same amount of time but needs space in Si(nk).1. change a character. We saw in Section 5.

Find an efficient algorithm to determine the minimum cost of a trip by canoe from each possible departure point i to each possible arrival point j. what is the computing time needed by your algorithm ? Problem 5. Problem 5.1. how much time does your algorithm take ? Problem 5. In terms of N. in which case you can return the first canoe at some post k between i and j and continue the journey in a second canoe. c } . At any of these posts you can rent a canoe to be returned at any other post downstream. it can happen that the cost of renting from i to j is higher than the total cost of a series of shorter rentals. b.8.8. the length of the string x.5. This algorithm works correctly in a country where there are coins worth 1. 5. 5.1. For instance. Thus ab = b. (This expression is not unique.) In terms of n. Consider the alphabet E = { a. In the introduction to Chapter 3 we saw a greedy algorithm for making change.4.8. have the multiplication table given in Table 5. and that tells us what these operations are.2. ba = c. However.Sec. your algorithm should return "yes" because (b (bb ))(ba) = a. How much time does your algorithm take as a function of the lengths of u and v ? Problem 5.8. For example.8. There is no extra charge if you change canoes in this way. but it does not always find an optimal solution if there .3. Note that the multiplication defined by this table is neither commutative nor associative. and so on.8. if x = bbbba. and 25 units. (b (b (b (ba )))) = a as well. Modify your algorithm from the previous problem so it returns the number of different ways of parenthesizing x to obtain a. TABLE 5.8 Supplementary Problems 165 Write a dynamic programming algorithm that finds the minimum number of operations needed to transform u into v. There are N Hudson's Bay Company posts on the River Koksoak. AN ABSTRACT MULTIPLICATION TABLE Right-hand symbol Left-hand symbol a b c a b b b c b a a c a c C Find an efficient algorithm that examines a string x = x 1x2 x of characters of E and decides whether or not it is possible to parenthesize x in such a way that the value of the resulting expression is a.) For each possible departure point i and each possible arrival point j the company's tariff gives the cost of a rental between i and j. 10. The elements of Y. (It is next to impossible to paddle against the current.

Give a dynamic programming algorithm to calculate A (m.1. Give a dynamic programming algorithm that calculates all the c. 4). relations "<" and "=". how much time does your algorithm take? iii. or cii _+00 if the amount j cannot be obtained using just these coins. T[2]. Let L be a bound on the sum we wish to obtain. ii. and A (4. n). ii.n)=A(m-1. 1) This function grows extremely rapidly.A(m. it is restricted to using a space in 0 (m). 1 <_ j <.7. and let T [I . ** Problem 5.8. Moreover. Give a greedy algorithm that can make change using the minimum number of coins for any amount M <_ L once the % have been calculated.L. We suppose that an unlimited number of coins of each value is available. . Your algorithm 0. Give a recurrence for cij .1 . 3). For example. Let n be the number of different coins that exist. let cij be the minimum number of coins required to obtain the sum j if we may only use coins of types T [1]. . A (m. A =B =C A =C <B C <A =B A =B <C B <A =C C <A <B A <B =C B <A <C C <B <A A <B <C A <C <B B <C <A B =C <A Give a dynamic programming algorithm that can calculate.n-l)) ifm. as a function of n. 5 also exists a coin worth 12 units (see Problem 3.8. the number of different possible orderings. Your algorithm may use only a single array of length L. Your algorithm should take a time in 0 (n 2) and space in 0 (n).n >0. 0) = A (m . i. Your algorithm must consist simply of two nested loops (recursion is not allowed). should take a time in 0 (n + provided You have n objects. 5). For 1 <. Calculate A (2.. A (3. n ] be an array giving the value of these coins.i <_ n and 1 <_ j <_ L.1).. including the initial conditions. T[i].Dynamic Programming 166 Chap. As a function of n and L.1.6. i. The general problem can be solved exactly using dynamic programming. 13 different orderings are possible with three objects. although some of these memory words can .. Ackermann's function is defined recursively as follows : A(0. which you wish to put in order using the * Problem 5.n)=n+1 ifm >0 A(m.

4. including the solution to Problem 5. comes from Gilbert and Moore (1959). and Lauriere (1979). The algorithm of Section 5.8. The solution to Problem 5.5.3 is described in Godbole (1973) . 1973). as shown in Figure 5. ind [i ]). grow quite large. can be found in Hu and Shing (1982. able to solve the problem of chained matrix multiplications in a time in 0 (n log n). Cutting a hexagon into triangles.5 for constructing optimal search trees. 1984). A solution to Problem 5.5. a hexagon can be cut in 14 different ways. polygon into n .9 167 References and Further Reading 0 Figure 5.1.. For example. Bellman and Dreyfus (1962). 5. The algorithm in Section 5. All these algorithms (with the exception of Fredman's) are unified in Tarjan (1981). this . m ] and ind [0. Floyd's algorithm for calculating all shortest paths is due to Floyd (1962). (Hint : use two arrays val [0.10 and 5.3).8.1. Catalan's numbers are discussed in many places.9 REFERENCES AND FURTHER READING Several books are concerned with dynamic programming.8. Both Floyd's and Warshall's algorithms are essentially the same as the one in Kleene (1956) to determine the regular expression corresponding to a given finite automaton (Hopcroft and Ullman 1979). m ] such that at every instant val [i ] = A (i.10 that is both simpler and more general is given by Yao (1980). the (n -1)st Catalan number (see Section 5.1). A theoretically more efficient algorithm is known : Fredman (1976) shows how to solve the problem in a time in O (n 3(log log n / log n) 1/3 ). a more efficient algorithm.2 triangles using diagonal lines that do not cross is T (n .8.5. 11 5. We mention only Bellman 1957..5. Nemhauser (1966). including Sloane (1973) and Purdom and Brown (1985).) 11 Prove that the number of ways to cut an n -sided convex Problem 5.9.2 is supplied by the algorithm in Warshall (1962).11 come from Knuth (1971.Sec. The improvements suggested by Problems 5.

12. Memory functions are introduced in Michie (1968) .8. Hopcroft and Ullman (1974).2.8 is discussed in Sloane (1973). Problem 5. Problem 5. which suggests how to avoid initial- izing a memory function.8.6 suggested itself to the authors one day when they set an exam including a question resembling Problem 2. Problem 5.5 is discussed in Wright (1975) and Chang and Korsh (1976). . The algorithm for the travelling salesperson problem given in Section 5.8.7 is based on Ackermann (1928). which takes cubic time to carry out the syntactic analysis of any context-free language (Hopcroft and Ullman 1979).12 in Aho.5.168 Dynamic Programming Chap.6 comes from Held and Karp (1962).2). A solution to Problem we were curious to know what proportion of all the possible answers was represented by the 69 different answers suggested by the students (see also Lemma 10. The optimal search tree for the 31 most common words in English is compared in Knuth (1973) with the tree obtained using the greedy algorithm suggested in Problem 5. Problem 5.7. Problem 5.8. An important dynamic programming algorithm that we have not mentioned is the one in Kasimi (1965) and Younger (1967). comes from Exercise 2.1. 5 paper gives a sufficient condition for certain dynamic programming algorithms that run in cubic time to be transformable automatically into quadratic algorithms. for further details see Marsh (1970).1 is given in Wagner and Fischer (1974).

the graph exists only implicitly. of the node we are in the process of visiting) and possibly representations of a few other positions. Up to now. For instance. we often need to look at all the nodes. it does not really exist in the memory of the machine. In this chapter we introduce some general techniques that can be used when no particular order of visits is required. we often use abstract graphs to represent games : each node corresponds to a particular position of the pieces on the board. Most of the time.1 INTRODUCTION A great many problems can be formulated in terms of graphs. When we explore such a graph. To solve such problems. the structure of the problem is such that we need only visit some of the nodes or edges.6 Exploring Graphs 6. to "find a neighbouring node" means to follow a pointer. We have seen. and the edges are represented by pointers. of a graph. and so on. We shall use the word "graph" in two different ways. the shortest edge. A graph may be a data structure in the memory of a computer. Sometimes. all we have is a representation of the current position (that is. and the fact that an edge exists between two nodes means that it is possible to get from the first to the second of these positions by making a single legal move. the shortest route problem and the problem of the minimal spanning tree. The operations to be carried out are quite concrete : to "mark a node" means to change a bit in memory. or all the edges. At other times. the algorithms we have seen have implicitly imposed an order on these visits : it was a case of visiting the nearest node. or to avoid arriving at 169 . the nodes are represented by a certain number of bytes. for instance. In this case to "mark a node" means to take any appropriate measures that enable us to recognize a position we have already seen. In this case. and so on.

do + c where d is a constant such that d ? 2c. However. g nodes are situated in the left-hand subtree. if we visit first the left-hand subtree. Then T(n) <- max (T(g) + T(n-g -1) + c) n > 0 O<g <n -1 We prove by mathematical induction that T (n) <.n < m . for some m > 0.2 TRAVERSING TREES We shall not spend long on detailed descriptions of how to explore a tree. For each of the six techniques mentioned. the time required is bounded above by some constant c. We simply remind the reader that in the case of binary trees three techniques are often used. then the right-hand subtree. 6 the same position twice . In this chapter we therefore do not distinguish the two cases. 0 <. Proof. we are traversing the tree in preorder . Preorder and postorder generalize in the obvious way to nonbinary trees. By the choice of c the hypothesis is true for n = 0. Suppose further that we are to explore a tree containing n nodes. and finally.T (0).1. and so on. n > 0. the node itself. These three techniques explore the tree from left to right. then the node itself. of which one node is the root. then all the nodes in the left-hand subtree. the techniques used to traverse it are essentially the same. then we are visiting the tree in postorder. +3c -d <. to "find a neighbouring node" means to change the current position by making a single legal move . that +c . If at each node of the tree we visit first the node itself. and n -g -1 nodes are in the right-hand subtree. Suppose that visiting a node takes a time in O(1). whether the graph is a data structure or merely an abstraction. the right-hand subtree. we may suppose that c >. It is obvious how to implement any of these techniques using recursion. and finally. the time T (n) needed to explore a binary tree containing n nodes is in O(n).2. Three corresponding techniques explore the tree from right to left. and if we visit first the left- hand subtree. Lemma 6. we are traversing the tree in inorder. Without loss of generality. and lastly.170 Exploring Graphs Chap. Now suppose that it is true for all n . all the nodes in the right-hand subtree. Then T (m) <_ <_ max (T (g) + T (m -g -1) + c) max (dg +c +d(m-g-1)+c +c) 0<g <m-1 0<g <m-1 <.

it is clear that T (n) is in c2(n) since each of the n nodes is visited. if there is another node adjacent to v that has not been visited. Problem 6. Therefore T (n) is in O(n). procedure search (G) for each v E N do mark [v] f.3 DEPTH-FIRST SEARCH : UNDIRECTED GRAPHS Let G = < N.visited for each node w adjacent to v do if mark [w] # visited then dfs (w ) . call the procedure recursively once again. Prove that both these techniques still run in a time in the order of the number of nodes in the tree to be traversed.1.2. and call the procedure yet again. This proves that T (n) 5 do + c for every n >_ 0.not-visited for each v E N do if mark [v] # visited then dfs (v ) procedure dfs (v : node ) [ node v has not been visited } mark [v] . To carry out a depth-first traversal of the graph. 6. choose any one of them as a new starting point. Next. the search starting at v is finished. Show how to generalize the concepts of preorder and postorder to arbitrary (nonbinary) trees. choose this node as a new starting point and call the depth-first search procedure recursively.3.5. 6. Mark this node to show that it has been visited.3 Depth-First Search : Undirected Graphs 171 so the hypothesis is also true for n = in. On return from the recursive call. Here is the recursive algorithm.2. On the other hand. choose any node v E N as the starting point.Sec.2. When all the nodes adjacent to v have been marked. if there is a node adjacent to v that has not yet been visited.9. choose this node as the next starting point. even when the nodes do not contain a pointer to their parents (otherwise the problem becomes trivial). Assume the trees are represented as in Figure 1. and hence T (n) is in O (n). Problem 6. Initially. Suppose that it is somehow possible to mark a node to indicate that it has already been visited.2. Continue in this way until all the nodes of G have been marked. a recursive implementation takes memory space in S2(n) in the worst case. A > be an undirected graph all of whose nodes we wish to visit. and so on. Prove that for any of the techniques mentioned. * Problem 6. no nodes are marked. Show how the preceding exploration techniques can be implemented so as to take only a time in 0(n) and space in 0(1). If there remain any nodes of G that have not been visited.

6 The algorithm is called depth-first search since it tries to initiate as many recursive calls as possible before it ever returns from a call. If the graph is represented in such a way as to make the lists of adjacent nodes directly accessible (type lisgraph of Section 1.3. Show how a depth-first search progresses through the graph in Figure 6.2). a depth-first search of the graph in Figure 6.3. this work is proportional to a in total. we look at the mark on each of its neighbouring nodes. there are n calls of the procedure dfs.1 progresses as follows : dfs (1) dfs (2) dfs (3) dfs (6) dfs (5) dfs (4) dfs (7) dfs (8) initial call recursive call recursive call recursive call recursive call . and that node 1 is the first starting point. When we visit a node.3.1 if the neighbours of a given node are examined in numerical order but the initial starting point is node 6.3. progress is blocked there are no more nodes to visit. At this point the recursion "unwinds" so that alternative possibilities at higher levels can be explored. progress is blocked a neighbour of node 1 has not been visited recursive call recursive call. n)). 0 Problem 6.1. An undirected graph.1. Figure 6.3. The recursivity is only stopped when exploration of the graph is blocked and can go no further. Example 6. If we suppose that the neighbours of a given node are examined in numerical order. The algorithm therefore takes a time in 0 (n) for the procedure calls and a time in 0 (a) to inspect the marks. How much time is needed to explore a graph with n nodes and a edges ? Since each node is visited exactly once. The execution time is thus in 0 (max(a.Exploring Graphs 172 Chap.9.1. .

The corresponding directed edges (1.3. Show how depth-first search can be used to find the connected components of an undirected graph.3. Problem 6. the tree and the numbering generated by a depth-first search in a graph are not unique.5).3. (3. A depth-first search also provides a way to number the nodes of the graph being visited : the first node visited (the root of the tree) is numbered 1.pnum + 1 prenum [v] E-.1. 6.1. To implement this numbering. (Continuation of Example 6. but rather a forest of trees.3.2. the second is numbered 2.1) The edges used in the depth-first search of Example 6. a depth-first search associates to it not merely a single tree.3. 14.1 numbers the nodes as follows : node prenum 1 1 2 2 3 3 4 6 5 5 6 4 The depth-first search illus- 7 7 8 8 11 Of course. 12.8).2).pnum where pnum is a global variable initialized to zero. we need only add the following two statements at the beginning of the procedure dfs : pnum F. Example 6.3. Edges that are not used in the traversal of the graph have no corresponding edge in the tree. the nodes of the associated tree are numbered in preorder.Sec.9. If the graph being explored is not connected.4.3.1 are (1. Example 6. 3).7) and (7. they are directed from the first node visited to the second. A depth-first traversal of a connected graph associates a spanning tree to the graph. 2).1) trated by Example 6. Exhibit the tree and the numbering generated by the search of .3.6).3 Depth-First Search : Undirected Graphs 173 What happens if the graph is represented by an adjacency Problem 6. The initial starting point of the exploration becomes the root of the tree.2) rather than by lists of adjacent nodes ? Problem 6. (1.3.3. (6.2. The edges of the tree correspond to the edges used to traverse the graph . matrix (type adjgraph of Section 1. but depend on the chosen starting point and on the order in which neighbours are visited. In other words. and so on. (2. form a spanning tree for the graph in Figure 6. one for each connected component of the graph. (Continuation of Example 6. The root of the tree is node 1. See Figure 6.3. and so on. Problem 6.

6.3. The values of lowest are calculated in postorder. 8) . say. and for each node v of the graph. 6 6.7. A graph G is biconnected (or unarticulated) if it is connected and has no articulation points.3. a.2.3. 3.61 and (4. These ideas are important in practice : if the graph G represents.5. there remain two connected components (2. and the value of lowest [v] to the right.3. and 6. Carry out a depth-first search in G. starting from any node. then the fact that it is biconnected assures us that the rest of the network can continue to function even if the equipment in one of the nodes fails . The search Example 6. and 1 successively. calculate lowest [v] as the minimum of i.2. Traverse the tree T in postorder.5. c.1 Articulation Points A node v of a connected graph is an articulation point if the subgraph obtained by deleting v and all the edges incident on v is no longer connected. Let T be the tree generated by the depth-first search.Exploring Graphs 174 Chap. let prenum [v ] be the number assigned by the search. 2.3. 6. The edges of G that have no corresponding edge in T are represented by broken lines. It is bicoherent (or isthmus free. For each node v visited.1.3. (Continuation of Examples 6. The articulation points of G are nodes 1 (by rule c(i)) and 4 (by rule c(ii)). or 2-edge-connected) if each articulation point is joined by at least two edges to each component of the remaining sub-graph. w } in G that has no corresponding edge in T W. we can be sure that all the nodes will be able to communicate with one another even if one transmission line stops working. the search at node 6. H. For example. that is.3. a telecommunications network. 8. prenum [v] ii. Articulation points are now determined as follows: i. prenum [w] for each node w such that there exists an edge [ v. Problem 6. A node v other than the root of T is an articulation point of G if and only if v has a child x such that lowest [x] >_ prenum [v]. lowest [x] for every child x of v in T.3. 4.3. for nodes 5.1 . 7. node 1 is an articulation point of the graph in Figure 6. if G is bicoherent. The value of prenum [v] appears to the left of each node v.1 generates the tree illustrated in Figure 6. The following algorithm finds the articulation points of a connected graph G .3) described in Example 6.4.3. The root of T is an articulation point of G if and only if it has more than one child. Verify that the same articulation points are found if we start 11 . b. if we delete it.

3.3.3. Show how to carry out the operations of steps (a) and (b) in parallel and write the corresponding algorithm. we can define lowest [v] by lowest [v] = min(prenum [w] I you can get to w from v by following down as many solid lines as you like and then going up at most one broken line) . On the other hand. there thus exists a chain of edges that joins x to the other nodes of the graph even if v is deleted.8. there is no chain joining lowest [x] >_ prenum [v]. .3. Problem 6. If x is a child of v and if lowest [x] < prenum [v]. x to the parent of v if v is not the root and if Problem 6.2.3. 6.3.Sec.10.3 Depth-First Search : Undirected Graphs Figure 6. Write an algorithm that decides whether or not a given con* Problem 6.6. 175 A depth-first search tree . given an undirected graph that is connected but not biconnected. Prove that an edge of G that has no corresponding edge in T Problem 6.7. Complete the proof that the algorithm is correct. nected graph is bicoherent.2) necessarily joins some node v to one of its ancestors in T. Your algorithm should find the smallest possible set of edges. finds a set of edges that could be added to make the graph biconnected.9. prenum on the left and lowest on the right. Write an efficient algorithm that. * Problem 6.3. (a broken line in Figure 6. Informally. Analyse the efficiency of your algorithm.

w) exists and (w .3 shows that the time taken by this algorithm is also in 0 (max(a.4 DEPTH-FIRST SEARCH : DIRECTED GRAPHS The algorithm is essentially the same as the one for undirected graphs. the algorithm progresses as follows : 1. then w is adjacent to v but v is not adjacent to w. If the neighbours of a given node are examined in numerical order. 6. The algorithm behaves quite differently. v) does not. ii. dfs (2) dfs (3) dfs (4) dfs (8) dfs (7) dfs (5) dfs (6) 4. the edges used to visit all the nodes of a directed graph G = <N. 6 Prove or give a counterexample : i. 9.1. In a directed graph.3. With this change of interpretation the procedures dfs and search from Section 6. dfs(1) 2. In this case.13. If a graph is bicoherent. 7.4. A > may form a forest of several trees even . Prove that a node v in a connected graph is an articulation point if and only if there exist two nodes a and b different from v such that every path joining a and b passes through v. If a graph is biconnected. Illustrate the progress of the algorithm if the neighbours of a Problem 6. w) exists.11. given node are examined in decreasing numerical order.3. 8. the difference being in the interpretation of the word "adjacent". initial call recursive call recursive call . Problem 6. and if the starting point is node 1. 6.4. then it is biconnected. Consider a depth-first search of the directed graph in Figure 6.1. biconnected graph. n)). there exist at least two chains of edges joining v and w that have no nodes in common (except the starting and ending nodes). 3.3. progress is blocked a neighbour of node 1 has not been visited recursive call recursive call . node w is adjacent to node v if the directed edge (v . 5. If (v .3 apply equally well in the case of a directed graph.Exploring Graphs 176 Problem 6. 0 An argument identical with the one in Section 6. however. Prove that for every pair of distinct nodes v and w in a Problem 6. progress is blocked new starting point recursive call . progress is blocked there are no more nodes to visit. however.12. then it is bicoherent. Chap.

In the case of a directed graph three kinds of edges can appear in A \ F (these edges are shown by the broken lines in Figure 6. Problem 6. This happens in our example : the edges used. and W.2) or (6.4. 4). (The numbers to the left of each node are explained in Section 6.4. (4. where the values of prenum are attributed as in Section 6. 1) or (7.1. i. 3).4 Depth-First Search : Directed Graphs Figure 6.6). Those like (3. A depth-first search forest. namely (1.3.Sec.2. form the forest shown by the solid lines in Figure 6. 6).4) that lead from a node to one of its ancestors .4. and if v is neither an ancestor nor a descendant of w in the forest.3) that join one node to another that is neither its ancestor nor its descendant. (1.4. Prove that if (v .2. (2. if G is connected. 8) that lead from a node to one of its descendants .2. ii. In the case of an undirected graph the edges of the graph with no corresponding edge in the forest necessarily join some node to one of its ancestors (Problem 6. and (5. 7). those like (5.4.) Let F be the set of edges in the forest.4.2. (8. 6. w) is an edge of the graph that has no corresponding edge in the forest. then prenum [v] > prenum [w]. 177 A directed graph. 2). those like (1. 8).3. Edges of this type are necessarily directed from right to left.2). . Figure 6.

d) . from the initial state to final completion. (What is the partial ordering in question?) Finally. directed graph G = <N. but is less general than the class of all directed graphs. 6 6.3.4. For example. Figure 6. from a node of G to one of its ancestors in the forest).4.4 illustrates part of another partial ordering defined on the integers. Another directed acyclic graph.1 Acyclic Graphs: Topological Sorting Directed acyclic graphs can be used to represent a number of interesting relations. This class of structures includes trees. Prove that G is acyclic if and only if A \ F includes no 11 edge of type (i) (that is. a directed acyclic graph can be used to represent the structure of an arithmetic expression that includes repeated subexpressions : thus Figure 6. Figure 6.4. A >. Figure 6.5 gives an example of this type of diagram. Let F be the forest generated by a depth-first search on a Problem 6.4. and the edges correspond to activities that have to be completed to pass from one state to another.Exploring Graphs 178 Chap.4.4.3 A directed acyclic graph.4. Figure 6. directed acyclic graphs are often used to represent the different stages of a complex project : the nodes are different states of the project.4. . Such graphs also offer a natural representation for partial orderings (such as the relation "smaller than" defined on the integers and the set-inclusion relation).3 represents the structure of the expression (a +b) (c +d) + (a +b) (c . Depth-first search can be used to detect whether a given directed graph is acyclic.

Despite the fact that there exist edges (1.4. If a directed graph is not strongly connected. as are several others.4 Depth-First Search : Directed Graphs 179 C make coffee drink coffee dress D get out G bring documents Figure 6. Yet another directed acyclic graph.4.2 Strongly Connected Components A directed graph is strongly connected if there exists a path from u to v and also a path from v to u for every distinct pair of nodes u and v. 4. Prove this.8). F. In the graph of Figure 6. 2.Sec. Each of these subgraphs is called a strongly connected component of the original graph. .4) and (1. Problem 6. D.8}. C.7. 12.4.4. the order A. For graphs as in Figure 6. Another component corresponds to the nodes {4. the natural order 1. Problem for instance. the numbers of the nodes will be printed in reverse topological order. For example. 6. 6. E.4. then i precedes j in the list. 3. 6. 8. we are interested in the largest sets of nodes such that the corresponding subgraphs are strongly connected. 3) and the corresponding edges form a strongly connected component. For the graph of Figure 6. 2. in our example. 8. what is the topological order obtained if the neighbours of a node are visited in numerical order and if the depth-first search begins at node I ? 6. 3. j ). The necessary modification to the procedure dfs to make it into a topological sort is immediate : if we add a supplementary line write v at the end of the procedure.5. it is not possible to merge these two strongly connected components into a single component because there exists no path from node 4 to node 1.5.1. 2.4.4. 4. A topological sort of the nodes of a directed acyclic graph is the operation of arranging the nodes in order in such a way that if there exists an edge (i. 24 is also acceptable. G will serve. a topological ordering of the states can be used to obtain a feasible schedule for the activities involved in the project. but the order 1.4. 12.5. 24 is adequate. for the graph of Figure 6. nodes 11. B.

we number each node at the moment when exploration of the node has been completed.6. 21 and '{ 4. the search reaches nodes 5 and 6. this time the search reaches nodes 1.1. 7. For each node v of the graph let postnum [v] be the number assigned during the search. 3. To do this.3 we number each node at the instant when exploration of the node begins.180 Exploring Graphs Chap.4. 7. and so on. it follows that postnum [w] = n. with postnum [1] = 6.2 shows to the left of each node the number thus assigned. 11.1) are the subgraphs corresponding to the sets of nodes (5. Prove that if two nodes u and v are in the same strongly connected component of G. (If G contains n nodes. 8 1. To each tree in the resulting forest there corresponds one strongly connected component of G. 3. 6 To detect the strongly connected components of a directed graph. and 8 are all reached. and 2. The graph G' is illustrated in Figure 6. We carry out a depth-first search starting from node 5. iii. The corresponding forest is illustrated in Figure 6.7. On the graph of Figure 6.4. For our second starting point.4. with the values of postnum shown to the left of each node. the nodes of the tree pro- duced are numbered in postorder.4.4.4.) If the search starting at w does not reach all the nodes. this time the remaining nodes 4. The strongly connected components of the original graph (Fig. Here.4. Begin this search at the node w that has the highest value of postnum. we must first modify the procedure dfs. 6. iv. Figure 6. Construct a new graph G' : G' is the same as G except that the direction of every edge is reversed. . Carry out a depth-first search of the graph starting from an arbitrary node. ii.1. The following algorithm finds the strongly connected components of a directed graph G . choose as the second starting point the node that has the highest value of postnum among all the unvisited nodes . i. For the third starting point we take node 4.4. Example 6. In Section 6. we choose node 1.nump where nump is a global variable initialized to zero. In other words.6. 6). the first depth-first search assigns the values of postnum shown to the left of each node in Figure 6.2. since postnum [5] = 8. we need only add at the end of procedure dfs the following two statements : nump F nump + 1 postnum [v] . then they are in the same tree when we carry out the depthfirst search of G'. * Problem 6. with postnum [4] = 5 . Carry out a depth-first search in G'.

But this is quite impossible.4. When we carried out the search in G.4. there exists at least one path from v to r in G.2). In the third case it would be necessary for the same reason that r be to the right of v. Since we chose r rather than v to be the root of the tree in question. since there exists a path from v to x in G. The forest of strongly connected components. any such path must go up the tree from v to a common ancestor (x. We should have postnum [x] > postnum [r] since x is an ancestor of r. When carrying out the search of G'. Let v be a node that is in the tree whose root is r when we carry out the search of G'. Before choosing r as . or r was neither an ancestor nor a descendant of v. and suppose v #r. Since in a depth-first search the edges not used by the search never go from left to right (Problem 6.Sec. as the root of a new tree) that node not yet visited with the highest value of postnum. Next. The second possibility is ruled out by the fact that postnum [r] > postnum [v]. 6. However.1. and then go down the tree to r. This implies that there exists a path from r to v in G' . 0 O O 0 O Figure 6. we always choose as a new starting point (that is. say) of v and r.4. r was a descendant of v . It is harder to prove the result the other way. thus there exists a path from v to r in G.4.6.4 Depth-First Search : Directed Graphs Figure 6.7. we have postnum [r] > postnum [v]. there would exist a path from x to v in G'. 181 Reversing the arrows in the graph of Figure 6. three possibilities seem to be open a priori: r was an ancestor of v .

Let stack be a data type allowing two operations. on the other hand. 6. We have thus proved that if node v is in the tree whose root is r when we carry out the search of G'. then there exist in G both a path from v to r and a path from r to v. If two nodes u and v are in the same tree when we search G'.5 BREADTH-FIRST SEARCH When a depth-first search arrives at some node v. this completes the proof that the algorithm works correctly. The function top denotes the element at the top of the stack. first served". This type represents a list of elements that are to be handled in the order "first come. it first visits all the neighbours of v. . 6 the root of a tree in the search of G'. and so on. Unlike depth-first search.visited push v on P while P is not empty do while there exists a node w adjacent to top (P) such that mark [w] # visited do mark [w] . by contrast.4. Only the first possibility remains : r was an ancestor of v when we searched G. This contradicts the hypothesis that v is in the tree whose root is r when we carry out the search of G. we begin by giving a nonrecursive formulation of the depth-first search algorithm. breadth-first search is not naturally recursive.4. it next tries to visit some neighbour of v. push and pop. The function first denotes the element at the front of the queue. procedure dfs'(v : node) P F empty-stack mark [v] t. Here is the modified depth-first search algorithm. With the result of Problem 6. then a neighbour of this neighbour. Estimate the time and space needed by this algorithm. first served".7.Exploring Graphs 182 Chap. we need a type queue that allows two operations enqueue and dequeue. and not until this has been done does it go on to look at nodes farther away. When a breadth-first search arrives at some node v. Here now is the breadth-first search algorithm.6. we would have already visited x (otherwise x rather than r would be chosen as the root of the new tree) and therefore also v. The type is intended to represent a list of elements that are to be handled in the order "last come.visited push w on P { w is now top (P) } pop top (P) For the breadth-first search algorithm. Problem 6. they are therefore both in the same strongly connected component of G since there exist paths from u to v and from v to u in G via node r. This implies that there exists a path from r to v in G. To underline the similarities and the differences between the two methods.

5. 7 8 8. n)). and if node I is used as the starting point. After a breadth-first search in an undirected graph G = <N. Node Visited Q 1.3.1 shows the tree generated by the search in Example 6. 4 5.1. Problem 6. but that neither u nor v is an ancestor of the other. procedure search (G) for each v E N do mark [v] . 6. 8 - As for depth-first search.visited enqueue v into Q while Q is not empty do u <--first (Q ) dequeue u from Q for each node w adjacent to u do if mark [w] # visited then mark [w] t.5 Breadth-First Search 183 procedure bfs (v : node ) Q . visited in numerical order.visited enqueue w into Q In both cases we need a main program to start the search.8 7. The edges of the graph that have no corresponding edge in the tree are represented by broken lines. 2 3. 3 4.6 3.1. we can associate a tree with the breadth-first search.6 4.5. Figure 6.8 5. 1 2.4. namely 0 (max(a. breadth-first search proceeds as follows.1.not-visited for each v E N do if mark [v] # visited then { dfs' or bfs } (v ) On the graph of Figure 6. v } E A \ F are such that u and v are in the same tree.empty-queue mark [v] F-.5.7. let F be the set of edges that have a corresponding edge in the forest of trees that is generated.5.6.8 6. It is easy to show that the time required by a breadth-first search is in the same order as that required by a depth-first search.Sec.5.7. Show that the edges { u . 6 7. A >.1. if the neighbours of a node are Example 6.3.4 2. If the appropriate . 5 6.5.

Relevant portions of the graph can thus be built as the search progresses.2.5. The economy in memory space is even more dramatic when nodes that have already been searched can be discarded. A breadth-first search tree. How many kinds of "broken" edges are possible ? (see Section 6. Problem 6.1.6 IMPLICIT GRAPHS AND TREES As mentioned at the outset of this chapter.1. For instance. If the graph contains a large number of nodes. Show how a breadth-first search progresses through the graph of Figure 6. various problems can be thought of in terms of abstract graphs. interpretation of the word "neighbouring" is used. it may be wasteful or infeasible to build it explicitly in computer memory before applying one of the search techniques we have encountered so far. One powerful application is in playing games of strategy by techniques known as minimax and alpha-beta . Often the original problem translates to searching for a specific node. An implicit graph is one for which we have available a description of its nodes and edges. assuming that the neighbours of a node are always visited in numerical order.3 (Continuation of Problem 6.4) Breadth-first search is most often used to carry out a partial exploration of certain infinite graphs or to find the shortest path from one point to another. the breadth-first search algorithm can be applied without modification to either directed or undirected graphs.2). we can use the nodes of a graph to represent configurations in a game of chess and edges to represent legal moves (see Section 6.5. making room for subsequent nodes to be explored.5.2) Sketch the corresponding forest and the remaining edges of the graph.4. path or pattern in the associated graph.5.6. 6. Therefore computing time is saved whenever the search succeeds before the entire graph has been constructed.Exploring Graphs 184 Chap. Problem 6. 6 Figure 6. and that necessary starting points are also chosen in numerical order. Backtracking is a basic search technique on implicit graphs.

4. since the number of positions we have to check would be [64 J = 4. These graphs are usually trees. we can write the algorithm very simply using eight nested loops. This approach is of no practical use. Consider the classic problem of placing eight queens on a chess-board in such a way that none of them threatens any of the others. each giving the position of the queen in the corresponding row. or at least they contain no cycles.1 Backtracking Backtracking algorithms use a special technique to explore implicit directed graphs.1.6. 2.6..299. To illustrate the general principle. 7) does not represent a solution since the queens in the third and the sixth rows are in the same column. We now discuss these notions.i2. . 6. although in fact the algorithm finds a solution and stops after considering only 1. Using this representation. 6.426. or on the same diagonals. Solve this problem without using a computer. program Queens 1 for ii-lto 8do for i2-lto 8do for i8-lto 8do try F-(iI.777.165. i8) if solution (try) then write try stop write "there is no solution" This time. 6. in the same column. For instance.. 216.6 Implicit Graphs and Trees 185 pruning. 8. looking for solutions to some problem. we shall. The first improvement we a than one queen in any given row. 6. the number of cases to be considered is reduced to 88 = 16.. 1. This might try consists of never putting more reduces the computer representation of the chess-board to a simple vector of eight elements. however. A backtracking algorithm carries out a systematic search.Sec. use a different example. checking each time to see whether a solution has been obtained. The first obvious way to solve this problem consists of trying systematically all the ways of placing eight queens on a chess-board..852 cases. (A queen threatens the squares in the same row.368.) Problem 6. the vector (3. and also two pairs of queens lie on the same diagonal. Some optimization problems can be handled by the more sophisticated branch-and-bound technique. At least one application of this technique dates back to antiquity : it allows one to find the way through a labyrinth without danger of going round and round in circles.

. procedure perm (i) if i = n then use (T) { T is a new permutation } else for j F.830 cases are in fact considered before the algorithm finds a solution. that is.6. . which prevents us from ever trying to put two queens in the same row. the informa0 tion just given should be of considerable help ! Once we have realized that the chess-board can be represented by a vector.6. as a function of n . only 2. easier in this case to verify whether a given position is a solution. The algorithm becomes program Queens 2 try initial permutation while try # final-permutation and not solution (try) do try F. n ] is a global array initialized to [ 1. Although it is more complicated to generate permutations rather than all the possible vectors of eight integers between 1 and 8. 0 line. Since we already know that two queens can neither be in the same row nor in the same column. for each of these leading values. it is natural to be equally systematic in our use of the columns. If the preceding algorithm is used to generate the permutations.i to n do exchange T [i] and T [ j ] perm (i +1) exchange T [i] and T [ j ] Here T [ 1 .2.3. on the other if solution (try) then write try else write "there is no solution" There are several natural ways to generate systematically all the permutations of the first n integers. all the permutations of the n -1 remaining elements.. If you have not yet solved the previous problem. show the result of calling perm (1) when n = 4. Hence we now represent the board by a vector of eight different numbers between 1 and 8. to execute the call perm (1) ? Now rework the problem 0 assuming that use (T) takes a time in 0(n).4.. This approach reduces the number of possible cases to 8! = 40.320. . it is. For instance. n ] and the initial call is If use (T) consists simply of printing the array T on a new Problem 6. perm (1). how much time is Problem 6. we might put each value in turn in the leading position and generate recursively. Assuming that use (T) takes constant time.Exploring Graphs 186 Chap. 6 Problem 6.6. it suffices to verify that they are not in the same diagonal. by a permutation of the first eight integers. 2. needed. .

Let N be the set of k-promising vectors. This check can be speeded up if we associate with each promising node the sets of columns. even the best of these algorithms makes 720 useless attempts to put the last six queens on the board when it has started by putting the first two on the main diagonal. we progressed first to a method that never puts two queens in the same row. V ) E A if and only if there exists an integer k. it seems at first sight that we have to check each of the 28 pairs of queens on the board. a vector V is k-promising if. of positive diagonals (at 45 degrees). 0. .. As a first step. On the other hand. 6. let us reformulate the eight queens problem as a tree searching problem. (k.Sec. . Let G = <N. we only need to check the last queen to be added.. For instance. to decide if some given permutation represents a solution. . V [k]) threatens any of the others. V [2]) . and of negative diago^. The solutions to the eight queens problem can be obtained by exploring this tree. k] of integers between I and 8 is k-promising.320. Mathematically. 0 5 k S 8. nodes are generated and abandoned during the course of the exploration. j . In fact. and then to a better method still where the only positions considered are those where we know that two queens can neither be in the same row nor in the same column.6 Implicit Graphs and Trees 187 Starting from a crude method that tried to put the queens absolutely anywhere on the chess-board. This graph is a tree. We say that a vector V [1 . 2. in order to decide whether a vector is k-promising.j . For k <. such that U is k -promising. Secondly. 0. the number of nodes in the tree is less than 8! = 40... k]. for 0S k 5 8. where try [I . This technique has two advantages over the previous algorithm that systematically tried each permutation. knowing that it is an extension of a (k -1)-promising vector. 0. where of course they threaten one another ! Backtracking allows us to do better than this. it is straightforward to count the nodes using a computer : #N = 2057.11s (at 135 degrees) controlled by the queens already placed. Solutions to the eight queens problem correspond to vectors that are 8-promising. 8] where it is impossible to place a queen in the next row without her threatening at least one of the queens already on the board. V [1]). all these algorithms share an important defect : they never test a position to see if it is a solution until all the queens have been placed on the board. Although it is not easy to calculate this number theoretically. it suffices to explore 114 nodes to obtain a first solution. Its root is the empty vector (k =0). Depth-first search is the obvious method to use. To print all the solutions to the eight queens problem. 0 <_ k < 8. However. 0). however : rather.1. 4. and U[i]=V[i] for every i E [1 . for every i #j between 1 and k. V is (k +l )-promising. if none of the k queens placed in positions (1. we have V[i] -V [j ] it (i . We do not generate the tree explicitly so as to explore it thereafter. (2. A > be the directed graph such that (U. particularly if we only require one solution. Its leaves are either solutions (k = 8) or else they are dead ends (k < 8) such as [1. call Queens (0. First.i ). 5. 8] is a global array. any vector V is k -promising.

the tree explored by the backtracking algorithm contains only 856. diag45 = {try[i]-i +l I 1 !5i !5k). col={try [i]I 1!5 i:5k}. ** Problem 6. k +11 is (k +1)-promising } Queens (k +1.6. the number of nodes in the tree of k -promising vectors.. k ] do backtrack (w [I . k ]) { v is a k -promising vector) if v is a solution then write v { otherwise } for each (k +1)-promising vector w such that w [1 .1 to 8 do if j Ocol and j -k Odiag45 and j +k Odiag 135 then try [k +I] F. diag 135) { try [ 1 .. For example. diag45. 6 procedure Queens (k. How does this number compare to n! ? Backtracking algorithms can also be used even when the solutions sought do not necessarily all have the same length.189 nodes. col u { j }.546.Exploring Graphs 188 Chap. and diag135={try [i]+i-1I 15i <-k} } if k = 8 then { an 8-promising vector is a solution } write try else { explore (k+1)-promising extensions of try for j F... As we might expect.5.001. .j } { try (L. k +1]) The otherwise should be present if and only if it is impossible to have two different solutions such that one is a prefix of the other. and the first solution to be found (using the generator given previously) corresponds to the 4. as a function of the number n of queens. and a solution is obtained when the 262nd node is visited. diag45 u { j -k }. Here is the general scheme.. for n = 12 there are 479. procedure backtrack (v [ 1 . Find a more interesting case than n = 2. Problem 6. k ] = v [I . col. Analyse mathematically.6. the advantage to be gained by using the backtracking algorithm instead of an exhaustive approach becomes more pronounced as n increases. on the other hand.044th position examined . diag 135 u { j +k)) It is clear that the problem generalizes to an arbitrary number of queens : how can we place n queens on an n x n "chess-board" in such a way that none of them threatens any of the others ? Show that the problem for n queens may have no solution.6.600 possible permutations to be considered. k ] is k -promising.

The graph is infinite if there is no a priori limit on the number of positions possible in the game. A node of the graph corresponds to a particular position in the game.9. green.Sec. we assume that the game is played by two players.6. The n-queens problem was solved using depth-first search in the corresponding tree. (This last constraint does not apply to the eight queens problem where each solution involves exactly the same number of pieces. Instant Insanity is a puzzle consisting of four coloured cubes. For instance. or white.2 Graphs and Games: An Introduction Most games of strategy can be represented in the form of directed graphs. and that chance plays no part in the outcome (the game is deterministic). 6. In this case. Some problems that can be formulated in terms of exploring an implicit graph have the property that they correspond to an infinite graph.10. . The ideas we present can easily be adapted to more general contexts. If the required change is impossible. The four cubes are to be placed side by side in such a way that each colour appears on one of the four top faces.7. it may be necessary to use breadth-first search to avoid the interminable exploration of some fruitless infinite branch. For simplicity.6 Implicit Graphs and Trees 189 Problem 6. and on one of the rear faces. 6. Problem 6. and hence some nodes in the graph have no successors : these are the terminal positions. your algorithm should say so rather than calculating forever.8.6. and an edge corresponds to a legal move between two positions. Problem 6.6.6. Each of the 24 faces is coloured blue. Show how to solve this problem using backtracking. Give an algorithm that determines the shortest possible series of manipulations needed to change one configuration of Rubik's Cube into another. 15 can be transformed into 4 using four function applications : 4 = gfgg (15). In particular. on one of the front faces. some positions in the game offer no legal moves. Breadth-first search is also appropriate if we have to find a solution starting from some initial position and making as few changes as possible. that the game is symmetric (the rules are the same for both players). each of whom moves in turn. Give other applications of backtracking.) The two following problems illustrate these ideas. Give an algorithm capable of transforming some initial integer n into a given final integer m by the application of the smallest possible number of transformations f(i) = 3i and g(i) = Li/21. red. on one of the four bottom faces.6. What does your algorithm do if it is impossible to transform n into m in this way ? * Problem 6. We further suppose that no instance of the game can last forever and that no position in the game offers an infinite number of legal moves to the player whose turn it is.

my opponent may take one. i. On the other hand. if at the outset I choose to remove a single match.1. There are seven matches on the table initially. except that he must take at least one and he must leave at least one. at least two matches are placed on the table between two players. but this is not necessarily the case (think of stalemate in chess).Exploring Graphs 190 Chap.4. we need only attach to each node of the graph a label chosen from the set win.1). if you find yourself in a terminal position. For most games. Example 6. On the other hand. or to remove more than two. or four. The player who has the first move in a game with seven matches is therefore cer- tain to win provided that he does not make an error. The labels assigned to terminal positions depend on the game in question. The labels are assigned systematically in the following way.11. If he takes only one match. A nonterminal position is a losing position if all of its successors are winning positions. I can in turn remove a single match. iv. . Problem 6. and he cannot prevent me from winning on my next turn. draw. then there is no legal move you can make. iii. you may verify that a player who has the first move in a game with eight matches cannot win unless his opponent makes an error.6. The player who removes the last match wins.6.6. leaving four matches on the table. Any remaining positions lead to a draw. If I take two of them. In the case of an acyclic finite graph (corresponding to a game that cannot continue for an indefinite number of moves). The label corresponds to the situation of a player about to move in the corresponding position. Initially.12. We illustrate these ideas with the help of a variant of Nim (also known as the Marienbad game). Grasp intuitively how these rules arise. 6 To determine a winning strategy for a game of this kind. ii. each player in turn must remove at least one match and at most twice the number of matches his opponent just took. If he takes more than one. then you may verify that my opponent has a winning strategy. two. assuming that neither player will make an error. and you have lost . I can remove all the matches that are left and win. lose. three. Thereafter. There are no draws. A nonterminal position is a winning position if at least one of its successors is a losing position. find a relationship between this method of labelling the nodes and topological sorting (Section 6. Can a player who finds himself in a winning position lose if his opponent makes an "error"? Problem 6. The first player removes as many matches as he likes.

