The Haskell Road to Logic, Math and Programming Kees Doets and Jan van Eijck March 4, 2004 Contents Preface 1 Getting Started Implementing a Prime Number Test . . . . . . 1.1 Starting up the Haskell Interpreter . . . . . 1.2 . . . . . . . . 1.3 Haskell Type Declarations . . . . . . . . 1.4 Identifiers in Haskell . . . . . . . . 1.5 Playing the Haskell Game . . . . . . . . 1.6 Haskell Types . . . . . 1.7 The Prime Factorization Algorithm . . 1.8 The map and filter Functions . . . . . . . . 1.9 Haskell Equations and Equational Reasoning . . . . . . . 1.10 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Talking about Mathematical Objects 2.1 Logical Connectives and their Meanings . . . . . 2.2 Logical Validity and Related Notions . . . . . . . 2.3 Making Symbolic Form Explicit . . . 2.4 Lambda Abstraction . . . . . . 2.5 Definitions and Implementations . . . . . . 2.6 Abstract Formulas and Concrete Structures . . . 2.7 Logical Handling of the Quantifiers . . . 2.8 Quantifiers as Procedures . . . . . . . . . . 2.9 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 The Use of Logic: Proof . . . . 3.1 Proof Style . . 3.2 Proof Recipes . . . . . 3.3 Rules for the Connectives . . . . . . 3.4 Rules for the Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i v 1 2 3 8 11 12 17 19 20 24 26 27 28 38 50 58 60 61 64 68 70 71 72 75 78 90 ii . . . 3.5 Summary of the Proof Recipes . . . . 3.6 Some Strategic Guidelines . . . . . . . 3.7 Reasoning and Computation with Primes . . . . . 3.8 Further Reading . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 . . . . . . . . 99 . . . . 103 . . . . 111 4 Sets, Types and Lists . . . . . . . 4.1 Let’s Talk About Sets . . . . . . . . 4.2 Paradoxes, Types and Type Classes . . . . . . . . . . . . . 4.3 Special Sets . . . . . . . . . . . . . . . . . 4.4 Algebra of Sets . . . . . . . . . . . 4.5 Pairs and Products . . . . . . . 4.6 Lists and List Operations . . . . . . . . . . 4.7 List Comprehension and Database Query . . . . 4.8 Using Lists to Represent Sets . . . . . . . . 4.9 A Data Type for Sets . . . . . . . . 4.10 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Relations . . . . . . . . 5.1 The Notion of a Relation . . . 5.2 Properties of Relations . . . . 5.3 5.4 5.5 Equivalence Relations . . . . . 5.6 Equivalence Classes and Partitions . . . . . . . . . Integer Partitions 5.7 . . . . 5.8 Further Reading . . . . . . . . . . . . . . . . . . . . . . Implementing Relations as Sets of Pairs . . Implementing Relations as Characteristic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Functions . . . . . . . . 6.1 Basic Notions . 6.2 Surjections, Injections, Bijections 6.3 Function Composition . . . . . . . . Inverse Function . . . . 6.4 . . . . 6.5 Partial Functions . . . . . . . 6.6 Functions as Partitions . . . . . . . . 6.7 Products . . . . . 6.8 Congruences . . . . . . . . 6.9 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Induction and Recursion 7.1 Mathematical Induction . . . . . . . . 7.2 Recursion over the Natural Numbers . 7.3 The Nature of Recursive Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 . . . . 114 . . . . 121 . . . . 125 . . . . 127 . . . . 136 . . . . 139 . . . . 145 . . . . 149 . . . . 153 . . . . 158 161 . . . . 162 . . . . 166 . . . . 175 . . . . 182 . . . . 188 . . . . 192 . . . . 202 . . . . 204 205 . . . . 206 . . . . 218 . . . . 222 . . . . 226 . . . . 229 . . . . 232 . . . . 234 . . . . 236 . . . . 238 239 . . . . 239 . . . . 246 . . . . 251 CONTENTS Induction and Recursion over Trees . 7.4 7.5 Induction and Recursion over Lists . . 7.6 Some Variations on the Tower of Hanoi 7.7 7.8 Further Reading . . . . . . . . . . . . . . . . Induction and Recursion over Other Data Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Working with Numbers . . . . . . . . . . . . . . 8.1 A Module for Natural Numbers . . . . 8.2 GCD and the Fundamental Theorem of Arithmetic Integers . . . . . . . . . . . . . 8.3 Implementing Integer Arithmetic . . . . . . . 8.4 . . . . . . . . 8.5 Rational Numbers . . . . . . . Implementing Rational Arithmetic . . 8.6 . . . . . . . . 8.7 . . Irrational Numbers . . . . . . . . 8.8 The Mechanic’s Rule . . . . . . . . . 8.9 Reasoning about Reals . . . . . . . . 8.10 Complex Numbers . . . . . . . . . . 8.11 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Polynomials 9.1 Difference Analysis of Polynomial Sequences . . . . . . . 9.2 Gaussian Elimination . . . . . . . . 9.3 Polynomials and the Binomial Theorem . . . . . . 9.4 Polynomials for Combinatorial Reasoning . . . . . . . . 9.5 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Corecursion . . . . 10.1 Corecursive Definitions . . . . . . . 10.2 Processes and Labeled Transition Systems . . . . . . . . 10.3 Proof by Approximation . . . . . . 10.4 Proof by Coinduction . . . . . . . . . . . 10.5 Power Series and Generating Functions . . . 10.6 Exponential Generating Functions . . . . . . . . . 10.7 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Finite and Infinite Sets 11.1 More on Mathematical Induction . . . . . . . . . . . 11.2 Equipollence . . . . . 11.3 Infinite Sets . . . . . . . . . . 11.4 Cantor’s World Implemented . . . . . . . . . 11.5 Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii . . . . 255 . . . . 265 . . . . 273 . . . . 281 . . . . 284 285 . . . . 286 . . . . 289 . . . . 293 . . . . 297 . . . . 299 . . . . 305 . . . . 309 . . . . 313 . . . . 315 . . . . 319 . . . . 329 331 . . . . 332 . . . . 337 . . . . 344 . . . . 352 . . . . 359 361 . . . . 362 . . . . 365 . . . . 373 . . . . 379 . . . . 385 . . . . 396 . . . . 398 399 . . . . 399 . . . . 406 . . . . 410 . . . . 418 . . . . 420 iv The Greek Alphabet References Index CONTENTS 423 424 428 Preface Purpose Long ago, when Alexander the Great asked the mathematician Menaechmus for a crash course in geometry, he got the famous reply “There is no royal road to mathematics.” Where there was no shortcut for Alexander, there is no shortcut for us. Still, the fact that we have access to computers and mature programming languages means that there are avenues for us that were denied to the kings and emperors of yore. The purpose of this book is to teach logic and mathematical reasoning in practice, and to connect logical reasoning with computer programming. The programming language that will be our tool for this is Haskell, a member of the Lisp family. Haskell emerged in the last decade as a standard for lazy functional programming, a programming style where arguments are evaluated only when the value is actu- ally needed. Functional programming is a form of descriptive programming, very different from the style of programming that you find in prescriptive languages like C or Java. Haskell is based on a logical theory of computable functions called the lambda calculus. Lambda calculus is a formal language capable of expressing arbitrary computable functions. In combination with types it forms a compact way to denote on the one hand functional programs and on the other hand mathematical proofs. [Bar84] Haskell can be viewed as a particularly elegant implementation of the lambda cal- culus. It is a marvelous demonstration tool for logic and math because its func- tional character allows implementations to remain very close to the concepts that get implemented, while the laziness permits smooth handling of infinite data struc- tures. v vi Haskell syntax is easy to learn, and Haskell programs are constructed and tested in a modular fashion. This makes the language well suited for fast prototyping. Programmers find to their surprise that implementation of a well-understood al- gorithm in Haskell usually takes far less time than implementation of the same algorithm in other programming languages. Getting familiar with new algorithms through Haskell is also quite easy. Learning to program in Haskell is learning an extremely useful skill. Throughout the text, abstract concepts are linked to concrete representations in Haskell. Haskell comes with an easy to use interpreter, Hugs. Haskell compilers, interpreters and documentation are freely available from the Internet [HT]. Every- thing one has to know about programming in Haskell to understand the programs in the book is explained as we go along, but we do not cover every aspect of the language. For a further introduction to Haskell we refer the reader to [HFP96]. Logic in Practice The subject of this book is the use of logic in practice, more in particular the use of logic in reasoning about programming tasks. Logic is not taught here as a mathematical discipline per se, but as an aid in the understanding and construction of proofs, and as a tool for reasoning about formal objects like numbers, lists, trees, formulas, and so on. As we go along, we will introduce the concepts and tools that form the set-theoretic basis of mathematics, and demonstrate the role of these concepts and tools in implementations. These implementations can be thought of as representations of the mathematical concepts. Although it may be argued that the logic that is needed for a proper understanding of reasoning in reasoned programming will get acquired more or less automatically in the process of learning (applied) mathematics and/or programming, students nowadays enter university without any experience whatsoever with mathematical proof, the central notion of mathematics. The rules of Chapter 3 represent a detailed account of the structure of a proof. The purpose of this account is to get the student acquainted with proofs by putting em- phasis on logical structure. The student is encouraged to write “detailed” proofs, with every logical move spelled out in full. The next goal is to move on to writing “concise” proofs, in the customary mathematical style, while keeping the logical structure in mind. Once the student has arrived at this stage, most of the logic that is explained in Chapter 3 can safely be forgotten, or better, can safely fade into the subconsciousness of the matured mathematical mind. PREFACE vii Pre- and Postconditions of Use We do not assume that our readers have previous experience with either program- ming or construction of formal proofs. We do assume previous acquaintance with mathematical notation, at the level of secondary school mathematics. Wherever necessary, we will recall relevant facts. Everything one needs to know about math- ematical reasoning or programming is explained as we go along. We do assume that our readers are able to retrieve software from the Internet and install it, and that they know how to use an editor for constructing program texts. After having worked through the material in the book, i.e., after having digested the text and having carried out a substantial number of the exercises, the reader will be able to write interesting programs, reason about their correctness, and doc- ument them in a clear fashion. The reader will also have learned how to set up mathematical proofs in a structured way, and how to read and digest mathematical proofs written by others. How to Use the Book Chapters 1–7 of the book are devoted to a gradual introduction of the concepts, tools and methods of mathematical reasoning and reasoned programming. Chapter 8 tells the story of how the various number systems (natural numbers, integers, rationals, reals, complex numbers) can be thought of as constructed in stages from the natural numbers. Everything gets linked to the implementations of the various Haskell types for numerical computation. Chapter 9 starts with the question of how to automate the task of finding closed forms for polynomial sequences. It is demonstrated how this task can be automated with difference analysis plus Gaussian elimination. Next, polynomials are imple- mented as lists of their coefficients, with the appropriate numerical operations, and it is shown how this representation can be used for solving combinatorial problems. Chapter 10 provides the first general textbook treatment (as far as we know) of the important topic of corecursion. The chapter presents the proof methods suitable for reasoning about corecursive data types like streams and processes, and then goes on to introduce power series as infinite lists of coefficients, and to demonstrate the uses of this representation for handling combinatorial problems. This generalizes the use of polynomials for combinatorics. Chapter 11 offers a guided tour through Cantor’s paradise of the infinite, while providing extra challenges in the form of a wide range of additional exercises. viii The book can be used as a course textbook, but since it comes with solutions to all exercises (electronically available from the authors upon request) it is also well suited for private study. Courses based on the book could start with Chapters 1–7, and then make a choice from the remaining Chapters. Here are some examples: Road to Numerical Computation Chapters 1–7, followed by 8 and 9. Road to Streams and Corecursion Chapters 1–7, followed by 9 and 10. Road to Cantor’s Paradise Chapters 1–7, followed by 11. Study of the remaining parts of the book can then be set as individual tasks for students ready for an extra challenge. The guidelines for setting up formal proofs in Chapter 3 should be recalled from time to time while studying the book, for proper digestion. Exercises Parts of the text and exercises marked by a * are somewhat harder than the rest of the book. All exercises are solved in the electronically avaible solutions volume. Before turning to these solutions, one should read the Important Advice to the Reader that this volume starts with. Book Website and Contact The programs in this book have all been tested with Hugs98, the version of Hugs that implements the Haskell 98 standard. The full source code of all programs is integrated in the book; in fact, each chapter can be viewed as a literate program [Knu92] in Haskell. The source code of all programs discussed in the text can be found on the website devoted to this book, at address http://www.cwi.nl/ ~jve/HR. Here you can also find a list of errata, and further relevant material. Readers who want to share their comments with the authors are encouraged to get in touch with us at email address jve@cwi.nl. PREFACE Acknowledgments ix Remarks from the people listed below have sparked off numerous improvements. Thanks to Johan van Benthem, Jan Bergstra, Jacob Brunekreef, Thierry Coquand (who found the lecture notes on the internet and sent us his comments), Tim van Erven, Wan Fokkink, Evan Goris, Robbert de Haan, Sandor Heman, Eva Hoog- land, Rosalie Iemhoff, Dick de Jongh, Anne Kaldewaij, Breannd´an ´O Nuall´ain, Alban Ponse, Vincent van Oostrom, Piet Rodenburg, Jan Rutten, Marco Swaen, Jan Terlouw, John Tromp, Yde Venema, Albert Visser and Stephanie Wehner for suggestions and criticisms. The beautiful implementation of the sieve of Eratos- thenes in Section 3.7 was suggested to us by Fer-Jan de Vries. The course on which this book is based was developed at ILLC (the Institute of Logic, Language and Computation of the University of Amsterdam) with finan- cial support from the Spinoza Logic in Action initiative of Johan van Benthem, which is herewith gratefully acknowledged. We also wish to thank ILLC and CWI (Centrum voor Wiskunde en Informatica, or Centre for Mathematics and Com- puter Science, also in Amsterdam), the home institute of the second author, for providing us with a supportive working environment. CWI has kindly granted permission to reuse material from [Doe96]. It was Krzysztof Apt who, perceiving the need of a deadline, spurred us on to get in touch with a publisher and put ourselves under contract. x Chapter 1 Getting Started Preview Our purpose is to teach logic and mathematical reasoning in practice, and to con- nect formal reasoning to computer programming. It is convenient to choose a programming language for this that permits implementations to remain as close as possible to the formal definitions. Such a language is the functional programming language Haskell [HT]. Haskell was named after the logician Haskell B. Curry. Curry, together with Alonzo Church, laid the foundations of functional computa- tion in the era Before the Computer, around 1940. As a functional programming language, Haskell is a member of the Lisp family. Others family members are Scheme, ML, Occam, Clean. Haskell98 is intended as a standard for lazy func- tional programming. Lazy functional programming is a programming style where arguments are evaluated only when the value is actually needed. With Haskell, the step from formal definition to program is particularly easy. This presupposes, of course, that you are at ease with formal definitions. Our reason for combining training in reasoning with an introduction to functional programming is that your programming needs will provide motivation for improving your reason- ing skills. Haskell programs will be used as illustrations for the theory throughout the book. We will always put computer programs and pseudo-code of algorithms in frames (rectangular boxes). The chapters of this book are written in so-called ‘literate programming’ style [Knu92]. Literate programming is a programming style where the program and its documentation are generated from the same source. The text of every chapter in 1 2 CHAPTER1. GETTINGSTARTED this book can be viewed as the documentation of the program code in that chapter. Literate programming makes it impossible for program and documentation to get out of sync. Program documentation is an integrated part of literate programming, in fact the bulk of a literate program is the program documentation. When writ- ing programs in literate style there is less temptation to write program code first while leaving the documentation for later. Programming in literate style proceeds from the assumption that the main challenge when programming is to make your program digestible for humans. For a program to be useful, it should be easy for others to understand the code. It should also be easy for you to understand your own code when you reread your stuff the next day or the next week or the next month and try to figure out what you were up to when you wrote your program. To save you the trouble of retyping, the code discussed in this book can be retrieved from the book website. The program code is the text in typewriter font that you find in rectangular boxes throughout the chapters. Boxes may also contain code that is not included in the chapter modules, usually because it defines functions that are already predefined by the Haskell system, or because it redefines a function that is already defined elsewhere in the chapter. Typewriter font is also used for pieces of interaction with the Haskell interpreter, but these illustrations of how the interpreter behaves when particular files are loaded and commands are given are not boxed. Every chapter of this book is a so-called Haskell module. The following two lines declare the Haskell module for the Haskell code of the present chapter. This mod- ule is called GS. module GS where 1.1 Starting up the Haskell Interpreter We assume that you succeeded in retrieving the Haskell interpreter hugs from the Haskell homepage www.haskell.org and that you managed to install it on your computer. You can start the interpreter by typing hugs at the system prompt. When you start hugs you should see something like Figure (1.1). The string Prelude> on the last line is the Haskell prompt when no user-defined files are loaded. You can use hugs as a calculator as follows: 1.2. IMPLEMENTINGAPRIMENUMBERTEST 3 ____ __ || || __ __ || || ___ || ||__ __ || ||___|| ||__|| ||__|| ||---|| ___|| || || || Version: November 2003 || __|| _________________________________________ Hugs 98: Based on the Haskell 98 standard Copyright (c) 1994-2003 World Wide Web: http://haskell.org/hugs Report bugs to: hugs-bugs@haskell.org _________________________________________ Haskell 98 mode: Restart with command line option -98 to enable extensions Type :? for help Prelude> Prelude> 2^16 65536 Prelude> Figure 1.1: Starting up the Haskell interpreter. The string Prelude> is the system prompt. 2^16 is what you type. After you hit the return key (the key that is often labeled with Enter or -), the system answers 65536 and the prompt Prelude> reappears. Exercise 1.1 Try out a few calculations using * for multiplication, + for addition, - for subtraction, ^ for exponentiation, / for division. By playing with the system, find out what the precedence order is among these operators. Parentheses can be used to override the built-in operator precedences: Prelude> (2 + 3)^4 625 To quit the Hugs interpreter, type :quit or :q at the system prompt. 1.2 Implementing a Prime Number Test Suppose we want to implement a definition of prime number in a procedure that recognizes prime numbers. A prime number is a natural number greater than 1 that has no proper divisors other than 1 and itself. The natural numbers are 0; 1; 2; 3; 4; : : : The list of prime numbers starts with 2; 3; 5; 7; 11; 13; : : : Except for 2, all of these are odd, of course. 4 CHAPTER1. GETTINGSTARTED Let n > 1 be a natural number. Then we use LD(n) for the least natural number greater than 1 that divides n. A number d divides n if there is a natural number a with a d = n. In other words, d divides n if there is a natural number a with n d = a, i.e., division of n by d leaves no remainder. Note that LD(n) exists for every natural number n > 1, for the natural number d = n is greater than 1 and divides n. Therefore, the set of divisors of n that are greater than 1 is non-empty. Thus, the set will have a least element. (cid:1) The following proposition gives us all we need for implementing our prime number test: Proposition 1.2 1. If n > 1 then LD(n) is a prime number. 2. If n > 1 and n is not a prime number, then (LD(n))2 6 n. In the course of this book you will learn how to prove propositions like this. Here is the proof of the first item. This is a proof by contradiction (see Chapter 3). Suppose, for a contradiction that c = LD(n) is not a prime. Then there are natural b, and also 1 < a and a < c. But then a divides n, numbers a and b with c = a and contradiction with the fact that c is the smallest natural number greater than 1 that divides n. Thus, LD(n) must be a prime number. (cid:1) (cid:1) in a For a proof of the second item, suppose that n > 1, n is not a prime and that a. Thus, a p = LD(n). Then there is a natural number a > 1 with n = p divides n. Since p is the smallest divisor of n with p > 1, we have that p 6 a, and therefore p2 6 p a = n, i.e., (LD(n))2 6 n. (cid:1) The operator b is a so-called infix operator. The operator is written between its arguments. If an operator is written before its arguments we call this prefix notation. The product of a and b in prefix notation would look like this: a b. (cid:1) (cid:1) (cid:1) In writing functional programs, the standard is prefix notation. In an expression op a b, op is the function, and a and b are the arguments. The convention is that function application associates to the left, so the expression op a b is interpreted as (op a) b. Using prefix notation, we define the operation divides that takes two integer expressions and produces a truth value. The truth values true and false are rendered in Haskell as True and False, respectively. The integer expressions that the procedure needs to work with are called the argu- ments of the procedure. The truth value that it produces is called the value of the procedure. 1.2. IMPLEMENTINGAPRIMENUMBERTEST 5 Obviously, m divides n if and only if the remainder of the process of dividing n by m equals 0. The definition of divides can therefore be phrased in terms of a predefined procedure rem for finding the remainder of a division process: divides d n = rem n d == 0 The definition illustrates that Haskell uses = for ‘is defined as’ and == for identity. (The Haskell symbol for non-identity is /=.) A line of Haskell code of the form foo t = ... (or foo t1 t2 = ..., or foo t1 t2 t3 = ..., and so on) is called a Haskell equation. In such an equa- tion, foo is called the function, and t its argument. Thus, in the Haskell equation divides d n = rem n d == 0, divides is the function, d is the first argument, and n is the second argument. Exercise 1.3 Put the definition of divides in a file prime.hs. Start the Haskell in- terpreter hugs (Section 1.1). Now give the command :load prime or :l prime, followed by pressing Enter. Note that l is the letter l, not the digit 1. (Next to :l, a very useful command after you have edited a file of Haskell code is :reload or :r, for reloading the file.) Prelude> :l prime Main> The string Main> is the Haskell prompt indicating that user-defined files are loaded. This is a sign that the definition was added to the system. The newly defined oper- ation can now be executed, as follows: Main> divides 5 7 False Main> The string Main> is the Haskell prompt, the rest of the first line is what you type. When you press Enter the system answers with the second line, followed by the Haskell prompt. You can then continue with: Main> divides 5 30 True 6 CHAPTER1. GETTINGSTARTED It is clear from the proposition above that all we have to do to implement a primal- ity test is to give an implementation of the function LD. It is convenient to define LD in terms of a second function LDF, for the least divisor starting from a given threshold k, with k 6 n. Thus, LDF(k)(n) is the least divisor of n that is > k. Clearly, LD(n) = LDF(2)(n). Now we can implement LD as follows: ld n = ldf 2 n This leaves the implementation ldf of LDF (details of the coding will be explained below): ldf k n | divides k n = k = n = ldf (k+1) n | k^2 > | otherwise n The definition employs the Haskell operation ^ for exponentiation, > for ‘greater than’, and + for addition. The definition of ldf makes use of equation guarding. The first line of the ldf definition handles the case where the first argument divides the second argument. Every next line assumes that the previous lines do not apply. The second line handles the case where the first argument does not divide the second argument, and the square of the first argument is greater than the second argument. The third line assumes that the first and second cases do not apply and handles all other cases, i.e., the cases where k does not divide n and k2 < n. The definition employs the Haskell condition operator | . A Haskell equation of the form foo t | condition = ... is called a guarded equation. We might have written the definition of ldf as a list of guarded equations, as follows: 1.2. IMPLEMENTINGAPRIMENUMBERTEST 7 ldf k n | divides k n = k = n ldf k n | k^2 > = ldf (k+1) n ldf k n n The expression condition, of type Bool (i.e., Boolean or truth value), is called the guard of the equation. A list of guarded equations such as foo t | condition_1 = body_1 foo t | condition_2 = body_2 foo t | condition_3 = body_3 = body_4 foo t can be abbreviated as foo t | condition_1 = body_1 | condition_2 = body_2 | condition_3 = body_3 = body_4 | otherwise Such a Haskell definition is read as follows: (cid:15) (cid:15) (cid:15) (cid:15) in case condition_1 holds, foo t is by definition equal to body_1, in case condition_1 does not hold but condition_2 holds, foo t is by definition equal to body_2, in case condition_1 and condition_2 do not hold but condition_3 holds, foo t is by definition equal to body_3, and in case none of condition_1, condition_2 and condition_3 hold, foo t is by definition equal to body_4. When we are at the end of the list we know that none of the cases above in the list apply. This is indicated by means of the Haskell reserved keyword otherwise. Note that the procedure ldf is called again from the body of its own definition. We will encounter such recursive procedure definitions again and again in the course of this book (see in particular Chapter 7). 8 CHAPTER1. GETTINGSTARTED Exercise 1.4 Suppose in the definition of ldf we replace the condition k^2 > n by k^2 >= n, where >= expresses ‘greater than or equal’. Would that make any difference to the meaning of the program? Why (not)? Now we are ready for a definition of prime0, our first implementation of the test for being a prime number. prime0 n | n < 1 = error "not a positive integer" = False | n == 1 | otherwise = ld n == n Haskell allows a call to the error operation in any definition. This is used to break off operation and issue an appropriate message when the primality test is used for numbers below 1. Note that error has a parameter of type String (indicated by the double quotes). The definition employs the Haskell operation < for ‘less than’. Intuitively, what the definition prime0 says is this: 1. the primality test should not be applied to numbers below 1, 2. if the test is applied to the number 1 it yields ‘false’, 3. if it is applied to an integer n greater than 1 it boils down to checking that LD(n) = n. In view of the proposition we proved above, this is indeed a correct primality test. Exercise 1.5 Add these definitions to the file prime.hs and try them out. Remark. The use of variables in functional programming has much in common with the use of variables in logic. The definition divides d n = rem n d == 0 is equivalent to divides x y = rem y x == 0. This is because the variables denote arbitrary elements of the type over which they range. They behave like universally quantified variables, and just as in logic the definition does not depend on the variable names. 1.3 Haskell Type Declarations Haskell has a concise way to indicate that divides consumes an integer, then another integer, and produces a truth value (called Bool in Haskell). Integers and 1.3. HASKELLTYPEDECLARATIONS 9 truth values are examples of types. See Section 2.1 for more on the type Bool. Section 1.6 gives more information about types in general. Arbitrary precision integers in Haskell have type Integer. The following line gives a so-called type declaration for the divides function. divides :: Integer -> Integer -> Bool Integer -> Integer -> Bool is short for Integer -> (Integer -> Bool). A type of the form a -> b classifies a procedure that takes an argument of type a to produce a result of type b. Thus, divides takes an argument of type Integer and produces a result of type Integer -> Bool, i.e., a procedure that takes an argument of type Integer, and produces a result of type Bool. The full code for divides, including the type declaration, looks like this: divides :: Integer -> Integer -> Bool divides d n = rem n d == 0 If d is an expression of type Integer, then divides d is an expression of type Integer -> Bool. The shorthand that we will use for d is an expression of type Integer is: d :: Integer. Exercise 1.6 Can you gather from the definition of divides what the type decla- ration for rem would look like? Exercise 1.7 The hugs system has a command for checking the types of expres- sions. Can you explain the following (please try it out; make sure that the file with the definition of divides is loaded, together with the type declaration for divides): Main> :t divides 5 divides 5 :: Integer -> Bool Main> :t divides 5 7 divides 5 7 :: Bool Main> 10 CHAPTER1. GETTINGSTARTED The expression divides 5 :: Integer -> Bool is called a type judgment. Type judgments in Haskell have the form expression :: type. In Haskell it is not strictly necessary to give explicit type declarations. For in- stance, the definition of divides works quite well without the type declaration, since the system can infer the type from the definition. However, it is good pro- gramming practice to give explicit type declarations even when this is not strictly necessary. These type declarations are an aid to understanding, and they greatly improve the digestibility of functional programs for human readers. A further advantage of the explicit type declarations is that they facilitate detection of pro- gramming mistakes on the basis of type errors generated by the interpreter. You will find that many programming errors already come to light when your program gets loaded. The fact that your program is well typed does not entail that it is correct, of course, but many incorrect programs do have typing mistakes. The full code for ld, including the type declaration, looks like this: ld :: Integer -> Integer ld n = ldf 2 n The full code for ldf, including the type declaration, looks like this: ldf :: Integer -> Integer -> Integer ldf k n | divides k n = k = n = ldf (k+1) n | k^2 > | otherwise n The first line of the code states that the operation ldf takes two integers and pro- duces an integer. The full code for prime0, including the type declaration, runs like this: 1.4. IDENTIFIERSINHASKELL 11 prime0 :: Integer -> Bool prime0 n | n < 1 = error "not a positive integer" = False | n == 1 | otherwise = ld n == n The first line of the code declares that the operation prime0 takes an integer and produces (or returns, as programmers like to say) a Boolean (truth value). In programming generally, it is useful to keep close track of the nature of the objects that are being represented. This is because representations have to be stored in computer memory, and one has to know how much space to allocate for this storage. Still, there is no need to always specify the nature of each data-type explicitly. It turns out that much information about the nature of an object can be inferred from how the object is handled in a particular program, or in other words, from the operations that are performed on that object. Take again the definition of divides. It is clear from the definition that an oper- ation is defined with two arguments, both of which are of a type for which rem is defined, and with a result of type Bool (for rem n d == 0 is a statement that can turn out true or false). If we check the type of the built-in procedure rem we get: Prelude> :t rem rem :: Integral a => a -> a -> a In this particular case, the type judgment gives a type scheme rather than a type. It means: if a is a type of class Integral, then rem is of type a -> a -> a. Here a is used as a variable ranging over types. In Haskell, Integral is the class (see Section 4.2) consisting of the two types for integer numbers, Int and Integer. The difference between Int and Integer is that objects of type Int have fixed precision, objects of type Integer have arbitrary precision. The type of divides can now be inferred from the definition. This is what we get when we load the definition of divides without the type declaration: Main> :t divides divides :: Integral a => a -> a -> Bool 1.4 Identifiers in Haskell In Haskell, there are two kinds of identifiers: 12 (cid:15) (cid:15) CHAPTER1. GETTINGSTARTED Variable identifiers are used to name functions. They have to start with a lower-case letter. E.g., map, max, fct2list, fctToList, fct_to_list. Constructor identifiers are used to name types. They have to start with an upper-case letter. Examples are True, False. Functions are operations on data-structures, constructors are the building blocks of the data structures themselves (trees, lists, Booleans, and so on). Names of functions always start with lower-case letters, and may contain both upper- and lower-case letters, but also digits, underscores and the prime symbol ’. The following reserved keywords have special meanings and cannot be used to name functions. case if let class import module data in newtype default infix of deriving infixl then do infixr type else instance where The use of these keywords will be explained as we encounter them. at the begin- all ning of a word is treated as a lower-case character. The underscore character by itself is a reserved word for the wild card pattern that matches anything (page 141). There is one more reserved keyword that is particular to Hugs: forall, for the defi- nition of functions that take polymorphic arguments. See the Hugs documentation for further particulars. 