It is also necessary to know the upper limit on the number of matches that it is permissible to remove on the next move. corresponding to the fact that such positions offer no winning move. but only < 0.6 Implicit Graphs and Trees 191 * Problem 6. 0 > is interesting : the nodes < i.j. < 7.6.6. Your characterization of n should be as simple as possible. 0 > for i > 0 are inaccessible. Part of a game graph. For n >.14. j > go to the j nodes < i -k. n 2.1. < 6. The square nodes represent losing positions and the round nodes are winning positions. min(2k. j >. give a necessary and sufficient condition on n to ensure that the player who has the first move in a game involving n matches have a winning strategy.1. whereas he does have such a strategy in the game with four matches. or five matches has no winning strategy.6.2. is < n . 6 >. i -k)>. Add nodes < 8. A position in this game is not specified merely by the number of matches that remain on the table. choose one of the heavy edges in order to win.Sec. There are no heavy edges leaving a losing position. Problem 6. 1 <_ j <. 1 k <. The nodes of the graph corresponding to this game are therefore pairs < i . The node corresponding to the initial position in a game with n matches. The edges leaving node < i .1 shows part of the graph corresponding to this game. . j > with j odd and j < i -1 cannot be reached starting from any initial position. Figure 6.13. nodes < i. In general.6. 6. three. 7 >. Figure 6. indicates that i matches remain on the table and that any number of them between 1 and j may be removed in the next move. All the nodes whose second component is zero correspond to terminal positions. The heavy edges correspond to winning moves : in a winning position. Prove your answer. < i . n -1 >. We observe that a player who has the first move in a game with two.6. 5 > and their descendants to the graph of Figure 6. Similarly.i . j >.

k + 1 G [i . rec (2.! while k < j and G [i -k. j ] F. min(2k.k < j. j > is a winning position. are there positions in which several different winning moves are available? Can this happen in the case of a winning initial position <n. G [i j] is set to true if and only if configuration < i. before calculating G [i .6. rec (5. 3) also calls rec (2. (If you want to work on this problem.Exploring Graphs 192 Chap. i -k)] do k E.i } for k F.1 to n do for j f. as well as all the values of G [i. we assume that 0 j <. j > is winning } G [0.j if it is a winning move to take away k matches.false for i f. 0] E. 1).17.15. min(2k.1 to i do k E. do not read the following paragraphs yet !) The first approach consists of using dynamic programming to create a Boolean array G such that G [i. rec (3.not G [i -k . 2) and rec (1. j ] = true if and only if < i.k <.5: it calculates the same value over and over. k ] for 1 <. Problem 6.I < i . 6 Problem 6. j ].n-1>? The obvious algorithm to determine whether a position is winning is the following.16. min(2k . i -k )] . 2) and rec (1. 1). This algorithm suffers from the same defect as the algorithm fib! in Section 1.6. For instance. calculating all the values of G [1. function rec (i. procedure dyn (n) {foreach1<-j <<-i n. 4) returns false having called successively rec (4.1 to j do if not rec (i -k. k ] for 1 <.7.k <. we proceed in a bottom-up fashion. Find two ways to remove this inefficiency. Can a winning position have more than one losing position among its successors ? In other words.6. 3). j > is winning . j ) { returns true if and only if the node < i. but rec (3. Modify this algorithm so that it returns an integer k such that k = 0 if the position is a losing position and 1 <. As usual with dynamic programming. 2). i -k)) then return true return false Problem 6.

is an algorithm for determining a winning strategy that is more efficient than any of those given previously.j <.7. j]. initialized to false..true return true G [i. n. 6.14 > is a winning position as soon as we discover that its second successor < 13. however.k < j. In fact. j] init [i. j] then return G [i. only 27 nodes are really useful when we calculate G [15. A solution that combines the advantages of both the algorithms consists of using a memory function (Section 5. although the dynamic programming algorithm looks at 121 of them.Sec. It is no longer of any interest to know whether the next successor < 12. k]. there is no particular reason to favour this approach over dynamic programming.6. j ). 1 <.. we know that < 15. 1 <. G [1. to avoid this initialization and to obtain a worthwhile gain in efficiency. n ]. The initial call of precond (n) is an application of the preconditioning technique to be discussed in the next chapter.6 Implicit Graphs and Trees 193 The preceding algorithm only _ises G [0. j] E. however. without explanation. In this context dynamic programming leads us to calculate wastefully some entries of the array G that are never needed.. 6 > is a winning or a losing position.i. n.l < i . Thereafter a call on whatnow (i . j] false return false At first sight. . function nim (i. n ] is global. The game we have considered up to now is so simple that it can be solved without really using the associated graph. since these nodes are never of interest. Here. it never calculates an unnecessary value.true for k . n ]. The recursive algorithm given previously is inefficient because it recalculates the same value several times. Because of its top-down nature. j] . 4 > is a losing position. k ] for 1 <. i -k)) then G [i.6]. 0. In an initial position with n matches. 14]. first call precond (n).18. This involves remembering which nodes have already been visited during the recursive computation using a global Boolean array init [0..1 to j do if not nim (i -k. Show how to improve its efficiency by also using the values of G [i . About half this work can be avoided if we do not calculate G [i . j ) if init [i. 0] and the values of Problem 6. For instance. j ] whenever j is odd and j < i -1. Using the technique suggested in Problem 5. but there is no "bottom-up" reason for not calculating G [12.7). determines in a time in O(1) the move to make in a situation where i matches remain on the table and the next player has the right to take at most j of them. min(2k.2 allows us.k <.. because in any case we have to take the time to initialize the whole array init [0. to calculate G [i. where n is an upper bound on the number of matches to be used. The array T [0. 0.

this graph allows us to play a perfect game of chess. j ) if j < T [i] then { prolong the agony ! } return 1 return T [i] * Problem 6. We also need to know whose turn it is to move. even with the fastest existing computers. the International Chess Federation has rules that prevent games dragging on forever : for example. Thanks to such rules. there are no cycles in the graph corresponding to chess. which rooks and kings have moved since the beginning of the game (to know if it is legal to castle). that is. However. Unfortunately (or perhaps fortunately for the game of chess). or a draw. Prove that this algorithm works correctly and that precond (n) takes a time in 0(n). say.) Adapting the general rules given at the beginning of this section. we can therefore label each node as being a winning position for White.2k do k F k + 1 T[i] . placed on a chess-board.19. Consider now a more complex game.6. Furthermore. 6 procedure precond (n) T [0] t-for i F 1 to n do k<-1 while T [i -k ] <. a winning position for Black.20. but also by an invisible item of information giving the number of matches that can be removed on the next move. At first sight. the graph contains so many nodes that it is quite out of the question to explore it completely. namely chess. as well as the older rule that makes a game a draw if the pieces return three times to exactly the same positions on the board. (For simplicity we ignore exceptions to the 50-move rule. For simplicity ignore the fact that certain positions are . a position is defined not merely by the number of matches on the table. Remember first that in the game we just looked at. a position in chess is not defined simply by the positions of the pieces on the board.Exploring Graphs 194 Chap. Thus we must include in our notion of position the number of moves made since the last irreversible action. to win whenever it is possible and to lose only when it is inevitable. the graph associated with this game contains cycles. Estimate the number of ways in which the pieces can be * Problem 6. a game is declared to be a draw after 50 moves in which no irreversible action (movement of a pawn. Once constructed. or a capture) took place. since if two positions u and v of the pieces differ only by the legal move of a rook. Similarly.k function whatnow (i. the king not being in check. then we can move equally well from u to v and from v to u.6. and whether some pawn has just been moved two squares forward (to know whether a capture en passant is possible). this problem disappears on closer examination.

It is customary to give values not too far from zero to positions where neither side has a marked advantage.W It is clear that this simplistic approach would not be very successful using the evaluation function suggested earlier.val then val . If the static evaluation function were perfect. that is. -oc if White has been mated. 5 points for each white rook. it underlies an important heuristic called minimax. and 10 points for each white queen . Since a complete search of the graph associated with the game of chess is out of the question. Although it does not allow us to be certain of winning. and large negative values to positions that favour Black. control of the centre. For example. Suppose it is White's turn to move from position u. count 1 point for each white pawn. and that both kings must be on the board). it would be easy to determine the best move to make. This is in a sense merely a systematic version of the method used by some human players that consists of looking ahead a small number of moves. the evaluation function should return +oo if Black has been mated. 31/4 points for each white bishop or knight. When applied to a terminal position. The first step is to define a static evaluation function eval that attributes some value to each possible position. Then we make the move that seems to cause our opponent the most trouble. an evaluation function that takes good account of the static aspects of the position but that is too simplistic to be of real use might be the following : for nonterminal configurations. Ignore also the possibility of 11 having promoted pawns. Here we give only an outline of the technique. The minimax principle. we want the value of eval (u) to increase as the position u becomes more favourable to White. This technique finds a move that may reasonably be expected to be among the best moves possible while exploring only a part of the graph starting from some given position. using one of several possible criteria. freedom of movement. they can never be obtained from the initial position by a legal series of moves (but take into account the fact that each bishop moves only on either white or black squares. val F-oo for each configuration w that is a successor of u do if eval(w) >. The best move would be to go to the position v that maximizes eval (v) among all the successors w of u. and the positions thus reached are evaluated heuristically. subtract a similar number of points for each black piece.eval(w) V f. In this situation the recursive approach comes into its own. and so on. Ideally.Sec. it is not practical to use a dynamic programming approach. Exploration of the graph is usually stopped before the leaves are reached. A compromise must be made between the accuracy of this function and the time needed to calculate it. since it would not hesitate to sacrifice a queen in order to take a pawn ! .6 Implicit Graphs and Trees 195 impossible. 6. and 0 if the game is a draw. This evaluation function must take account of many factors : the number and the type of pieces remaining on both sides.

val then val f.Black (w. n -1) 1 x is a successor of w } function White (x. .196 Exploring Graphs Chap. On the other hand. n) if n = 0 or x has no successor then return eval(x) else return max { Black (w. it is preferable to look several moves ahead.W where the functions Black and White are the following : function Black (w.-°° for each configuration w that is a successor of u do if Black (w. n) >. and White. (Ideally.min{eval(x) I x is a successor of w } if valw >. To add more dynamic aspects to the static evaluation provided by eval. which of course may be exactly the wrong rule to apply if it prevents White from finding the winning move : maybe if he looked further ahead the gambit would turn out to be profitable. n -1) 1 w is a successor of x } We see why the technique is called minimax : Black tries to minimize the advantage he allows to White.) We are now looking half a move ahead. we are sure to avoid moves that would allow Black to mate immediately (provided we can avoid this). a better strategy for White is to assume that Black will reply with the move that minimizes the function eval.val then val F. White should move to the position v given by val <.eval(w) else valw . val --°° for each configuration w that is a successor of u do if w has no successor then valw . n) if n = 0 or w has no successor then return eval(w) else return min { White (x. tries to maximize the advantage he obtains from each move. 6 If the evaluation function is not perfect. n) V f. To look n half-moves ahead from position U. since the smaller the value taken by this function.valw v t-W There is now no question of giving away a queen to take a pawn. on the other hand. he would like a large negative value. the better the position is supposed to be for him.

6. he will choose the second of the three possible moves. besides the fact that it would take far too long to calculate in practice? Justify your answer.3. If A plays so as to maximize his advantage. Example 6. Alpha-beta pruning. This assures him of a value of at least 10. Look back at Figure 6. the exploration of certain branches can be abandoned early if the information we have about them is already sufficient to show that they cannot possibly influence the values of nodes farther up the tree.6 Implicit Graphs and Trees Player Rule A max Figure 6.Sec.6.21.2. Similarly. 12800). Let < i.6. 1 <_ j 5 18. 6.2 shows part of the graph corresponding to some game.6.6. it may be worthwhile to explore the most promising moves in greater depth. In the example we suppose that player A is trying to maximize the evaluation function and that player B is trying to minimize it.2. This second type of improvement is generally known as alpha-beta pruning. Figure 6. Example 6. Let u correspond to the initial position of the pieces. 1 > starting from the values calculated by the function eval for the leaves < 4. visiting the successors of a given node from left to right.2. To do this. j > represent the j th node in the i th row of the tree. What can you say about White (u. the values for the other nodes can be calculated using the minimax rule. We give just one simple example of the technique. 197 The minimax principle. For example. We want to calculate the value of the root < 1. If the values attached to the leaves are obtained by applying the function eval to the corresponding positions. The basic minimax technique can be improved in a number of ways. j >. we carry out a bounded depth-first search in the tree.6. Problem 6. .

3. we visit only 19 of the 31 nodes in the tree.6.6. whereas node < 2.23.2>. 1> (a node that maximizes eval) has value at least -7. Thus as soon as the first leaf <4.Exploring Graphs 198 Chap. (This has already been done !) ** Problem 6. 3 > cannot have any influence on the value of node < 2.22. 3 > has value at most 1.6. 1> has value -7 and that <3. < 3. 1 > (a node that minimizes eval) has value at most 5. we say that the corresponding branches of the tree have been pruned. Similarly.6. 11>. Since we already know that the value of < 1. Node < 2. game of strategy. Write a program capable of playing brilliantly your favourite 11 ** Problem 6.6. Write a program that can beat the world backgammon champion.2> has value 5. 1 >. To establish that the value of the root <1.4> at the situation illustrated in Figure 6.3>. I> is at least 10.24. we have to transmit immediately to the higher levels of the tree any information obtained by evaluating a leaf. 1> is evaluated. . and < 2. 6 If we want to abandon the exploration of certain branches because it is no longer useful. after evaluation of the leaf <4. there is no need to evaluate the other children of node < 2. we know that <4. After evaluation of the second leaf <4.3.6. 3 >. Since node <3. ** Problem 6. the exact value of node < 3.1> is 10. 3 > has value at least 10. we know that <4. 1 > has value 5. It is therefore unnecessary to evaluate the other descendants of node <3. 1 > has value at most -3. we arrive after evaluation of the leaf <4. we are in the situation shown in Figure 6. -3 10 ? ? Alpha-beta pruning.4. What modifications should be made to the principles set out in this section to take account of those games of strategy in which chance plays a certain part ? What about games with more than two players ? Player Rule A max B min A max B eva! -7 5 Figure 6. Continuing in this way.

branch-and-bound is a technique for exploring an implicit directed graph.Sec.6). An example illustrates the technique. 6. This time. More often. we are looking for the optimal solution to some problem. At each node we calculate a bound on the possible value of any solutions that might happen to be farther on in the graph. a breadth-first search finishes exploring nodes in the order of their creation.6 Implicit Graphs and Trees Player Rule A max 199 It is not necessary to explore these branches B eval Figure 6. as we have just explained.2 and 5. If the bound shows that any such solution must necessarily be worse than the best solution we have found so far.6. In the simplest version. More alpha-beta pruning. We return to the travelling salesperson problem (see Sections . 3. this graph is usually acyclic or even a tree. In general terms we may say that a depth-first search finishes exploring nodes in inverse order of their creation.3 Branch-and-Bound Like backtracking. Again.4. using a stack to hold those nodes that have been generated but not yet explored fully . to prune certain branches of a tree or to close certain paths in a graph. and serves only. Branch-and-bound uses auxiliary computations to decide at each instant which node should be explored next.4.4. 6. Example 6. the calculated bound is used not only to close off certain paths. but also to choose which of the open paths looks the most promising.6. calculation of these bounds is combined with a breadthfirst or a depth-first search. using this time a queue to hold those that have been generated but not yet explored (see Section 6. then we do not need to go on exploring this part of the graph. so that it can be explored first. however.5). and a priority list to hold those nodes that have been generated but not yet explored.6.

For instance. 10/2.2). a complete tour must include a departure from node 1. or 5: minimum 7/2 A visit to 3 that neither comes from I nor leaves for 2: minimum 11/2 A similar visit to 4: minimum 3 A similar visit to 5: minimum 3 A return to 1 from 3.5. For instance. a visit to each of the nodes 2. namely. In Figure 6. or 5: minimum 2 . Its length is therefore at least 2+6+4+3+3+2= 20. The bound for node (1. the minimum of 14/2. Returning to node 1 costs at least 2. and (1.3.Exploring Graphs 200 Chap.3) corresponds to two complete tours : (1. 5. For instance. 2) must include The trip 1-2 : 14 (formally. 4. 4. and the other half when we arrive at j. 4.3). suppose that half the distance between two points i and j is counted at the moment we leave i. 1) and (1. 4/2. (This bound on the root of the implicit tree serves no purpose in the algorithm . The nodes in the implicit graph correspond to partially specified paths.4. To obtain a bound on the length of a path.6.5).) Our search begins by generating (as though for a breadth-first search) the four possible successors of the root. this arbitrary choice of a starting point does not alter the length of the shortest tour. 6 Let G be the complete graph on five points with the following distance matrix : 014 4 10 20 14 0 7 8 7 4 5 0 7 16 11 7 9 0 2 18 7 17 4 0 We are looking for the shortest tour starting from node 1 that passes exactly once through each other node before finally returning to node 1. At each node we calculate a lower bound on the length of the corresponding complete tours. nodes (1. 2. leaving I for 2 and arriving at 2 from 1: 7+7 ) A departure from 2 toward 3. To calculate this bound. visiting node 2 costs us at least 6 (at least 5/2 when we arrive and at least 7/2 when we leave).2. and 20/2. 3. We have just calculated the lower bound shown for this node. for example. is calculated as follows. 4/2.5 the root of the tree specifies that the starting point for our tour is node 1. A tour that begins with (1.4. and 18/2. Obviously. and 5 (not necessarily in this order) and a return to 1. it was computed here for the sake of illustration. 1). Notice that this calculation does not imply the existence of a solution that costs only 20. namely the lowest of the values 14/2. 11/2. 2). 3. The successors of a given node correspond to paths in which one additional node has been specified. (1. 4. node (1. it suffices to add elements of this kind. (1.4). Similarly. leaving node I costs us at least 2.

3). 2 :Z Bound 31 1.4) and (1.5) are generated. . Next.6 Implicit Graphs and Trees 201 1 Bound 20 1.3. 2). We find that the length of the tour (1.5) of this node are therefore generated. as node (1. 5. 4. for instance. corresponds to exactly one complete tour (1. 5 Bound 41 Figure 6. Its two children (1. the most promising node seems to be (1.5.3. The three children (1.2. which gives a total length of at least 24.3.2. 2) as follows : The trip 1.2.3. and (1.4.2). (1. 6.2. 1).4).3. 2. The length of such a tour is therefore at least 31. we do not need to continue exploration of the nodes (1. Branch-and-bound. To give just one example.3. 2). 1) is 31.5.Sec. we do not need to calculate a lower bound since we may calculate immediately its length 37. we calculate the bound for node (1. The most promising node is now (1. 3. This time.6. If we are only concerned to find one optimal solution. whose bound is 24. The other bounds are calculated similarly.4). 3.3 -2: 9 A departure from 2 toward 4 or 5: minimum 7/2 A visit to 4 that comes from neither 1 nor 3 and that leaves for neither 2 nor 3: minimum 3 A similar visit to 5: minimum 3 A return to 1 from 4 or 5: minimum 11/2. 3.3.

6.Exploring Graphs 202 Chap. In the worst case it may turn out that even an excellent bound does not allow us to cut any branches off the tree. (ii) using a breadth-first search. The heap is an ideal data structure for holding this list. Nevertheless. (Why?) There remains only node (1.3. situated in all the levels of the tree and preferably sorted in order of the corresponding bounds. it almost always pays to invest the necessary. however. we have looked at merely 15 of the 41 nodes that are present in a complete tree of the type illustrated in Figure 6. we find that the tour (1. There is always a compromise to be made concerning the quality of the bound to be calculated : with a better bound we look at less nodes. 3. Problem 6. the optimal solution does not come from there. 3. 1). After looking at the two complete tours (1. To obtain our answer. which cannot possibly lead to a better solution. 3) was the most promising node. but on the other hand. graph is in fact a tree (i) using a depth-first search .7. for problems of the size encountered in applications.25.6. 5). no elegant recursive formulation of branch-and-bound is available to the programmer. 3.5. we shall most likely spend more time at each one calculating the corresponding bound. 5. Problem 6. 2. The need to keep a list of nodes that have been generated but not yet completely explored. * Problem 6.4) is pointless. 5) and (1.27.6. Implement this algorithm on a computer and test it on our example. How much time do your algorithms take? .4. Unlike depth-first search and its related techniques.6. 1) and (1. 5. 2.7 SUPPLEMENTARY PROBLEMS Write algorithms to determine whether a given undirected Problem 6. time in calculating the best possible bound (within reason). 6. Solve the same problem using the method of Section 5.6. In practice.5). one finds applications such as integer programming handled by branch-and-bound. the technique is sufficiently powerful that it is often used in practical applications. 2.26. For instance.4) to explore. the bound at each node being obtained by solving a related problem in linear programming with continuous variables. This example illustrates the fact that although at one point (1. 1) of length 30 is optimal. and all the extra work we have done is wasted. It is next to impossible to give any idea of how well the technique will perform on a given problem using a given bound. The only child to offer interesting possibilities is (1. 3. makes branch-and-bound quite hard to program. 6 (1. 4. 4. 4.1. 5. Even exploration of the node (1. Show how to solve the same problem using a backtracking algorithm that calculates a bound as shown earlier to decide whether or not a partially defined path is promising.

it is permissible to go to adjacent points in the same row or in the same column.4. general.7. if M [i..7. and prints the path if so. j] is true. j ).) A Boolean array M [ 1 . ii. A > is called a sink if for every node v E N . you may use statements such as "for each point v that is a neighbour of x do "). 6. and if so. * Problem 6.3.7. Write an algorithm that determines whether or not a given graph has an Euler path. Operations are executed from left to right. 1) to (n.7.6.7. then you may pass through point (i. Euler's problem.2.4 for a directed graph. How much time does your algorithm take? *Problem 6.7. p) exists. An Euler path in a finite undirected graph is a path such that every edge appears in it exactly once. .7. Without being completely formal (for instance. Without giving all the details of the algorithm. Give a backtracking algorithm that finds a path. We want to express 13 in this way. indicate how to solve this problem by branch-and-bound. The value 1 is available. To construct other values. Your algorithm should accept the graph represented by its adjacency matrix (type adjgraph of Section 1.7. (This approach is used in a number of compilers. Problem 6. Write an algorithm to determine whether a given directed graph is in fact a rooted tree.. whereas the edge (p . i. Repeat Problem 6. For instance 10= 1x2x2x2x2/3x2.1 gives an example. from (1. Show how the problem can be expressed in terms of exploring a graph and find a minimum-length solution. j ). A node p of a directed graph G = <N. Figure 6. Write an algorithm that can detect the presence of a sink in G in a time in O (n). then you may not pass through point (i. How much time does your algorithm take? * Problem 6.7 Supplementary Problems 203 Problem 6. n ] represents a square maze. if one exists. v) does not exist.7. j] is false. v #p. n).5.9. Notice that a running time in 0 (n) for this problem is remarkable given that the instance takes a space in Q(n 2) merely to v:'rite down.8.7. starting from a given point. * Problem 6. the edge (v . any resulting fraction being dropped). to find the root. you have available the two operations x2 (multiplication by 2) and /3 (division by 3. If M [i. your algorithm must be clear and precise. I . In Problem 6.Sec. Show how the problem of carrying out a syntactic analysis of a programming language can be solved in top-down fashion using a backtracking algorithm. n.2).

1. Lawler (1976). Other algorithms based on depth-first search appear in Aho. Several applications of depth-first search are taken from Tarjan (1972) and Hopcroft and Tarjan (1973).7. and Ullman (1974. and Deo (1977).8 REFERENCES AND FURTHER READING There exist a number of books concerning graph algorithms or combinatorial problems that are often posed in terms of graphs.2 is given in Robson (1973). Gondran and Minoux (1979). A linear time algorithm for testing the planarity of a graph is given in Hopcroft and Tarjan (1974). Backtracking is described in Golomb and Baumert (1965) and techniques for analysing its efficiency are given in Knuth (1975a). Nievergelt. The branch-andbound technique is explained in Lawler and Wood (1966). The use of this technique to solve the travelling salesperson problem is described in Bellmore and Nemhauser (1968). The latter is analysed in Knuth (1975b). and alpha-beta pruning. Hopcroft. 6 17 Figure 6. and Tarjan (1983). 6. Papadimitriou and Steiglitz (1982). A solution of problem 6. Even (1980). the minimax technique. The book by Nilsson (1971) is a gold mine of ideas concerning graphs and games.23) is given in Deyong (1977). 1983). 1970). The mathematical notion of a graph is treated at length in Berge (1958.2. . We mention in chronological order Christofides (1975). Some algorithms for playing chess appear in Good (1968).Exploring Graphs 204 -+ 11 .3. Problem 6. Reingold. consult Berliner (1980). For a more technical description of this feat. -+ Chap.6. A lively account of the first time a computer program beat the world backgammon champion (Problem 6. A maze.10 is solved in Rosenthal and Goldner (1977).

1. This is preconditioning. A preconditioning algorithm for this problem is an algorithm A that accepts as input some element j of J and produces as output a new algorithm Bj . Example 7. 7.1 PRECONDITIONING 7.1. it is sometimes worthwhile to invest some time in calculating auxiliary results that can thereafter be used to speed up the solution of each instance. Suppose each instance jet can be separated into two components j E J and k E K (that is. Pascal. k > of the original problem. The general problem is to know whether a given program is syntactically correct with respect to some given language. In this case I is the set of instances of the type "Is kEK a valid program in the language defined by the grammar j E J ? ".1. Even when there is only one instance to be solved. Simula. For example. precomputation of auxiliary tables may lead to a more efficient algorithm. J might be a set of grammars in Backus-Naur form for such languages as Algol. Let J be a set of grammars for a family of programming languages. This algorithm Bj must be such that if k E K and < j. Let K be a set of programs. I c J x K ). k > E I.1 Introduction Let I be the set of instances of a given problem. then the application of Bj on k gives the solution to the instance < j .7 Preconditioning and Precomputation If we know that we shall have to solve several similar instances of the same problem. 205 . and so on.

then. Let J be a set of sets of keywords. 1 pour.=1 with preconditioning.) if we work without preconditioning. for example. . it generates a compiler Bj for the language in question. Example 7.. . be possible to calculate and store ahead of time #J preconditioned algorithms. .) .. Thereafter. to know whether k E K is a program in language j. Obviously. k > directly It is usually the case that bj (k) <_ t( j. { si.1. b. Let a( j) = the time required to produce Bj given j b j (k) = the time required to apply Bj to k t ( j. k). we simply apply the O compiler Bj to k. to stop a runaway reactor. a. we are wasting our time using preconditioning if bj (k) > t ( j. k2>. k) = the time required to solve < j. In this case it is sometimes impractical to calculate and store ahead of time the #1 solutions to all the relevant instances. kn > with the same j. for example to ensure a sufficiently fast response time for a real-time application. k) 5 a ( j) + bb (k). and t2=a(j)+ n bj (k. In this case the time taken to solve all the instances is ti = t(j. on the other hand. 7 One possible preconditioning algorithm for this example is a compiler generator : applied to the grammar j E J . jusqu'a. else. k > from scratch consists of producing Bj from j and then applying it on k. k 1 >. <j. <j. for example J = { { if. and on the other hand. Whenever n is sufficiently large.2. sinon. We need to be able to solve any instance i E I very rapidly. { for. pas 1 1. It may.Preconditioning and Precomputation 206 Chap. Such an application of preconditioning may be of practical importance even if only one crucial instance is solved in the whole lifetime of the system : this may be just the instance that enables us. by 1. Preconditioning can be useful in two situations. it often happens that t2 is much smaller than t I. finsi } .k. endif 1. The time you spend studying before an exam may also be considered as an example of this kind of preconditioning. one way of solving < j. We have to solve a series of instances <j. alors. to.

we have t(j. If we expect to have several systems of equations to solve with the same matrix A but different vectors b. Example 7. If we solve each instance directly. w > of nodes. Calculating the necessary values cj is an example of preconditioning that allows us subsequently to make change quickly every time this is required. recursively. function}. For a given pair k = < v. We have to solve a large number of instances of the type "Is the keyword kEK a member of the set j E J ? ".k)E8(nj) in the worst case.8. (k) E O( log nj ) for the search If there are many instances to be solved for the same j. then the second technique is clearly preferable. if we start by sorting j (this is the preconditioning).1. On the other hand. 7. We are to solve the system of equations Ax = b . 7. k > by a binary search algorithm. so that we can subsequently solve any particular instance in a time in 0(1).5. then we can subsequently solve < j.2 Ancestry in a rooted tree Let J be the set of all rooted trees. w > and a given rooted tree j we want to know whether node v is an ancestor of node w in tree j.Sec.) If the tree j contains n nodes. jusqu'A. where A is a non-singular square matrix and b is a column vector. then it is probably worthwhile to calculate the inverse of A once and for all. and let K be the set of pairs < v .1.5) is a further example of preconditioning. Problem 5. Example 7. however. . Problem 7. a (j) E O(nj log nj) for the sort bi .1. and to multiply this inverse by each b. the ancestor of all the nodes of which its children are ancestors.1. (By definition. Creating an optimal search tree (Section 5.3. Example suggests how to obtain an efficient greedy algorithm for making change. where nj is the number of elements in the set j. for example K = (Si. begin. possible to precondition the tree in a time in O(n). Why? It is. any direct solution of this instance takes a time in S2(n) in the worst case. every node is its own ancestor and.1.1 Preconditioning 207 Let K be a set of keywords.

and let postnum [v] be the number assigned during the traversal in postorder. (Notice.1. Show. There exist several similar ways of preconditioning a tree so as to be able thereafter to verify rapidly whether one node is an ancestor of another. respectively.9. Thus prenum [v] <_ prenum [w] b v is an ancestor of w or v is to the left of w in the tree. that this can be done using a traversal in preorder followed by a traversal in inverted preorder. It follows that prenum [v] prenum [w] and postnum [v] >_ postnum [w] t* v is an ancestor of w. A rooted tree with preorder and postorder numberings. that traversal of a rooted tree in inverted preorder requires more work than traversal in postorder if the representation of rooted trees suggested in Figure 1.5 is used. D 1 2 B 5 4 E 3 I 2 C 6 F 4 8 12 . which visits first a node and then its subtrees from right to left. In preorder we first number a node and then we number its subtrees from left to right. numbering the nodes sequentially as we visit them. 7 We illustrate this approach using the tree in Figure 7.208 Preconditioning and Precomputation Chap.1.1.2. for example.G6 10 J 9 711 K H 8121 L 913 M 10 Figure 7.2). and then we number the node itself. however.1. let prenum [v] be the number assigned to the node when we traverse the tree in preorder.1. In Figure 7. we traverse it first in preorder and then in postorder (see Sec- tion 6. Problem 7.) 11 . Let v and w be two nodes in the tree. Once all the values of prenum and postnum have been calculated in a time in 0(n).1. the required condition can be checked in a time in 0(1).1 these two numbers appear to the left and the right of the node. For a node v. It contains 13 nodes. In postorder we first number the subtrees of a node from left to right. To precondition the tree. Thus postnum [v] >_ postnum [w] a v is an ancestor of w or v is to the right of w in the tree.

) In general. we first express it in the form p(x) = (x(n+t)12+a)q(x) + r(x). x7.17.I -1. If we evaluate p (x) as p (x) = ((((((x -5)x +4)x .rt=-3. Equating the coefficients of x6. 7.6. and let K be the set of values this variable can take. . . Consider the polynomial p(x)=x7-5x6+4x5.3 Repeated Evaluation of a Polynomial Let J be the set of polynomials in one variable x. where i is a power of 2. A naive method for evaluating this polynomial is to calculate first the series of values x2.a =2. . where a is a constant and q (x) and r (x) are monic polynomials of degree 2k . 5x6 and finally p(x)... . we apply the same procedure recursively to q (x) and r (x). x3. (A typical 7th degree monic polynomial would require 5 multiplications as is the case here. It is easy to do better.1 209 7. if p (x) is a monic polynomial of degree n = 2k -1. . . we can calculate p(x)=(x4+2)[(x2+3)(x-5)+(x+2)]+[(x2-4)x + (x +9)] using only 5 multiplications (of which two are to calculate x2 and x4) plus 9 additions. x5.1. We use the number of integer multiplications that have to be carried out as a barometer to measure the efficiency of an algorithm. Finally. from which we can obtain 5x.. 1. we restrict ourselves even further and consider only monic polynomials (the leading coefficient is 1) of degree n = 2k . In the preceding example we first express p (x) in the form (x4+a)(x3+82x2+qlx+qo)+(x3+r2x2+rlx+ro) .Preconditioning Sec. The problem consists of evaluating a given polynomial at a given point. Example 7.. qo=-13. Better still. we restrict ourselves to polynomials with integer coefficients. p (x) is expressed entirely in terms of polynomials of the form x` + c. we obtain successively q2 = -5. For simplicity.r2=0. Initially. Next.andro=9. Thus .13x4+3x3. taking no account of the size of the operands involved nor of the number of additions and subtractions.I for some integer k ? 1.1. lOx2.10x2+5x .10)x +5)x -17 we need only 6 multiplications and 7 additions. but 10 additions. This method requires 12 multiplications and 7 additions (counting the subtractions as additions). q I = 4.13)x +3)x .. evaluated at integer values of x.

Problem 7. In other words. x (n + 1)/2 We obtain the recurrence equa- 0 k=1 2M(k -1) + 1 k>2. Analysis of the method.7.3x + 9 = (x2-4)x + (x +9) to arrive finally at the expression for p (x) given in Example 7.4. Let M(k) = M(k) . are not monic. we find x3-5x2+4x . Further generalize the . tion M(k) = . Prove that if the monic polynomial p (x) is given by its coefficients.k + 1 be the number of multiplications required if we do not count those used in the calculation of x 2. Express p (x) ax7 in preconditioned form. (Continuation of Problem 7. Problem 7.210 Preconditioning and Precomputation Chap. x 4..13=(x2+3)(x-5)+(x+2) x3 . In other words.1. Generalize this method of preconditioning to polynomials that Problem 7.7) method to polynomials of any degree. the time invested in preconditioning the polynomial allows us to evaluate it subsequently using essentially half the number of multiplications otherwise required.3.1.8. Similarly. .1. Problem 7. with no risk of rounding error due to the use of floating-point arithmetic. This expression is the preconditioned form of the polynomial. Problem 7. Show that evaluation of a preconditioned polynomial of degree n = 2k . and hence M(k)=2k -1+k-2. Your generalization must give an exact answer.1.1.1. there does not exist an algorithm that can calculate p (x) using less than n -1 multiplications in the worst case.1. (n -3)/2+Ig(n + 1) multiplications are sufficient to evaluate a preconditioned polynomial of degree n = 2k -1.1. a preconditioned monic polynomial of degree n = 2k -1. * Problem 7. Consequently. Express p(x)=x7+2x6-5x4+2x3-6x2+6x -32 in preconditioned form.6.6..1 requires (3n . 7 p(x) = (x4+2)(x3-5x2+4x -13) + (x3-3x +9). Let M(k) be the number of multiplications required to evaluate p (x).5.1. M(k) = 2k ' -1 for k >.1)/2 additions in the worst case.

Is the method appropriate for polynomials involving real coefficients and real variables ? Justify your answer. whereabouts in S it occurs. which is in Q(mn) if n is much larger than in.) The total number of comparisons to be made is therefore in S1(m (n -m )).). i = 1. m).1. if it occurs in S. In the worst case it makes m comparisons at each position to see whether or not P occurs there.1 =Pi . and that the pattern P. It returns r if the first occurrence of P in S begins at position r (that is. S. S = s is 2 s .1 Signatures Suppose that the target string S can be decomposed in a natural way into substrings. S = S 1S 2 . information retrieval systems. if the Si are the lines in a text file S and we are searching for the lines in the file that contain P. and a pattern consisting of m characters..2. we want to know whether P is a substring of S. and if so. .. P=p1p2 p.m do ifp[j]#s[i+j] then ok <--false else j F j + 1 if ok then return i + 1 return 0 The algorithm tries to find the pattern P at every position in S. Can we do better ? 7. .2 211 Problem 7. 7. for example. Given a target string consisting of n characters. macroprocessors. . r is the smallest integer such that Sr+i . for i E. we use the number of comparisons between pairs of characters as a barometer to measure the efficiency of our algorithms.. (Think of S = "aaa aab".. In the analyses that follow.9. 7. . Show using an explicit example that the method described here does not necessarily give an optimal solution (that is. must occur entirely within one of these substrings (thus we exclude the possibility that P might straddle several consecutive substrings).O to n -m do ok F. Problem 7. and it returns 0 if P is not a substring of S.m.1.1 while ok and j <. P = "aaaab". This situation occurs.Precomputation for String-Searching Problems Sec. 2.true j *. etc.10.2 PRECOMPUTATION FOR STRING-SEARCHING PROBLEMS The following problem occurs frequently in the design of text-processing systems (editors. it does not necessarily minimize the required number of multiplications) even in the case of monic polynomials of degree n = 2k -1. The following naive algorithm springs immediately to mind. Suppose without loss of generality that n >.

c 2) = (27 val (c 1) + val (c 2)) mod 32. Define val ("a") = 0. z and other are equiprobable.. . For each pattern P we are given we need a further time in O (m) to calculate its signature.1. This gives us the function T we need : T (P .. define . . cr as a 32-bit word where M. .. Only seven bits are set to 1 in the signature because B ("e". o") = 27 x 2 + 14 mod 32 = 4. B ("r". Si) is false. B (c 2.. * Problem 7. Suppose too that we are working on a computer with 32-bit words.1. ii. Signatures offer a simple way of implementing such a function. i. Si) is true. If Si contains the pattern P. where the and operator represents the bitwise conjunction of two whole words. we calculate B ("c". c 2). Define the signature sig (C) of a string C = c 1c2 the bits numbered B (c 1. then P cannot be a substring of Si . If T (P . if T (P... 7 The basic idea is to use a Boolean function T (P. B (c 1.. n = 40. Si) that can be calculated rapidly to make a preliminary test.. . T can be computed very rapidly once all the signatures have been calculated. Calculating the signatures for S takes a time in O (n). val ("b") = 1. The improvement actually obtained in practice depends on the judicious choice of a method for calculating signatures. cr) are set to 1 and the other bits are 0. If signatures are calculated as described. m = 5). however. val ("z") = 25.Preconditioning and Precomputation 212 Chap. B (cr_1. . If C is the string "computers". Si) = [(sig (P) and sig (Si )) = sig (P )J. . using the naive algorithm given earlier). B ("o". and if the characters a. then all the bits that are set to 1 in the signature of P are also set to 1 in the signature of Si .. . x. where we have lumped all the non-alphabetic characters together. c. b. Example 7. We calculate a signature for each substring Si and for the pattern P. but we have to carry out a detailed check to verify this (for instance. but from then on we hope that the preliminary test will allow us to speed up the search for P.2.2. This is yet another example of preconditioning. . z. r") = B ("r". b. what is the probability that the signature of a random string of n characters contains a 1 in all the bit positions that contain a 1 in the signature of another random string of m characters ? Calculate the numerical value of this probability for some plausible values of m and n (for instance. the signature of this string is the word 0000 1110 0100 0001 0001 0000 0000 0100 . m") = 27 x 14 + 12 mod 32 = 6. . y. s") = 27 x 17 + 18 mod 32 = 29. If c1 and c2 are characters. Here is one common way of defining a signature. . it is possible that P might be a substring of Si .. c 3). other) . If the bits of a word are numbered from 0 (on the left) to 31 (on the right). Suppose that the character set used for the strings S and P is { a. val (other) = 26. s") = 29.

Is this to be recommended ? If not.2.2. and so on. c 2) = (128 val (c 1) + val (c 2)) mod 32. The arrows show the comparisons carried out before we find a character that does not match. The number of bits in the signature can also be changed. is it useful ? Problem 7. In the preceding example the function B takes two consecutive characters of the string as parameters.4.2. we try the following configuration : S P babcbabcabcaabcabcabcacabc abcabcacab T We check the characters of P from left to right. Si) is true with probability E > 0 even if Si does not contain P. 7.2. If the character set contains the 128 characters of the ASCII code.3. we might define B by B (c 1. If T (P. After this failure we try S P babcbabcabcaabcabcabcacabc abcabcacab TTTT .2.2 The Knuth-Morris-Pratt Algorithm We confine ourselves to giving an informal description of this algorithm (henceforth : the KMP algorithm). To find P in S we slide P along S from left to right. what is the order of the number of operations required in the worst case to find P in S or to confirm that it is absent ? Many variations on this theme are possible. looking at the characters that are opposite one another.2.5. Example 7.2 213 Problem 7. Problem 7. It is easy to invent such functions based on three consecutive characters. Initially.2. Can we define a function B based on a single character? If this is possible. what do you suggest instead ? 7.2. and if the computer in use has 32-bit words. In this case there is only one comparison.Precomputation for String-Searching Problems Sec. Is the method illustrated of interest if the target string is very long and if it cannot be divided into substrings ? Problem 7. Let S = "babcbabcabcaabcabcabcacabc" and P = "abcabcacab". which finds the occurrences of P in S in a time in 0 (n).

however moving it three places might work. So let us try sliding P four characters along. To implement this algorithm. we know that the last eight characters examined in S are abcabcax where x # "c". we have proceeded exactly as in the naive algorithm. we can conclude that it is useless to slide P one. and this time a three-place movement is necessary. (A three-place movement is not enough: we know that the last characters examined in S are ax. However we now know that the last four characters examined in S are abcx where x #"a". S P babcbabcabcaabcabcabcacabc abcabcacab TTTTTTTT Following this mismatch. 7 This time the first three characters of P are the same as the characters opposite them in S.) S P babcbabcabcaabcabcabcacabc abcabcacab TTTTTTTT Yet again we have a mismatch. It suffices to start checking at the current position of the pointer. We must instead line up P with the first character of S what . This array tells us to do when a mismatch occurs at position j in the pattern. Without making any more comparisons with S.Preconditioning and Precomputation 214 Chap. or three characters along : such an alignment cannot be correct. Up to now. where x is not a "b". but the fourth does not match. If next [ j ] = 0. S P babcbabcabcaabcabcabcacabc abcabcacab T There is no need to recheck the first four characters of P : we chose the movement of P in such a way as to ensure that they necessarily match. two. S P babcbabcabcaabcabcabcacabc abcabcacab TTTTTT We complete the verification starting at the current position time the correspondence between the target string and the pattern is complete. m ]. Sliding P one or two places along cannot be right . sliding P four places along might work. In this case we have a second mismatch in the same position. This time.. we need an array next[ I. it is useless to compare further characters of the pattern to the target string at the current position.

which can be neglected since m <_ n. including the search for a single pattern in a single target. The pointer and P can each be moved a maximum of n times.6. or the variable k in the algorithm) or the pattern P.2. Problem 7.2 Precomputation for String-Searching Problems 215 that has not yet been examined and start checking again at the beginning of P. preconditioning does not apply if several distinct patterns are sought in a given target string.2.m else return 0 It returns either the position of the first occurrence of P in S. In both cases we slide P along j . After each comparison of two characters. It is correct to talk of preconditioning in this case only if the same pattern is sought in several distinct target strings.8.7.1 while j <_ m and k 5 n do while j > O ands [k] # p [j ] do j'.next [ j ] characters to the right with respect to S. m ] in a time in Problem 7. in ] can be carried out in a time in 0(m). On the other hand. 7.2. Modify the KMP algorithm so that it finds all the occurrences of P in Sin a total time in 0 (n).2. If next [ j ] = i > 0.. 0 (M). Find a way to compute the array next [ 1 . The time required by the algorithm is therefore in O (n). which does happen in some applications.2. In the preceding example we have j 1 p[j] 2 3 4 5 8 9 10 b c a b 6 c 7 a a c a b next[ j] 0 1 1 0 1 1 0 5 0 1 Once this array has been calculated. Precomputation of the array next [ 1 . we should align the i th character of P on the current character of S and start checking again at this position. the execution time is thus in O (n).k <.next[j] k-k+1 j-j+1 if j >m then return k .. Follow the execution of this algorithm step by step using the strings from Example 7. or else 0 if P is not a substring of S. * Problem 7. here is the algorithm for finding P in S. In all cases. Overall. we move either the pointer (the arrow in the diagrams. function KMP j.Sec. it is correct to talk of precomputation. .

The BM algorithm. and again there is an immediate mismatch. As with the KMP algorithm. Furthermore. If we have a mismatch immediately after moving P. we try . we slide P along S from left to right. Example 7. We know that c #p [m].3. S P This is a delicate topic cat T Again we examine P from right to left. increases. we align the latter just after the occurrence of c in the target string. since the KMP algorithm examines every character of the string S at least once in the case when P is absent. 7 7. In the best case the BM algorithm finds all the occurrences of P in S in a time in O (m + n /m). There is an immediate mismatch in the position shown by the arrow. The character opposite p [m] is "i". the BM algorithm tends to become more efficient as m.2. the number of characters in the pattern P. is often sublinear : it does not necessarily examine every character of S. This time. we slide the latter along in such a way as to align the last occurrence of c in the pattern with the character c in the target string. the algorithm due to Boyer and Moore (henceforth : the BM algorithm) finds the occurrences of P in S in a time in O (n) in the worst case. Since the pattern does not include this character. Let S = "This is a delicate topic" and P = "cat". and the number of comparisons carried out can be less than n. Since "i" does not appear in the pattern. We use two rules to decide how far we should move P after a mismatch. If c appears elsewhere in the pattern. ii. then we use this partial knowledge of S (just as in the KMP algorithm) to slide P along to a new position compatible with the information we possess.3 The Boyer-Moore Algorithm Like the KMP algorithm. however. checking corresponding characters. let c be the character opposite p [m]. on the other hand. i. If a certain number of characters at the end of P correspond to the characters in S. it makes at least n comparisons.216 Preconditioning and Precomputation Chap. the characters of P are checked from right to left after each movement of the pattern. we slide the pattern just to the right of the arrow. However. If c does not appear in the pattern. The target string and the pattern are initially aligned as follows : S This is a delicate topic P cat T We examine P from right to left.2.

In this example we have found P without ever using rule (ii). We slide P along to align the "c" found in S with the last "c" in P. If we slide P five places right.) S P babcbabcabcaabcabcabcacabc abcabcacab T Unlike the KMP algorithm. P is correctly aligned. . We have made only 9 comparisons between a character of P and a character of S. A final check.2 S 217 This is a delicate topic P cat T There is once again an immediate mismatch. Some unnecessary checks (corresponding to the underscored characters in P) may be made at times. S P Consider the same strings as in Example 7. We know that starting at this position S contains the characters xcab where x #"a". S This is a delicate topic P cat T Now. 7. We slide P one place along to align the occurrences of the letter "a". there is an immediate mismatch. (Underscores show which characters were aligned.2: babcbabcabcaabcabcabcacabc abcabcacab TTTT We examine P from right to left. always from right to left. The left-hand arrow shows the position of the first mismatch.Precomputation for String-Searching Problems Sec. but this time the character "a" that appears opposite p[m] also appears in P. In our example when we start over checking P from right to left. and start checking again (at the right-hand end of P).2.4. when we slide P along one position to align the "a" in the target string with the "a" in the pattern. S This is a delicate topic P cat T After two more immediate mismatches we are in this situation. will confirm this. this information is not contradicted. Example 7. we check all the positions of P after moving the pattern.2.

A second application of rule (ii) gives us S P babcbabcabcaabcabcabcacabc abcabcacab T We apply rule (i) once to align the letters "a": S P babcbabcabcaabcabcabcacabc abcabcacab T and one last time to align the letters "c": S P babcbabcabcaabcabcabcacabc abcabcacab TTTTTTTTTT We have made 21 comparisons in all to find P. the former to implement rule (i) and the latter for rule (ii). Suppose further that at some moment during our search for the string P in S we have a mismatch in position p [7]. m ] then m elsem -max{i I p[i]=c] .5. is easy to compute. This is the distance to move P according to rule (i) when we have an immediate mismatch. but only an example. It is more complicated to compute d2. We shall not give the details here. For every character c dl[c] F. indexed by the character set we are using. Example 7. Since we always examine the characters of P from right to left. we know that starting at the position of the mismatch the characters of S are xs. To implement the algorithm.if c does not appear in p [ 1 . The interpretation of d2 is the following : after a mismatch at position i of the pattern.. Suppose the pattern is P = "assesses". begin checking again at position m (that is.. we need two arrays d1 [ { character set } ] and d2[1 . we again have a mismatch.2. at the right-hand end) of the pattern and d2[i] characters further along S. where x # "e" : .218 Preconditioning and Precomputation S P Chap. 7 babcbabcabcaabcabcabcacabc abcabcacab TTTT After four comparisons between P and S (of which one is unnecessary). m -1]. carried out as usual from right to left. The array d 1.

10 characters further on in S than the previous comparison: thus d2[6] = 10. the characters of S are xes. As a third instance.2 S P ??????xs???????? assesses 219 x "e" TT The fact that x * "e" does not rule out the possibility of aligning the "s" in p [6] with the "s" we have found in S. and we must slide P all the way to the right under the characters of S that we have not yet examined : S P ?????xes???????? assesses X "s T We start checking again at the end of P. 7. Starting from the position of the mismatch. Starting from the position of the mismatch. that is.Precomputation for String-Searching Problems Sec. suppose we have a mismatch at position p [4]. It is therefore impossible to align P under these characters. Similarly. 3 characters further on in S than the previous comparison: thus d2[7] = 3. where x # "s" : S P ?????xes???????? assesses TTT X "s" The fact that x # "s" rules out the possibility of aligning the "e" and the "s" in p [4] and p [5] with the "e" and the "s" found in S. the characters of S are xsses. suppose now that we have a mismatch at position p [6]. It may therefore be possible to align P as follows : S P ??????xs???????? assesses X #"e T We start checking again at the end of P. that is. where x # "e": S P ???xsses???????? assesses TTTTT x "e" In this case it may be possible to align P with S by sliding it three places right : S P ???xsses???????? assesses T xe .

Preconditioning and Precomputation 220 Chap.m while k <.1 if j # 0 then if j = m then k <-. the choice between using rule (i) and rule Problem 7.2. Problem 7.4. because an immediate mismatch is impossible at a position where S contains "s".2. function BM j. 7 characters further on in S than the previous comparison.10. Calculate d 1 and d2 for P = "abracadabraaa".9. so k is increased by d2[71 = 3 to obtain .2.k-1 j *. Problem 7. consider the following situation : S P ??????ts???????? assesses TT The failure of the match between "t" and "e" is of the second kind. it is possible that rule (i) might allow k to advance more than rule (ii). so d2[4] = 7. In this algorithm. Note that d 1 ["s"] has no significance. Calculate d 1 and d2 for the pattern in Example 7. d 1 ["a"] = 7 and d 1 [any other character ] = 8.n and j > 0 do while j > 0 and s [k] = p [j ] do k+. However. Here finally is the BM algorithm.5. (ii) depends on the test "j = m ? ".k F.k + d 1[s [k]] elsek k+d2[j] jFm if j = 0 then return k + 1 else return 0 It returns either the position of the first occurrence of P in S. Continuing from Example 7.j . or else 0 if P is not a substring of S.11. For this example we find i 1 2 3 4 5 6 7 8 p[i] a s s e s s e s d2[i] 15 14 13 7 11 10 3 We also have d 1 ["s"] = 0. even if j < m . d 1 ["e"] =1. 7 Now we start checking at the end of P.2.2.

m by k <. ??????te???????? assesses S P T Show that the algorithm is still correct if we replace if j= m then k E. 7.k+d2[j] j E. they repeat comparisons previously made) ? * Problem for String-Searching Problems Sec. Show the progress of the algorithm if we search (unsuccessfully. the fact that "t" does not appear in P should have allowed us to increase k directly by d 1["t"] = 8 positions. For a character set of reasonable size (say.k + max(d I [s [k] ]. ** Problem 7. It is easy to see intuitively why the algorithm is often more efficient for longer patterns. provided we define d 2[m] = 1 and d 1[ p [m]] = 0. Thus we look at approximately one character out of every m in the target string. This modification corresponds to the algorithm usually known by the name Boyer-Moore (although these authors also suggest other improvements). Find a way to calculate the array d2 in a time in 0 (m). d I [c] is equal to m for most characters c.2 221 ??????ts???????? assesses S P T However.2. and how many of these comparisons are redundant (that is. d2[ j ]) j4-m . of course) for the pattern P = "assesses" in S = "I guess you possess a dress fit for a princess".2.14.13. ten figures and about a dozen other characters) and a pattern that is not too long. Prove that the total execution time of the algorithm (computation of d I and d2 and search for P) is in 0 (n). Problem 7. 52 letters if we count upper. How many comparisons are made altogether before the failure to match is discovered.15.and lowercase separately. Modify the BM algorithm so that it will find all the occurrences of P in S in a time in 0 (n). As long as m stays small . Problem 7.k+ d I [s [k]] else k<.

For a probabilistic string-searching algorithm (see Chapter 8). read Karp and Rabin (1987). and Pratt (1977) and Boyer and Moore (1977). 7 compared to the size of the character set. Finite automata. Signatures are discussed in Harrison (1971). For an efficient algorithm capable of finding all the occurrences of a finite set of patterns in a target string. Rytter (1980) corrects the algorithm given in Knuth.222 Preconditioning and Precomputation Chap. Morris. see.3 come from Knuth. Boyer and Moore give some empirical results : if the target string S is a text in English. . only 15% of the characters in S are examined.2 and 7. Morris. for example. when m = 12. the number of characters examined goes down as m goes up. as described for instance in Hopcroft and Ullman (1979). consult Aho and Corasick (1975). 7. Baase (1978). The KMP and BM algorithms of Sections 7.3 REFERENCES AND FURTHER READING Preconditioning polynomials for repeated evaluation is suggested in Belaga (1961). can be used to introduce the KMP algorithm in an intuitively appealing way. and Pratt (1977) for calculating the array d2 to be used in the Boyer-Moore algorithm.2. about 20% of the characters are examined when m = 6.2.

An elf offers to show you how to decipher the map if you pay him the equivalent of the treasure that the dragon can carry away in three nights. you can expect to come home with x . and that y is the value of the treasure carried off every night by the dragon. A treasure is hidden at a place described by a map that you cannot quite decipher. Remembering that it will take you five days to reach the hiding-place.1. you can do better. you can set out 223 . Suppose further that x > 9y.8 Probabilistic Algorithms 8. Suppose that x is the value of the treasure remaining today. You estimate that it will take four more days' computation to solve the mystery of the map and thus to know with certainty where the treasure is hidden. however. or to travel from one of them to the other. It takes five days to get to either of the possible hiding-places. which are.1.9y if you wait four days to finish deciphering the map. Leaving out of consideration the possible risks and costs of setting off on a treasure-hunting expedition. however. If you were at one or the other of these two places. If you accept the elf's offer. Problem 8. you would immediately know whether it was the right one. but if you set out on a journey you will no longer have access to your computer. If you are willing to take a calculated risk.1 INTRODUCTION Imagine that you are the hero (or the heroine) of a fairy tale. a considerable distance apart. should you accept the elf's offer? Obviously it is preferable to give three nights' worth of treasure to the elf rather than allow the dragon four extra nights of plunder. You have managed to reduce the search to two possible hiding-places. The problem is complicated by the fact that a dragon visits the treasure every night and carries part of it away to an inaccessible den in the mountains.

The latter.8y left.5y. 8 immediately and bring back x . This fable can be translated into the context of algorithmics as follows : when an algorithm is confronted by a choice. not the time incurred if the worst possible probabilistic choices are unfortunately taken. The average execution time of a deterministic algorithm was discussed in section 1. It is always possible that bad luck will force the algorithm to explore many unfruitful possibilities. but choosing the pseudomedian still takes quite some time. We make an important distinction between the words "average" and "expected".6 describes an algorithm that can find the k th small- est of an array of n elements in linear time in the worst case. journeying on to the other if you find you have decided wrong.1. and one chance out of two of coming home with x -10y .5y. of which 3y will go to pay the elf. This avoids the infinite recursion. By contrast. Despite this. the more efficient the algorithm will be. This is like buying a ticket for a lottery that has a positive expected return. Clearly. without making the algorithm catastrophically bad for the worst-case instances. It refers to the average time taken by the algorithm when each possible instance of a given size is considered equally likely. refers to the expected time taken by the worst possible instance of a given size.1. Example 8.3). for instance.1 that choosing the pivot randomly gives a substantial improvement in the expected execution time as compared to the algorithm using the pseudomedian. Recall that this algorithm begins by partitioning the elements of the array on either side of a pivot. the probabilistic algorithm can only be more efficient with respect to its expected execution time. the expected execution time of a probabilistic algorithm is defined on each individual instance : it refers to the mean time that it would take to solve the same instance over and over again. you will thus have x . and that it then calls itself recursively on the appropriate section of the array if need be. This gives you one chance out of two of coming home with x -5y. One fundamental principle of the divide-and-conquer technique suggests that the nearer the pivot is to the median of the elements. We once asked the students in an algorithmics course to implement the selection algorithm of their choice. Thus we choose a suboptimal so-called pseudomedian.4. Your expected profit is therefore x . it is sometimes preferable to choose a course of action at random.6. rather than to spend time working out which alternative is the best. Section 4. but at the price of a quadratic worst case : it simply decides to use the first element of the array as the pivot. compared to the time that will be saved on the average by making this optimal choice.7. This makes it meaningful to talk about the average expected time and the worst-case expected time of a probabilistic algorithm. We shall see in Section 8. there is no question of choosing the exact median as the pivot because this would cause an infinite recursion (see Problem 4. The only algorithms they had seen were those in Sec- . A better strategy is to toss a coin to decide which possible hidingplace to visit first.4.Probabilistic Algorithms 224 Chap. we saw another algorithm that is much faster on the average. Such a situation arises when the time required to determine the optimal choice is prohibitive. On the other hand.

2 describes an efficient probabilistic algorithm to solve this problem provided that one is willing to accept an arbitrarily small probability of error. Show how the effect of uniform (i .3. a < b .2.6. b) is available. For this reason. Section 6.. returns an element chosen randomly." Of course. where i and j are integers.1 describes a systematic way of exploring an implicit tree to solve the eight queens problem. we may obtain different correct solutions on different occasions. a number of results are cited without proof in the following sections. If we are content with finding one solution rather than all of them.1. Example 8. statistics. Nevertheless.2. j ).1 Introduction 225 tion 4. thought of using a probabilistic approach. uniform (X).1.6.Sec. A call on uniform (a. This idea allowed them to beat their colleagues hands down : their programs took an average of 300 milliseconds to solve the trial instance. They are sometimes used to solve problems that allow several correct solutions. and the function returns an integer k chosen randomly. Throughout this chapter we suppose that we have available a random number generator that can be called at unit cost. Problem 8. The distribution of x is Iniform on the interval. and independently in the interval i 5 k S j.1. whereas the majority of the deterministic algorithms took between 1500 and 2600 milliseconds. where X is a nonempty finite set. For more details. Similarly. and independently among the elements of X.1. uniformly. Example 8. Using the same probabilistic algorithm on the same instance. Section 8.5. b) returns a real number x chosen randomly in the interval a <_ x < b . we extend the notation to include uniform (i . This problem has important applications in cryptology (Section 4. j) can be obtained if only uniform (a. uniformly. but in this case the choice of algorithm determines uniquely which solution will be obtained whenever the algorithm is applied on any given instance. and successive calls on the generator yield independent values of x. . The analysis of probabilistic algorithms is often complex. Let a and b. No known deterministic algorithm can decide in a reasonable Example 8. Section 8. requiring an acquaintance with results in probability.8).6.2 raises an important consideration concerning probabilistic algorithms.. be real numbers. consult the references suggested in the last section. we can improve the backtracking technique by placing the first few queens at random. time whether a given integer with several hundred decimal digits is prime or composite. To generate random integers. i <_ j. none of them took the risk of using a deterministic algorithm with a quadratic worst case.1 goes into this more thoroughly. however. For another example of the same phenomenon consider the problem : "Find a nontrivial factor of a given composite integer. Since the students did not know which instances would be used to test their programs (and suspecting the worst of their professors). 8. such problems can also be handled by deterministic algorithms. and number theory beyond the scope of this book. Three students.

independent choice of an element of X. and independently. However. Truly random generators are not usually available in practice. a seed that depends on the date or time. For example. function draw (a . where X is a sufficiently large set and Y is the domain of pseudorandom values to be generated.1. the index of 2 modulo 31 is 5. that of 3 is 30. an index modulo p always divides p -1 exactly. Most programming languages include such a generator. p ) j F..8 } Problem 8. y 1 . The same seed always gives rise to the same sequence. defined by yi = R (xi ). the theoretical results obtained in this chapter concerning the efficiency of different algorithms can generally be expected to hold. However. Suggestions for further reading are given at the end of the chapter.uniform(1.Probabilistic Algorithms 226 Chap. The index of a modulo p is the smallest strictly positive integer i such that a i = 1 (mod p ). uniform. Give other examples of sets in which there is an efficient way to choose an element randomly. This suggests one way of making a random. and even the result obtained. i >_ 0. Using the function S. Y2. and the sequence may be for most practical purposes statistically indistinguishable from a truly random sequence of elements of Y. if S and R (and sometimes g) are chosen properly. may vary considerably . Most generators are based on a pair of functions S : X -+ X and R : X -> Y.1. Let p be a prime number. 8. 8 Example 8. To start a sequence. a probabilistic algorithm leaves some of its decisions to chance. the period can be made very long.4. It is thus the cardinality of X = { a i mod p I j > 1 } . p) { Section 4. Using a good pseudorandom generator. .. We shall not worry about the fact that such a concept conflicts with the definition of "algorithm" given at the beginning of the first chapter. we may choose. Most of the time pseudorandom generators are used instead : these are deterministic procedures that are able to generate long sequences of values that appear to have the properties of a random sequence.1) for i > 0.3. but a simple example will illustrate the general idea. Finally. and let a be an integer such that 1 5 a < p . j. By Fermat's theorem.2 CLASSIFICATION OF PROBABILISTIC ALGORITHMS By definition. This sequence is necessarily periodic. the impractical hypothesis that a genuinely random generator is available is crucial when we carry out the analysis.p-1) return dexpoiter (a . The theory of pseudorandom generators is complex. The execution time. so to obtain different sequences.. although some implementations should be used with caution. The fundamental characteristic of these algorithms is that they may react differently if they are applied twice to the same instance. the function R allows us to obtain the pseudorandom sequence YO. uniformly. for example. this seed defines a sequence : x0 = g and xi = S (xi . we must supply an initial value called a seed. with a period that cannot exceed #X. Let g r= X be a seed. and that of 5 is 3.

Monte Carlo algorithms. Simulation can be used. but sometimes they do not find an answer at all. perhaps because of uncertainties in the experimental data to be used. or maybe because a digital computer can only handle binary or decimal values while the answer to be computed is irrational.2 Classification of Probabilistic Algorithms 227 from one use to the next. for example. the probability of success increases as the time available to the algorithm goes up. the probability of success (that is. Sometimes the answer is given in the form of a confidence interval. such as the simplex algorithm for linear programming. For other problems. These algorithms should not be confused with those. it is hard to see what an "approximation" might be. Las Vegas. but its expected precision improves as the time available to the algorithm increases. and in particular for those we call "numerical". computation of an exact solution is not possible even in principle. In the case of a decision problem. Similarly. but the answer is not necessarily right. for example. to estimate the mean length of a queue in a system so complex that it is impossible to get closed-form solutions or to get numerical answers by deterministic methods. on the other hand.) For certain real-life prob- lems. The answer obtained by such a probabilistic algorithm is always approximate. but catastrophic for a few instances. Some authors use the term "Monte Carlo" for any probabilistic algorithm. any answer that is obtained is necessarily correct. (The error is usually inversely proportional to the square root of the amount of work performed. Randomness was first used in algorithmics for the approximate solution of numerical problems. The principal disadvantage of such algorithms is that it is not in general possible to decide efficiently whether or not the answer given is correct. of getting a correct answer) increases as the time available to the algorithm goes up. Monte Carlo. and Sherwood. if we are trying to factorize an integer. the probability of failure can be made arbitrarily small by repeating the same algorithm enough times on this instance. They are used when some known deterministic algorithm to solve a particular problem runs much faster on the average than in the worst case. but rather of breaking the link between the occurrence of such behaviour and the particular . it is of little interest to know that such-and-such a value is "almost a factor". A Monte Carlo algorithm always gives an answer. 8. Las Vegas algorithms never return an incorrect answer. Finally.Sec. Thus a certain doubt will always exist. that are extremely efficient for the great majority of instances to be handled. and sometimes even to eliminate. and the answer is always correct. It is not a case of preventing the occasional occurrence of the algorithm's worst-case behaviour. a precise answer exists but it would take too long to figure it out exactly. this difference between good and bad instances. Sherwood algorithms always give an answer. Incorporating an element of randomness allows a Sherwood algorithm to reduce. Probabilistic algorithms can be divided into four major classes : numerical. As with Monte Carlo algorithms. are used when there is no question of accepting an approximate answer. Whatever the instance to be salved. A way to put down seven queens on the chess-board is little help in solving the eight queens problem. However. since only two answers are possible. and only an exact solution will do.

5. each one independently of all the others. A problem is well-characterized if it is always possible to Problem 8. 8.2. Since it reacts more uniformly than the deterministic algorithm. the average number of toothpicks expected to fall across a crack can be calculated : it is almost exactly 113. .2. This suggests a probabilistic "algorithm" for estimating the value of it by spilling a sufficiently large number of toothpicks onto the floor. Contrariwise. do you think ? NUMERICAL PROBABILISTIC ALGORITHMS Remember that it is a question of finding an approximate answer for a numerical problem. If you know that there were 355 toothpicks in the box. do you think the problem of finding the smallest nontrivial factor of a composite integer is well-characterized? Problem 8. and this uncertainty is typical of probabilistic algorithms. Show that the problem of finding a nontrivial factor of a composite integer (Section 8. Show how to obtain a Las Vegas algorithm to solve a wellcharacterized problem given that you already have a Monte Carlo algorithm for the same problem.1. show how to obtain a Monte Carlo algorithm for any problem whatsoever given that you already have a Las Vegas algorithm for the same problem. Why 113 ? Prove it. You should realize this in no way implies that the problem is easy to solve. 8 instance to be solved.1 Buffon's Needle You spill a box of toothpicks onto a wooden floor.4). each toothpick has one chance in it of falling across a crack. However. Problem 8. how many toothpicks will fall across a crack between two planks ? Clearly any answer between 0 and 355 is possible. Why Buffon ? In fact. as Georges Louis Leclerc showed. Intuitively.Probabilistic Algorithms 228 Chap. verify efficiently the correctness of a proposed solution for any given instance.3. this method is not used in practice since better methods of calculating the decimal expansion of it are known.1. a Sherwood algorithm is less vulnerable to an unexpected probability distribution of the instances that some particular application might give it to solve (see the end of Section is well-characterized. Why "Sherwood". Needless to say. The toothpicks spread out on the ground in random positions and at random angles. Problem 8. and that each one is exactly half as long as the planks in the floor are wide (we realize that this gets unlikelier every minute ! ).

except that it only throws darts into the upper right quadrant of the target. then its area is nr 2. type discussed previously.required.000. give an algorithm Buffon (n) that simulates the experiment of dropping n toothpicks. .2.of each toothpick that falls. how many of the latter should you drop in order to obtain with probability at least 90% an estimate of it whose absolute error does not exceed 0. length of the toothpicks.1 illustrates the experiment. and return n / k as its estimate of it. so the average proportion of the darts that fall inside the circle is nr2/4r2 = it/4.3. using a pseudorandom generator. What are your estimates of it ? (It is likely that you will need the value of it during the simulation to generate the random angle . Supposing that the width of the planks is exactly twice the * Problem 8. At AF r M r r At Si r 4 r Figure 8.3. In our example. 8.Sec. But then 0 nobody said this was a practical method !) Consider next the experiment that consists of throwing n darts at a square target and counting the number k that fall inside a circle inscribed in this square. the precision of your estimate of it would be limited by the precision of the ratio of the length of the toothpicks to the width of the planks.1. we are not surprised to find 21 of them inside the circle. whereas that of the square target is 4r2. Your algorithm should count the number k of toothpicks that fall across a crack. We suppose that every point in the square has exactly the same probability of being hit by a radians . Figure 8.3.2. This allows us to estimate it = 4k/n.2. The following algorithm simulates this experiment.001 ? Supposing that you have available a random generator of the Problem 8. (It is much easier to simulate this experiment on a computer than to find a darts-player with exactly the degree of expertise-or of incompetence . where 28 darts have been thrown.3 Numerical Probabilistic Algorithms 229 Furthermore.) If the radius of the inscribed circle is r. where we expect to see on average 28n/4 = 22. / Throwing darts to compute it. Try your algorithm on a computer with n = 1000 and n = 10.

I I < E ] > 1. and the line x =1 is given by if (x)dx 0 To estimate this integral.3.1)" by "x t.uniform (0. What value is estimated if we replace "x . 1) .3.4. . Therefore it is sufficient to use n = 11/4c281 (because I (1-I) <-'/a) to reduce below 8 the probability of an absolute error exceeding E. then Prob[ I h . 8 function darts (n) k -0 for i F.uniform (0. we could throw a sufficient number of darts at the unit square and count how many of them fall below the curve.1) y <.230 Probabilistic Algorithms Chap. y F. Consider two real constants a and S strictly between 0 and 1.uniform (0.1 to n do x F-.1 to n do x t.1) if y <-f(x)then k Fk+1 return k In Thus the algorithm using darts to estimate it is equivalent to the evaluation of 45(1-x 2)Zdx 0 by hit-or-miss Monte Carlo.S whenever the number n of iterations is at least I (1-I)/E2S. 1] -* [0. the y-axis.1) y F uniform (0.uniform (0. (This name is unfortunate. function hitormiss (f .uniform (0. the x-axis.3. Prove that if ! is the correct value of the integral and if h is the value returned by the preceding algorithm.1) if x2+y2<-lthen k-k+1 return 4k In Problem 8.) Recall that if f : [0. y F x " in this algorithm ? 8. n) k+-0 for i E.2 Numerical Integration This brings us to the best known of the numerical probabilistic algorithms : Monte Carlo integration. * Problem 8.uniform (0. because in our terminology it is not an example of a Monte Carlo algorithm.5. then the area of the surface bounded by the curve y = f (x).1) .1] is a continuous function.

Sec. c. b) sum .n.(f (a)+f (b))/2 for x . the variance of the estimate calculated by this algorithm is inversely proportional to the number of points generated randomly. and d be four real numbers such that a < b and c S d.(b -a)l(n -1) sum <-.a +delta step delta to b -delta do sum<-sum+f(x) return sum x delta . a.sum + f (x) return (b -a) x (sum/n) fh fab Provided f (x) dx and f 2(x) dx exist. As presented thus far. A better estimate of the integral can generally be obtained by various deterministic methods. and let f : [a. one of the simplest of which is the trapezoidal algorithm. function trapezoidal (f . besides f . Generalize the preceding algorithm to estimate h Jf(x)dx . however.3 Numerical Probabilistic Algorithms 231 Notice that this is not very good : one more decimal digit of precision requires one hundred times more computation. which hit-or-miss can do without by proceeding as in darts. One should not immediately conclude.1 to n do x F. the number n of iterations to make and the values of c and d. Monte Carlo integration is of little practical use. a Your algorithm should accept as parameters. 8.[c.6. each iteration of crude requires the computation of a square root. a. n . function crude(f . d ] be a continuous function.3.b) sum E-0 for i . Moreover.a. b ) we assume n >-21 delta .uniform (a. The simplest consists of generating a number of points randomly and uniformly inside the interval concerned. Usually more efficient probabilistic methods of estimating the value of a definite integral exist. and the distribution of the estimate is approximately normal when n is large. b. for any fixed number of iterations. because hit-or-miss can sometimes make more iterations than crude in a given amount of time. Let a. its variance is never worse than that of the hit-or-miss algorithm. Problem 8. If both are used to compute it as previously described. b I . that crude always outperforms hit-or-miss. The estimated value of the integral is obtained by multiplying the width of the interval by the arithmetic mean of the values of the function at these points. and b. for instance.

even the most sophisticated.7. although the amount of work for each iteration is likely to increase slightly with the dimension.101 returns the value zero. Consider for example the function f (x) = sin2((100!)itx). In general.1) with 2 5 n <.Probabilistic Algorithms 232 Problem 8. Why is it called the trapezoidal algorithm ? Problem 8. even when f is a thoroughly ordinary function). Let X be a finite set.4). uniformly.1. and independently (see Example 8. numeric probabilistic algorithms are used to approximate a real number. The same technique can also be used to estimate the value of an integer.3. 8. In practice. Any call on trapezoidal (f . We would like to know the cardinality of X. Monte Carlo integration is of interest when we have to evaluate a multiple integral. 10. However. the number of sample points needed to achieve a given precision grows exponentially with the dimension of the integral to be evaluated. to every deterministic integration algorithm. a technique not discussed here (but Section 8. . on the other hand. This is typical of most of the natural functions that we may wish to integrate. then it will probably be necessary to use all the points of a 100 x 100 grid. the dimension of the integral generally has little effect on the precision obtained. Compare experimentally the trapezoidal algorithm and the two probabilistic algorithms we have seen. there correspond continuous functions that can be constructed expressly to fool the algorithm. In each case. 8 Try to grasp intuitively why this algorithm works.3 Probabilistic Counting In the preceding examples. the trapezoidal algorithm needs many less iterations than does Monte Carlo integration to obtain a comparable degree of precision. If the dimension is fixed. that we are able to choose an element from X randomly. estimate the value of it by calculating fo 4(1-x2)zdx.8. one million points will be needed for a triple integral. but the number of elements is too large for it to be practical simply to count them one by one. to achieve the same precision when a double integral is evaluated. Monte Carlo integration is used to evaluate integrals of dimension four or higher. No function can play this kind of trick on the Monte Carlo integration algorithm (although there is an extremely small probability that the algorithm might manage to make a similar kind of error. Chap. In practice.3. it may even be preferable to use quasi Monte Carlo integration. In Monte Carlo integration.000 points. even though the true value of this integral is z . that is. The precision of the answer can be improved using hybrid techniques that are partly systematic and partly probabilistic.7 gives a reference for further reading). and so on. Suppose. 0.3. If 100 points are needed to evaluate a simple integral. on the other hand. n . A classic brain-teaser helps to explain how this ability to choose random elements from X allows us to estimate its cardinality. If a deterministic integration algorithm using some systematic method to sample the function is generalized to several dimensions.

12. show that the expected value of k tends to where (3 = = 1. it is when k z a4n-.3. Let k be the number of choices before the occurrence of the first repetition. More generally.Sec. uniformly and independently choose elements with replacement. Let X be a set of n elements from which we randomly.. Show that n!/(n -k )!n k = e-k'/2n * Problem 8. the probability that you would win your bet is greater than 56%. n! E I (n le )" [1 + l/12n + O(n-2)] ln(l+x) Ex -x2/2+x3/3-8(x4) when -l <x < 1 to conclude that n /(n-k) nk E e-k(k-1)12n-k'/6n'±0(max(k'/n'.3. Calculate 365!/340!36525 to four significant figures.3.3.) The intuitive answer to the preceding question is almost invariably "of course not". The calculation in problem 8.3 Numerical Probabilistic Algorithms 233 A room contains 25 randomly chosen people. and the approximation ln(l+x) Z Stirling's approximation. ** Problem 8.3. where a = 21n 2 z 1. It is harder to determine the average value of k corresponding to the first repetition. . When n is large. Would you be Problem 8.9. Use the more accurate formulas . because births are not uniformly distributed through the year. Since there are n k different ways of choosing k objects if repetitions are allowed.10 does not correspond exactly to the puzzle in problem 8. Problem 8. taking into account the order in which they are chosen.10. 8. that the probability of having a repetition exceeds 50%. Problem 8. allow us to estimate this probability. Does this make it more or less likely that you would win your bet? Justify your answer intuitively. What about leap years ? (n le )" . Nevertheless. there are n!/(n-k)! different ways of choosing k distinct objects from among n objects.9. n! = x _X2 / 2 when x is near zero. the probability that k objects chosen randomly and uniformly from n (with repetition allowed) are all distinct is n!/(n -k )!n k .3.k'/n')) provided that 1 << k << n. Problem 8. willing to bet that at least two of them share the same birthday? (Do not read the following paragraphs if you wish to think about this.3. This suggests the following probabilistic algorithm for estimating the number of elements in a set X .253.11. In particular.

hence k = q +p.15. through X. Because the walk is pseudorandom. x 1.t <. Show also that t 0 (mod p). We have not merely to choose an ele- ment at random from X.1r. X2. i > 0. with t = q +p only possible if q = 0. * Problem 8. and hence deterministic. . since this corresponds precisely to the first repetition.. function of the value of n. 6) that returns an estimate N of the number n of elements in X such that Prob[ 11.Probabilistic Algorithms 234 Chap.o a <--uniform (X ) repeat k F-k+1 S <-S u (a} a .O defined by yt =X2. The space can be reduced to a constant with only a linear increase in the execution time by using a pseudorandom generator. The following exercise shows how to obtain both q and p in constant space and in a time in 0 (k).16. Let f : X -* X be a pseudorandom function and let xoE X be a randomly chosen starting point. and let k be the smallest integer larger than q such that xq = xk . and more generally. e. is such that q <.q and j -i = 0 (mod p ). The algorithm count (X) estimates the number n of elements in X in an expected time and space that are both in O(X ). Incorporate all these ideas into a simple algorithm that is capable of finding the values of q and p.1. This is one of the rare instances where using a truly random generator would be a hindrance rather than a help. but also to step through X in a pseudorandom. Consider the sequence {y. We are interested in computing the value of k. and thus 2k2/it is our estimate on #X. and deduce that the smallest integer j such that xj = x1++ is precisely q.uniform (X) until a E S return 2k2/1t Carry out a statistical analysis of the random variable k2 as a ** Problem 8. = x. }1. way. For this reason {xi } Q_o is called the tail of the walk and {xi } 4 q is its period. where xi = f (xi _ 1).N In I < c ] >. 8 function count (X :set) kF-0 S F. Let q be the smallest integer such that xq appears more than once in the sequence. Show that the smallest integer t > 0 such that y.S.q +p.3. the length of the tail.q.3. This quantity of space can be prohibitive if n is large. hence in an expected time in 0 ( X ). This defines a walk xo. provided operations on the set S are counted at unit cost. we also have xq+i = xk+i for every nonnegative integer i.. Let p stand for k . the sequence {x. . must eventually repeat itself. xi = xj whenever j >_ i >. . Does the function count provide an unbiased estimator of n? Give an algorithm count 2(X. Because X is a finite set. in a time in 0 (k) and in constant space.

keys. although this may not imply a uniform distribution on Xm . be used unchanged to estimate a lower bound on the number of elements in X.3.(Ek.(m)) I k I .4 More Probabilistic Counting You have the complete works of Shakespeare on a magnetic tape. Let N be the total number of words on the tape. that is.k2EK)(3k3EK)(b'mEM)[Ek. Probabilistic counting is used to estimate a lower bound on the cardinality of Xm . and so on) as distinct items? Two obvious solutions to this problem are based on the techniques of sorting and searching. Such a system is closed (an undesirable property) if (dki. 8. if some elements of X are favoured to the detriment of others. (Even 256 microseconds is more than two millennia. if the system is not closed.. counting different forms of the same word (plurals. it is reasonable to hope that #Xm >> #K provided that #M >> #K. Let m be chosen randomly from M. where a << b . see Problem 8. How can you determine the number of different words he used. the probabilistic algorithm was able to arrive at this conclusion in less than a day..3 Numerical Probabilistic Algorithms 235 Problem 8. = { Ek (Ek. It suffices to choose k I and k2 randomly in K and to calculate Ek (Ek.17. (Continuation of Problem 8.(m))=Ek. possessives.3. It is clear that #Xm <_ #K if the system is closed. which rules out any possibility of an exhaustive verification of the hypothesis that #X. In this application #K = 256 and #M = 264.(m)) in order to choose a random element from Xm .3. The first approach might be to sort the words on the tape so as to . The variance of the estimate obtained from this algorithm is unfortunately too high for most practical applications (unless the solution to Problem 8. > 256. On the other hand.) It is improbable that the system is closed if this estimate is significantly greater than the cardinality of K. k 2 E K }. A similar approach was used to demonstrate that the American Data Encryption Standard is almost certainly not closed.1.3.17. and let n be the number of different words. (We can only estimate a lower bound since there is no reason to believe that elements are chosen uniformly from Xm .3. nevertheless.Sec. a finite set M of messages and two permutations Ek : M -* M and Dk : M -* M for each key k E K such that Dk (Ek (m)) = m for every m E M and k E K. For every m r= M consider the set X. 8.(m)] . All this suggests a probabilistic approach to testing whether or not a cryptosystem is closed.11) The probabilistic counting algorithm no longer works if the generation of elements in X is not uniform.15 is used).3. An endomorphic cryptosystem consists of a finite set K of Example 8. Show that it can. The following example shows that it can nonetheless be useful if we simply need to know whether X contains less than a elements or more than b .) Implemented using specialized hardware.

8 bring identical forms together.1 } . but there is no word x4 such that h (x4) begins with 0001. the smallest i such that y [i ] =b.6% = e-1. there are words x 1. "jacko'lantern". Moreover. that the value returned by this algorithm is 4. denote by y [i] the i th bit of y. Prob[ k = 41 n =16 ] = 313/4 %.a -eh'' . Consider the following algorithm. If we are willing to tolerate some imprecision in the estimate of n. This method takes a time in O(N log N) but requires a relatively modest amount of space in central memory if a suitable external sorting technique is used. denote by 7t(y.) Let U be the set of such sequences. since the probability that a random binary string begins with 001 is 2-3. ). in fact.1) y[i]<-l return 7t(y. function wordcnt { initialization } y .) The second approach consists of making a single pass over the tape and constructing in central memory a hash table (see Section 8. (The probability that n(h (xi). 1) # 4 for 16 different values of xi is (15/16)16 = 35. and if we already know an upper bound M on the value of n (or failing this. x2 and x3 on the tape such that h (x1) begins with 1. then there exists a probabilistic algorithm for solving this problem that is efficient with respect to both time and space. but also whether we want to count such sequences as "jack-rabbit". This means that the final y begins with 1110. If y is a string of bits of length k. and 001. 0) Suppose. Consequently. (This may depend not only on the character set we are using. ) This crude rea- .4) holding a single occurrence of each form so far encountered. for example. 1 5i 5k . or four words. Prob[ k = 41 n = 4 ] = 183/4 %. (Such techniques are not covered in this book. b E { 0.1 }'" be a hash function that transforms a sequence from U in a pseudorandom way into a string of bits of length m.Probabilistic Algorithms 236 Chap. which will most likely prove prohibitive. We must first define what sequences of characters are to be considered as words. three. and then to make a sequential pass over the sorted tape to count the number of different words. The required time is thus in 0 (N) on the average. but it is in S2(Nn) in the worst case. on the value of N ). two. it is unlikely that there could be more than 16 distinct words on the tape. (The probability that it(h (x. or k+1 if none of the bits of y is equal to b. 01. in fact. respectively. Let m be a parameter somewhat larger than lg M (a more detailed analysis shows that m = 5 + Fig MI suffices).4.4 % = 1. this second method requires a quantity of central memory in S2(n). 1) = 3 for at least one value of xi among 4 different values is 1-(7/8)4 = 41. assuming h has sufficiently random behaviour . it is unlikely that there could be less than 4 distinct words on the tape. b ).string of (m + 1) bits set to zero { sequential passage through the tape } for each word x on the tape do i -7t(h(x). Since the probability that a random binary string begins with 0001 is 2-4. and "jack-in-the-box" as one. Let k be the value returned by a call on wordcnt. ) Conversely. Let h : U -* {0.

the standard deviation of R shows that this estimate may be in error by a factor of of m bits. Prove further that the standard deviation of R fluctuates around 1.62950 when n is sufficiently large. typically requiring that 1-A have only small eigenvalues. in about 1/n2 of the time they would require to compute the whole inverse. This offers a first approach for estimating the number of different words : calculate k using the algorithm wordcnt and estimate n as 2k/1. for any given 1 <_ i < n and 1 < j <. when the tape contains n different words and the function h : U ---)10. Consider. Unfortunately. your hash function should produce strings of m +lg t bits. 8. for instance. matrix inversion. Among those are matrix multiplication. ** Problem 8. 1 }' is randomly chosen with uniform distribution among all such functions (this last assumption is not reasonable in practice).64) .3. We do not discuss any of them in detail here because it is only for very specialized applications that they perform better than the obvious deterministic algorithms. By contrast. Classic deterministic inversion algorithms compute its inverse B as a whole or perhaps column by column.--. (Hint : by using t strings y 1 . you can obtain a relative precision of about 0.18. which is unacceptable. the solution of a system of simultaneous linear equations.78/-J provided t is sufficiently large (t >. and the computation of eigenvalues and eigenvectors. It is far from obvious how to carry out a more precise analysis of the unbiased estimate of n given by k.12127. The reader is referred to the literature for further discussion. Y 22 . . Prove that the expected value of R is in Ig n + O(1).Sec. where I stands for the identity matrix.5 Numerical Problems in Linear Algebra Many classic problems in linear algebra can be handled by numerical probabilistic algorithms.) 0 Notice that this approximate counting algorithm is completely insensitive to the order in which the words appear on the tape and to the number of repetitions of each of them. 8. Let R be the random variable returned by this algorithm ** Problem 8.3 Numerical Probabilistic Algorithms 237 soning indicates that it is plausible to expect that the number of distinct words on the tape should lie between 2k-2 and 2"`. Show how to obtain an arbitrarily precise estimate by using a little more space but with no appreciable increase in execution time. An intriguing feature of these probabilistic algorithms is their ability to compute independently the various entries in the result. where the hidden constant in 0(1) fluctuates around 0.3. provided n is sufficiently large.54703.n . an n x n nonsingular matrix A. .19. These algorithms are only applicable if the matrices concerned are well conditioned. there are probabilistic algorithms that are capable of estimating the value of B.3.

We wish to obtain a probabilistic algorithm B such that tB (x) = 1A (n) + s (n) for every instance x of size n. tA(x)/#X xEX This in no way rules out the possibility that there exists an instance x of size n such that to (x) >> FA (n). independently of the specific instance x to be solved. A hypothesis that is correct for a given application of the algorithm may prove disastrously wrong for a different application. but only worst-case executions.Probabilistic Algorithms 238 Chap.1 Selection and Sorting We return to the problem of finding the kth smallest element in an array T of n elements (Section 4. that quicksort (Section 4. for example. If we define Fe(n) = I te(a)l#X xex' the average expected time taken by algorithm B on a random instance of size n. but this fortuitous behaviour is only due to the probabilistic choices made by the algorithm. Analysis of this sorting method shows that it takes an average time in O(n log n) to sort n items provided that the instances to be sorted are chosen randomly. Suppose.4 SHERWOOD ALGORITHMS Section 1.1). it is clear that to (n) = tA(n)+s (n).5) is used as a subalgorithm inside a more complex algorithm. Thus there are no longer worst-case instances.4 mentions that analysing the average efficiency of an algorithm may sometimes give misleading results. This analysis no longer bears any relation to reality if in fact we tend to give the algorithm only instances that are already almost sorted. Algorithm B may occasionally take more time than tA(n)+s(n) on an instance x of size n. 8 8.1. Sherwood algorithms free us from the necessity of worrying about such situations by evening out the time required on different instances of a given size. be the set of instances of size n. The heart of this algorithm is the choice of a . Supposing that every instance of a given size is equiprobable. The reason is that any analysis of the average case must be based on a hypothesis about the probability distribution of the instances to be handled. Let A be a deterministic algorithm and let to (x) be the time it takes to solve some instance x.6 and Example 8. The Sherwood algorithm thus involves only a small increase in the average execution time if s (n) is negligible compared to to (n). 8. where tB (x) is the expected time taken by algorithm B on instance x and s (n) is the cost we have to pay for this uniformity. For every integer n let X.4. the average time taken by the algorithm to solve an instance of size n is tA(n)= Y.

n2 >> cp n >_ tp(n. the permutation of the first n integers that corresponds to the relative order of the elements of the array. a) and t. More precisely.)(3n3E IN)(Vn >C. and that we are looking for the median. For the execution time to be independent of the permutation a. function selectionRH (T [ 1 . n ]. Suppose that the elements of T are distinct. (n) = I t . we can express it as a function of both n and a. the simpler algorithm has the advantage of a much smaller hidden constant on account of the time that is saved by not calculating the pseudomedian. a). 8. (n . using the first element of the array as the pivot assures us of a linear execution time on the average. be the set of n! permutations of the first n integers. with the risk that the algorithm will take quadratic time in the worst case (Problems 4. n ] but (3c.j f. the simplified algorithm is sometimes disastrous : t. but only on their relative order. If we decide to aim for speed on the average thanks to the simpler deterministic algorithm.6). (n . Despite this prohibitive worst case.6 do not depend on the values of the elements of the array. (Y) for most values of a. a) < tp (n .6. Let tp (n . let S.a)/n! .239 Sherwood Algorithms Sec.4 pivot around which the other elements of the array are partitioned. k) { finds the kth smallest element in array T . we assume that 1 <_ k <_ n } iE. Using the pseudo- median as the pivot assures us of a linear execution time in the worst case. t. The simplified algorithm is generally faster : for every n . (n . Define F.ElN)(`dn?n1)(Vo (acs << cp)(3n2E lNT)(Vn ? n2)[ts(n) <_ c. On the other hand. The execution times of the algorithms in Section 4.5 and 4.. Rather than express this time as a function solely of n. The fact that we no longer calculate a pseudomedian simplifies the algorithm and avoids recursive calls. The decision whether it is more important to have efficient execution in the worst case or on the average must be taken in the light of the particular application. We have the following equations : 6ESn (3cp)(3n.3. On the other hand. we must make sure that the instances to be solved are indeed chosen randomly and uniformly.a) is occasionally much greater than tp (n . even though finding this pivot is a relatively costly operation. (5) be the times taken by the algorithm that uses the pseudomedian and by the simplified algorithm.6. respectively. (n .l. it suffices to choose the pivot randomly among the n elements of the array T. The resulting algorithm resembles the iterative binary search of Section 4. which forces us to distinguish between the worst case and an average case.n .(Y)] .

v ] are equal to m. Stochastic preconditioning allows us to obtain a Sherwood algorithm without changing the deterministic algo- . and. after this operation the elements of T [i . m.Probabilistic Algorithms 240 Chap.k return T[i] Here partition (T.1).no)(VaESf)[tRH(n. The probabilistic nature of the algorithm ensures that tRH (n .v) ifk < u <-else if k > v then i -v+1 else i. Notice that quicksort must first be modified along 0 the lines of Problem 4.. this probability is independent of the instance concerned. A similar analysis to that of Problem 4.4. occasions when we are given a deterministic algorithm efficient on the average but that we cannot reasonably expect to modify.6.1. (Y) be the average time taken by the Sherwood algorithm to determine the median of an array of n elements arranged in the order specified by a. independently of the instance to be solved. var v) pivots the elements of T [i . There are. j .. we have transformed this algorithm into a Sherwood algorithm that is efficient (with high probability) whatever the instance considered. j i. Using the probabilistic approach. It is always possible that some particular execution of the algorithm will take quadratic time. those of T [u .i. Its simplicity ensures that (3noEN)(f/n >. to repeat.4.T [uniform (i . j ] around the value m . and those of T [v +1 . badly documented software package. Problem 8. for instance. u -I] are less than m .u. var u. Thus its efficiency is not affected by the peculiarities of the application in which the algorithm is used. i .. j ] are greater than m.m.a)] To sum up. 8. 8 while i < j do m <-.5.j.(Y) < tp(n.6. (Y) is independent of a. but the probability that this will happen becomes increasingly negligible as n gets larger. This happens.. if the algorithm is part of a complicated. The values of u and v are calculated and returned by the pivoting algorithm (see Problem 4.. Show how to apply the Sherwood style of probabilistic approach to quicksort (Section 4.5). however.5 shows that the expected time taken by this probabilistic selection algorithm is linear.4. Let tRH (n .2 Stochastic Preconditioning The modifications we made to the deterministic algorithms for sorting and for selection in order to obtain Sherwood algorithms are simple. we started with an algorithm that is excellent when we consider its average execution time on all the instances of some particular size but that is very inefficient on certain specific instances. j)] partition (T.

Denote the discrete logarithm of x modulo p to the base g by .. )[f (x) = v(r. and H.Y such that i. procedure shuffle (T [I . Assume random sampling with uniform distribution is possible efficiently within A . and let A be a set with the same number of elements. Simply call the following procedure before the deterministic sorting or selection algorithm. f(u(x.uniform (i n) interchange T [i] and T [ j ] Example 8. let X be the set of instances of size n. and then to deduce the solution to the original instance.1. to use the given deterministic algorithm to solve this random instance. We thus obtain the following Sherwood algorithm. sorting is the same : it is simply a question of randomly shuffling the elements of the array in question. the first property ensures that this instance is transformed into an instance y chosen randomly and uniformly from all those of the same size.1 to n -1 do j .u (x.f (y) { solved by the deterministic algorithm } return v (r. Stochastic preconditioning consists of a pair of functions u : X x A -* X and v : A x Y . The stochastic preconditioning required for selection or for Example 8. n ] ) for i f. Suppose for the purposes of illustration that someone were to discover an algorithm efficient on the average but prohibitively slow in the worst case. Recall that no efficient algorithm is known for calculating discrete logarithms (Section function RH (x) { computation of f (x) by Sherwood algorithm } let n be the size of x r .uniform (A. 8.4 Sherwood Algorithms 241 rithm. the solution to this random instance allows us to recover the solution of the original instance x. iii. No posttreatment (function v) is needed to recover the solution in these cases. s ) Whatever the instance x to be solved. r) { random instance of size n } s .) y E. the functions u and v can be calculated efficiently in the worst case. Thanks to the second property. Suppose the problem to be solved consists of the computation of some function f : X -i Y for which we already have an algorithm that is efficient on the average.r)))].Sec.4. For every integer n. Let A be the union of all the A . The trick is to transform the instance to be solved into a random instance.

logg..1). 8.p a { using the assumed algorithm } return (s . logg. For instance. . and so on. Assume that you would like to compute f (x) for some instance x but that you lack the computing power or the efficient algorithm to do so. ii. 13. perhaps for a fee.3 Searching an Ordered List A list of n keys sorted into ascending order is implemented using two arrays val [ 1 . 8 logg.bx mod p s f. 8. r . then ptr [i] gives the index of the following key.. here is one way to represent the list 1. Stochastic preconditioning offers an intriguing possibility : computing with an encrypted instance. The rank of a key is the number of keys in the list that are less than or equal to the given key. furthermore. Find other problems that can benefit from stochastic precondi- tioning.p x + logg. x . logs. Here is the Sherwood algorithm. r) is independent of x as long as r is chosen randomly with uniform probability. that some other party is capable of carrying out this computation and willing to do so for you.p(gr modp)=r for 05r Sp-2. Problem 8.4.2.p y) mod (p . 3. This process yields no information on your actual request. 2. 5. The end of the list is marked by ptr [i] = 0. p ) r <--uniform (0. The following equations allow us to transform our hypothetical algorithm into a Sherwood algorithm : i.p (xy mod p) = (logg. because the probability distribution of u (x. The smallest key is in val [head ].p x.4. and then use the function v to deduce f (x). function dlogRH (g .3. n ] and an integer head. have f (y) computed for you..Probabilistic Algorithms 242 Chap.8) a F. Why does the algorithm dlogRH work? Point out the functions corresponding to u and v. except for its size. 21 .4. p -2) b Fdexpoiter (g. p) { Section 4. n ] and ptr [ 1 .r) mod (p -1) 11 Problem 8. Assume. In general. What should you do if you are unwilling to divulge your actual request x? The solution is easy if stochastic preconditioning applies to the computation of f : use the function u to encrypt x into some random y. if val [i] is not the largest key. the next smal- lest is in val [ptr [head ]].

4 Sherwood Algorithms i val [i] ptr [i] 243 1 2 3 4 5 6 7 2 2 3 13 1 5 21 8 5 6 1 7 0 3 In this example head = 4 and the rank of 13 is 6. wA (n) and mA (n) denote its worst-case and its mean time.}. 1 5 i <_ n . . Prove the preceding assertion. respectively. Thus WA(n)=max{tA(n. Whether the algorithm is deterministic or probabilistic. If A is any deterministic algorithm.4. As usual. k. there exists a deterministic algorithm that is capable of carrying out such a search in an average time in 0 (I ). Problem 8. We want a deterministic algorithm B such that mB (n)E 0 ('/n) and a Sherwood algorithm C such that we (n) = mB (n). (Y) denotes the time taken by this algorithm to find the key of rank k among the n keys in the list when the order of the latter in the array val is specified by the permutation a.) Despite this inevitable worst case.4 implies that wA (n) E S2(n) for every deterministic algorithm A. The following algorithm finds a key x starting from some position i in the list.k. We can use binary search to find a key efficiently in a sorted array. From this we can obtain a Sherwood algorithm whose expected execution time is in 0 (') whatever the instance to be solved. Here. however.(Y) nxn! vES k=I Problem 8.Sec. Any instance can be characterized by a permutation a of the first n integers and by the rank k of the key we are looking for. provided that x ? val [i] and that x is indeed present. the problem is thus to find that index i. and mA(n)= E EtA(n.4. In fact. 8. k . Given a key x. which would correspond to the first step in binary search. such that val [i] = x. Let Sn be the set of all n! permutations. but it does not have such a thing as a worst-case instance. and that all the elements of the list are distinct. any deterministic algorithm takes a time in Q (n) in the worst case to find a key in this kind of list. Suppose for the moment that the required key is always in fact present in the list.k.4.(Y)I 1 <_k <_n and aeS. In the case of a probabilistic algorithm. the Sherwood algorithm is no faster on the average than the corresponding deterministic algorithm. there is no obvious way to select the middle of the list. tA (n . to (n . (Hint : show how a worst-case instance can be constructed systematically from the probes made into the list by any given deterministic algorithm. (Y) denotes the expected value of this time.

Determine 1A (n. and m.max .y return search (x.val [i ] case x < y : return search (x. i ) while x > val [i] do i <. head ) Let iA (n.val [i] for j f.4.4.ptr [i ] return i Here is the obvious deterministic search. Show. however. (See Problem 8.) Define WA (n) and mA (n) similarly. and m introduced in the previous probProblem 8. Here is a first probabilistic algorithm. The following deterministic algorithm is efficient on the average.) The quantities F.8. k) if and only if k > f (n).5. Determine wD (n) and mD (n) for every integer n. Determine WA (n) and mA (n) for every integer n . w. lems facilitate our analysis. why should we choose to execute the for loop . k) for every integer n and for every k between 1 and n.4. val made by the algorithm A to find the key of rank k in a list of n keys. w.Probabilistic Algorithms 244 Chap. function D (x) i .l to L'u J do y <--val[j] ifmax<y <_x theni F j. that they do not tell the whole story by exhibiting a deterministic algorithm E such that WE (n) E 0 (log n).head max <. 8 function search (x. Determine 1D (n . k) be the exact number of references to the array Problem 8. (The order a of the keys is irrelevant for this algorithm. thus apparently contradicting Problem 8.4.7. k) for every integer n and for every k between 1 and n.4. function A (x) return search (x. k) ? Compare wD (n) and mA (n).uniform (L.6. function B (x) i .4. times in algorithm B ? Intuitively. head ) x > y : return search (x .4. ptr [i ]) otherwise : return i Problem 8. k) < iA (n . As a function of n.5 for the definition of i . what values of k maximize t D (n. i ) Problem 8. Give explicitly a function f (n) such that ?D (n . n) y .