1.5 Playing the Haskell Game This section consists of a number of further examples and exercises to get you acquainted with the programming language of this book. To save you the trouble of keying in the programs below, you should retrieve the module GS.hs for the present chapter from the book website and load it in hugs. This will give you a system prompt GS>, indicating that all the programs from this chapter are loaded. In the next example, we use Int for the type of fixed precision integers, and [Int] for lists of fixed precision integers. Example 1.8 Here is a function that gives the minimum of a list of integers: 1.5. PLAYINGTHEHASKELLGAME 13 mnmInt :: [Int] -> Int mnmInt [] = error "empty list" mnmInt [x] = x mnmInt (x:xs) = min x (mnmInt xs) This uses the predefined function min for the minimum of two integers. It also uses pattern matching for lists . The list pattern [] matches only the empty list, the list pattern [x] matches any singleton list, the list pattern (x:xs) matches any non-empty list. A further subtlety is that pattern matching in Haskell is sensitive to order. If the pattern [x] is found before (x:xs) then (x:xs) matches any non-empty list that is not a unit list. See Section 4.6 for more information on list pattern matching. It is common Haskell practice to refer to non-empty lists as x:xs, y:ys, and so on, as a useful reminder of the facts that x is an element of a list of x’s and that xs is a list. Here is a home-made version of min: min’ :: Int -> Int -> Int min’ x y | x <= y = x | otherwise = y You will have guessed that <= is Haskell code for 6. Objects of type Int are fixed precision integers. Their range can be found with: Prelude> primMinInt -2147483648 Prelude> primMaxInt 2147483647 Since 2147483647 = 231 1, we can conclude that the hugs implementation uses four bytes (32 bits) to represent objects of this type. Integer is for arbitrary pre- cision integers: the storage space that gets allocated for Integer objects depends on the size of the object. (cid:0) Exercise 1.9 Define a function that gives the maximum of a list of integers. Use the predefined function max. 14 CHAPTER1. GETTINGSTARTED Conversion from Prefix to Infix in Haskell A function can be converted to an infix operator by putting its name in back quotes, like this: Prelude> max 4 5 5 Prelude> 4 ‘max‘ 5 5 Conversely, an infix operator is converted to prefix by putting the operator in round brackets (p. 21). Exercise 1.10 Define a function removeFst that removes the first occurrence of an integer m from a list of integers. If m does not occur in the list, the list remains unchanged. Example 1.11 We define a function that sorts a list of integers in order of increas- ing size, by means of the following algorithm: (cid:15) (cid:15) an empty list is already sorted. if a list is non-empty, we put its minimum in front of the result of sorting the list that results from removing its minimum. This is implemented as follows: srtInts :: [Int] -> [Int] srtInts [] = [] srtInts xs = m : (srtInts (removeFst m xs)) where m = mnmInt xs Here removeFst is the function you defined in Exercise 1.10. Note that the second clause is invoked when the first one does not apply, i.e., when the argument of srtInts is not empty. This ensures that mnmInt xs never gives rise to an error. Note the use of a where construction for the local definition of an auxiliary func- tion. Remark. Haskell has two ways to locally define auxiliary functions, where and let constructions. The where construction is illustrated in Example 1.11. This can also expressed with let, as follows: 1.5. PLAYINGTHEHASKELLGAME 15 srtInts’ :: [Int] -> [Int] srtInts’ [] = [] srtInts’ xs = let m = mnmInt xs in m : (srtInts’ (removeFst m xs)) The let construction uses the reserved keywords let and in. 2 (cid:1)(cid:1)(cid:1) k +nk Example 1.12 Here is a function that calculates the average of a list of integers. The average of m and n is given by m+n , the average of a list of k integers n1; : : : ; nk is given by n1+ . In general, averages are fractions, so the result type of average should not be Int but the Haskell data-type for floating point numbers, which is Float. There are predefined functions sum for the sum of a list of integers, and length for the length of a list. The Haskell operation for division / expects arguments of type Float (or more precisely, of Fractional type, and Float is such a type), so we need a conversion function for converting Ints into Floats. This is done by fromInt. The function average can now be written as: average :: [Int] -> Float average [] = error "empty list" average xs = fromInt (sum xs) / fromInt (length xs) Again, it is instructive to write our own homemade versions of sum and length. Here they are: sum’ :: [Int] -> Int sum’ [] = 0 sum’ (x:xs) = x + sum’ xs 16 CHAPTER1. GETTINGSTARTED length’ :: [a] -> Int length’ [] = 0 length’ (x:xs) = 1 + length’ xs Note that the type declaration for length’ contains a variable a. This variable ranges over all types, so [a] is the type of a list of objects of an arbitrary type a. We say that [a] is a type scheme rather than a type. This way, we can use the same function length’ for computing the length of a list of integers, the length of a list of characters, the length of a list of strings (lists of characters), and so on. The type [Char] is abbreviated as String. Examples of characters are ’a’, ’b’ (note the single quotes) examples of strings are "Russell" and "Cantor" (note the double quotes). In fact, "Russell" can be seen as an abbreviation of the list [’R’,’u’,’s’,’s’,’e’,’l’,’l’]. Exercise 1.13 Write a function count for counting the number of occurrences of a character in a string. In Haskell, a character is an object of type Char, and a string an object of type String, so the type declaration should run: count :: Char -> String -> Int. Exercise 1.14 A function for transforming strings into strings is of type String -> String. Write a function blowup that converts a string a1a2a3 (cid:1) (cid:1) (cid:1) to a1a2a2a3a3a3 (cid:1) (cid:1) (cid:1) blowup "bang!" should yield "baannngggg!!!!!". (Hint: use ++ for string concatenation.) : Exercise 1.15 Write a function srtString :: [String] -> [String] that sorts a list of strings in alphabetical order. Example 1.16 Suppose we want to check whether a string str1 is a prefix of a string str2. Then the answer to the question prefix str1 str2 should be either yes (true) or no (false), i.e., the type declaration for prefix should run: prefix :: String -> String -> Bool. Prefixes of a string ys are defined as follows: 17 1.6. HASKELLTYPES 1. [] is a prefix of ys, 2. if xs is a prefix of ys, then x:xs is a prefix of x:ys, 3. nothing else is a prefix of ys. Here is the code for prefix that implements this definition: prefix :: String -> String -> Bool prefix [] ys = True prefix (x:xs) [] = False prefix (x:xs) (y:ys) = (x==y) && prefix xs ys The definition of prefix uses the Haskell operator && for conjunction. Exercise 1.17 Write a function substring :: String -> String -> Bool that checks whether str1 is a substring of str2. The substrings of an arbitrary string ys are given by: 1. if xs is a prefix of ys, xs is a substring of ys, 2. if ys equals y:ys’ and xs is a substring of ys’, xs is a substring of ys, 3. nothing else is a substring of ys. 1.6 Haskell Types The basic Haskell types are: Int and Integer, to represent integers. Elements of Integer are un- bounded. That’s why we used this type in the implementation of the prime number test. Float and Double represent floating point numbers. The elements of Double have higher precision. Bool is the type of Booleans. (cid:15) (cid:15) (cid:15) 18 (cid:15) Char is the type of characters. CHAPTER1. GETTINGSTARTED Note that the name of a type always starts with a capital letter. To denote arbitrary types, Haskell allows the use of type variables. For these, a, b, . . . , are used. New types can be formed in several ways: (cid:15) (cid:15) (cid:15) (cid:15) By list-formation: if a is a type, [a] is the type of lists over a. Examples: [Int] is the type of lists of integers; [Char] is the type of lists of characters, or strings. By pair- or tuple-formation: if a and b are types, then (a,b) is the type of pairs with an object of type a as their first component, and an object of type b as their second component. Similarly, triples, quadruples, . . . , can be formed. If a, b and c are types, then (a,b,c) is the type of triples with an object of type a as their first component, an object of type b as their second component, and an object of type c as their third component. And so on (p. 139). By function definition: a -> b is the type of a function that takes arguments of type a and returns values of type b. By defining your own data-type from scratch, with a data type declaration. More about this in due course. Pairs will be further discussed in Section 4.5, lists and list operations in Section 4.6. Operations are procedures for constructing objects of a certain types b from ingre- dients of a type a. Now such a procedure can itself be given a type: the type of a transformer from a type objects to b type objects. The type of such a procedure can be declared in Haskell as a -> b. If a function takes two string arguments and returns a string then this can be viewed as a two-stage process: the function takes a first string and returns a transformer from strings to strings. It then follows that the type is String -> (String -> String), which can be written as String -> String -> String, because of the Haskell convention that -> associates to the right. Exercise 1.18 Find expressions with the following types: 1. [String] 1.7. THEPRIMEFACTORIZATIONALGORITHM 19 2. (Bool,String) 3. [(Bool,String)] 4. ([Bool],String) 5. Bool -> Bool Test your answers by means of the Hugs command :t. Exercise 1.19 Use the Hugs command :t to find the types of the following pre- defined functions: 1. head 2. last 3. init 4. fst 5. (++) 6. flip 7. flip (++) Next, supply these functions with arguments of the expected types, and try to guess what these functions do. 1.7 The Prime Factorization Algorithm Let n be an arbitrary natural number > 1. A prime factorization of n is a list of pj = n. We will show prime numbers p1; : : : ; pj with the property that p1 (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) that a prime factorization of every natural number n > 1 exists by producing one by means of the following method of splitting off prime factors: WHILE n = 1 DO BEGIN p := LD(n); n := n p END Here := denotes assignment or the act of giving a variable a new value. As we have seen, LD(n) exists for every n with n > 1. Moreover, we have seen that LD(n) is always prime. Finally, it is clear that the procedure terminates, for every round through the loop will decrease the size of n. 6 20 CHAPTER1. GETTINGSTARTED So the algorithm consists of splitting off primes until we have written n as n = pj, with all factors prime. To get some intuition about how the procedure p1 (cid:1) (cid:1) (cid:1) works, let us see what it does for an example case, say n = 84. The original as- signment to n is called n0; successive assignments to n and p are called n1; n2; : : : and p1; p2; : : :. n0 = 84 = 1 p1 = 2 n1 = 84=2 = 42 = 1 p2 = 2 n2 = 42=2 = 21 = 1 p3 = 3 n3 = 21=3 = 7 = 1 p4 = 7 n4 = 7=7 = 1 n0 6 n1 6 n2 6 n3 6 n4 = 1 This gives 84 = 22 3 7, which is indeed a prime factorization of 84. (cid:1) (cid:1) The following code gives an implementation in Haskell, collecting the prime fac- tors that we find in a list. The code uses the predefined Haskell function div for integer division. factors :: Integer -> [Integer] factors n | n < 1 = error "argument not positive" = [] | n == 1 | otherwise = p : factors (div n p) where p = ld n If you load the code for this chapter, you can try this out as follows: GS> factors 84 [2,2,3,7] GS> factors 557940830126698960967415390 [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71] 1.8 The map and filter Functions Haskell allows some convenient abbreviations for lists: [4..20] denotes the list of integers from 4 through 20, [’a’..’z’] the list of all lower case letters, "abcdefghijklmnopqrstuvwxyz". The call [5..] generates an infinite list of integers starting from 5. And so on. If you use the Hugs command :t to find the type of the function map, you get the following: 1.8. THE MAP AND FILTER FUNCTIONS 21 Prelude> :t map map :: (a -> b) -> [a] -> [b] The function map takes a function and a list and returns a list containing the results of applying the function to the individual list members. If f is a function of type a -> b and xs is a list of type [a], then map f xs will return a list of type [b]. E.g., map (^2) [1..9] will produce the list of squares [1, 4, 9, 16, 25, 36, 49, 64, 81] You should verify this by trying it out in Hugs. The use of (^2) for the operation of squaring demonstrates a new feature of Haskell, the construction of sections. Conversion from Infix to Prefix, Construction of Sections If op is an infix operator, (op) is the prefix version of the operator. Thus, 2^10 can also be written as (^) 2 10. This is a special case of the use of sections in Haskell. In general, if op is an infix operator, (op x) is the operation resulting from ap- plying op to its right hand side argument, (x op) is the operation resulting from applying op to its left hand side argument, and (op) is the prefix version of the operator (this is like the abstraction of the operator from both arguments). Thus (^2) is the squaring operation, (2^) is the operation that computes powers of 2, and (^) is exponentiation. Similarly, (>3) denotes the property of being greater than 3, (3>) the property of being smaller than 3, and (>) is the prefix version of the ‘greater than’ relation. The call map (2^) [1..10] will yield [2, 4, 8, 16, 32, 64, 128, 256, 512, 1024] If p is a property (an operation of type a -> Bool) and xs is a list of type [a], then map p xs will produce a list of type Bool (a list of truth values), like this: Prelude> map (>3) [1..9] [False, False, False, True, True, True, True, True, True] Prelude> The function map is predefined in Haskell, but it is instructive to give our own version: 22 CHAPTER1. GETTINGSTARTED map :: (a -> b) -> [a] -> [b] map f [] = [] map f (x:xs) = (f x) : (map f xs) Note that if you try to load this code, you will get an error message: Definition of variable "map" clashes with import. The error message indicates that the function name map is already part of the name space for functions, and is not available anymore for naming a function of your own making. Exercise 1.20 Use map to write a function lengths that takes a list of lists and returns a list of the corresponding list lengths. Exercise 1.21 Use map to write a function sumLengths that takes a list of lists and returns the sum of their lengths. Another useful function is filter, for filtering out the elements from a list that satisfy a given property. This is predefined, but here is a home-made version: filter :: (a -> Bool) -> [a] -> [a] filter p [] = [] filter p (x:xs) | p x = x : filter p xs | otherwise = filter p xs Here is an example of its use: GS> filter (>3) [1..10] [4,5,6,7,8,9,10] Example 1.22 Here is a program primes0 that filters the prime numbers from the infinite list [2..] of natural numbers: 1.8. THE MAP AND FILTER FUNCTIONS 23 primes0 :: [Integer] primes0 = filter prime0 [2..] This produces an infinite list of primes. (Why infinite? See Theorem 3.33.) The list can be interrupted with ‘Control-C’. Example 1.23 Given that we can produce a list of primes, it should be possible now to improve our implementation of the function LD. The function ldf used in the definition of ld looks for a prime divisor of n by checking k n for all k with j n for the primes p with 2 6 p 6 pn. 2 6 k 6 pn. In fact, it is enough to check p j Here are functions ldp and ldpf that perform this more efficient check: ldp :: Integer -> Integer ldp n = ldpf primes1 n ldpf :: [Integer] -> Integer -> Integer ldpf (p:ps) n | rem n p == 0 = p = n = ldpf ps n | p^2 > n | otherwise ldp makes a call to primes1, the list of prime numbers. This is a first illustration of a ‘lazy list’. The list is called ‘lazy’ because we compute only the part of the list that we need for further processing. To define primes1 we need a test for primality, but that test is itself defined in terms of the function LD, which in turn refers to primes1. We seem to be running around in a circle. This circle can be made non-vicious by avoiding the primality test for 2. If it is given that 2 is prime, then we can use the primality of 2 in the LD check that 3 is prime, and so on, and we are up and running. 24 CHAPTER1. GETTINGSTARTED primes1 :: [Integer] primes1 = 2 : filter prime [3..] prime :: Integer -> Bool prime n | n < 1 = error "not a positive integer" = False | n == 1 | otherwise = ldp n == n Replacing the definition of primes1 by filter prime [2..] creates vicious circularity, with stack overflow as a result (try it out). By running the program primes1 against primes0 it is easy to check that primes1 is much faster. Exercise 1.24 What happens when you modify the defining equation of ldp as follows: ldp :: Integer -> Integer ldp = ldpf primes1 Can you explain? 1.9 Haskell Equations and Equational Reasoning The Haskell equations f x y = ... used in the definition of a function f are genuine mathematical equations. They state that the left hand side and the right hand side of the equation have the same value. This is very different from the use of = in imperative languages like C or Java. In a C or Java program, the statement y have the same value, but rather it is a x = x*y does not mean that x and x command to throw away the old value of x and put the value of x y in its place. (cid:3) It is a so-called destructive assignment statement: the old value of a variable is destroyed and replaced by a new one. (cid:3) Reasoning about Haskell definitions is a lot easier than reasoning about programs that use destructive assignment. In Haskell, standard reasoning about mathemat- ical equations applies. E.g., after the Haskell declarations x = 1 and y = 2, the Haskell declaration x = x + y will raise an error "x" multiply defined. Be- cause = in Haskell has the meaning “is by definition equal to”, while redefinition 1.9. HASKELLEQUATIONSANDEQUATIONALREASONING 25 is forbidden, reasoning about Haskell functions is standard equational reasoning. Let’s try this out on a simple example. a = 3 b = 4 f :: Integer -> Integer -> Integer f x y = x^2 + y^2 To evaluate f a (f a b) by equational reasoning, we can proceed as follows: f a (f a b) = f a (a2 + b2) = f 3 (32 + 42) = f 3 (9 + 16) = f 3 25 = 32 + 252 = 9 + 625 = 634 The rewriting steps use standard mathematical laws and the Haskell definitions of a, b, f . And, in fact, when running the program we get the same outcome: GS> f a (f a b) 634 GS> Remark. We already encountered definitions where the function that is being defined occurs on the right hand side of an equation in the definition. Here is another example: g :: Integer -> Integer g 0 = 0 g (x+1) = 2 * (g x) Not everything that is allowed by the Haskell syntax makes semantic sense, how- ever. The following definitions, although syntactically correct, do not properly define functions: 26 CHAPTER1. GETTINGSTARTED h1 :: Integer -> Integer h1 0 = 0 h1 x = 2 * (h1 x) h2 :: Integer -> Integer h2 0 = 0 h2 x = h2 (x+1) The problem is that for values other than 0 the definitions do not give recipes for computing a value. This matter will be taken up in Chapter 7. 1.10 Further Reading The standard Haskell operations are defined in the file Prelude.hs, which you should be able to locate somewhere on any system that runs hugs. Typically, the file resides in /usr/lib/hugs/libraries/Hugs/. In case Exercise 1.19 has made you curious, the definitions of these example func- tions can all be found in Prelude.hs. If you want to quickly learn a lot about how to program in Haskell, you should get into the habit of consulting this file regularly. The definitions of all the standard operations are open source code, and are there for you to learn from. The Haskell Prelude may be a bit difficult to read at first, but you will soon get used to the syntax and acquire a taste for the style. Various tutorials on Haskell and Hugs can be found on the Internet: see e.g. [HFP96] and [JR+]. The definitive reference for the language is [Jon03]. A text- book on Haskell focusing on multimedia applications is [Hud00]. Other excellent textbooks on functional programming with Haskell are [Tho99] and, at a more ad- vanced level, [Bir98]. A book on discrete mathematics that also uses Haskell as a tool, and with a nice treatment of automated proof checking, is [HO00]. Chapter 2 Talking about Mathematical Objects Preview To talk about mathematical objects with ease it is useful to introduce some sym- bolic abbreviations. These symbolic conventions are meant to better reveal the structure of our mathematical statements. This chapter concentrates on a few (in fact: seven), simple words or phrases that are essential to the mathematical vo- cabulary: not, if, and, or, if and only if, for all and for some. We will introduce symbolic shorthands for these words, and we look in detail at how these building blocks are used to construct the logical patterns of sentences. After having isolated the logical key ingredients of the mathematical vernacular, we can systematically relate definitions in terms of these logical ingredients to implementations, thus building a bridge between logic and computer science. The use of symbolic abbreviations in specifying algorithms makes it easier to take the step from definitions to the procedures that implement those definitions. In a similar way, the use of symbolic abbreviations in making mathematical statements makes it easier to construct proofs of those statements. Chances are that you are more at ease with programming than with proving things. However that may be, in the chapters to follow you will get the opportunity to improve your skills in both of these activities and to find out more about the way in which they are related. 27 28 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS module TAMO where 2.1 Logical Connectives and their Meanings Goal To understand how the meanings of statements using connectives can be described by explaining how the truth (or falsity) of the statement depends on the truth (or falsity) of the smallest parts of this statement. This understanding leads directly to an implementation of the logical connectives as truth functional procedures. In ordinary life, there are many statements that do not have a definite truth value, for example ‘Barnett Newman’s Who is Afraid of Red, Yellow and Blue III is a beautiful work of art,’ or ‘Daniel Goldreyer’s restoration of Who is Afraid of Red, Yellow and Blue III meets the highest standards.’ Fortunately the world of mathematics differs from the Amsterdam Stedelijk Mu- seum of Modern Art in the following respect. In the world of mathematics, things are so much clearer that many mathematicians adhere to the following slogan: every statement that makes mathematical sense is either true or false. The idea behind this is that (according to the adherents) the world of mathematics exists independently of the mind of the mathematician. Doing mathematics is the activity of exploring this world. In proving new theorems one discovers new facts about the world of mathematics, in solving exercises one rediscovers known facts for oneself. (Solving problems in a mathematics textbook is like visiting famous places with a tourist guide.) This belief in an independent world of mathematical fact is called Platonism, after the Greek philosopher Plato, who even claimed that our everyday physical world is somehow an image of this ideal mathematical world. A mathematical Platonist holds that every statement that makes mathematical sense has exactly one of the two truth values. Of course, a Platonist would concede that we may not know which value a statement has, for mathematics has numerous open problems. Still, a Platonist would say that the true answer to an open problem in mathematics like ‘Are there infinitely many Mersenne primes?’ (Example 3.40 from Chapter 3) is 2.1. LOGICALCONNECTIVESANDTHEIRMEANINGS 29 either ‘yes’ or ‘no’. The Platonists would immediately concede that nobody may know the true answer, but that, they would say, is an altogether different matter. Of course, matters are not quite this clear-cut, but the situation is certainly a lot In the first place, it may not better than in the Amsterdam Stedelijk Museum. be immediately obvious which statements make mathematical sense (see Example 4.5). In the second place, you don’t have to be a Platonist to do mathematics. Not every working mathematician agrees with the statement that the world of mathe- matics exists independently of the mind of the mathematical discoverer. The Dutch mathematician Brouwer (1881–1966) and his followers have argued instead that the mathematical reality has no independent existence, but is created by the work- ing mathematician. According to Brouwer the foundation of mathematics is in the intuition of the mathematical intellect. A mathematical Intuitionist will therefore not accept certain proof rules of classical mathematics, such as proof by contra- diction (see Section 3.3), as this relies squarely on Platonist assumptions. Although we have no wish to pick a quarrel with the intuitionists, in this book we will accept proof by contradiction, and we will in general adhere to the practice of classical mathematics and thus to the Platonist creed. In mathematical reasoning, it is usual to employ shorthands for if Connectives (or: if. . . then), and, or, not. These words are called connectives. The word and is called the conjunction symbol. is used to form conjunctions, its shorthand is called the disjunction The word or is used to form disjunctions, its shorthand symbol. The word not is used to form negations, its shorthand is called the negation symbol. The combination if. . . then produces implications; its shorthand is the implication symbol. Finally, there is a phrase less common in everyday ) conversation, but crucial if one is talking mathematics. The combination . . . if and only if . . . produces equivalences, its shorthand is called the equivalence symbol. These logical connectives are summed up in the following table. , ^ : _ and or not if—then if, and only if symbol ^ _ : ) , name conjunction disjunction negation implication equivalence Remark. Do not confuse if. . . then ( ) on one hand with thus, so, therefore on the other. The difference is that the phrase if. . . then is used to construct conditional ) 30 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS statements, while thus (therefore, so) is used to combine statements into pieces of mathematical reasoning (or: mathematical proofs). We will never write when we want to conclude from one mathematical statement to the next. The rules of inference, the notion of mathematical proof, and the proper use of the word thus are the subject of Chapter 3. ) In mathematical English it is usual to abbreviate if, and only if to iff. We will as a symbolic abbreviation. Sometimes the phrase just in case is used Iff. also use with the same meaning. , The following describes, for every connective separately, how the truth value of a compound using the connective is determined by the truth values of its compo- nents. For most connectives, this is rather obvious. The cases for have some peculiar difficulties. and ) _ The letters P and Q are used for arbitrary statements. We use t for ‘true’, and f for ‘false’. The set is the set of truth values. t; f g f Haskell uses True and False for the truth values. Together, these form the type Bool. This type is predefined in Haskell as follows: data Bool = False | True Negation An expression of the form P (not P , it is not the case that P , etc.) is called the negation of P . It is true (has truth value t) just in case P is false (has truth value f). : In an extremely simple table, this looks as follows: P t f P : f t This table is called the truth table of the negation symbol. The implementation of the standard Haskell function not reflects this truth table: 2.1. LOGICALCONNECTIVESANDTHEIRMEANINGS 31 not not True not False :: Bool -> Bool = False = True This definition is part of Prelude.hs, the file that contains the predefined Haskell functions. Conjunction The expression P and Q are called conjuncts of P Q are both true. ^ ^ Q ((both) P and Q) is called the conjunction of P and Q. P Q is true iff P and Q. The conjunction P ^ Truth table of the conjunction symbol: Q P Q P t t f t t f f f ^ t f f f This is reflected in definition of the Haskell function for conjunction, && (also from Prelude.hs): :: Bool -> Bool -> Bool (&&) False && x && x True = False = x What this says is: if the first argument of a conjunction evaluates to false, then the conjunction evaluates to false; if the first argument evaluates to true, then the conjunction gets the same value as its second argument. The reason that the type declaration has (&&) instead of && is that && is an infix operator, and (&&) is its prefix counterpart (see page 21). 32 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Disjunction The expression P _ are the disjuncts of P Q (P or Q) is called the disjunction of P and Q. P and Q Q. _ The interpretation of disjunctions is not always straightforward. English has two disjunctions: (i) the inclusive version, that counts a disjunction as true also in case both disjuncts are true, and (ii) the exclusive version either. . . or, that doesn’t. Remember: The symbol _ will always be used for the inclusive version of or. Even with this problem out of the way, difficulties may arise. Example 2.1 No one will doubt the truth of the following: for every integer x, x < 1 or 0 < x. However, acceptance of this brings along acceptance of every instance. E.g., for x := 1:1 1 < 1 or 0 < 1: Some people do not find this acceptable or true, or think this to make no sense at all since something better can be asserted, viz., that 0 < 1. In mathematics with the inclusive version of , you’ll have to live with such a peculiarity. _ The truth table of the disjunction symbol now looks as follows. _ Q P Q P t t f t t f f f _ t t t f Here is the Haskell definition of the disjunction operation. Disjunction is rendered as || in Haskell. :: Bool -> Bool -> Bool (||) False || x || x True = x = True 1:= means: ‘is by definition equal to’. 2.1. LOGICALCONNECTIVESANDTHEIRMEANINGS 33 What this means is: if the first argument of a disjunction evaluates to false, then the disjunction gets the same value as its second argument. If the first argument of a disjunction evaluates to true, then the disjunction evaluates to true. Exercise 2.2 Make up the truth table for the exclusive version of or. Implication An expression of the form P of P and Q. P is the antecedent of the implication and Q the consequent. Q (if P , then Q; Q if P ) is called the implication ) is perhaps the only surprising one. However, it can be moti- The truth table of vated quite simply using an example like the following. No one will disagree that for every natural number n, ) ) Therefore, the implication must hold in particular for n equal to 2, 4 and 6. But then, an implication should be true if 5 < n 3 < n: both antecedent and consequent are false (n = 2), antecedent false, consequent true (n = 4), and both antecedent and consequent true (n = 6). (cid:15) (cid:15) (cid:15) And of course, an implication should be false in the only remaining case that the antecedent is true and the consequent false. This accounts for the following truth table. Q P Q P t t f t t f f f ) t f t t If we want to implement implication in Haskell, we can do so in terms of not and ||. It is convenient to introduce an infix operator ==> for this. The number 1 in the infix declaration indicates the binding power (binding power 0 is lowest, 9 is highest). A declaration of an infix operator together with an indication of its binding power is called a fixity declaration. 34 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS infix 1 ==> (==>) :: Bool -> Bool -> Bool x ==> y = (not x) || y It is also possible to give a direct definition: (==>) :: Bool -> Bool -> Bool True False ==> x = True ==> x = x Trivially True Implications. Note that implications with antecedent false and those with consequent true are true. For instance, because of this, the following two sentences must be counted as true: if my name is Napoleon, then the decimal expansion of (cid:25) contains the sequence 7777777, and: if the decimal expansion of (cid:25) contains the sequence 7777777, then strawberries are red. Implications with one of these two properties (no matter what the values of param- eters that may occur) are dubbed trivially true. In what follows there are quite a number of facts that are trivial in this sense that may surprise the beginner. One is that the empty set is included in every set (cf. Theorem 4.9 p. 126). ; Remark. The word trivial is often abused. Mathematicians have a habit of calling things trivial when they are reluctant to prove them. We will try to avoid this use of the word. The justification for calling a statement trivial resides in the psycho- logical fact that a proof of that statement immediately comes to mind. Whether a proof of something comes to your mind will depend on your training and experi- ence, so what is trivial in this sense is (to some extent) a personal matter. When we are reluctant to prove a statement, we will sometimes ask you to prove it as an exercise. Implication and Causality. The mathematical use of implication does not al- ways correspond to what you are used to. In daily life you will usually require a certain causal dependence between antecedent and consequent of an implication. (This is the reason the previous examples look funny.) In mathematics, such a 2.1. LOGICALCONNECTIVESANDTHEIRMEANINGS 35 causality usually will be present, but this is quite unnecessary for the interpreta- tion of an implication: the truth table tells the complete story. (And in this section in particular, causality usually will be absent.) However, in a few cases, natural language use surprisingly corresponds with truth table-meaning. E.g., I’ll be dead if Bill will not show up must be interpreted (if uttered by someone healthy) as strong belief that Bill will indeed turn up.2 Converse and Contraposition. The converse of an implication P ) P ; its contraposition is P . The converse of a true implication does not need to be true, but its contraposition is true iff the implication is. Cf. Theo- rem 2.10, p. 45. Q is Q ) : ) Q : Necessary and Sufficient Conditions. The statement P is called a sufficient condition for Q and Q a necessary condition for P if the implication P Q holds. ) Q can be expressed in a mathematical text in a number of An implication P ways: ) 1. if P , then Q, 2. Q if P , 3. P only if Q, 4. Q whenever P , 5. P is sufficient for Q, 6. Q is necessary for P . Equivalence , Q (P iff Q) is called the equivalence of P and Q. P and Q The expression P are the members of the equivalence. The truth table of the equivalence symbol is unproblematic once you realize that an equivalence P Q amounts to the con- junction of two implications P P taken together. (It is sometimes Q.) The outcome is that an equivalence must convenient to write Q ) be true iff its members have the same truth value. Q and Q P as P , ( ) ) Table: 2‘If Bill will not show up, then I am a Dutchman’, has the same meaning, when uttered by a native speaker of English. What it means when uttered by one of the authors of this book, we are not sure. 36 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Q P Q P t t f t t f f f , t f f t From the discussion under implication it is clear that P is called a condition that is both necessary and sufficient for Q if P Q is true. , There is no need to add a definition of a function for equivalence to Haskell. The type Bool is in class Eq, which means that an equality relation is predefined on it. But equivalence of propositions is nothing other than equality of their truth values. Still, it is useful to have a synonym: infix 1 <=> (<=>) :: Bool -> Bool -> Bool x <=> y = x == y Example 2.3 When you are asked to prove something of the form P iff Q it is often convenient to separate this into its two parts P Q. The ‘only ) ( Q means the same as P only if’ part of the proof is the proof of P ) if Q), and the ‘if’ part of the proof is the proof of P Q means the same as Q P , which in turn means the same as P , if Q). Q (for P Q (for P Q and P ( ) ( ) Exercise 2.4 Check that the truth table for exclusive or from Exercise 2.2 is equiv- Q). Conclude that the Haskell implementation of the alent to the table for function <+> for exclusive or in the frame below is correct. (P , : infixr 2 <+> (<+>) :: Bool -> Bool -> Bool x <+> y = x /= y The logical connectives are written in infix notation. Their Haskell coun- terparts, && and || are also infix. Thus, if p and q are expressions of type Bool, and ^ _ 2.1. LOGICALCONNECTIVESANDTHEIRMEANINGS 37 then p && q is a correct Haskell expression of type Bool. If one wishes to write this in prefix notation, this is also possible, by putting parentheses around the op- erator: (&&) p q. Although you will probably never find more than 3–5 connectives occurring in one mathematical statement, if you insist you can use as many connectives as you like. Of course, by means of parentheses you should indicate the way your expression was formed. For instance, look at the formula P : ^ ((P ) Q) (Q , : P )): ^ : Using the truth tables, you can determine its truth value if truth values for the components P and Q have been given. For instance, if P has value t and Q has value f, then P ): t; P ): f, and the displayed expression thus has value f. This Q) (P calculation can be given immediately under the formula, beginning with the values given for P and Q. The final outcome is located under the conjunction symbol , ^ which is the main connective of the expression. Q becomes f, Q P has f, P : (Q P : f; , : ^ : ^ : ^ : (Q ) ) : P t ^ ((P t : f ) f Q) f , : (Q f P )) t ^ : f f t f f In compressed form, this looks as follows: P t : f ((P t Q) f ) f ^ f , : t f (Q f ^ : f f P )) t Alternatively, one might use a computer to perform the calculation. 38 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS p = True q = False formula1 = (not p) && (p ==> q) <=> not (q && (not p)) After loading the file with the code of this chapter, you should be able to do: TAMO> formula1 False TAMO> Note that p and q are defined as constants, with values True and False, respec- tively, so that the occurrences of p and q in the expression formula1 are evaluated as these truth values. The rest of the evaluation is then just a matter of applying the definitions of not, &&, <=> and ==>. 