4.4. N ] of lists in which T [i ] is the list of those identifiers x found in the program such that h (x) = i. 8. To do this.14. Is this better than 0 (n log n) ? Justify your answer. is used in just about every compiler to implement the symbol table. Such a function is a good choice if it efficiently disperses all the probable identifiers. 2.4.5. (Continuation of Problem 8. [Hint : Let M1.4.4. Other ways to handle collisions.4. Let X be the set of possible identifiers in the language to be compiled. n }.11. uniformly.4 Universal Hashing Hash coding.13. Show that the when expected value of M1. Suggest a few. N }. . k. . give a Sher- wood algorithm C such that wC (n) E O Show more precisely Problem 8. are legion. . Starting from the deterministic algorithm B. A hash function is a function h : X --* { 1. 2.. Show.Sec.4. the table takes space in O(N+n) and the average length of the lists is a. Prove that m8 (n) E 0 (' )..4. When x * y but h (x) = h (y). which is unavoidable from Problem 8... Give an efficient Sherwood algorithm that takes into account the possibility that the key we are seeking may not be in the list and that the keys may not all be distinct. if h (x) # h (y) for most of the pairs x * y that are likely to be found in the same program. (The ratio (x may well be greater than 1..) If we suppose that every identifier and every pointer occupies a constant amount of space. be the random variable that corresponds to the minimum of I integers chosen randomly.4. Thus we see that increasing the value of N reduces the average search time but increases the space occupied by the table. however.. where n is the number of distinct identifiers in the table. Analyse your algorithm. Problem 8.. where the meaning of w is given in Problem 8. and let N be a parameter chosen to obtain an efficient system.11) that wC (n) E 2' + O(1). is about n /(l + 1) when 1 is a constant and about Problem 8. The load factor of the table is the ratio a = n IN. . .9..4. 8.4. Find a link between this random variable and the average-case analysis of algorithm B. or simply hashing.4. 0) E U(n). Sherwood Algorithms 245 * Problem 8.. that w8 (n) E S2(n). that is. Use the structure and the algorithms we have just seen to obtain a Sherwood sorting algorithm that is able to sort n elements in a worst-case expected time in 0 (n 312). Problem 8. The hash table is an array T [I . give explicitly a permutation a and a rank k such that tB (n . Problem 8. Problem 8. and independently with replacement from the set ( 1. besides the use of a table of lists as outlined here. we say that there is a collision between x and y .

These programs will compile slowly every time they are submitted.16. there are f a r too many functions from X into { 1. In a sense they are paying the price for all other programs to compile quickly..2. that #X is very much greater than N. This technique is very efficient provided that the function h disperses the identifiers properly.4. the probability of a collision between x and y is therefore at most 1/N.9 before answering.B are there if the cardinalities of A and B are a and b. Let S c X be a set of n identifiers already in the table. Let H be a universa12 class of functions f r o m X to { 1. 2. .4. does it suffice to choose N rather larger than this bound to ensure that the probability of a collision is negligible ? (Hint : solve Problem 8. 8 Problem 8. A Sherwood approach allows us to retain the efficiency of hashing on the average. N } for it to be reasonable to choose one at random.4. Prove that the average number of collisions between x and the elements of S (that is. . We content ourselves with mentioning just one. y E A such that x #y.. respectively ? This difficulty is solved by universal hashing. . the average length of the list T [h (x)]) is less than or equal to the load factor a. N } . Unfortunately. (If you have not yet solved Problem 8. and let x and y be any two distinct identifiers.. Let xEX \ S be a new identifier. . * Problem 8.) Problem 8. How many functions f :A .18. however. Several efficient universal2 classes of functions are known.246 Probabilistic Algorithms Chap. 2.3.4. The following problem generalizes this situation. What do you think of the "solution" that consists of ignoring the problem? If we are given an a priori upper bound on the number of identifiers a program may contain.17.. it is inevitable that certain programs will cause a large number of collisions. . without arbitrarily favouring some programs at the expense of others. Show that n calls on the symbol table can take a total time in f2(n2) in the worst case. now is the time to think some more about it!) The basic idea is to choose the hash function randomly at the beginning of each compilation. If we suppose.3. Problem 8.19. If we choose a hash function h randomly and uniformly in H. Prove further that the probability that the number of colli- sions will be greater than t a is less than 1/t for all t >_ 1. A program that causes a large number of collisions during one compilation will therefore probably be luckier next time it is compiled. By definition a class H of functions from A to B is universal2 if # (h E H (h (x) = h (y)) < #H/#B for every X.

(x) = ((mx + n) mod p) mod N. 8. and let p be a prime number * Problem 8. Better still is the existence of some constant 8 > 0 such that p (x) >. success).20. Now consider the following algorithm. not less than N. It may be able to solve in practice certain problems for which no efficient deterministic algorithm is known even on the average. However.n < p E is a universal2 class of functions. Las Vegas algorithms usually have a return parameter success. .. whatever happens. respectively. y . even though the expected time required for each instance may be small and the probability of encountering an excessive execution time is negligible.4. on the other hand.. sometimes for every instance. The distinguishing characteristic of Las Vegas algorithms is that now and again they take the risk of making a random decision that renders it impossible to find a solution.4.) I * Problem 8. Thus these algorithms react by either returning a correct solution or admitting that their random decisions have led to an impasse. 1 5 m < p and 0 <.1) never takes a time in excess of 0 (n2) to sort n elements. .. Let s (x) and e (x) be the expected times taken by the algorithm on instance x in the case of success and of failure. there is no upper bound on the time that may be required to obtain a solution.. 2. 8.S for every instance x. (Remarks : In practice we take N to be a power of 2 so that the second mod operation can be executed efficiently. Find applications of universal hashing that have nothing to do with compilation..5 Las Vegas Algorithms 247 Let X be { 0... In the latter case it suffices to resubmit the same instance to the same algorithm to have a second. The overall probability of success therefore increases with the amount of time we have available. For example. 1. .5 LAS VEGAS ALGORITHMS Although its behaviour is more uniform. the Sherwood version of quicksort (Problem 8. . The typical call to solve instance x is LV (x . where we are able to predict the maximum time needed to solve a given instance. allows us to obtain an increase in efficiency..Sec.21. we require that p (x) > 0 for every instance x. a -11. Let p (x) be the probability of success of the algorithm each time that it is asked to solve the instance x. nor even with management of a symbol table. independent chance of arriving at a solution.. For an algorithm to be correct. which is set to true if a solution is obtained and false otherwise. l . a Sherwood algorithm is no faster on the average than the deterministic algorithm from which it arose. N-1 ) by h. A Las Vegas algorithm. Define hm n :X -* (0. Let m and n be two integers. where y is a return parameter used to receive the solution thus obtained whenever success is set to true. Prove that H = { hm. It is also more efficient to carry out all the computations in a Galois field whose cardinality is a power of 2.4. . Contrast this to a Sherwood algorithm.

5. that the queens placed on the board do not threaten one another. it may be preferable to accept a smaller probability of success if this also decreases the time required to know that a failure has occurred. . success) until success return y Let t (x) be the expected time taken by the algorithm obstinate to find an exact solution to the instance x. but the algorithm does not take into account one important fact : there is nothing systematic about the positions of the queens in most of the solutions. 8.. however. for any t >.1 The Eight Queens Problem Revisited The eight queens problem (Section 6.( )x) WW e (x) There is a compromise to be made between p (x). which still takes an expected time t (x). and e (x) if we want to minimize t (x). The recurrence is easily solved to yield t (x) = s (x) + 1 . The resulting algorithm is not recursive.5. 8 function obstinate (x) repeat LV (x . For example.6. With probability 1. This observation suggests a greedy Las Vegas algorithm that places queens randomly on successive rows. Suppose that s (x) and e (x) are not just expected times.248 Probabilistic Algorithms Chap. Using this technique. before starting all over again to solve the instance. . Recall that the backtracking technique used involves systematically exploring the nodes of the implicit tree formed by the k-promising vectors. e (x) and p (x). On the contrary. This is not bad.1.. s (x). Neglecting the time taken by the control of the repeat loop. What is the probability that the algorithm obstinate will find a correct solution in a time not greater than t. Problem 8. s (x). y .057 nodes in the tree. we obtain the first solution after examining only 114 of the 2. taking an expected time e (x). The algorithm ends either successfully if it manages to place all the queens on the board or in failure if there is no square in which the next queen can be added.1) provides a nice example of this kind of algorithm.s (x) ? Give your answer as a function of t.p (x) it first makes an unsuccessful attempt to solve the instance. we obtain the following recurrence : t (x) = p (x) s (x) + (1.p (x)) (e (x) + t (x)).) in the case of success and of failure. the queens seem more to have been positioned haphazardly. respectively. but that they are in fact the exact times taken by a call on LV (x . taking care. thus taking an expected time s (x). This follows because the algorithm succeeds at the first attempt with probability p (x).

it is column j that has been chosen (with probability 1 / nb) } try [k +I] '. the average number s of nodes that it explores in the case of success. less than half the number of nodes explored by the systematic backtracking technique..2. Show that each position has. nb) = 1 then ( maybe try column i } j4-i if nb > 0 then ( amongst all nb possibilities for the (k + 1)st queen. diag 135 f.927 .5. diag45. and e = 6.Sec.diag 135 u { j + k } { try [ 1 . When there is more than one position open for the (k + I )st Problem 8. the same probability of being chosen. nevertheless. We can do better still.colu{j} diag45 .971 time out of eight by proceeding in a completely random fashion ! The expected number of nodes explored if we repeat the algorithm until a success is finally obtained is given by the general formula s + (1-p )e/p = 55. The Las Vegas algorithm is too defeatist : as soon as it detects a failure it starts all over again from the beginning.I to 8 do if i 0 col and i -k e diag45 and i +k 0 diag 135 then ( column i is available for the (k + I )st queen } nb -nb+1 if uniform (1 .0 for i E. the algorithm QueensLV chooses one at random without first counting the number nb of possibilities.diag45 u { j . we need to determine its probability p of success. k ] is k-promising } nb .0 k -0 repeat { try [1 .5 Las Vegas Algorithms 249 procedure QueensLV (var success) { if success = true at the end.(nb > 0) To analyse the efficiency of this algorithm.. then try [1 . k + 1 ] is (k + 1)-promising } k-k+1 until nb = 0 or k =8 success . queen.j col .. and the average number e of nodes that it explores in the case of failure.. The backtracking .k } diag 135 F. 8] (a global array) contains a solution to the eight queens problem } col. Clearly s = 9 (counting the 0-promising empty vector). Using a computer we can calculate A solution is therefore obtained more than one p = 0. 8.1293 .

An unfortunate random choice of the positions of the first few queens can make it impossible to add the others. The pure backtracking algorithm finds the first solution in 40 milliseconds.00 9. and then uses backtracking to try and add the remaining queens. makes a systematic search for a solution that we know has nothing systematic about it.1293 0. This is a fraction more than .00 39. and the expected number t = s +(I -p)e/p of nodes explored if the algorithm is repeated until it eventually finds a solution.10 8.1624 0. where 1 < stopVegas S 8 indicates how many queens are to be placed randomly before moving on to the backtracking phase. however. on the other hand.0000 1.50 55.29 6. without. success) else success *. col. reconsidering the positions of the queens that were placed randomly. for instance.4931 0.false . if the first two queens are placed in positions 1 and 3. takes on the average 23 milliseconds to find a solution.92 53. stopVegas p s 0 1.1357 0. The case stopVegas = 0 corresponds to using the deterministic algorithm directly.79 7.97 6.33 9.98 6. diag45. which places all the queens in a random way.63 28.31 9.97 114. 8 algorithm. the expected number e of nodes explored in the case of failure.10 46. The resulting algorithm is similar to QueensLV.0000 0. The more queens we place randomly.01 35. The original greedy algorithm QueensLV. except that the last two lines are replaced by until nb = 0 or k = stopVegas if nb > 0 then backtrack (k .67 15. diag 135. The following table gives for each value of stopVegas the probability p of suc- cess. but the greater is the probability of a failure.2618 0.48 1 2 3 4 5 6 7 8 10. respectively. whereas an average of 10 milliseconds is all that is needed if the first two or three queens are placed at random.93 55.1 except that it has an extra parameter success and that it returns immediately after finding the first solution if there is one.93 -- We tried these different algorithms on a CYBER 835.00 e t 39. This happens.1293 114. the expected number s of nodes explored in the case of success.00 39.05 9.53 13.8750 0. The latter looks like the algorithm Queens of Section 6.63 22.6.20 29.Probabilistic Algorithms 250 Chap. A judicious combination of these two algorithms first places a number of queens on the board in a random way. the smaller is the average time needed by the subsequent backtracking stage.

this can be understood intuitively in terms of the lack of regularity in the solutions (at least when the number of queens is not 4k + 2 for some integer k).5 Las Vegas Algorithms 251 half the time taken by the backtracking algorithm because we must also take into account the time required to make the necessary pseudorandom choices of position.4] are explored to no effect. Using the probabilistic approach and placing the first ten queens at random. Thus the probabilistic algorithm turned out to be almost 1. For the eight queens problem. 8. is that a solution can be obtained more rapidly on the average if several queens are positioned randomly before embarking on the backtracking phase. the same corner is a better than average starting point for the problems with five or twelve queens. What is significant. (Try it!) This unlucky characteristic of the upper left-hand corner is nothing more than a meaningless accident.39 222. For instance.5. and then try again. 36 different solutions were found in about five and a half minutes. On the other hand. This is one reason why it is more efficient to place the first queen at random rather than to begin the systematic search immediately. e.00 33.4. If we want a solution to the general n queens problem.0465 262.00 - 47. however. this time placing the first five queens randomly.0000 0.000 times faster per solution than the deterministic algorithm.71.20 262.5. a systematic search for a solution beginning with the first queen in the first column takes quite some time. If you are still not convinced of the value of this technique. First.11 On the CYBER 835 the Las Vegas algorithm that places the first five queens randomly before starting backtracking requires only 37 milliseconds on the average to find a solution. s. An empirical study of the twenty queens problem was also carried out using an Apple II personal computer. and t for a few values of stopVegas in the case of the twelve queens problem. Once again. Here are the values of p.21 and [1. ** Problem 8.3. First the trees below the 2-promising nodes [1. we waste time with [1.00 80. The deterministic backtracking algorithm took more than 2 hours to find a first solution.88 13. it wastes so much time making its pseudorandom choices of position that it requires essentially the same amount of time as the pure backtracking algorithm. t stopVegas p s e 0 5 12 1. whereas the pure backtracking algorithm takes 125 milliseconds. it is obviously silly to analyse exhaustively all the possibilities so as to discover the optimal .5. Problem 8.5039 0. Even when the search starting from node [1. As for the greedy Las Vegas algorithm.5] begins. we suggest you try to solve the twelve queens problem without using a computer. try to solve the problem systematically.23 10. a systematic search that begins with the first queen in the fifth column is astonishingly quick.Sec.5.3] and [1.

63 is a square root of 55 modulo 103. (Hint : assuming that a 2 ° h 2 (mod p). Prove further that x is a quadratic residue modulo p if and only if x(P.2 Square Roots Modulo p Let p be an odd prime.6. An integer x is a quadratic residue modulo p if 1 <. on the other hand.5. Such a y is a square root o x modulo p provided 1 :.b 2. An integer z is a quadratic nonresidue modulo p if 1 < z <.x <. Prove or disprove : the n queens problem can be solved for every n >.7. 8 value of stopVegas. (Hint: one direction follows immediately from Fermat's theorem : x n = 1 (mod p ).) Problem 8.5.8 to calculate x(p .p .2py + y 2 = y 2 (mod p). that no quadratic residue has more than two distinct square roots. can you find a constant S > 0 such that the probability of success of the Las Vegas algorithm to solve the n queens problem is at least S for every n ? 8.4.5. Any quadratic residue has at least two distinct square roots since (p _Y)2 = p 2 . If no solution exists.1)/2 mod p.5. In fact. Conclude from the preceding results that exactly half the integers between 1 and p -1 are quadratic residues modulo p.x < p -1 and if there exists an integer y such that x = y2 (mod p).5. value of stopVegas to be determined rapidly as a function of n. Technically.9. determining the optimal value of stopVegas takes longer than a straightforward search for a solution using backtracking.5. and then try to place the queens on the board) can only be considered to be a Las Vegas algorithm if its probability of success is strictly positive for every n. * Problem 8.p .p . the obstinate proba- bilistic algorithm will loop forever without realizing what is happening. but not necessarily optimal. Problem 8. consider a 2 . (We needed more than 50 minutes computation on the CYBER to establish that stopVegas = 5 is the optimal choice for the twelve queens problem !) Find an analytic method that enables a good. For instance. Problem 8.1.5. the number of queens. ** Problem 8. Given an odd prime p and a quadratic residue x .) The preceding problem suggests an efficient algorithm for testing whether x is a quadratic residue modulo p : it suffices to use the fast exponentiation algorithm of Section 4.y <. Prove that x(P-1)/2 =± 1 (mod p) for every integer 1<.y # y and that 1 <_ p . Prove that p . the other direction requires some 0 knowledge of group theory.I and z is not a quadratic residue modulo p. Combining this with Problem 8.p -l and every odd prime p. Prove.1.4.5.Probabilistic Algorithms 252 Chap.8. This is the case if and only if there exists at least one solution.y <. the general algorithm obtained using the previous problem (first determine stopVegas as a function of n. and then to apply the corresponding Las Vegas algorithm.1)/2 = + 1 (mod p).

b. But 7 has two square roots modulo p. This modulo p. . carry out the symbolic calProblem 8. (1-1) mod 53 is also a quadratic nonresidue modulo 53. Subtracting them. Suppose.9).I (mod 53). an efficient Las Vegas algorithm to solve this problem when p = 1 (mod 4). it is possible to carry out the symbolic multiplication of a +b-6.2 (mod 53).3 (mod 4) and let x be a quadratic residue Problem 8. where a. modulo p. and hence d / . however. Let p = 53 = 1 (mod 4) and x = 7.5 Las Vegas Algorithms 253 modulo p.5.and c + d are integers between 0 and p . we find 2d/ . and d product is ((ac +bdx) mod p) + ((ad + bc) mod p Note the similarity to a product of complex numbers. Even if the value of is unknown.1 (mod 53) and c =.11. What happens if we calculate symbolically (a + / )26 == c +dI (mod 53) in a case when one of (a +') mod 53 and (a -i7) mod 53 is a quadratic residue modulo p and the other is not? for instance.3 (mod 4).F7-) =0+421 (1+ 41 )26 = -1 (mod 53). Adding these two equations. A preliminary computation shows that x is a quadratic residue modulo p since 726 = 1 (mod 53). 8. (All the following calculations are modulo 53. c. Suppose that p .F7_ =(22+101)(22+101)=18+16I (1+J)12=(18+161)(18+l61)=49+461 (1+x)13=(1+x)(49+46-. c -d Using Example 8.) (1 + 7)2 = (1 + -0) (1 + 17) =8+ (1+V)3 =(1+1)(8+2J) =22+10.1)/2.1 (mod 53).1. and the symbolic calculation that we just carried out is valid regardless of which of them we choose to call '. There exists.Sec. Let us decide arbitrarily to denote by the smaller of the two square roots of x. Let us calculate symbolically (1 mod 53.1 (mod 4). but no efficient deterministic algorithm is known to solve this problem when p =.1 as a guide. we obtain 2c = 0 (mod 53) and hence c = 0 since 0 <.1. Example 8.5. Prove that ±x(p+ 1)/4 mod p are the two square roots of x modulo p. The symbolic exponentiation (a +b-6.10. that c +d 1 _. culation (2+I7)26 = 0+41J (mod 53) in detail.)" can be calculated efficiently by adapting the algorithms of Section 4. Consequently.5. does there exist an efficient algorithm for calculating the two square roots of x modulo p ? The problem is easy when p . we conclude that (1+') mod 53 is a quadratic nonresidue modulo 53 (Problem 8.c < 52. Calculate 5526 mod 103 and verify that its square modulo 103 is indeed 55.5.8. Since 26 = (p .5.

and let x be a quadratic residue * Problem 8. a. [Hint : Suppose without loss of generality that u >. If p is prime and 1 a <. p .52 and 41y . Otherwise. i.22 = 31.) By mathematical induction.4). An integer a.uniform (I.y < p -1 and dy = l (mod p ) It remains to determine the probability of success of this algorithm. prove that there exists a unique y such that 1 <. Let u and v be two positive integers.1 (mod 4). the proof is trivial (a = 0 and b = 1). Give an efficient iterative algorithm for calculating d. Then we need only take a = b' and b = a' -b' Lu l v j.1. The other square root is 53 .true compute y such that 1 <. Prove that . 8 To obtain a square root of 7.11). I (mod p). and let d be their greatest common divisor.true y Fa else compute c and d such that 0 <-c p-1. and x is a quadratic residue modulo p } a F.p -1 and ay given p and a. now let a' and b' be such that a'v + b'w = d. This is indeed a square root of 7 modulo 53 since 222 = 7 (mod 53).Probabilistic Algorithms 254 Chap. p -1) if a 2 = x (modp) { very unlikely } then success f. This suggests the following Las Vegas algorithm for calculating square roots. Your algorithm should not calculate d before starting to work on a and b. This can be done efficiently using a modification of Euclid's algorithm for calculating the greatest common divisor (Section 1. the preceding problem shows that we need only find the unique integer y such that 1 <.7. var success) { may find some y such that y 2 = x (modp ) assuming p is a prime.13.5..] ii. (This is the heart of Euclid's algorithm. iii.y <. let w = u mod v.1 (mod 53).a p -1. 1 <.12.1 (mod 53). we find y = 22 because 41 x 22 . modulo p. If v = d. and b from u and v. var y.y <. procedure rootLV (x . Let p = 1 (mod 4) be prime. Prove that there exist integers a and b such that au + by = d.5. Give an efficient algorithm for calculating y In our example (following Problem 8.p . First show that d is also the greatest common divisor of v and w.0 <d p-1 and (a +')(r-1)12-c+dL (modp) if d = 0 then success F false else { c = 0) success f. gives the key to ' if (a2-x) mod p is not a quadratic residue modulo P.5. p. * Problem 8.v.

Sec. 8.5

Las Vegas Algorithms


i. The Las Vegas algorithm finds a square root of x if and only if it randomly
chooses an a that gives the key to ' ; and
ii. Exactly (p + 3)/2 of the p -1 possible random choices for a give the key to x_.
[Hint : Consider the function








, p ,p-


defined by the equation (a - NFx_ ) f (a) = a +L (mod p ). Prove that this function is one-to-one and that f (a) is a quadratic residue modulo p if and only if a


does not give the key to '.]

This shows that the Las Vegas algorithm succeeds with probability somewhat
greater than one half, so that on the average it suffices to call it twice to obtain a
square root of x. In view of the high proportion of integers that give a key to ', it is
curious that no known efficient deterministic algorithm is capable of finding even one
of them with certainty.

The previous problem suggests a modification to the algoProblem 8.5.14.
rithm rootLV : only carry out the symbolic calculation of (a +' )(p- 1)12 if
(a2-x) mod p is a quadratic nonresidue. This allows us to detect a failure more
rapidly, but it takes longer in the case of a success. Give the modified algorithm explicitly. Is it to be preferred to the original algorithm ? Justify your answer.
* Problem 8.5.15.
if p

1 (mod 8)

The following algorithm increases the probability of success


procedure rootLV 2(x, p, var y, var success)
I assume that p is a prime and p = 1 (mod 4) }

a F-uniform(1..p-1)
if a 2 = -x (mod p)

I very unlikely and unfortunate }

then success - false
else let odd t and k >- 2 be such that p = 2k t + 1

compute c and d such that 0 <c 5p-1,0<-d <p-1
)t =-c + d

and (a +

(mod p)

ifc =0ord =0
then success - false
else while c 2

d 2x (mod p) do

b F-(c2-d2x)modp
d E-- 2cd mod p

c -h

compute y such that 1 5 y 5 p- 1 and yd = 1 (mod p)

y - cy mod p
success F- true
(1/2)k-1 and that the
Prove that the probability of failure of this algorithm is exactly
while loop is executed at most k - 2 times, where k is specified in the algorithm.

Probabilistic Algorithms


Chap. 8

Problem 8.5.16.
An even more elementary problem for which no efficient
deterministic algorithm is known is to find a quadratic nonresidue modulo p where
p = 1 (mod 4) is a prime.
i. Give an efficient Las Vegas algorithm to solve this problem.

ii. Show that the problem is not more difficult than the problem of finding an
efficient deterministic algorithm to calculate a square root. To do this, suppose
that there exists a deterministic algorithm root 2(x , p) that is able to calculate
efficiently ' mod p, where p =- 1 (mod 4) is prime and x is a quadratic residue
modulo p. Show that it suffices to call this algorithm less than L19 P I times to
be certain of obtaining, in a way that is both deterministic and efficient, a quadratic nonresidue modulo p. (Hint : Let k be the largest integer such that 2k

divides p -1 exactly. Consider the sequence x i = p - 1, xi = x7 mod p for
2<- i <- k. Prove that xi is a quadratic residue modulo p for 1 <- i < k , but that
xk is not.)

** Problem 8.5.17.
The converse of Problem 8.5.16. Give an efficient deterministic algorithm rootDET (x , p , z) to calculate a square root of x modulo p, provided

p is an odd prime, x is a quadratic residue modulo p, and z is an arbitrary quadratic
nonresidue modulo p.
The two preceding problems show the computational equivalence between the
efficient deterministic calculation of square roots modulo p and the efficient deterministic discovery of a quadratic nonresidue modulo p. This is an example of the
technique called reduction, which we study in Chapter 10.

8.5.3 Factorizing Integers
Let n be an integer greater than 1. The problem of factorizing n consists of finding the
unique decomposition n = p 1 ' p 2
such that m I , M2, ... , Mk are positive
integers and p I < P 2 <
< p k are prime numbers. If n is composite, a nontrivial
factor is an integer x, 1 < x < n , that divides n exactly. Given composite n, the
problem of splitting n consists of finding some nontrivial factor of n.

Problem 8.5.18.
Suppose you have available an algorithm prime (n), which
tests whether or not n is prime, and an algorithm split(n), which finds a nontrivial
factor of n provided n is composite. Using these two algorithms as primitives, give an
algorithm to factorize any integer.

Section 8.6.2 concerns an efficient Monte Carlo algorithm for determining primality. Thus the preceding problem shows that the problem of factorization reduces to
the problem of splitting. Here is the naive algorithm for the latter problem.

Sec. 8.5

Las Vegas Algorithms


function split (n)
{ finds the smallest nontrivial factor of n if n is composite
or returns 1 if n is prime }

for i F- 2 to LJ j do
if (n mod i) = 0 then return i
return I
Problem 8.5.19.

Why is it sufficient to loop no further than ' ?

The preceding algorithm takes a time in Q(J) in the worst case to split n. It is
therefore of no practical use even on medium-sized integers : it could take more than 3

million years in the worst case to split a number with forty or so decimal digits,
counting just
microsecond for each trip round the loop. No known algorithm,
whether deterministic or probabilistic, can split n in a time in 0 (p (m)) in the worst
case, where p is a polynomial and m = [log(l +n)l is the size of n. Notice that
Z 10"'12, which is not a polynomial in m. Dixon's probabilistic algorithm is

nevertheless able to split n in a time in 0 (20( m '09 - ))

Problem 8.5.20.
Prove that 0 (mk ) c O (2014"' kg n )) C O
the values of the positive constants k and b.


The notion of quadratic residue modulo a prime number (Section 8.5.2) generalLet n be any positive integer. An integer x,
1 5 x S n -1, is a quadratic residue modulo n if it is relatively prime to n (they have
no nontrivial common factor) and if there exists an integer y, 1 <- y < n - 1, such that
izes to composite numbers.

x = y 2 (mod n). Such a y is a square root of x modulo n. We saw that a quadratic
residue modulo p has exactly two distinct square roots modulo p when p is prime.
This is no longer true modulo n if n has at least two distinct odd prime factors. For
instance, 82 = 132 = 222 = 272 = 29 (mod 35).

* Problem 8.5.21.
Prove that if n = pq, where p and q are distinct odd primes,
then each quadratic residue modulo n has exactly four square roots. Prove further that

exactly one quarter of the integers x that are relatively prime to n and such that

<- x <- n - 1 are quadratic residues modulo n.

Section 8.5.2 gave efficient algorithms for testing whether x is a quadratic residue

modulo p, and if so for finding its square roots. These two problems can also be
solved efficiently modulo a composite number n provided the factorization of n is
given. If the factorization of n is not given, no efficient algorithm is known for either

of these problems. The essential step in Dixon's factorization algorithm is to find two
integers a and h relatively prime to n such that a2 = b2 (mod n) but a ±b (mod n).
This implies that a2 - h2 = (a - b)(a + b) = 0 (mod n). Given that n is a divisor nei-

ther of a +h nor of a -b, it follows that some nontrivial factor x of n must divide
a + b while nix divides a -b. The greatest common divisor of n and a + b is thus a

Probabilistic Algorithms


Chap. 8

nontrivial factor of n. In the previous example, a = 8, b = 13, and n = 35, and the
greatest common divisor of a +b = 21 and n = 35 is x = 7, a nontrivial factor of 35.
Here is an outline of Dixon's algorithm.

procedure Dixon (n, var x, var success)
{ tries to find some nontrivial factor x of composite number n }

if n is even then x <- 2, success - true
else for i - 2 to Llog3n J do
if n 11' is an integer then x <- n
success - true
{ since n is assumed composite, we now know that it has
at least two distinct odd prime factors }
a, b <- two integers such that a 2 = b 2 (mod n)

if a °±b (mod n) then success <- false
else x <- gcd(a+b, n) { using Euclid's algorithm }
success F- true

So how we can find a and b such that a2 = b2 (mod n)? Let k be an integer to
be specified later. An integer is k -smooth if all its prime factors are among the first k
prime numbers. For instance, 120 = 23 x 3 x 5 is 3-smooth, but 35 = 5 x 7 is not.
When k is small, k-smooth integers can be factorized efficiently by an adaptation of
the naive algorithm split (n) given earlier. In its first phase, Dixon's algorithm chooses
integers x randomly between 1 and n - 1. A nontrivial factor of n is already found if
by a lucky fluke x is not relatively prime to n . Otherwise, let y = x 2 mod n . If y is
k-smooth, both x and the factorization of y are kept in a table. The process is repeated
until we have k + 1 different integers for which we know the factorization of their
squares modulo n.

Let n = 2,537 and k = 7. We are thus concerned only with
Example 8.5.2.
the primes 2, 3, 5, 7, 11, 13, and 17. A first integer x = 1,769 is chosen randomly.
We calculate its square modulo n : y = 1,240. An attempt to factorize
1,240 = 23 x 5 x 31 fails since 31 is not divisible by any of the admissible primes.
A second attempt with x = 2,455 is more successful : its square modulo n is
y = 1,650 = 2 x 3 x 52 x 11. Continuing thus, we obtain
x,= 2 , 455 y I= 1 ,650 = 2 x 3 x 52 x 11

y2=2,210 =2x5x 13x 17
X3=1,105 y3= 728 =23x7x13

x2= 970

X4 = 1,458

y4 = 2,295 = 33 x 5 x 17

x5= 216

y5= 990 =2x32x5x11

X6 =


y6= 1,326 = 2 x 3 x 13 x 17

X7 = 1,844

y7= 756 =22x33x7

xs =

ys= 2,288 = 24 x 11 x 13


Sec. 8.5

Las Vegas Algorithms


Problem 8.5.22. Given that there are 512 integers x between 1 and 2,536 such
that x2 mod 2537 is 7-smooth, what is the average number of trials needed to obtain
eight successes like those in Example 8.5.2 ?

The second phase of Dixon's algorithm finds a nonempty subset of the k + I
equations such that the product of the corresponding factorizations includes each of the
k admissible prime numbers to an even power (including zero).

Example 8.5.3.

There are seven possible ways of doing this in Example 8.5.2,


y1y2y4y8=26x34x54x70 x112x132x172
Y1Y3Y4Y5Y6Y7=28 x310x54x72x 112x 132x 172.

Problem 8.5.23.

Find the other five possibilities.

Problem 8.5.24. Why is there always at least one solution ? Give an efficient
algorithm for finding one. [Hint : Form a (k + 1) x k binary matrix containing the parities of the exponents. The rows of this matrix cannot be independent (in arithmetic
modulo 2) because there are more rows than columns. In Example 8.5.3, the first
dependence corresponds to
(1,1,0,0,1,0,0) + (1,0,1,0,0,1,1) + (0,1,1,0,0,0,1) + (0,0,0,0,1,1,0)

(0,0,0,0,0,0,0) (mod 2) .

Use Gauss-Jordan elimination to find a linear dependence between the rows.]

This gives us two integers a and b such that a2 = b2 (mod n). The integer a is
obtained by multiplying the appropriate xi and the integer b by halving the powers of
the primes in the product of the y; If a 4t±b (mod n), it only remains to calculate the
greatest common divisor of a + b and n to obtain a nontrivial factor. This occurs with
probability at least one half.

Example 8.5.4.
The first solution of Example 8.5.3 gives
a = x1x2x4x8 mod n = 2,455 x 970 x 1,458 x 433 mod 2,537 = 1,127 and

b = 23 x 32 x 52 x 11 x 13 x 17 mod 2,537 = 2,012 *±a (mod n).
The greatest common divisor of a +b = 3,139 and n = 2,537 is 43, a nontrivial factor
of n. On the other hand, the second solution gives
a =xlx3x4x5x6x7 mod n = 564 and

b=24x35x52x7x11x13x17mod 2,537=1,973=-a (mod n),
which does not reveal a factor.

Problem 8.5.25.
Why is there at least one chance in two that
a t ± b (mod n)? In the case when a
b (mod n), can we do better than simply
starting all over again ?

Probabilistic Algorithms


Chap. 8

It remains to be seen how we choose the best value for the parameter k. The
larger this parameter, the higher the probability that x 2 mod n will be k -smooth when
x is chosen randomly. On the other hand, the smaller this parameter, the faster we can
carry out a test of k -smoothness and factorize the k -smooth values y; , and the fewer
such values we need to be sure of having a linear dependence. Set L = e
let h E IR+. It can be shown that if k = L b , there are about L 112h failures for every success when we try to factorize x2 mod n. Since each unsuccessful attempt requires k
divisions and since it takes k + 1 successes to end the first phase, the latter takes an
average time that is approximately in 0 (L 2h +1/2h ), which is minimized by b = z . The
second phase takes a time in O (k 3) = 0(L31) by Problem 8.5.24 (it is possible to do
better than this), which is negligible when compared to the first phase. The third phase
can also be neglected. Thus, if we take k = E , Dixon's algorithm splits n
with probability at least one half in an approximate expected time in 0(L2)
) and in a space in O (L).
= O (e 2 I
Several improvements make the algorithm more practical. For example, the proI

bability that y will be k-smooth is improved if x is chosen near [I 1, rather than
being chosen randomly between I and n - 1. A generalization of this approach,
known as the continued fraction algorithm, has been used successfully. Unlike
Dixon's algorithm, however, its rigorous analysis is unknown. It is therefore more
properly called a heuristic. Another heuristic, the quadratic sieve, operates in a time in
0 (L 9is) and space in 0 (LD). In practice, we would never implement Dixon's algorithm because the heuristics perform so much better. More recently, H. W. Lenstra Jr.
has proposed a factorization algorithm based on the theory of elliptic curves.

Problem 8.5.26.

Let n = 1040 and L = e 1° n In Inn


Compare L 9/8, L 2, and

microseconds. Repeat the problem with n = 10s0p
8.5.4 Choosing a Leader

A number of identical synchronous processors are linked into a network in the shape of

a ring, as in Figure 8.5.1. Each processor can communicate directly with its two
immediate neighbours. Each processor starts in the same state with the same program
and the same data in its memory. Such a network is of little use so long as all the processors do exactly the same thing at exactly the same time, for in this case a single
processor would suffice (unless such duplication is intended to catch erratic behaviour
from faulty processors in a sensitive real-time application). We seek a protocol that

allows the network to choose a leader in such a way that all the processors are in
agreement on its identity. The processor that is elected leader can thereafter break the
symmetry of the network in whatever way it pleases by giving different tasks to different processors.
No deterministic algorithm can solve this problem, no matter how much time is
available. Whatever happens, the processors continue to do the same thing at the same
time. If one of them decides that it wants to be the leader, for example, then so do all

Sec. 8.5

Las Vegas Algorithms

Figure 8.5.1.


A ring of identical processors.

the others simultaneously. We can compare the situation to the deadlock that arises
when two people whose degree of courtesy is exactly equal try to pass simultaneously
through the same narrow door. However, if each processor knows in advance how
many others are connected in the network, there exists a Las Vegas algorithm that is
able to solve this problem in linear expected time. The symmetry can be broken on the
condition that the random generators used by the processors are independent. If the
generators are in fact pseudorandom and not genuinely random, and if each processor
starts from the same seed, then the technique will not work.

Suppose there are n processors in the ring. During phase zero, each processor
initializes its local variable m to the known value n, and its Boolean indicator active to
the value true. During phase k, k > 0, each active processor chooses an integer randomly, uniformly, and independently between 1 and m. Those processors that chose 1
inform the others by sending a one-bit message round the ring. (The inactive processors continue, nevertheless, to pass on messages.) After n -1 clock pulses, each processor knows the number I of processors that chose 1. There are three possibilities. If
/ = 0, phase k produces no change in the situation. If l > 1, only those processors that
chose 1 remain active, and they set their local variable m to the value 1. In either case
phase k + I is now begun. The protocol ends when l = 1 with the election of the
single processor that just chose 1.
This protocol is classed as a Las Vegas algorithm, despite the fact that it never
ends by admitting a failure, because there is no upper bound on the time required for it

to succeed. However, it never gives an incorrect solution: it can neither end after
electing more than one leader nor after electing none.

Let 1(n) be the expected number of phases needed to choose a leader among n
processors using this algorithm (not counting phase zero, the initialization). Let
p (n , j) _ [ ] n-i (1- lln)" -i be the probability that j processors out of n randomly
choose the value I during the first stage. With probability p (n, 1) only a single phase
is needed ; with probability p (n , 0) we have to start all over again ; and with

but it finds a correct solution with high probability whatever the instance considered.p(n. This is not the same as saying that it works correctly on a majority of instances.n)) i=2 * Problem 8. j=2 A little manipulation gives n-I I(n)=(I+ F. Problem 8.5. A Monte Carlo algorithm occasionally makes a mistake. only failing now and again in some special cases. Prove that the following algorithm decides correctly whether or not an integer is prime in more than 80% of the cases.27.0)-p(n.j)l(j). 30030) = 1 then return true else return false ( using Euclid's algorithm } . the preceding problems show that the choice of a leader also takes expected linear time. function wrong (n) if gcd(n . Since each phase thus takes linear time. ** Problem 8. that it is not a Monte Carlo algorithm by exhibiting an instance on which the algorithm systematically gives a wrong answer.28.5.6 MONTE CARLO ALGORITHMS There exist problems for which no efficient algorithm is known that is able to obtain a correct solution every time.5.718 for every n >_ 2. on the other hand.j)l(j))/(1-p(n.0)l(n)+J:p(n. Show that l (n) < e = 2. 8 probability p (n . n l(n)=1+p(n. Nevertheless. give a Monte Carlo algorithm that is able to determine n exactly with a probability of error less than p whatever the number of processors. Prove that no protocol (not even a Las Vegas protocol) can solve the problem of choosing a leader in a ring of n identical processors if they are not given the value n . j ).442. No warning is usually given when the algorithm makes a mistake. n>_2. n-)oo 11 One phase of this protocol consists of single bit messages passed round the ring. 8. Consequently.29.1.) Show. Show that lim I (n) < 2. whether it be deterministic or probabilistic. 2 <_ j <_ n.6. given an arbitrary parameter p > 0. (It performs much better on small integers.Probabilistic Algorithms 262 Chap. * Problem 8. l (j) subsequent phases will be necessary on the average.

030 to bring the proportion of successes above 85% even if very large integers are considered? Let p be a real number such that 1/2 < p < 1. we need only call it several times and choose the most frequent answer. Problem 8. Its error probability is therefore at most m-1 E Prob[ i correct answers in n tries ] i=0 < M-1 1: i=0 n []piqnl i z+e.Monte Carlo Algorithms Sec. p-correct algorithm. function MC 3(x) MC (x) . however small it may be. More generally. Consider the following algorithm.4c2 ). let a and S be two positive real numbers such that e + S < -1. . The time taken by such algorithms is then expressed as a function of both the size of the instance and the reciprocal of the acceptable error probability. whatever the instance considered. Let c. A Monte Carlo algorithm is p-correct if it returns a correct solution with probability not less than p. Some Monte Carlo algorithms take as a parameter not only the instance to be solved but also an upper bound on the probability of error that is acceptable. were not consistent. hence 84%-correct. p = and q = I -p = z -e. This increases our confidence in the result in a way similar to the world series calculation of Section 5. The advantage of such an algorithm is p -'/2. u .cE lg 1/8 be the number of times Let m = Ln / 2j+ 1. Show on the other hand that it might not even be 71%-correct if MC. though 75%-correct. The repetitive algorithm finds the correct answer if the latter is that the ('+F-)-correct algorithm is called. Let MC(x) be a consistent. = -2 / Ig (1. The algorithm is consistent if it never gives two different correct solutions to the same instance. To increase the probability of success of a consistent. so as to obtain a new algorithm whose error probability is as small as we choose. It suffices to call MC (x) at least [cElg 1/81 times and to return the most frequent answer (ties can be broken arbi- trarily) to obtain an algorithm that is consistent and (1. To prove the preceding claim. v .6.2. 75%-correct Monte Carlo algo- rithm.S)-correct.2. This allows us to amplify the advantage of the algorithm. Let MC (x) be a Monte Carlo algorithm that is consistent and (z + e)-correct.MC (x) t if t = u or t = v then return t return v Prove that this algorithm is consistent and 27/32-correct. Let x be some instance to be solved. let n >.6 263 What constant should be used instead of 30. 8. obtained at least m times.MC (x) .

Prove that cE < (In 2)/2E2. and that repeating it 600 times yields an algorithm that is better that 99%-correct. A more precise calculation shows that it is enough to repeat the original algorithm 269 times. [This is because some of the inequalities used in the proof are rather crude. we wish to go from a 55%-correct algorithm to a 95%-correct algorithm). suppose we have a consistent Monte Carlo algorithm whose advantage is 5% and we wish to obtain an algorithm whose probability of error is less than 5% (that is. Fortunately.1/(2SI) and then set m = Fx/4E21.4E2 < 1 =6 since aliisa = 2 for every a > 0. 8 -i m-1 since q /p < 1 and 2. most Monte Carlo algorithms that occur in practice are such that we can increase our confidence in the result obtained much more rapidly. If we repeat such an algorithm several times to increase our confidence in the final result. where m-1 S= 2-E v 2i i =0 i (4 -£2)i < 1 (1-4E2)m 4E nm The first part of this formula can be used efficiently to find the exact number of repetitions required to reduce the probability of error below any desired threshold S. the resulting algorithm is (1-S)-correct.Probabilistic Algorithms 264 =(Pq)ni2 I. Assume for simplicity that we are dealing with a decision problem and that the original Monte Carlo algorithm is biased in the sense that it is always correct whenever it returns the answer true.0 i=0 (pq)n/2 ` (nl (Pq)ni22n = 2)n i2 _ (4pq )n i2 = (1-4F (1. it suffices to repeat such an . The preceding theorem tells us that this can be achieved by calling the original algorithm about 600 times on the given instance. The probability of success of the repetitive algorithm is therefore at least 1.] Repeating an algorithm several hundred times to obtain a reasonably small probability of error is not attractive.S. it would be silly to return the most frequent answer : a single true outweighs any number of falses. n i=0 (pq )n12 (qlp) Chap. Problem 8.4E2)( = 2-1g 1/6 12) lg 1/8 since 0 < 1. A more complicated argument shows that if a consistent (Z + E)-correct Monte Carlo algorithm is repeated 2m -1 times.3. errors being possible only when it returns the answer false.i i >. For example. As we shall see shortly. Alternatively.6. a good upper bound on this number of repetitions is quickly obtained from the second part: find x such that e'x .

and let y be the solution returned by MC (x). the restriction that the original algorithm be p-correct for some p > i no longer applies : arbitrarily high confidence can be obtained by repeating a biased p-correct algorithm enough times. but the algorithm may not always return the correct solution to these instances. A Monte Carlo algorithm is yo-biased if there exists a subset X of the instances such that i. if there exist i # j such that y. y is indeed correct . it is not required that an efficient test be known for membership in X. the algorithm always returns the correct solution. Now suppose that we call MC (x) k times and that the answers obtained are y. and if x EX. the correct solution is necessarily yo. 8.yj . but the probability of such an occurrence is at most (1. the solution returned by the algorithm is always correct whenever the instance to be solved is not in X. and if x E X. Moreover. even if p < z (as long as p > 0). the preceding argument shows that this is indeed the correct solution . In both cases. . and ii. the probability of such an error occurring is not greater than 1-p given that the algorithm is p-correct. = y # yo for all the i. What happens if on the other handy # y o ? If x e X. the only possible explanation is that x EX (because the algorithm is consistent by assumption). let us return to an arbitrary problem (not necessarily a decision problem). and therefore the correct solution is yo. and let yo be some distinguished answer. Let x be an instance. Yk- If there exists an i such that y. so yo is indeed correct . and if y. it is still possible that the correct solution is yo and that the algorithm has made a mistake k times in succession on x E X. the algorithm has made a mistake since the correct solution is y o .Monte Carlo Algorithms Sec. the correct solution to all the instances that belong to X is yo. we may conclude that yo is a correct solution. = yo. . yo-biased and p-correct.p )k . or 6 times to obtain a 99%-correct algorithm. More formally. Although the distinguished answer yo is known explicitly. Let MC be a Monte Carlo algorithm that is consistent. :t.6 265 algorithm 4 times to improve it from 55%-correct to 95%-correct. The following paragraph shows that this definition is tuned precisely to make sure that the algorithm is always correct when it answers yo. What can we conclude if y =yo? If x e X.

2.true while ans and i < k do if-i+1 ans F. k repetitions of a consistent. This last remark may appear trivial. Prob[ MC (x) = true ] = I for each instance x E S and Prob[ MC (x) = true ] <_ q for each instance x 0 S.6. k) i-0 ans E. It suffices to repeat the algorithm at most 20 times to be either sure that the correct solution is yo (if either of the first two cases previously described is observed). The probability in question is therefore either 0 or 1. given any x E I. if you systematically quiz the algorithm with instances for which the correct solution is yo. Consider the following algorithms. or not. whether x E S.(1-p )k )-correct and still consistent and yo-biased. Indeed. Let q =1-p .Probabilistic Algorithms 266 Chap. function repeatMC (x . for example. p-correct. but it is in fact crucial if the probabilistic algorithm is used to generate with high probability some random instance x on which the correct answer is a specific Y #Yo To illustrate this situation. however. It is important to understand how to interpret such behaviour correctly. yo-biased algorithm yield an algorithm that is (1 . but it causes no problems with a biased algorithm). or extremely confident that the solution obtained on every one of the trials is correct (since otherwise the probability of obtaining the results observed is less than one chance in a million). my proportion of errors should not significantly exceed (1-p )k ". The correct interpretation is as follows : "I believe that y is the correct answer and if you quiz me enough times on different instances. The "proportion of errors" in question. despite the fact that we cannot tell for sure which it is. p-correct. Assume your consistent. it will always be wrong whenever it "believes" otherwise. In general. It may be tempting to conclude that "the probability that y is an incorrect answer is at most (1-p )k ".MC (x) return ans . that p = s (once again. p-correct Monte Carlo algorithm to decide. we may be interested in generating a random prime of a given length . By definition of a false-biased algorithm. (As a practical example. yo-biased Monte Carlo algorithm has yielded k times in a row the same answer y # yo on some instance x. not only over those occurrences in which the algorithm actually provides such probabilistic answers. this is not allowed for general Monte Carlo algorithms. Such a conclusion of course makes no sense because either the correct answer is indeed y. 8 Suppose.see Section 8. We are interested in generating a random member of some subset S S 1. is averaged over the entire sequence of answers given by the algorithm. consider a nonempty finite set I of instances.) Let MC be a false-biased.

5. It is correct that Prob[X I Y ] <_ q k. var y. They all involve the solution of a decision problem. Thus the section continues with some examples of biased Monte Carlo algorithms. k) return x It is tempting to think that genrand(k) returns a random member of S with a probability of failure at most q k . where X stands for "repeatMC (x . Give an efficient Las Vegas algorithm LV (x . var success) to solve the same problem. What is the best value of r you can deduce so that your Las Vegas algorithm succeeds with probability at least r on each instance ? El .6. k) returns true" and Y stands for "x 0 S ". Let A and B be two efficient Monte Carlo algorithms for Problem 8. we need an a priori probability that a call on uniform (I) returns a member of S. Prove that the probability that a call on genrand (k) erroneously returns some x 0 S is at most 1 1+ 1r q-' r k In particular. 8. Algorithm A is p-correct and true-biased.6. the only possible answers that they can return are true and false. whereas algorithm B is q-correct and false-biased. * Problem 8. This is wrong in general.4. We are not aware of any unbiased Monte Carlo algorithm sufficiently simple to feature in this introduction.uniform (I) until repeatMC (x. This situation illustrates dramatically the difference between the error probability of genrand and that of repeatMC. To calculate this pro- bability. then the probability that a call on genrand (k) returns some x 0 S is about q k /r if q k<< r<< 1. and nearly 1 if r << g k This can be significant when the confidence one gets in the belief that x belongs to S from running MC several times must be weighed against the a priori likelihood that a randomly selected x does not belong to S. solving the same decision problem. that is.6 267 Monte Carlo Algorithms function genrand (k) repeat x .Sec. if the error probability of MC on instances not in S is exactly q (rather than at most q). but we are in fact interested in Prob[ Y I X ]. Let r denote the probability that x E S given that x is returned by a call on uniform (I). The problem is that we must not confuse the conditional probabilities Prob[ X I Y ] and Prob[ Y I X ]. about z2 if q k= r << 1.