2.2 Logical Validity and Related Notions Goal To grasp the concepts of logical validity and logical equivalence, to learn how to use truth tables in deciding questions of validity and equivalence, and in the handling of negations, and to learn how the truth table method for testing validity and equivalence can be implemented. Logical Validities. There are propositional formulas that receive the value t no matter what the values of the occurring letters. Such formulas are called (logically) valid. Examples of logical validities are: P P , P ) P , and P _ : (Q ) ) P ). Truth Table of an Expression. If an expression contains n letters P; Q; : : :, then there are 2n possible distributions of the truth values between these letters. The 2n-row table that contains the calculations of these values is the truth table of the expression. If all calculated values are equal to t, then your expression, by definition, is a validity. 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 39 Example 2.5 (Establishing Logical Validity by Means of a Truth Table) The following truth table shows that P P ) is a logical validity. (Q ) ) P t t f f ) t t t t (Q t f t f ) t t f t P ) t t f f To see how we can implement the validity check in Haskell, look at the imple- mentation of the evaluation formula1 again, and add the following definition of formula2: formula2 p q = ((not p) && (p ==> q) <=> not (q && (not p))) To see the difference between the two definitions, let us check their types: TAMO> :t formula1 formula1 :: Bool TAMO> :t formula2 formula2 :: Bool -> Bool -> Bool TAMO> The difference is that the first definition is a complete proposition (type Bool) in itself, while the second still needs two arguments of type Bool before it will return a truth value. In the definition of formula1, the occurrences of p and q are interpreted as con- stants, of which the values are given by previous definitions. In the definition of formula2. the occurrences of p and q are interpreted as variables that represent the arguments when the function gets called. A propositional formula in which the proposition letters are interpreted as variables can in fact be considered as a propositional function or Boolean function or truth function. If just one variable, say p occurs in it, then it is a function of type Bool -> Bool (takes a Boolean, returns a Boolean). If two variables occur in it, say p and q, then it is a function of type Bool -> Bool -> Bool (takes Boolean, 40 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS then takes another Boolean, and returns a Boolean). If three variables occur in it, then it is of type Bool -> Bool -> Bool -> Bool, and so on. In the validity check for a propositional formula, we treat the proposition letters as arguments of a propositional function, and we check whether evaluation of the function yields true for every possible combination of the arguments (that is the essence of the truth table method for checking validity). Here is the case for propo- sitions with one proposition letter (type Bool -> Bool). valid1 :: (Bool -> Bool) -> Bool valid1 bf = (bf True) && (bf False) The validity check for Boolean functions of type Bool -> Bool is suited to test functions of just one variable. An example is the formula P P that expresses the principle of excluded middle (or, if you prefer a Latin name, tertium non datur, for: there is no third possibility). Here is its implementation in Haskell: _ : excluded_middle :: Bool -> Bool excluded_middle p = p || not p To check that this is valid by the truth table method, one should consider the two cases P := t and P := f, and ascertain that the principle yields t in both of these cases. This is precisely what the validity check valid1 does: it yields True precisely when applying the boolean function bf to True yields True and applying bf to False yields True. And indeed, we get: TAMO> valid1 excluded_middle True Here is the validity check for propositional functions with two proposition let- ters, Such propositional functions have type Bool -> Bool -> Bool), and need a truth table with four rows to check their validity, as there are four cases to check. 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 41 valid2 :: (Bool -> Bool -> Bool) valid2 bf = True) False) (bf True && (bf True && (bf False True) && (bf False False) -> Bool Again, it is easy to see that this is an implementation of the truth table method for validity checking. Try this out on P P , and discover that the bracketing matters: P ) and on (P (Q Q) ) ) ) ) form1 p q = p ==> (q ==> p) form2 p q = (p ==> q) ==> p TAMO> valid2 form1 True TAMO> valid2 form2 False The propositional function formula2 that was defined above is also of the right argument type for valid2: TAMO> valid2 formula2 False It should be clear how the notion of validity is to be implemented for propositional functions with more than two propositional variables. Writing out the full tables becomes a bit irksome, so we are fortunate that Haskell offers an alternative. We demonstrate it in valid3 and valid4,. 42 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS valid3 :: (Bool -> Bool -> Bool -> Bool) -> Bool valid3 bf = and [ bf p q r | p <- [True,False], q <- [True,False], r <- [True,False]] valid4 :: (Bool -> Bool -> Bool -> Bool -> Bool) -> Bool valid4 bf = and [ bf p q r s | p <- [True,False], q <- [True,False], r <- [True,False], s <- [True,False]] The condition p <- [True,False], for “p is an element of the list consisting of the two truth values”, is an example of list comprehension (page 118). The definitions make use of Haskell list notation, and of the predefined function and for generalized conjunction. An example of a list of Booleans in Haskell is [True,True,False]. Such a list is said to be of type [Bool]. If list is a list of Booleans (an object of type [Bool]), then and list gives True in case all mem- bers of list are true, False otherwise. For example, and [True,True,False] gives False, but and [True,True,True] gives True. Further details about working with lists can be found in Sections 4.6 and 7.5. Leaving out Parentheses. We agree that and P . Thus, for instance, P (Q R)). ) Q ^ and R stands for (P ^ _ bind more strongly than ) R (and not for Q) ) ^ , ^ ) In Haskell, the convention is not quite the Operator Precedence in Haskell same, for || has operator precedence 2, && has operator precedence 3, and == has operator precedence 4, which means that == binds more strongly than &&, which in turn binds more strongly than ||. The operators that we added, ==> and <=>, follow the logic convention: they bind less strongly than && and ||. Logically Equivalent. Two formulas are called (logically) equivalent if, no mat- ter the truth values of the letters P; Q; : : : occurring in these formulas, the truth values obtained for them are the same. This can be checked by constructing a truth table (see Example (2.6)). Example 2.6 (The First Law of De Morgan) 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 43 (P t t f f : f t t t Q) t f t f ^ t f f f ( : f f t t P t t f f _ f t t t : f t f t Q) t f t f The outcome of the calculation shows that the formulas are equivalent: note that Q. the column under the Q) coincides with that under the (P of of P : : ^ _ : _ : Notation: (cid:8) can say that the truth table of Example (2.6) shows that (cid:9) indicates that (cid:8) and (cid:9) are equivalent3. Using this notation, we Q). Q) (P (cid:17) P : ^ ( : (cid:17) _ : Example 2.7 (De Morgan Again) The following truth table shows that which establishes that Q) (P : ^ (cid:17) (P t t f f : f t t t Q) t f t f ^ t f f f (P P : ( : , t t t t P ( : Q) is a logical validity, _ : Q) , Q). ^ _ : ( : f f t t P t t f f _ f t t t : f t f t Q) t f t f Example 2.8 A pixel on a computer screen is a dot on the screen that can be either on (i.e., visible) or off (i.e., invisible). We can use 1 for on and 0 for off. Turning pixels in a given area on the screen off or on creates a screen pattern for that area. The screen pattern of an area is given by a list of bits (0s or 1s). Such a list of bits can be viewed as a list of truth values (by equating 1 with t and 0 with f), and given two bit lists of the same length we can perform bitwise logical operations on them: the bitwise exclusive or of two bit lists of the same length n, say L = [P1; : : : ; Pn] denotes and K = [Q1; : : : ; Qn], is the list [P1 (cid:8) exclusive or. Qn], where Q1; : : : ; Pn (cid:8) (cid:8) In the implementation of cursor movement algorithms, the cursor is made visible on the screen by taking a bitwise exclusive or between the screen pattern S at the cursor position and the cursor pattern C. When the cursor moves elsewhere, the original screen pattern is restored by taking a bitwise exclusive or with the cursor pattern C again. Exercise 2.9 shows that this indeed restores the original pattern S. 3The Greek alphabet is on p. 423. 44 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Exercise 2.9 Let cise 2.2, that (P (cid:8) (cid:8) Q) Q is equivalent to P . (cid:8) stand for exclusive or. Show, using the truth table from Exer- In Haskell, logical equivalence can be tested as follows. First we give a procedure for propositional functions with 1 parameter: logEquiv1 :: logEquiv1 bf1 bf2 = (Bool -> Bool) -> (Bool -> Bool) -> Bool (bf1 True <=> bf2 True) && (bf1 False <=> bf2 False) What this does, for formulas (cid:8); (cid:9) with a single propositional variable, is testing the formula (cid:8) (cid:9) by the truth table method. , We can extend this to propositional functions with 2, 3 or more parameters, us- ing generalized conjunction. Here are the implementations of logEquiv2 and logEquiv3; it should be obvious how to extend this for truth functions with still more arguments. logEquiv2 :: (Bool -> Bool -> Bool) -> (Bool -> Bool -> Bool) -> Bool logEquiv2 bf1 bf2 = and [(bf1 p q) <=> (bf2 p q) | p <- [True,False], q <- [True,False]] logEquiv3 :: (Bool -> Bool -> Bool -> Bool) -> (Bool -> Bool -> Bool -> Bool) -> Bool logEquiv3 bf1 bf2 = and [(bf1 p q r) <=> (bf2 p q r) | p <- [True,False], q <- [True,False], r <- [True,False]] Let us redo Exercise (2.9) by computer. formula3 p q = p formula4 p q = (p <+> q) <+> q 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 45 Note that the q in the definition of formula3 is needed to ensure that it is a function with two arguments. TAMO> logEquiv2 formula3 formula4 True We can also test this by means of a validity check on P follows: ((P Q) (cid:8) (cid:8) , Q), as formula5 p q = p <=> ((p <+> q) <+> q) TAMO> valid2 formula5 True (cid:17) and (cid:9) ex- Warning. Do not confuse . If (cid:8) and (cid:9) are formulas, then (cid:8) , (cid:9) is presses the statement that (cid:8) and (cid:9) are equivalent. On the other hand, (cid:8) just another formula. The relation between the two is that the formula (cid:8) (cid:9) is logically valid iff it holds that (cid:8) (cid:9). (See Exercise 2.19.) Compare the dif- ference, in Haskell, between logEquiv2 formula3 formula4 (a true statement about the relation between two formulas), and formula5 (just another formula). (cid:17) , , (cid:17) The following theorem collects a number of useful equivalences. (Of course, P , Q and R can be arbitrary formulas themselves.) Theorem 2.10 1. P P (law of double negation), P (laws of idempotence), (cid:17) :: P (cid:17) Q; _ _ Q, ^ : P P ; P 2. P 3. (P ^ (cid:17) Q) ) (P ) P P (cid:17) : Q) (cid:17) : 4. ( P Q) (Q : (P ) : Q) Q) ) : P ) ( : 5. (P ) P ); P ); ) : Q P ) ) Q) (cid:17) (Q (cid:17) ( (cid:17) : ((P Q) , (cid:17) ((P Q) ) ^ (cid:17) (Q ( : ) P ^ _ (laws of contraposition), P )) Q)), ^ : 46 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Q P ; P ^ Q Q (cid:17) _ P (laws of commutativity), _ Q; Q _ : ^ : (P 6. P 7. ^ (P (P : : 8. P P 9. P P ^ _ ^ _ Q ^ _ (Q (Q (Q (Q (cid:17) Q) Q) (cid:17) : (cid:17) : R) R) R) R) P P (cid:17) (cid:17) (cid:17) (cid:17) ^ _ _ ^ Q) Q) Q) Q) ^ _ _ ^ R; R (P (P R); R) ^ _ ^ _ ^ _ (P (P (P (DeMorgan laws). (laws of associativity), (distribution laws), Equivalence 8 justifies leaving out parentheses in conjunctions and disjunctions of three or more conjuncts resp., disjuncts. Non-trivial equivalences that often are used in practice are 2, 3 and 9. Note how you can use these to re-write negations: a negation of an implication can be rewritten as a conjunction, a negation of a conjunction (disjunction) is a disjunction (conjunction). Exercise 2.11 The First Law of De Morgan was proved in Example 2.6. This method was implemented above. Use the method by hand to prove the other parts of Theorem 2.10. We will now demonstrate how one can use the implementation of the logical equiv- alence tests as a check for Theorem 2.10. Here is a question for you to ponder: does checking the formulas by means of the implemented functions for logical equivalence count as a proof of the principles involved? Whatever the answer to this one may be, Figure 2.1 defines the tests for the statements made in Theorem 2.10, by means of lambda abstraction The expression \ p -> not (not p) is the Haskell way of referring to the lambda term (cid:21)p: p, the term that denotes the operation of performing a double negation. See Section 2.4. :: If you run these tests, you get result True for all of them. E.g.: TAMO> test5a True The next theorem lists some useful principles for reasoning with that is always true; the Haskell counterpart is True) and always false; the Haskell counterpart of this is False). ? (the proposition (the proposition that is > Theorem 2.12 1. ; :> (cid:17) ? , :? (cid:17) > 2. P P , ) ? (cid:17) : 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 47 = logEquiv1 id (\ p -> not (not p)) test1 test2a = logEquiv1 id (\ p -> p && p) test2b = logEquiv1 id (\ p -> p || p) test3a = logEquiv2 (\ p q -> p ==> q) (\ p q -> not p || q) test3b = logEquiv2 (\ p q -> not (p ==> q)) (\ p q -> p && not q) test4a = logEquiv2 (\ p q -> not p ==> not q) (\ p q -> q ==> p) test4b = logEquiv2 (\ p q -> p ==> not q) (\ p q -> q ==> not p) test4c = logEquiv2 (\ p q -> not p ==> q) (\ p q -> not q ==> p) test5a = logEquiv2 (\ p q -> p <=> q) test5b = logEquiv2 (\ p q -> p <=> q) (\ p q -> (p ==> q) && (q ==> p)) (\ p q -> (p && q) || (not p && not q)) test6a = logEquiv2 (\ p q -> p && q) (\ p q -> q && p) test6b = logEquiv2 (\ p q -> p || q) (\ p q -> q || p) test7a = logEquiv2 (\ p q -> not (p && q)) (\ p q -> not p || not q) test7b = logEquiv2 (\ p q -> not (p || q)) (\ p q -> not p && not q) test8a = logEquiv3 (\ p q r -> p && (q && r)) (\ p q r -> (p && q) && r) test8b = logEquiv3 (\ p q r -> p || (q || r)) (\ p q r -> (p || q) || r) test9a = logEquiv3 (\ p q r -> p && (q || r)) (\ p q r -> (p && q) || (p && r)) test9b = logEquiv3 (\ p q r -> p || (q && r)) (\ p q r -> (p || q) && (p || r)) Figure 2.1: Defining the Tests for Theorem 2.10. 48 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS ^ ? (cid:17) ? P ^ > (cid:17) 3. P 4. P 5. P 6. P ; P _ > (cid:17) > P ; P _ ? (cid:17) P _ : P ^ : (cid:17) > (cid:17) ? (dominance laws), (identity laws), (law of excluded middle), (contradiction). Exercise 2.13 Implement checks for the principles from Theorem 2.12. Without proof, we state the following Substitution Principle: If (cid:8) and (cid:9) are equivalent, and (cid:8)0 and (cid:9)0 are the results of substituting (cid:4) for every occurrence of P in (cid:8) and in (cid:9), respectively, then (cid:8)0 and (cid:9)0 are equivalent. Example 2.14 makes clear what this means. Example 2.14 From follows that (P : Q) P (cid:17) ) Q plus the substitution principle it ^ : ( : : P for P ), but also that ) P Q) P (cid:17) : Q ^ : (by substituting : (a = 2b 1 a is prime) a = 2b 1 a is not prime (cid:0) : (by substituting a = 2b ^ (cid:17) ) 1 for P and a is prime for Q). (cid:0) (cid:0) Exercise 2.15 A propositional contradiction is a formula that yields false for every combination of truth values for its proposition letters. Write Haskell definitions of contradiction tests for propositional functions with one, two and three variables. Exercise 2.16 Produce useful denials for every sentence of Exercise 2.31. (A de- nial of (cid:8) is an equivalent of (cid:8).) : Exercise 2.17 Produce a denial for the statement that x < y < z (where x; y; z R). 2 Exercise 2.18 Show: 1. ((cid:8) (cid:9)) , (cid:8) 2. ( (cid:8) ( : ((cid:8) (cid:17) (cid:9)) : , (cid:17) (cid:9)), , : (cid:9)). , : Exercise 2.19 Show that (cid:8) (cid:9) is true iff (cid:8) (cid:17) , (cid:9) is logically valid. 2.2. LOGICALVALIDITYANDRELATEDNOTIONS 49 Exercise 2.20 Determine (either using truth tables or Theorem 2.10) which of the following are equivalent, next check your answer by computer: 1. 2. 3. P : P : P : ) ) ) Q and P Q and Q Q, ) : P , ) : Q and Q : ) P , 4. P 5. P (Q (Q ) ) ) ) R) and Q (P ) ) R), R) and (P Q) R, ) ) 6. (P Q) ) ) P and P , 7. P Q ) _ R and (P R) ) ^ (Q ) R). Exercise 2.21 Answer as many of the following questions as you can: 1. Construct a formula (cid:8) involving the letters P and Q that has the following truth table. P Q (cid:8) t t t t f t f t f t f f 2. How many truth tables are there for 2-letter formulas altogether? 3. Can you find formulas for all of them? 4. Is there a general method for finding these formulas? 5. And what about 3-letter formulas and more? 50 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS 2.3 Making Symbolic Form Explicit In a sense, propositional reasoning is not immediately relevant for mathematics. Few mathematicians will ever feel the urge to write down a disjunction of two statements like 3 < 1 1 < 3. In cases like this it is clearly “better” to only write down the right-most disjunct. _ Fortunately, once variables enter the scene, propositional reasoning suddenly be- comes a very useful tool: the connectives turn out to be quite useful for combining open formulas. An open formula is a formula with one or more unbound variables in it. Variable binding will be explained below, but here is a first example of a formula with an unbound variable x. A disjunction like 3 < x x < 3 is (in some cases) a useful way of expressing that x = 3. _ Example. Consider the following (true) sentence: Between every two rational numbers there is a third one. (2.1) The property expressed in (2.1) is usually referred to as density of the rationals. We will take a systematic look at proving such statements in Chapter 3. Exercise 2.22 Can you think of an argument showing that statement (2.1) is true? A Pattern. There is a logical pattern underlying sentence (2.1). To make it visi- ble, look at the following, more explicit, formulation. It uses variables x; y and z for arbitrary rationals, and refers to the ordering < of the set Q of rational numbers. For all rational numbers x and z, if x < z, then some rational number y exists such that x < y and y < z. (2.2) You will often find ‘x < y and y < z’ shortened to: x < y < z. Quantifiers Note the words all (or: for all), some (or: for some, some. . . exists, there exists. . . such that, etc.). They are called quantifiers, and we use the symbols and as shorthands for them. 9 8 With these shorthands, plus the shorthands for the connectives that we saw above, Q for the property of being a rational, we arrive at the and the shorthand : : : following compact symbolic formulation: 2 Q x 8 2 z 8 2 Q ( x < z y ) 9 2 Q ( x < y ^ y < z ) ): (2.3) 6 2.3. MAKINGSYMBOLICFORMEXPLICIT 51 We will use example (2.3) to make a few points about the proper use of the vocab- ulary of logical symbols. An expression like (2.3) is called a sentence or a formula. Note that the example formula (2.3) is composite: we can think of it as constructed out of simpler parts. We can picture its structure as in Figure (2.2). x 8 2 Q z 8 2 Q(x < z y ) 9 2 Q(x < y ^ y < z)) Q(x < z z 8 2 y ) 9 2 Q(x < y y < z)) ^ (x < z y ) 9 2 Q(x < y y < z)) ^ x < z Q(x < y y 9 2 y < z) ^ (x < y y < z) ^ x < y y < z Figure 2.2: Composition of Example Formula from its Sub-formulas. As the figure shows, the example formula is formed by putting the quantifier prefix Q in front of a Q in front of the result of putting quantifier prefix x z 8 2 8 simpler formula, and so on. 2 The two consecutive universal quantifier prefixes can also be combined into Q. This gives the phrasing x; z 8 2 x; z 8 2 Q( x < z y ) 9 2 Q ( x < y ^ y < z ) ): x 2 2 Q ^ 8 Q(x < z Q(x < y y ) 9 between the two quantifiers is incorrect, however. In other words, the y < z)) is considered x ^ Q is itself not a formula, Putting an expression z 2 ^ 8 ungrammatical. The reason is that the formula part but a prefix that turns a formula into a more complex formula. The connective can only be used to construct a new formula out of two simpler formulas, so cannot serve to construct a formula from Q and another formula. 8 The symbolic version of the density statement uses parentheses. Their function is to indicate the way the expression has been formed and thereby to show the scope of operators. The scope of a quantifier-expression is the formula that it combines with to form a more complex formula. The scopes of quantifier-expressions and connectives in a formula are illustrated in the structure tree of that formula. Figure ^ ^ 2 2 x 8 52 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS 2.2 shows that the scope of the quantifier-expression Q is the formula x 8 2 2 Q(x < z z 8 Q is the formula y ) 9 Q (x < y y < z)); ^ 2 the scope of z 8 2 and the scope of y 9 2 Q is the formula (x < y ^ (x < z y ) 9 2 Q(x < y y < z)); ^ y < z). Exercise 2.23 Give structure trees of the following formulas (we use shorthand notation, and write A(x) as Ax for readability). 1. 2. 3. x(Ax 8 ) (Bx ) Cx)). x(Ax 9 ^ Bx). xAx 9 xBx. ^ 9 , is called The expression for all (and similar ones) and its shorthand, the symbol the universal quantifier; the expression there exists (and similar ones) and its short- hand, the symbol , is called the existential quantifier. The letters x, y and z that have been used in combination with them are variables. Note that ‘for some’ is equivalent to ‘for at least one’. 8 9 Unrestricted and Restricted Quantifiers, Domain of Quantification Quanti- fiers can occur unrestricted: A(x > 0), B(y < a) (where A and B are sets). x(y > x), and restricted: x(x > 0), y 9 2 x 8 8 8 In the unrestricted case, there should be some domain of quantification that often is implicit in the context. E.g., if the context is real analysis, x may mean for all reals x. . . , and 8 f may mean for all real-valued functions f . . . . y 9 2 8 Example 2.24 R is the set of real numbers. The fact that the R has no greatest element can be expressed with restricted quantifiers as: x R y 9 R(x < y): 8 2 If we specify that all quantifiers range over the reals (i.e., if we say that R is the domain of quantification) then we can drop the explicit restrictions, and we get by with y(x < y). 2 x 8 9 2.3. MAKINGSYMBOLICFORMEXPLICIT 53 The use of restricted quantifiers allows for greater flexibility, for it permits one to indicate different domains for different quantifiers. Example 2.25 x 8 2 R y 8 2 R(x < y z ) 9 2 Q (x < z < y)): 9 x(Ax : : :) one can write Instead of A(: : :). The advantage when all 9 quantifiers are thus restricted is that it becomes immediately clear that the domain This can make the logical is subdivided into different sub domains or types. translation much easier to comprehend. ^ 2 x Remark. We will use standard names for the following domains: N for the natural numbers, Z for the integer numbers, Q for the rational numbers, and R for the real numbers. More information about these domains can be found in Chapter 8. Exercise 2.26 Write as formulas with restricted quantifiers: 1. 2. 3. y(x x 9 x(x 9 8 x(x 8 2 R Z Q ^ ) 9 y 2 y(y Q 2 ^ R x < y). x < y)). ^ m; n(m ) 9 N n ^ 2 2 N ^ x = m n)). (cid:0) 2 2 Exercise 2.27 Write as formulas without restricted quantifiers: 1. 2. x 8 x 8 2 2 Q m; n Z(n = 0 x = m=n). 9 F y 8 2 2 D(Oxy ) ^ Bxy). Bound Variables. Quantifier expressions y,. . . (and their restricted com- panions) are said to bind every occurrence of x, y,. . . in their scope. If a variable occurs bound in a certain expression then the meaning of that expression does not change when all bound occurrences of that variable are replaced by another one. x, 9 8 6 54 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS 2 Example 2.28 y 9 This indicates that y is bound in Q(x < z). 2 Q(x < y) and Q(z < y) have different meanings, for the first asserts that there exists a ra- y 9 tional number greater than some given number x, and the second that there exists a rational number greater than some given z. Q(x < y) has the same meaning as y 9 Q(x < y). But z 9 2 y 9 2 2 Universal and existential quantifiers are not the only variable binding operators used by mathematicians. There are several other constructs that you are probably familiar with which can bind variables. 5 i=1 i is nothing but Example 2.29 (Summation, Integration.) The expression a way to describe the number 15 (15 = 1 + 2 + 3 + 4 + 5), and clearly, 15 does in no way depend on i. Use of a different variable does not change the meaning: P 5 k=1 k = 15. Here are the Haskell versions: P Prelude> sum [ i | i <- [1..5] ] 15 Prelude> sum [ k | k <- [1..5] ] 15 Similarly, the expression R 1 0 xdx denotes the number 1 2 and does not depend on x. Example 2.30 (Abstraction.) Another way to bind a variable occurs in the ab- , cf. (4.1), p. 118. The Haskell counterpart to this straction notation x is list comprehension: A 2 P g f j [ x | x <- list, property x ] The choice of variable does not matter. The same list is specified by: [ y | y <- list, property y ] The way set comprehension is used to define sets is similar to the way list compre- hension is used to define lists, and this is similar again to the way lambda abstrac- tion is used to define functions. See 2.4. 2.3. MAKINGSYMBOLICFORMEXPLICIT 55 Bad Habits. as follows. It is not unusual to encounter our example-statement (2.1) displayed For all rationals x and y, if x < y, then both x < z and z < y hold for some rational z. Note that the meaning of this is not completely clear. With this expression the true z < y)) could be meant, but statement that (x < z z ^ 9 what also could be meant is the false statement that z 9 (x < z Q (x < y Q (x < y z < y)). x; y x; y ) ) Q Q 2 2 2 2 8 8 ^ Putting quantifiers both at the front and at the back of a formula results in ambigu- ity, for it becomes difficult to determine their scopes. In the worst case the result is an ambiguity between statements that mean entirely different things. It does not look too well to let a quantifier bind an expression that is not a variable, such as in: for all numbers n2 + 1, . . . Although this habit does not always lead to unclarity, it is better to avoid it, as the result is often rather hard to comprehend. If you insist on quantifying over complex terms, then the following phrasing is suggested: for all numbers of the form n2 + 1; : : : Of course, in the implementation language, terms like n + 1 are important for pattern matching. Translation Problems. It is easy to find examples of English sentences that are hard to translate into the logical vernacular. E.g., in between two rationals is a third one it is difficult to discover a universal quantifier and an implication. Also, sometimes the English has an indefinite article where the meaning is clearly universal. Consider the sentence a well-behaved child is a quiet child. The indef- inite articles here may suggest existential quantifiers; however, the reading that is clearly meant has the form x 8 2 C (Well-behaved(x) Quiet(x)): ) A famous example from philosophy of language is: if a farmer owns a donkey, he beats it. Again, in spite of the indefinite articles, the meaning is universal: x y((Farmer(x) 8 8 Donkey(y) ^ ^ Own(x; y)) ) Beat(x; y)): 56 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS In cases like this, translation into a formula reveals the logical meaning that re- mained hidden in the original phrasing. In mathematical texts it also occurs quite often that the indefinite article a is used to make universal statements. Compare Example (2.43) below, where the following universal statement is made: A real function is continuous if it satisfies the "-(cid:14)- definition. Exercise 2.31 Translate into formulas, taking care to express the intended mean- ing: 1. The equation x2 + 1 = 0 has a solution. 2. A largest natural number does not exist. n for ‘d divides n’). 3. The number 13 is prime (use d j 4. The number n is prime. 5. There are infinitely many primes. Exercise 2.32 Translate into formulas: 1. Everyone loved Diana. (Use the expression L(x; y) for: x loved y, and the name d for Diana.) 2. Diana loved everyone. 3. Man is mortal. (Use M (x) for ‘x is a man’, and M’(x) for ‘x is mortal’.) 4. Some birds do not fly. (Use B(x) for ‘x is a bird’ and F (x) for ‘x can fly’.) Exercise 2.33 Translate into formulas, using appropriate expressions for the pred- icates: 1. Dogs that bark do not bite. 2. All that glitters is not gold. 3. Friends of Diana’s friends are her friends. 4.*The limit of 1 n as n approaches infinity is zero. 2.3. MAKINGSYMBOLICFORMEXPLICIT 57 Expressing Uniqueness. If we combine quantifiers with the relation = of iden- tity, we can make definite statements like ‘there is precisely one real number x with the property that for any real number y, xy = y’. The logical rendering is (assuming that the domain of discussion is R): x( y(x y = y) z( y(z y = y) z = x)): 9 8 8 The first part of this formula expresses that at least one x satisfies the property y = y), and the second part states that any z satisfying the same property is y(x ^ 8 ) (cid:1) (cid:1) 8 identical to that x. (cid:1) The logical structure becomes more transparent if we write P for the property. This gives the following translation for ‘precisely one object has property P ’: x(P x 9 z(P z ^ 8 ) z = x)): Exercise 2.34 Use the identity symbol = to translate the following sentences: 1. Everyone loved Diana except Charles. 2. Every man adores at least two women. 3. No man is married to more than one woman. Long ago the philosopher Bertrand Russell has proposed this logical format for the translation of the English definite article. According to his theory of description, the translation of The King is raging becomes: x(King(x) 9 y(King(y) ^ 8 y = x) ^ ) Raging(x)): Exercise 2.35 Use Russell’s recipe to translate the following sentences: 1. The King is not raging. 2. The King is loved by all his subjects. (use K(x) for ‘x is a King’, and S(x; y) for ‘x is a subject of y’). Exercise 2.36 Translate the following logical statements back into English. 1. x 9 2 R(x2 = 5). 58 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS N(n < m). N(1 < d < (2n + 1) n 8 n 8 2 2 n N N N m 9 :9 m 2 d 2 2. 3. 4. 5. (2n + 1)). d j ^ N(p 6 n m 6 p)). N(n < m p 2 9 R+ 8 " 8 j positive reals; a; a0; a1; : : : refer to real numbers .) ^ 8 N(m > n 2 n a ( j am ) m (cid:0) N _ 2 2 2 2 9 8 6 ")). (R+ is the set of Remark. Note that translating back and forth between formulas and plain English involves making decisions about a domain of quantification and about the predi- cates to use. This is often a matter of taste. For instance, how does one choose between P (n) for ‘n is prime’ and the spelled out n > 1 d 2 ^ :9 N(1 < d < n n); d j ^ which expands the definition of being prime? Expanding the definitions of mathe- matical concepts is not always a good idea. The purpose of introducing the word prime was precisely to hide the details of the definition, so that they do not bur- den the mind. The art of finding the right mathematical phrasing is to introduce precisely the amount and the kind of complexity that are needed to handle a given problem. Before we will start looking at the language of mathematics and its conventions in a more systematic way, we will make the link between mathematical definitions and implementations of those definitions. 2.4 Lambda Abstraction The following description defines a specific function that does not depend at all on x: The function that sends x to x2. Often used notations are x lambda term. 7! x2 and (cid:21)x:x2. The expression (cid:21)x:x2 is called a If t is an expression of type b and x is a variable of type a then (cid:21)x:t is an expression of type a b, i.e., (cid:21)x:t denotes a function. This way of defining functions is called lambda abstraction. ! 2.4. LAMBDAABSTRACTION 59 Note that the function that sends y to y2 (notation y same function as (cid:21)x:x2. 7! y2, or (cid:21)y:y2) describes the In Haskell, function definition by lambda abstraction is available. Compare the following two definitions: square1 :: Integer -> Integer square1 x = x^2 square2 :: Integer -> Integer square2 = \ x -> x^2 In the first of these, the parameter x is given as an argument. In the second, the function is defined as a lambda abstract. The Haskell way of lambda abstraction goes like this. The syntax is: \ v -> body, where v is a variable of the argu- ment type and body an expression of the result type. It is allowed to abbreviate \ v -> \ w -> body to \ v w -> body. And so on, for more than two vari- ables. E.g., both of the following are correct: m1 :: Integer -> Integer -> Integer m1 = \ x -> \ y -> x*y m2 :: Integer -> Integer -> Integer m2 = \ x y -> x*y And again, the choice of variables does not matter. Also, it is possible to abstract over tuples. Compare the following definition of a function that solves quadratic equations by means of the well-known ‘abc’-formula b x = (cid:0) (cid:6) pb2 2a 4ac : (cid:0) 60 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS solveQdr :: (Float,Float,Float) -> (Float,Float) solveQdr = \ (a,b,c) -> if a == 0 then error "not quadratic" else let d = b^2 - 4*a*c in if d < 0 then error "no real solutions" else ((- b + sqrt d) / 2*a, (- b - sqrt d) / 2*a) . x (cid:0) , and To solve the equation x2 1 = 0, use solveQdr (1,-1,-1), and you will (cid:0) get the (approximately correct) answer (1.61803,-0.618034). Approximately correct, for 1:61803 is an approximation of the golden ratio, 1+p5 0:618034 is an approximation of 1 p5 (cid:0) 2 One way to think about quantified expressions like or binations of a quantifier expression The lambda abstract (cid:21)x:P x denotes the property of being a P . The quantifier 8 is a function that maps properties to truth values according to the recipe: if the property holds of the whole domain then t, else f. The quantifier is a function that maps properties to truth values according to the recipe: if the property holds of anything at all then t, else f. This perspective on quantification is the basis of the Haskell implementation of quantifiers in Section 2.8. yP y is as com- and a lambda term (cid:21)x:P x or (cid:21)y:P y. xP x and (cid:0) 8 9 9 8 9 2 2.5 Definitions and Implementations Here is an example of a definition in mathematics. A natural number n is prime if n > 1 and no number m with 1 < m < n divides n. We can capture this definition of being prime in a formula, using m n for ‘m j divides n’, as follows (we assume the natural numbers as our domain of discourse): n > 1 ^ :9 m(1 < m < n n): m j ^ Another way of expressing this is the following: n > 1 m((1 < m < n) ^ 8 n): m j ) : (2.4) (2.5) If you have trouble seeing that formulas (2.4) and (2.5) mean the same, don’t worry. We will study such equivalences between formulas in the course of this chapter. 2.6. ABSTRACTFORMULASANDCONCRETESTRUCTURES 61 If we take the domain of discourse to be the domain of the natural numbers N = , then formula (2.5) expresses that n is a prime number. 0; 1; 2; : : : g f We can make the fact that the formula is meant as a definition explicit by introduc- ing a predicate name P and linking that to the formula:4 P (n) : (cid:17) n > 1 m((1 < m < n) ^ 8 n): m j ) : (2.6) One way to think about this definition is as a procedure for testing whether a natural number is prime. Is 83 a prime? Yes, because none of 2; 3; 4; : : : ; 9 divides 83. Note that there is no reason to check 10; : : :, for since 10 10 > 83 any factor m of 83 with m > 10 will not be the smallest factor of 83, and a smaller factor should have turned up before. (cid:2) The example shows that we can make the prime procedure more efficient. We only have to try and find the smallest factor of n, and any b with b2 > n cannot be the smallest factor. For suppose that a number b with b2 > n divides n. Then there is b = n, and therefore a2 6 n, and a divides n. Our definition a number a with a can therefore run: (cid:2) P (n) : (cid:17) n > 1 m((1 < m ^ 8 m2 6 n) ^ n): m j ) : In Chapter 1 we have seen that this definition is equivalent to the following: The Haskell implementation of the primality test was given in Chapter 1. P (n) : (cid:17) n > 1 ^ LD(n) = n: (2.7) (2.8) 2.6 Abstract Formulas and Concrete Structures The formulas of Section 2.1 are “handled” using truth values and tables. Quan- tificational formulas need a structure to become meaningful. Logical sentences involving variables can be interpreted in quite different structures. A structure is a domain of quantification, together with a meaning for the abstract symbols that occur. A meaningful statement is the result of interpreting a logical formula in a certain structure. It may well occur that interpreting a given formula in one structure yields a true statement, while interpreting the same formula in a differ- ent structure yields a false statement. This illustrates the fact that we can use one logical formula for many different purposes. Look again at the example-formula (2.3), now displayed without reference to Q and using a neutral symbol R. This gives: y ( xRy = ) 9 8 4:(cid:17) means: ‘is by definition equivalent to’. x 8 z ( xRz zRy ) ): ^ (2.9) 62 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS It is only possible to read this as a meaningful statement if 1. it is understood which is the underlying domain of quantification, and 2. what the symbol R stands for. Earlier, the set of rationals Q was used as the domain, and the ordering < was employed instead of the —in itself meaningless— symbol R. In the context of Q and <, the quantifiers z in (2.9) should be read as: for all rationals x . . . , resp., for some rational z . . . , whereas R should be viewed as standing for <. In that particular case, the formula expresses the true statement that, between every two rationals, there is a third one. x and 8 9 However, one can also choose the set N = of natural numbers as domain and the corresponding ordering < as the meaning of R. In that case, the formula expresses the false statement that between every two natural numbers there is a third one. 0; 1; 2; : : : g f A specification of (i) a domain of quantification, to make an unrestricted use of the quantifiers meaningful, and (ii) a meaning for the unspecified symbols that may occur (here: R), will be called a context or a structure for a given formula. As you have seen here: given such a context, the formula can be “read” as a meaningful assertion about this context that can be either true or false. Open Formulas, Free Variables, and Satisfaction. tifier expression x from the example formula (2.9), then the following remains: If one deletes the first quan- 8 y ( xRy = 8 ) 9 z ( xRz zRy ) ): ^ (2.10) Although this expression does have the form of a statement, it in fact is not such a thing. Reason: statements are either true or false; and, even if a quantifier domain and a meaning for R were specified, what results cannot be said to be true or false, as long as we do not know what it is that the variable x (which no longer is bound by the quantifier x) stands for. 8 However, the expression can be turned into a statement again by replacing the variable x by (the name of) some object in the domain, or —what amounts to the same— by agreeing that x denotes this object. For instance, if the domain consists of the set N of natural numbers together with all rationals between 0 and 1, and the meaning of R is the 0 < q < 1 q [ f Q 2 g j 2.6. ABSTRACTFORMULASANDCONCRETESTRUCTURES 63 usual ordering relation < for these objects, then the expression turns into a truth upon replacing x by 0:5 or by assigning x this value. We say that 0:5 satisfies the formula in the given domain. However, (2.10) turns into a falsity when we assign 2 to x; in other words, 2 does not satisfy the formula. Of course, one can delete a next quantifier as well, obtaining: xRy = z ( xRz ) 9 zRy ): ^ Now, both x and y have become free, and, next to a context, values have to be assigned to both these variables in order to determine a truth value. An occurrence of a variable in an expression that is not (any more) in the scope of a quantifier is said to be free in that expression. Formulas that contain free variables are called open. An open formula can be turned into a statement in two ways: (i) adding quantifiers that bind the free variables; (ii) replacing the free variables by (names of) objects in the domain (or stipulating that they have such objects as values). Exercise 2.37 Consider the following formulas. 1. 2. 3. 4. 5. x x x y(xRy), 8 y(xRy). 9 y(xRy). 8 x y(x = y 8 8 8 9 9 x y(xRy 9 8 xRy). _ z(xRz ^ ^ :9 zRy)). Are these formulas true or false in the following contexts?: a. Domain: N = f b. Domain: N; meaning of R: >, ; meaning of R: <, 0; 1; 2; : : : g c. Domain: Q (the set of rationals); meaning of R: <, d. Domain: R (the set of reals); meaning of xRy: y2 = x, e. Domain: set of all human beings; meaning of R: father-of, f. Domain: set of all human beings; meaning of xRy: x loves y. 64 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Exercise 2.38 In Exercise 2.37, delete the first quantifier on x in formulas 1–5. Determine for which values of x the resulting open formulas are satisfied in each of the structures a–f. 2.7 Logical Handling of the Quantifiers Goal To learn how to recognize simple logical equivalents involving quantifiers, and how to manipulate negations in quantified contexts. Validities and Equivalents. Compare the corresponding definitions in Section 2.2. 1. A logical formula is called (logically) valid if it turns out to be true in every structure. 2. Formulas are (logically) equivalent if they obtain the same truth value in every structure (i.e., if there is no structure in which one of them is true and the other one is false). Notation: (cid:8) equivalent. (cid:17) (cid:9) expresses that the quantificational formulas (cid:8) and (cid:9) are Exercise 2.39 (The propositional version of this is in Exercise 2.19 p. 48.) Argue (cid:9) is valid. that (cid:8) and (cid:9) are equivalent iff (cid:8) , Because of the reference to every possible structure (of which there are infinitely many), these are quite complicated definitions, and it is nowhere suggested that you will be expected to decide on validity or equivalence in every case that you may encounter. In fact, in 1936 it was proved rigorously, by Alonzo Church (1903– 1995) and Alan Turing (1912–1954) that no one can! This illustrates that the complexity of quantifiers exceeds that of the logic of connectives, where truth tables allow you to decide on such things in a mechanical way, as is witnessed by the Haskell functions that implement the equivalence checks for propositional logic. Nevertheless: the next theorem already shows that it is sometimes very well possi- ble to recognize whether formulas are valid or equivalent — if only these formulas are sufficiently simple. Only a few useful equivalents are listed next. Here, (cid:9)(x), (cid:8)(x; y) and the like denote logical formulas that may contain variables x (or x; y) free. 2.7. LOGICALHANDLINGOFTHEQUANTIFIERS 65 Theorem 2.40 1. 2. y(cid:8)(x; y) y(cid:8)(x; y) x x 8 9 8 9 y (cid:17) 8 y (cid:17) 9 x(cid:8)(x; y); 8 x(cid:8)(x; y), 9 x(cid:8)(x) :8 x(cid:8)(x) :9 :8 :9 x x : : (cid:8)(x) (cid:8)(x) (cid:17) 9 : (cid:17) 8 x x (cid:8)(x); (cid:8)(x); : x(cid:8)(x); (cid:17) 9 x(cid:8)(x), (cid:17) 8 3. x((cid:8)(x) 8 x((cid:8)(x) 9 ^ _ (cid:9)(x)) (cid:9)(x)) (cid:17) (cid:17) ( 8 ( 9 x(cid:8)(x) x(cid:8)(x) x(cid:9)(x)); ^ 8 x(cid:9)(x)). _ 9 Proof. There is no neat truth table method for quantification, and there is no neat proof here. You just have to follow common sense. For instance (part 2, first item) common sense dictates that not every x satisfies (cid:8) if, and only if, some x does not satisfy (cid:8). Of course, common sense may turn out not a good adviser when things get less simple. Chapter 3 hopefully will (partly) resolve this problem for you. Exercise 2.41 For every sentence (cid:8) in Exercise 2.36 (p. 57), consider its nega- tion (cid:8) by working the negation symbol through the quantifiers. (cid:8), and produce a more positive equivalent for : : Order of Quantifiers. Theorem 2.40.1 says that the order of similar quantifiers (all universal or all existential) is irrelevant. But note that this is not the case for quantifiers of different kind. On the one hand, if you know that x(cid:8)(x; y) (which states that there is one 8 y such that for all x, (cid:8)(x; y) holds) is true in a certain structure, then a fortiori y(cid:8)(x; y) will be true as well (for each x, take this same y). However, if y(cid:8)(x; y) holds, it is far from sure that x(cid:8)(x; y) holds as well. y 9 x x 8 8 9 9 y 9 8 Example 2.42 The statement that y(x < y) is true in N, but the statement x(x < y) in this structure wrongly asserts that there exists a greatest natural x 9 8 y 9 8 number. 66 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Restricted Quantification. You have met the use of restricted quantifiers, where the restriction on the quantified variable is membership in some domain. But there are also other types of restriction. Example 2.43 (Continuity) According to the “"-(cid:14)-definition” of continuity, a real function f is continuous if (domain R): x " > 0 (cid:14) > 0 y ( 8 8 9 This formula uses the restricted quantifiers compact formulation here. (cid:0) 8 j 8 x j y < (cid:14) = f (x) ) j " > 0 and (cid:0) (cid:14) > 0 that enable a more f (y) j < " ): 9 Example 2.44 Consider our example-statement (2.3). Here it is again: 8 This can also be given as y 8 x(x < y = z(x < z ) 9 z < y)) ^ y 8 x < y 8 z < y(x < z); 9 but this reformulation stretches the use of this type of restricted quantification probably a bit too much. Remark. If A is a subset of the domain of quantification, then A (cid:8)(x) means the same as x(x 8 2 A ) (cid:8)(x)), x 8 whereas 2 A (cid:8)(x) is tantamount with x 9 2 x(x 9 2 A ^ (cid:8)(x)). Warning: The restricted universal quantifier is explained using existential quantifier is explained using ! ^ , whereas the ) Example 2.45 ‘Some Mersenne numbers are prime’ is correctly translated as P x). The translation P x) is wrong. It is much too weak, for it ex- presses(in the domain N) that there is a natural number x which is either not a Mersenne number or it is a prime. Any prime will do as an example of this, and so will any number which is not a Mersenne number. x(M x ) 9 9 x(M x ^ x R(P x In the same way, ‘all prime numbers have irrational square roots’ is translated as Q) is wrong. This 8 time we end up with something which is too strong, for this expresses that every real number is a prime number with an irrational square root. Q). The translation px = 2 px = 2 R(P x ) 2 2 ^ x 8 2.7. LOGICALHANDLINGOFTHEQUANTIFIERS 67 x Restricted Quantifiers Explained. There is a version of Theorem 2.40 that em- ploys restricted quantification. This version states, for instance, that A (cid:8) is equivalent to (cid:8), and so on. The equivalence follows immediately from A the remark above. We now have, e.g., that x(x 2 (cid:8)(x)), A hence to (and here the implication turns into a conjunction — cf. Theorem 2.10) x (cid:8)(x)), which in turn is equivalent to (Theorem 2.40) A(cid:8)(x) is equivalent to A x (cid:8)(x)), and, finally, to :8 ) (cid:8)(x). x(x :8 :8 ) (x A A 2 : 2 2 : 2 x x 9 9 9 2 ^ : 9 2 : Exercise 2.46 Does it hold that A (cid:8)(x)? If your answer is ‘yes’, give a proof, if ‘no’, then you should show this by giving a simple refutation (an example of formulas and structures where the two formulas have different truth values). A (cid:8)(x) is equivalent to :9 62 2 x x 9 Exercise 2.47 Is 62 answer is ‘yes’, and a refutation otherwise. (cid:8)(x) equivalent to A : x 9 A x 9 2 (cid:8)(x)? Give a proof if your : Exercise 2.48 Produce the version of Theorem 2.40 (p. 65) that employs restricted quantification. Argue that your version is correct. Example 2.49 (Discontinuity Explained) The following formula describes what it means for a real function f to be discontinuous in x: " > 0 : 8 (cid:14) > 0 9 y ( 8 x j y j (cid:0) < (cid:14) = f (x) f (y) j (cid:0) < " ): ) j Using Theorem 2.40, this can be transformed in three steps, moving the negation over the quantifiers, into: " > 0 9 (cid:14) > 0 8 y 9 ( x j : y j (cid:0) < (cid:14) = f (x) f (y) j (cid:0) < " ): ) j According to Theorem 2.10 this is equivalent to " > 0 9 (cid:14) > 0 8 y ( 9 x j y j (cid:0) < (cid:14) f (x) ^ : j f (y) j (cid:0) < " ); i.e., to " > 0 9 (cid:14) > 0 8 y ( 9 x j y j (cid:0) < (cid:14) f (x) ^ j f (y) j (cid:0) > " ): What has emerged now is a clearer “picture” of what it means to be discontinuous in x: there must be an " > 0 such that for every (cid:14) > 0 (“no matter how small”) a y > "; i.e., there are numbers can be found with f (x) j y “arbitrarily close to x” such that the values f (x) and f (y) remain at least " apart. < (cid:14), whereas f (y) j x j (cid:0) (cid:0) y j 68 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Different Sorts. Several sorts of objects, may occur in one and the same context. (For instance, sometimes a problem involves vectors as well as reals.) In such a situation, one often uses different variable naming conventions to keep track of the differences between the sorts. In fact, sorts are just like the basic types in a functional programming language. Just as good naming conventions can make a program easier to understand, nam- ing conventions can be helpful in mathematical writing. For instance: the letters n; m; k; : : : are often used for natural numbers, f; g; h; : : : usually indicate that functions are meant, etc. The interpretation of quantifiers in such a case requires that not one, but several domains are specified: one for every sort or type. Again, this is similar to providing explicit typing information in a functional program for easier human digestion. Exercise 2.50 That the sequence a0; a1; a2; : : : limn itive equivalent for the statement that the sequence a0; a1; a2; : : : converge. R converges to a, i.e., that 2 m > n( < (cid:14)). Give a pos- a j R does not an = a, means that (cid:14) > 0 am !1 (cid:0) 2 n 8 9 8 j 2.8 Quantifiers as Procedures One way to look at the meaning of the universal quantifier is as a procedure to test whether a set has a certain property. The test yields t if the set equals the whole is a procedure that maps domain of discourse, and f otherwise. This means that the domain of discourse to t and all other sets to f. Similarly for restricted universal quantification. A restricted universal quantifier can be viewed as a procedure that takes a set A and a property P , and yields t just in case the set of members of A that satisfy P equals A itself. 8 8 can be In the same way, the meaning of the unrestricted existential quantifier specified as a procedure. takes a set as argument, and yields t just in case the argument set is non-empty. A restricted existential quantifier can be viewed as a procedure that takes a set A and a property P , and yields t just in case the set of members of A that satisfy P is non-empty. 9 9 If we implement sets as lists, it is straightforward to implement these quantifier procedures. In Haskell, they are predefined as all and any (these definitions will be explained below): 2.8. QUANTIFIERSASPROCEDURES 69 any, all any p all p :: (a -> Bool) -> [a] -> Bool = or . map p = and . map p The typing we can understand right away. The functions any and all take as their first argument a function with type a inputs and type Bool outputs (i.e., a test for a property), as their second argument a list over type a, and return a truth value. Note that the list representing the restriction is the second argument. To understand the implementations of all and any, one has to know that or and and are the generalizations of (inclusive) disjunction and conjunction to lists. (We have already encountered and in Section 2.2.) They have type [Bool] -> Bool. Saying that all elements of a list xs satisfy a property p boils down to: the list map p xs contains only True (see 1.8). Similarly, saying that some element of a list xs satisfies a property p boils down to: the list map p xs contains at least one True. This explains the implementation of all: first apply map p, next apply and. In the case of any: first apply map p, next apply or. The action of applying a function g :: b -> c after a function f :: a -> b is performed by the function g . f :: a -> c , the composition of f and g. See Section 6.3 below. The definitions of all and any are used as follows: Prelude> any (<3) [0..] True Prelude> all (<3) [0..] False Prelude> The functions every and some get us even closer to standard logical notation. These functions are like all and any, but they first take the restriction argument, next the body: every, some :: [a] -> (a -> Bool) -> Bool every xs p = all p xs xs p = any p xs some 70 CHAPTER2. TALKINGABOUTMATHEMATICALOBJECTS Now, e.g., the formula as a test, as follows: x 8 1; 4; 9 y g9 2 f 2 f g 1; 2; 3 x = y2 can be implemented TAMO> every [1,4,9] (\ x -> some [1,2,3] (\ y -> x == y^2)) True But caution: the implementations of the quantifiers are procedures, not algorithms. A call to all or any (or every or some) need not terminate. The call every [0..] (>=0) will run forever. This illustrates once more that the quantifiers are in essence more complex than the propositional connectives. It also motivates the development of the method of proof, in the next chapter. Exercise 2.51 Define a function unique :: (a -> Bool) -> [a] -> Bool that gives True for unique p xs just in case there is exactly one object among xs that satisfies p. Exercise 2.52 Define a function parity :: [Bool] -> Bool that gives True for parity xs just in case an even number of the xss equals True. Exercise 2.53 Define a function evenNR :: (a -> Bool) -> [a] -> Bool that gives True for evenNR p xs just in case an even number of the xss have property p. (Use the parity function from the previous exercise.) 2.9 Further Reading Excellent books about logic with applications in computer science are [Bur98] and [HR00]. A good introduction to mathematical logic is Ebbinghaus, Flum and Thomas [EFT94]. Chapter 3 The Use of Logic: Proof Preview This chapter describes how to write simple proofs. Section 3.1 is about style of presentation, while Section 3.2 gives the general format of the proof rules. Sec- tions 3.3 and 3.4 describe the rules that govern the use of the connectives and the quantifiers in the proof process. The recipes are summarized in Section 3.5. Section 3.6 gives some strategic hints for handling proof problems. Section 3.7 applies the proof recipes to reasoning about prime numbers; this section also illus- trates how the computer can be used (sometimes) as an instrument for checking particular cases, and therefore as a tool for refuting general claims. Representation and proof are two sides of the same coin. In order to handle the stuff of mathematics, we start out from definitions and try to find meaningful rela- tions. To check our intuitions and sometimes just to make sure that our definitions accomplish what we had in mind we have to provide proofs for our conjectures. To handle abstract objects in an implementation we have to represent them in a concrete way. Again, we have to check that the representation is faithful to the original intention. It turns out that proofs and implementations have many things in common. In the first place, variables are a key ingredient in both. Variables are used to denote members of certain sets. In our example proofs, we will used x; y for rational numbers and real numbers, and m; n for integers. Similarly, the variables used in the implementation of a definition range over certain sets, indicated by means of typings for the variables. 71 72 CHAPTER3. THEUSEOFLOGIC:PROOF The main purpose of this chapter is to spur you on to develop good habits in setting up mathematical proofs. Once acquired, these habits will help you in many ways. Being able to ‘see structure’ in a mathematical proof will enable you to easily read and digest the proofs you find in papers and textbooks. When you have learned to see the patterns of proof, you will discover that they turn up again and again in any kind of formal reasoning. Because it takes time to acquire new habits, it is not possible to digest the contents of this chapter in one or two readings. Once the recommended habits are acquired, but only then, the contents of the chapter have been properly digested, and the chapter will have served its purpose. The module containing the code of this chapter depends on the module containing the code of the previous chapter. In the module declaration we take care to import that module. This is done with the reserved keyword import. module TUOLP where import TAMO 3.1 Proof Style The objects of mathematics are strange creatures. They do not exist in physical space. No one ever saw the number 1. One can consistently argue that mathematics has no subject at all, or maybe that the subject matter of mathematics is in the mind. But what does that mean? For sure, different mathematicians have the same objects in mind when they investigate prime numbers. But how can that be, if mathematics has no subject matter? As to the method of mathematics, there is no disagreement: the heart of the matter is the notion of proof. A proof is an argument aimed at convincing yourself and others of the truth of an assertion. Some proofs are simple, but others can be pieces of art with aesthetic and intellectual qualities. In daily life, people argue a lot, but these efforts are not always capable of convinc- ing others. If you have a row with your partner, more often than not it is extremely difficult to assess who is right. (If this were an easy matter, why have a row in 3.1. PROOFSTYLE 73 the first place?) By contrast, in mathematical matters it usually is very clear who is right and who is wrong. Doing mathematics is a sports where the rules of the game are very clear. Indeed, mathematical proofs go undisputed most of the time. This remarkable phenomenon is probably due to the idealized character of mathe- matical objects. Idealization is a means by which mathematics provides access to the essence of things, and grasping the essence of something is a hallmark of the creative mind. It remains a source of wonder that the results are so often applicable to the real world. The mathematical content of most of the proofs discussed in this chapter is minute; later on you will call such proofs “routine”. For the time being, the attention is focused on the form of proofs. However, to begin with, here follows a list of stylistic commandments. A proof is made up of sentences. Therefore: 1 Write correct English, try to express yourself clearly. If you are not a native speaker you are in the same league as the authors of this book. Still, English is the lingua franca of the exact sciences, so there is no way around learning enough of it for clear communication. Especially when not a native speaker, keep your style simple. Write with a reader in mind, and do not leave your reader in the dark concerning your proof plans. 2 Make sure the reader knows exactly what you are up to. Applying the rules of proof of the next section in the correct way will usually take care of this. In particular, when you feel the need to start writing symbols —often, variables— inform the reader what they stand for and do not let them fall out of the blue. 3 Say what you mean when introducing a variable. Section 3.2 again shows you the way. Use the symbolic shorthands for connectives and quantifiers sparingly. Something like the continuity definition on p. 66 with its impressive sequence of four quanti- fiers is about the maximum a well-educated mathematician can digest. 4 Don’t start a sentence with symbols; don’t write formulas only. 74 CHAPTER3. THEUSEOFLOGIC:PROOF It is best to view symbolic expressions as objects in their own right. E.g., write: The formula (cid:8) is true, or: It holds that (cid:8). Note that a proof that consists of formulas only is not a suitable way to inform the reader about what has been going on in your brain. 5 Use words or phrases like ‘thus’, ‘therefore’, ‘hence’, ‘it follows that’,etc. to link up your formulas. Be relevant and succinct. Do not use the implication symbol in cases that really require “thus” or “there- fore”. Beginner’s proofs can often be recognized by their excessive length. Of course, it will be difficult at first to keep to the proper middle road between com- mandments (4) and (5). ) When embarking on a proof, it is a good idea to write down exactly (i) what can be used in the proof (the Given), (ii) and what is to be proved. It is very helpful to use a schema for this. 6 When constructing proofs, use the following schema: Given: . . . To be proved: . . . Proof: . . . N.B.: A proof using Mathematical Induction offers some extra structure; cf. Sec- tion 7.1 below and p. 400. In the course of this chapter you will discover that proofs are highly structured pieces of texts. The following guideline will help you keep track of that structure. 7 Use layout (in particular, indentation) to identify subproofs and to keep track of the scopes of assumptions. The general shape of a proof containing subproofs will be discussed in Section 3.2. 8 Look up definitions of defined notions, and use these definitions to re-write both Given and To be proved. In particular, elimination of defined notions from the sentence to be proved will show you clearly what it is that you have to prove. Do not expect that you will be able to write down a correct proof at the first try. Do not expect to succeed at the second one. You will have to accept the fact that, for 3.2. PROOFRECIPES 75 the time being, your efforts will not result in faultless proofs. Nevertheless: keep trying! 9 Make sure you have a sufficient supply of scratch paper, make a fair copy of the end-product — whether you think it to be flawless or not. Finally: before handing in what you wrote, try to let it rest for at least one night. And only then, 10 Ask yourself two things: Is this correct? Can others read it? The honest answers usually will be negative at first. . . Fortunately, you did not finish your supply of scratch paper. And surely, if you have understood properly, you must be able to get it down correctly eventually. (Apply to this the law of contraposition in Theorem 2.10, p. 45.) 3.2 Proof Recipes Goal Develop the ability to apply the proof rules in simple contexts. The proof rules are recipes that will allow you to cook up your own proofs. Try to distill a recipe from every rule, and and make sure you remember how to apply these recipes in the appropriate situation. It is completely normal that you get stuck in your first efforts to prove things. Of- ten, you will not even know how to make a first move. This section then may come to your help. It provides you with rules that govern the behaviour of the logical phrases in mathematical proof, and thus with recipes to use while constructing a proof. In fact, constructing proofs has a lot in common with writing computer programs. In structured programming, layout can be used to reveal the building blocks of a program more clearly. We will also use layout as a tool to reveal structure. The most important structure principle is that a proof can contain subproofs, just like a program may contain procedures which have their own sub-procedures, and so on. We will indicate subproofs by means of indentation. The general structure of a proof containing a subproof is as follows: 76 CHAPTER3. THEUSEOFLOGIC:PROOF Given: A, B, . . . To be proved: P Proof: . . . Suppose C . . . To be proved: Q Proof: . . . . . . Thus Q . . . Thus P To be sure, a subproof may itself contain subproofs: Given: A, B To be proved: P Proof: . . . Suppose C To be proved: Q Proof: . . . Suppose D To be proved: R Proof: . . . . . . Thus R . . . Thus Q . . . Thus P 3.2. PROOFRECIPES 77 The purpose of ‘Suppose’ is to add a new given to the list of assumptions that may be used, but only for the duration of the subproof of which ‘Suppose’ is the head. If the current list of givens is P1; : : : ; Pn then ‘Suppose Q’ extends this list to P1; : : : ; Pn; Q. In general, inside a box, you can use all the givens and assumptions of all the including boxes. Thus, in the innermost box of the example, the givens are A; B; C; D. This illustrates the importance of indentation for keeping track of the ‘current box’. There are some 15 rules discussed here for all seven notions of logic. At a first encounter, this may seem an overwhelming number. However, only some of these are really enlightening; several are so trivial that you won’t even notice using them. An example is: Given: P , Q Q. Thus P ^ What is really remarkable is this: together these 15 rules are sufficient to tackle every possible proof problem. Of course, this does not mean that the process of proving mathematical facts boils down in the end to mastery of a few simple rules. Think of it as chess: the rules of the game are extremely simple, and these rules can be used to play very compli- cated and beautiful games. To be able to play you must know the rules. But if your knowledge does not extend beyond the rules, you will only be capable of playing games that are rather dull. Learning to play chess does not stop with learning the rules, and what is beautiful in a game of chess cannot be found in the rules. Sim- ilarly, the rules of proof cannot teach you more than how to prove the simplest of things. (And, in the beginning, that is difficult enough.) It is ideas that make a proof interesting or beautiful, but the rules keep silent about these things. Classification of Rules. Every logical symbol has its own rules of use. There are basically two ways in which you can encounter a logical symbol: it can either appear in the given or in the statement that is to be proved. In the first case the rule to use is an elimination rule, in the second case an introduction rule. Elimination rules enable you to reduce a proof problem to a new, hopefully simpler, one. Introduction rules make clear how to prove a goal of a certain given shape. All rules are summarized in Section 3.5. Safety. As we go along, we will provide arguments that try to explain why a certain rule is safe. A rule is safe if it will never allow you to prove something false on the basis of statements that are true. Obviously, this is a requirement that 78 CHAPTER3. THEUSEOFLOGIC:PROOF proofs should fulfill. That the rules for connectives are safe is due to truth tables. Safety of two of the quantifier rules is obvious; the remaining ones are tough nuts. Don’t worry when these explanations in some cases appear mystifying. Eventually, you’ll understand! 3.3 Rules for the Connectives Implication Here come a complicated but important introduction rule and a trivial one for elim- ination. Introduction The introduction rule for implication is the Deduction Rule. It en- (cid:9). Instead, it ables you to reduce the problem of proving an implication (cid:8) prescribes to assume (cid:8) as an additional new Given, and asks you to derive that (cid:9). ) Given: . . . To be proved: (cid:8) Proof: (cid:9) ) Suppose (cid:8) To be proved: (cid:9) Proof: . . . (cid:9). Thus (cid:8) ) Safety. In case (cid:8) is false, the implication (cid:8) (cid:9) will be true anyway. Thus, the case that is left for you to consider is when (cid:8) is true, meaning that you can assume it as given. But then of course you should show that (cid:9) is true as well (otherwise, (cid:8) (cid:9) would be false). ) ) Example 3.1 We show that the implication P given P ) (p. 74) for the first time: Q and Q R is provable on the basis of the R. Thus, employing the schema in the 7th commandment ) ) Q, Q Given: P To be proved: P Proof: ) ) ) R R Given: P To be proved: R ) Q, Q R (old), P (new) ) 3.3. RULESFORTHECONNECTIVES 79 Proof: From P Next, from Q Q and P , conclude Q. R and Q, conclude R. ) ) Thus, according to the deduction rule, P R ) Here is a slightly more concise version: Q, Q Given: P To be proved: P Proof: ) ) ) R R Suppose P To be proved: R Proof: From P Next, from Q ) ) Thus P R ) Q and P , conclude Q. R and Q, conclude R. Detailed vs. Concise Proofs. The proof just given explains painstakingly how the Deduction Rule has been applied in order to get the required result. However, in practice, you should write this in a still more concise way: Q, Q Given: P To be proved: P Proof: ) ) ) R R Suppose P From P Next, from Q ) Q and P , conclude Q. R and Q, conclude R. ) Note that the final concise version does not mention the Deduction Rule at all. The actual application of this rule should follow as a last line, but that is left for the reader to fill in. This is not considered an incomplete proof: from the situation it is perfectly clear that this application is understood. Several other examples of detailed proofs and their concise versions are given in what follows. Here is the key to the proof of an implication: If the ‘to be proved’ is an implication (cid:8) following Obligatory Sentence: ) (cid:9), then your proof should start with the Suppose that (cid:8) holds. 80 CHAPTER3. THEUSEOFLOGIC:PROOF The obligatory first sentence accomplishes the following things (cf. the 2nd com- mandment on p. 73). (cid:15) (cid:15) (cid:15) It informs the reader that you are going to apply the Deduction Rule in order to establish that (cid:8) (cid:9). ) The reader also understands that it is now (cid:9) that you are going to derive (instead of (cid:8) (cid:9)). ) Thus, starting with the obligatory sentence informs the reader in an efficient way about your plans. Reductive Character. givens P Deduction Rule, to the problem of showing that from the givens P R; P it can be proved that R. This requires a subproof. In Example (3.1) the problem of showing that from the R was reduced, using the R it can be proved that P Q; Q ) ) Q; Q ) ) ) Marking Subproofs with Indentation. Note that only during that subproof you are allowed to use the new given P . Some logic texts recommend the use of markers to indicate beginning and end of such a subproof, between which the extra given is available. In our example proofs in this chapter, we indicate the beginnings and ends of subproofs by means of indentation. In more colloquial versions of the proofs, identifying the beginning and end of a subproof is left to the reader. In this particular case, the reduced proof problem turned out to be easy: P and R together produce R, as P desired. (Of course, in general, such a subproof may require new reductions.) Q together produce Q; whereas Q and Q ) ) The last two steps in this argument in fact apply the Elimination Rule of implica- tion, which is almost too self-evident to write down: Elimination. This rule is also called Modus Ponens. In words: from (cid:8) (cid:8) you can conclude that (cid:9). ) (cid:9) and In a schema: Given: (cid:8) Thus (cid:9). ) (cid:9), (cid:8) Safety. Immediate from truth tables: if (cid:8) (cid:9) and (cid:8) are both true, then so is (cid:9). ) 3.3. RULESFORTHECONNECTIVES 81 The two proof rules for implication enable you to handle implications in all types of proof problems. You may want to use more obvious properties of implication, but fact is that they usually can be derived from the two given ones. Example (3.1) is a case in point, as is Exercise 3.2. (We will not bother you with the exceptions; one is in Exercise 3.9.) Exercise 3.2 Apply both implication rules to prove P Q; P R). (Q ) ) R from the givens P ) ) Conjunction The conjunction rules are almost too obvious to write down. Introduction. A conjunction follows from its two conjuncts taken together. In a schema: Given: (cid:8); (cid:9) (cid:9). Thus (cid:8) ^ Elimination. From a conjunction, both conjuncts can be concluded. Schematically: Given: (cid:8) Thus (cid:8). ^ (cid:9) Given: (cid:8) Thus (cid:9). ^ (cid:9) Suppose we want to check whether the sum of even natural numbers always is even. It is not difficult to implement this check in Haskell, using the built-in func- tion even and the list of even numbers. evens = [ x | x <- [0..], even x ] Formulating the check is easy, but of course we won’t get an answer, as the check takes an infinite amount of time to compute. 82 CHAPTER3. THEUSEOFLOGIC:PROOF TUOLP> forall [ m + n | m <- evens, n <- evens ] even If we want to check a statement about an infinite number of cases, we can either look for a counterexample, or we can attempt to give a proof. In the present case, the latter is easy. Example 3.3 Assume that n; m n is even) m + n is even. N. 2 ) To show: (m is even ^ Detailed proof: n even). -elimination) m and n are both even. Assume that (m even Then ( ^ For instance, p; q Then m + n = 2p + 2q = 2(p + q) is even. ^ 2 N exist such that m = 2p, n = 2q. The result follows using the Deduction Rule. Concise proof: Assume that m and n are even. For instance, m = 2p, n = 2q, p; q Then m + n = 2p + 2q = 2(p + q) is even. N. 2 Exercise 3.4 Assume that n; m Show: (m is odd n is odd) ^ ) N. 2 m + n is even. Equivalence An equivalence can be thought of as the conjunction of two implications. Thus, . the rules follow from those for ^ and ) Introduction. In order to prove (cid:8) example 2.3): , (cid:9), you have to accomplish two things (cf. ( ( ) add (cid:8) as a new given, and show that (cid:9); ) ) add (cid:9) as a new given, and show that (cid:8). ( If you can do this, (cid:8) (cid:9) has been proved. , 3.3. RULESFORTHECONNECTIVES 83 Concise Proof Schema. Given: . . . To be proved: (cid:8) Proof: (cid:9) , Suppose (cid:8) To be proved: (cid:9) Proof: . . . Suppose (cid:9) To be proved: (cid:8) Proof: . . . Thus (cid:8) (cid:9). , , Instead of you may also encounter ‘iff’ or ‘if and only if’. A proof of a statement of the form ‘(cid:8) iff (cid:9)’ consists of two subproofs: the proof of the ‘only if’ part and (cid:9), and the proof of the ‘if’ part. Caution: the ‘only if’ part is the proof of (cid:8) the ‘if’ part is the proof of (cid:8) (cid:9)’, and ‘(cid:8) if (cid:9)’ means ‘(cid:8) ) (cid:9). This is because ‘(cid:8) only if (cid:9)’ means ‘(cid:8) (cid:9)’. Thus, we get: ( ) ( Given: . . . To be proved: (cid:8) iff (cid:9) Proof: Only if: Suppose (cid:8) If: To be proved: (cid:9) Proof: . . . Suppose (cid:9) To be proved: (cid:8) Proof: . . . Thus (cid:8) iff (cid:9). Elimination. Schematically: Given: (cid:8) Thus (cid:9) , (cid:9), (cid:8), . . . Given: (cid:8) Thus (cid:8) , (cid:9), (cid:9), . . . Exercise 3.5 Show: 84 CHAPTER3. THEUSEOFLOGIC:PROOF 1. From P 2. From P , , Q it follows that (P Q it follows that (R R) P ) (Q (R ) ) , , ) ) R), Q). Negation General Advice. In no matter what concrete mathematical situation, before ap- plying any of the negation rules given below: whether you want to prove or use a negated sentence, you should first attempt to convert into something positive. If this succeeds, you can turn to the other rules. Theorems 2.10 (p. 45) and 2.40 (p. 65) contain some tools for doing this: you can move the negation symbol inward, across quantifiers and connectives. If this pro- cess terminates, the “mathematics” of the situation often allows you to eliminate b 6 x) can be rewritten as the negation symbol altogether. E.g., x(x < a x < b). For a more complicated example, cf. (2.49) on p. 67. :8 _ x(a 6 x 9 ^ Remark. The general advice given above does not apply to the exercises in the present chapter. Firstly, many of the exercises below are purely logical: there is no mathematical context at all. Secondly, all exercises here are designed to be solved using the rules only; possible shortcuts via the results of Chapter 2 will often trivialize them. Introduction. If (cid:8) is to be proved, do the following. Assume (cid:8) as a new given, and attempt to prove something (depending on the context) that is evidently false. : This strategy clearly belongs to the kind of rule that reduces the proof problem. Schematically: Given: . . . To be proved: Proof: (cid:8) : Suppose (cid:8) To be proved: Proof: . . . (cid:8). : ? Thus Evidently False. Here, ? stands for the evidently false statement. 