Probabilistic Algorithms 268 Chap.11.6. 8 8. the answer returned is false. The general technique for biased Monte Carlo algorithms allows us to reduce this probability efficiently to any arbitrary value. If the array does have a majority element. and if one of its elements is chosen at random. In sum.11. the probability that the first call of maj (T) will return true is p > z .1 Majority Element in an Array The problem is to determine whether an array T [I . in which case maj 2(T) also returns true.0 forj f. 4.6.. the element chosen is a majority element. n ]) i F uniform (1 . the second call of maj (T) may still with probability p return true. with probability > z T has no majority element = maj (T) = false. since majority elements occupy more than half the array. which happens with probability 1-p . consider function maj 2(T) if maj (T) then return true else return maj (T) If the array does not have a majority element. First.. Consider the following algorithm. and hence trivially there is a majority element in T. An error probability of 50% is intolerable in practice. function maj (T [1. too. The algorithm maj2 is therefore also true-biased. each call of maj(T) is certain to return false. if the answer returned by maj (T) is false. the probability that maj 2(T) will return true if the array T has a majority element is p+(I-p)p=I-(1-p)2>3/4.11. with certainty . if the first call of maj (T) returns false. On the other hand.. Therefore.5. but 3/4-correct. although in this case the element chosen randomly is in a minority. on the other hand. and then checks whether this element forms a majority in T. If.7). and in this case maj 2(T) returns true. hence so does maj 2(T ). and 4. it is nonetheless possible that T contains a majority element. If the array does indeed contain a majority element. z that is T has a majority element = maj (T) = true. The probability of error decreases because the successive calls of maj (T) are independent : the fact that . the probability of choosing an element that is in a minority is less than one-half.ltondoifT[jI=x thenk '-k +1 return (k > n / 2) We see that maj (T) chooses an element of the array at random. n ] has a majority element (see Problems 4. Summing up. If the answer returned is true. n) x -T[i] k F. we may reasonably suspect that T does indeed have no majority element. this algorithm is true-biased and 4correct.

11. . For larger values of n the situation gets worse. and we can be certain that n is composite. In 98% of calls the algorithm will inform us incorrectly that n is prime. but it is still unsatisfactory.I to k do if maj (T) then return true return false The algorithm takes a time in 0 (n log(1/E)).5 and 4. 8. E) k rlg(1/E)1 for i <-. we can be certain that T contains a majority element. (It is currently possible to establish with certainty the primality of numbers up to 213 decimal digits within approximately 10 minutes of computing time on a CDC CYBER 170/750. return false is less than 2-k if T contains a majority element. the answer true is returned with high probability even if n is in fact composite. The algorithm chooses an integer randomly between 2 and 51.uniform (2 .6. This is interesting only as an illustration of a Monte Carlo algorithm since a linear time deterministic algorithm is known (Problems 4. Show that the probability that k successive calls of maj(T) all Problem 8.6 Monte Carlo Algorithms 269 maj (T) has returned false on an array with a majority element does not change the probability that it will return true on the following call on the same instance. function majMC (T.6). On the other hand. 8.Sec. Thus there is only a meagre 2% probability that it will happen on d = 43 and hence return false.623 = 43 x 61. The problem is to decide whether a given integer is prime or composite. No deterministic or Las Vegas algorithm is known that can solve this problem in a reasonable time when the number to be tested has more than a few hundred decimal digits. using Euclid's algorithm. The algorithm can be improved slightly by testing whether n and d are relatively prime. El The following Monte Carlo algorithm solves the problem of detecting the presence of a majority element with a probability of error less than E for every E > 0.6. where n is the number of elements in the array and E is the acceptable probability of error.) A first approach to finding a probabilistic algorithm might be function prime (n) d .. LJ J) return ((n mod d) # 0) If the answer returned is false. Unfortunately. Consider for example n = 2.11.6. the algorithm has been lucky enough to find a nontrivial factor of n. as soon as any call returns true.2 Probabilistic Primality Testing This classic Monte Carlo algorithm recalls the algorithm used to determine whether or not an array has a majority element.

This Monte Carlo algorithm for deciding primality is falsebiased and 3/4-correct. The theorem assures us that if n is composite. 1582"9 = 1312 = 110 (mod 289). Give an efficient algorithm for testing whether n is a strong pseudoprime to the base a. and 1588 9 = 2512 =_ I (mod 289). and let s and t be positive integers such that n -1 = 2s t.6. it cannot be a strong pseudoprime to more than (n . whereas 737 does not even have one. n -2) if n is strongly pseudoprime to the base a then return true else return false For an odd integer n > 4 the theorem ensures that n is composite if Rabin (n) returns false. function Rabin (n) { this algorithm is only called if n > 4 is odd } a <--uniform (2. Let a be an integer such that 2 5 a <. Problem 8.8. 289 has only 14 false witnesses of primality. For instance. Your algorithm should not take significantly longer (and sometimes even be faster) than simply calculating a (" . For example. Contrariwise. where 4(n) < n -1 is Euler's totient function. finally.7. This suggests the following algorithm. 1589 = 131 (mod 289).1)/2 mod n with dexpoiter from Section 4. there is at least a 75% chance that n will not be a strong pseudoprime to the base a. 8 To obtain an efficient Monte Carlo algorithm for the primality problem. If n is composite. In both cases. There exist however composite numbers that are strong pseudoprimes to some bases.2. an integer a is chosen at random. This test of primality has several points in common with the Las Vegas algorithm for finding square roots modulo p (Section 8. it is a strong pseudoprime to any base.2 different bases.6.. If n is prime. The theorem is generally pessimistic. we need a theorem whose proof lies beyond the scope of this book. 1584"9 = 1102 = 251 (mod 289). This certainty is not accompanied however by any indication of what are the nontrivial factors of n.5. where t is odd. 158 is a false witness of primality for 289 because 288 = 9 x 25.i < s and art _ -1 (mod n).n . in which case we obtain the correct answer with cer- . ** Problem 8. Prove this theorem.8. we may begin to suspect that n is prime if Rabin (n) returns true. We say that n is a strong pseudoprime to the base a if a' _ 1 (mod n) or if there exists an integer i such that 0 <.2). Such a base is then a false witness of primality for this composite number. The situation is even better if n is composed of a large number r of distinct prime factors : in this case it cannot be a strong pseu- doprime to more than 4(n)/2r-' .9)/4 different bases.Probabilistic Algorithms 270 Chap. Let n be an odd integer greater than 4.

Find a true-biased Monte Carlo algorithm for primality testing whose running time is polynomial in the logarithm of the integer being tester. At any given moment the question "Does Si = Si ?" may be asked.6. The basic operation consists of adding x to the set Si . A philosophical remark is once again in order: the algorithm does not reply "this number is prime with probability 1-E". regardless of how long the program is allowed to run.6.9. all of which are empty at the outset. this can be due either to the fact that n is indeed prime or to the fact that a is a false witness of primality for the composite number n. (Note : this figure of 99% is very conservative as it would still hold even if Rabin (n) had a flat 25% chance of failure on each composite integer. the probability of error can be made arbitrarily small by repeating the algorithm. program print primes print 2. One might also expect composite numbers to be produced erroneously once in a while. where xE U \ Si and I S i _< n.2. otherwise I have observed a natural phenomenon whose probability of occurrence was not greater than E". On the other hand. Prove that this is unlikely to happen. . not necessarily disjoint. but rather.3 A Probabilistic Test for Set Equality We have a universe U of N elements. Llg n j) then print n n -n + 2 ad museum Clearly. every prime number will eventually be printed by this program. "I believe this number to be prime . The difference can also be explained using Problem 8.10.6. * Problem 8.) ** Problem 8. 8. if n is a strong pseudoprime to the base a.6.3 n 5 repeat if repeatRahin (n.2. Notice that this combines with the false-biased algorithm described previously to yield a Las Vegas algorithm (by Problem 8.5).Monte Carlo Algorithms Sec.6 271 tainty. As usual. More precisely. prove that the probability is better than 99% that not even a single composite number larger than 100 will ever be produced. Consider the following nonterminating program. We suppose that N is quite large while n is small. and a collection of n sets. Similarly there is better than a 50% chance that a will provide a key for finding . 8. since every integer larger than I is either prime or composite. The first reply would be nonsense. Nevertheless the algorithm for testing primality is only a Monte Carlo algorithm whereas the one for finding square roots is Las Vegas. This difference is explained by the fact that the Las Vegas algorithm is able to detect when it has been unlucky : the fact that a does not provide a key for is easy to test.

if indeed it is not in S2(k log k ). 1/E))]. 8 The naive way to solve this problem is to keep the sets in arrays. The table is used to implement a random function rand : U -+ (0. j ) ifv[i]=v[jI then return true else return false It is obvious that Si # Sj if v [i] # v [ j ]. For any e > 0 fixed in advance.1 }k as follows. let . each test of equality will take a time in S2(k).7. and two calls with different arguments are independent. The algorithm never makes an error when Si = S/ . To each set Si we associate a variable v [i] initialized to the binary string composed of k zeros. Here is the algorithm for adding an element x to the set Si .1 }k . search trees. We suppose that x is not already a member of Si . Each call of rand (x) returns a random string chosen with equal probability among all the strings of length k.Probabilistic Algorithms 272 Chap. For a set S c U.v[i]®rand(x) The notation t ® u stands for the bit-by-bit exclusive-or of the binary strings t and u. procedure add (i . The Monte Carlo algorithm first chooses a function at random in this class and then initializes a hash table that has U for its domain. Let H be a universal2 class of functions from U into (0. x) v[i] F. there exists a Monte Carlo algorithm that is able to handle a sequence of m questions in an average total time in 0 (m). lists. or hash tables. Let e > 0 be the error probability that can be tolerated for each request to test the equality of two sets. Whatever structure is chosen.4. the set of k -bit strings. where k is the cardinality of the larger of the two sets concerned. each call of rand (x) takes constant expected time.some random k -bit string add x to the table and associate y to it return y Notice that this is a memory function in the sense of Section 5. Let Si = Si \ (x0}. Thanks to the use of universal hashing. Two different calls with the same argument return the same value.4). This algorithm provides an interesting application of universal hashing (Section 8. The algorithm to test the equality of Si and S1 is: function test (i . Let k = Ilg(max(m. function rand (x) if x is in the table then return its associated value y E. in the opposite case its probability of error does not exceed E. What is the probability that v [i ] = v [ j ] when Si # Sj ? Suppose without loss of generality that there exists an xo a Si such that xa 9 S/ .

Sec. j ). . (Hint: reread Problem 5. x 9 Si u SS .6 Monte Carlo Algorithms 273 XOR (S) be the exclusive-or of the rand (x) for every x E S . Suppose that a call of uniform (u . which removes the element x from the set Si . 2. which decides without ever making an error whether x E Si .. x). Your implementation should satisfy each request in constant time in the worst case. N 1. let N be an integer and let U = { 1. Show how to implement a random permutation. Moreover. A call of init initializes a new permutation tt : U -* U. Si = Sj u {x }.2. x) is only permitted when x is already in Si . More precisely. You may use memory space in O (N). v) takes constant time for 1 <_ u <_ v <_ N.1. Problem 8. may be troublesome for other applications. but no request may consult or modify more than a constant number of memory locations : thus it is not possible to create the whole permutation when init is called. This Monte Carlo algorithm differs from those in the two previous sections in that our confidence in an answer "Si = Si " cannot be increased by repeating the call of test (i . Notice the similarity to the use of signatures in Section 7.6. You must accept two kinds of request : init and p (i) for 1 <_ i <_ N. Modify the algorithm so that it will work correctly (with probability of error c) even if a call of add (i. rand : U -*{O.13. For instance. and if an application of the algorithm replies incorrectly that Si = Sj .14.6. The possibility that rand (x 1) = rand (X2) even though x I * X2. Let yo = XOR (S.2 and Example 8.7.) and v[j] = XOR (SS ).. the different tests of equality are not independent. then it will also reply incorrectly that Sk=S1. A call of elim (i. The fact that v [i ] = v [ j ] implies that rand (xo) = yo. ) ®XOR (Si ). A sequence of m requests must still be handled in an expected time in O (m). whatever happens. which does not worry us when we want to test set equality.6. What happens with this algorithm if by mistake a call of Problem 8. . v [i] = XOR (Si) = rand (xo) ®XOR (S. Sk = Si u {x }. Universal hashing allows us to implement a random function ** Problem 8. Two such calls separated by a call of init should on the other hand be independent. Show how you could also implement a procedure elim (i.1}".4. x) is made when xE Si .6.12. if Si # Sj .. A call of p (i) returns the value 7t(i) for the current permutation. Two calls p (i) and p Q) that are not separated by a call of init should therefore return two different answers if and only if i # j. x) is made when x is already in Si ? Problem 8. It is only possible to increase our confidence in the set of answers obtained to a sequence of requests by repeating the application of the algorithm to the entire sequence. add (i. Also implement a request member (i . x). the probability of this happening is only 2-k since the value of rand (xo) is chosen independently of those values that contribute to yo. 8.1). By definition. .11.

n and 2n. and C and you would like to decide whether AB = C. and r (x) of degrees n. we must compute ABX as A times BX. correct Monte Carlo algorithm that is capable of solving this problem in a time in d (n2).1 ton do X [i ] <. function goodproduct (A. Compare this with the fastest i known deterministic algorithm to compute the product AB (Section 4. Given three polynomials p (x). n) array X [ 1 .17.6.uniform ({ -1.3. C . providing a dramatic example of the topic discussed in Section 5. but no such algorithm is known that only takes a time in 0 (n). Recall that it is often used to describe any probabilistic algorithm. (Hint: consider the columns of the matrix AB -C and show that at least half the ways to add and subtract them yield a nonzero column vector. see Anon.6. B .15.16. 8 8. The term "Monte Carlo" was introduced into the literature by Metropolis and Ulam (1949). The term "Las Vegas" was introduced by Babai' (1979) to distinguish probabilistic algorithms that occasionally make a mistake from those that reply correctly if they reply at all. It is no doubt the earliest recorded probabilistic algorithm. It is obvious that goodproduct (A . Here is an intriguing false-biased. q (x).5.) Given two n x n matrices A and B.6. Problem 8. .. (In the next chapter we shall see a deterministic algorithm that is capable of computing the symbolic product of two polynomials of degree n in a time in 0 (n log n ).7 REFERENCES AND FURTHER READING The experiment devised by Leclerc (1777) was carried out several times in the nineteenth century. The term "Sherwood" is our own.2. whenever AB = C. decide probabilistically whether B is the inverse of A.6. C. (1495). respectively.Probabilistic Algorithms 274 Chap.4 Matrix Multiplication Revisited You have three n x n matrices A. adapt this algorithm to Problem 8. B. in particular in Los Alamos. and with the probabilistic algorithm mentioned in Section 8. correct Monte Carlo algorithm to decide whether r (x) is the symbolic product of p z (x) and q (x).376). which takes a time in 0 (n 2. For the solution to Problem 8. see for instance Hall (1873).3.9). n ] { to be considered as a column vector) for i F. B .11) if ABX = CX then return true else return false In order to take a time in 0 (n 2). New Mexico.3. n) returns true Problem 8. give a false-biased.) 8. provided 11 AB #C. Prove that it returns false with probability at least z whenever AB # C. which only computes the product approximately. but it was already in use in the secret world of atomic research during World War 11. Your algorithm should run in a time in 0 (n ).

15. the same reference gives the statement of Problem 8. The application of probabilistic counting to the Data Encryption Standard (Example 8. More references on this subject can be found in Brassard (1988).3.5.Sec. and Ullman (1974).4.3. and Shub (1986).4 follows Flajolet and Martin (1985).6. many solutions to Problem 8. Stanat.12) is given in Bentley.21. this article and the one by Yao (1982) introduce the notion of an unpredictable generator.14 is solved in Klamkin and Newman (1967).3. An early (1970) linear expected time probabilistic median finding algorithm is attributed to Floyd : see Exercise 5. Under the assumption that it is infeasible to factorize large numbers. and Kilian (1987). is due to Peralta (1986). For solutions to Problem 8.4. which can pass any statistical test that can be carried out in polynomial time. consult Aho. including Problem 8. A guide to simulation is provided by Bratley. A probabilistic algorithm that is capable of finding the i th smallest among n elements in an expected number of comparisons in n +i +O(J) is given in Rivest and Floyd (1973).4. It predates the classic worst-case linear time deterministic algorithm described in Section 4. Rabin (1980a) gives an efficient probabilistic algorithm for com- puting roots of arbitrary polynomials over any finite field. Brassard. A more interesting generator from a cryptographic point of view is given by Blum and Micali (1984). The algorithm of Section 8. Hammersley and Handscomb (1965). and Robert (1988).4. Hopcroft. consult Sobol' (1974). Section 8. Problem 8.2 for finding square roots modulo a prime number. 1987) to cope with generators that are only "semirandom". Computation with encrypted instances (end of Section 8. The solution to .5. including the one from Problem 8.7 References and Further Reading 275 Two encyclopaedic sources of techniques for generating pseudorandom numbers are Knuth (1969) and Devroye (1986). and Steele (1981). and Bennett. Numeric probabilistic algorithms designed to solve problems from linear algebra are discussed in Curtiss (1956).1) is described in Kaliski. The former includes tests for trying to distinguish a pseudorandom sequence from one that is truly random. Blum.2) is an idea originating in Feigenbaum (1986) and developed further in Abadi.3. Classic hash coding is described in Knuth (1968). Rivest. a technique known as quasi Monte Carlo.4. General techniques are given in Vazirani (1986. For a solution to Problem 8. The probabilistic approach to the eight queens problem was suggested to the authors by Manuel Blum. several universal2 classes are described there. For more information on numeric probabilistic algorithms. An analysis of this technique (Problem 8.3. As a generalization. Fox. The technique for searching in an ordered list and its application to sorting (Problem 8. a more efficient unpredictable pseudorandom generator is proposed in Blum. The point is made in Fox (1986) that pure Monte Carlo methods are not specially good for numerical integration with a fixed dimension : it is preferable to choose your points systemat- ically so that they are well spaced.15 appear there.4. see Wegman and Carter (1981). and Carasso (1971). Vickery (1956). Universal hashing was invented by Carter and Wegman (1979). and Sherman (1988).14) come from Janko (1976).20. Feigenbaum. Early algorithms to solve this problem are given in Lehmer (1969) and Berlekamp (1970).13 in Knuth (1973). and Schrage (1983).5. The experiments on the twenty queens problem were carried out by Pierre Beauchemin.

17 are given in Freivalds (1979). Amplification of the advantage of an unbiased Monte Carlo algorithm is used to serve cryptographic ends in Goldwasser and Micali (1984).17 is given by the algorithm of Shanks (1972) and Adleman. Cre peau.6. The integer factorization algorithm of Pollard (1975) has a probabilistic flavour. The solution to Problem 8. For an anthology of probabilistic algorithms. A Monte Carlo algorithm is given in Schwartz (1978) to decide whether a multivariate polynomial over an infinite domain is identically zero and to test whether two such polynomials are identical. Rabin (1976) gives an algorithm that is capable of finding the closest pair in expected linear time (contrast this with Problem 4. Rabin (1980a) gives an efficient probabilistic algorithm for factorizing polynomials over arbitrary finite fields. Brassard. consult Williams (1978). and Miller (1977). which allows us to decide efficiently whether a given integer is a perfect number and whether a pair of integers is amicable. We close by mentioning a few of them.5. read Valois (1987).6.5. Consult Zippel (1979) for sparse polynomial interpolation probabilistic algorithms.6.14). 1980b). Goutier.14 is in Brassard and Kannan (1988). they also give a cryptographic application of universal hashing. Kranakis (1986).276 Probabilistic Algorithms Chap. Miller. The Monte Carlo algorithm to verify matrix multiplication (Section 8. and Pomerance (1988). which also gives a fast probabilistic splitting algorithm whose probability of success on any given composite integer is at least as large as the probability of failure of Rabin's test on the same integer. Lenstra (1982). also read Freivalds (1977).2. Given the cartesian coordinates of points in the plane. The implication of Problem 8. Our favourite unbiased Monte Carlo algorithm for a decision problem. A theoretical solution to Problem 8. including Problem 8. Pomerance.11.6. The test of Solovay and Strassen (1977) was discovered independently. The probabilistic test of primality presented here is equivalent to the one in Rabin (1976. For more information on tests of primality and their implementation. The algorithm based on elliptic curves is discussed in Lenstra (1986). . Several interesting probabilistic algorithms have not been discussed in this chapter. The expected number of false witnesses of primality for a random composite integer is investigated in Erdos and Pomerance (1986) .29. and for finding irreducible polynomials. consult Montgomery (1987). Adleman. and Cohen and Lenstra (1987). The algorithm for electing a leader in a network. The probabilistic test for set equality comes from Wegman and Carter (1981). see also Monier (1980). More information on number theory can be found in the classic Hardy and Wright (1938).10 is given in Goldwasser and Kilian (1986) and Adleman and Huang (1987). The probabilistic integer factorization algorithm discussed in Section 8. An efficient probabilistic algorithm is given in Karp and Rabin (1987) to solve the stringsearching problem discussed in Section 7. and Rumely (1983). Manders. 8 Problem 8. is described in Bach. comes from Itai and Rodeh (1981).5.4) and the solution to Problem 8.4 for the generation of random numbers that are probably prime is explained in Beauchemin. refer to Pomerance (1982). for a comparison with other methods. For efficiency considerations in factorization algorithms.6. and Shallit (1986).3 originated with Dixon (1981) .

a(xt ))) xt in the domain D . Since the computation of a and a1 would take more time than the original multiplication. If you were asked..1 illustrates this principle. an invertible transformation function a : D -+ R and a transformed function g : R' -> R such that PX1 .1. Let f :Dt -4 D be a function to be calculated. and an exponentiation. for example. this idea is only of interest when tables of logarithms are .--- Example 9. for all x 1 . a(u) = In u and g (x . an addition. x 2 . f (u . (You would thus use an algorism. and if the transformations a and 6-1 can also be computed efficiently. y) =x +y.. This allows a multiplication to be replaced by the calculation of a logarithm. v) = u x v . let D be the domain of objects to be manipulated in order to solve a given problem. with this word's original meaning!) More generally. R = R. to multiply two large numbers given in Roman figures. Figure 9. . In this case. The most important transformation used before the advent of computers resulted from the invention of logarithms by Napier in 1614. An algebraic transformation consists of a transformed domain R. x22 . Kepler found this discovery so useful that he dedicated his Tabulae Rudolphinae to Napier.9 Transformations of the Domain 9.1.1 INTRODUCTION It is sometimes useful to reformulate a problem before trying to solve it.1. Such a transformation is of interest if g can be calculated in the transformed domain more rapidly than f can be calculated in the original domain.--C02). you would probably begin by translating them into Arabic notation. . D = IN+ or R+. xt) = a' ' (g (a(x 1).

1. 34) and q = ( . You want to perform a symbolic multiplication of two poly- nomials. for example. this does not allow us to recover the coefficients of r (x) because a polynomial of degree 6 is not uniquely defined by its value at only four points. 2. . The new representation suffices to domain is still define the original polynomials because there is exactly one polynomial of degree at most 3 that passes through any four given points. 53. d.6. 22. Let r(x) = 3x6-17x5+37x4-31x3+8x -2 be the product of p(x) and q(x).d. 3. Most computers that handle numerical data read these data Example 9. but its meaning has changed. calculating the transformed function (carrying out the pointwise multiplication) is much quicker than straightforwardly calculating the original function . thus furnish a historical example of preconditioning (Chapter 7). We would have needed to use seven points from the outset. we can carry out the multiplication in a time in O (d) since r (i) is calculated straightforwardly as p (i) q (i) for 0 5 i <.1.. . 1.2. 106). calculated once and for all.2.Transformations of the Domain 278 Chap. it takes a time in O(d 2) if the scalar operations are taken to be elementary. and print the results in decimal but carry out their computations in binary. Example 9.4. f o r instance.2. In this example. 1. Transformation of the domain. An alternative way to represent the same polynomials is to give their values at d+ 1 distinct points. 7.109.1.. However. These polynomials are represented by their coefficients.2. . 463) and q = (. taking p = (1. The naive algorithm for multiplying these polynomials resembles the classic algorithm for integer multiplication .2. Such tables. 246.2. 1. p(x) = 3x3-5x2-x +I and q(x) =x3-4x2+6x -2. where d is the degree of the polynomials involved. computed beforehand. if the computation of r were to be carried out correctly using this representation. It is often useful to transform between Cartesian and polar coordinates. Example 9. 34.1. 3. 9 I D Dt o` Rt Figure 9. The original domain of the problem is therefore Zd + 1. Using the transformed representation. 2. 2. Thus we obtain r = (. 0.3.1. . The transformed Zd+I. The polynomials in our example would be represented by p = (1.2. 7). 238)..

They are assumed to be executed at unit cost unless it is explicitly stated otherwise. a i . n -1 ] (n is a power of2andwn'2=-1 ) array A [0.n / 2 arrays b.(a) is calculated using the divide-and-conquer technique. a' = 1 and co' _ -1.. Clearly. a naive implementation of Lagrange's algorithm to carry out the final interpolation would take a time in S2(d 3).7. Pa (w')=pb((X')+w'pC((X').1.. function FFT (a [0. B. .2 THE DISCRETE FOURIER TRANSFORM For the rest of this chapter all the arithmetic operations involved are carried out either modulo some integer m to be determined or in the field of complex numbers . Worse still. The Fourier transform Fu. this is only useful if the transformation function (evaluation of the polynomials) and its inverse (interpolation) can be calculated rapidly. a`12 = (m2)t/2 = m' = wn/2 = -1. . an _ 1).(a) = (pa (1).2 The Discrete Fourier Transform 279 (performing a symbolic multiplication of two polynomials given by their coefficients). it seems that the computation of the values p (i) and q (i) for 0 <_ i <. This algorithm is vitally important for a variety of applications. particularly in the area of signal processing (Section 1. an_3. . co) : array [0. The discrete Fourier transform of a with respect to m is the n-tuple Fu.. an-4. In particular.a [0] else t .6). c.4.w' pC (c'). which appears to remove any interest this approach may have had.. .. Consider an n-tuple a = (ao. We shall see in Section 9. for example. We denote by co some constant such that 012 =-1. it appears at first glance that the number of scalar operations needed to calculate this transform is in S2(n2).. n -1] [ the answer is computed in this array ) if n = 1 then A [0] <. As in Example 9. a3. Suppose that n > 2 and set t = n / 2. The t-tuples b and c defined by b = (ao.or more generally in any commutative ring. this is not in fact the case. hence a`+' = a` and w`+` d. This defines in the natural way a polynomial pa(x)=an_1Xn-I+an_2Xn-2+ "' +alx+ao of degree less than n. thanks to an algorithm known as the Fast Fourier Transform (FFT).. Let n > I be a power of 2. so it is legitimate to talk about Fa.(b) and Fa(c ). .. where a=w2. so that pa (w`+') = Pb ((X') .. C [0. pa (w).Sec. and w= (1+i )/J is a possible value in the field of complex numbers. Furthermore. an-2) and c = (al . pa (wn .4 that this is not in fact the case. t -1 ] ( intermediate arrays ) ( creation of the sub-instances ) fori -0tot-1 dob[i] 4-a[2i] c[i] E-a[2i+1] . 9. an_j) are such that pa(X)=ph(x2)+xpc(x2)..1)).. pa ((02).. 9.2d must take a time in S2(d2). n -1 ]. then w= 4 is a possible value if we are using arithmetic modulo 257. At first glance. However. However. . If n = 8. a 2 . . . provided that the points at which we evaluate the polynomials are chosen judiciously.

yield B = F 16(b) = (38.1. where n is a power of 2 greater than the degree of the product . and an iterative algorithm. it is preferable in practice to use an iterative version. 200.2..w2) C -FFT(c.127. Using the fast Fourier transform allows us to evaluate the polynomials p (x) and q (x) at the points 1. .3) and c = (8. 9 { recursive Fourier transform computation of the sub-instances } B t-FFT(b.240.170+43co = 85 A [4] <. (x x1) be a polynomial of degree t such that x1 are arbitrary distinct values.9+7w3 = 200 32 .0 to t-1 do (a= w' I A[i] F-B[i]+aC[i] A [t+i ] E.aC [i] a F-ato return A Show that the execution time of this algorithm Problem 9. Show that p (xi) = r (x. A[ 11 E-. & -1 in a time in 0 (n log n).0).4. 32. 255.) for 15 i <. 226. 85. q(x1) = q (x 2) = = q (xr) = 0.] 9. 9) and C = F 16(c) = (217. 170. which use t = 4 and w2 = 16. let q (x) = (x-x1 )(x-x2) .3 THE INVERSE TRANSFORM Despite its importance in signal processing. 240. Example 9. Our principal aim is to save the idea proposed in Example 9.2. 43. X2.1. Despite the conceptual simplicity of this recursive algorithm. 78. where w = 4.9-70 = 75 The final result is therefore A = (255.22to2 = 194 A [7] . 194. 22.I for i F. First a is split into b = (255. Let us calculate FW(a) in arithmetic modulo m = 257.(o 2) { Fourier transform computation of the original instance } af. 75).217 = 78 A [5] F 170-43w = 255 A [2] A [6] A [0] 38 + 217 = 255 32 + 22002 = 127 A [3] F. 7).226..t. Let n = 8 and a = (255. the remainder of the symbolic division of p (x) by the monomial (x-x. Give such [Hint: Let p (x) be an arbitrary polynomial.-- Let r (x) be the remainder of the symbolic division of p (x) by q (x).) is the constant polynomial whose value is p (x1). The recursive calls.2.3.37.B [i] . We combine these results to obtain A. .38 . our concern here is to use the discrete Fourier transform as a tool for transforming the domain of a problem. In particular. cue.0.8.0. is in O(n log n).37. 0).-1 (mod 257).Transformations of the Domain 280 Chap.1.. This can be done because 44 =. where x 1 . ** Problem 9. w.2.

wn -' are all distinct . is a principal n th root of unity in the field of . Prove that e 2i 1/. Thus we have in fact to invert the Fourier transformation. each summing to zero. Problem 9.1. Show that co-" = .3.Sec. wn -1 is the multiplicative inverse of co in the ring in question . iv. and let w be an element of the ring such that con/2 = . With this in view we say that w is a principal n th root of unity if it fulfills three conditions : 1. Let s =2k-1. j=0 When n is a power of 2. let n = 2k and decompose p = 2n v where u and v are integers.1. complex numbers. w# 1 2. Hints : i.3. Then i. let us use " n " also to denote the element in our ring obtained by adding n ones together (the "1" from our ring). The pointwise multiplication can be carried out in a time in O (n).3 The Inverse Transform 281 polynomial. w is a principal n th root of unity. iv. To obtain the final result. Notice that (w 1)n i2 = -1. . Use p = n/2 in condition (3). w 1 is also a principal n th root of unity. Assume co' = w j for 0 <. iii.2.1. we shall denote it by col. ii.1. and v is odd. and use the existence of n-1.p < n .i. w' .1. wn = 1. Conclude by splitting y o' wjP into 2n sub sums of 2s elements.3. wn 12 = -1 is a consequence of w being a principal n th root of unity. Prove Theorem 9. Consider any commutative ring. ii. as we have assumed already. it turns out that these conditions are automatically satisfied whenever con/2 = . Conditions (1) and (2) are obviously fulfilled.3. v. E or" = 0 for every I <. Use condition (3) and n * 0 to obtain a contradiction. As a slight abuse of notation. Assuming the existence of a multiplicative inverse n-1 for n in our ring.p < n . v. Then wP = 1 and 1 <. 1 = w0 . Theorem 9. Assuming this n is not zero. we still have to interpolate the unique polynomial r (x) of degree less than n that passes through these points. w2 . More generally we use W-' to denote (0-1)' for any integer i. Let n > 1 be a power of 2.i < j < n and let p = j . Obvious. 9.they are called the n th roots of unity. and n-I 3.1. * Problem 9. To show condition (3).. iii.(dj+s )P for every integer j.

since 1 <. n -'Pa (w (n -1))) Problem 9.a < m and 0 <. When m is of the form 2" + 1. . Let A and B be the matrices just defined.1) be an n-tuple.j <.. assume that w is a principal n th root of unity and that n-1 exists in the algebraic structure considered. If i < j.2 justifies the following definition. 9 Problem 9.(m -1)/n . set d =i-j.b < m .3. More generally. provided that n-1 is either known or easily calculable. n .j = n'o'. namely the matrix B defined by B. Then AB = I. then co('-j)k=co0=1. Let a and b be two integers such that 0 <. ... Now C.3. By definition. Prove that 0 <. If i > j. n -'Pa ((02 ).(a) = aA.j = n-1 Y.j = di for 0 <. by showing that n-1 = m . Let C = AB.k Bki . multiplications modulo m can be carried out on a binary computer without using division operations. Problem 9. k=0 n-1 H. Proof.3. Theorem 9. Prove that Fw' (Fu. let p = i -j. and let m = w"12 + 1. show how x mod m can be calculated in a time in p 0 (u) when m = 2" + 1.d < m and ab = d (mod m). Let a = (a o.3.i < n and 0 <.n-I ik -o w(i-j)k There are three cases to consider. .3.4. . Let A be the n x n matrix defined by A.j . Decompose c into two blocks of u bits : c = 2" j + i. The main theorem asserts that A has an inverse.2. as in Problem 9.3.i.i < 2' and 0 <.2". and otherwise set d = m + i .Ifi=j. we wish to obtain their product modulo m.I iii.j < n . let p = j .(a)) = Fw(FW' (a)) = a for every a.j =n(o)1'P = 0 because co 1 is also a k=0 principal n th root of unity by Theorem 9. Theorem 9.1(iii). (The only possibility for j = 2" is when a = b = m -1.3. where 0 <. Prove that w is a principal n th root of unity in arithmetic modulo m.) If i > j. OP = 0 by property (3) of a principal k=0 n th root of unity. the n x n identity matrix.Transformations of the Domain 282 Chap.3.p < n . From now on. n -1pa (W-'). Let c = ab be the ordinary product obtained from a multiplication. n i. Now C.. an . This matrix A provides us with another equivalent definition of the discrete Fourier transform : F".(22u + 2u) < x < 22u +2u+'. a I .. provided that .andsoC11=n-1 nxn-1=1.' (a) = aB = (n -'pa (1).. Cu = Jk _o A. . Prove further that n-1 exists modulo m.5. Let n and w be positive powers of 2. The inverse Fourier transform can be calculated efficiently by the following algorithm. The inverse Fourier transform of a with respect to w is the n-tuple F.3.

where w = 4. the fact that w is a power of 2 means that multiplications in the FFT algorithm can be replaced by shifts. 255. these results give A = (241.Sec.8. which is consistent with Example 9. On the other hand.(B ± C T (3) mod m.65.3.w"-') for i <-Oton-1 do F[i]Fn-'F[i] return F Example 9.143.. 39. Since the number of operations of this type is in 0 (n log n). Consequently . x x 2Y.127.200. co) : array [0.75). where w' = co7 = 193..12. The heart of the algorithm consists of executing instructions of the form A . There remains the mul- tiplication by n-' = m -(m -1)/n = 225 (Problem 9.3. a is decomposed into b=(255. 0). 24. The value of the shift never exceeds (z .0). If the Fourier transform is calculated in the field of complex numbers (Problem 9.9. 85. 127.226. Combined. we supply the base 2 logarithm of w. 121.3. We already know (Problem 9. 9.2. w is indeed a principal n th root of unity.B < m and 0 <.4.w2)_(101. All the arithmetic is carried out modulo m = w"'2+ I using Problem 9.w" -' + w" '2. The calls with yield B =FFT(b.w" -' <. of ' ).194) 0-2 = 241 recursive and c = (85. where x Ty denotes the value of x shifted left y binary places.240. Let us calculate Fj' (a) in arithmetic modulo m = 257. First we calculate FFT (a . which means that it can be reduced modulo m in a time in O (log m) = O (n log (o) by Problem 9. n -1 ] { n is a power of 2 and w" = 1 } array F [0. Let n = 8 and a = (255.3.0. For the rest of this section we no longer suppose that arithmetic operations can be performed at unit cost : the addition of two numbers of size 1 takes a time in O (1).3.11. the complete computation of the Fourier . if the transform is calculated modulo m (Problem 9. 194.3. where 0 <.B[i]-C[i]TP RF-R+y.31). instead of giving w as the second argument.3). By Problem 9.3. thanks to the particular form chosen for m.1. To do this.B ± C T Q <. 78.133) and C =FFT(c. it may be necessary to handle large integers.4.0.3.C < m .1) Ig w. Secondly. that is. 200. denoted by y.4) that reductions modulo m can be carried out in a time in O (log m). even when the recursive calls are taken into account.. First. Furthermore. For this it is convenient to modify the algorithm slightly. n . The final result is thus F = (255.0 for i f-0tot-1 do I R= iy} A [i] <-B[i]+C[i]TR A[t+i]<-. the recursive calls are made with 2y rather than w2 as the second argument. The final loop becomes R <-.2).3.255.3).78.37.w2)_(140.1.3. 75). rounding errors may occur on the computer.221.3 The Inverse Transform 283 function FFTinv (a [0.64. n -1 ] F <-FFT(a.

1) lg co.(x) = 2x + 1 and p2(x) = 5x + 1 in the ring of integers modulo 9. p.3. By Problem 9. since -n-1 = (o"t2/n is a power of 2.. cd .. as. . In order to implement this algorithm for the symbolic multiplication of polynomials. and C = Fu. . 0 .. (Consider. where n is the smallest power of 2 greater than d.) Problem 9.. causes shifts that can go up to (2 -1) (n . (From a practical point of view.(c). bt..4 can no longer be SYMBOLIC OPERATIONS ON POLYNOMIALS We now have available the tools that are necessary to finish Example 9.6.3.(b). c 1. corresponding to the use of w = co' as a principal root of unity. Let p(x) =a. Putting all this together. . The easiest approach is to use the complex number field and Problem 9.. b. Give explicitly the algorithm we have just sketched.) Let A = Fu. B.. (The algorithm FFTinv has to be modified. Let n be the smallest power of 2 greater than d.xs-1+ +alx+ao and q(x) = btxt+bt_lxt-1+ +b1x+bo be two polynomials of degrees s and t. Therefore C is the pointwise product of A and B. .1. It is somewhat surprising that efficient symbolic multiplication of polynomials with integer .5. provided that a principal n th root of unity and the multiplicative inverse of n are both easily obtainable.. . 1(Fw(a) x Fjb)). 0). for instance. respectively.xs +aS_.3.. which means that Problem 9. the coefficients of the product polynomial r (x) are given by the first d + 1 entries in c = F. 0). 9 Transformations of the Domain transform modulo m can be carried out in a time in 0 (n 2 log n log co).0. respectively. the final multiplication by n-1 can profitably be replaced by a multiplication by -n-1 followed by a change of sign.. (Padding c with zeros is unnecessary if d -1 is a power of 2. and c = (c0 . .284 Chap. and this is fortunate because unique interpolation does not always hold when the arithmetic is performed in a ring rather than in a field.O. and c be the n-tuples defined by a =(ao.) Problem 9. the algorithm takes a time in 0 (n log n).2.) 9. By definition of the Fourier transform.4. respectively. Both of these degree 1 polynomials evaluate to 1 and 7 at the points 0 and 3. Let a. B = F".1) lg co. b =(bo. we need to be able to calculate efficiently a principal n th root of unity. Notice that this reasoning made no use of the classic unique interpolation theorem. 0 . We want to calculate symbolically the product polynomial r(x)=cdxd+cd_IXd-1+ +clx+co=p(x)q(x) of degree d = s + t.0. . C..4. if m is sufficiently small that arithmetic modulo m can be considered to be elementary. . 1 1 Similarly. 0).. Show that the inverse transform modulo m = co"t2+ I can also be computed in a time in 0 (n 2 log n log co). Otherwise a direct call on the new FFT with y = (n . Show that it can be used to multiply two polynomials whose product is of degree d with a number of scalar operations in 0 (d log d).0.(a). and let co be a principal n th root of unity. c = Fw 1(C ). = r (co') = p (w`) q (co`) = A. .3.1.

3.3 depends on the degrees s and t of the polynomials to be multiplied and on the size of their coefficients. If an exact answer is required.255. Since the product is of degree 6.5.3. t ]) : array[0 . all the coefficients of the product polynomial r (x) = p (x) q (x) lie between -120 and 120.3. 0. (Continuation of Example 9.3.2.. 0).1.4. 226. The pointwise product of these two transforms. and 240 correspond to -2. This may require the use of multipleprecision arithmetic. 3. By Problem 9. 37. Let u be the maximum of the degrees of the two polynomials.4. the algorithm multiplies p (x) and q (x) symbolically in a time in 0 (d log d). Two applications of the algorithm FFT yield A = F. For this reason it may be more attractive to carry out the arithmetic modulo a sufficiently large number (Problem 9. it becomes necessary to take precautions against the possibility of rounding errors on the computer. 22.Sec. 9. Since all the coefficients of r (x) lie between -120 and Among other things.(a) = (255. b [0 .85.3 and 9. so no coefficient of r(x) can exceed 120 in absolute value.4. If the latter are sufficiently small that it is reasonable to consider operations modulo m to be elementary.4.(b) _ (1. 0) and b =(-2..2. 240. s+t ] that carries out symbolic multiplication of polynomials with integer coefficients. a = 5.1. still working modulo 257.) Q The analysis of the algorithm of Problem 9. . 226.4. 244.4. and n -' = The final answer is therefore r(x) = 3x6-17x5+37x4-31x3+8x -2 .194. By Example 9.4) We wish to multiply sym- bolically the polynomials p(x) = 3x3-5x2-x +l and q(x) =x3-4x2+6x -2.3. Generalize this idea : give explicitly an algorithm mul (a [0 .0. 188) . 193.200. 0. 8. respectively.1.199.4. (In Example 9. 82. it suffices to take n = 8. Problem 9.(c) = C is c = (255.1. and to use Problem 9..) Example 9. 3. and u = 3. the integers 255. thus it suffices to calculate them modulo in = 257. Prove that no coefficient of the product polynomial p(x)q(x) exceeds ab (u+l ) in absolute value. respectively.3 to obtain a principal n th root of unity. b = 6. By Problem 9.109. -1. Let a = (1. the vector c such that F. and m. 0.6. In this case a more thorough analysis of the possible build-up of rounding errors is needed.4 285 Symbolic Operations on Polynomials coefficients should require operations on complex numbers.O.251. your algorithm should determine suitable values for n.75). Let p (x) and q (x) be two polynomials with integer coefficients. w.127. . Let a and b be the maxima of the absolute values of the coefficients of p (x) and q (x). (Use Problems 9. 179.133) and B = F. is C =(255. s ]. and -17. 103. 0. -31.29.0).1.2 for this. Problem 9. w = 4 is a principal n th root of unity in arithmetic modulo 257.2).-4.

2. the initial computation of the Fourier transforms and the final calculation of the inverse take a time in 0 (d 2 log d log (o). whereas the algorithm using divide-and-conquer requires only a time in 0 (n 1. x be n distinct points. x 2.7. Give an efficient algorithm to calculate the coefficients of the unique polynomial p (x) of degree less than n such that p (xi) = yi for every 1 <_ i <_ n.--.6.2 is relevant here too. x 22 . Let x 1 .5. The naive algorithm would have taken a time in 0 (st ). Since M (1) n O(1 log l log log 1) with the best-known algorithm for integer multiplication (Section 9. . the naive algorithm takes a time in 0 (s t M (1)). Problem 9. distinct points. where 1 is the size of the largest coefficient in the polynomials to be multiplied. Suppose for simplicity that n is a power of 2 (nonsignificant leading zeros are added at the left of the operands if necessary). In every case. Let a and b be two n-bit integers whose product we wish to calculate.7. The classic algorithm takes a time in f (n2). (The polynomial is monic if the coefficient of x" is 1.5 MULTIPLICATION OF LARGE INTEGERS We return once more to the problem of multiplying large integers (Sections 1.) * Problem 9. The naive algorithm is therefore preferable to the "fast" algorithm if d is very large and I is reasonably small. x2 2 .11. where M (1) is the time required to multiply two integers of size 1. and the intermediate pointwise multiplication of the transforms takes a time in 0 (d M (d log co)).4. . 1. .4.. The total time is therefore in 0 (d M (d log co)).Transformations of the Domain 286 Chap.59). Your algorithm should take a time in 0 (n log2n ).5). then the symbolic product of these polynomials is obtained using the discrete Fourier transform.4.7). where Co = 2 suffices if none of the coefficients of the polynomials to be multiplied exceeds 2' 14/ 2(1 + max(s . y 1. Give an efficient algorithm to calculate each y. t )) in absolute value.4.--. We can do better than this thanks to a double transformation of the domain. not necessarily distinct. the algorithm that uses 0) = e 2fi /" can multiply approximately the two polynomials in a time in 0 (d log d). if we are obliged to use multiple-precision arithmetic. (Hint : see Problem 4. The original integer domain is first transformed into the domain of polynomials represented by their coefficients . (Hint : and let yn be n values. 9 where d = s + t. the first term in this analysis can be neglected. (Remember that n is the smallest power of 2 greater than d.2. Your algorithm should take the hint to Problem 9. On the other hand. and 4. It is possible for this time to be in 0 (st) in practice.x a time be n in 0 (n log2n). and let x 1 . Y2.1.2.) Your algorithm should take a time in 0 (n log2n) provided all the necessary operations are taken to be elementary. Let p (x) be a polynomial of degree n .) ** Problem 9. Give an efficient algorithm to calculate the coefficients of the unique monic polynomial p (x) of degree n such that p (xi) = 0 for every 1 <_ i <_ n.8). . 9.) By comparison.x be n distinct points. Let x 1 . = p (x1) for 1 <_ i <_ n. or even in 0 (n") for any a > 1 (Problem 4.. if arithmetic can be carried out on integers of size l at unit cost.