3.3. RULESFORTHECONNECTIVES 85 In a mathematical context, this can be anything untrue, such as 1 = 0. For a more complicated falsehood, cf. the proof that limits are unique on p. 317. In the logical examples below, may consist of the occurrence of a statement together with its negation. In that case, one statement contradicts the other. For instance, you might derive a sentence (cid:9), thereby contradicting the occurrence of (cid:9) among the given (cid:0). Cf. Examples (3.8) and and (3.30). ? : Example 3.6 The proof of theorem 3.33 (the number of primes is not finite) is an example of the method of negation introduction. Another example can be found in the proof of theorem 8.14 (the square root of 2 is not rational). Exercise 3.7 Produce proofs for: 1. Given: P Q. To show: Q : P , ) : ) 2. Given P Q. To show: P : Q. , : , Safety. Suppose that from (cid:0); (cid:8) it follows that Then (cid:8) cannot be true. (Otherwise, your proof would show the evidently false to be true as well.) Thus, , and that all given (cid:0) are satisfied. (cid:8) must be true. ? ? : Elimination. When you intend to use the given (cid:8), you can attempt to prove, on the basis of the other given, that (cid:8) must hold. In that case, the elimination rule declares the proof problem to be solved, no matter what the statement To be proved! : Schematically: Given: (cid:8), Thus (cid:9). (cid:8) : Safety. The rule cannot help to be safe, since you will never find yourself in a (cid:8) are both true. (Remarkably, this nevertheless is a useful situation where (cid:8) and rule!) : There is one extra negation rule that can be used in every situation: Proof by Contradiction, or Reductio ad Absurdum. 86 CHAPTER3. THEUSEOFLOGIC:PROOF Proof by Contradiction. In order to prove (cid:8), add to deduce an evidently false statement. (cid:8) as a new given, and attempt : In a schema: Given: . . . To be proved: (cid:8) Proof: (cid:8) Suppose To be proved: Proof: . . . : ? Thus (cid:8). Safety. The argument is similar to that of the introduction rule. Advice. Beginners are often lured into using this rule. The given (cid:8) that comes in free looks so inviting! However, many times it must be considered poisoned, making for a cluttered bunch of confused givens that you will not be able to disen- tangle. It is a killer rule that often will turn itself against its user, especially when that is a beginner. Proof by Contradiction should be considered your last way out. Some proof problems do need it, but if possible you should proceed without: you won’t get hurt and a simpler and more informative proof will result. : Comparison. Proof by Contradiction looks very similar to the introduction rule (both in form and in spirit), and the two are often confused. Indeed, in ordinary mathematical contexts, it is usually “better” to move negation inside instead of applying -introduction. : : Example 3.8 From Q : P it follows that P ) : Q. ) P Q Given: ) : : To be proved: P Detailed proof: Q ) Suppose P To be proved: Q Proof: : Suppose Q To be proved: Proof: From : From P and Q and ? : Q : ) : P derive P derive P . : . ? 3.3. RULESFORTHECONNECTIVES 87 Thus, Q, by contradiction. Thus, P Q, by the Deduction Rule. ) Concise proof: Assume that P . If Q, then (by Contradiction. : Q : P ) it follows that ) : P . : Exercise 3.9* Show that from (P Q) ) ) P it follows that P . Hint. Apply Proof by Contradiction. (The implication rules do not suffice for this admittedly exotic example.) Disjunction Introduction. A disjunction follows from each of its disjuncts. Schematically: Given: (cid:8) Thus (cid:8) _ Given: (cid:9) Thus (cid:8) _ (cid:9). (cid:9). Safety is immediate from the truth tables. Elimination. You can use a given (cid:8) (cid:8), and one employing (cid:9). _ (cid:9) by giving two proofs: one employing In a proof schema: (cid:9), . . . Given: (cid:8) To be proved: (cid:3) Proof: _ Suppose (cid:8) To be proved: (cid:3) Proof: . . . Suppose (cid:9) To be proved: (cid:3) 88 CHAPTER3. THEUSEOFLOGIC:PROOF Proof: . . . Thus (cid:3). Example 3.10 We show that from P Q; _ P it follows that Q. : Given: P Q; : To be proved: Q. Proof: _ P . Suppose P . Then from P and Suppose Q. Then Q holds by assumption. P we get Q. : Therefore Q. Exercise 3.11 Assume that A; B; C and D are statements. 1. From the given A B the previous example.) ) C and B _ A, derive that A ) : C. (Hint: use ) 2. From the given A D. B ) B C ) _ _ D, C ) A, and B A, derive that ) : Example 3.12 Here is a proof of the second DeMorgan law (Theorem 2.10): (P Given: _ To be proved: : Detailed proof: : Q). P Q. ^ : Assume P . By This contradicts the given (i.e., we have an evident falsity here). -introduction, we get -introduction, it follows that P Q. _ _ : By In a similar way, By : -introduction, it follows that : Q can be derived. P P . ^ ^ : The concise version (no rules mentioned): Assume P . Then a fortiori P : Q. Thus, Thus, P . Similarly, P Q follows. : : : ^ : _ Q is derivable. Q holds, contradicting the given. 3.3. RULESFORTHECONNECTIVES 89 Example 3.13 The following example from the list of equivalences in Theorem (2.10) is so strange that we give it just to prevent you from getting entangled. Q) (P Given: ) : To be proved: P Proof: By Q ^ : ^ -introduction, it suffices to prove both P and Q. : To be proved: P Proof: (by contradiction) Suppose that P . : Then if P holds, Q follows by Thus (Deduction Rule), we get that P However, this contradicts the given. : -elimination. Q. ) To be proved: Proof: (by Q : introduction) : Assume that Q. Then, by a trivial application of the Deduction Rule P (Trivial, since we do not need P at all to conclude Q.) Again, this contradicts the given. Q follows. ) Note that the rule for implication introduction can be used for reasoning by cases, P is a logical truth, it can always be added to the list of as follows. Because P P both yield conclusion Q, then this proves givens. If the two sub cases P and Q. Here is the schema: _ : : Given: . . . To be proved: Q Proof: Suppose P . To be proved: Q. Proof: . . . P . Suppose To be proved: Q. Proof: . . . : Thus Q. This pattern of reasoning is used in the following examples. Example 3.14 90 CHAPTER3. THEUSEOFLOGIC:PROOF Let n Proof: 2 N. To be proved: n2 n is even. (cid:0) Assume n even. Then n = 2m, so n2 and therefore n2 Assume n odd. Then n = 2m + 1, so n2 and therefore n2 (cid:0) (cid:0) (cid:0) n is even. (cid:0) Thus n2 (cid:0) n is even. n = (n 1)n = (2m 1)2m, (cid:0) (cid:0) n is even. n = (n 1)n = 2m(2m + 1), (cid:0) Exercise 3.15 Show that for any n 1. 2 N, division of n2 by 4 gives remainder 0 or Example 3.16 Let R be the universe of discourse, and let P (x) be the following property: In other words, x has property P iff x is irrational and xp2 is rational. We will show that either p2 or p2 p2 has property P . Q x = 2 ^ xp2 Q: 2 Q. Q (Theorem 8.14), we know that p2 has P . p2 (cid:1) p2 = p2 2 = 2 2 Q, we know that p2 p2 has P . p2 Suppose p2 2 Then, since p2 = 2 Thus, P (p2) P (p2 p2). _ p2 = Q. Suppose p2 2 p2)p2 = p2 Then, since (p2 p2). Thus, P (p2) P (p2 _ Therefore P (p2) P (p2 p2). _ Exercise 3.17 Prove the remaining items of Theorem 2.10 (p. 45). To prove (cid:8) (cid:9) means (i) to derive (cid:9) from the given (cid:8) and (ii) to derive (cid:8) from the given (cid:9). (cid:17) 3.4 Rules for the Quantifiers The rules for the quantifiers come in two types: for the unrestricted, and for the restricted versions. Those for the restricted quantifiers can be derived from the others: see Exercise 3.32 p. 102. 3.4. RULESFORTHEQUANTIFIERS 91 Universal Quantifier Introduction. When asked to prove that Sentence: x E(x), you should start a proof by writing the Obligatory 8 Suppose that c is an arbitrary object. And you proceed to show that this object (about which you are not supposed to assume extra information; in particular, it should not occur earlier in the argument) has the property E in question. Schematic form: Given: . . . To be proved: Proof: xE(x) 8 Suppose c is an arbitrary object To be proved: E(c) Proof: . . . Thus xE(x) 8 Here is the Modification suitable for the restricted universal quantifier: A E(x) is to be proved, you should start proof this time by the, again, obliga- x If 8 tory: 2 Suppose that c is any object in A. And you proceed to show that this object (about which you only assume that it belongs to A) has the property E in question. Schematic form: 92 CHAPTER3. THEUSEOFLOGIC:PROOF Given: . . . To be proved: Proof: x 8 2 A E(x) Suppose c is any object in A To be proved: E(c) Proof: . . . Thus x 8 2 A E(x) Arbitrary Objects. You may wonder what an arbitrary object is. For instance: what is an arbitrary natural number? Is it large? small? prime? etc. What exactly is an arbitrary object in A when this set happens to be a singleton? And: are there objects that are not arbitrary? Answer: the term ‘arbitrary object’ is only used here as an aid to the imagination; it indicates something unspecified about which no special assumptions are made. Imagine that you allow someone else to pick an object, and that you don’t care what choice is made. ‘Suppose c is an arbitrary A’ is the same as saying to the reader: ‘Suppose you provide me with a member c from the set A; the choice is completely up to you.’ Often, a universal quantifier occurs in front of an implication. Therefore, you may find the following rule schema useful. Given: . . . To be proved: Proof: x(P (x) 8 ) Q(x) Suppose c is any object such that P (c) To be proved: Q(c) Proof: . . . Thus x(P (x) 8 ) Q(x)) This rule is derivable from the introduction rules for similar to the recipe for restricted -introduction. Cf. Exercise 3.18. and ) 8 . Note that it is very 8 Elimination. Schematic form: Given: Thus E(t). 8 x E(x) Here, t is any object of your choice. 3.4. RULESFORTHEQUANTIFIERS 93 That this rule is safe is obvious: if every thing satisfies E, then in particular so must t. Modification suitable for the restricted universal quantifier, in schematic form: Given: x Thus E(t). 8 2 A E(x), t A 2 -introduction and Deduction Rule: if from (cid:0); P (c) it Exercise 3.18 Show, using follows that Q(c) (where c satisfies P , but is otherwise “arbitrary”), then from (cid:0) it follows that x(P (x) Q(x)). 8 8 ) Existential Quantifier Introduction. In order to show that which E(t) holds. x E(x), it suffices to specify one object t for 9 Schematic form: Given: E(t) Thus, x E(x). 9 Here, t can be anything: any example satisfying E can be used to show that xE(x). 9 Modification suitable for the restricted existential quantifier, in schematic form: Given: E(t); t Thus, x A 2 A E(x). 9 2 That these rules are safe goes without saying. Example-objects. An object t such that E(t) holds is an example-object for E. Thus, the introduction rule concludes xE(x) from the existence of an example- object. 9 However, it is not always possible or feasible to prove an existential statement by exhibiting a specific example-object, and there are (famous) proofs of existential statements that do not use this rule. Example 3.19 A transcendent real is a real which is not a root of a polynomial with integer coefficients (a polynomial f (x) = anxn +an 1 +: : :+a1x+a0, where the ai are integers and an = 0). 1xn (cid:0) (cid:0) By an argument establishing that the set of reals must be strictly larger than the set of roots of polynomials with integer coefficients, it is relatively easy to show 6 94 CHAPTER3. THEUSEOFLOGIC:PROOF that transcendent reals exist. (Compare the reasoning about degrees of infinity in Appendix 11.) Still, it is not immediately clear how to get from this argument at an example- transcendent real. In particular, it is hard to see that e and (cid:25) are transcendent reals. Example 3.20 There is a proof that, in chess, either white has a winning strategy, or black has a strategy with which he cannot lose. However, the proof neither informs you which of the two cases you’re in, nor describes the strategy in a usable way. Example 3.21 For a logical example, look at the proof given in Example 3.30 (p. 101). What is proved there is introduction is used somewhere) no example-x for (cid:8) is exhibited. And, on the basis of the given (cid:8)(x), but (although : x 9 9 x(cid:8)(x), it is unrealistic to expect such an example. :8 Sometimes, it is known only that an example-object must be present among the members of a certain (finite) set but it is impossible to pinpoint the right object. See the following example. Example 3.22 Given: P (a) To be proved: Proof: By P (b). But this is immediate, by xP (x). _ _ P (b) 9 -elimination, it is sufficient to prove introduction. 9 xP (x) from both P (a) and 9 Example 3.23 To make example (3.22) more concrete, consider the following question. Is there an irrational number (cid:11) with the property that (cid:11)p2 is rational? In Example 3.16 we established that either p2 or p2 p2 has this property. Thus, the answer to the question is ‘yes’. But the reasoning does not provide us with an example of such a number. It is a general feature of Proofs by Contradiction of existential statements that no example objects for the existential will turn up in the proof, and this is one reason to stay away from this rule as long as you can in such cases. Elimination. (Note the similarity with the -rule.) _ 3.4. RULESFORTHEQUANTIFIERS 95 When you want to use that Sentence: xE(x) in an argument to prove (cid:3), you write the Obligatory 9 Suppose that c is an object that satisfies E. However, this is all that you are supposed to assume about c. Now, you proceed to prove (cid:3) on the basis of this assumption. Schematic form: xE(x), . . . Given: 9 To be proved: (cid:3) Proof: Suppose c is an object that satisfies E To be proved: (cid:3) Proof: . . . Thus (cid:3) Modification suitable for the restricted existential quantifier: When you want to use that x 9 2 A E(x) in an argument to prove (cid:3), you write Suppose that c is an object in A that satisfies E. Again, this is all that you are supposed to assume about c. Subsequently, you proceed to prove (cid:3) on the basis of this assumption. Schematic form: xE(x), . . . A; Given: c 9 To be proved: (cid:3) Proof: 2 Suppose c is an object in A that satisfies E To be proved: (cid:3) Proof: . . . Thus (cid:3) 96 CHAPTER3. THEUSEOFLOGIC:PROOF 3.5 Summary of the Proof Recipes Here is a summary of the rules, with introduction rules on the left hand side, and elimination rules on the right hand side. Proof by Contradiction has been put in the Elimination-column. The Recipes for ) Introduction and Elimination Given: . . . To be proved: P Proof: Q ) Suppose P To be proved: Q Proof: . . . Q. Thus P ) Given: P , P Thus Q ) Q, . . . The Recipes for , Introduction and Elimination Given: . . . To be proved: P Proof: Q , Suppose P To be proved: Q Proof: . . . Suppose Q To be proved: P Proof: . . . Thus P Q. , Given: P , P Thus Q , Q, . . . Given: Q, P Thus P , Q, . . . 3.5. SUMMARYOFTHEPROOFRECIPES 97 The Recipes for : Introduction and Elimination Given: . . . To be proved: Proof: P : Suppose P To be proved: Proof: . . . P . : ? Thus Given: . . . To be proved: P Proof: P Suppose To be proved: Proof: . . . : ? Thus P . Given: P , Thus Q. P : The Recipes for ^ Introduction and Elimination Given: P , Q Q. Thus P ^ Given: P Thus P . Q ^ Given: P Thus Q. Q ^ The Recipes for _ Introduction and Elimination Given: P Thus P _ Given: Q Thus P _ Q, . . . Given: P To be proved: R Proof: _ Q. Q. Suppose P To be proved: R Proof: . . . Suppose Q To be proved: R Proof: . . . Thus R. 98 CHAPTER3. THEUSEOFLOGIC:PROOF The Recipes for 8 Introduction and Elimination Given: . . . To be proved: Proof: xE(x) 8 Suppose c is an arbitrary object To be proved: E(c) Proof: . . . Thus xE(x) 8 Given: . . . To be proved: Proof: x 8 2 A E(x) Suppose c is any object in A To be proved: E(c) Proof: . . . Thus x 8 2 A E(x) xE(x), . . . Given: Thus E(t). 8 Given: x Thus E(t). 8 2 A E(x), t A, . . . 2 The Recipes for 9 Introduction and Elimination Given: E(t), . . . Thus xE(x). 9 xE(x), . . . Given: 9 To be proved: P Proof: Suppose c is an object that satisfies E To be proved: P Proof: . . . Thus P 3.6. SOMESTRATEGICGUIDELINES 99 Given: t Thus x 9 2 2 A, E(t), . . . A E(x). A E(x), . . . x Given: 9 To be proved: P Proof: 2 Suppose c is an object in A that satisfies E To be proved: P Proof: . . . Thus P 3.6 Some Strategic Guidelines Here are the most important guidelines that enable you to solve a proof problem. 1. Do not concentrate on the given, by trying to transform that into what is to be proved. 2. Instead, concentrate on (the form of) what is to be proved. 3. A number of rules enable you to simplify the proof problem. For instance: (cid:15) (cid:15) When asked to prove P (Deduction Rule). ) Q, add P to the givens and try to prove Q. When asked to prove ( -introduction). 8 x E(x), prove E(c) for an arbitrary c instead 8 4. Only after you have reduced the problem as far as possible you should look at the givens in order to see which of them can be used. (cid:15) (cid:15) When one of the givens is of the form P Q, and R is to be proved, make a case distinction: first add P to the givens and prove R, next add Q to the givens and prove R. _ When one of the givens is of the form x E(x), and P is to be proved, give the object that satisfies E a name, by adding E(c) to the givens. Next, prove P . 9 5. It is usually a good idea to move negations inward as much as possible before attempting to apply -introduction. : 6. Stay away from Proof by Contradiction as long as possible. 100 CHAPTER3. THEUSEOFLOGIC:PROOF Example 3.24 To show: from x(P (x) Q(x)); xP (x) it follows that xQ(x). ) Concise proof: Using the second given, assume that x is such that P (x) holds. Applying the first given to this x, it follows that P (x) Q(x). Thus, we have that Q(x). Conclusion: xQ(x). ) 9 8 9 9 Exercise 3.25 Show: 1. from 2. from x(P (x) 8 x(P (x) 9 ) ) Q(x)); xP (x) it follows that 8 xQ(x), 8 Q(x)); xP (x) it follows that 8 xQ(x). 9 What about: from x(P (x) 9 ) Q(x)); xP (x) it follows that 9 xQ(x) ? 9 Exercise 3.26 From the given x 8 y(xRy); 9 8 x(xRx). 8 derive that x y(xRy 8 yRx); ) x y 8 z(xRy 8 ^ yRz ) xRz); 8 Exercise 3.27 Give proofs for the following: 1. From x 8 yRx), : y 8 z(xRy 8 yRz ^ ) xRz); xRx it follows that x 8 : x y(xRy ) 8 8 2. From 3. From x 8 x = y), x 8 y(xRy 8 y(xRy 8 yRx) it follows that ) : xRx, x 8 : x = y ^ yRx) it follows that ) : x y(xRy 8 8 yRx ^ ) 4. From x 8 it follows that : xRx; x 8 x :9 y(xRy 8 y(xRy). 9 yRx); x y 8 z(xRy 8 ^ yRz ) xRz) 8 ) The following exercise is an example on how to move a quantifier in a prefix of quantifiers. Exercise 3.28 Show that from y 8 z 9 xP (x; y; z) it follows that 8 x y 8 zP (x; y; z). 9 8 6 3.6. SOMESTRATEGICGUIDELINES 101 Example 3.29 That a real function f is continuous means that (cf. p. 66), in the domain R, x 8 " > 0 8 (cid:14) > 0 9 y ( 8 x j y j (cid:0) < (cid:14) f (x) ) j f (y) j (cid:0) < " ): Uniform continuity of f means j y 8 9 8 x (cid:0) y ( x j " > 0 (cid:14) > 0 < (cid:14) = f (y) j 8 x has moved two places. Ac- Compared with the first condition, the quantifier 8 cording to Exercise 3.28 (where P (x; "; (cid:14)) is < x y( j ")), continuity is implied by uniform continuity. (But, as you may know, the im- plication in the other direction does not hold: there are continuous functions that are not uniformly so.) f (y) j < " ): ) j f (x) f (x) ) j < (cid:14) (cid:0) (cid:0) (cid:0) 8 y j Example 3.30 :8 x(cid:8)(x). x Given: To be proved: Proof: We apply Proof by Contradiction. (cid:8)(x). : 9 Assume, for a contradiction, that We will show that x(cid:8)(x). x (cid:8)(x). :9 : 8 Suppose x is arbitrary. To show that (cid:8)(x), we apply Proof by Contradiction again. Assume Then x Thus (cid:8)(x). 9 : : (cid:8)(x). (cid:8)(x), and contradiction with :9 x (cid:8)(x). : Therefore x 8 (cid:8)(x). Thus 9 : x(cid:8)(x), and contradiction with x(cid:8)(x). :8 The concise and more common way of presenting this argument (that leaves for the reader to find out which rules have been applied and when) looks as follows. x : :9 (cid:8)(x) and let x be arbitrary. : (cid:8)(x) is true, then so is Proof. Assume that If : Thus, (cid:8)(x) holds; and, since x was arbitrary, we conclude that However, this contradicts the first given; which proves that, in fact, be true. (cid:8)(x), contradicting the assumption. x 8 9 9 x(cid:8)(x). x : (cid:8)(x) must Exercise 3.31 Prove the other equivalences of Theorem 2.40 (p. 65). (To prove (cid:9) means (i) deriving (cid:9) from the given (cid:8) and (ii) deriving (cid:8) from the given (cid:8) (cid:17) (cid:9).) 102 CHAPTER3. THEUSEOFLOGIC:PROOF Exercise 3.32 Derive the rules for the restricted quantifiers from the others, using the facts (cf. the remark preceding Example (2.45), page 66) that A E(x) x 8 is equivalent with A E(x) is equivalent with A E(x)), and 8 E(x)). x(x x(x ) A 2 2 2 x 9 9 2 ^ With practice, you will start to see that it is often possible to condense proof steps. Here are some examples of condensed proof recipes: Given: . . . To be proved: Proof: x(A(x) 8 ) B(x)). Suppose c is an arbitrary object such that A(c). To be proved: B(c). Proof: . . . Thus x(A(x) 8 ) B(x)). Given: . . . To be proved: Proof: x y A(x; y). 8 8 Suppose c and d are arbitrary objects. To be proved: A(c; d). Proof: . . . Thus x y A(x; y). 8 8 Given: . . . To be proved: Proof: x 8 2 A y 8 B R(x; y). 2 Suppose c, d are arbitrary objects such that A(c) and B(d). To be proved: R(c; d). Proof: . . . Thus x 8 2 A y 8 B R(x; y). 2 3.7. REASONINGANDCOMPUTATIONWITHPRIMES 103 Given: . . . To be proved: Proof: x y(R(x; y) 8 8 S(x; y)). ) Suppose c, d are arbitrary objects such that R(c; d). To be proved: S(c; d). Proof: . . . Thus x y(R(x; y) 8 8 S(x; y)). ) 3.7 Reasoning and Computation with Primes In this section we will demonstrate the use of the computer for investigating the theory of prime numbers. For this, we need the code for prime that was given in Chapter 1. It is repeated here: prime :: Integer -> Bool prime n | n < 1 = error "not a positive integer" | n == 1 = False | otherwise = ldp n == n where = ldpf primes ldp ldpf (p:ps) m | rem m p == 0 = p = m = ldpf ps m | p^2 > m | otherwise primes = 2 : filter prime [3..] Euclid (fourth century B.C.) proved the following famous theorem about prime numbers. Theorem 3.33 There are infinitely many prime numbers. Proof. Suppose there are only finitely many prime numbers, and p1; : : : ; pn is a pn) + 1. Note that m is list of all primes. Consider the number m = (p1p2 (cid:1) (cid:1) (cid:1) pn and remainder 1. not divisible by p1, for dividing m by p1 gives quotient p2 (cid:1) (cid:1) (cid:1) Similarly, division by p2; p3; : : : always gives a remainder 1. Thus, we get the following: 104 (cid:15) (cid:15) CHAPTER3. THEUSEOFLOGIC:PROOF LD(m) is prime, For all i 1; : : : n 2 f , LD(m) g = pi. Thus, we have found a prime number LD(m) different from all the prime numbers in our list p1; : : : ; pn, contradicting the assumption that p1; : : : ; pn was the full list of prime numbers. Therefore, there must be infinitely many prime numbers. j f n 4n + 3 N (See Example 5.90 below). Show that Exercise 3.34 Let A = A contains infinitely many prime numbers. (Hint: any prime > 2 is odd, hence of the form 4n + 1 or 4n + 3. Assume that there are only finitely many primes of the form 4n + 3, say p1; : : : ; pm. Consider the number N = 4p1 (cid:1) (cid:1) (cid:1) 1 = 1) + 3. Argue that N must contain a factor 4q + 3, using the fact 4(p1 (cid:1) (cid:1) (cid:1) that (4a + 1)(4b + 1) is of the form 4c + 1.) pm pm (cid:0) (cid:0) 2 g Use filter prime [ 4*n + 3 | n <- [0..] ] to generate the primes of this form. Euclid’s proof suggests a general recipe for finding bigger and bigger primes. Finding examples of very large primes is another matter, of course, for how do you know whether a particular natural number is a likely candidate for a check? Example 3.35 A famous conjecture made in 1640 by Pierre de Fermat (1601– 1665) is that all numbers of the form n 22 + 1 + 1 = 22 + 1 = 5, 222 are prime. This holds for n = 0; 1; 2; 3; 4, for we have: 220 221 prime, and 224 as Fermat got. + 1 = 21 + 1 = 3, + 1 = 28 + 1 = 257, which is + 1 = 216 + 1 = 65537, which is prime. Apparently, this is as far + 1 = 24 + 1 = 17, 223 Our Haskell implementation of prime allows us to refute the conjecture for n = 5, using the built-in function ^ for exponentiation. We get: TUOLP> prime (2^2^5 + 1) False This counterexample to Fermat’s conjecture was discovered by the mathematician L´eonard Euler (1707–1783) in 1732. The French priest and mathematician Marin Mersenne (1588–1647; Mersenne was a pen pal of Descartes) found some large prime numbers by observing that Mn = 2n 1 sometimes is prime when n is prime. (cid:0) 6 3.7. REASONINGANDCOMPUTATIONWITHPRIMES 105 Exercise 3.36 It is not very difficult to show that if n is composite, Mn = 2n (cid:0) composite too. Show this. (Hint: Assume that n = ab and prove that xy = 2n 1 and y = 1 + 2b + 22b + for the numbers x = 2b + 2(a 1)b). 1 is 1 (cid:0) (cid:0) (cid:0) But when n is prime, there is a chance that 2n 1 = 7, 25 22 1 = 3, 23 (cid:0) (cid:0) (cid:0) (cid:0) (cid:1) (cid:1) (cid:1) 1 = 31. Such primes are called Mersenne primes. 1 is prime too. Examples are Example 3.37 Let us use the computer to find one more Mersenne prime. Put the procedure prime in a file and load it. Next, we use ^ for exponentiation to make a new Mersenne guess, as follows: TUOLP> prime 5 True TUOLP> prime (2^5-1) True TUOLP> 2^5-1 31 TUOLP> prime (2^31-1) True TUOLP> 2^31-1 2147483647 TUOLP> It may interest you to know that the fact that 231 Euler in 1750. Using a computer, this fact is a bit easier to check. (cid:0) 1 is a prime was discovered by We have already seen how to generate prime numbers in Haskell (Examples 1.22 and 1.23). We will now present an elegant alternative: a lazy list implementation of the Sieve of Eratosthenes. The idea of the sieve is this. Start with the list of all natural numbers > 2: 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; : : : In the first round, mark 2 (the first number in the list) as prime, and mark all mul- tiples of 2 for removal in the remainder of the list (marking for removal indicated by over-lining): 2 ; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; : : : 106 CHAPTER3. THEUSEOFLOGIC:PROOF In the second round, mark 3 as prime, and mark all multiples of 3 for removal in the remainder of the list: 2 ; 3 ; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; : : : 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; : : : In the third round, mark 5 as prime, and mark all multiples of 5 for removal in the remainder of the list: 2 ; 3 ; 4; 5 ; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; : : : And so on. A remarkable thing about the Sieve is that the only calculation it involves is counting. If the 3-folds are to be marked in the sequence of natural numbers starting from 3, walk through the list while counting 1; 2; 3 and mark the number 6, next walk on while counting 1; 2; 3 and mark the number 9, and so on. If the 5-folds are to be marked in the sequence the natural numbers starting from 5, walk on through the sequence while counting 1; 2; 3; 4; 5 and mark the number 10, next walk on while counting 1; 2; 3; 4; 5 and mark the number 15, and so on. In the Haskell implementation we mark numbers in the sequence [2..] for re- moval by replacing them with 0. When generating the sieve, these zeros are skipped. sieve :: [Integer] -> [Integer] sieve (0 : xs) = sieve xs sieve (n : xs) = n : sieve (mark xs 1 n) where mark :: [Integer] -> Integer -> Integer -> [Integer] 0 : (mark ys m) mark (y:ys) k m | k == m y : (mark ys (k+1) m) = | otherwise = 1 primes :: [Integer] primes = sieve [2..] This gives: TUOLP> primes 3.7. REASONINGANDCOMPUTATIONWITHPRIMES 107 [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79, 83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163, 167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251, 257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349, 353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443, 449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557, 563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647, 653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757, 761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863, 877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983, 991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063, {Interrupted!} Does this stream ever dry up? We know for sure that it doesn’t, because of Euclid’s proof. It is possible, by the way, to take a finite initial segment of an infinite Haskell list. This is done with the built in function take, as follows: TUOLP> take 100 primes [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79, 83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163, 167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251, 257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349, 353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443, 449,457,461,463,467,479,487,491,499,503,509,521,523,541] TUOLP> Exercise 3.38 A slightly faster way to generate the primes is by starting out from the odd numbers. The stepping and marking will work as before, for if you count k positions in the odd numbers starting from any odd number a = 2n + 1, you will move on to number (2n + 1) + 2k, and if a is a multiple of k, then so is a + 2k. Implement a function fasterprimes :: [Integer] using this idea. The odd natural numbers, starting from 3, can be generated as follows: oddsFrom3 :: [Integer] oddsFrom3 = 3 : map (+2) oddsFrom3 Still faster is to clean up the list at every step, by removing multiples from the list as you go along. We will come back to this matter in Section 10.1. 108 CHAPTER3. THEUSEOFLOGIC:PROOF Exercise 3.39 Write a Haskell program to refute the following statement about pk) + 1 is a prime numbers: if p1; : : : ; pk are all the primes < n, then (p1 (cid:2) (cid:1) (cid:1) (cid:1) (cid:2) prime. A computer is a useful instrument for refuting guesses or for checking particular cases. But if, instead of checking a guess for a particular case, you want to check the truth of interesting general statements it is of limited help. You can use the function mersenne to generate Mersenne primes, but the computer will not tell you whether this stream will dry up or not . . . mersenne = [ (p,2^p - 1) | p <- primes, prime (2^p - 1) ] This is what a call to mersenne gives: TUOLP> mersenne [(2,3),(3,7),(5,31),(7,127),(13,8191),(17,131071),(19,524287), (31,2147483647) If you are interested in how this goes on, you should check out GIMPS (“Great Internet Mersenne Prime Search”) on the Internet. To generate slightly more in- formation, we can define: notmersenne = [ (p,2^p - 1) | p <- primes, not (prime (2^p-1)) ] This gives: TUOLP> notmersenne [(11,2047),(23,8388607),(29,536870911),(37,137438953471), (41,2199023255551),(43,8796093022207),(47,140737488355327), (53,9007199254740991),(59,576460752303423487) The example may serve to illustrate the limits of what you can do with a computer when it comes to generating mathematical insight. If you make an interesting mathematical statement, there are three possibilities: 3.7. REASONINGANDCOMPUTATIONWITHPRIMES 109 (cid:15) (cid:15) (cid:15) You succeed in proving it. This establishes the statement as a theorem. You succeed in disproving it (with or without the help of a computer). This establishes the statement as a refuted conjecture. Neither of the above. This may indicate that you have encountered an open problem in mathematics. It may also indicate, of course, that you haven’t been clever enough. Example 3.40 Here is an example of an open problem in mathematics: Are there infinitely many Mersenne primes? It is easy to see that Euclid’s proof strategy will not work to tackle this problem. The assumption that there is a finite list p1; : : : ; pn of Mersenne primes does yield a larger prime, but nothing guarantees that this larger prime number is again of the form 2m 1. (cid:0) Mersenne primes are related to so-called perfect numbers. A perfect number is a number n with the curious property that the sum of all its divisors equals 2n, or, in other words, the sum of all proper divisors of n equals n (we call a divisor d of n proper if d < n). The smallest perfect number is 6, for its proper divisors are 1; 2 and 3, and 1 + 2 + 3 = 6, and it is easy to check that 1; 2; 3; 4 and 5 are not perfect. 1 is prime, then 2n Euclid proved that if 2n (cid:0) perfect numbers found by Euclid’s recipe are: 2 24 1) = 496. (25 (cid:0) 1(2n (22 (cid:1) (cid:1) (cid:0) 1) is perfect. Examples of (23 1) = 6, 22 1) = 28, (cid:1) (cid:0) (cid:0) (cid:0) Exercise 3.41 How would you go about yourself to prove the fact Euclid proved? Here is a hint: if 2n 1) are 1 is prime, then the proper divisors of 2n (cid:0) 1); : : : ; 2n 1; 2n 1); 22(2n 1; 2(2n (cid:0) 1(2n 2(2n (cid:0) 1; 2; 22; : : : ; 2n (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 1): (cid:0) Here is a function for generating the list of proper divisors of a natural number. This is not an efficient way to generate proper divisors, but never mind. pdivisors :: Integer -> [Integer] pdivisors n = [ d | d <- [1..(n-1)], rem n d == 0 ] 110 CHAPTER3. THEUSEOFLOGIC:PROOF With this it is easy to check that 8128 is indeed a perfect number: TUOLP> pdivisors 8128 [1,2,4,8,16,32,64,127,254,508,1016,2032,4064] TUOLP> sum (pdivisors 8128) 8128 Even more spectacularly, we have: TUOLP> prime (2^13 -1) True TUOLP> 2^12 * (2^13 -1) 33550336 TUOLP> pdivisors 33550336 [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8191,16382,32764, 65528,131056,262112,524224,1048448,2096896,4193792,8387584, 16775168] TUOLP> sum [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8191,16382, 32764,65528,131056,262112,524224,1048448,2096896,4193792,8387584, 16775168] 33550336 TUOLP> Prime pairs are pairs (p; p + 2) where both p and p + 2 are prime. Prime pairs can be generated as follows: primePairs :: [(Integer,Integer)] primePairs = pairs primes where pairs (x:y:xys) | x + 2 == y = (x,y): pairs (y:xys) | otherwise = pairs (y:xys) This gives: TUOLP> take 50 primePairs take 50 primePairs [(3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61),(71,73), (101,103),(107,109),(137,139),(149,151),(179,181),(191,193), (197,199),(227,229),(239,241),(269,271),(281,283),(311,313), (347,349),(419,421),(431,433),(461,463),(521,523),(569,571), (599,601),(617,619),(641,643),(659,661),(809,811),(821,823), 3.8. FURTHERREADING 111 (827,829),(857,859),(881,883),(1019,1021),(1031,1033), (1049,1051),(1061,1063),(1091,1093),(1151,1153),(1229,1231), (1277,1279),(1289,1291),(1301,1303),(1319,1321),(1427,1429), (1451,1453),(1481,1483),(1487,1489)] TUOLP> Does this stream ever dry up? We don’t know, for the question whether there are infinitely many prime pairs is another open problem of mathematics. Exercise 3.42 A prime triple is a triple (p; p + 2; p + 4) with p; p + 2; p + 4 all prime. The first prime triple is (3; 5; 7). Are there any more? Note that instructing the computer to generate them is no help: primeTriples :: [(Integer,Integer,Integer)] primeTriples = triples primes where triples (x:y:z:xyzs) | x + 2 == y && y + 2 == z = (x,y,z) : triples (y:z:xyzs) | otherwise = triples (y:z:xyzs) We get: TUOLP> primeTriples [(3,5,7) Still, we can find out the answer . . . How? Exercise 3.43 Consider the following call: TUOLP> filter prime [ p^2 + 2 | p <- primes ] [11 Can you prove that 11 is the only prime of the form p2 + 2, with p prime? 