For instance. in order that using the discrete Fourier transform should be attractive. Thus pa (x) = 2x3 +3 X2 + 1 and Pb (x) = x3+9x +5. let pa (x) therefore denote the polynomial whose coefficients are given by the successive digits of a.1. Let a = 2301 and b = 1095. For instance.1. This polynomial is thus of degree less than n / 2. The algorithm is recursive because one of the stages in the symbolic multiplication of polynomials consists of a pointwise multiplication of Fourier transforms. . where d = 2n is a power of 2 greater than the degree of the product polynomial. 0 The recursive nature of this algorithm obliges us to refine the analysis of symbolic multiplication of polynomials given following Problem 9. 9. This time the degree of the product polynomial r (x) is less than n.519. To obtain the product of the integers a and b. The central step in the symbolic multiplication of Pa (x) and Pb (x) therefore only requires n multiplications of integers less than m = w"i2+ 1.) Taking into account also the time spent in computing the two initial Fourier transforms and the final inverse transform. The central step in the symbolic multiplication of two polynomials of degree less than n consists of d multiplications f integers less than wd12+ 1. even if this means increasing the size of their coefficients. Unfortunately. pa (2) = a for every integer a. so that pa (10) = a. these integers are of silt. The symbolic product is r (x) = pa (X)pb (x) = 2x6+3x5 + 18x4+38x3 + 15x2+9x +5 and we obtain the desired product ab as r (10) = 2. where n is a power of 2. and the final answer is obtained by evaluating r (4). the symbolic multiplication takes a time in n M (-'n Ig (o) + 0 (n 2log n logo)). Let M (n) be the time required to multiply two n-bit integers.7 consists of representing each integer by a polynomial of degree 1 whose two coefficients lie between 0 and 2'12 . we need only calculate symbolically the polynomial r (x) = p (x) q (x) using the fast Fourier transform (Section 9. Clearly. The original multiplication of two integers of size n therefore requires 2n multiplications of slightly larger integers ! To correct this. (The second term is added in case certain operands are exactly w" 2.5 Multiplication of Large Integers 287 We denote by pa (x) the polynomial of degree less than n whose coefficients are given by the successive bits of the integer a. the polynomial r (x) = Pa (x)pb (x) is calculated symbolically using the Fourier transform. the polynomials considered must have a sufficiently high degree. However.595.3. This can be carried out in a time in n M (2 n Ig w) + 0 (n 2). its coefficients lie between 0 and 3. For the purpose of this example only.Sec. we perform the computation Example 9. and pa (4) = a. As an extreme case.5.4). suppose we redefine pa (x) to be the polynomial whose coefficients are given by the successive figures of the representation of a in base 4. As before. we must reduce the degree of the polynomials used to represent the integers to be multiplied. To make the illustration simpler. p53(x) = X5+X4+X2 + I because 53 in binary is 00110101. even if we take co = 2. and then evaluate r (2).4. since this integer of length 1 + i n Ig (o has to be treated specially. here in decimal rather than in binary. 1 + d Ig w = n + 1. the algorithm of Section 4.

we must lower the degree of the polynomials still further. Note that 1 and k are powers of 2. This time. t(1) = 1 n =2 k . To calculate the product of the integers a and b. / = 4.260.155. Let k = n /I. We thus obtain the asymptotic recurrence M (n) e n M (n / 2) + 0 (n 2 log n). and then evaluate r (21). When n is a power of 2. Since the coefficients of the polynomials pa (x) and pb (x) lie between 0 and 3.5. r(x) =pa(x)pb(x)=10x6+36x5+63x4+116x3+111x2+108x + 156 and the final evaluation yields r (16) = 210. comes from the decomposition into four blocks of the binary representation of a 0010 0110 1001 1101. Problem 9.2. namely the evaluation of r (4).100 = 9885 x 21. The choice to = 2 is adequate provided n >: 16. even if we do not take into account the time required to compute the Fourier transforms ! This is explained by the fact that we used the "fast" algorithm for multiplying two polynomials in exactly the circumstances when it should be avoided : the polynomials are of high degree and their coefficients are small. that is. Thus we have that pa (21) = a. consists of n shifts and n additions of integers whose size is not more than lg (9n / 2). for instance. The last stage of the multiplication of a and b. using Fourier transforms. Let d = 2k. and the degree of these poly- .288 Transformations of the Domain Chap. and since these polynomials are of degree less than n / 2. The first of these polynomials. 9 Before analysing the new multiplication algorithm. denote by pa (x) the polynomial of degree less than k whose coefficients correspond to the k blocks of I successive bits in the binary representation of a. we need only calculate symbolically the polynomial r (x) = Pa (X)Pb (x). l = I or l = 2n . prove that t(n) = the value of the constant k. 1! Ign1 Let l = 2 2 .260. Since the coefficients of the polynomials pa (x) and Pb (x) lie between 0 and 21 .4. depending on whether lg n is even or odd. which takes a negligible time in 0 (n log n).1. We form the polynomials pa (x) = 2x3+6x2+9x + 13 and Pb (x) = 5x3 + 3x2 + 12. To correct this. The symbolic product is Example 9. k E lN+ . Show that t(n) 0 (nk ).2). we must choose a principal n th root of unity tv. As the computations are carried out modulo m = o)'12 + 1.1. This time we need to choose a principal d th root of unity t. Let a = 9885 and b = 21. it is thus sufficient that 9n / 2 < (0/2 .5. whatever The preceding problem shows that the modified algorithm is still bad news.3. and k = 4. a power of 2 greater than the degree of the product polynomial r (x). so that n = 16. The symbolic computation of r (x) therefore takes a time in n M (n / 2) + 0 (n 2 log n). Consider the recurrence t (n) = nt (n / 2). the largest coefficient possible in r (x) is 9n/2 (Problem 9.

and w = 32 is sufficient when Ig n is odd. Prove that 0 (n (log n)2. we split the n-bit integers to be multiplied into k blocks of I bits.)/ = 2 + (Ig J )/W W. we obtain Ig w ? 4 + (Ig' )/ . This is possible provided we increase slightly the size of the coefficients of the polynomials used in order to decrease their degree. that is Ig co >. for some constants S.(2' + IgFn.3. d = 2n and w = 32. this algorithm can multiply two n-bit integers in a time in M (n) = nt (n) E 0 (n (log n)2.lg(5/3) provided that n >_ (310 for every real constant 0 > Multiplication of Large Integers 289 nomials is less than k. the largest coefficient possible in r (x) is k (21 -1)2.59 ). [Hints: For the second case use the fact that Iglg(1J) (Iglgn) . Is this algorithm optimal ? To go still faster using a similar approach. which is negligible. The multiplication of two n-bit integers is thus carried out using a symbolic multiplication of two polynomials.p Ig n . This algorithm therefore outperforms all those we have seen previously. In the case when Ig n is even I = k = dl 2 = I and Ig w >. for all real constants Q ? 1 and y > 2. For the third case prove by constructive induction that t (n) 5 S [(lg n)1gT .11 .5. which arises here as the max- imum of 2 x 3 and ' x i . It suffices therefore that k 221 < m = wd 12 + 1. When n is odd. Consequently.2. 9. More precisely. Consequently. and p that you must determine and for n sufficiently n)(197). this can easily be carried out in a time in 0 (d 2 log w). Prove that t(n)E if y < 2 O (log n) O (log n log log n) if y= 2 O ((log n)19 Y) if y > 2 . * Problem 9. yt. Also use the fact that (lg1')Ig1< T(lgn)IST+2lgylg1(lg1`r)(IgY)-1 provided n y21g R. where . hence. which gives M (n) e 24W M (3') + O (n log n). and t(n)E5t(52 )+O(logn) when n is odd. When n is even. By Problem 9.(21 + Ig k)/(dl 2). Let y > 0 be a real constant.5. Problem 9. when n is odd.59) c n°C whatever the value of the real constant a > 1.2 suggests that we should reduce the constant y = 6. which takes a time in d M (2 d lg w) + O (d 2 log d log w). The equations obtained earlier for M (n) lead to t(n)E6t(3N/n-)+O(logn) when lgn is even. w = 8 suffices to guarantee that the computation of the coefficients of r (x) will be correct when Ig n is even.5. provided n is sufficiently large. Problem 9. As far as the final evaluation of r(21) is concerned. d = 2J and w = 8. Simi- larly. M (n) E 2n M (2 I) + 0 (n log n). t (n) E O ((log n)1 g 6) c 0 ((log n)2-59).] Let t(n) = M (n)/n . and let t(n) be a function satisfying the asymptotic recurrence t(n) E yt(O (' )) + 0 (log n).

it is then possible to obtain the coefficients of r (x) by the Chinese remainder theorem. ignl and k = n/I for an arbitrary constant i >_ 0. This is still not optimal. and Welch (1967). 9. This approach allowed Schonhage and Strassen to obtain y = 2. An integer multiplication algorithm based on the fast Fourier transform has been used by the Japanese to calculate it to 10 million decimal places. Their approach requires care to avoid problems due to rounding errors but gives rise to a simpler algorithm. The corresponding y is thus 1 y=max(21-'(2'+1+2-'). Lewis. at least asymptotically. The complexity of this algorithm is such that it is of theoretical interest only. an algorithm that takes a time in 0 (n log n log log n). and 2-' 2n recursive calls on integers of size (2' + + 2-' if Ig n is odd provided n is sufficiently large. we have to redefine the notion of the "product" of two polynomials so as to avoid doubling the degree of the result obtained. 2-' f(2'++2-i`1)J)=4+21 2r The algorithm obtained takes a time in O(n(log n)a). Even more decimals of it have been calculated since.6 REFERENCES AND FURTHER READING The first published algorithm for calculating discrete Fourier transforms in a time in O (n log n) is by Danielson and Lanczos (1942). but the algorithms that are even faster are too complicated to be described in detail here. it is astonishing that the existence of a fast algorithm remained almost entirely unknown until its rediscovery nearly a quarter of a century later by Cooley and Tukey (1965). thus avoiding a costly conversion when printing out the result. Detailed analysis shows that recursive calls on integers of size (2' + 1 + 2-' if Ig n is this gives rise to 21-' even. . We mention simply that it is possible to obtain y=4 by calculating the coefficients of the polynomial r (x) modulo 221 + 1 (using the Fourier transform and proceeding recursively) and modulo k (using the algorithm for integer multiplication of Section 4. they used a variant involving operations in the complex number field. In view of the great practical importance of Fourier transforms. To go faster still. Because 221 + 1 and k are coprime. where a=2+Ig(1+2-1-21) <2+2-1-2'/ln2 can be reduced arbitrarily close to 2. read Cooley. Rather than resorting to modulo computations in a finite ring.Transformations of the Domain 290 Chap. These authors mention that the source of their method goes back to Runge and Konig (1924). 9 I _ 2r+f. increasing the parameter i reduces the exponent a at the expense of increasing the hidden constant in the asymptotic notation. that is. An efficient implementation and numerous applications are suggested in Gentleman and Sande (1966). For a more complete account of the history of the fast Fourier transform. Needless to say. The outcome is an algorithm that is capable of multiplying two n-bit integers in a time in 0 (n log2n). and finally to evaluate r(2 21 ). It also allows the computation to be carried out directly in decimal.7).

Tamura.6 is given in Aho. Further ideas concerning the symbolic manipulation of polynomials.2. The algorithm used by the Japanese to com- pute it to 10 million decimal places is described in Kanada. The empire struck back shortly thereafter when the Japanese computed 134 million decimals. 1975). Pollard (1971) studies the computation of Fourier transforms in a finite field. Monet. The nonrecursive algorithm suggested in Problem 9.2 is described in several references. and Turk (1982). 9.5 and The algorithm that is able to multiply two integers of size n in a time in O (n log2n) is attributed to Karp and described in Borodin and Munro (1975). and Ullman (1974). Also read Turk (1982). but does not explain the algorithm used. The details of the algorithm by Schonhage and Strassen (1971) are spelled out in Brassard. Hopcroft. Yoshino and Ushiro (1986). which is the world record at the time of this writing. Cray Research (1986) mentions an even more precise computation of the decimals of it. evalua- tion. for example Aho. The solution to Problems 9. A practical algorithm for the rapid multiplication of integers with up to 10 thousand decimal digits is given in Pollard (1971). and Zuffellato (1986). read Gleick (1987). Hopcroft. The book by Brigham (1974) is also worth mentioning. The second edition of Knuth (1969) includes a survey of algorithms for integer multiplication. and Ullman (1974). .6 References and Further Reading 291 and Rabiner and Gold (1974). and interpolation can be found in Borodin and Munro (1971. although the solution given there to Problem 9. And the saga goes ever on. Horowitz and Sahni (1978).3.3.3 is unnecessarily complicated in the light of Theorem 9.Sec.

In this chapter we introduce only a few of the principal techniques and concepts used in the study of computational complexity. considers globally the class of all algorithms that are able to solve a given problem. 10. 292 . we can prove.1 DECISION TREES This technique applies to a variety of problems that use the concept of comparisons between elements. Computational complexity. unfortunately. ignoring those that may be made to control the loops in our program. that a certain problem can be solved in a time in 0 (f (n)) for some function f (n) that we aim to reduce as much as possible. We illustrate it with the sorting problem. Our satisfaction is complete when f (n) e O(g(n)). to solve some given problem. Using complexity. we try to find a function g(n) as large as possible and to prove that any algorithm that is capable of solving our problem correctly on all of its instances must necessarily take a time in 92(g(n)).10 Introduction to Complexity Up to now. In this case we say that the complexity of the problem is known exactly . Consider first the following algorithm. Thus we ask the following question : what is the minimum number of comparisons that are necessary to sort n elements? For simplicity we count only comparisons between the elements to be sorted. a field of study that runs in parallel with algorithmics. by giving an explicit algorithm. Using algorithmics. this does not happen often. we have been interested in the systematic development and analysis of specific algorithms. since then we know that we have found the most efficient algorithm possible (except perhaps for changes in the hidden multiplicative constant). each more efficient than its predecessors.

if max(T) . 2. This difference resembles that between binary search and hash coding.min(T). If the answer is "yes". directed binary tree. 9.j]F-0 fork F. it becomes impractical.. A decision tree is a labelled. The trip ends when it reaches a leaf . 3. 6. Each internal node contains a comparison between two of the elements to be sorted.1). the algorithm provides an efficient and practical way of sorting an array in linear time.max(T) array C[i.Sec.1.1. In this book we pay no further attention to algorithms for sorting by transformation. how many comparisons between elements are made? Coming back to the question we asked at the beginning of this section : what is the minimum number of comparisons that are necessary in any algorithm for sorting n elements by comparison ? Although the theorems set out in this section still hold even if we consider probabilistic sorting algorithms (Section 8. Problem 10. the number of elements to be sorted. 10.1 to ndo C[T[k]]*-C[T[k]]+1 k F--1 for p F. variants such as radix sort or lexicographic sort (not discussed here) can sometimes be used to advantage. 10] containing the values 3.1 Decision Trees 293 procedure countsort(T [1. The most important characteristic of countsort and its variations is that they work using transformations : arithmetic operations are carried out on the elements to be sorted.1 to C [p ] do T[k]*-p k -k+l Problem 10. For example. j F. n ]) i <. Show exactly how countsort can be said to carry out arithmetic operations on the elements to be sorted. 5. 5. 1. this algorithm on an array This algorithm is very efficient if the difference between the largest and the smallest values in the array to be sorted is not too large.i to j do for q <. when the difference between the elements to be sorted is large. Simulate the operation of T [ 1 . only in rare applications will these algorithms prove preferable to quicksort or heapsort. all the sorting algorithms considered in the preceding chapters work using comparisons : the only operation allowed on the elements to be sorted consists of comparing them pairwise to determine whether they are equal and.. A decision tree for sorting n elements is valid if to each possible order relation between the elements it associates a verdict that is . if not. this leaf contains the verdict associated with the order relation used. However.min(T) = #T . 1. which is the greater. a trip through the tree consists of starting from the root and asking oneself the question that is found there. otherwise it continues recursively in the right-hand subtree. However. Given a total order relation between the elements. 4.1.4. On the other hand. Each leaf contains an ordering of the elements. In this case.2. As a function of n. we shall for simplicity confine our discussion to deterministic algorithms. the trip continues recursively in the left-hand subtree . on account of both the memory and the time it requires..

Verify that the decision tree given in Figure 10. (The annotations on the trees are intended to help follow the progress of the corresponding algorithms.B. Figures 10.1. and C.T [2].T [3] if A < B then if B < C then { already sorted } else if A < C then T . C.3.2 and 10..C.B. for each value of n.T [1]. .A else T <.4 and Problem 2.1.A. a decision tree that is valid for sorting n elements.2.<A<B B<A < C B<C<A A valid decision tree for sorting three elements.1. respectively.) Notice that heapsort C<B<A A<B<C 1 A<C<B Figure 10.A. The following problem will help you grasp these notions. C <. when three elements are to be sorted.1.C.Introduction to Complexity 294 Chap. procedure adhocsort3(T [l .A Similarly.B else if B < C then if A <C then T .1 there corresponds the following algorithm.C. For example.1.A.3 give the trees corresponding to the insertion sorting algorithm (Section 1.1. 10 compatible with this relation.B.C. Problem 10. a decision tree is pruned if all its leaves are accessible from the root by making some consistent sequence of decisions.9.3). Every valid decision tree for sorting n elements gives rise to an ad hoc sorting algorithm for the same number of elements.1.4) and to heapsort (Section 1. B.C else T . to every deterministic algorithm for sorting by comparison there corresponds. Finally. to the decision tree of Figure 10.1 is valid for sorting three elements A.B else T . 3]) A. B.

but it.1. The three element insertion sort decision tree.4). sometimes makes unnecessary comparisons. nonetheless. 10.4.<C<B ABC A<B.Decision Trees Sec. the decision tree of Figure 10.5) for the case of three elements. the distance from the root to the most distant leaf. and to quicksort (Section 4. gives the number of comparisons carried . that is. It would now be possible to establish the correct verdict. so that every leaf can be reached by some consistent sequence of decisions. Problem 10.1 295 ABC BAC ABB BAA x=C CBA I C<B<A x=C BAC BCA B. the tree is pruned : the leaf that would correspond to a contradictory answer "yes" to the third question has been removed.1.5.4) and by merging (Section 4. This situation does not occur with the decision tree of Figure 10.<C<A I ACB A. and then whether C > A (answer: yes). but beware of appearances : it occurs even more frequently with the insertion sorting algorithm than with heapsort when the number of elements to be sorted increases.2.3 first tests whether B >A (answer : no).1. if B <A A < C. For instance. In the two latter cases do not stop the recursive calls until there remains only a single element to be "sorted".1. 0 Problem 10.2.) Thus heapsort is not optimal insofar as the number of comparisons is concerned.1. (Despite this. (You will need a big 0 piece of paper !) The following observation is crucial: the height of the pruned decision tree corresponding to any algorithm for sorting n elements by comparison. Give the pruned decision trees corresponding to the algorithms for sorting by selection (Section 1. asks again whether B > A before reaching its conclusion.C Figure 10. Give the pruned decision trees corresponding to the insertion sorting algorithm and to heapsort for the case of four elements.

The lemma follows immediately. a possible worst case for sorting three elements by insertion is encountered if the array is already sorted into descending order (C < B < A ). The decision trees we have seen for sorting three elements are all of height 3. n! leaves. C < A ?. 10 yes C> B C A rBA BC C A BC A C BC AB BA A C BC A C A B BA BC AB CB AC C B BA CA A<B<C C<A<B A< C<B B<A<C C < B < A B<C<A I Figure (It may have more than n! leaves if it is not pruned or if some of the leaves can only be reached when some keys are equal.2.1. Can we find a valid decision tree for sorting three elements whose height is less ? If so. Any valid decision tree for sorting n elements contains at least Lemma 10. out by this algorithm in the worst case. we shall have an ad hoc algorithm for sorting three elements that is more efficient in the worst case.1. It is easy to show (by mathematical induction on the total number of nodes in the tree) that any binary tree with k leaves must have at least k -1 internal nodes. a binary tree with t nodes in all cannot have more than It /21 leaves. this time on the height of the tree).) . Any binary tree with k leaves has a height of at least [lg k 1. and C < B ? situated on the path from the root to the appropriate verdict in the decision tree all have to be made.Introduction to Complexity 296 Chap. and hence it has at most 2" leaves. The upper limit on the number of leaves of any pruned decision tree can be computed with Problem 5. in this case the three comparisons B < A ?. Lemma 10. We now prove more generally that such a tree is impossible. Proof. For example. To say the same thing differently.8. The three element heapsort decision tree. Try it : you will soon see that this cannot be done.1. Now a binary tree of height h can have at most 2h+1-1 nodes in all (by another simple argument using mathematical induction.

A valid tree must be able to produce at least one verdict corresponding to each of the n! possible orderings of the n elements to be sorted. whereas heapsort makes 59 (of which the first 18 are made during construction of the heap).1. Prove further that if n is a power of 2.1. and yet rlg(12!)] = 29. if any.2. Give a lower bound on the number of comparisons required in the worst case to handle n elements. the algorithm takes a time in S2(n log n) in the worst case. 10. the insertion sorting algorithm makes 66 comparisons when sorting 12 elements.1. and that sorting by merging almost attains the lower bound.6.1.1. A = B. This certainly does not mean that it is always possible to sort n elements with as few as [lg(n!)1 comparisons in the worst case. Rework this problem assuming that there are three possible outcomes of a comparison between A and B : A < B .C is not acceptable : the algorithm must answer either A < B < C or A < B = C. a verdict such as A< B <. How well do these algorithms do when compared to the lower bound Fig (n!)] for n = 50 ? ** Problem 10. This proof shows that any deterministic algorithm for sorting by comparison must make at least Iig(n!)l comparisons in the worst case when sorting n elements. Problem 10. By the crucial observation that precedes Lemma 10. a time in S2(n log n) to sort n elements in the worst case. Any deterministic algorithm for sorting by comparison takes Theorem 10.1. n ? 2. are equal. it has been proved that 30 comparisons are necessary and sufficient in the worst case for sorting 12 elements. This tree contains at least n! leaves by Lemma 10. Proof. (Some modifications of heapsort come very close to being optimal for the worst-case number of comparisons. To every deterministic algorithm for sorting by comparison there corresponds a pruned valid decision tree for sorting n elements.Sec. In the worst case. orA >B.1.7.) Problem 10. Give exact formulas for the number of comparisons carried out in the worst case by the insertion sorting algorithm and by the selection sorting algorithm when sorting n elements. is never greater than 2n ign. Suppose we ask our sorting algorithm not merely to determine the order of the elements but also to determine which ones.17). Since each comparison takes a time in 82(1).1 Decision Trees 297 Proof. then mergesort makes n Ig n . What can you say about sorting by merging in the general case ? More precise analysis shows that [lg(n!)1 E n lgn -O(n).n +I comparisons in the worst case when sorting n elements.1. the algorithm thus needs at least [lg(n!)l comparisons in the worst case to sort n elements. Its height is therefore at least [lg(n!)1 by Lemma 10. For example. In fact. The previous problem therefore shows that heapsort is optimal to within a factor of 2 as far as the number of comparisons needed in the worst case is concerned.1. . 0 Prove that the number of comparisons carried out by heap- sort on n elements.1.1.8. and lg(n!)eS2(n logn) (Problem 2.

Let T [I . At first sight this recurrence is not well founded since it defines h (k) in terms of itself (when we take i =0 or i =k in the minimum. In the first case the root is the only leaf in the tree and H (T) = 0. the single child is the root of a subtree A.1. In particular.1. for some 1 <. h (1) = 0.i S n and T [i ] <. Suppose for simplicity that the n elements are all distinct.x). Since the dis- tance from each leaf to the root of A is one less than the distance from the same leaf to the root of T. If we define h (0) = 0. we see that there is little difference between the worst case and the average.1 has an average height (2+3+3+3+3+2)/6=8/3. The root of T can have 0. n ] be an array sorted into ascending order. or 2 children. In the second case. rather than to 0 determine its position ? Decision trees can also be used to analyse the complexity of a problem on the average rather than in the worst case.o0 and T [n + 1 ] = + oo. then 8/3 is the average number of comparisons made by the sorting algorithm associated with this tree. In the third case the tree T is composed of a root and of two subtrees B and C with i and k .Ig (k -x). Lemma 10.k -1. For example.x) lg(k . We can thus reformulate the recurrence that defines h (k).3. Let T be a binary tree with k leaves. which corresponds to the root having only one child).i leaves.. 1. Calculating the derivative gives g' (x) =1g x . Define H (T) as the sum of the depths of the leaves. By definition.3. respectively. the average height of T is H (T )/k . the decision tree of Figure 10. H (T) = 16 for the tree in Figure 10. which is .) Proof.1. 10 Problem 10. we have H (T) = H (A) + k. Any binary tree with k leaves has an average height of at least lgk.1. 0 if k<-1 h(k) = k +min[h(i)+h(k-i) 11 <-i <-k-1 } ifk > 1 Now. By a similar argument we obtain this time H (T) = H (B) +H (C) + k. For k >-1.1.9. This difficulty disappears because it is impossible that h (k) = h (k) + k.x < T [i + 11.1. (By comparison with Lemma 10. For example. define h (k) as the smallest value possible for H (X) for all the binary trees X with k leaves. Define the average height of T as the sum of the depths of all the leaves divided by the number of leaves.i < k .i ) + k 10<-i <-k } for every k > 1. consider the function g (x) = x Igx + (k . and let x be some element. How many comparisons between elements are needed in the worst case to locate x in the array ? As in Section 4. Let T be a binary tree. the problem is to find an index i such that 0 <. How does binary search compare to this lower bound ? What lower bound on the number of comparisons do you obtain using the decision tree technique if the problem is simply to determine whether x is in the array. where x e R is such that 1 <.x <. with the logical convention that T [0] = . If each verdict is equally likely.298 Introduction to Complexity Chap.1. which also has k leaves. the preceding discussion and the principle of optimality used in dynamic programming lead to h (k) = min I h (i) + h (k .

Prove that this also implies that h (k) >. it is therefore at least Ig k. both in the worst case and on the average.k lg k for every integer k ? 1 now follows by mathematical induction. h(k)=k +min{h(i)+h(k-i) I 1 <i <-k-1 }.1. n ] be an array and k <.n an integer. where h (k) is the function used in the proof of Lemma 10. Conclude that it must take a time in Q(k log n).2 and 10.1 Decision Trees 299 zero if and only if x = k -x. The problem consists of returning in descending order the k largest elements of T. it follows that h (k) > k + min {g (x) I < x 5 k -1. Let k > 1. By definition. This shows that H (T) >. The base k = 1 is immediate. if x = k12. Follows immediately from Lemmas 10.3.10. 10. i E IN }.2. Prove that h (k) = k t + 21 .12. How do these values compare to the number of comparisons performed by these algorithms in the worst case? * Problem 10. h (k) ? k + min { g (i) I 1 < i 5 k -1. . Let t = Llg k and 1 =k-2. Let T [1 ..3. * Problem 10.1.k lg k. Justify your analysis of the time and space used by your algorithm. It therefore takes an average time in Q (n log n). Proof.1. Suppose by the induction hypothesis that h (j) >.j lg j for every strictly positive integer j < k -1. By the induction hypothesis. Determine the number of comparisons performed on the average by the insertion sorting algorithm and by the selection sorting algorithm when sorting n elements. Any deterministic algorithm for sorting by comparison makes at least lg(n!) comparisons on the average to sort n elements.) Theorem 10. Using the result obtained previously.1.11. Since the second derivative is positive.k. g (x) attains its minimum at x = k / 2.k lg k for every tree T with k leaves. that is.1. Prove that any deterministic algorithm that solves this problem using comparisons between the elements must make at least (k / 2) lg (n / 2) comparisons. The proof that h (k) >. This minimum is g (k / 2) = (k lg k) .Sec. Your algorithm should make no more than one sequential pass through the array T . Problem 10.1. On the other hand. (Optionally : give an intuitive interpretation of this formula in the context of the average height of a tree with k leaves. it is therefore efficient even if n is large and if the array is supplied on a magnetic tape. give an algorithm able to solve this problem in a time in 0 (n log k) and a space in O (k) in the worst case. The average height of T being H (T )lk . Because min X ? min Y for any two nonempty sets X and Y such that X c Y. x E 1R } .1. we have h (k) ? k + g (k / 2) = k lg k.

it does not often happen in the present state of our knowledge that the bounds derived from algorithmics and complexity meet so satisfactorily. when A =' B we can be sure that they are the same. Suppose we are able to prove that a certain number of problems are equivalent in the sense that they have about the same complexity. A and B are linearly equivalent.2. for any function t(n). then the fact that the problems are equivalent makes it even more unlikely that such an algorithm exists. implies that there exists an algorithm for A that works in a time in 0 (t(0 (n))).2. we content ourselves with comparing the efficiency of algorithms for A on instances of size n with the efficiency of algorithms for B on instances of size in O (n). Problem 10. if the existence of an algorithm for B that works in a time in 0(t(n)). denoted A <_ 1 B. a more efficient algorithm for all the others. both on the average and in the worst case. write A <g B if the existence of an algorithm for B that works in a time in O(t(n)). at least in theory. implies that there exists an algorithm for A that also works in a time in 0 (t (n)). Let A and B be two solvable problems. Even if we are not able to determine the complexities of A and B exactly.1. we often have to be content to compare the relative difficulty of different problems. the question of the complexity of sorting by comparison is therefore settled : a time in O(n log n) is both necessary and sufficient for sorting n elements. if these problems have all been studied independently in the past. For the purposes of this problem only.3 goes into this second motivation in more detail. . On the other hand. A is linearly reduDefinition 10. Except for the value of the multiplicative constant.2. Section 10. where p (n) is a polynomial. cible to B . Show with the help of an explicit example that the notions A <_! B and A <g B are not equivalent even if there exists an algorithm for B that works in a time in O (p (n )). 10 10. for any function t(n). In the remainder of this section we shall see a number of examples of reduction from a variety of application areas. A less restrictive definition of linear reduction is obtained if * Problem 10. There are two reasons for doing this.2.Introduction to Complexity 300 Chap. and if all the efforts to find an efficient algorithm for any one of them have failed.1. denoted A = 1 B. Prove that the relations <_ 1 and = 1 are transitive. Unfortunately.2 REDUCTION We have just shown that any algorithm for sorting by comparison takes a minimum time in SZ(n log n) to sort n elements. When A 51 B and B S ' A both hold. From a negative point of view. Because it is so difficult to determine the exact complexity of most of the problems we meet in practice. Any algorithmic improvement in the method of solution of one of these problems now automatically yields. we know that heapsort and mergesort both solve the problem in a time in O (n log n).

strongly and supra linear functions are defined similarly. counting the multiplications at unit cost.5) to algorithms and problems.8.20. this algorithm is smooth because it takes a time in O(log n ). which cannot be smooth.3. A problem is smooth if any reasonable algorithm that solves it is smooth.2 Reduction 301 Let us extend the notion of smooth functions (introduced in Section 2. for any smooth function t(n). (Hint: apply Problem 2. and logn is smooth. Most theorems in this section are stated conditionally on a "reasonable" assumption.8. These notions extend to algorithms and problems as in the case of smooth functions.1. it takes longer to compute a 31st power than a 32nd. Even though a smooth function must be eventually nondecreasing by definition. it is supra quadratic if it is eventually nondecreasing and if there exists an CEIR+ such that t(an) >_ a2+et(n) for every positive integer a and every sufficiently large integer n. A function t : IN -* It* is at least quadratic if t(n) E c (n2). We have seen (Problem 4.Sec. assuming B is smooth". Prove that any strongly quadratic function is at least quadratic. At least. i. No problem could be smooth without this restriction to reasonable algorithms because any problem that can be solved at all can be solved by an algorithm that takes a time in Q(2'). this does not imply that the actual time taken by a specific implementation of the algorithm must also be given by an eventually nondecreasing function.5) that the time it takes to compute a"mod m is a linear function both of lg n and the number of Is in the binary representation of n . Show that n 2log n is strongly quadratic but not supra quadratic. implies that there exists an algorithm for A that also works in a time in 0 (t(n)). From a more practical point of view it also means that the existence of an algorithm for B that works in a time in 0 (t(n)). we mean an algorithm that does not purposely waste time. such as "A <_ ' B. 10. . It is strongly at least quadratic (strongly quadratic for short) if it is eventually nondecreasing and if t(an) >_ a2t(n) for every positive integer a and every sufficiently large integer n. An algorithm is smooth if it takes a time in O(t(n)) for some smooth function t. In particular. Consider for instance the modular exponentiation algorithm dexpo of Section 4. Give an explicit example of an eventually nondecreasing function that is at least quadratic but not strongly quadratic. Moreover. Nonetheless.1. This can be interpreted literally as meaning that A <_ 1 B under the assumption that B is smooth. The actual time taken by any reasonable implementation of this algorithm is not an eventually nondecreasing function of the exponent.2. 0 iii. Problem 10.) ii. all these theorems are constructive : the algorithm for B follows from the algorithm for A and the proof of the corresponding theorem. By "reasonable". Finally.

MQ <_ ' MT.1 Reductions Among Matrix Problems An upper triangular matrix is a square matrix M whose entries below the diagonal are all zero.2. but the proof of this is much more difficult. Is it possible that multiplication of upper triangular matrices could be carried out significantly faster than the multiplication of two arbitrary square matrices ? From another point of view. by MQ. We shall show under reasonable assumptions that MQ I MT =_1 IT. referring to an algorithm that runs in a time in O(n 2) as quadratic. Theorem 10. this is incorrect because the running time should be given as a function of the size of the instance. Suppose there exists an algorithm that is able to multiply two n x n upper triangular matrices in a time in 0 (t (n)). multiplication of upper triangular matrices. contrary to the intuition that may suggest that this problem will inevitably require a time in S2(n3).2. = 0 when i > j. Let A and B be two arbitrary n x n matrices to be multiplied.2. more efficient algorithm for inverting arbitrary nonsingular matrices (at least in theory). so that a time in O(n2) is really linear. where t (n) is a smooth function. that is. In particular.) Once again this means that any new algorithm that allows us to multiply upper triangular matrices more efficiently will also provide us with a new. Proof.2. 10 10. No confusion should arise from this. and inversion of nonsingular upper triangular matrices. (The problem of inverting an arbitrary nonsingular matrix is also linearly equivalent to the three preceding problems (Problem 10. and the resulting algorithm is numerically unstable.9). We denote these three problems. Theorem 10. In what follows we measure the complexity of algorithms that manipulate n x n matrices in terms of n.9 that a time in O (n 2. Proof.Introduction to Complexity 302 Chap. that is. Notice that the problems considered are at least quadratic in the worst case because any algorithm that solves them must look at each entry of the matrix or matrices concerned. M. MT <_ I MQ.1. We saw in Section 4.81) (or even 0 (n 2-376)) is sufficient to multiply two arbitrary n x n matrices.376). it implies that we can invert any nonsingular n x n matrix in a time in 0 (n 2. experience might well lead us to believe that inverting nonsingular upper triangular matrices should be an operation inherently more difficult than multiplying them.2. multiplication of arbitrary square matrices. Any algorithm that can multiply two arbitrary square matrices can be used directly for multiplying upper triangular matrices. Formally speaking. MT. it requires a slightly stronger assumption. assuming MT is smooth. Consider the following matrix product: . respectively. and IT.

IT <' MQ. Suppose there exists an algorithm that is able to multiply two arbitrary n x n matrices in a time in 0 (t (n)). weaker but less natural hypothesis suffices : it is enough that MQ be supra linear. where t (n) is strongly quadratic.' IT. By the smoothness of t(n). (In fact. MQ <. plus 0 (t (3n)) for the multiplication of the two upper triangular matrices. Let A and B be two arbitrary n x n matrices to be multiplied. assuming IT is smooth. Now consider the following product : BC 0D X B-1 -B-'CD-' 0 D-' 10 01 This product shows us how to obtain A-' by first calculating B-1. This product shows us how to obtain the desired result AB by inverting the first of the 3n x 3n upper triangular matrices. The upper triangular matrices B and D.4.2. this operation takes a time in 0 (t(n)). Proof.2 Reduction 303 0 A 0 000 0 0 AB 0 0 0 x 0 0 0 00B = 0 0 0 00 0 0 0 0 where the "0" are n x n matrices all of whose entries are zero. Consider the following matrix product : I AO 0IB x 0 0/ 1 -A AB 0 1 -B 00 1 100 = 010 001 where I is the n x n identity matrix. By Problem 10.3. As in the proof of the previous theorem. a Theorem 10.5. and D each of size n / 2 x n / 2 such that Cl A= I B0DJ where B and D are upper triangular and C is arbitrary.2.) If n = 1. the matrices B and D are nonsingular. inversion is trivial. Let A be a nonsingular n x n upper triangular matrix to be inverted. Consequently. C.) Proof. C. (See Problem 10. Suppose for simplicity that n is a power of 2. assuming MQ is strongly quadratic. Suppose there exists an algorithm that is able to invert a nonsingular n x n upper triangular matrix in a time in O (t(n)). the total time required to obtain the product AB is in 0 (t(n)). Theorem 10. Otherwise decompose A into three submatrices B.2.4 for the general case.2. and D-'. and D-'. n 2 E O (t (n)). 10. This product shows us how to obtain the desired result AB by multiplying two upper triangular 3n x 3n matrices. The time required for this operation is in 0 (n 2) for the preparation of the two big matrices and the extraction of AB from their product.Sec. Because t(n) is at least quadratic. and then multiplying the matrices B-'. where t (n) is a smooth function. . t (3n) E O (t (n)).

An infinite cost represents the absence of a direct route.7.2. and if t(n) is strongly quadratic.6. with the natural conventions that x +(+oo) =+oo and min(x.3. by MS the problem of multiplying symmetric matrices.2.3(iii). Prove that if g(n) E 2g(n / 2) +O (t (n)) when n is a power of Problem 10.5. Prove that MS =' MQ under suitable assumptions. Problem 10. 2. Denote by . note that it is enough to assume that t(n) is supra linear. using the assumption that t(n) is strongly quadratic (or at least supra linear).) 10. Prove that IQ =' MQ. Denote by SU the problem of squaring a unitary upper triangular MQ under suitable assumptions.see Problem t7 10. diagonal are 1. Let f : X x Y -4 IR°° and g : Yx Z -4 IR°° be two functions representing the cost of going directly from one node to another. The fact that t(n)ES2(n2) and the assumption that t(n) is eventually nondecreasing (since it is strongly quadratic) yield g(n) e 2g(n / 2) +O (t(n)) when n is a power of 2.2.2.Introduction to Complexity 304 Chap. Let X. Let IT2 be the problem of inverting nonsingular upper triangular matrices whose size is a power of 2.6. then B and D are nonsingular. and Z be three sets of nodes.2. What assumptions do you need? Denote by IQ the problem of inverting an arbitrary non** Problem 10. (Note : this reduction would not go through should an algorithm that is capable of multiplying n x n matrices in a time in 0 (n 2 log n) exist . Complete the proof that IT <_' MQ. (Hint : apply Problem 2. are smaller than the original matrix A. Using the divide-and-conquer technique suggests a recursive algorithm for inverting A in a time in O(g(n)) where g(n) a 2g(nl2)+2t(nl2)+O(n2). Assume that both IQ and MQ are smooth and supra quadratic. and if B and D are defined as in the proof of theorem 10.8. Prove that if A is a nonsingular upper triangular matrix whose size is even. Denote Problem 10.2 Reductions Among Graph Problems In this section IR°° denotes IR*u I+-}.2.4 really shows is that IT2 <_' MQ.2. Prove that SU need ? A matrix A is symmetric if Aid = Aji for all i and j. this implies that g(n) e 0 (t (n) I n is a power of 2). Problem 10.2.4. 10 which we now have to invert. All that the proof of theorem 10.2.13(iv).+oo) = x for all x e R-.9.) An upper triangular matrix is unitary if all the entries on its Problem 10.4. What assumptions do you matrix. Y. By Problem 10.2. then g(n) E O (t(n) I n is a power of 2).2. singular matrix.

This calculation is nothing other than the calculation of f * . which we shall write f 2. computing f * for a given function f seems to need more time than calculating a simple product fg. we do not discuss them here. f never takes negative values.Sec. coincide. At first sight. Thus it is possible to get away with a time in O(n 3) after all. too). there is no obvious way of adapting to this problem Strassen's algorithm for ordinary matrix multiplication (Section 4. Nevertheless. Y. without increasing the cost of the resulting path. namely Floyd's algorithm. but do not confuse this operation with the composition of functions. z) I y E Y }.2 305 Reduction fg the function h : X x Z -* IR°° defined for every x E X and z (=Z by h (x . f2) gives the minimum cost of going from one node of X to another either directly or by passing through exactly one intermediate node. which we write f*.) So the definition of f * can be taken to give us a direct algorithm for calculating its value in a time in O(n4). Notice the analogy between this definition and ordinary matrix multiplication (where addition and multiplication are replaced by the minimum operation and addition. and Z. is therefore f * = min { f' I i >_ 0 1.9). f ° represents the cost of going from one node to another while staying in the same place. and also the functions f and g. y) + g (y. We thus have that f * = min ( f' I 0 <_ i < n . By analogy. there exist more efficient algorithms for this problem. However. This represents the minimum cost of going from x to z passing through exactly one node in Y.4 a dynamic programming algorithm for calculating shortest paths in a graph. gives the minimum cost of going from one node of X to another (possibly the same) while passing through exactly one intermediate node (possibly the same. z) = min { f (x. Because they are quite complicated and have only theoretical advantages. Unfortunately. The preceding notation becomes particularly interesting when the sets X. Similarly. we saw in Section 5. Any path that passes twice through the same node can therefore be shortened by taking out the loop thus formed. In this case ff . Could it be that the problems of calculating fg and f * are of the same complexity ? The following two theorems show that this is indeed the case : these two problems are linearly equivalent. } There is no equivalent to this operation in the present context since taking the minimum is not a reversible operation. (The intuitive reason is that Strassen's algorithm does subtractions. min(f . it suffices to consider only those paths whose length is less than the number of nodes in X. respectively). This definition is not practical because it apparently implies an infinite computation . However. Let this number be n. The meaning of f' is similar for any i > 0. 10. The existence of algorithms asymptotically more . so that ° f (x' y) 0 ifx =y +00 otherwise. The minimum cost of going from one node to another without restrictions on the number of nodes on the path. Consequently. The straightforward algorithm for calculating fg takes a time in O(n3) if the three sets of nodes concerned are of cardinality n. it is not even immediately clear that f * is well defined.

and let f : X x Y .& + oo when y E Y is to have u EX.Introduction to Complexity 306 Chap. v) I y c: Y ). y) + h (y. v) # +oo when y E Y is to have v EZ. assuming TRC is smooth. Define the function h : W x W. y) = f (u . Therefore h (u. v) = +. Suppose there exists an algorithm that is able to calculate h* in a time in O(t(n)). n 2 and n 3. Denote by MUL and TRC the problems consisting of calculating fg and f*. w) + which implies that h 3(U.v) ifuEY andvEZ +00 otherwise . h 2) is given by the following equation. The calculation of h 3(u . Notice in particular that h (x.0 for every WE W. w) + h 2(w . respectively. 10 efficient than O(n3) for solving the problem of calculating fg therefore implies that Floyd's algorithm for calculating shortest routes is not optimal. As in the previous section. for a smooth function t(n). v) I w E W I.v)= g(u. Suppose without loss of generality that X. z) _ + oo when x e X and z E Z. v) to conclude that h 2(U. v) = g (y. Let X. Y. for instance. v) is easier. v) =min { h (u . Theorem 10. v) = h 2(w . By definition. MUL <_ ' TRC. time complexities will be measured as a function of the number of nodes in the graphs concerned. v). h 2(U.1R°° and g : Y x Z . Again. In the case when u EX. y) :. we have 2 h (u. But the only way to have h (u. where n is the cardinality of the set W such that h : W x W . at least in theory. w) + h (w. respectively. v) I W E W 1. If u 0 X or if v 0 Z. The conclusion from all this is that h* = min(h °. v) = min ( h (u. v) = hh 2(u . v E W. Proof.v) if u EX and veZ + otherwise . Let W = X u Y u Z. By the definition of h. min ( h (u . v) # +oo simultaneously unless w c= Y.1R°°. v) I y E Y 1.1R°° be two functions for which we wish to calculate fg.1R°° as follows : f(u. By definition. and Z are disjoint. Summing up. the problems considered are at least quadratic in the worst case because any algorithm that solves them must look at each edge concerned. h 3(u . and Z be three sets of nodes of cardinality n . which is precisely the definition of f g (u. and the only way to have h (y.2. v) = min { f (u . But h (u . and v e Z it suffices to note that h (u .= v) fg(u. v) = +o. h . let us find the value of h 2(u . Y. v) _ + oo for all u. y) + g (y. v) = + 00 when w e X. w) # +oo and h (w. Now. v). y) and h (y. The same holds for h ` (u .5.v) ifuEX andvEY h(u. would be considered quadratic even though it is linear in the number of edges (for dense graphs). y E Y. v) for all i>3. . it is impossible that h (u. Consequently. w) _ + oo when w E X whereas h 2(W. it therefore follows that h 2(u . An algorithm such as Dijkstra's. h 2(U.

the time g(n) required by this approach is characterized by the equation g(n) E 2g(n / 2)+0 (t(n)). Let e : J x J .1R.v) +oo otherwise Therefore the restriction of h* to X x Z is precisely the product fg we wished to calculate. e*(u. v) if u E K and V E J (min(c*. The other restrictions are obtained similarly. c : K x K -* 1R°°. Otherwise split H into two disjoint subsets J and K. it is obvious that h* (u . The desired result is then obtained after a number of instances of the MUL problem.2. each of size n / 2. (In the restrictions of h to the corresponding subdomains. Using the algorithm for calculating h* thus allows us to compute fg in a time in t (n) + O (n 2)C O (t(3 max(n i . Y. which implies that g(n) E 0 (t(n) I n is a power of 2). each containing half the nodes. Suppose for simplicity that n is a power of 2 (see Problem 10. n 2. n 3))). Notice that e (u . n 2. Theorem 10.and g : Y x Z -* 1R°°. Let H be a set of cardinality n and let h : H x H . If n = 1. however. e* is the restriction to J x J of the h* that we wish to calculate. TRC <_' MUL. Define a : J x J -> IR°°.2. and Z such that f : X x Y -+ 1R. for u and v in J. assuming MUL is strongly quadratic. 10. u) = 0 for the unique u EH.) Proof. v) ifu E K and v E K To calculate h* using divide-and-conquer. The minimum cost for going from u to v with no restrictions on the path is therefore e* (u. represents the minimum cost for going from u to v without passing through any node in J (not counting the endpoints) . v ). in order to obtain c* and e*. v) when u and v are both in J. a weaker but less natural hypothesis suffices again : it is enough that MUL be supra linear. Assume t (n) is strongly quadratic. n 2.Sec. As in theorem 10. and n 3 are the cardinalities of the sets X. v) if u E J and v E K c*de* (u. n 3))) because t(n) is smooth and at least quadratic.v) if uEX and vEZ g( a cost function for which we wish to calculate h*.6.v) = < if ueJ and veJ e*bc* (u. .11 for the general case).1R°° be the function given by e = min(a .v) h*(u. Let n = n I +n 2+n 3 be the cardinality of W. In other words. c*de*bc* )) (u .1R°°. we therefore solve recursively two instances of the TRC problem.v) = if uEX andvEY ifuEY andvEZ fg(u. Suppose there exists an algorithm that is able to calculate fg in a time in 0 (t (max(n I . where n I . n 3))) + O (n 2)g 0 (t (max(n I .2 Reduction 307 0 ifu =v f(u.v) h*(u.4. be*d ). b : J x K . the nodes in K can be used as often as necessary.2. and d : KxJ -* R. or not used at all if this is preferable. n 2.

+00 }.81). where p is any prime number larger than n. We saw that it is possible to multiply two integers of size n in a time in 0 (n 1. restricted cost functions. We saw that Warshall's algorithm (Problem 5.2.12. Conclude that MULB <_ 1 MQ. when the cost functions are restricted in this way. show how to transform the problem of calculating fg into the computation of an ordinary arithmetic matrix multiplication. v E X . Chap.7. Let TRC2 be the problem of calculating h* for h : X x X . MULBS. v) = f (v . Problem 10. No algorithm is known that can solve MULB faster than MQ. TRCS. Show that the arithmetic can be done modulo p. Let f : X x Y -* 10. calculating f * comes down to determining for each pair of nodes whether or not there is a path joining them. that using Strassen's algorithm requires a number of arithmetic operations in 0 (n 2-81 ) . u) for every u . All the proof of theorem 10.2.3 Reductions Among Arithmetic and Polynomial Problems We return to the problems posed by the arithmetic of large integers (sections 1. When the range of the cost functions is restricted to 10. A cost function f : X x X -4R. and 9. Call these four problems MULS. Note. What can we say about integer division and taking square roots ? Our everyday experience leads us to believe that the second of .2. respectively. * Problem 10. the time in 0 (n 3) taken by Warshall's algorithm counts only Boolean operations as elementary.1R°° when the cardinality of X is a power of 2.10.+-l be two Problem 10. Complete the proof that TRC 5 ' MUL.2) solves this problem in a time in 0(n3). 10 Prove formally that the preceding formula for h* is correct. Strassen's algorithm can therefore be used to solve the problems MULB and TRCB in a time in 0 (n symmetric if f (u .2. This is interesting because MULB <_ 1 MQ.2. and TRCBS. where MQ is the problem of multiplying arbitrary arithmetic square matrices (Problem 10.2.5). Assuming we count arithmetic operations at unit cost.+-) and g: Y x Z -* (0.2. does one of these two problems appear strictly more difficult than the other? Which one? Justify your answer.11.13. the proof that MUL =/ TRC can easily be adapted to show that MULB =' TRCB. 10.2. 4. regardless of the cost of the path. Unlike the case of arbitrary cost functions. Do you believe that MULBS=1 TRCBS? If not.7.59) and even in 0 (n log n log log n).12). Furthermore. Prove that MULBS=1 MULB. Each of the four problems discussed earlier has a symmetric version that arises when the cost functions involved are symmetric.Introduction to Complexity 308 * Problem 10. thus showing that Warshall's algorithm is not optimal.4. It is clear that MULB <_ 1 MUL and TRCB <_ 1 TRC since the general algorithms can also be used to solve instances of the restricted problems. Let MULB and TRCB be the problems consisting of calculating fg and h*.6 really shows is that TRC2 5 1 MUL. however.

of multiplying two integers of size n.22). we take a moment to prove that SQR MLT.2. The following formula enables us to obtain their product by carrying out two squaring operations of integers of size at most n + 1. since squaring is only a special case of multiplication.2 Reduction 309 these problems. suppose there exists an algorithm that is able to square an integer of size n in a time in 0 (t(n)). is inapplicable in the case of division unless the two polynomials concerned divide one another exactly with no remainder. MLT.7. and DIV be the problems consisting of squaring an integer of size n. 10. To show that MLT <_' SQR. SQR 51 MLT. Once again this turns out not to be true.2. Notice that a direct approach using discrete Fourier transforms. assuming these three problems are smooth and MLT is strongly linear (weaker but more complicated assumptions would suffice). too.2. Assume without loss of generality that x >. For this reason we content ourselves in the rest of this section with showing the equivalence of these operations in the "cleaner" domain of polynomial arithmetic. is genuinely more difficult than multiplication. and probably the first one. where t(n) is smooth (it is enough to assume that t (n + 1) E O (t (n)) ). and of determining the quotient when an integer of size 2n is divided by an integer of size n. In this . For example. Its conceptual beauty is also defaced in places by the necessity of using an inordinate number of ad hoc tricks to circumvent the problems caused by integer truncation (see Problem 10. we can solve MLT in a time in 2t(n +1)+O(n)c O(t(n)) because t(n) is smooth and t (n) E Q(n). Proof outline. which works so well for multiplying two polynomials. this choice is not critical : the time taken by the various algorithms would be in the same order if given as a function of the size of their operands in decimal digits or computer words. We now show that the problem of polynomial division is linearly equivalent to that of multiplication.) Theorem 10. however. let p(x)=x3+3x2+x+2 and d(x) = x2+x +2. This is the case precisely because we assume all these algorithms to be smooth. (For simplicity we measure the size of integers in bits. Let x and y be two integers of size n to be multiplied. SQR =' MLT =' DIV. Clearly.y . Clearly. The quotient of the division of p (x) by d (x) is q (x) =x+2. The full proof of this theorem is long and technical. respectively. provided that the necessary arithmetic operations on the coefficients can be counted at unit cost. Let SQR. and a division by 4: xy =((x+y)2-(x-y)2)/4. assuming SQR is smooth (a weaker assumption would do). Nonetheless. We have seen in Section 9. these problems are at least linear because any algorithm that solves them must take into account every bit of the operands involved. a few additions.4 how to multiply two polynomials of degree n in a time in 0 (n log n). As mentioned in Section 1. Since the additions and the division by 4 can be carried out in a time in O (n).Sec.7.

17. Show that if both p (x) and d (x) are monic polynomials (the coefficient of highest degree is 1) with integer coefficients then both q (x) and r (x) have integer coefficients and q (x) is monic (unless q (x) = 0). Prove the existence and the uniqueness of the quotient and Problem 10. then p2(x) _ p i(x)d 2(x) ± p2(x)d 1(x) d i(x)d2(x) ± d2(x) in particular. (There is a very simple proof. LLp(x)/di(x)j/d2(x)J = Lp(x)/(di(x)d2(x))]. Notice that p (x) and p* (x) are always of the same degree. the remainder. The inverse of p (x). We also need the notion of an inverse. Recall that p (x) = E.14.3 x 2 + 8x .3x .2. and let d (x) be a nonzero polynomial of degree m. For example.15. and if d (x). and d* (x). Let p(x) = x 3 +x 2 + 5 x + 1 Problem 10. q (x) = p* (x) then q* (x) = p (x). which we denote p* (x). We even have that p (1) = 7 is not divisible by d (1) = 4. the quotient and the remainder of the division of p (x) by d (x). d i(x) and d2(x) are three nonzero polynomials. and iii.2. 11 Consider the four following problems : SQRP consists of squaring a polynomial of degree n. MLTP of multiplying two polynomials of degree at most n. if p(x)=x3+3x2+x+2. 10 case p (2) = 24 and d (2) = 8. Then there exists a unique polynomial r (x) of degree strictly less than m and a unique polynomial q (x) such that p (x) = q (x) d (x) + r (x). then p*(x)= x 3 . p I (x) and p 2(x) are three arbitrary poly* Problem 10.2.16. L p (x)/d (x)]. and d(x)=x-2. ii. otherwise q (x) = 0.'_ 0 ai x' is a polynomial of degree n provided that an # 0. By analogy with the integers. the quotient is denoted by q(x) = Lp(x)Id(x)J. it is possible to determine the quotient and the remainder produced when a polynomial of degree 2n is divided by a polynomial of degree n in a time in 0 (n log n) by reducing these problems to a certain number of polynomial multiplications calculated using the Fourier transform. Let p (x) be a nonzero polynomial of degree n. INVP of determining the inverse of a polynomial of degree n and DIVP of calculating the . p* (x).m if n > m -1 . but q (2) = 4 # 24/8. nomials. Lp1(x)/d(x)] ± Lp2(x)/d(x)] = L(p1(x)±p2(x))/d(x)]. This is all due to the remainder of the division r (x) = . We call q (x) and r (x). Calculate Prove that if p (x) is a nonzero polynomial and if Problem 10. The polynomial q (x) is of degree n . Despite this difficulty. provided that the degree of p1(x) P I(x) d i(x) is not more than twice the degree of LP 2(X)ld (x)j. respectively.23. By convention the polynomial p (x) = 0 is of degree -1. Let p (x) be a polynomial of degree n.) Prove that if p (x).2.2.Introduction to Complexity 310 Chap. Lp1(x)/Lp2(x)/d(x)Jj = L(p1(x)d(x))/p2(x)]. is defined by p*(x)= Lx 2nIp(x)].

Sec. 10.2



quotient of the division of a polynomial of degree at most 2n by a polynomial
of degree n. We now prove under suitable assumptions that these four problems
linearly equivalent using the following chain of reductions : MLTP <-'
SQRP <' INVP <-' MLTP and INVP:-' DIVP <-' INVP. Again, all these problems
are at least linear. We assume arithmetic operations can be carried out at unit cost on

the coefficients.

MLTP <-' SQRP, assuming SQRP is eventually nonde-

Theorem 10.2.8.

Proof. Essentially the same as the proof that MLT <-1 SQR given in theorem
10.2.7. There is no need for a smoothness assumption this time because the sum or

difference of two polynomials of degree n cannot exceed degree n.

Theorem 10.2.9.


SQRP <- 1 INVP, assuming INVP is smooth.

Proof. The intuitive idea is given by the following formula, where x is a
nonzero real number :



A direct attempt to calculate the square of a polynomial p (x) using the analogous formula (p* (x) - (p (x) + 1)* )* -p (x) has no chance of working : the degree of this
expression cannot be greater than the degree of p (x). This failure is caused by truncation errors, which we can, nevertheless, eliminate using an appropriate scaling factor.

Suppose there exists an algorithm that is able to calculate the inverse of a polynomial of degree n in a time in 0 (t(n)), where t(n) is a smooth function. Let p (x) be a
polynomial of degree n > 1 whose square we wish to calculate. The polynomial
x In p (x) is of degree 3n, so
[X2np(x)]* = Lx6n/x2np(x)J = LX4n/p(x)J


[x2n (p(x)+1)]* = Lx4n/(p(x)+l)J
By Problem 10.2.17
[x2n p(x)]* -[x2n (p(x)+l)]* = Lx4n/p(x)]- Lx4n/(p(x)+1)]


(p(x)+1)-x 4np(x))l(p(x)(p(x)+1))j

= Lx4nl(p2(x)+p(x))j

= [p2(x)+p(x)]*


The last equality follows from the fact that p 2(x) + p (x) is of degree 2n. By Problem
10.2.16, we conclude finally that
p2(x) _

[[x2n p(x)]* -[x2n (p(x)+l)]*]* -p(x)


Introduction to Complexity

Chap. 10

This gives us an algorithm for calculating p2(x) by performing two inversions of polynomials of degree 3n, one inversion of a polynomial of degree 2n, and a few operations (additions, subtractions, multiplications by powers of x) that take a time in O (n).
This algorithm can therefore solve SQRP in a time in

2t(3n)+t(2n)+0(n)c 0(t(n))
because t(n) is smooth and at least linear.

Theorem 10.2.10.


Proof. To calculate p* (x), where p (x) is a polynomial of degree n, we evaluate
Lx2n /p (x)j, an instance of size n of the problem of polynomial division.
Theorem 10.2.11.

DIVP :- I INVP, assuming INVP is smooth.

Proof. The intuitive idea is given by the following formula, where x and y are
real numbers and y # 0:

x/y = xy-I
If we try to calculate the quotient of a polynomial p (x) divided by a nonzero polynomial d (x) using directly the analogous formula p (x) d* (x), the degree of the result is
too high. To solve this problem we divide the result by an appropriate scaling factor.
Suppose there exists an algorithm that is able to calculate the inverse of a polynomial of degree n in a time in 0 (t(n)), where t(n) is a smooth function. Let p (x) be

a polynomial of degree less than or equal to 2n, and let d (x) be a polynomial of
degree n. We wish to calculate LP (x)/d (x)j. Let r (x) be the remainder of the division
of x 2n by d (x), which is to say that d* (x) = Lx 2" /d (x) j = (x 2n - r (x)) /d (x) and that
the degree of r (x) is strictly less than n . Now consider
d* (x)p (x)
X 2"


x 2n p (x) - r (x)p (x)
x2n d(x)
x 2n p (x)



r (x) p (x)


by Problem 10.2.17(i). But the degree of r (x) p (x) is strictly less than 3n , whereas the
degree of x 2" d (x) is equal to 3n , and so L(r (x) p (x))/(x 2" d (x))] = 0. Consequently,
L(d* (x)p (x ))/x2n] = L p (x) Id (x)], which allows us to obtain the desired quotient by
performing an inversion of a polynomial of degree n, the multiplication of two polynomials of degree at most 2n, and the calculation of the quotient from a division by a

power of x. This last operation corresponds to a simple shift and can be carried
out in a time in O(n). The multiplication can be performed in a time in O(t(2n))
thanks to theorems 10.2.8 and 10.2.9 (using the assumption that t(n) is smooth).

Sec. 10.2



The calculation of Lp(x)/d(x)] can therefore be carried out in a time in
t(n)+O(t(2n))+O(n)c O(t(n)) because t(n) is smooth and at least linear.
Theorem 10.2.12.
INVP 5' MLTP, assuming MLTP is strongly linear
(a weaker assumption will do - see the hint of Problem 10.2.19).
Proof. This reduction is more difficult than the previous ones. Once again, we
appeal to an analogy with the domain of real numbers, namely, Newton's method for
finding the zero of f (w) = I - xw. Let x be a positive real number for which we want
to calculate x-1. Let y be an approximation to x-1 in the sense that xy = I -S, for
-1 < S < 1. We can improve the approximation y by calculating z = 2y -y2x.
Indeed, xz = x (2y - y 2x) = xy (2 -xy) _ (1- S) (1 + S) = l _ S2. From our assumption
on S, S2 is smaller than 6 in absolute value, so that z is closer than y to x-1. To calculate the inverse of a polynomial, we proceed similarly, first finding a good approximation to this inverse and then correcting the error.
Suppose there exists an algorithm able to multiply two polynomials of degrees
less than or equal to n in a time in 0 (t (n)) where t(n) is strongly linear. Let p (x) be a
nonzero polynomial of degree n whose inverse we wish to calculate. Suppose for simplicity that n + 1 is a power of 2 (see Problem 10.2.18 for the general case).
If n =0, the inverse p* (x) is easy to calculate. Otherwise let k = (n + 1)/2.
During the first stage of the polynomial inversion algorithm we find an approximation h (x) to p* (x) such that the degree of x In - p (x) h (x) is less than 3k -1. (Note
that the degree of x In - p (x) p* (x) can be as high as n -1 = 2k - 2.) The idea is to rid
ourselves provisionally of the k coefficients of lowest degree in the polynomial p (x) by
dividing the latter by x'. Let h (x) = x k LP (x)/x k J * . Note first that the degree of
LP(x)/xk] is n -k = k -1, so Lp(x)/xk j* = Lx2k-2/LP(x)/xk]] = LX 3k -21p (x)] by
Problem 10.2.17(iii). Let r (x) be the polynomial of degree less than n such that
Lx 3k -2/p (x) J = (x 3k-2 - r (x)) /p (x). Then we have
x2n-p(x)h(x)=x4k-2 _p(x)xk(x3k-2-r(x))/p(x)=xkr(x)

which is indeed a polynomial of degree less than 3k - 1.
During the second stage, we improve the approximation h (x) in order to obtain
p* (x) exactly. Taking into account the appropriate scaling factor, the analogy introduced at the beginning of this proof suggests that we should calculate
q (x) = 2h (x) - Lh2(x)p (x)1x2n j. Let s (x) be the polynomial of degree less that 2n

such that Lh2(x)p(x)/x2n ] _ (h2(x)p(x)-s(x))/x2". Now calculate
p (x) q (x) = 2p (x) h (x) _ (p 2(x) h 2(x) _ P (x) s (x)) /x


= [(P(x)h(x))(2x2n -p(x)h(x))+P(x)s(x)]Ix In
_xkr(x))(xIn +xkr(x))+p(x)s(x)]/xIn
= [(x2n
= [xa" _x2kr2(x)+P(x)s(x)]lx2"

=x2n +(P(x)s(x)-x2kr'2(x))/x2"

Introduction to Complexity


Chap. 10

It remains to remark that the polynomials p (x) s (x) and x 2k r 2(x) are of degree at most
3n -1 to conclude that the degree of x 2n - p (x) q (x) is less than n , hence
q (x) = p* (x), which is what we set out to calculate.
Combining these two stages, we obtain the following recursive formula :

p*(x) = 2xk LP(x)lxkJ* - Lv(x)[Lp(x)lxk J*]2/xn-1 J
Let g(n) be the time taken to calculate the inverse of a polynomial of degree n by the
divide-and-conquer algorithm suggested by this formula. Taking into account the
recursive evaluation of the inverse of Lp(x)/xk J, the two polynomial multiplications
that allow us to improve our approximation, the subtractions, and the multiplications

and divisions by powers of x, we see that

g(n)Eg((n -1)/2)+t((n -1)/2)+t(n)+O(n)c g((n -1)/2)+O(t(n))
because t(n) is strongly
g(n) E O (t (n))


Using Problem 10.2.19, we conclude that

Problem 10.2.18.
Let INVP2 be the problem of calculating p* (x) when p (x)
is a polynomial of degree n such that n + 1 is a power of 2. All that the proof of
theorem 10.2.12 really shows is that INVP2 <-' MLTP. Complete the proof that

Problem 10.2.19.
Prove that if g(n)Eg((n -1)/2)+O(t(n)) when n +1 is a
power of 2, and if t(n) is strongly linear, then g(n) E O (t (n) I n +1 is a power of 2).
(Hint : apply Problem 2.3.13(iv) with the change of variable T (n) = g(n -1) ; note
that it is enough to assume the existence of a real constant a> 1 such that
t (2n) ? a t (n) for all sufficiently large n - strong linearity of t (n) would unnecessarily impose a=2.)
Problem 10.2.20.
Let p (x) = x 3 +x 2 + 5 x + 1. Calculate p* (x) using the
approach described in the proof of theorem 10.2.12. You may carry out directly the
intermediate calculation of the inverse of Lp (x)/x2J rather than doing so recursively.
Compare your answer to the one obtained as a solution to Problem 10.2.15.
* Problem 10.2.21.
We saw in Section 9.4 how Fourier transforms can be used
to perform the multiplication of two polynomials of degrees not greater than n in a
time in O (n log n). Theorems 10.2.11 and 10.2.12 allow us to conclude that this time
is also sufficient to determine the quotient obtained when a polynomial of degree at
most 2n is divided by a polynomial of degree n. However, the proof of theorem
10.2.11 depends crucially on the fact that the degree of the dividend is not more than
double the degree of the divisor. Generalize this result by showing how we can divide

a polynomial of degree m by a polynomial of degree n in a time in 0 (m log n).
** Problem 10.2.22.

Following the general style of theorems 10.2.9 to 10.2.12,

complete the proof of theorem 10.2.7. You will have to define the notion of

Sec. 10.3

Introduction to NP-Completeness


inverse for an integer : if i is an n-bit integer (that is, 2n-1 < i <- 2" -1), define
i* = L22n - '/i J. Notice that i* is also an n-bit integer, unless i is a power of 2. The
problem INV is defined on the integers in the same way as INVP on the polynomials.
The difficulties start with the fact that (i* )* is not always equal to i, contrary to
Problem 10.2.16. (For example, 13* = 9 but 9* = 14.) This hinders all the proofs.
For example, consider how we prove that DIV <- I INV. Let i be an integer of size 2n

and let j be an integer of size n ; we want to calculate

Li I j J.

If we define

z = Li j * / 22n - 1 ] by analogy with the calculation of L(p (x) d* (x )) /x 2n j in the proof

of theorem 10.2.11, we no longer obtain automatically the desired result z = Li / j].
Detailed analysis shows, however, that z <- Li / j J < z + 2. The exact value of Li / j J
can therefore be obtained by a correction loop that goes around at most three times.
z E- Li j* /22n


t - (z +1)x j
while t

<- i

t Ft +j


return z

The other proofs have to be adapted similarly.
** Problem 10.2.23.
Let SQRT be the problem of computing the largest integer
less than or equal to the square root of a given integer of size n. Prove under suitable

assumptions that SQRT=/ MLT. What assumptions do you need? (Hint: for the
reduction SQRT <-' MLT, follow the general lines of theorem 10.2.12 but use
Newton's method to find the positive zero of f (w) = w 2 -x ; for the inverse reduction,
use the fact that

x++- x l- x- x+l+ + 1=1/x.)
Let MOD be the problem of computing the remainder when
Problem 10.2.24.
an integer of size 2n is divided by an integer of size n . Prove that MOD <_ 1 MLT.
Let GCD be the problem of computing the greatest common
** Problem 10.2.25.
divisor of two integers of size at most n . Prove or disprove GCD = MLT. (Warning:
at the time of writing, this is an open problem.)


There exist many real-life, practical problems for which no efficient algorithm is
known, but whose intrinsic difficulty no one has yet managed to prove. Among these
are such different problems as the travelling salesperson (Sections 3.4.2, 5.6, and
6.6.3), optimal graph colouring (Section 3.4.1), the knapsack problem, Hamiltonian cir-

cuits (Example 10.3.2), integer programming, finding the longest simple path in a


Introduction to Complexity

Chap. 10

graph (Problem 5.1.3), and the problem of satisfying a Boolean expression. (Some of
these problems are described later.) Should we blame algorithmics or complexity ?
Maybe there do in fact exist efficient algorithms for these problems. After all, computer science is a relative newcomer: it is certain that new algorithmic techniques
remain to be discovered.
This section presents a remarkable result : an efficient algorithm to solve any one
of the problems we have listed in the previous paragraph would automatically provide
us with efficient algorithms for all of them. We do not know whether these problems
are easy or hard to solve, but we do know that they are all of similar complexity. The
practical importance of these problems ensured that each of them separately has been
the object of sustained efforts to find an efficient method of solution. For this reason it
is widely conjectured that such algorithms do not exist. If you have a problem to solve
and you are able to show that it is equivalent (see Definition 10.3.1) to one of those

mentioned previously, you may take this result as convincing evidence that your
problem is hard (but evidence is not a proof). At the very least you will be certain that
nobody else claims to be able to solve your problem efficiently at the moment.

10.3.1 The Classes P and NP

Before going further it will help to define what we mean by an efficient algorithm.
Does this mean it takes a time in O (n log n) ? O (n 2) ? O (n'-") ? It all depends on
the problem to be solved. A sorting algorithm taking a time in O(n2) is inefficient,
whereas an algorithm for matrix multiplication taking a time in 0 (n 2 log n) would be
an astonishing breakthrough. So we might be tempted to say that an algorithm is
efficient if it is better than the obvious straightforward algorithm, or maybe if it is the
best possible algorithm to solve our problem. But then what should we say about the
dynamic programming algorithm for the travelling salesperson problem (Section 5.6)
or the branch-and-bound algorithm (Section 6.6.3) ? Although more efficient than an
exhaustive search, in practice these algorithms are only good enough to solve instances
of moderate size. If there exists no significantly more efficient algorithm to solve this
problem, might it not be reasonable to decide that the problem is inherently intractable ?

For our present purposes we answer this question by stipulating that an algorithm
is efficient (or polynomial-time) if there exists a polynomial p (n) such that the algorithm can solve any instance of size n in a time in 0 (p (n)). This definition is
motivated by the comparison in Section 1.6 between an algorithm that takes a time in
O(2') and one that only requires a time in O (n 3 ), and also by sections 1.7.3, 1.7.4,
and 1.7.5. An exponential-time algorithm becomes rapidly useless in practice, whereas

generally speaking a polynomial-time algorithm allows us to solve much larger
instances. The definition should, nevertheless, be taken with a grain of salt. Given
two algorithms requiring a time in O(ntg'g" ) and in O(n 10), respectively, the first, not
being polynomial, is "inefficient". However, it will beat the polynomial algorithm on
all instances of size less than 10300 assuming that the hidden constants are similar. In
fact, it is not reasonable to assert that an algorithm requiring a time in O(n 10) is

Sec. 10.3

Introduction to NP-Completeness


efficient in practice. Nonetheless, to decree that O(n3) is efficient whereas S2(n4) is
not, for example, seems rather too arbitrary.

In this section, it is crucial to avoid pathological algorithms and analyses such as
those suggested in Problem 1.5.1. Hence no algorithm is allowed to perform arithmetic operations at unit cost on operands whose size exceeds some fixed polynomial in
the size of the instance being solved. (The polynomial may depend on the algorithm
but not of course on the instance.) If the algorithm needs larger operands (as would be
the case in the solution of Problem 1.5.1), it must break them into sections, keep them

in an array, and spend the required time to carry out multiprecision arithmetic.
Without loss of generality, we also restrict all arrays to contain a number of elements
at most polynomial in the size of the instance considered.
The notion of linear reduction and of linear equivalence considered in Section
10.2 is interesting for problems that can be solved in quadratic or cubic time. It is,
however, too restrictive when we consider problems for which the best-known algorithms take exponential time. For this reason we introduce a different kind of reduction.

Definition 10.3.1.
Let X and Y be two problems. Problem X is polynomially
reducible to problem Y in the sense of Turing, denoted X <_ T Y, if there exists an algo-

rithm for solving X in a time that would be polynomial if we took no account of the
time needed to solve arbitrary instances of problem Y. In other words, the algorithm
for solving problem X may make whatever use it chooses of an imaginary procedure
that can somehow magically solve problem Y at no cost. When X <_T Y and Y <_T X
simultaneously, then X and Y are polynomially equivalent in the sense of Turing,
denoted X -T Y. (This notion applies in a natural way to unsolvable problems - see
Problem 10.3.32.)

Example 10.3.1.
Let SMALLFACT(n) be the problem of finding the smallest
integer x > 2 such that x divides n (for n >_ 2), let PRIME(n) be the problem of determining whether n >_ 2 is a prime number, and let NBFACT(n) be the problem of counting
the number of distinct primes that divide n. Then both PRIME <_ T SMALLFACT and
NBFACT <_ T SMALLFACT. Indeed, imagine solutions to the problem SMALLFACT can be

obtained at no cost by a call on SolveSF ; then the following procedures solve the
other two problems in a time polynomial in the size of their operand.

function DecidePRiME(n)
we assume n _ 2}
if n = SolveSF (n) then return true
else return false
function SolveNBF (n)
nb - 0

while n > I do nh - nh + I
x - SolveSF (n)
while x divides n don f- n /x
return nh

2. When no confusion can arise. P is the class of decision problems that can be solved by a 0 polynomial-time algorithm. Let X and Y be two decision problems. Let X and Y be two problems such that X <T Y. X is many-one polynomially reducible to problem Y. The restriction to decision problems allows us to introduce a simplified notion of polynomial reduction. Definition 10. Prove that if X <_ T Y and Y E P. then X E P. the existence of an algorithm to solve problem Y in polynomial time implies that there also exists a polynomial-time algorithm to solve problem X. Problem 10.Introduction to Complexity 318 Chap. nondecreasing function. For technical reasons we confine ourselves from now on to the study of decision problems. or "the set of all possible graphs".3. Problem 10. 10 Notice that SolveNBF works in polynomial time (counting calls of SolveSF at no cost) because no integer n can have more than Llg n ] prime factors.3.3. We also assume that the instances can be coded efficiently in the form of strings of bits. known as the reduction function between X and Y. Let X and Y be two problems such that X <_ T Y Y. Let X c I and Y c J be two decision problems. whereas "find the smallest prime factor of n " is not. denoted X = m Y. A decision problem can be thought of as defining a subset X of the set I of all its instances. or none of them can. "Is n a prime number?" is a decision problem. Problem Definition 10. such as IN. . given some x e 1. if there exists a function f :I . denoted by X <m Y. such that (VXEI)[xEX f (x)EY] . The usefulness of this definition is brought out by the two following exercises. When X <_.J computable in polynomial time.° Y and Y <_ m X both hold. In particular. where t(n) is a nonzero. even taking repetitions into account.3. whether or not x EX. we may sometimes omit to state explicitly the set of instances for the decision problem under consideration. Prove that there exist a polynomial p (n) and an algorithm that is able to solve problem X in a time in O (p (n) t(p (n))). For example. We generally assume that the set of all instances is easy to recognize. the equivalence mentioned in the introduction to this section implies that either all the problems listed there can be solved in polynomial time. Then the problem consists of deciding. Prove that Problem 10. then X and Y are many-one polynomially equivalent.3. Suppose there exists an algorithm that is able to solve problem Y in a time in O (t(n)).

then xSTY.2 shows that the restriction to decision problems is not a severe constraint.4. are transi- tive. An instance of HAM is a directed graph.3. c. El T .3. It is also the case that TSPD m HAM. Lemma 10.3. To prove that HAM 5 m TSPD. A > be a directed graph for which you would like to decide if it has a Hamiltonian circuit. function DecideX (x) y -f (X) if DecideY (y) then return true else return false * Problem 10. NxN >. Then the following procedure solves X in polynomial time. 10. the cost function c(u . and the question is to decide whether there exists a circuit in the graph passing exactly once through each node n (with no optimality constraint). An instance of TSPD consists of a directed graph with costs on the edges. Problem 10.6 and 6. L > E TSPD. Prove that the converse of Lemma 10.5. Let TSPD and HAM be the travelling salesperson decision problem and the Hamiltonian circuit problem. let G = < N. If X and Y are two decision problems such that X <_m Y.1.3. In fact. Imagine solutions to problem Y can be obtained at no cost by a call on DecideY and let f be the polynomial-time computable reduction function between X and Y. v) _ I if (u.3 Introduction to NP-Completeness 319 Example 10. Proof.2. most optimization problems are polynomially equivalent in the sense of Turing to an analogous decision problem. as the following exercise illustrates. respectively. together with some bound L used to turn the travelling salesperson optimization problem (as in Sections 5.v)EA 2 otherwise and the bound L = #N. and whose cost does not exceed L.Sec. the number of nodes in G. but this is significantly harder to prove. Define f (G) as the instance for TSPD consisting of the complete graph H = <N. <_ m .6.3. The introduction of TSPD in Example 10. Clearly. Prove that the relations <_ T .3. after having visited each of the other nodes exactly once. G E HAM if and only if < H.3) into a decision problem : the question is to decide whether there exists a tour in the graph that begins and ends at some node. and = R.1 does not necessarily hold by giving explicitly two decision problems X and Y for which you can prove that X <_ T Y whereas it is not the case that X <_ m Y. .

Introduction to Complexity


Chap. 10

* Problem 10.3.6.
Let G = < N, A > be an undirected graph, let k be an
integer, and c : N -* (1, 2, ... , k } a function. This function is a valid colouring of G
if there do not exist nodes u , v E N such that (u , v ) E A and c (u) = c (v) (Section
3.4.1). The graph G can be coloured with k colours if there exists such a valid

colouring. The smallest integer k such that G can be coloured with k colours is called
the chromatic number of G, and in this case a colouring with k colours is called an
optimal colouring. Consider the three following problems.
COLD: Given a graph G and an integer k, can G be coloured with k colours ?
COLO: Given a graph G, find the chromatic number of G.
COLC: Given a graph G, find an optimal colouring of G.

Conclude that there exists a polynomial-time
algorithm to determine the chromatic number of a graph, and even to find an optimal
Prove that COLD =T COLO =T COLC.

colouring, if and only if COLD E P.

These graph colouring problems have the characteristic that although it is perhaps

difficult to decide whether or not a graph can be coloured with a given number of
colours, it is easy to check whether a suggested colouring is valid.

Let X be a decision problem. Let Q be a set, arbitrary for
Definition 10.3.4.
the time being, which we call the proof space for X. A proof system for X is a subset

FcXxQ such that (`dxEX)(3gEQ)[<x,q>EF]. Any q such that <x,q >EF
is known as a proof or a certificate that x E X. Intuitively, each true statement of the
type x E X has a proof in F, whereas no false statement of this type has one (because if
x E X, there does not exist a q E Q such that < x, q > (=- F ).
Example 10.3.3.
Let I = N and COMP = { n I n is a composite integer }. We
can take Q = 1N as the proof space and F = ( < n, q > I 1 < q < n and q divides n
exactly } as the proof system. Notice that some problems may have more than one
natural proof system. In this example we could also use the ideas of Section 8.6.2 to


< n, q > I (n is even and n > 2) or
(1 < q < n and n is not a strong pseudoprime to the base q

which offers a large number of proofs for all odd composite numbers.

Example 10.3.4.
Consider the set of instances I = { < G , k > I G is an
undirected graph and k is an integer } and the problem COLD = { < G, k > E 1 I G
coloured with k colours }. As proof space we may take Q =
(c : N -* (1, 2, ... , k} I N is a set of nodes and k is an integer). Then a proof

can be

system is given by

Sec. 10.3


Introduction to NP-Completeness


< < G, k >, c > I G = < N, A > is an undirected graph,
k is an integer,
c : N -4 { 1, 2, ... , k } is a function and

(`du,vEN)[{u,v}EA =c(u)#c(v)] }


Problem 10.3.7. Let G = < N, A > be an undirected graph. A clique in G is
a set of nodes K E_ N such that { u, v } E A for every pair of nodes u, v K. Given a
graph G and an integer k, the CLIQUE problem consists of determining whether there
exists a clique of k nodes in G. Give a proof space and a proof system for this decision problem.

Definition 10.3.5.
NP is the class of decision problems for which there exists
a proof system such that the proofs are succinct and easy to check. More precisely, a
decision problem X is in NP if and only if there exist a proof space Q, a proof system
F c X x Q, and a polynomial p (n) such that

i. (VxEX)(agEQ)[<x,q>eFand IqI <p(Ix1)],
where I q I and I x I denote the sizes of q and x, respectively ; and
ii. F E P.

We do not require that there should exist an efficient way to find a proof of x when
x c X, only that there should exist an efficient way to check the validity of a proposed
short proof.

The conceptual distinction between P and NP is best grasped
Example 10.3.5.
with an example. Let comp be the problem in Example 10.3.3. In order to have
comp E P, we would need an algorithm
function DecideCOMP (n)
{ decides whether n is a composite number or not }

return true

return false

whose running time is polynomial in the size of n. No such algorithm is currently
known. However, to show that coMP E NP, we need only exhibit the following
(obvious) polynomial-time algorithm.

Introduction to Complexity


Chap. 10

function VerifyCOMP (n, q)

if 1 < q < n and q divides n then return true
else return false
By definition of NP, any run of VerifyCOMP (n, q) that returns true is a proof that n is

composite, and every composite number has at least one such proof (but prime
numbers have none). However, the situation is not the same as for a probabilistic
algorithm (Chapter 8) : we are content even if there exist very few q (for some
composite n) such that VerifyCOMP (n , q) is true and if our chance of hitting one at
random would be staggeringly low.

Problem 10.3.8.
Let X be a decision problem for which there exists a
polynomial-time true-biased Monte Carlo algorithm (section 8.6). Prove that X E NP.
(Hint : the proof space is the set of all sequences of random decisions possibly taken
by the Monte Carlo algorithm.)
Problem 10.3.9.
Prove that P E_ NP. (Hint : Let X be a decision problem
in P. It suffices to take Q = { 0 } and F = (< x, 0 > I x e X } to obtain a system of
"proofs" that are succinct and easy to check. This example provides an extreme illustration of the fact that the same proof may serve for more than one instance of the
same problem.)

Example 10.3.6.
The problems COLD and CLIQUE considered in example
10.3.4 and Problem 10.3.7 are in NP.
Although COLD is in NP and COLO =T COLD, it does not appear that NP contains
the problem of deciding, given a graph G and an integer k, whether k is the chromatic
number of G. Indeed, although it suffices to exhibit a valid colouring to prove that a
graph can be coloured with a given number of colours (Example 10.3.4), no one has
yet been able to invent an efficient proof system to demonstrate that a graph cannot be
coloured with less than k colours.

Definition 10.3.6.
Let X S I be a decision problem. Its complementary
problem consists of answering "Yes" for an instance x E I if and only if x 0 X. The
class co-NP is the class of decision problems whose complementary problem is in NP.
For instance, the preceding remark indicates that we do not know whether COLD E
co-NP. Nonetheless, we know that COLD E co-NP if and only if NP = co-NP (Problems 10.3.27 and 10.3.16). The current conjecture is that NP # co-NP, and therefore
that COLD


Problem 10.3.10.

Let A and B be two decision problems. Prove that if

A 5 m B and B E NP, then A E NP.

Problem 10.3.11.
Let A and B be two decision problems. Do you believe that
if A 5 T B and B E NP, then A E NP ? Justify your answer.

Sec. 10.3

Introduction to NP-Completeness


Show that HAM, the Hamiltonian circuit problem defined in
Problem 10.3.12.
Example 10.3.2, is in NP.


In 1903, two centuries after Mersenne claimed without proof
Example 10.3.7.
is a prime number, Frank Cole showed that
267_ 1

267 -1 = 193,707,721 x 761,838,257, 287


It took him "three years of Sundays" to discover this factorization. He was lucky that
the number he chose to attack is indeed composite, since this enabled him to offer a
proof of his result that is both short and easy to check. (This was not all luck : Lucas
had already shown in the nineteenth century that 267-1 is composite, but without
finding the factors.)
The story would have had quite a different ending if this number had been prime.
In this case the only "proof" of his discovery that Cole would have been able to produce would have been a thick bundle of papers covered in calculations. The proof
would be far too long to have any practical value, since it would take just as long to
check as it did to produce in the first place. (A similar argument may be advanced
concerning the "proof' by computer of the famous four colour theorem.) This results
from a phenomenon like the one mentioned in connection with the chromatic number
of a graph : the problem of recognizing composite numbers is in NP (Example 10.3.3),

but it seems certain at first sight not to be in co-NP, that is, the complementary
problem of recognizing prime numbers seems not to be in NP.

However, nothing is certain in this world except death and taxes : this problem
too is in NP, although the notion of a proof (or certificate) of primality is rather more
subtle than that of a proof of nonprimality. A result from the theory of numbers shows
that n, an odd integer greater than 2, is prime if and only if there exists an integer x
such that

0<x <n
x"- I = 1 (mod n ), and
xI" - I)lP 4P 1 (mod n) for each prime factor p of n -1

A proof of primality for n therefore consists of a suitable x, the decomposition of n -1
into prime factors, and a collection of (recursive) proofs that each of these factors is
indeed prime. (More succinct proof systems are known.)

Complete the proof sketched in Example 10.3.7 that the
* Problem 10.3.13.
problem of primality is in NP. It remains to show that the length of a recursive proof
of primality is bounded above by a polynomial in the size (that is, the logarithm) of the
integer n concerned, and that the validity of such a proof can be checked in polynomial

Introduction to Complexity


Chap. 10

Problem 10.3.14.
Let F = { < x , y > I x , y E N and x has a prime factor less
than y }. Let FACT be the problem of decomposing an integer into prime factors.
Prove that

i. F E NP n co-NP; and
ii. F = T FACT.

If we accept the conjecture that no polynomial-time factorization algorithm exists, we
can therefore conclude that F E (NP n co-NP) 1 P.

10.3.1 NP-Complete Problems

The fundamental question concerning the classes P and NP is whether the inclusion
P9 NP is strict. Does there exist a problem that allows an efficient proof system but
for which it is inherently difficult to discover such proofs in the worst case ? Our intuition and experience lead us to believe that it is generally more difficult to discover a
proof than to check it: progress in mathematics would be much faster were this not so.
In our context this intuition translates into the hypothesis that P # NP. It is a cause of
considerable chagrin to workers in the theory of complexity that they can neither prove
nor disprove this hypothesis. If indeed there exists a simple proof that P # NP, it has
certainly not been easy to find !

On the other hand, one of the great successes of this theory is the demonstration

that there exist a large number of practical problems in NP such that if any one of
them were in P then NP would be equal to P. The evidence that supports the
hypothesis P # NP therefore also lends credence to the view that none of these problems can be solved by a polynomial-time algorithm in the worst case. Such problems
are called NP-complete.
Definition 10.3.7.

A decision problem X is NP-complete if

I. X E NP ; and
ii. for every problem Y E NP, Y S T X.

Some authors replace the second condition by Y

n, X or by other (usually stronger)

kinds of reduction.

Prove that there exists an NP-complete problem X such that
Problem 10.3.15.
X E P if and only if P = NP.

Prove that if there exists an NP-complete problem X such
* Problem 10.3.16.
that X E co-NP, then NP = co-NP.

Sec. 10.3

Introduction to NP-Completeness


Prove that if the problem X is NP-complete and the

Problem 10.3.17.
problem Z E NP, then

i. Z is NP-complete if and only if X <T Z ;
ii. if X n, Z, then Z is NP-complete.

Be sure to work this important problem. It provides the fundamental tool for
proving NP-completeness. Suppose we have a pool of problems that have already
been shown to be NP-complete. To prove that Z is NP-complete, we can choose an
appropriate problem X from the pool and show that X is polynomially reducible to Z
(either many-one or in the sense of Turing). We must also show that Z E NP by exhibiting an efficient proof system for Z. Several thousand NP-complete problems have
been enumerated in this way.
This is all well and good once the process is under way, since the more problems

there are in the pool, the more likely it is that we can find one that can be reduced
without too much difficulty to some new problem. The trick, of course, is to get the
ball rolling. What should we do at the outset when the pool is empty to prove for the
very first time that some particular problem is NP-complete? (Problem 10.3.17 is then
powerless.) This is the tour de force that Steve Cook managed to perform in 1971,
opening the way to the whole theory of NP-completeness. (A similar theorem was
discovered independently by Leonid Levin.)

10.3.2 Cook's Theorem
Definition 10.3.8.
A Boolean variable takes its values in the set
B = { true, false }. Boolean variables are combined using logical operators (not, and,
or, t:* , r , and so on) and parentheses to form Boolean expressions. It is customary
to represent disjunction (or) in such expressions by the symbol "+" and conjunction
(and) by simply juxtaposing the operands (as for arithmetic multiplication). Negation
is often denoted by a horizontal bar above the variable or the expression concerned. A
Boolean expression is satisfiable if there exists at least one way of assigning values to
its variables so as to make the expression true. A Boolean expression is a tautology if
it remains true whatever values are assigned to its variables. A Boolean expression is a
contradiction if it is not satisfiable, that is, if its negation is a tautology. We denote by

SAT, TAUT and CONT, respectively, the problems of deciding, given a Boolean
expression, whether it is satisfiable, whether it is a tautology, and whether it is a con-



Example 10.3.8.
ables p and q

Here are three Boolean expressions using the Boolean vari-


i. (p+q)=pq
ii. (p b q) (p +q)(p +q )
iii. p (p +q)q

Introduction to Complexity


Chap. 10

Expression (i) is satisfiable because it is true if p = true and q = true, but it is
not a tautology because it is false if p = true and q = false . Verify that expression
(ii) is a tautology and that expression (iii) is a contradiction.
To prove that a Boolean expression is satisfiable, it suffices to produce an assignment that satisfies it. Moreover, such a proof is easy to check. This shows that
SATE NP. It is not apparent that the same is true of the two other problems : what
short and easy proof can one give in general of the fact that a Boolean expression is a
tautology or a contradiction? These three problems are, nevertheless, polynomially
equivalent in the sense of Turing.
Prove that

Problem 10.3.18.

i. SAT = T TAUT - T CONT ; and even

ii. TAUT -m CONT.
It is possible in principle to decide whether a Boolean expression is satisfiable by
working out its value for every possible assignment to its Boolean variables. However
this approach is impractical when the number n of Boolean variables involved is large,
since there are 2" possible assignments. No efficient algorithm to solve this problem is

Definition 10.3.9.
A literal is either a Boolean variable or its negation.
A clause is a literal or a disjunction of literals. A Boolean expression is in conjunctive
normal form (CNF) if it is a clause or a conjunction of clauses. It is in k-CNF for
some positive integer k if it is composed of clauses, each of which contains at most k
literals (some authors say : exactly k literals).

Example 10.3.9.

Consider the following expressions.

L (p +q +r)(p +q +r)q r
ii. (p +qr)(p +q (p +r))

W. (p =* q)b(p+q)
Expression (i) is composed of four clauses. It is in 3-CNF (and therefore in CNF), but

not in 2-CNF. Expression (ii) is not in CNF since neither p +qr nor p +q (p +r) is a
clause. Expression (iii) is also not in CNF since it contains operators other than conjunction, disjunction and negation.

*Problem 10.3.19.

i. Show that to every Boolean expression there corresponds an equivalent expression in CNF.

ii. Show on the other hand that the shortest equivalent expression in CNF can be
exponentially longer than the original Boolean expression.

algorithm A accepts at least one instance of size 5 if and only if T5(A) is satisfiable. all these problems are in NP or in co-NP.10. Thus it remains to prove that X <_ T SAT-CNF for every problem X E NP. (The number of additional Boolean variables is polynomial in the size of the instance because the algorithm runs in polynomial time and because we have assumed without loss of generality that none of its variables or arrays can ever occupy more than a polynomial number of bits of memory. SAT-k-CNF is the restriction of SAT-CNF to Boolean expressions in k-CNF..3 Introduction to NP-Completeness 327 Definition 10. polynomial time. Proof. -The Boolean expression is constructed so that there exists a way to satisfy it by choosing the values of its other Boolean variables if and only if algorithm A accepts the instance corresponding to the Boolean value of the x variables. algorithm A accepts the instance 10010 if and only if the expression xix2x3xax5`P5(A) is satisfiable. x i . 10. such as the Turing machine. El The interest of Boolean expressions in the context of NP-completeness arises from their ability to simulate algorithms. Theorem 10. SAT-CNF is the restriction of the SAT problem to Boolean expressions in CNF. For any positive integer k. the clauses of 'P (A) force these Boolean variables to simulate the step-by-step execution of the algorithm on the corresponding instance.1.20.3.--. For example. Polynomial-time algorithms are known for a few of them. among which x correspond in a natural way to the bits of instances of size n for A. For every integer n there exists a Boolean expression 'P (A) in CNF that can be obtained efficiently (in a time polynomial in n. It usually requires a formal model of computation beyond the scope of this book. x2 2 . and for each unit t of time taken by this computation.3. SAT-CNF is NP-complete. Once the variables x I . This Boolean expression contains a large number of variables. where the polynomial may depend on the algorithm A . Suppose that the size of the instances is measured in bits. X2. We content ourselves with mentioning that the expression 'P (A) contains among other things a distinct Boolean variable b.) We are finally in a position to state and prove the fundamental theorem of the theory of NP-completeness. Consider an arbitrary decision problem that can be solved by a polynomial-time algorithm A. the size of 'Y (A) is also polynomial in n). Prove that SAT-2-CNF and TAUT-CNF can be solved in * Problem 10. More interestingly.Sec.3. We already know that SAT-CNF is in NP. Let p (n) be the polynomial (given by the definition of NP) such . The problems TAUT-(k-)CNF and 11 CONT-(k-)CNF are defined similarly.x are fixed. The proof that this Boolean expression exists and that it can be constructed efficiently poses difficult technical problems. for each bit i of memory that algorithm A may need to use when solving an instance of size n. Let Q be a proof space and F an efficient proof system for X. Clearly.

q > E F. (To be precise. one technical detail is worth mentioning : to each algorithm A. 10 that to every x E X there corresponds a q E Q whose length is bounded above by p (I x I) such that < x. where 0 <_ i <_ p (n). function DecideX (x) let n be the size of x (in bits) for i . q > implies that algorithm Ax accepts q.3. We illustrate this principle with several examples. (q ) if A (< x. let Ax be an algorithm whose purpose is to verify whether a given q E Q is a valid proof that x E X. The fact that algorithm A accepts < x.Introduction to Complexity 328 Chap. which implies that DecideX (x) will find no integer i such that Ti (Ax) is satisfiable. if x 0 X.3 Some Reductions We have just seen that SAT-CNF is NP-complete. whether q is a proof that x E X (that is. and let q e Q be a proof of size i that x E X. That such an algorithm exists is ensured by the fact that F E P. there corresponds a two variable polynomial r such that the Boolean formula Ti (Ax) can be constructed in a time in 0 (r (I x I. Let A be a polynomial-time algorithm able to decide. Let X E NP be some other decision problem. We already know that SAT E NP. if we imagine that answers concerning the satisfiability of certain Boolean expressions in CNF can be obtained at no cost by a call on DecideSATCNF. and therefore Ax accepts no inputs. which is part of the definition of X E NP. Prove that in fact X <_ m SAT-CNF for any decision problem 10. as DecideX (x) will discover. q >) then return true else return false Here finally is an algorithm to solve problem X in polynomial time. For each x.17). function Ax.3. q > E F. To show that X too is NP-complete.3. This completes the proof that X <_ T SAT-CNF. we need only prove that SAT CNF <_ T X (Problem 10. to show that Y E NP is NP-complete. given < x.21. i )) and such that its size is bounded similarly. Thereafter.3. It therefore remains to show that SAT-CNF <_ T SAT. This is an instance of size i.10. q > belongs to F ). whether < x. then there exists no q e Q such that < x. we have the choice of proving SAT-CNF <_ T Y or X <_ T Y.O to p (n) do if DecideSATCNF (Ti (Ax)) then return true return false Let X E X be of size n.) Problem 10. The reduction is immediate because Boolean expressions in CNF . SAT is NP-complete. XENP. hence the Boolean expression Ti (Ax) is satisfiable. q > as input. Conversely. Example 10.

10. which is therefore a disjunction of k literals. Let u be a new Boolean variable. Let u 1 .3. in 3-CNF that is satisfiable if and only if P is satisfiable. We have already seen that SAT-3-CNF is in NP. More precisely. u2 . . so it cannot be computed in polynomial time in the worst case. However. . i.19(i) to obtain the algorithm function DecideSAT(Y) let E be a Boolean expression in CNF equivalent to Y if DecideSATCNF (E..3.22. l3 .12..3. 12. 1 1 =(11+12+u1)(u1+13+u2)(u2+14+u3) (uk_3+lk-1+lk). Consider first how to proceed if `P contains only one clause. let l . and 14 be the literals such that `P is l 1 + 12 + 13 + l4 ..Sec. which is already in 3-CNF. SAT-3-CNF is NP-complete.3 Introduction to NP-Completeness 329 are simply a special case of general Boolean expressions. Let P be a Boolean expression in CNF. here is a polynomial-time algorithm for solving SAT-CNF. can be exponentially longer than expression T.11) If `P=(p+q+r+s)(r+s)(p+s+x+v+u) we obtain E=(p+q+u1)(u1+r+s)('+s)(p+s+u2)(u2+x+u3)(u3+v+w) .) Example 10. . Example 10. function DecideSATCNF(`Y) if P is not in CNF then return false if DecideSAT (P) then return true else return false Prove that SAT <_ T SAT-CNF. Problem 10.3. If k = P. lk be the literals such that `P is +lk .. set If k = 4. uk_3 be new Boolean variables. It remains to show that SAT CNF <_ T SAT-3-CNF. Take = (l 1 + 12 + u) (i + 13 +14)iii. Take l 1 +12+ ii. resist the temptation to use Problem 10. 3.3. 12. if we imagine that the satisfiability of Boolean expressions can be decided at no cost by a call on DecideSAT.10. More generally.3. if k >_ 4. let 1 . (Hint : This problem has a very simple solution.11. Our problem is to construct efficiently a Boolean expression i. .3. (Continuation of Example 10. If the expression `P consists of several clauses. Using Example 10. .) then return true else return false because Problem 10. This time let us show that SAT-CNF <_ m SAT-3-CNF. conclude that SAT= T SAT-CNF. treat each of them independently (using different new variables for each clause) and form the conjunction of all the expressions in 3-CNF thus obtained.19(ii) shows that expression i.