3.8 Further Reading The distinction between finding meaningful relationships in mathematics on one hand and proving or disproving mathematical statements on the other is drawn 112 CHAPTER3. THEUSEOFLOGIC:PROOF very clearly in Polya [Pol57]. A good introduction to mathematical reasoning is [Ecc97]. More detail on the structure of proofs is given by Velleman [Vel94]. Automated proof checking in Haskell is treated in [HO00]. An all-time classic in the presentation of mathematical proofs is [Euc56]. Chapter 4 Sets, Types and Lists Preview The chapter introduces sets, not by means of a definition but by explaining why ‘set’ is a primitive notion, discusses the process of set formation, and explains some important operations on sets. Talking about sets can easily lead to paradox, but such paradoxes can be avoided either by always forming sets on the basis of a previously given set, or by imposing certain typing constraints on the expressions one is allowed to use. Section 4.2 explains how this relates to functional programming. The end of the Chapter discusses various ways of implementing sets using list representations. The chapter presents ample opportunity to exercise your skills in writing implementations for set operations and for proving things about sets. module STAL where import List import DB 113 114 CHAPTER4. SETS,TYPESANDLISTS 4.1 Let’s Talk About Sets Remarkably, it is not possible to give a satisfactory definition of the notion of a set. There are several axiomatic approaches to set theory; the standard one (that is implicitly assumed throughout mathematics) is due to Zermelo and Fraenkel and dates from the beginning of the 20th century. Axioms vs. Theorems, Primitive vs. Defined Notions. The truth of a mathemati- cal statement is usually demonstrated by a proof. Most proofs use results that have been proved earlier. It follows that some truths must be given outright without proof. These are called axioms. The choice of axioms often is rather arbitrary. Criteria can be simplicity or intuitive evidence, but what must be considered an axiom can also be dictated by circumstance (for instance, a proof cannot be given within the given context, but in another context a proof would be possible). In the present context, we shall accept some of the Zermelo-Fraenkel axioms, as well as some fundamental properties of the number systems, such as the principle of mathematical induction for N (see Sections 7.1 and 11.1). Statements that have been proved are called theorems if they are considered to be of intrinsic value, and lemmas if they are often used in proving theorems. Notions are split up in two analogous categories. Their meaning may have been explained by definitions in terms of other notions. However, the process of defining notions must have a beginning somewhere. Thus, the need for notions that are primitive, i.e., undefined. For instance, we shall consider as undefined here the notions of set and natural number. Given the notion of a set, that of a function can be defined. However, in a context that is not set-theoretic, it could well be an undefined notion. Georg Cantor (1845-1915), the founding father of set theory, gave the following description. The Comprehension Principle. A set is a collection into a whole of definite, dis- tinct objects of our intuition or of our thought. The objects are called the elements (members) of the set. Usually, the objects that are used as elements of a set are not sets themselves. To handle sets of this kind is unproblematic. But it is not excluded at all that members are sets. Thus, in practice, you can encounter sets of sets of . . . sets. Notation. sometimes by A If the object a is member of the set A, this is denoted by a 2 a. If a is not a member of A, we express this as a = 2 3 A, or A, or 4.1. LET’STALKABOUTSETS 115 a. A 63 Example: 0 N, 1 2 62 N, 1 2 2 Q. 2 A set is completely determined by its elements: this is the content of the following Principle of Extensionality. Sets that have the same elements are equal. Symbolically, for all sets A and B, it holds that: x(x 8 2 A x 2 , B) = ) A = B: The converse of this (that equal sets have the same elements) is trivial. The Prin- ciple of Extensionality is one of Zermelo’s axioms. Figure 4.1: A set with a proper subset. Subsets. The set A is called a subset of the set B, and B a superset of A; notations: A A, if every member of A is also a member of B. In symbols: B, and B (cid:18) (cid:19) x (x 8 2 A = ) B): x 2 If A (cid:18) B and A = B, then A is called a proper subset of B. For instance, proper subset of the set of all even integers. 0; 2 g f is a proper subset of N, and the set of all multiples of 4 is a B and B Note that A = B iff A have to find an object c with c or an object c with c = 2 A = B will in general have the following form: (cid:18) A; c = 2 A; c (cid:18) 2 A. To show that A B (in this case c is a witness of A 2 B (in this case c is a witness of B = B we therefore either B), A). A proof of 6(cid:18) 6(cid:18) 6 6 116 CHAPTER4. SETS,TYPESANDLISTS Given: . . . To be proved: A = B. Proof: : Let x be an arbitrary object in A. : Let x be an arbitrary object in B. (cid:18) (cid:18) B. 2 To be proved: x Proof: . . . Thus x B. 2 A. 2 To be proved: x Proof: . . . Thus x A. 2 Thus A = B. Warning. Sometimes, A B (cid:26) to indicate that A is a proper subset of B. In this book we will stick to , and to express that A is properly included in B we will always use the conjunction of A B. Other authors use A B is written as A B and A = B. (cid:18) (cid:26) (cid:18) (cid:18) (cid:18) versus . Beginners often confuse , but these relations are very B implies that A and B are both sets, whereas B only implies that B is a set. Assuming that numbers are not sets, we have (provided the latter makes sense); whereas 2 different. For instance, A a 2 that 1 1 0; 1; 2 0; 1; 2 and 1 1 g , and 0; 1; 2 0; 1; 2 2 f and (cid:18) (cid:18) 2 6(cid:18) f g 62 f g . g g (cid:18) f f Theorem 4.1 For all sets A; B; C, we have that: g f 1. A 2. A 3. A A B B (cid:18) (cid:18) (cid:18) Proof. 1. B B A = ) C = ) (cid:18) (cid:18) A = B A C (cid:18) ^ ^ To be proved: A Proof: (cid:18) A, i.e., x(x 8 A x 2 ) 2 A). (reflexivity), (antisymmetry), (transitivity). Suppose c is any object in A. Then c x A), i.e., A x(x A 2 A. A. Therefore 8 2 ) 2 (cid:18) 6 4.1. LET’STALKABOUTSETS 117 2. This is Extensionality — be it in a somewhat different guise. 3. To be proved: A Proof: B B ^ (cid:18) (cid:18) C ) A (cid:18) C. Suppose A To be proved: A Proof: (cid:18) B and B C. (cid:18) C. (cid:18) Suppose c is any object in A. Then by A x(x B C), i.e., A C. (cid:18) x A B. c A C B, and by B C. (cid:18) 2 ) ) 2 (cid:18) 2 (cid:18) Thus A Thus B (cid:18) 8 ^ C, c C. 2 (cid:18) Remark. Note that the converse of antisymmetry also holds for if A; B are sets, then A = B iff A of the subproofs mentioned above: the proof of A . In other words, (cid:18) A. It is because of antisymmetry relation that a proof that two sets A; B are equal can consist of the two B and the proof of B B and B (cid:18) (cid:18) (cid:18) (cid:18) A. (cid:18) Exercise 4.2 Show that the superset relation also has the properties of Theorem is reflexive, antisymmetric and transitive. 4.1, i.e., show that (cid:19) Enumeration. A set that has only few elements a1; : : : ; an can be denoted as a1; : : : ; an f : g Extensionality ensures that this denotes exactly one set, for by extensionality the set is uniquely determined by the fact that it has a1; : : : ; an as its members. iff x = a1 _ (cid:1) (cid:1) (cid:1) _ Note that x a1; : : : ; an x = an. 2 f g Example 4.3 have that 3 0; 2; 3 f 0; 2; 3 2 f is the set the elements of which are 0, 2 and 3. We clearly g , and that 4 g 0; 2; 3 62 f . g Note that elements. Thus: 0; 2; 3 f g = f 2; 3; 0 = 3; 2; 2; 0 . Indeed, these sets have the same g f g Order and repetition in this notation are irrelevant. Exercise 4.4 Show, that 1; 2 ; g ff 0 ; g f f 2; 1 gg = 0 ; g ff f 1; 2 . gg 118 CHAPTER4. SETS,TYPESANDLISTS An analogue to the enumeration notation is available in Haskell, where [n::m] can be used for generating a list of items from n to m. This presupposes that n and m are of the same type, and that enumeration makes sense for that type. (Technically, the type has to be of the class Ord; see 4.2 below.) Sets that have many elements, in particular, infinite sets, cannot be given in this way, unless there is some system in the enumeration of their elements. For in- stance, N = is the set 0; 1; 2; : : : g f of even natural numbers. is the set of natural numbers, and 0; 2; 4; : : : g f Abstraction. If P (x) is a certain property of objects x, the abstraction g denotes the set of things x that have property P . f j x P (x) (4.1) Thus, for every particular object a, the expression is equivalent with a x j 2 f P (x) g P (a): By Extensionality, our talking about the set of x such that P (x) is justified. The abstraction notation binds the variable x: the set pends on x; P (x) P (y) = x f j g y f j . g x f j P (x) g in no way de- Usually, the property P will apply to the elements of a previously given set A. In that case j denotes the set of those elements of A that have property P . For instance, the set of even natural numbers can be given by 2 f g x A P (x) N n 2 f j n is even : g This way of defining an infinite set can be mimicked in functional programming by means of so-called list comprehensions, as follows: naturals = [0..] evens1 = [ n | n <- naturals , even n ] 4.1. LET’STALKABOUTSETS 119 N, which is of course in- Note the similarity between n <- naturals and n tended by the Haskell design team. The expression even n implements the prop- erty ‘n is even’, so we follow the abstraction notation from set theory almost to the letter. Here is the implementation of the process that generates the odd numbers: 2 odds1 = [ n | n <- naturals , odd n ] Back to the notation of set theory. A variation on the above notation for abstraction looks as follows. If f is an operation, then j denotes the set of things of the form f (x) where the object x has the property P . For instance, g f f (x) P (x) is yet another notation for the set of even natural numbers. 2n n j 2 f N g And again, we have a counterpart in functional programming. Here it is: evens2 = [ 2*n | n <- naturals ] Still, the similarity in notation between the formal definitions and their imple- mentations should not blind us to some annoying divergences between theory and practice. The two notations and n2 f j n 0; : : : ; 999 2 f gg ^ are equivalent. They are merely two ways of specifying the set of the first 1000 square numbers. But the Haskell counterparts behave very differently: 2 g j n2 f n N n < 1000 small_squares1 = [ n^2 | n <- [0..999] ] 120 CHAPTER4. SETS,TYPESANDLISTS A call to the function small_squares1 indeed produces a list of the first thousand square numbers, and then terminates. Note the use of [0..999] to enumerate a finite list. Not so with the following: small_squares2 = [ n^2 | n <- naturals , n < 1000 ] The way this is implemented, n <- naturals generates the infinite list of natu- ral numbers in their natural order, and n < 1000 tests each number as it is gen- erated for the property of being less than 1000. The numbers that satisfy the test are squared and put in the result list. Unlike the previous implementation small_squares1, the function small_squares2 will never terminate. Example 4.5 (*The Russell Paradox) It is not true that to every property E there of all objects that have E. The simplest example corresponds a set was given by Bertrand Russell (1872–1970). Consider the property of not having yourself as a member. E(x) x f g j Most sets that you are likely to consider have this property: the set of all even natural numbers is itself not an even natural number, the set of all integers is itself not an integer, and so on. Call such sets ‘ordinary’. The corresponding abstraction is R = x x x f j 62 . g It turns out that the question whether the set R itself is ordinary or not is impossible to answer. For suppose R R, that is, suppose R is an ordinary set. Ordinary sets are the sets that do not have themselves as a member, so R does not have itself as a member, i.e., R = R, that is, R 2 is an extraordinary set. Extraordinary sets are the sets that have themselves as a member, so R has itself as a member, i.e., R R. Suppose, on the contrary, that R = 2 R. 2 2 If R were a legitimate set, this would unavoidably lead us to the conclusion that which is impossible. R R 2 () R ; R 62 You do not have to be afraid for paradoxes such as the Russell paradox of Example 4.5. Only properties that you are unlikely to consider give rise to problems. In particular, if you restrict yourself to forming sets on the basis of a previously given set A, by means of the recipe A x 2 j f E(x) ; g 4.2. PARADOXES,TYPESANDTYPECLASSES 121 no problems will ever arise. Example 4.6 There is no set corresponding to the property F (x) : there is no infinite sequence x = x0 3 : : :. To see this, assume to the contrary x2 3 x1 3 that F is such a set. Assume F F . This implies F F : : :, so by the F 2 3 F . Then by the defining property of defining property of F , F = F . Assume F = 2 2 : : :. Now take the infinite F , there is an infinite sequence F = x0 3 x1 3 F , contradicting sequence x1 3 x1. F = x0 3 Exercise 4.7* Assume that A is a set of sets. Show that : : :. By the defining property of F , x1 = 2 3 x2 3 x2 3 A. (cid:17) A 3 F x x x f 2 j 62 g 62 It follows from Exercise (4.7) that every set A has a subset B Take B = A x x f 2 x = 2 j . g A with B = 2 (cid:18) A. 4.2 Paradoxes, Types and Type Classes It is a well-known fact from the theory of computation that there is no general test for checking whether a given procedure terminates for a particular input. The halting problem is undecidable. Intuitively, the reason for this is that the existence of an algorithm (a procedure which always terminates) for the halting problem would lead to a paradox very similar to the Russell paradox. Here is a simple example of a program for which no proof of termination exists: run :: Integer -> [Integer] run n | n < 1 = error "argument not positive" | n == 1 = [1] | even n = n: run (div n 2) | odd n = n: run (3*n+1) This gives, e.g.: STAL> run 5 [5,16,8,4,2,1] STAL> run 6 [6,3,10,5,16,8,4,2,1] STAL> run 7 [7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1] 122 CHAPTER4. SETS,TYPESANDLISTS We say that a procedure diverges when it does not terminate or when it aborts with an error. Stipulating divergence in Haskell is done by means of the predeclared function undefined, which causes an error abortion, just like error. In fact, Haskell has no way to distinguish between divergence and error abortion. Now suppose halts can be defined. Then define the procedure funny, in terms of halts, as follows (the part of a line after -- is a comment; in this case a warning that funny is no good as Haskell code): funny x | halts x x = undefined | otherwise = True -- Caution: this -- will not work What about the call funny funny? Does this diverge or halt? Suppose funny funny does not halt. Then by the definition of funny, we are in the first case. This is the case where the argument of funny, when applied to itself, halts. But the argument of funny is funny. Therefore, funny funny does halt, and contradiction. Suppose funny funny does halt. Then by the definition of funny, we are in the second case. This is the case where the argument of funny, when applied to itself, does not halt. But the argument of funny is funny. Therefore, funny funny does not halt, and contradiction. Thus, there is something wrong with the definition of funny. The only peculiarity of the definition is the use of the halts predicate. This shows that such a halts predicate cannot be implemented. It should be clear that funny is a rather close analogue to the Russell set x = 2 keeping track of the types of all objects and operations. j . Such paradoxical definitions are avoided in functional programming by x x g f As we have seen, new types can be constructed from old. Derived types are pairs of integers, lists of characters (or strings), lists of reals, and so on. How does this type discipline avoid the halting paradox? Consider the definition of funny. It makes a call to halts. What is the type of halts? The procedure halts takes as first argument a procedure, say proc, and as second argument an argument to that procedure, say arg. This means that the two arguments of halts have types a -> b and a, respectively. and that (proc arg), the result of applying proc to arg, has type b. But this means that the application halts x x in the definition of funny is ill-formed, for as we have seen the types of the two arguments to halts must be different, so the arguments themselves must be different. 4.2. PARADOXES,TYPESANDTYPECLASSES 123 For another example, take the built-in elem operation of Haskell which checks whether an object is element of a list. This operation is as close as you can get in Haskell to the ‘ ’ relation of set theory. The operation expects that if its first argument is of certain type a, then its second argument is of type ‘list over a’. 2 Thus, in Haskell, the question whether R R does not make sense, for any R: wit- 2 ness the following interaction. In the transcript, Prelude> is the Haskell prompt when no user-defined files are loaded. Prelude> elem ’R’ "Russell" True Prelude> elem ’R’ "Cantor" False Prelude> elem "Russell" "Cantor" ERROR: Type error in application *** expression *** term *** type *** does not match : Char : "Russell" ‘elem‘ "Cantor" : "Russell" : String Prelude> You would expect from this that elem has type a -> [a] -> Bool, for it takes an object of any type a as its first argument, then a list over type a, and it returns a verdict ‘true’ or ‘false’, i.e., an object of type Bool. Almost, but not quite. The snag is that in order to check if some thing x is an element of some list of things l, one has to be able to identify things of the type of x. The objects that can be identified are the objects of the kinds for which equality and inequality are defined. Texts, potentially infinite streams of characters, are not of this kind. Also, the Haskell operations themselves are not of this kind, for the Haskell operations denote computation procedures, and there is no principled way to check whether two procedures perform the same task. For suppose there were a test for equality on procedures (implemented functions). Then the following would be a test for whether a procedure f halts on input x (here /= denotes inequality): halts f x = f /= g where g y | y == x = undefined | otherwise = f y -- Caution: this -- will not work 124 CHAPTER4. SETS,TYPESANDLISTS The where construction is used to define an auxiliary function g by stipulating that g diverges on input x and on all other inputs behaves the same as f. If g is not equal to f, then the difference must be in the behaviour for x. Since we have stipulated that g diverges for this input, we know that f halts on x. If, on the other hand, g and f are equal, then in particular f and g behave the same on input x, which means that f diverges on that input. The types of object for which the question ‘equal or not’ makes sense are grouped into a collection of types called a class. This class is called Eq. Haskell uses == for equality and /= for inequality of objects of types in the Eq class. Using the hugs command :t to ask for the type of a defined operation, we get for elem: Prelude> :t elem elem :: Eq a => a -> [a] -> Bool Prelude> In the present case, the type judgment means the following. If a is a type for which equality is defined (or, if a is in the Eq class), then a -> [a] -> Bool is an appropriate type for elem. In other words: for all types a in the Eq class it holds that elem is of type a -> [a] -> Bool. This says that elem can be used to check whether an integer is a member of a list of integers, a character is a member of a string of characters, a string of characters is a member of a list of strings, and so on. But not whether an operation is a member of a list of operations, a text a member of a list of texts, and so on. Ord is the class of types of things which not only can be tested for equality and inequality, but also for order: in addition to == and /=, the relations < and <= are defined. Also, it has functions min for the minimal element and max for the maximal element. The class Ord is a subclass of the class Eq. Classes are useful, because they allow objects (and operations on those objects) to be instances of several types at once. The numeral ‘1’ can be used as an integer, as a rational, as a real, and so on. This is reflected in Haskell by the typing: Prelude> :t 1 1 :: Num a => a Prelude> All of the types integer, rational number, real number, complex number, and so on, are instances of the same class, called Num in Haskell. The class Num is a subclass of the class Eq (because it also has equality and inequality). For all types in the class Num certain basic operations, such as + and *, are defined. As we will see in 4.3. SPECIALSETS 125 Chapter 8, addition has different implementations, depending on whether we oper- ate on N; Z; Q; : : : and depending on the representations we choose. Still, instead of distinguishing between add, add1, add2, and so on, one could use the same name for all these different operations. This is standard practice in programming language design, and it is called operator overloading. Exercise 4.8 Explain the following error message: Prelude> elem 1 1 ERROR: [a] is not an instance of class "Num" Prelude> 4.3 Special Sets Singletons. Sets that have exactly one element are called singletons. The set whose only element is a is ; this is called the singleton of a. Note that it a g follows from the definition that x iff x = a. a f 2 f g Warning. Do not confuse a singleton a with its element a. In most cases you will have that a we have that = 0; 1 g 6 1. On the other hand, 0; 1 0; 1 f ff ff gg gg g f . For instance, in the case that a = a g 0; 1 = f . For, f has only one element: the set , g has two elements: the numbers 0 and 0; 1 0; 1 g f . g f has solutions (or, more Remark. The question whether the equation a = generally, whether sets a exist such that a a) is answered differently by different axiomatizations of set theory. A set satisfying this equation has the shape a = , but of course this is unofficial notation. For the mathematical 2 a f g fff(cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) ggg content of set theory this problem is largely irrelevant. An example of a case where it would be useful to let sets have themselves as members would be infinite streams, like an infinite list of ‘1’s. Such an object is easily programmed in Haskell: ones = 1 : ones 6 126 CHAPTER4. SETS,TYPESANDLISTS If you load and run this, and endless stream of 1’s will cover the screen, and you will have to kill the process from outside. Still, the process specified by ‘first generate a ‘1’ and then run the same process again’ is well-defined, and it can plausibly be said to have itself as its second member. To be sure, the order does matter here, but any set theory can encode ordered pairs: see Section 4.5 below. A set of the form = a; b a f g f g g is, in fact, a singleton. f a; b is called an (unordered) pair. Of course, if a = b, then Empty Set. Finally, there is a set without any elements at all: the empty set. This curious object can be a source of distress for the beginner, because of its unexpected properties. (A first one is Theorem 4.9.) The notation for the empty set is Note that there is exactly one set that is empty: this is due to Extensionality. : ; Theorem 4.9 For every set A, we have that A: ; (cid:18) Proof. Suppose x is an arbitrary object with x Then ? Therefore x x 2 ; ) Thus x(x 8 A. A), i.e., 2 2 A. ; (cid:18) . 2 ; (contradiction with the fact that has no members). ; Exercise 4.10 Show: iff a = b, b g 1. 2. a = f g f a1; a2g f = b1; b2g f iff: a1 = b1 ^ a2 = b2, or a1 = b2 ^ a2 = b1. Exercise 4.11 Explain that = ; 6 f;g . And that = f;g 6 . ff;gg 4.4. ALGEBRAOFSETS 127 Remark. Abstraction and Restricted Quantification. Note that Similarly, x 8 2 x 9 2 A (cid:8)(x) is true iff x f 2 A j (cid:8)(x) g = A: A (cid:8)(x) is true iff x f 2 A j (cid:8)(x) g 6 = : ; 4.4 Algebra of Sets Definition 4.12 (Intersection, Union, Difference.) Assume that A and B are sets. Then: 1. A \ 2. A [ and B = B = x x x x j j f f 2 2 A A ^ _ x x 2 2 B B g g is the intersection of A and B, is their union, 3. A B = x x j f 2 A ^ x 62 B g (cid:0) their difference. and Types of Set Theoretic Expressions Often, the symbols with the connectives _ clear. However, their functioning is completely different: two sets A and B, new sets A B (thus, between sets only), whereas statements (cid:8) are confused . From Definition 4.12, their intimate connection is produce, given can be written combine two statements (cid:8) and (cid:9) into new B resp. A (cid:9) resp. (cid:8) and and \ and \ \ ^ (cid:9). [ [ and \ [ _ [ ^ ^ _ In the idiom of Section 4.2: the operations on the other have different types. The typing underlying the notation of set theory is much simpler than that of functional programming. To start with, we just distinguish between the types s (for set), t (for a proposition, i.e., an expression with a truth value), (for a family of sets, see below). (for anything at all), and on one hand and ; \ ; ^ _ [ s g f t is like Bool in Haskell, but s has no Haskell analogue. In set theory, sets like N, Z, Q, R are all of the same type, namely s, whereas in Haskell much finer distinctions are drawn. The simplicity of the typing underlying set theory is an advantage, for it makes the language of set theory more flexible. To find the types of the set theoretic operations we use the same recipe as in func- s, for they take tional programming. Thus, t, tor they two set arguments and produce a new set. t both have type s and ! have type t and [ \ s ^ _ ! ! ! 128 CHAPTER4. SETS,TYPESANDLISTS Figure 4.2: Set union, intersection, difference, and complement. 4.4. ALGEBRAOFSETS 129 t, take two propositions and produce a new proposition. for it takes anything at all as its first argument, a set as its second argument, and produces a proposition. has type ! ! 2 s Exercise 4.13 What are the types of the set difference operator clusion operator ? (cid:18) and of the in- (cid:0) Exercise 4.14 Give the types of the following expressions: . g 1. x x E(x) 2 f x j E(x) . g 2. j f 3. (A B) \ (cid:18) C. 4. (A B) [ x(x 5. 8 2 6. A = B. \ A C. x 2 ) B). 7. a A a 2 , 2 B. The relationships between ^ equalities from Definition 4.12 are written in the form of equivalences: and between and and [ \ _ become clearer if the 1. x 2. x 3. x A A A B B B \ [ (cid:0) 2 2 2 () () () x x x A A A x x x ^ _ ^ 2 2 2 2 2 62 B, B, B. Disjointness. Sets A and B are called disjoint if A B = . ; \ Example 4.15 For A = 3 A and A B = f B = 1; 2; 3 g 1; 2 \ f g (cid:0) f 3; 4 and B = . A and B are not disjoint, for 3 g , we have: A g [ f B = 2 1; 2; 3; 4 B. f A \ , g Theorem 4.16 For all sets A, B and C, we have the following: 1. A A \ ; [ ; = ; ; = A, 130 CHAPTER4. SETS,TYPESANDLISTS 2. A A 3. A A 4. A A 5. A A \ [ \ [ \ [ \ [ A = A; A = A B = B B = B A; A \ [ (B (B (B (B \ [ [ \ C) = (A C) = (A C) = (A C) = (A B) B) B) B) C; C (A (A \ [ [ \ \ [ \ [ C); C) \ [ (idempotence), (commutativity), (associativity), (distributivity). By part 4.16.4, we can omit parentheses in intersections and unions of more than two sets. , these laws all reduce to Theorem 2.10 Proof. Using the definitions of (cid:0) (p. 45). Still, it is instructive to give a direct proof. Here is one for the first distribution law: , \ and [ : (cid:18) Let x be any object in A A and either x Then x 2 Suppose x Suppose x Thus in either case x B. Then x C. Then x (A 2 2 \ 2 C). (B [ B or x A A 2 2 B) C. B, so x C, so x C). 2 \ \ (A Thus A (B C) \ [ 2 (A \ B) [ (A \ C). (cid:18) \ [ \ B) (A C). : (cid:18) Let x be any object in (A Then either x A \ A 2 Suppose x Then x 2 Suppose x Then x Thus in either case x 2 (A \ B) A 2 \ [ Therefore (A \ B or x [ (A 2 B, i.e., x 2 \ A and either x 2 C, i.e., x \ C). A and x \ 2 B or x 2 A and x 2 \ A and either x (B A \ C) 2 B or x C). (B 2 [ A 2 C). (cid:18) \ [ By Extensionality, the required equality follows. (A (A B) B) (A (A [ [ \ \ \ \ 2 2 C). C). 2 B. C, i.e., x C. C, i.e., x 2 A A \ \ 2 2 (B (B [ [ C). C). Finally, a third form of proof: using the definitions, the distribution law reduces, as in the above, to the equivalence x A (x B x _ 2 2 ^ 2 C) (x A x B) (x A x ^ 2 2 _ 2 ^ 2 C); () 4.4. ALGEBRAOFSETS 131 which involves the three statements x out true always can be checked using an 8-line truth table in the usual way. B and x C. That this comes A, x 2 2 2 Exercise 4.17 A, B and C are sets. Show: 1. A 2. A B , B = A 6(cid:18) \ A (cid:0) (A (cid:0) = . ; B). B (cid:0) Example 4.18 We show that A C (cid:0) (cid:18) (A (cid:0) B) [ (B (cid:0) C). Given: x (A To be proved: x Proof: 2 (cid:0) 2 C): (A B) x _ 2 (cid:0) (B (cid:0) C). Suppose x 2 Therefore x Suppose x = 2 Therefore x B. From x B) (A 2 (cid:0) B. From x B) (A 2 _ x 2 (cid:0) (B (cid:0) (A 2 x _ 2 (A 2 x _ C). 2 (cid:0) (B (cid:0) (B Thus x (A (cid:0) 2 B) C) we get that x = 2 C). (cid:0) C) we get that x C, so x A, so x 2 (B (cid:0) C). (A (cid:0) B). 2 2 C). (cid:0) Exercise 4.19 Express (A B) (C \ [ [ D) as a union of four intersections. Complement. Fix a set X, of which all sets to be considered are subsets. The complement Ac of a set A X is now defined by (cid:18) Clearly, we have that for all x Ac := X A: (cid:0) X: Ac 2 A: x 62 , 2 x Theorem 4.20 1. (Ac)c = A; X c = ; ; c = X, ; 6 132 CHAPTER4. SETS,TYPESANDLISTS 2. A A Ac = X; Ac = , ; [ \ 3. A (cid:18) 4. (A (A [ \ Bc B , B)c = Ac B)c = Ac Ac, (cid:18) Bc; Bc \ [ (DeMorgan laws). Figure 4.3: Symmetric set difference. Symmetric Difference. The symmetric difference of two sets A and B, notation . This is the set of all objects that A x A are either in A or in B, but not in both. B, is the set given by (cid:8) (cid:8) B 2 2 x x f g j Exercise 4.21 Show that A B = (A B) (B [ (cid:0) (cid:0) (cid:8) A) = (A B) (A \ (cid:0) [ B). 1; 2; 3 2; 3 2 g g g f f f ; 1; 2 1 g g f f 8 >>< >>: Figure 4.4: The power set of 1; 3 3 g g f f 9 >>= >>; 1; 2; 3 . g f Definition 4.22 (Power Set) The powerset of the set X is the set }(X) = X of all subsets of X. g A f A j (cid:18) By Theorem 4.9 and 4.1.1 we have that instance, }( ; 1 ) = ; 1 1 ; ; f; g f; f;g f f; gg ; g }(X) and X ; 2 . Note that X 2 }(A) 2 }(X). So, for X A. (cid:18) , 4.4. ALGEBRAOFSETS 133 Exercise 4.23 Let X be a set with at least two elements. Then by Theorem (4.1), the relation on }(X) has the properties of reflexivity, antisymmetry and transi- tivity. The relation 6 on R also has these properties. The relation 6 on R has the further property of linearity: for all x; y on }(X) lacks this property. R, either x 6 y or y 6 x. Show that (cid:18) (cid:18) 2 Figure 4.5: Generalized set union and intersection. Definition 4.24 (Generalized Union and Intersection) Suppose that a set Ai has been given for every element i of a set I. 1. The union of the sets Ai is the set I Ai. i 2 Notation: x f i j 9 2 I(x . Ai) g 2 2. The intersection of the sets Ai is the set S Notation: I Ai i 2 x f i j 8 2 I(x . Ai) g 2 If the elements of I are sets themselves, and Ai = i (i the union of I; I), then 2 I i is called i 2 T The short notation for this set is I. Similarly, S I i is written as i 2 I Ai can also be written as A0 [ I. , resp., A0 \ In the case that I = N, A2 [ (cid:1) (cid:1) (cid:1) A set of sets is sometimes called a family of sets or a collection of sets. If family of sets, I Ai and S i 2 . A2 \ (cid:1) (cid:1) (cid:1) are sets. For example, if A1 \ S and i 2 T T T A1 [ is a F F S F 1; 2; 3; 4; 5 ff and F T = g 1; 2; 3 2; 3; 4 ; g 3; 4; 5 f ; gg ; g = f 3 f . g F T then = F S f 134 CHAPTER4. SETS,TYPESANDLISTS Example 4.25 For p mp j all natural numbers that have p as a factor. N, let Ap = 2 f m N; m > 1 . Then Ap is the set of g 2 Ai is the set of all natural numbers of the form n 2;3;5;7 i 2f least one of (cid:11); (cid:12); (cid:13); (cid:14) > 0. S g g Ai is the set of all natural numbers of the form n 2;3;5;7 i 2f of (cid:11); (cid:12); (cid:13); (cid:14) > 0, which is the set A210. T 2(cid:11)3(cid:12)5(cid:13)7(cid:14), with at (cid:1) 2(cid:11)3(cid:12)5(cid:13)7(cid:14), with all (cid:1) Let and be collections of sets. To check the truth of statements such as F F (cid:18) it is often useful to analyze their logical form by means of a translation in G G terms of quantifiers and the relation T . The translation of 2 S becomes: x( y(y 8 9 2 F ^ y) x 2 z(z ) 8 2 G ) F (cid:18) G T z)): S x 2 Exercise 4.26 Give a logical translation of using only the relation . 2 F (cid:18) G T S Types of Generalized Union and Intersection Again, it is useful to consider the types of the operations of generalized union and intersection. These operations is the take families of sets and produce sets, so their type is type of a family of sets. s g ! f s, where s g f be a family of sets. Show that there is a set A with the Exercise 4.27 Let F following properties: 1. F (cid:18) }(A), 2. For all sets B: if }(B) then A B. (cid:18) F (cid:18) , then Remark. If I = I Ai = i 2 This last fact is an example of a trivially true implication: if I = statement i I is false, hence the implication i everything is member of I Ai is the collection of all sets. i 2 , then every ; Ai true, and , and ) S T 2 2 2 x ; ; I x f i j 8 2 I(x Ai) g 2 = x f i(i j 8 I 2 x : Ai) g 2 ) Therefore, the notation I Ai usually presupposes that I i 2 = . ; T 6 4.4. ALGEBRAOFSETS 135 Example 4.28 For x x 2 2 B)c (A [ X, we have that: , : , : (cid:3) , : x , x (x (x x 2 A A [ _ 2 2 B) x A 2 Ac Ac ^ : x 2 ^ Bc: B) B 2 x 2 Bc 2 Step ( ) is justified by propositional reasoning. See 2.10.9. Extensionality allows (cid:3) us to conclude the first DeMorgan law: , \ (A [ B)c = Ac Bc: \ Exercise 4.29 Prove the rest of Theorem 4.20. Exercise 4.30 Answer as many of the following questions as you can. ))). )) and }(}(}( ), }(}( 1. Determine: }( ; ; ; ) = }(}(}(}(}( 2. How many elements has }5( ; 3. How many elements has }(A), given that A has n elements? )))))? ; Exercise 4.31 Check whether the following is true: if two sets have the same subsets, then they are equal. I.e.: if }(A) = }(B), then A = B. Give a proof or a refutation by means of a counterexample. Exercise 4.32 Is it true that for all sets A and B: 1. }(A 2. }(A \ [ B) = }(A) B) = }(A) }(B)? }(B)? \ [ Provide either a proof or a refutation by counter-example. Exercise 4.33* Show: 1. B ( \ I Ai) = i 2 I (B i 2 \ Ai), S S CHAPTER4. SETS,TYPESANDLISTS 136 2. B 3. ( 4. ( S T [ ( Ai), I Ai) = i 2 I Ai)c = T i 2 I Ai)c = i 2 I (B i 2 I Ac i , assuming that T i 2 I Ac i 2 i , assuming that T [ i 8 i 8 2 2 I Ai I Ai (cid:18) (cid:18) X, X. S Exercise 4.34* Assume that you are given a certain set A0. Suppose you are as- signed the task of finding sets A1; A2; A3; : : : such that }(A1) A1, }(A3) fail, that is: hit a set An for which no An+1 exists such that }(An+1) (cid:18) A2,. . . Show that no matter how hard you try, you will eventually An. (I.e., A0, }(A2) (cid:18) (cid:18) An.) ; 62 Hint. Suppose you can go on forever. Show this would entail }( (cid:18) N Ai) i 2 (cid:18) T N Ai. Apply Exercise 4.7. i 2 T Exercise 4.35* Suppose that the collection tion: K of sets satisfies the following condi- 2 K Show that every element of 8 A (A = B (A = }(B))): ; _ 9 2 K has the form }n( ) for some n ; K N. (N.B.: }0( ; 2 ) = .) ; 4.5 Ordered Pairs and Products Next to the unordered pairs immaterial ( b; a The ordered pair of objects a and b is denoted by of Section 4.3, in which the order of a and b is a; b f ), there are ordered pairs in which order does count. g a; b = g g f f (a; b) : Here, a is the first and b the second coordinate of (a; b). Ordered pairs behave according to the following rule: (a; b) = (x; y) = a = x b = y: (4.2) ^ ) This means that the ordered pair of a and b fixes the objects as well as their order. Its behaviour differs from that of the unordered pair: we always have that = a; b g , whereas (a; b) = (b; a) only holds — according to (4.2) — when a = b. g f Warning. If a and b are reals, the notation (a; b) also may denote the open interval x . The context should tell you which of the two is meant. a < x < b b; a R f f 2 j g 4.5. PAIRSANDPRODUCTS 137 Defining Ordered Pairs. Defining ; g allows you to prove (4.2). Cf. Exercise 4.41. (a; b) = ff a a; b f gg Definition 4.36 (Products) The (Cartesian) product of the sets A and B is the set of all pairs (a; b) where a B. In symbols: A and b 2 B = 2 (a; b) f A (cid:2) a j 2 A ^ b 2 B : g Instead of A (cid:2) Example 4.37 A one usually writes A2. 