.3. The graph G that we are about to build contains 3 + 2t + 6k nodes and x2 2 . 3-COL is NP-complete. Prove that SAT-CNF <_ m SAT-3-CNF still holds even if in the definition of SAT-3-CNF we insist that each clause should contain exactly three literals. Prove that the construction of Example 10. Figure 10. We still have to add 6 nodes and 12 edges for each clause in T. imagine that the colours assigned to T and F represent the Boolean values true and false.-y. of `P the graph contains two nodes yj and zi that are linked to each other and to the control node C.24).3. we shall prove this time that SAT-3-CNF <_ m 3-COL. and C. One copy of this widget is added to the graph for each clause in T. See Figure 10. Suppose further without loss of generality that the Boolean variables appearing in `P are x t .1. Given a Boolean expression `P in 3-CNF. If y.3. For each Boolean variable x. In any colouring of G in three colours. corresponds to the literal x. -x.--3 + 3t + 12k edges. In every case node y.6). F. we have to construct efficiently a graph G that can be coloured in three colours if and only if T is satisfiable. . Figure 10. . X2 . Chap. corresponds to an assignment of Boolean values to x t .11. To show that 3-COL is NP-complete. think of this intuitively as an assignment of the value true to the Boolean variable x.23. Let G be an undirected graph and k an integer constant. These are added in such a way that the graph will be colourable with three colours if and only if the choice of colours f o r y l . Suppose for simplicity that every clause of the expression `P contains exactly three literals (see Problem 10.3. Let k be the number of clauses in T.24. respectively. If t = 3. this forces yj to be the same colour as either T or F and zi to be the complementary colour. and node zi corresponds to the literal x. When the time comes to colour G in three colours.3.1. It is easy to see that 3-COL E NP. Contrariwise. The control triangle. Y2.3.13. The colour used for node C will be a control colour. This can be accomplished thanks to the widget illustrated in Figure 10.2 shows the part of the graph that we have constructed up to now.Introduction to Complexity 330 Problem 10.3. is the same colour as T. This reduction is considerably more complex than those we have seen so far. 10 is satisfiable if and only if P is satisfiable in Problem 10. x. Three distinguished nodes of this graph are linked in a triangle : call them T. The problem k-COL consists of determining whether G can be coloured with k colours (see Problem 10.3. . xi is false if it is zi that is the same colour as T.3. for example. that satisfies every clause.3. Example 10.3.

A widget. since the colour assigned to node T represents true. 2. Each widget is linked to five other nodes of the graph : nodes C and T of the control triangle. Because these input nodes shows that the widget can be coloured with the colours assigned to C.3. the widget simulates the disjunction of the three literals represented by the nodes to which it is joined.2. 2 and 3 cannot be the same colour as C. .3. 10. and three nodes chosen from the y. so as to correspond to the three literals of the clause concerned. 331 Graph representation of three Boolean variables. In other words. and 3 is coloured with the same colour as node T. Problem 10.Sec. and F if and only if at least one of the nodes 1. T Figure 10. T.3 Introduction to NP-Completeness Figure 10. and z.

3. We conclude that SAT-3-CNF <_ m 3-COL. Problem 10.3. Prove that COLD (Problem 10. The name NP arose from this other definition: it represents the class of problems that can be solved by a Non-deterministic algorithm in Polynomial time.3.3. Prove that CLIQUE (Problem 10.which we do not do here).26. * Problem 10. Prove that 3-COL is still NP-complete even if we restrict ourselves to planar graphs of degree not greater than 4.31.3. T.3. on the other hand. and therefore that 3-COL is NP-complete. 10. ** Problem 10. Prove that the problem of the Hamiltonian circuit (Example 10. Problem 10.25. A function f : IN -* IN is polynomially bounded if the size of its value on any argument is polynomially bounded by the size of its argument.3. Show. The notion of reducibility extends in a natural way to unsolvable problems such as the halting problem (although it is usual to drop the polynomial-time aspect of the reductions . Prove that the problem of computing any polynomially bounded computable function is polynomially reducible to the halting problem in the sense of Turing.332 Introduction to Complexity Chap. The classic definition involves the notion of non-deterministic algorithms.2 and Problem 10.7) is NP-complete. Verify that the colours attributed to nodes C.27.3 if and only if at least one of the input nodes is coloured with the same colour as node T (knowing that the input nodes cannot be coloured with the same colour as node C).3.12) is NP-complete.29. and F suffice to colour the widget of Figure 10.) ** Problem 10. * Problem 10.3.6) is NP-complete. The halting problem consists of deciding. 10 This ends the description of the graph G. It is clear that the graph can be constructed efficiently starting from the Boolean expression `I' in 3-CNF. which can be coloured with three colours if and only if `i is satisfiable. that 2-COL is in P.3. * Problem 10. Prove however that there exist decision problems that are not polynomially reducible to the halting problem in the sense of Turing.32. (Hint : prove that SAT-3-CNF <_ m CLIQUE. .4 Non-determinism The class NP is usually defined quite differently. although the definitions are equivalent.3. which we only sketch here.3. whether the latter will ever halt when started.30.3. Problem 10. Give a simple proof that 4-COL is NP-complete.28.3. given any program as instance.

The algorithm is consistent if two different computations on any given instance always lead to the same result. (This explains why. this does not imply that P = NP. otherwise no solution is obtained. we write return success . If the algorithm sets success to true.) To avoid confusion with probabilistic algorithms. j) as in Chapter 8.. Definition 10. then y is a correct solution to instance x .bool return . It is even allowable that for some instances the algorithm will never set success to true. but instead.bool as an abbreviation for success . 10. success). whose effect is to set n to some value between i and j inclusive. although nondeterministic algorithms can solve NP-complete problems in polynomial time. we use a special instruction choose n between i and j .11.5). we do not denote nondeterministic choices by calls on uniform (i . The effect of the algorithm is determined by the existence or the nonexistence of sequences of non-deterministic choices that lead to the production of a result. y .3 Introduction to NP-Completeness 333 On the surface. We are not concerned with how such sequences could be determined efficiently or how their nonexistence could be established.Sec. Notice that there is no limit on how long a polynomial-time non-deterministic algorithm can run if the "wrong" non-deterministic . hence to returning a solution. A non-deterministic algorithm runs in polynomial time if the time it takes is bounded by some polynomial in the size of its instance provided that the instance is in the domain of the algorithm.3. The time taken by a non-deterministic algorithm on a given instance of its domain is defined as the shortest possible time that any computation can cause it to run on this instance . a non-deterministic algorithm resembles a Las Vegas probabilistic algorithm (Section 8. nor is it subject to the laws of probability. it computes a well-defined function on the instances of its domain. where y and success are return parameters. the time is undefined on instances that are not in the domain of the algorithm. For simplicity. The difference between non-deterministic and Las Vegas algorithms is that for the former we do not set any bound on the probability of success. The domain of the algorithm is the set of instances on which it has at least one computation. The actual value assigned to n is not specified by the algorithm. A computation of the algorithm is a sequence of nondeterministic choices that leads to setting success to true. When a non-deterministic algorithm is consistent. For this reason non-deterministic algorithms are only a mathematical abstraction that cannot be used directly in practice : we would not program such an algorithm in the hope of running it successfully and efficiently on a computer. The algorithm is total if its domain is the set of all possible instances. we call ND (x . To solve instance x of some problem X.

7. n .m /p else return success f.3.7.3 or when n is even.3.n < 3) return success f. n) # 1 then return success f. n) then while p divides m do m .true else success . n >. 10 choices are made.3.1.5. When n is prime. suc) if suc and pr and p divides m and dexpo (x. (n -1)/p . Example 10. the algorithm also has a computation that consists of choosing guess = 1.true 1 When n is an odd composite number.false m f-n -1 while m > 1 do choose p between 2 and m { the guess is that p is a new prime divisor of n -1 } primeND (p. var success) { non-deterministically determines whether n is prime } if n < 3 or n is even then prime F.7 } if dexpo (x. the algorithm also has a computation when n <.false choose m between 2 and n -1 if m divides n then success . and guessing successively each prime divisor of n -1.14.false success f.true choose x between 1 and n -1 { the guess is that x is as in Example 10. It is also consistent by the same theorem from Example 10. procedure primeND (n . Consider the following total consistent non-deterministic primality testing algorithm (recall from Section 4.let's guess a proof ! } prime . Notice again that it would be pointless to attempt implementing this algorithm on the computer by replacing the choose instructions by random choices in the same interval : the probability of success would be infinitesimal. choosing x in accordance with the theorem mentioned in example 10. var prime. n) computes x' mod n efficiently).(2<. . the algorithm has a computation that consists of choosing guess =0 and non-deterministically setting m to some nontrivial factor of n. pr. i . n >.true choose guess between 0 and 1 if guess = 0 then { the guess is that n is composite .false else { the guess is that n is prime .let's guess a proof ! } prime .8 that dexpo (x . The algorithm primeND is therefore total.3. It is even possible for a computation to be arbitrarily long. provided that the same instance also admits at least one polynomially bounded computation. Clearly.9.Introduction to Complexity 334 Chap.

3. and let p be the polynomial that bounds the length of a proof as a function of the length of the corresponding positive instance. For instance.true else success . var success) n .2 Every decision problem in NP is the domain of some polynomial-time non-deterministic algorithm.3. Although the authors prefer the definition based on proof systems.34.3 335 * Problem 10. that there are sequences of non-deterministic choices that can cause the algorithm to run for a time exponential in the size of its instance. The following polynomialtime non-deterministic algorithm has a computation on instance x if and only if x E X .1 to l do choose b between 0 and 1 append bit b to the right of q if < x. this problem belongs to NP. Prove that a decision problem can be solved by a total con- sistent polynomial-time non-deterministic algorithm if and only if it belongs to NP c co-NP. Show.false 11 Problem 10.8 and 10. q > E F then ans . (Hint: use the set of all possible computations as proof space.9 completely obvious. and the corresponding parameter may be ignored altogether (there is no point in algorithm XND setting ans to true when it finds a q such that <x. Prove that the non-deterministic algorithm primeND of Example 10.33.3. var ans.3. 10. let Q be its proof space.35 Prove the converse of Theorem 10.Introduction to NP-Completeness Sec. .size of x choose l between 0 and p (n) q .true.3.) The preceding theorem and problem suggest the alternative (and more usual) definition for NP: it is the class of decision problems that are the domain of some polynomial-time non-deterministic algorithm. procedure XND (x. it makes Problems 10.q > EF ). success . the actual result returned by the algorithm in case of success is irrelevant.3.empty binary string for i . Let X E NP be a decision problem. Assume for simplicity that Q is the set of binary strings. however.2: whenever a decision problem is the domain of some polynomial-time non-deterministic algorithm. we are only concerned with the existence or nonexistence of computations (usually called accepting computations in this context) . Theorem 10.3. let F c X x Q be its efficient proof system. (The algorithm can easily be made consistent but it will not in general be total.) Proof. * Problem 10. In this case.14 runs in a time polynomial in the size of its instance. it is sometimes easier to show that a problem belongs to NP with this other definition.3.

9) comes from Bunch and Hopcroft (1974). consult Aho. more succinct primality certificates can be found in Pomerance (1987).11. Arlazarov. Pippenger (1978) describes a method similar to decision trees for determining a lower bound on the size of logic circuits such as those presented in Problems 4.12 are optimal. 10. Hopcroft.3.2.2. the lexicographic sorting algorithm can sort n elements in a time in O(n +m). comes from Cook and Aanderaa (1969). see Hopcroft and Ullman (1979).14.16 are from Brassard (1979). and Ullman (1974). To be historically exact. This chapter has done little more than sketch the theory of computational complexity. it should be noted that TAUT-DNF is probably not NP-complete since otherwise NP = co-NP (Problem 10. and Problems 10. Kronrod. however.12. and Ullman (1974). Sahni and Horowitz (1978).28 can be found in Stockmeyer (1973). A good introduction can also be found in Hopcroft and Ullman (1979).22).2.3. The fact that the set of prime numbers is in NP (Examples 10.2. However in this case we have to be content with heuristics and approximations as described in Garey and Johnson (1976).2. Problem and 10. then an algorithm that is asymptotically more efficient than the naive algorithm for calculating fg is given in Fredman (1976). and Karp (1972) underlines the importance of this notion by presenting a large number of NP-complete problems. In his tutorial article. Hopcroft. Problems 10.11. and Ullman (1974).33. Borodin and .11. consult Gonnet and Munro (1986) and Carlsson (1986. The uncontested authority in matters of NP-completeness is Garey and Johnson (1979).4 REFERENCES AND FURTHER READING For an overview of algorithms for sorting by transformation consult Aho.12 is solved in Fischer and Meyer (1971). Part of the solution to Problem was discovered by Pratt (1975). In particular. In particular.16). In the case of cost functions whose range is restricted to (0. Several important techniques have gone unmentioned. this technique shows that the required solutions for Problems 4.5 is due to Fischer and Meyer (1971) and Theorem 10. and 10. 1987).14 and 10.+-). For further information concerning the topics of Section 10.8 to 4. A similar theory was developed independently by Levin (1973).6 is due to Furman (1970). For modifications of heapsort that come close to being optimal for the worstcase number of comparisons. 10 10.3.3. The reduction IQ <_/ MQ (Problem 10. Dinic. the original statement from Cook (1971) is that X <_ T TAUT-DNF for every X E NP. In practice. and Faradzev (1970) present an algorithm to calculate fg using a number of Boolean operations in 0 (n 3/log n).3. the fact that a problem is NP-complete does not make it go away.9 and 4. To find out more about non-determinism. where TAUT-DNF is concerned with tautologies in disjunctive normal form .3. which is crucial in the proof of Theorem 10. The theory of NP-completeness originated with two fundamental papers : Cook (1971) proves that SAT-CNF is NP-complete. An algebraic approach to lower bounds is described in Aho.3.11. Hopcroft. where m is the sum of the sizes of the elements to be sorted. The reduction INV <_/ MLT (Problem 10.336 Introduction to Complexity Chap. and Horowitz and Sahni (1978).2. Theorem 10. If f and g are two cost functions as in section 10.

whatever the resources available. Hopcroft. as described in Aho. consult Horowitz and Sahni (1978). even if it is allowed to take a time comparable to the age of the universe and as many bits of memory as there are elementary particles in the known universe (Stockmeyer and Chandra 1979).4).32. and Brassard and Monet (1982). there exist problems that are intrinsically difficult. there also exist problems that cannot be solved by any algorithm. read Turing (1936). For an introduction to adversary arguments (Problem 8. As mentioned in Problem 10. but it can be proved that no algorithm can solve them in practice when the instances are of moderate size. These can be solved in theory.4. Gardner and Bennett (1979).4 References and Further Reading 337 Munro (1975). 10. Hopcroft and Ullman (1979). and Ullman (1974). . Although we do not know how to prove that there are no efficient algorithms for NP-complete problems. and Winograd (1980).3.Sec.

n factorial (0! = 1 and n! =n x (n -1)!. cardinality of set T. interval of integers : { k E IN I i <_ k <_ j } arithmetic quotient and modulo . n! > 1) number of combinations of k elements chosen among n 338 . and b.3 arithmetic and matrix multiplication . respectively basis of the natural logarithm : 2. [lg(1+x)l if x is an integer. 2. In..Table of Notation #T i j div. lg.. mod x T E- var x return return v IxI number of elements in array T . log. Cartesian product pointer assignment return parameter of a procedure or function dynamic end of a procedure dynamic end of a function with value v returned size of instance x .2. e. logb e n! absolute value of x logarithm in basis 10. extended to polynomials in Section 10.7182818.

. P (x) asymptotic notation (see Section 2. reals. b > (in particular : edge of a directed graph) .1. open interval {x E IR I a <x < b [ closed interval {xeIR I a <-x <_b } set of real numbers larger than or equal to a denotes comments in algorithms .3 ceiling of x : smallest integer larger than or equal to x iterated logarithm (see page 63) randomly and uniformly selected integer between i and j x is congruent to y modulo n (n divides x . and Booleans (see Section 2.1) f is a function from A to B there exists an x such that P (x) there exists one and only one for all x.Table of Notation <a.2. set of elements such that set inclusion (allowing equality) strict set inclusion set union: AuB={x IxEA or xEB { set intersection : A nB = { x I x E A and x E B I set membership set nonmembership set difference : A\ B={ x IxEA and x 0 B empty set sets of integers.b > (a.y exactly) exclusive-or of x and y for bits or Booleans . b) 339 ordered pair consisting of elements a and b same as < a. j) x =y (mod n) xO+y floor of x : largest integer less than or equal to x extended to polynomials in Section 10.1) implies if and only if summation integral plus or minus derivative of the function f (x) 00 infinity limit of f (x) when x goes to infinity rx lg* uniform (i . bit-by-bit exclusive-or for bit strings .

Table of Notation 340 and.(a) Fw' (a) A B A B A .B A is many-one polynomially reducible to B A PB A is many-one polynomially equivalent to B P NP co-NP choose class of decision problems that can be solved in polynomial time class of decision problems that have an efficient proof system class of decision problems whose complementary problem is in NP instruction for non-deterministic choice . or A <_ T B Boolean conjunction and disjunction Boolean complement of x Fourier transform of vector a with respect to co inverse Fourier transform A is linearly reducible to B A is linearly equivalent to B A is polynomially reducible to B in the sense of Turing A =T B A is polynomially equivalent to B in the sense of Turing x F(.

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G. 118. 204. 178-179.. S.L. 301 Atomic research. 20 Articulation point. 276. 141. 174-176 Asymptotic notation. 263-266. 3. 276. 185. 35 Baase... 204 backtrack. 27 Amplification of stochastic advantage. E. 2-3 tree Barometer.. 275. 141. 344 Abadi. 7-9. 37-51 with several arguments. 336. 168.Index Aanderaa.. 47 At least quadratic. 78. 137 Adel'son-Vel'skii. 294.. E. 122. 342 Bachmann. 199 Ad hoc sorting. 35. 198. 207 Ancestry in a rooted tree. 342 Array. 342 Backgammon. 336 Arlazarov. 78.. 222. 207-208 APL. 341 adjgraph. M. 25. 222. K. 342 Bellmore. 275. 138 Baumert. 22.. 277 Algol. 54. L. 227. 342 Belaga. L. L. 204. S. 337 Aho. 6.M. 7-9 Ancestor. A. 5. 346 Beauchemin. V. 35. 275. 35 Alpha-beta pruning. xiii. 342 Bach.G. 106 Batcher. 173. 185-189. 205 Algorithm.. 4 Apple II. 296 Adder. 167.. 35. 5 hybrid. 274 Average height..M. R. 205 Balanced tree. 22. 276. Complexity AVL tree. 63. 291. Nested asymptotic notation. 342 Babaf. I Algorithmics. W..E. 188 Backtracking. Operations on asymptotic notation Asymptotic recurrence. 342 353 . 276 Analysis of algorithms: on average. 36. 2. 23. 224.H.O. 5 in worst case. 342 Bellman. 341 Ackermann.. 341 Ackermann's function.. 341 Algebraic transformation. 204 alter-heap. 336.. See Analysis of algorithms. 222. See AVL tree. 203 Adleman. 18 theoretical. 204. P. 168.V. 184. 313. 248-252 Backus-Naur form.C. 190.. 342 Batcher's sorting circuit. P.. 274. 336. 166 Acyclic graph. 238 empirical. 341. 251 Approximation. M.. 197-199. 204. 341 Ajtai. 43 See also Conditional asymptotic notation. M. 298 Average. 350 Adversary argument. 52 Basic subalgorithm. 377. 140.

48-51 Context-free language... 80. Tally circuit. 14-17. 199-204 Brassard. 69. 344 Chromatic number. 235. 336. 276. Batcher's sorting circuit. 35. 10. 275. 146-150. 342 Berliner. 275. 304-308 of large integer arithmetic. N. 292-337 on average. 225. 0. 344 Cook's theorem. 260 Contradiction. 276 choose. 342. 164 Binomial queue. 6 Curtiss. 325. 293 CRAY. 250-252. M. 71 Cheriton. 333 Christofides... 78. 182-184 Brigham. 204 Child.. 343 Bottom-up technique. C. 174-176 binary-node. S. 128-132. 323. A... 336. 2 Comparison sort.. 138 of graph problems. 246 Black. 332 CNF. F.. A. 104. 19.. 140. 275. 35. See Graph colouring Comment. 291. 343 Boyer-Moore algorithm. 78.L. 291 Crepeau. Telephone switching Classic.H. See p-correct probabilistic algorithm correction loop. 204. Merge circuit. 292-299 Compound interest. See Fibonacci heap Birthday surprise. 143. 275. 222. 326 Connected component. D. 140. J. 325 Borodin. 104. 343 Bratley.H.. 350 Branch-and-bound. C. 275. 140.. 337. 78 Conditional probability. 322 Consistent probabilistic algorithm. 35. 315 count. 104.J. 275. Hamiltonian circuit. 231 Cryptographic key. 323 Collision.B. 344 Coppersmith. 344 Coprime. 124. 350 Chang. 72-75 Characteristic equation. J. 25 Binary search. 298-299 of circuits. 315-335 of polynomial arithmetic. 290. 274 Bunch. 342 Berge. 314-315 of matrix problems. 220 Boolean expression. 344 CYBER. 78. 222. 342. 65. 216. 184.. 228-230.. P.. 194-198. 339 Certificate of primality. 36. 142 Boyer. 309-314 of sorting. 233. 325 Cook. 206 Complementary problem. C. 343 Brown. 35. 45-47. J. S. 207 Change of variable. 167 Chandra.. 35. 293.L. 336 Chained matrix multiplication. 216-222. 336. 291. L. 276.A. H.. 291. 351 Catalan number... 308-309. 325-328 Cooley. 52. 342. 293 Compiler generator. R. 4.J. 148. D. 128. 35. 349 Buffon's needle. 204. C. 196 Blum. 78.R. 24 Binomial coefficient. 337. 320 Circuit.. 140. 275. J. E. 343 Carter. 234 . 326 Clique. See Conjunctive normal form Cohen. 243.S. 341 Correct. H. 227 Conjunctive normal form. 165. 31 Carasso. 235 Cryptology. 345 Bentley.O.. 167. 343 Chess. 65. P. 265 Bicoherent... 343 Breadth-first search. 342 crude. 342 bfs. 23 Chinese remainder theorem. 234 countsort. 245 Colouring. 276. C. 343. See Adder. 344 Cole. 260-262.. 276. 106 Conditional asymptotic notation. 336.. 155. 35. 302-304 of NP-complete problems. 15 Corasick. 343 Candidate. 104. 167 Ceiling. 298 Binary tree. 133.A. See Multiplication Clause. 275.. 343 BM... 263 Constructive induction. 290 Choice of leader. J. 168. 337. 321.. 275.. 276 Cubic algorithm. 141.Index 354 Bennett. M.K. 269 Cycle detection. 174-176 Biconnected. 343 Carlsson. 275. 279-290 Complexity. 109-115. 168 Continued fraction algorithm. 173 Co-NP. 325 Boolean variable.M. 267 Confidence interval. 343 Bordvka. 336. 204. E. 322 Complex number. 343 Change making. 343.. 342 Berlekamp. 183 Biased probabilistic algorithm. G. 343 Buneman. 275. 79 Canonical object. 342 Blum. L.

91. See Dicrete Fourier transform Fox. 136 find. 276. 167. R. 35. 270. 344 Dense graph.. 344 Dixon's algorithm. 34 See also Disjoint sets find-max. 56-57 Devroye. 142. 168. 294 trip. 30. 318 sense of Turing. 151. 9-11 Exact order of (e). 78. 104. 2 delete-max. 345 Fermat's theorem. 58.C. See Double ended heap Decision problem. Polynomial arithmetic Dixon.K. 242. 276 Faradzev. 135.W. 19. 241 Disjoint sets. 203 Euler's function. 140. 167. 350 Floyd's algorithm. 87-92. 342 Fast Fourier transform. J. 6 Expected time. 185-188.. 210 Floor. 262. 342 Directed graph. 344 Dijkstra. 290. 298 Efficient algorithm. 235 Data Encryption Standard. 138 Depth of node. 343. 270 Evaluation of polynomial: with fast Fourier transform. 103. 35. 226. N. 24 Depth-first search. 104 Fibonacci sequence. S. 344 Euclid. 134-136.. P. 300 many-one.. R. 19. 76. 104 Euler path.Index 355 Danielson. 66. 275 Elementary operation. 19. 129.. 344 Dewdney. 16. 279. 143 See also De Moivre's formula fib 1. 269 extended. 334 dexpoiler.J. 13 Four colour theorem. 344 De Moivre's formula. 345.L.A.. 224 Exponential algorithm. 349 Depth of circuit. I1-12. 306 Deo. 323 Fourier transform. 301. 128.A. 18. 140. 222 Even. See Fast Fourier transform Fibonacci. 316 Eigenvalue. 143. 276. 104. 339 Floyd. 171-182.G.. 237 Eigenvector. 204. 270 Deyong. 130. 16.E. 36. 78. 293 valid. 36. 17-18. 342 Dromey. 275.W.G. 351 Flajolet. 209-211.. 140. 237 Eight queens problem. 336. 79 Feigenbaum. 28 Demars. 275 Endomorphic cryptosystem. 17-18. 343. 204 Determinant of matrix. 58. 11. 248-252. 35. 78. 345. 46 pruned. 344 Denning. 167. 176-182 Discrete Fourier transform. 20-34 de Bruijn. 35. 124. 293 Declaration. 6 Exponentiation. E. 16. 67 Deadline. 10. 317. 30-34. 344 Dijkstra's algorithm. M. B. 286. 150-153. 151 Dinic. 336 div. 258. 19. G. 60-63.. 10 Fibonacci heap. 291 with preconditioning. 256-260 dlogRH.L.. 344 dexpo. 222 Fischer. 242 Double ended heap. 344 darts. 336. 167. See Large integer arithmetic Factor. 204. 279. 226. 317 Erdo s.E. 242. 45. See Large integer arithmetic.... 293 verdict. 275. C. W. 292-299 Equivalence: linear. 19. 131. J.. 275 Data Strutures. 57 Euclid's algorithm. I.. 28 Finite state automaton. L. 142-168. 16. 171 dfs'. 11. 35. 256-260... 57. 230 Encrypted instance. P. 192-196. 204. 14-15. 341. 140 Execution time. 345 Floating point. N. 225.. 104. 276 False witness of primality. 306 FORTRAN.. 21. 36 Dreyfus. 336. 275. E. 192 fib2.R. 136. 167. 345 . 344 Dynamic programming. 261 Deap. 305. L. 279-280 Feasible. 17. See Nontrivial factor Factorization. 18. 344 Eventually nondecreasing function.D. M. 345 Fredman. 3 Divide-and-conquer. 336. 17.. 105-141. 318 polynomial. 182 Diffie. 30.. 344 dfs. A. 127. 59. 252 FFT. 254 Euclidean travelling salesperson. D. 78. 1. 318 Decision tree. 58-59 fib3. 275. 35. 95-100 Deadlock. 302-315 Division. S. 87. 98-100 Disjunctive normal form. 84. 267. 16. 42 Exchange of two sections in an array. 275.. 235. 309 Discrete logarithm..

S. 13. 35. 254 of a matrix. 91-92. 26. 78. 344. 346 Good. M. 248-252 Greene.. 336 heapsort. 36. 128. 280-284 modulo p. 316. 336. 347 Kaliski.B. 336. 20. M.. See Adder Furman. C. 274 Goutier.. 336..Index 356 Freivalds. 140. 189 Integration. 347 Kasimi. 202 See also Double ended heap. 237. 277 . 65-68 Hopcroft.. 291.. 341. 274. 347 Kanada.. 347 Huang. 68-72 Inorder. 336. 347 Karp.. 295-299 insert-node. 345 God. 184.. A. 101-104.. 346 See also Quicksort Hode.. A. 195. 347 Kepler. 310 of an integer. 115. 36. W. 35. 104. 315.. 346 Haken. D. 346 Halting problem. 332 Hamiltonian path. 276. D.. E. 274. 67 Goldner. 345 Golomb. 4 Instant Insanity. 342 Graham. R. 184. 104. 275. 275. 78. 350 Hu. J. E. S.. 36.. 336. K. 135. 337. J. 302-304 of a polynomial. 276. 342 Hall. xiv. 273 Generator of cyclic group. Fibonacci heap Heap property. 257. V. 291.. 204. 140. 140. 150-153. Chess. R. 346 Homer's rule.. D. 222. 84. See Backgammon. 56 Height. 337. 346 Heuristic...A.. S. 24 See also Average height Held. 291. 140. 350 Goldwasser. 291. 346. 346 Handscomb. Y. 336 Hidden constant. 169-204. 25-30. 345 Kahn. 63. 101-102.. 54 Harrison. 35-36. See Principle of invariance Inverse: Fourier transform. M. 290. 116. M. 347 Jensen.R. P. 346 Graph. 15.. 209 Horowitz. 260. 275.L.C..J. 346 Harmonic series. 204. 276. G. 336. 345 Gilbert. 291.. 347. 345 Gauss-Jordan elimination. 168. 30... 85 Heapsort. 115. 140. 128. 276. 347 Johnson. 345 Gold. 104. 330-332 Greatest common divisor. 35. 346 Gondran. W. 341 Hudson's Bay. 304-308 Graph colouring. A.C. 55-56.S. 165 Huffman code. 170 insert. 7. 100-104. 185-202 Indentation. 64 Godbole. 315 Isthmus-free. 100 Janko.A. B. 319.E. 167. 42. See Greatest common divisor Generation of permutation.-D. 346 Gries. A. 28 Instance. 167. 20.. 343 Karatsuba. 320. 81-92. M. 79-104. 347 Johnson. 349 Golden ratio.. Nim Gardner.H.S.. 345 Game graph. 230 Hoare. 346 Hashing. 346 Gonnet. 204 Games. 347 Iterated logarithm. 3 Infinite graph. 323. 15. D. T. M. See Numeric integration Internal node.. 346 Harel. 346 Hellman.. 186. 13. 346 goodproduct. 343. 141.R. 140.E. 85. 168. 168. 151. 78. 13. 159. 347 Kannan.. 276. 16.. 222. I.N. M. 337. 291. 21-22. S. 167. 43. 276. 104 Hammersley. J. 267 Gentleman. 276. 336. 168. 115. 36... 337 Inhomogeneous recurrence. 35. 245-247 Heap. 293-298. 204. 53-54.C.. 56..M. 25. 345 Full adder. 167. 104. 140. 332 Hamiltonian circuit. 189 Inherently difficult problem. 275. 276.. G. xiii. 120. M.... 104. 189-198. C... 104 Implicit graph. T. 35. 128 genrand.E... See Robyn Hode Homogeneous recurrence. 6 hitormiss. B.. 7. R. J. 345 Garey. 204. 35. 204. 275.M. 346 Hell. 345 Gleick. 346 Hanoi.. See Towers of Hanoi Hardy. D. 166.H. 347 Jarm'k. 119. 315 Greedy algorithm. 259 gcd. D. D. 35. xiv. 117 Insertion sort. 23 Invariance. See Bicoherent Itai.. D.H.

275. 7. 87. 24 Levin. 346.. 301 equations. 301 greatest common divisor. 275.. A. 195-197. 348 Lewis. 241 Label. 121-124. 168. 248-251 Kranakis. 297 mergesort. 55 Manders. Inverse of a matrix. 348. 227 reduction. 325. 165 Koml6s. 104 k-smooth integer. H. See: Choice of leader Leaf.. 124 division. 19. 275. E. See Canonical object Labyrinth. 272 Memory space. 141. 315 Knight's tour problem. 204. 344 Landis.. 7 Merge circuit. 348 Leader. G. D. 140. 168. 336. 115 Mersenne.R. 345 Matrix. J. 276. 308-315 addition. C. See Equivalence. 315 exponentiation. 140.H. 346 Key for' . 286-290.Index Kerr. D. 346. 117. 238-240. 137. 351 Micali. 184. 140. 78. 104 Knuth. D. K... L. 13-14. 78. 336 L'Hopital's rule. 309 Las Vegas algorithm.. H. 204 Minimization of waiting time. 315 multiplication. 349.. 348 Median. Jr. 344. 269 Majority element. 5.C. 342.. 275. 168.. 33 See also Disjoint sets Merge sort. 36. 300 lisgraph. 271 Kilian. 204. 274. 188. Multiplication of matrices. 228. 344 Lexicographic sorting. 268 maj 2. H. 343. 3 Modular arithmetic. 167. 103. 1-4. 254. L. 120.. Unitary matrix McDiarmid. 345 Klamkin.K. 168. 348 Leclerc.. 35. 35. L... 35. 309. 35. 216. 275. 349 Memory function. 104. 258 k th smallest element. Reduction Marienbad.. E.R. 237 equivalence. 276.A. 268-269 make-heap. 342 Kruskal. E. 275. 215 Knapsack.L. 222. 348 Levy. 293. 308-315 square root. 92-94 Minimum spanning tree. 140. 344 Lenstra.. 341. 347. 341 Many-one. 300 programming. 187. 260. 144. 119. 20-21 Literal. 341. 23 Lecarme. 13-14.. 277 See also Discrete logarithm. 308. S. 308 reductions. 346 Levin. 315 Los Alamos. Determinant of matrix.. 227. 290. 275 algorithm. G. 128. 124-132. Symmetric matrix. M. 138... 167.B. 185. Jr. 349 Miller. 204. 140. 222 Koksoak. 82-85. 140. 128. 343 Lewis. 193. 347 Minimax principle.. See Chained matrix multiplication. 247-262. 336.. K. 19. 128. 276. 351 Level. 203-204 Lanczos. Strassen's algorithm. 348 Lawler. 237. Triangular matrix. 276. F. 0. 168. 162-164. 345 Michie. 104.. 276. 40. 128-131. 348. 346 mod. 202. 140. 347 Kleene. 120-124.N. M. 291. 348 Knuth-Morris-Pratt's algorithm.L. 348. 78 Limit. G. 141.. M. 350 Korsh. 35. 167. 348 Left-hand child. 211 maj. IN. 275 Melhom. 228. 345. 40. 78.E.R. 213.. C. 290. J. 213-215. 343 k-promising. J. 6.H.A. 78 357 Linear: algebra. M. 35.. 349 Meyer. See Nim Marsh.M. 24 Lehmer. 315 squaring. 286-290. G. 336.. P. 341 Konig. 347 KMP. 274.. 139 merge.. 124-128. 35. 91. 276. 29.. A. 323 Lueker. 140. D. 35. 128-132. 92. 15. 323 Metropolis. 276. 348 Martin.. 238-240. G. 13. 341 Large integer arithmetic. 348 Lenstra. 175 Lucas. 115-116. 245 Logarithm. 169 Minoux. 308.E. 224. S. See Atomic research Lower bound theory. 348 Macroprocessor. 336 lowest. Reductions among martix problems...S. J.W. Iterated logarithm Longest simple path. 252-256. 268 majMC. 224. 348 Kruskal's algorithm.. 326 Load factor.. 290.. 290 . 348 Kronrod. 172 List.. 81-87. 333 LauriBre..S..A. 252-253. 22. J-L.J. 274.. 92..

276. 2. 290-291 Pippenger. J. 140. 6. 310 Monier. R.C. 208 Preorder. 349 Moore. 349 Pivot. 332 Pohl. 35. 30. 5. 222. 2-4. 23 Pascal (computer language). 349 Nim. 348 Nemhauser. 124 classic... 348. 349 Pointer. 337. 154-159. 291. 1. Sherwood algorithm. See Theory of numbers Numeric integration: Monte Carlo. 231 Numerical probabilistic algorithm. Unbiased probabilistic algorithm Probabilistic counting. 225. 4 Program. 35. Consistent probabilistic algorithm.. 278. J. 25. 278 prenum. 321. 209-211. N. 180. 337. 79 See also k-promising Proof space.M. 247. 170. 275 trapezoidal. N. 291 interpolation. 227.. 281. 79-80 obstinate.-L.. 316 Polynomial arithmetic: division. 276. 333 Problem. 146-150.. 228. 308-314 Polynomial.. Y. 208 Pratt. See Principle of optimality Order of (0). 51. 332-335. 35. 13-14. D. 240-242. 166. 256. See Generation of Permutation. 291. 153. Monte Carlo algorithm. 336. 275 Objective function. R. 248 Ofman. 224. 334 Principal root of unity. 27. 256. 330-332 Optimal search tree.. Reduction Pomerance. S. Polynomial arithmetic a la russe. 344. 205-211. 204..190-194 Node. 36. 38 Principle of optimality. 102. 140. 349 Monte Carlo algorithm. 349 Polynomial algorithm. 124-132. 215. V. 35. C. 278 of matrices..J. 332-335 NP-complete. 274. 204. 106. 320 . 275.. 207 Optimality. 347 Newton's binomial. 193. 21 node. 10 Newton's method. 348. 263. J. 333 Nontrivial factor. 136.. 216. 167-168.R. L. 342. 302-308 Munro. 204.H. 346 Napier..S. 1. 141. P. 28-30. 3. 13-14. 6. 309-314 reductions. 230-232. 222. 284-286. 117 Planar graph. 349 postnum. 170. 133. 336. 349 Papadimitriou. 27. 62 Pattern. 140. See also Large integer arithmetic. 154-159. J. 263 Peralta. 41 One-way equality. 347 Omega (a). J. 232. 3. 167. 204. 334 NP.. 37 P. 136.. Numerical probabilistic algorithm. 276. 124. 276. 2 Programming language. 315. 275 Probability of success. 286. 349 Percolate. 320. 334 See also Certificate of primality primeND. 20. 315-335. 286 multiplication. 237. 322 See also Numeric integration. 275. 143.. 34. 140. 2 Promising. 146 Path compression.. 8-17. 349 parent. 21 Nondecreasing. 205. 336. 209-211. 343 Monic polynomial. 336-337 Number theory. 228-237. 222. 222. 313. 336. 136.Index 358 Monet. 269-271.. 116. 173. 348 Multiplication.. 140. 204. 174. See Equivalence. 232-237. 343. 35. E. 78 Operations on asymptotic notation. 276. 35.I. C.. J. 208 Prim. 132-133. 85-87.C. 281 Principle of invariance. 239-240 pivot. 24 Nebut. 212-222 p-correct probabilistic algorithm. 345 Moore. 6. 103. 26. 144-159. 324-332 NP-completeness. 349 Precomputation. 342. 275 multiple.. V. 298 Priority list. 211-222 Preconditioning. 20 Pollard. Random permutation Pi (it). 211. 120. 228-234. 33. 92 Primality testing. 262-274. 318 Pan. Quasi Monte Carlo Montgomery. 208 Postorder. 167. 9. 35. J. 143. 309-314 evaluation. p-correct probabilistic algorithm. 227. 205 Pascal's triangle. G. 45 Newman. 315 Nievergelt.F. 104. 91 percolate.L. 291. 27 Permutation. See Biased probabilistic algorithm. 336 Non-deterministic algorithm. 343. 349 Nilsson. 199 Probabilistic algorithm. See Eventually nondecreasing function Non-determinism. 277 n-ary-node.H.. 341.. Las Vegas algorithm. 43-45 Optimal graph colouring. 286. 349 Nested asymptotic notation. 343 Morris. 349 Prim's algorithm.

349 Rabin's algorithm. 277 rootLV. 204. 207 Sedgewick. 8. 247. 139-140 Schonhage... 140.. 170-171. 239 Pseudoprime. 260-262 Rivest. 80 selectionRH. 273 Range transformation. 347.E. 260 Quasi Monte Carlo. 226. 350 Sahni. 24 359 Ring of processors. 347 Roman figures. 140. 276. 275. 65-78 return. 320 Pruned. 238-247. See Alpha-beta pruning pseudomed. 240. 290. See Decision tree Pruning. L. 350 Rubik's Cube. 275 Queens. 79. M.. 226. 250 QueensLV. S. 276. 347. 274 Shing. A. Jr. 154-159. 227. 35. 34. 348 Reflexive transitive closure. 171-184 in a sorted array. 318 polynomial. 320 Pseudorandom generation. 120 Selection function.. 25.. 275 Pseudorandom walk. 44 Set equality. 13. 276. M. 78. 291... M. 254 rootLV2. 167. 290 Schrage. 255 Root of a tree. 7. 15. 276 Shakespeare. J.R. 23-25 Rosenthal. 308-309.. 204. 276. See Satisfiability Satisfiability.. 342 Robson. 272 Random permutation. 302-304 NP-complete problems. 271-273. G. 275. J. 235 Shallit. 6. 35. 350 Rytter. R. 347 . I. J. 350 Seed of a pseudorandom generator. 60-63.. Inhomogeneous recurrence. B. 309-314 Reed. 349 Quadratic: algorithm. 252. 242-245 record. See Asymptotic recurrence.R. 224. 290. E. 290. See coprime repeatMC. 35.. 110 series. 342 Scaling factor.M. 222.. 257 sieve. 19. 336 Saxe. 54. Resolution of recurrences Recursivity. 275. 153. 234 Purdom P. 317 Reductions among: arithmetic problems.. 350 Schwartz. 75 Rank. C. 78. L. J. 304-308 matrix problems. 242-245 in a tree. 275. 122 Pseudomedian. 249 Queue.W.S. 350 Search tree. 204. 316. 341 Runge.. 104. See Binary search in a sorted list. 154-159.A. 350 Schonhage-Strassen algorithm. 199. 122-124. A. 116-119. 207 Searching: for a string. 342 Shamir. 325. 266 Resolution of recurrences. 261 select.. 7 Selection. 304-308 Regular expression. 227 See also Binomial queue Quicksort. 350 Robert. 343 Schwartz. 276. R. 23 Rooted tree. W.L.S. 276. 337.. 188. 35. 314-315 graph problems.O. 106 Reduction: linear... 211-222 in a graph. 342 Shanks.. 146 Set difference. 349 Relatively prime. 291. 291. 300 many-one. 350 Rumely.-M.. 78. AT. 317.. 295-299 Sequential search... 270. 140. 239 Selection sort. 182. See kth smallest element selection. 92-100. 293 Random function. W. 350 Robyn Hode. 293-295 quicksort. 3 RH..B. 270. 345 SAT.Index Proof system. 241 Right-hand child. 276 Rabiner. 232. 52. 350 Sherman. Homogeneous recurrence. 141. 189 Rudin. 301 nonresidue. 347. 13. 343. 318 sense of Turing.. 350 Shamos. 167 Reingold. 104. E. A.. 3 Recurrence. 252 residue. 167.. J.M. 228. R. D. 347 Sherwood algorithm. 336. 228 Rodeh. 275. 326-332. 238. 117 Rabin. 25. 350 Sande. 349 Radix sort. 311-313 Scheduling. 36.. 315-335 polynomial problems. 115.

R. 37. 153. 350 Statistical test.. 107.. 102. 276. D. 139-140. 290. 141. 342 Steiglitz. xiv. 35. Decision tree.. 30. R. J. 159-162. Rooted tree. 69. 55-56. 204. 350 Stone. 153 Shub. 213-222 Tarjan. 144. 351 See also Equivalence. Minimum spanning tree. 336 Telephone switching. D. 293. 347 Target string. 179-182 Strongly quadratic. 46. See Telephone switching Symbolic manipulation of polynomials.. K.R.. 167. 35. 23 Sift-down. 87. 56 Smooth algorithm.. 36. 204. 320 Strongly connected.. 39. countsort. See also Large integer arithmetic modulo n..A. See Decision tree TRS-80. 137. 252-256 Stack. 319 See also Euclidean travelling salesperson Traversal of tree. See Searching Strong pseudoprime. 91. 150 Sparse graph. Reduction Turing machine. 275 Sink... 104. 91 Switch.M. 5 Sloane. Binary tree. 168. 233 Stochastic advantage. 88-90 Splitting. 86. I. Selection sort. 22 See also Ancestry in a rooted tree. 19. 276 Threshold for asymptotic notation. 28. 132. V. 349. 140.. 302 Trip. 275 Steele. Comparison sort.. 78. 277 sort. See Numeric integration Traub. 227. 301 Supra quadratic. 327 Turk. Insertion sort. 276. 103. 344 Turing.F... 276 Square root. 277-291 function. 106 determination. 351 2-edge-connected. 167. 342 Standish. 109-115. See Inorder. 211 Theory of numbers. 350. 168. 19.. 132-133. 301 Smooth integer. See Percolate Signal processing. 140. 277 Trapezoidal algorithm.W.. 28-30. Quicksort. 257 modulo p. Preorder. 337. 140. M. 351 Travelling salesperson.. 342 shuffle. 275. 35.M. Optimal search tree. 325. 87. 123 Tijdeman. 270. A. 336 Text editor. Searching Tree.M. Heapsort. H. 351 Timetable. 78 Transformation: of the domain. 103 Triangular matrix. 348. 199-202. 128-132. 349 Stinson. 35 . See Polynomial arithmetic Symmetric matrix. 211-213. E. 275. 151 Special path. Batcher's sorting circuit. 304 Syntactic analysis. See k-smooth integer Smooth problem. 226. 28. Postorder. 315. 290. 301 tablist. 121. 178-179. 23 Triangle inequality. 308 String. 87. 301 Sobol'. 87-92. 142 Topological sorting.J. 153. T. 21. 137 Tamura. 336 Transformed domain. Balanced tree. 277 Transformed function. 203. 291. 345. 182. 141. Y. 103-104. Transformation sort Source. 291. 346. See Amplification of stochastic advantage Stochastic preconditioning. 144-146 Towers of Hanoi. 305. 315 Simplex. 341 sift-down. 350 Stirling's formula. 35. 168. 279. 27. R.M. 336. 211 Text processing.Index 360 Shortest path. 301 Smooth function. 140. Radix sort. 138 Size of instance. 240-242 Stockmeyer. 291. 25. 203 Size of circuit.J. 343. 349. 27 sift-up. L. 64. J. 139-140 Top-down technique. 350 Strassen's algorithm. Topological sorting.F. 204.W. 20 Tally circuit. 106. 150-153. 256. 108. 350 Sort. See Ad hoc sorting. 227 Simplification. 345. 351 Tautology. 241 Sibling. 107-109 ultimate. 15.S. N.. 104. 350 Solovay. J.. 43 Threshold for divide-and-conquer. 167. 205 Szemeredi. 304-308 Shortest simple path. 74. 280 Signature. 350 slow-make-heap. See Bicoherent 2-3 tree. 134-136 Simula. 35. Searching in a tree. 273 Simple path. J.E. 275. Lexicographic sorting. Traversal of tree treenode.A. 35. 22.. 350 Strassen. 275. 326-327. 205 Simulation. 337. 35 Tukey. 199 Stanat. 222. 30. 211. 190 Tournament.

S. 168. S. 168..A. U. P. 291. H. 351 Wirth. 140. C. 144-146. 291. 167. 274.W.J. 351 Warshall. A. 168. 276. 245-247. 304 Universal hashing.M. 337 Undirected graph.. 348 wordcnt. Y.. R. 228 White.. 104. 272. 222. 78.. 342 Urbanek. 276. 153. 351 Zuffellato. N. 204. D. 331 Williams. 167. 275.. 225 Unit cost. 351 Zippel. 133.W.W.N. 275. R.. 343.H. 167. 104.. 35. 196 Widget. 140.F. 290. 276.. 164.D. 276. 351 Virtual initialization. D.. 347 Valid. 351 VAX. 276 Undecidable problem.J. 343 .. 8. See Biconnected Unbiased probabilistic algorithm. S.. 351 Yao. 56. D. Heapsort Wilson's theorem.. 236 World Series. 13..E. J. 337..V. 341.C. J. 78. 351 Williams.. 275. 168. 10 Winograd. 171-176 Uniform random generation. 351 361 Warusfel.. See Elementary operation Unitary matrix. 344. 9. 351 Ushiro. 36. 317. 275. 291. 35.W. 21. 13 Vazirani. E.. 275-276 Unpredictable pseudorandom generator.. 351 Yoshino. 275. 344 Well-characterized problem.. A. 273 Wagner. 204. 349 Ullman.Index Ulam.D. J. 263 Worst case. 351 Welch.. 291. 337. 168. 351 Verdict. F. 347 Younger.. 347 Wood. 35. See Decision tree Vickery. 351 Wegman. 346 Ultimate. D. 193. See Analysis of algorithms Wright. 35.. 346 Wright.. M.. 351 Warshall's algorithm. See Decision tree Valois. See Threshold for divide-and-conquer Unarticulated. 351 See also Heap. 332. 276. S. 141. 275. 36. 308 Yao. F. 336.

numerical analysis. However. artificial intelligence. cryptography. Presents real-life applications for most algorithms. PRENTICE HALL.11fl Mcpo)[PT TIPMT0T( GILLES BRASSARD PAUL BRATLEY The computer explosion has made calculations once thought impossible seem routine. Details each technique in full. computing in the humanities. among others. Now. NJ 07632 ISBN 0-13-023243-2 . Englewood Cliffs. symbolic computation. this innovative new book gives readers the basic tools they need to develop their own algorithms--in whatever field of application they may be required! CONTENT HIGHLIGHTS: Concentrates on the techniques needed to design and analyze algorithms. Contains approximately 500 exercises-many of which call for an algorithm to be implemented on a computer so that its efficiency may be measured experimentally and compared to the efficiency of alternative solutions. operations research. linear algebra. Illustrates each technique with concrete examples of algorithms taken from such different applications as optimization. another factor has had an even more important effect in extending the frontiers of feasible computation: the use of efficient algorithms.