0; 1 f g (cid:2) f g f 1; 2; 3 = (0; 1); (0; 2); (0; 3); (1; 1); (1; 2); (1; 3) . g When A and B are real intervals on the X resp., the Y -axis in two-dimensional space, A B can be pictured as a rectangle. (cid:2) Ran(R) R Dom(R) Theorem 4.38 For arbitrary sets A; B; C; D the following hold: D) = (A D) 1. (A 2. (A 3. (A 4. (A B) B) B) B) \ (cid:2) (cid:2) (cid:2) (cid:2) [ \ [ (C (cid:2) C = (A (C (C \ [ C) [ (cid:2) D) = (A D) = (A 5. [(A C) B] [A [ (cid:2) (cid:2) (B (cid:0) D)] (cid:18) (cid:2) (cid:2) (cid:0) (cid:2) (B (cid:2) C) (C \ (cid:2) C); (A (B (cid:2) \ B), B) \ D), (cid:2) C = (A C) (B \ (cid:2) (cid:2) C), C) [ (A D) (B C) (cid:2) (A [ B) (cid:2) (C [ D). (cid:2) (cid:0) (cid:2) (B (cid:2) D), 138 CHAPTER4. SETS,TYPESANDLISTS Proof. As an example, we prove the first item of part 2. A B and c 2 C exist such that p = (a; c). [ (A (A C) C) (B (B [ [ (cid:2) (cid:2) (cid:2) (cid:2) 2 2 C). C). B) [ (cid:2) C = (A C) (B [ (cid:2) (cid:2) C): To be proved: (A : (cid:18) Suppose that p Thus (i) a 2 2 (A [ A or (ii) a (i). In this case, p (ii). Now p (A 2 C) B B) C. Then a 2 (cid:2) B. 2 C, and hence p A 2 C, and hence again p (cid:2) (B C). (cid:2) (cid:2) B) [ C [ (cid:2) (cid:18) (cid:2) (A C) (B [ (cid:2) (cid:2) C). Thus p Therefore, (A 2 2 2 C or (ii) p : (cid:18) Conversely, assume that p Thus (i) p A (i). In this case a 2 a fortiori, a A [ B and c (ii). Now b a fortiori b A (A (cid:2) Thus p Therefore, (A 2 2 2 2 B) [ C) (cid:2) [ C. (cid:2) (B [ (cid:2) (cid:2) C). (B (A B 2 A and c B and hence p C) [ C. C exist such that p = (a; c); B) (cid:2) 2 C. (cid:2) (A C exist such that p = (b; c); B) (A (cid:2) [ 2 B and hence, again, p C. 2 2 [ (cid:2) The required result follows using Extensionality. C) (A B) C. (cid:18) [ (cid:2) Exercise 4.39 Prove the other items of Theorem 4.38. Exercise 4.40 B (cid:2) A. Show that A = B. 1. Assume that A and B are non-empty and that A B = (cid:2) 2. Show by means of an example that the condition of non-emptiness in 1 is necessary. (Did you use this in your proof of 1?) Exercise 4.41* To show that defining (a; b) as a ; g ff f a; b gg works, prove that 1. 2. a; b = a; c g f b = c, = ) g f a ; g ff f a; b gg = x ; g ff x; y f = ) gg a = x ^ b = y. 4.6. LISTSANDLISTOPERATIONS 139 If we assume the property of ordered pairs (a; b) = (x; y) = b = y, we can define triples by (a; b; c) := (a; (b; c)). For suppose (a; b; c) = (x; y; z). Then by definition, (a; (b; c)) = (a; b; c) = (x; y; z) = (x; (y; z)), and by the property of ordered pairs, we have that a = x and (b; c) = (y; z). Again, by the property of ordered pairs, b = y and c = z. This shows that (a; b; c) = (x; y; z) (a = x c = z). a = x b = y ) ) ^ ^ ^ Let us go about this more systematically, by defining ordered n-tuples over some N. We proceed by recursion. base set A, for every n 2 Definition 4.42 (n-tuples over A) , 1. A0 := f;g 2. An+1 := A An. (cid:2) In Haskell, ordered pairs are written as (x1,x2), and there are predefined func- tions fst to get at the first member and snd to get at the second member. Ordered triples are written as (x1, x2, x3), and so on. If x1 has type a and x2 has type b, then (x1,x2) has type (a,b). Think of this type as the product of the types for x1 and x2. Here is an example: (1, ’A’) Prelude> :t (1,’A’) :: Num a => (a,Char) Prelude> 4.6 Lists and List Operations Assuming the list elements are all taken from a set A, the set of all lists over A N An. We abbreviate this set as A(cid:3). For every list L is the set A(cid:3) there is 2 some n 2 An we say that list L has length n. An. If L n N with L S 2 2 2 Standard notation for the (one and only) list of length 0 is []. A list of length n > 0 1]. looks like (x0; (x1; ( A one element list is a list of the form (x; []). In line with the above square bracket notation, this is written as [x]. )). This is often written as [x0; x1; : : : ; xn ; xn (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) 1) (cid:0) (cid:0) If one uses lists to represent sets, there is a difficulty. The list [a; b; b] is different from the list [a; b], for [a; b; b] has length 3, and [a; b] has length 2, but the sets are identical. This shows that sets and lists have different a; b; b and a; b f g identity conditions. f g This is how the data type of lists is (pre-)declared in Haskell: 140 CHAPTER4. SETS,TYPESANDLISTS data [a] = [] | a : [a] deriving (Eq, Ord) To grasp what this means, recall that in functional programming every set has a type. The data declaration for the type of lists over type a, notation [a] specifies that lists over type a are either empty or they consist of an element of type a put in front of a list over type a. Haskell uses : for the operation of putting an element in front of a list. The operation : combines an object with a list of objects of the same type to form a new list of objects of that type. This is reflected by the type judgment: Prelude> :t (:) (:) :: a -> [a] -> [a] The data declaration for lists also tells us, by means of deriving (Eq,Ord), that if equality is defined on the objects of type a (i.e., if the type a belongs to the class Eq) then this relation carries over to lists over type a, and if the objects of type a are ordered, i.e., if the type a belongs to the class Ord (e.g., the objects of type Z are ordered by <), then this order carries over to lists over type a. If we have an equality test for members of the set A, then it is easy to see how equality can be defined for lists over A (in fact, this is all taken care of by the predefined operations on lists). Lists are ordered sets, so two lists are the same if (i) they either are both empty, or (ii) they start with the same element (here is where the equality notion for elements from A comes in), and their tails are the same. In Haskell, this is implemented as follows (this is a definition from Prelude.hs): instance Eq a => Eq [a] where [] = == [] (x:xs) == (y:ys) = = == _ _ True x==y && xs==ys False This says that if a is an instance of class Eq, then [a] is so too. This specifies how the equality notion for objects of type a carries over to objects of type [a] (i.e., lists over that type). As we have seen, Haskell uses == for equality and = for the definition of operator values. Part (i) of the definition of list equality is taken care of by the statement that [] == [] is true, part (ii) is covered by: 4.6. LISTSANDLISTOPERATIONS 141 (x:xs) == (y:ys) = x==y && xs==ys This says: the truth value of (x:xs) == (y:ys) (equality of two lists that are both non-empty) is given by the conjunction of the first elements are the same (x == y), and the remainders are the same (xs == ys). (cid:15) (cid:15) The final line _ == _ = False states that in all other cases the lists are not equal. This final line uses _ as a ‘don’t care pattern’ or wild card. Exercise 4.43 How does it follow from this definition that lists of different length are unequal? A type of class Ord is a type on which the two-placed operation compare is de- fined, with a result of type Ordering. The type Ordering is the set , LT; EQ; GT g f where LT; EQ; GT have the obvious meanings. Suppose a is a type of class Ord, i.e., compare is defined on the type a. What does a reasonable ordering of lists over type a look like? A well-known order on lists is the lexicographical order: the way words are ordered in a dictionary. In this ordering, the empty list comes first. For non-empty lists L1; L2, we compare their first elements, using the function compare for objects of type a. If these are the same, the order is determined by the remainders of the lists. If the first element of L1 comes before the first element of L2 then L1 comes before L2, otherwise L2 comes before L1. The following piece of Haskell code implements this idea, by defining the function compare for lists over type a. instance Ord a => Ord [a] where compare [] compare [] compare (_:_) compare (x:xs) (y:ys) = primCompAux x y (compare xs ys) (_:_) [] [] = LT = EQ = GT This specifies how the ordering on type a carries over to an ordering of lists over type a. The first line says that the empty list [] is less than any non-empty list. 142 CHAPTER4. SETS,TYPESANDLISTS The second line says that the empty list is equal to itself. The third line says that any non-empty list is greater than the empty list. This fully determines the relation between [] and any list. The last line uses an auxiliary function primCompAux to cover the case of two non-empty lists. This function is defined by: primCompAux primCompAux x y o = :: Ord a => a -> a -> Ordering -> Ordering case compare x y of EQ -> o; LT -> LT; GT -> GT The type declaration of primCompAux says that if a is an ordered type, then primCompAux expects three arguments, the first one of type a, the second one of type a and the third one an element of the set (or type) Ordering, i.e., the set . The result is again a member of the type Ordering. LT; EQ; GT f g The definition of the operation primCompAux uses a case construction, using the reserved keywords case and of, and the arrow -> to point at the results for the various cases. . It says that in case the first two arguments are equal, the function returns the element of Ordering which is its third argument, in case the first ar- gument is less than the second argument, the function returns LT, in case the first argument is greater than the second argument, the function returns GT. Exercise 4.44 Another ordering of lists is as follows: shorter lists come before longer ones, and for lists of the same length we compare their first elements, and if these are the same, the remainder lists. Give a formal definition of this ordering. How would you implement it in Haskell? In list processing, fundamental operations are checking whether a list is empty, accessing the first element of a non-empty list, determining what remains after removal of the first element from a list (its tail). The operations for accessing the head or the tail of a list are implemented in Haskell as follows: head head (x:_) :: [a] -> a = x tail tail (_:xs) :: [a] -> [a] = xs 4.6. LISTSANDLISTOPERATIONS 143 The type of the operation head: give it a list over type a and the operation will return an element of the same type a. This is specified in head :: [a] -> a. The type of the operation tail is: give it a list over type a and it will return a list over the same type. This is specified in tail :: [a] -> [a]. Note that these operations are only defined on non-empty lists. (x:_) specifies the pattern of a non-empty list with first element x, where the nature of the tail is irrelevant. (_:xs) specifies the pattern of a non-empty list with tail xs, where the nature of the first element is irrelevant. Accessing the last element of a non-empty list is done by means of recursion: the last element of a unit list [x] is equal to [x]. The last element of a non-unit list is equal to the last element of its tail. Here is the Haskell implementation: last last [x] last (_:xs) :: [a] -> a = x = last xs Note that because the list patterns [x] and (_:xs) are tried for in that order, the pattern (_:xs) in this definition matches non-empty lists that are not unit lists. Exercise 4.45 Which operation on lists is specified by the Haskell definition in the frame below? init init [x] init (x:xs) :: [a] -> [a] = [] = x : init xs It is often useful to be able to test whether a list is empty or not. The following operation accomplishes this: null null [] null (_:_) :: [a] -> Bool = True = False 144 CHAPTER4. SETS,TYPESANDLISTS Exercise 4.46 Write your own definition of a Haskell operation reverse that re- verses a list. Exercise 4.47 Write a function splitList that gives all the ways to split a list of at least two elements in two non-empty parts. The type declaration is: splitList :: [a] -> [([a],[a])] The call splitList [1..4] should give: STAL> splitList [1..4] [([1],[2,3,4]),([1,2],[3,4]),([1,2,3],[4])] STAL> ] An operation on lists that we will need in the next sections is the operation of re- moving duplicates. This is predefined in the Haskell module List.hs as nub (‘nub’ means essence), but here is a home-made version for illustration: nub :: (Eq a) => [a] -> [a] nub [] = [] nub (x:xs) = x : nub (remove x xs) where remove y [] remove y (z:zs) | y == z = [] = remove y zs | otherwise = z : remove y zs What this says is, first, that if a is any type for which a relation of equality is defined, then nub operates on a list over type a and returns a list over type a. In Haskell, strings of characters are represented as lists, and the shorthand "abc" is allowed for [’a’,’b’,’c’]. Here is an example of an application of nub to a string: STAL> nub "Mississippy" "Mispy" Of course, we can also use nub on lists of words: STAL> nub ["Quentin Tarantino","Harrison Ford","Quentin Tarantino"] ["Quentin Tarantino","Harrison Ford"] 4.7. LISTCOMPREHENSIONANDDATABASEQUERY 145 4.7 List Comprehension and Database Query To get more familiar with list comprehensions, we will devote this section to list comprehension for database query, using the movie database module DB.hs given in Figure 4.6. The database that gets listed here is called db, with type DB, where DB is a synonym for the type [WordList], where Wordlist is again a synonym for the type [String]. The reserved keyword type is used to declare these type synonyms. Notice the difference between defining a type synonym with type and declaring a new data type with data. The database can be used to define the following lists of database objects, with list comprehension. Here db :: DB is the database list. characters = nub [ x | ["play",_,_,x] <- db ] = movies [ x | ["release",x,_] <- db ] = nub [ x actors <- db ] | ["play",x,_,_] = nub [ x directors | ["direct",x,_] <- db ] | ["release",_,x] <- db ] = nub [ x dates = nub (characters++actors++directors++movies++dates) universe Next, define lists of tuples, again by list comprehension: direct act play release | ["direct",x,y] = [ (x,y) | ["play",x,y,_] = [ (x,y) = [ (x,y,z) | ["play",x,y,z] = [ (x,y) <- db ] <- db ] <- db ] | ["release",x,y] <- db ] Finally, define one placed, two placed and three placed predicates by means of lambda abstraction. 146 CHAPTER4. SETS,TYPESANDLISTS module DB where type WordList = [String] type DB = [WordList] db :: DB db = [ ["release", "Blade Runner", "1982"], ["release", "Alien", "1979"], ["release", "Titanic", "1997"], ["release", "Good Will Hunting", "1997"], ["release", "Pulp Fiction", "1994"], ["release", "Reservoir Dogs", "1992"], ["release", "Romeo and Juliet", "1996"], {- ... -} ["direct", "Brian De Palma", "The Untouchables"], ["direct", "James Cameron", "Titanic"], ["direct", "James Cameron", "Aliens"], ["direct", "Ridley Scott", "Alien"], ["direct", "Ridley Scott", "Blade Runner"], ["direct", "Ridley Scott", "Thelma and Louise"], ["direct", "Gus Van Sant", "Good Will Hunting"], ["direct", "Quentin Tarantino", "Pulp Fiction"], {- ... -} ["play", "Leonardo DiCaprio", "Romeo and Juliet", "Romeo"], ["play", "Leonardo DiCaprio", "Titanic", "Jack Dawson"], ["play", "Robin Williams", "Good Will Hunting", "Sean McGuire"], ["play", "John Travolta", "Pulp Fiction", "Vincent Vega"], ["play", "Harvey Keitel", "Reservoir Dogs", "Mr White"], ["play", "Harvey Keitel", "Pulp Fiction", "Winston Wolf"], ["play", "Uma Thurman", "Pulp Fiction", "Mia"], ["play", "Quentin Tarantino", "Pulp Fiction", "Jimmie"], ["play", "Quentin Tarantino", "Reservoir Dogs", "Mr Brown"], ["play", "Sigourney Weaver", "Alien", "Ellen Ripley"], {- ... -} Figure 4.6: A Database Module. 4.7. LISTCOMPREHENSIONANDDATABASEQUERY 147 charP actorP movieP directorP dateP actP releaseP directP playP -> elem x characters = \ x -> elem x actors = \ x -> elem x movies = \ x -> elem x directors = \ x -> elem x dates = \ x -> elem (x,y) act = \ (x,y) -> elem (x,y) release = \ (x,y) -> elem (x,y) direct = \ (x,y) = \ (x,y,z) -> elem (x,y,z) play We start with some conjunctive queries. ‘Give me the actors that also are directors.’ q1 = [ x | x <- actors, directorP x ] ‘Give me all actors that also are directors, together with the films in which they were acting.’ q2 = [ (x,y) | (x,y) <- act, directorP x ] ’Give me all directors together with their films and their release dates.’ The fol- lowing is wrong. q3 = [ (x,y,z) | (x,y) <- direct, (y,z) <- release ] The problem is that the two ys are unrelated. infinite list. This can be remedied by using the equality predicate as a link: In fact, this query generates an q4 = [ (x,y,z) | (x,y) <- direct, (u,z) <- release, y == u ] 148 CHAPTER4. SETS,TYPESANDLISTS ‘Give me all directors of films released in 1995, together with these films.’ q5 = [ (x,y) | (x,y) <- direct, (u,"1995") <- release, y == u ] ‘Give me all directors of films released after 1995, together with these films and their release dates.’ q6 = [ (x,y,z) | (x,y) <- direct, (u,z) <- release, y == u, z > "1995" ] ‘Give me the films in which Kevin Spacey acted.’ q7 = [ x | ("Kevin Spacey",x) <- act ] ‘Give me all films released after 1997 in which William Hurt did act.’ q8 = [ x | (x,y) <- release, y > "1997", actP ("William Hurt",x) ] Yes/no queries based on conjunctive querying: ‘Are there any films in which the director was also an actor?’ q9 = q1 /= [] ‘Does the database contain films directed by Woody Allen?’ q10 = [ x | ("Woody Allen",x) <- direct ] /= [] 4.8. USINGLISTSTOREPRESENTSETS 149 Or simply: q10’ = directorP "Woody Allen" Disjunctive and negative queries are also easily expressed, since we have predi- cates and Boolean operators. Exercise 4.48 Translate the following into a query: ‘Give me the films in which Robert De Niro or Kevin Spacey acted.’ Exercise 4.49 Translate the following into a query: ‘Give me all films with Quentin Tarantino as actor or director that appeared in 1994.’ Exercise 4.50 Translate the following into a query: ‘Give me all films released after 1997 in which William Hurt did not act.’ 4.8 Using Lists to Represent Sets Sets are unordered, lists are ordered, but we can use lists to represent finite (or countably infinite) sets by representing sets as lists with duplicates removed, and by disregarding the order. If a finite list does not contain duplicates, its length gives the size of the finite set that it represents. Even if we gloss over the presence of duplicates, there are limitations to the rep- resentation of sets by lists. Such representation only works if the sets to be repre- sented are small enough. In Chapter 11 we will return to this issue. To removing an element from a list without duplicates all we have to do is remove the first occurrence of the element from the list. This is done by the predefined function delete, also part of the Haskell module List.hs. Here is our home-made version: delete :: Eq a => a -> [a] -> [a] delete x [] = [] delete x (y:ys) | x == y = ys | otherwise = y : delete x ys 150 CHAPTER4. SETS,TYPESANDLISTS As we have seen, the operation of elem for finding elements is built in. Here is our demo version, re-baptized elem’ to avoid a clash with Prelude.hs. elem’ :: Eq a => a -> [a] -> Bool elem’ x elem’ x [] (y:ys) | x == y = False = True | otherwise = elem’ x ys Further operations on sets that we need to implement are union, intersection and difference. These are all built into the Haskell module List.hs. Our version of union: union :: Eq a => [a] -> [a] -> [a] union [] ys union (x:xs) ys = x : union xs (delete x ys) = ys Note that if this function is called with arguments that themselves do not contain duplicates then it will not introduce new duplicates. Here is an operation for taking intersections: intersect :: Eq a => [a] -> [a] -> [a] intersect [] intersect (x:xs) s | elem x s s = [] = x : intersect xs s intersect xs s = | otherwise Note that because the definitions of union and intersect contain calls to delete or elem they presuppose that the type a has an equality relation defined on it. This is reflected in the type declarations for the operations. Exercise 4.51 The Haskell operation for list difference is predefined as \\ in List.hs. Write your own version of this. 4.8. USINGLISTSTOREPRESENTSETS 151 The predefined versions of the functions elem and notElem for testing whether an object occurs in a list or not use the functions any and all: elem, notElem elem notElem :: Eq a => a -> [a] -> Bool = any . (==) = all . (/=) Be cautious with this: elem 0 [1..] will run forever. Let’s turn to an operation for the list of all sublists of a given list. For this, we first need to define how one adds a new object to each of a number of lists. This can be done with the Haskell function map . Adding an element x to a list l that does not contain x is just a matter of applying the function (x:) (prefixing the element x) to l. Therefore, the following simple definition gives an operation that adds an element x to each list in a list of lists. addElem :: a -> [[a]] -> [[a]] addElem x = map (x:) Note the type declaration: [[a]] denotes the type of lists over the type of lists over type a. The operation addElem is used implicitly in the operation for generating the sub- lists of a list: powerList powerList powerList :: [a] -> [[a]] [] = [[]] (x:xs) = (powerList xs) ++ (map (x:) (powerList xs)) Here is the function in action: STAL> powerList [1,2,3] [[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]] 152 CHAPTER4. SETS,TYPESANDLISTS Example 4.52 For a connection with Exercise 4.30, let us try to fool Haskell into generating the list counterparts of }2( ) and }4( ), }3( ). Some fool- f;g ; ; ; ing is necessary, because Haskell is much less flexible than set theory about types. :: s, for these are all sets. In Haskell, In set theory, we have: ff;gg the type of the empty list is polymorphic, i.e., we have [] :: [a]. Thus, the list counterpart of }2( ; , can be consistently typed: ; ) = ) = }( f;gg :: s, :: s, f;g f; ; STAL> :t [[],[[]]] [[],[[]]] :: [[[a]]] What happens is that the first occurrence of [] gets type [[a]], and the second occurrence type [a], so that [[],[[]]] is a list containing two objects of the same type [[a]], and is itself of type [[[a]]]. However, Haskell refuses to display a list of the generic type [a]. The empty list will only be displayed in a setting where it is clear what the type is of the objects that are being listed: STAL> [ x | x <- [1..10], x < 0 ] [] This is OK, for it is clear from the context that [] is an empty list of numbers. STAL> [] ERROR: Cannot find "show" function for: *** Expression : [] *** Of type : [a] STAL> [[],[[]]] ERROR: Cannot find "show" function for: *** Expression : [[],[[]]] *** Of type : [[[a]]] These are not OK, for the context does not make clear what the types are. The following data type declaration introduces a data type S containing a sin- gle object Void. You should think of this as something very mysterious, for the Void is the unfathomable source of an infinite list-theoretic universe built from empty :: [S]. Void is used only to provide empty with a type. data S = Void deriving (Eq,Show) empty :: [S] empty = [] 4.9. ADATATYPEFORSETS 153 Here are the first stages of the list universe: STAL> powerList empty [[]] STAL> powerList (powerList empty) [[],[[]]] STAL> powerList (powerList (powerList empty)) [[],[[[]]],[[]],[[],[[]]]] STAL> powerList (powerList (powerList (powerList empty))) [[],[[[],[[]]]],[[[]]],[[[]],[[],[[]]]],[[[[]]]],[[[[]]],[[],[[]]]], [[[[]]],[[]]],[[[[]]],[[]],[[],[[]]]],[[]],[[],[[],[[]]]],[[],[[]]], [[],[[]],[[],[[]]]],[[],[[[]]]],[[],[[[]]],[[],[[]]]], [[],[[[]]],[[]]],[[],[[[]]],[[]],[[],[[]]]]] STAL> Exercise 4.53 Write functions genUnion and genIntersect for generalized list union and list intersection. The functions should be of type [[a]]-> [a]. They take a list of lists as input and produce a list as output. Note that genIntersect is undefined on the empty list of lists (compare the remark about the presupposition of generalized intersection on page 134). 4.9 A Data Type for Sets The representation of sets as lists without duplicates has the drawback that two finite lists containing the same elements, but in a different order, e.g., [1,2,3] and [3,2,1], are unequal as lists, but equal as sets. The Haskell equality operator == gives the wrong results when we are interested in set equality. For instance, we get: Prelude> [1,2,3] == [3,2,1] False Prelude> This can be remedied by defining a special data type for sets, with a matching definition of equality. All this is provided in the module SetEq.hs that is given in Figs. 4.7 and 4.8. If a is an equality type, then Set a is the type of sets over a. The newtype declaration allows us to put Set a as a separate type in the Haskell type system. Equality for Set a is defined in terms of the subSet relation: 154 CHAPTER4. SETS,TYPESANDLISTS module SetEq (Set(..),emptySet,isEmpty,inSet,subSet,insertSet, deleteSet,powerSet,takeSet,list2set,(!!!)) where import List (delete) infixl 9 !!! newtype Set a = Set [a] instance Eq a => Eq (Set a) where set1 == set2 = subSet set1 set2 && subSet set2 set1 instance (Show a) => Show (Set a) where showsPrec _ (Set s) str = showSet s str showSet [] showSet (x:xs) str = showChar ’{’ (shows x (showl xs str)) str = showString "{}" str where showl [] str = showChar ’}’ str showl (x:xs) str = showChar ’,’ (shows x (showl xs str)) emptySet emptySet = Set [] :: Set a :: Set a -> Bool isEmpty isEmpty (Set []) = True isEmpty = False _ inSet inSet x (Set s) = elem x s :: (Eq a) => a -> Set a -> Bool subSet :: (Eq a) => Set a -> Set a -> Bool _ subSet (Set []) subSet (Set (x:xs)) set = (inSet x set) && subSet (Set xs) set = True insertSet :: (Eq a) => a -> Set a -> Set a insertSet x (Set ys) | inSet x (Set ys) = Set ys | otherwise = Set (x:ys) Figure 4.7: A Module for Sets as Unordered Lists Without Duplicates. 4.9. ADATATYPEFORSETS 155 deleteSet :: Eq a => a -> Set a -> Set a deleteSet x (Set xs) = Set (delete x xs) list2set :: Eq a => [a] -> Set a list2set [] = Set [] list2set (x:xs) = insertSet x (list2set xs) powerSet :: Eq a => Set a -> Set (Set a) powerSet (Set xs) = Set (map (\xs -> (Set xs)) (powerList xs)) powerList powerList powerList :: [a] -> [[a]] [] = [[]] (x:xs) = (powerList xs) ++ (map (x:) (powerList xs)) takeSet :: Eq a => Int -> Set a -> Set a takeSet n (Set xs) = Set (take n xs) (!!!) :: Eq a => Set a -> Int -> a (Set xs) !!! n = xs !! n Figure 4.8: A Module for Sets as Unordered Lists Without Duplicates (ctd). instance Eq a => Eq (Set a) where set1 == set2 = subSet set1 set2 && subSet set2 set1 The instance declaration says that if a is in class Eq, then Set a is also in that class, with == defined as specified, in terms of the procedure subSet, that in turn is defined recursively in terms of inSet. This gives: SetEq> Set [2,3,1] == Set [1,2,3] True SetEq> Set [2,3,3,1,1,1] == Set [1,2,3] True SetEq> The module SetEq.hs gives some useful functions for the Set type. In the first place, it is convenient to be able to display sets in the usual notation. 156 CHAPTER4. SETS,TYPESANDLISTS instance (Show a) => Show (Set a) where showsPrec _ (Set s) = showSet s showSet [] showSet (x:xs) str = showChar ’{’ ( shows x ( showl xs str)) str = showString "{}" str where showl [] str = showChar ’}’ str showl (x:xs) str = showChar ’,’ (shows x (showl xs str)) This gives: SetEq> Set [1..10] {1,2,3,4,5,6,7,8,9,10} SetEq> powerSet (Set [1..3]) {{},{3},{2},{2,3},{1},{1,3},{1,2},{1,2,3}} SetEq> The empty set emptySet and the test isEmpty are implemented as you would expect. Useful functions for operating on sets are insertSet and deleteSet. The function insertSet is used to implement the translation function from lists to sets list2set. The function powerSet is implemented by lifting powerList to the type of sets. Exercise 4.54 Give implementations of the operations unionSet, intersectSet and differenceSet, in terms of inSet, insertSet and deleteSet. Exercise 4.55 In an implementation of sets as lists without duplicates, the imple- mentation of insertSet has to be changed. How? ), V4 = }4( ), V3 = }3( ), V2 = }2( , V1 = }( ; ; ; ; ; Figure 4.9 gives a module for generating the first five levels of the set theoretic hierarchy: V0 = ), V5 = }5( ). Displaying V5 in full takes some time, but here are the first few items. ; This uses the operator !!!, defined in the SetEq module as a left-associative infix operator by means of the reserved keyword infixl (the keyword infixr can be used to declare right associative infix operators). Hierarchy> v5 !!! 0 4.9. ADATATYPEFORSETS 157 module Hierarchy where import SetEq data S = Void deriving (Eq,Show) empty,v0,v1,v2,v3,v4,v5 :: Set S empty = Set [] v0 v1 v2 v3 v4 v5 = empty = powerSet v0 = powerSet v1 = powerSet v2 = powerSet v3 = powerSet v4 Figure 4.9: The First Five Levels of the Set Theoretic Universe. Hierarchy> display 88 (take 1760 (show v5)) {{},{{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}},{{}}}},{{{},{{{}}},{{}}},{{},{{{}}},{{}},{{ },{{}}}}},{{{},{{{}}},{{},{{}}}}},{{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{ },{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}},{{},{{{}}} ,{{}},{{},{{}}}}},{{{},{{{}}}}},{{{},{{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}},{{ },{{{}}},{{}}}},{{{},{{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{{}}}},{ {},{{{}}},{{},{{}}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{ },{{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{{}}}},{{},{{{}}},{{},{{}}}},{{} ,{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{} ,{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{} ,{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}}},{{ {},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}} },{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}},{{},{{}}}},{{} ,{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}}},{{{},{{}},{{ },{{}}}},{{},{{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{ {}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}} },{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}} }},{{},{{{}}},{{},{{}}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{ },{{{}}},{{},{{}}}},{{},{{{}}},{{}}}},{{{},{{}},{{},{{}}}},{{},{{{}}}},{{},{{{}}},{{},{{ }}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}},{{{},{{}}}},{{{},{{}}},{{},{{{}}},{{}} ,{{},{{}}}}},{{{},{{}}},{{},{{{}}},{{}}}},{{{},{{}}},{{},{{{}}},{{}}},{{},{{{}}},{{}},{{ Figure 4.10: An initial segment of }5( ). ; 158 CHAPTER4. SETS,TYPESANDLISTS {} Hierarchy> v5 !!! 1 {{{},{{{}}},{{}},{{},{{}}}}} Hierarchy> v5 !!! 2 {{{},{{{}}},{{}}}} Hierarchy> v5 !!! 3 {{{},{{{}}},{{}}},{{},{{{}}},{{}},{{},{{}}}}} Hierarchy> Figure 4.10 displays a bit more of the initial segment of }5( ), with the help of the ; following function: display :: Int -> String -> IO () display n str = putStrLn (display’ n 0 str) where display’ _ _ [] = [] display’ n m (x:xs) | n == m | otherwise = = ’\n’: display’ : display’ x 0 n n (m+1) (x:xs) xs Exercise 4.56 What would have to change in the module SetEq.hs to get a repre- sentation of the empty set as 0? Exercise 4.57* 1. How many pairs of curly braces in the representation where fg appears as ; ? fg occur in the expanded notation for }5( ), ; 2. How many copies of 0 occur in the expanded notation for }5( ), in the ; representation where appears as 0 (Exercise 4.56)? ; 3. How many pairs of curly braces occur in the expanded notation for }5( in the representation where appears as 0? ; ), ; 4.10 Further Reading Russell’s paradox, as stated by Russell himself, can be found in [Rus67]. A further introduction to sets and set theory is Doets, Van Dalen and De Swart [DvDdS78]. 4.10. FURTHERREADING 159 A good introduction to type theory is given in [Hin97]. There are many books on database query an database design; a lucid introduction is [SKS01]. SQL (Stan- dard Query Language) is explained in the documentation of all state of the art relational databases. Implementations of the relational database model are freely available on the Internet. See, e.g., www.postgresql.org or www.mysql.com. Logical foundations of database theory are given in [AHV95]. 160 CHAPTER4. SETS,TYPESANDLISTS Chapter 5 Relations Preview The first section of this chapter explains the abstract notion of a relation. Sec- tion 5.2 discusses various properties of relations. Next, in Sections 5.3 and 5.4, we discuss various possible implementations of relations and relation processing. Sections 5.5 and 5.6 focus on an often occurring type of relation: the equivalence. The following declaration turns the code in this chapter into a module that loads the List module (to be found in same directory as Prelude.hs, under the name List.hs), and the SetOrd module (see below, in Figs. 5.3 and 5.4). module REL where import List import SetOrd 161 162 CHAPTER5. RELATIONS 5.1 The Notion of a Relation Although you probably will not be able to explain the general notion of a relation, you are definitely familiar with a couple of instances, such as the usual ordering relation < between natural numbers, or the subset relation between sets. Non- mathematical examples are the different family relationships that exist between humans. For instance, the father-of relation holds between two people if the first one is the father of the second. (cid:18) N, the statement n < m is either true or false. E.g., For every two numbers n; m 3 < 5 is true, whereas 5 < 2 is false. In general: to a relation you can “input” a pair of objects, after which it “outputs” either true or false. depending on whether these objects are in the relationship given. 2 In set theory, there is a clever way to reduce the notion of a relation to that of a set. Consider again the ordering 6 on N. With it, associate the set R of ordered pairs (n; m) of natural numbers for which n 6 m is true: R6 = (n; m) N2 n 6 m : g 2 Note that a statement n 6 m now has become tantamount with the condition, that (n; m) R6. Thus, (3; 5) R6, and (5; 2) R6. f j 2 2 62 This connection can be converted into a definition. That is, the ordering relation 6 of N is identified with the set R6. The following definition puts it bluntly. Definition 5.1 (Relations, Domain, Range) A relation is a set of ordered pairs. Instead of (x; y) R(x; y), or Rxy. 2 R — where R is a relation — one usually writes xRy, or The set dom (R) = of pairs in R, is called the domain of R and ran(R) = second coordinates of pairs in R, its range. y ( xRy ) , i.e., the set consisting of all first coordinates , the set of x ( xRy ) y f j 9 j 9 x g f g Example 5.2 If A and B are sets, then A trivially is a relation (for, all its members are ordered pairs). (cid:2) B is a relation. The empty set ; ) = ran( dom ( ; ; thus dom (A (cid:2) ; empty). ) = ) = , dom (A ; ), and ran(A ; (cid:2) (cid:2) B) = A (provided B is non-empty: A , ; B) = B (analogously: provided A is non- (cid:2) ; = Definition 5.3 (From . . . to, Between, On) The relation R is a relation from A to B or between A and B, if dom (R) A and ran(R) B. (cid:18) A relation from A to A is called on A. (cid:18) 5.1. THENOTIONOFARELATION 163 If R is a relation on A, then A is called the underlying set (of the structure that consists of the domain A and the relation R). Example 5.4 R = and it also is a relation on (1; 4); (1; 5); (2; 5) 1; 2; 4; 5; 6 f g is a relation from . Furthermore, dom (R) = g f 1; 2; 3 to g 1; 2 , 4; 5; 6 g f , ran(R) = g f f 4; 5 . g f Example 5.5 If A is a set, then is a relation on }(A). (cid:18) 2 1 0 -1 -2 2 1 0 -1 -2 -2 -1 0 1 2 -2 -1 0 1 2 Figure 5.1: The relations 6 and > on R2. Example 5.6 The relations 6 and > on R are subsets of the real plane R2, and can be pictured as in Figure 5.1. Definition 5.7 (Identity and Inverse) 1. (cid:1)A = (a; b) f identity on A. 2 A2 j a = b = g f (a; a) a j 2 A g is a relation on A, the 2. If R is a relation between A and B, then R(cid:0) 1 = of R, is a relation between B and A. (b; a) aRb j f , the inverse g Example 5.8 The inverse of the relation ‘parent of’ is the relation ‘child of’. 164 CHAPTER5. RELATIONS 2 1 0 -1 -2 2 1 0 -1 -2 -2 -1 0 1 2 -2 -1 0 1 2 Figure 5.2: The relations (cid:1)R and R R on R2. (cid:2) Example 5.9 B is the biggest relation from A to B. (cid:2) is the smallest relation from A to B. 1. A 2. ; 3. For the usual ordering < of R, <(cid:0) 1 = >. 4. (R(cid:0) 1)(cid:0) 1 1 = R; (cid:1)(cid:0) A = (cid:1)A; 1 = (cid:0) ; ; and (A (cid:2) B)(cid:0) 1 = B A. (cid:2) In practice, you define a relation by means of a condition on the elements of or- dered pairs. This is completely analogous to the way you define a set by giving a condition its elements should fulfill. Example 5.10 If R is the set of pairs (n; m) (n; m is a divisor of m, the definition of R may look as follows: 2 Z) such that (condition) n2 For n; m 2 Z: nRm : (cid:17) n2 m. j In Haskell this is implemented as: \ n m -> rem m n^2 == 0. 5.1. THENOTIONOFARELATION 165 N; ab = n; a 6 is a relation on N. This relation gives all the divisor pairs of n. Here is its N, the set of pairs (a; b) a; b 2 2 f j Example 5.11 For all n b g implementation: divisors :: Integer -> [(Integer,Integer)] divisors n = [ (d, quot n d) | d <- [1..k], rem n d == 0 ] where k = floor (sqrt (fromInteger n)) Example 5.12 We can use the relation divisors of the previous example for yet another implementation of the primality test: prime’’ :: Integer -> Bool prime’’ = \n -> divisors n == [(1,n)] Also, here are the list of divisors of a natural number, the list of all proper divisors of a natural number, and a test for being a perfect natural number: divs :: Integer -> [Integer] divs n = (fst list) ++ reverse (snd list) where list = unzip (divisors n) properDivs :: Integer -> [Integer] properDivs n = init (divs n) perfect :: Integer -> Bool perfect n = sum (properDivs n) == n ‘There is a relation between a and b.’ In daily life, a statement like ‘there is a relation between a and b’ (or: ‘a and b are having a relation’) may not sound 166 CHAPTER5. RELATIONS unusual, but in the present context this should be avoided. Not because these are false statements, but because they are so overly true, and, therefore, uninformative. Cf. Exercise 5.13. Of course, a statement like ‘a stands in relation R to b’ or ‘the relation R subsists between a and b’ can very well be informative, when R is some specific relation. Exercise 5.13 Show that some relation.”) x 8 y 8 R (xRy). (“Between every two things there exist 9 In everyday situations, by saying that a and b are related we usually mean more than that it is possible to form the ordered pair of a and b. We usually mean that there is some good reason for considering the pair (a; b), because there is some specific link between a and b (for instance, you can make Mrs. a blush by mentioning Mr. b in conversation). 5.2 Properties of Relations In this section we list some useful properties of relations on a set A. Further on, in Sections 5.3 and 5.4, we will illustrate how tests for these properties can be implemented. A relation R is reflexive on A if for every x A: xRx. 2 Example 5.14 On any set A, the relation (cid:1)A is reflexive. Note that (cid:1)A is the smallest reflexive relation on A: it is a subset of any reflexive relation on A. In other words, a relation R is reflexive on A iff (cid:1)A R. (cid:18) Example 5.15 The relation 6 on N is reflexive (for every number is less than or equal to itself). A relation R on A is irreflexive if for no x A: xRx. 2 Example 5.16 The relation < on N is irreflexive. Exercise 5.17 Show that a relation R on A is irreflexive iff (cid:1)A R = . ; \ There are relations which are neither reflexive nor irreflexive (the reader is urged to think up an example). A relation R on A is symmetric if for all x; y A: if xRy then yRx. 2 5.2. PROPERTIESOFRELATIONS 167 Example 5.18 The relation ‘having the same age’ between people is symmetric. Unfortunately, the relation ‘being in love with’ between people is not symmetric. Exercise 5.19 Show the following: 1. A relation R on a set A is symmetric iff x; y 8 2 A(xRy , yRx). 2. A relation R is symmetric iff R (cid:18) R(cid:0) 1, iff R = R(cid:0) 1. A relation R on A is asymmetric if for all x; y A: if xRy then not yRx. The relation < on N is asymmetric. relation R on A is asymmetric iff R R(cid:0) which are neither symmetric nor asymmetric. \ It is immediate from the definition that a . Note that there are relations 1 = 2 ; Exercise 5.20 Show that every asymmetric relation is irreflexive. A relation R on A is antisymmetric if for all x; y x = y. 2 A: if xRy and yRx then n (m is a divisor of n) on N is antisymmetric. If m Example 5.21 The relation m j is a divisor of n and n is a divisor of m, then m and n are equal. The relation in example 5.15 is antisymmetric. Exercise 5.22 Show from the definitions that an asymmetric relation always is antisymmetric. The converse of the statement in Exercise 5.22 is not true: the relation 6 on N provides a counterexample. A relation R on A is transitive if for all x; y; z A: if xRy and yRz then xRz. 2 Examples of transitive relations are < and 6 on N. If F is the relation ‘friendship’ on a set of people E, and every member of E endorses the laudable principle “The friends of my friends are my friends”, then F is transitive. Exercise 5.23 Show that a relation R on a set A is transitive iff x; z 8 A( y 9 2 2 A(xRy yRz) ^ ) xRz): 168 CHAPTER5. RELATIONS A relation R on A is intransitive if for all x; y; z xRz. 2 A: if xRy and yRz then not Example 5.24 The relation ‘father of’ on the set of all human beings is intransi- tive. Again: there are relations that are neither transitive nor intransitive (think up an example). A relation R on A is a pre-order (or quasi-order) if R is transitive and reflexive. Example 5.25 Let L be the set of all propositional formulas built from a given set of atomic propositions. Then the relation = given by j P = Q iff P j ) Q is logically valid ) ) Q and Q P is logically valid, so R are logically valid, then P is a pre-order on L. To check this, note that for every propositional formula P , = is reflexive. Also, for all propositional formulas P j R is logically P; Q; R, if P ) valid. This can be checked as follows, by contraposition. Suppose P R is not logically valid. Then there is a valuation for the atomic propositions that makes P true and R false. Now there are two possibilities: this valuation makes Q true or it makes Q false. If it makes Q true, then we know that Q R is not logically valid. If it makes Q false, then we know that P Q is not logically valid. So either P R is not logically valid. This shows that Q or Q ) ) ) ) = is transitive. j ) ) on L given by P Example 5.26 Let L be the set of all propositional formulas built from a given set of atomic propositions. Then the relation Q iff there is a proof of Q from the given P (this relation was defined in Chapter 3) is a pre-order. See example 3.1 for the transitivity of coincides with = of the previous example. The safety checks on the proof rules for the relation j the logical connectives that were performed in Chapter 3 guarantee that =. This fact is called the soundness of the proof system. The inclusion in the other direction, , is called the completeness of the proof system. It is outside the scope of this book, but it is proved in [EFT94]. . In fact, the relation ‘ (cid:18) j (cid:18) ‘ = j ‘ ‘ ‘ ‘ A relation R on A is a strict partial order if R is transitive and irreflexive. A relation R on A is a partial order if R is transitive, reflexive and antisymmetric. Note that the relation las P P ^ Q and Q Q and Q ^ ^ ^ = from example 5.25 is not antisymmetric. Take the formu- j Q, but P . Then we have P P and Q Q P = Q j ^ = P j = is not a partial order. j ^ ^ P are different formulas. Thus, ^ 5.2. PROPERTIESOFRELATIONS 169 Example 5.27 The relation < on N is a strict partial order and the relation 6 on N is a partial order. Theorem 4.1 states that for any set A, the relation on }(A) is a partial order. (cid:18) Exercise 5.28 Show that every strict partial order is asymmetric. Exercise 5.29 Show that every relation which is transitive and asymmetric is a strict partial order. Exercise 5.30 Show that if R is a strict partial order on A, then R order on A. (So every strict partial order is contained in a partial order.) [ (cid:1)A is a partial Exercise 5.31 Show that the inverse of a partial order is again a partial order. A relation R on A is linear (or: has the comparison property) if for all x; y A: xRy or yRx or x = y. A partial order that is also linear is called a total order. All Haskell types in class Ord a are total orders. A set A with a total order on it is called a chain. 2 Exercise 5.32 Let S be a reflexive and symmetric relation on a set A. A path is a finite sequence a1; : : : ; an of elements of A such that for every i, 1 6 i < n, we have that aiSai+1. Such a path connects a1 with an. Assume that for all a; b A there is exactly one path connecting a with b. A. Define the relation 6 on A by: a 6 b iff a is one of the elements in the 2 Fix r path connecting r with b. 2 Show the following: 1. 6 is reflexive, 2. 6 is antisymmetric. 3. 6 is transitive, 4. for all a 2 5. for every a A, r 6 a, then b 6 c or c 6 b. 2 A, the set Xa = x f 2 A x 6 a g j is finite and if b; c Xa 2 (A structure (A; 6; r) with these five properties is called a tree with root r. The directory-structure of a computer account is an example of a tree. Another example is the structure tree of a formula (see Section 2.3 above). These are examples of finite trees, but if A is infinite, then the tree (A; 6; r) will have at least one infinite branch.) 170 CHAPTER5. RELATIONS Here is a list of all the relational properties of binary relations on a set A that we discussed: reflexivity irreflexivity symmetry asymmetry antisymmetry transitivity intransitivity linearity yRx). yRx). : x xRx. xRx. x xy (xRy ) xy (xRy ) : xy (xRy ^ xyz (xRy xyz (xRy xy (xRy yRx yRz ^ yRz ^ yRx _ 8 8 8 8 8 8 8 8 x = y). xRz). ) ) ) : x = y). _ xRz). Because of their importance, relations that are transitive, reflexive and symmetric (equivalence relations) deserve a section of their own (see Section 5.5). pre-order strict partial order partial order total order equivalence p p irrefl refl asymm antisymm symm trans p p p p p p p p p p p p p linear p Exercise 5.33 Consider the following relations on the natural numbers. Check their properties (some answers can be found in the text above). The successor relation is the relation given by . The divisor relation is (cid:17) GCD(n; m) = 1, i.e., the only factor of n that divides m is 1, and vice versa (see Section 8.2). . The coprime relation C on N is given by nCm : n divides m g n + 1 = m (n; m) (n; m) f g f j j < 6 successor divisor coprime irreflexive reflexive asymmetric antisymmetric symmetric transitive linear Definition 5.34 If - O closure of a relation R is the smallest relation S that includes R and that has all the properties in is a set of properties of relations on a set A, then the O . O 5.2. PROPERTIESOFRELATIONS 171 The most important closures are the reflexive closure, the symmetric closure, the transitive closure and the reflexive transitive closure of a relation. Remark. To show that R is the smallest relation S that has all the properties in , show the following: O 1. R has all the properties in O 2. If S has all the properties in , O , then R S. (cid:18) Exercise 5.35 Suppose that R is a relation on A. 1. Show that R 2. Show that R [ [ (cid:1)A is the reflexive closure of R. R(cid:0) 1 is the symmetric closure of R. Exercise 5.36 Let R be a transitive binary relation on A. Does it follow from (cid:1)A is also the transitivity of R that its symmetric reflexive closure R transitive? Give a proof if your answer is yes, a counterexample otherwise. 1 R(cid:0) [ [ To define the transitive and reflexive transitive closures of a relation we need the concept of relation composition. Composing Relations. Suppose that R and S are relations on A. The composition R S of R and S is the relation on A that is defined by (cid:14) x(R S)z : y (cid:17) 9 (cid:14) 2 A(xRy ySz): ^ N; n > 1 we define Rn by means of R1 := R, Rn+1 := 2 Furthermore, for n Rn R. (cid:14) Example 5.37 In imperative programming, the meaning of a command can be viewed as a relation on a set of machine states, where a machine state is a set of pairs consisting of machine registers with their contents. If s = (r1; 4); (r2; 6); : : : f then executing r2 := r1 in state s gives a new state s0 = . g (r1; 4); (r2; 4); : : : g f If C1 and C2 are commands, then we can execute them in sequential order as C1; C2 or as C2; C1. If the meaning of command C1 is R and the meaning of command C2 is S, then the meaning of C1; C2 is R S and the meaning of C2; C1 is S R. (cid:14) (cid:14) 172 CHAPTER5. RELATIONS Exercise 5.38 Determine the composition of the relation “father of” with itself. Determine the composition of the relations “brother of” and “parent of” Give an example showing that R R can be false. S = S (cid:14) (cid:14) Exercise 5.39 Consider the relation R = (0; 2); (0; 3); (1; 0); (1; 3); (2; 0); (2; 3) f g on the set A = 0; 1; 2; 3; 4 f . g 1. Determine R2, R3 and R4. 2. Give a relation S on A such that R (S (cid:14) [ R) = S. Exercise 5.40 Verify: 1. A relation R on A is transitive iff R R (cid:14) (cid:18) R. 2. Give an example of a transitive relation R for which R R = R is false. (cid:14) Exercise 5.41 Verify: (R S) = (Q 1. Q (cid:14) 2. (R (cid:14) S)(cid:0) (cid:14) 1 1 = S(cid:0) S. (Thus, the notation Q R) (cid:14) 1. R(cid:0) (cid:14) (cid:14) R (cid:14) (cid:14) S is unambiguous.) We will show that for any relation R on A, the relation R+ = transitive closure of R. S R+. We still have to prove Because R = R1 that R+ is transitive, and moreover that it is the smallest transitive relation that contains R. n>1 Rn, we know that R S (cid:18) (cid:18) n>1 Rn is the To see that R+ is transitive, assume xR+y and yR+z. From xR+y we get that there is some k > 1 with xRky. From yR+z we get that there is some m > 1 with yRmz. It follows that xRk+mz, and therefore xR+z, proving that R+ is transitive. 5.2. PROPERTIESOFRELATIONS 173 Proposition 5.42 R+ is the smallest transitive relation that contains R. Proof. The proposition can be restated as follows. If T is a transitive relation on T . We prove this fact by induction (see Chapter A such that R 7). T , then R+ (cid:18) (cid:18) Basis If xRy then it follows from R T that xT y. (cid:18) Induction step Assume (induction hypothesis) that Rn T . We have to show T . Consider a pair (x; y) with xRn+1y. Then, by the definition that Rn+1 of Rn+1, there is a z such that xRnz and zRy. Because R T we have zT y, and the induction hypothesis yields xT z. By transitivity of T it now follows that xT y. (cid:18) (cid:18) (cid:18) Example 5.43 If A = a; b; , then c ggg f + A; b f b; f c gg 2 f f + A; a 2 + A; 2 c g 2 f + A and c + A: 2 Example 5.44 On the set of human beings, the transitive closure of the ‘parent of’ relation is the ‘ancestor of’ relation, and the transitive closure of the ‘child of’ relation is the ‘descendant of’ relation. Exercise 5.45 Show that the relation < on N is the transitive closure of the relation R = (n; n + 1) N n f j 2 . g Exercise 5.46 Let R be a relation on A. Show that R+ transitive closure of R. [ (cid:1)A is the reflexive The reflexive transitive closure of a relation R is often called the ancestral of R, notation R(cid:3). Note that R(cid:3) is a pre-order. Exercise 5.47 Give the reflexive transitive closure of the following relation: R = (n; n + 1) f n N : g 2 j Exercise 5.48* 174 CHAPTER5. RELATIONS 1. Show that an intersection of arbitrarily many transitive relations is transitive. 2. Suppose that R is a relation on A. Note that A2 is one example of a transitive relation on A that extends R. Conclude that the intersection of all transitive relations extending R is the least transitive relation extending R. In other words, R+ equals the intersection of all transitive relations extending R. Exercise 5.49* 1. Show that (R(cid:3))(cid:0) 1 = (R(cid:0) 1)(cid:3). 2. Show by means of a counter-example that (R be false. R(cid:0) 1)(cid:3) = R(cid:3) 1 R(cid:0) (cid:3) may [ [ 3. Prove: if S R (cid:14) (cid:18) R (cid:14) S, then (R S)(cid:3) R(cid:3) (cid:14) (cid:18) S(cid:3). (cid:14) \ \ R R (cid:18) (cid:18) (cid:18) (cid:18) S, i.e., R S, so (cid:1)A R and (cid:1)A Exercise 5.50 Suppose that R and S are reflexive relations on the set A. Then S is reflexive as well. We say: (cid:1)A reflexivity is preserved under intersection. Similarly, if R and S are reflexive, then S is reflexive. Reflexivity is preserved under union. If R S, so R (cid:1)A [ 1 is reflexive. Reflexivity is preserved under 1, so R(cid:0) is reflexive, then (cid:1)A R(cid:0) inverse. If R and S are reflexive, (cid:1)A S is reflexive. Reflexivity is R preserved under composition. Finally, if R on A is reflexive, then the complement of R, i.e., the relation A2 R, is irreflexive. So reflexivity is not preserved under complement. These closure properties of reflexive relations are listed in the table below. S, so R [ (cid:18) (cid:18) (cid:0) (cid:14) (cid:14) We can ask the same questions for other relational properties. Suppose R and S is symmetric? That S are symmetric relations on A. Does it follow that R 1 is symmetric? That A2 S is symmetric? That R(cid:0) R is symmetric? That R S is symmetric? Similarly for the property of transitivity. These questions are R summarized in the table below. Complete the table by putting ‘yes’ or ‘no’ in the appropriate places. \ (cid:0) [ (cid:14) property preserved under preserved under preserved under inverse? preserved under complement? preserved under composition? ? \ ? [ reflexivity yes yes yes no yes symmetry ? ? ? ? ? transitivity ? ? ? ? ? 5.3. IMPLEMENTINGRELATIONSASSETSOFPAIRS 175 5.3 Implementing Relations as Sets of Pairs Our point of departure is a slight variation on the module SetEq.hs of the previ- ous Chapter. This time, we represent sets a ordered lists without duplicates. See Figs. 5.3 and 5.4 for a definition of the module SetOrd.hs. The definition of deleteList inside Figure 5.3 employs a Haskell feature that we haven’t encountered before: ys@(y:ys’) is a notation that allows us to refer to the non-empty list (y:ys’) by means of ys. If the item to be deleted is greater then the first element of the list then the instruction is to do nothing. Doing nothing boils down to returning the whole list ys. This possibility to give a name to a pattern is just used for readability. Next we define relations over a type a as sets of pairs of that type, i.e., Rel a is defined and implemented as Set(a,a). type Rel a = Set (a,a) domR gives the domain of a relation. domR :: Ord a => Rel a -> Set a domR (Set r) = list2set [ x | (x,_) <- r ] ranR gives the range of a relation. ranR :: Ord a => Rel a -> Set a ranR (Set r) = list2set [ y | (_,y) <- r ] idR creates the identity relation (cid:1)A over a set A: 176 CHAPTER5. RELATIONS module SetOrd (Set(..),emptySet,isEmpty,inSet,subSet,insertSet, deleteSet,powerSet,takeSet,(!!!),list2set) where import List (sort) {-- Sets implemented as ordered lists without duplicates --} newtype Set a = Set [a] deriving (Eq,Ord) instance (Show a) => Show (Set a) where showsPrec _ (Set s) str = showSet s str showSet [] showSet (x:xs) str = showChar ’{’ ( shows x ( showl xs str)) str = showString "{}" str where showl [] str = showChar ’}’ str showl (x:xs) str = showChar ’,’ (shows x (showl xs str)) emptySet emptySet = Set [] :: Set a :: Set a -> Bool isEmpty isEmpty (Set []) = True = False isEmpty _ inSet :: (Ord a) => a -> Set a -> Bool inSet x (Set s) = elem x (takeWhile (<= x) s) subSet :: (Ord a) => Set a -> Set a -> Bool subSet (Set []) _ subSet (Set (x:xs)) set = (inSet x set) && subSet (Set xs) set = True insertSet :: (Ord a) => a -> Set a -> Set a insertSet x (Set s) = Set (insertList x s) Figure 5.3: A Module for Sets as Ordered Lists Without Duplicates. 5.3. IMPLEMENTINGRELATIONSASSETSOFPAIRS 177 insertList x [] = [x] insertList x ys@(y:ys’) = case compare x y of GT -> y : insertList x ys’ EQ -> ys _ -> x : ys deleteSet :: Ord a => a -> Set a -> Set a deleteSet x (Set s) = Set (deleteList x s) deleteList x [] = [] deleteList x ys@(y:ys’) = case compare x y of GT -> y : deleteList x ys’ EQ -> ys’ _ -> ys list2set :: Ord a => [a] -> Set a list2set [] = Set [] list2set (x:xs) = insertSet x (list2set xs) -- list2set xs = Set (foldr insertList [] xs) powerSet :: Ord a => Set a -> Set (Set a) powerSet (Set xs) = Set (sort (map (\xs -> (list2set xs)) (powerList xs))) powerList powerList powerList :: [a] -> [[a]] [] = [[]] (x:xs) = (powerList xs) ++ (map (x:) (powerList xs)) takeSet :: Eq a => Int -> Set a -> Set a takeSet n (Set xs) = Set (take n xs) infixl 9 !!! (!!!) :: Eq a => Set a -> Int -> a (Set xs) !!! n = xs !! n Figure 5.4: A Module for Sets as Ordered Lists Without Duplicates (ctd). 178 CHAPTER5. RELATIONS idR :: Ord a => Set a -> Rel a idR (Set xs) = Set [(x,x) | x <- xs] The total relation over a set is given by: totalR :: Set a -> Rel a totalR (Set xs) = Set [(x,y) | x <- xs, y <- xs ] invR inverts a relation (i.e., the function maps R to R(cid:0) 1). invR :: Ord a => Rel a -> Rel a invR (Set []) = (Set []) invR (Set ((x,y):r)) = insertSet (y,x) (invR (Set r)) inR checks whether a pair is in a relation. inR :: Ord a => Rel a -> (a,a) -> Bool inR r (x,y) = inSet (x,y) r The complement of a relation R R. The operation of relational complementation, relative to a set A, can be implemented as follows: A is the relation A (cid:2) (cid:0) (cid:2) (cid:18) A A complR :: Ord a => Set a -> Rel a -> Rel a complR (Set xs) r = Set [(x,y) | x <- xs, y <- xs, not (inR r (x,y))] A check for reflexivity of R on a set A can be implemented by testing whether (cid:1)A R: (cid:18) 5.3. IMPLEMENTINGRELATIONSASSETSOFPAIRS 179 reflR :: Ord a => Set a -> Rel a -> Bool reflR set r = subSet (idR set) r A check for irreflexivity of R on A proceeds by testing whether (cid:1)A R = : ; \ irreflR :: Ord a => Set a -> Rel a -> Bool irreflR (Set xs) r = all (\ pair -> not (inR r pair)) [(x,x) | x <- xs] A check for symmetry of R proceeds by testing for each pair (x; y) (y; x) R: 2 R whether 2 symR :: Ord a => Rel a -> Bool symR (Set []) = True symR (Set ((x,y):pairs)) | x == y = symR (Set pairs) | otherwise = inSet (y,x) (Set pairs) && symR (deleteSet (y,x) (Set pairs)) A check for transitivity of R tests for each couple of pairs (x; y) whether (x; v) R if y = u: 2 R; (u; v) R 2 2 transR :: Ord a => Rel a -> Bool transR (Set []) = True transR (Set s) = and [ trans pair (Set s) | pair <- s ] where trans (x,y) (Set r) = and [ inSet (x,v) (Set r) | (u,v) <- r, u == y ] Now what about relation composition? This is a more difficult matter, for how do Szy)? The key to the implementation is the following we implement z(Rxz 9 ^ 180 CHAPTER5. RELATIONS procedure for composing a single pair of objects (x; y) with a relation S, simply by forming the relation . This is done by: (x; z) (z; y) S f j 2 g composePair :: Ord a => (a,a) -> Rel a -> Rel a composePair (x,y) (Set []) = Set [] composePair (x,y) (Set ((u,v):s)) | y == u | otherwise = composePair (x,y) (Set s) = insertSet (x,v) (composePair (x,y) (Set s)) For relation composition we need set union (Cf. Exercise 4.54): unionSet :: (Ord a) => Set a -> Set a -> Set a unionSet (Set []) set2 unionSet (Set (x:xs)) set2 = = insertSet x (unionSet (Set xs) (deleteSet x set2)) set2 Relation composition is defined in terms of composePair and unionSet: compR :: Ord a => Rel a -> Rel a -> Rel a compR (Set []) _ = (Set []) compR (Set ((x,y):s)) r = unionSet (composePair (x,y) r) (compR (Set s) r) Composition of a relation with itself (Rn): repeatR :: Ord a => Rel a -> Int -> Rel a repeatR r n | n < 1 = error "argument < 1" = r | n == 1 | otherwise = compR r (repeatR r (n-1)) 5.3. IMPLEMENTINGRELATIONSASSETSOFPAIRS 181 Example 5.51 Let us use the implementation to illustrate Exercise 5.39. r = Set [(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)] r2 = compR r r r3 = repeatR r 3 r4 = repeatR r 4 This gives: REL> r {(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)} REL> r2 {(0,0),(0,3),(1,2),(1,3),(2,2),(2,3)} REL> r3 {(0,2),(0,3),(1,0),(1,3),(2,0),(2,3)} REL> r4 {(0,0),(0,3),(1,2),(1,3),(2,2),(2,3)} REL> r == r2 False REL> r == r3 True REL> r == r4 False REL> r2 == r4 True Also, the following test yields ‘True’: s = Set [(0,0),(0,2),(0,3),(1,0),(1,2),(1,3),(2,0),(2,2),(2,3)] test = (unionSet r (compR s r)) == s Exercise 5.52 Extend this implementation with a function restrictR :: Ord a => Set a -> Rel a -> Rel a 182 CHAPTER5. RELATIONS that gives the restriction of a relation to a set. In the type declaration, Set a is the restricting set. The union of two relations R and S is the relation R for this we can use unionSet. [ S. Since relations are sets, Exercise 5.53 Use unionSet to define procedures rclosR for reflexive closure and sclosR for symmetric closure of a relation. As background set for the reflex- ive closure you can take the union of domain and range of the relation. Exercise 5.54 Define a function tclosR :: Ord a => Rel a -> Rel a to compute the transitive closure of a relation R, for relations implemented as Ord a => Rel a. Hint: compute the smallest relation S with the property that S = R R2 [ [ (cid:1) (cid:1) (cid:1) [ Rk (for some k) is transitive. Use transR for the transitivity test. 5.4 Implementing Relations as Characteristic Func- tions A characteristic function is a function of type A , for some set A. Char- g acteristic functions are so-called because they characterize subsets of a set A. The function f : A A. Characteristic functions implemented in Haskell have type a -> Bool, for some type a. characterizes the set B = f (a) = 1 ! f ! f g (cid:18) 0; 1 0; 1 A 2 a g f j From the fact that a binary relation r is a subset of a product A B, you would expect that a binary relation is implemented in Haskell as a function of type (a,b) -> Bool. Given a pair of objects (x,y), with x of type a and y of type b, the function proclaims the verdict True if (x,y) in the relation, False otherwise. (cid:2) Standard relations like == (for identity, or equality) and <= (for 6) are represented in Haskell in a slightly different way. They take their arguments one by one. Let us check their types: 5.4. IMPLEMENTINGRELATIONSASCHARACTERISTICFUNCTIONS183 Prelude> :t (==) (==) :: Eq a => a -> a -> Bool Prelude> :t (<=) (<=) :: Ord a => a -> a -> Bool Prelude> What this means is: if a is a type in the class Eq, then == takes a first argument of that type and a second argument of that type and proclaims a verdict True or False, and similarly for <=, except that now the arguments have to be of a type in the class Ord. Another example of a relation in Haskell is the following implementation divides Z with y (‘x divides y’) on the integers (x divides y if there a q of the relation x j y = xq). 2 divides :: Integer -> Integer -> Bool divides d n | d == 0 = error "divides: zero divisor" | otherwise = (rem n d) == 0 Switching back and forth between types a -> a -> Bool and (a,a) -> Bool (or, more generally, between types a -> b -> c and (a,b) -> c), can be done by means of the procedures for currying and uncurrying a function. The proce- dures refer to the logician H.B. Curry who helped laying the foundations for func- tional programming. (The initial H stands for Haskell; the programming language that we use in this book is also named after him.) If f is of type (a,b) -> c, then currying f means transforming it into a function that takes its arguments one by one, i.e., a function of type a -> b -> c. The procedure curry is predefined in Haskell as follows: curry curry f x y :: ((a,b) -> c) -> (a -> b -> c) = f (x,y) If f is of type a -> b -> c, then uncurrying f means transforming it into a func- tion that takes its arguments as a pair, i.e., a function of type (a,b) -> c. The procedure uncurry is predefined in Haskell as follows: 184 CHAPTER5. RELATIONS uncurry uncurry f p :: (a -> b -> c) -> ((a,b) -> c) = f (fst p) (snd p) As an example, here are some definitions of relations. eq :: Eq a => (a,a) -> Bool eq = uncurry (==) lessEq :: Ord a => (a,a) -> Bool lessEq = uncurry (<=) If a relation is implemented as a procedure of type (a,b) -> Bool it is very easy to define its inverse: inverse :: ((a,b) -> c) -> ((b,a) -> c) inverse f (x,y) = f (y,x) This gives: REL> inverse lessEq (3,4) False REL> inverse lessEq (4,3) True REL> Can we do something similar for procedures of type a -> b -> c? Yes, we can. Here is the predefined procedure flip: flip flip f x y :: (a -> b -> c) -> b -> a -> c = f y x Here it is in action: 5.4. IMPLEMENTINGRELATIONSASCHARACTERISTICFUNCTIONS185 REL> flip (<=) 3 4 False REL> flip (<=) 4 3 True REL> The procedure flip can be used to define properties from relations. Take the property of dividing the number 102. This denotes the set d N+ d divides 102 = 1; 2; 3; 6; 17; 34; 51; 102 g It is given in Haskell by (‘divides‘ 102), which in turn is shorthand for 2 f f j : g flip divides 102 Trying this out, we get: REL> filter (‘divides‘ 102) [1..300] [1,2,3,6,17,34,51,102] We will now work out the representation of relations as characteristic functions. To keep the code compatible with the implementation given before, we define the type as Rel’, and similarly for the operations. type Rel’ a = a -> a -> Bool emptyR’ :: Rel’ a emptyR’ = \ _ _ -> False list2rel’ :: Eq a => [(a,a)] -> Rel’ a list2rel’ xys = \ x y -> elem (x,y) xys idR’ creates the identity relation over a list. idR’ :: Eq a => [a] -> Rel’ a idR’ xs = \ x y -> x == y && elem x xs invR’ inverts a relation. 186 CHAPTER5. RELATIONS invR’ :: Rel’ a -> Rel’ a invR’ = flip inR’ checks whether a pair is in a relation. inR’ :: Rel’ a -> (a,a) -> Bool inR’ = uncurry Checks whether a relation is reflexive, irreflexive, symmetric or transitive (on a domain given by a list): reflR’ :: [a] -> Rel’ a -> Bool reflR’ xs r = and [ r x x | x <- xs ] irreflR’ :: [a] -> Rel’ a -> Bool irreflR’ xs r = and [ not (r x x) | x <- xs ] symR’ :: [a] -> Rel’ a -> Bool symR’ xs r = and [ not (r x y && not (r y x)) | x <- xs, y <- xs ] transR’ :: [a] -> Rel’ a -> Bool transR’ xs r = and [ not (r x y && r y z && not (r x z)) | x <- xs, y <- xs, z <- xs ] Union, intersection, reflexive and symmetric closure of relations: 5.4. IMPLEMENTINGRELATIONSASCHARACTERISTICFUNCTIONS187 unionR’ :: Rel’ a -> Rel’ a -> Rel’ a unionR’ r s x y = r x y || s x y intersR’ :: Rel’ a -> Rel’ a -> Rel’ a intersR’ r s x y = r x y && s x y reflClosure’ :: Eq a => Rel’ a -> Rel’ a reflClosure’ r = unionR’ r (==) symClosure’ :: Rel’ a -> Rel’ a symClosure’ r = unionR’ r (invR’ r) Relation composition: compR’ :: [a] -> Rel’ a -> Rel’ a -> Rel’ a compR’ xs r s x y = or [ r x z && s z y | z <- xs ] Composition of a relation with itself: repeatR’ :: [a] -> Rel’ a -> Int -> Rel’ a repeatR’ xs r n | n < 1 = error "argument < 1" = r | n == 1 | otherwise = compR’ xs r (repeatR’ xs r (n-1)) Exercise 5.55 Use the implementation of relations Rel’ a as characteristic func- tions over type a to define an example relation r with the property that unionR r (compR r r) is not the transitive closure of r. Exercise 5.56 If a relation r :: Rel’ a is restricted to a finite list xs, then we can calculate the transitive closure of r restricted to the list xs. Define a function 188 CHAPTER5. RELATIONS transClosure’ :: [a] -> Rel’ a -> Rel’ a for this. Hint: compute the smallest relation S with the property that S = R R2 [ [ (cid:1) (cid:1) (cid:1) [ Rk (for some k) is transitive. Use transR xs for the transitivity test on domain xs. 5.5 Equivalence Relations Definition 5.57 A relation R on A is an equivalence relation or equivalence if R is transitive, reflexive on A and symmetric. Example 5.58 On the set of human beings the relation of having the same age is an equivalence relation. Example 5.59 The relation R = equivalence relation on N. (n; m) f n; m j 2 N and n + m is even is an g The equivalence test can be implemented for relations of type Ord a => Rel a as follows: equivalenceR :: Ord a => Set a -> Rel a -> Bool equivalenceR set r = reflR set r && symR r && transR r For relations implemented as type Rel’ a the implementation goes like this: equivalenceR’ :: [a] -> Rel’ a -> Bool equivalenceR’ xs r = reflR’ xs r && symR’ xs r && transR’ xs r Example 5.60 The next table shows for a number of familiar relations whether they have the properties of reflexivity, symmetry and transitivity. Here: 5.5. EQUIVALENCERELATIONS 189 is the empty relation on N, (cid:15) ; (cid:1) = (cid:1)N = (n; n) n N is the identity on N, j f N is the biggest relation on N, 2 g N2 = N (cid:2) < and 6 are the usual ordering relations on N, Suc is the relation on N defined by Suc(n; m) n + 1 = m, (cid:17) (cid:15) (cid:15) (cid:15) (cid:15) is the inclusion relation on }(N). (cid:15) (cid:18) property: reflexive (on N resp. }(N)) symmetric transitive yes: (cid:1), N2, 6, , (cid:1), N2 ; , (cid:1), N2, <, 6, ; (cid:18) (cid:18) no: , <, Suc ; <, 6, Suc, Suc (cid:18) The table shows that among these examples only (cid:1) and N2 are equivalences on N. Example 5.61 The relation is reflexive on the set , but on no other set. ; ; ; is (trivially) symmetric and transitive. The relation is an equivalence on the set , but on no other set. Thus: ; ; Example 5.62 Let A be a set. (cid:1)A is the smallest equivalence on A. A2 is the biggest equivalence on A. Example 5.63 The relation defined by ~a ~b (cid:24) r (cid:17) 9 2 (cid:24) between vectors in 3-dimensional space R3 that is R+ (~a = r~b) is an equivalence. n on Z is given by m Example 5.64 For any n iff m and k have the same remainder when divided by n. More precisely, m (or: m = 0, the relation k (mod n)) iff Z, n (cid:17) 2 (cid:17) n k n k (cid:17) (cid:17) m = qn + r, with 0 6 r < n, k = q0n + r0, with 0 6 r0 < n, r = r0. (cid:15) (cid:15) (cid:15) When m called the (mod n) relation. (cid:17) n k we say that m is equivalent to k modulo n. The relation n is also (cid:17) 6 190 To show that is useful. (cid:17) CHAPTER5. RELATIONS n from Example 5.64 is an equivalence, the following proposition Proposition 5.65 m n k iff n m j (cid:0) k. (cid:17) Proof. and 0 6 r0 < n and r = r0. Thus, m n k. Then m = qn + r and k = q0n + r0 with 0 6 r < n k. q0)n, and it follows that n : Suppose m k = (q ) m (cid:17) : Suppose n n < r m k. Then n (cid:0) r0 < n, this implies r j ( (cid:0) (cid:0) j (cid:0) (cid:0) (cid:0) (qn + r) (cid:0) r0 = 0, so r = r0. It follows that m (q0n + r0), so n (cid:0) r j j (cid:0) r0. Since n k. (cid:17) From this we get that the following are all equivalent: n k. m (cid:17) m n j k. (cid:15) (cid:15) (cid:0) Z : an = m k. (cid:0) Z : m = k + an. Z : k = m + an. a (cid:15) 9 a (cid:15) 9 a (cid:15) 9 2 2 2 Exercise 5.66 Show that for every n equivalence on Z. 2 Z with n = 0 it holds that n is an (cid:17) Example 5.67 Here is a Haskell implementation of the modulo relation: modulo :: Integer -> Integer -> Integer -> Bool modulo n = \ x y -> divides n (x-y) Example 5.68 The relation that applies to two finite sets in case they have the same number of elements is an equivalence on the collection of all finite sets. The corresponding equivalence on finite lists is given by the following piece of Haskell code: 6 5.5. EQUIVALENCERELATIONS 191 equalSize :: [a] -> [b] -> Bool equalSize list1 list2 = (length list1) == (length list2) Abstract equivalences are often denoted by or . (cid:25) (cid:24) Exercise 5.69 Determine whether the following relations on N are (i) reflexive on N, (ii) symmetric, (iii) transitive: 1. 2. (2; 3); (3; 5); (5; 2) f ; g (n; m) f n j j m j > 3 . g (cid:0) Exercise 5.70 A = . Can you guess how many relations there are on this small set? Indeed, there must be sufficiently many to provide for the following questions. 1; 2; 3 f g 1. Give an example of a relation on A that is reflexive, but not symmetric and not transitive. 2. Give an example of a relation on A that is symmetric, but not reflexive and not transitive. 3. Give examples (if any) of relations on A that satisfy each of the six remain- ing possibilities w.r.t. reflexivity, symmetry and transitivity. Exercise 5.71 For finite sets A (0, 1, 2, 3, 4 and 5 elements and n elements gener- ally) the following table has entries for: the number of elements in A2, the number of elements in }(A2) (that is: the number of relations on A), the number of rela- tions on A that are (i) reflexive, (ii) symmetric and (iii) transitive, and the number of equivalences on A. 192 CHAPTER5. RELATIONS A A2 }(A2) 0 1 2 3 4 5 n 0 1 ? ? ? ? ? 1 2 ? ? ? ? ? reflexive 1 1 ? ? ? ? ? symmetric 1 2 ? ? ? ? ? transitive 1 2 13 — — — — equivalence 1 1 — — — — — Give all reflexive, symmetric, transitive relations and equivalences for the cases that A = (1 element). Show there are exactly 13 ; , and give the 3 that are not transitive. Put numbers on transitive relations on the places with question marks. (You are not requested to fill in the —.) (0 elements) and A = 0; 1 g f 0 f g Example 5.72 Assume relation R on A is transitive and reflexive, i.e, R is a pre- order. Then consider the relation yRx. on A given by: x is an equivalence relation on A. Symmetry is immediate from the The relation (cid:24) definition. y and is reflexive because R is. (cid:24) yRz, by transitivity of z. Then xRy y R, xRz, and from zRy zRy, and from xRy ^ yRx, by transitivity of R, zRx; thus x is transitive, for assume x yRx xRy yRz y : z. (cid:24) (cid:24) (cid:24) (cid:24) (cid:24) (cid:17) ^ ^ ^ ^ (cid:24) Exercise 5.73 Suppose that R is a symmetric and transitive relation on the set A such that A(xRy). Show that R is reflexive on A. A x 8 2 y 9 2 Exercise 5.74 Let R be a relation on A. Show that R is an equivalence iff (i) (cid:1)A R and (ii) R = R R(cid:0) 1. (cid:18) ^ (cid:14) 5.6 Equivalence Classes and Partitions Equivalence relations on a set A enable us to partition the set A into equivalence classes. Definition 5.75 Suppose R is an equivalence relation on A and that a set bRa R = equivalence class of a modulo R. A. The is called the R-equivalence class of a, or the = A 2 2 a a g f b j j j j j Elements of an equivalence class are called representatives of that class. Example 5.76 (continued from example 5.62) The equivalence class of a modulo (cid:1)A is The only equivalence class modulo A2 is A itself. . g a f A 2 5.6. EQUIVALENCECLASSESANDPARTITIONS 193 Example 5.77 (continued from example 5.63) The equivalence class of (1; 1; 1) R3 modulo g (not including the origin). (A “direction”.) 2 : half a straight line starting at the origin is the set (r; r; r) r > 0 (cid:24) f j Example 5.78 (continued from example 5.67) The equivalence class of 2 in Z (mod 4) is the set 6; : : : ; 2 + 4n = Z n f j 2 . g 2; 2; 6; 10; 14; : : : g f The implementation yields: (cid:0) (cid:0) REL> filter (modulo 4 2) [-15..15] [-14,-10,-6,-2,2,6,10,14] Example 5.79 (continued from example 5.68) The equivalence class of 0; 1; 2 g modulo the equivalence of having the same number of elements is the collection of all three-element sets. f *(According to the Frege-Russell definition of natural number, this is the number three.) Lemma 5.80 Suppose that R is an equivalence on A. If a; b A, then: 2 Proof. b a 2 j : Note that a ) R, i.e., aRb. j 2 j j a R = a j b j R j , aRb: R (for, R is reflexive). Thus, if j R = a j j b j j R, then : Assume that aRb. Then also bRa (R is symmetric.) R signifies xRa, and hence xRb follows (R transitive); j a (cid:18) j R: x b ( b a R j therefore, x a b 2 j j R. 2 j j R: similarly. j Extensionality completes the proof. (cid:18) j R j j j Lemma 5.81 Let R be an equivalence on A. 1. Every equivalence class is non-empty, 2. every element of A belongs to some equivalence class, 3. different equivalence classes are disjoint. A: a Proof. 1/2. Since R is reflexive on A, we have, for every a a j 3. Suppose that . Then we have both j cRa and cRb. Since R is symmetric, it follows that aRc. Thus, aRb (R transitive). Therefore, follows, using Lemma 5.80. are not disjoint. Say, c j \ j and 2 j 2 j 2 b a a b j j j j . = a b j j j j 194 CHAPTER5. RELATIONS Exercise 5.82 Use the implementation of relations Rel’ a as characteristic func- tions over type a to implement a function raccess that takes a relation r, an object x, and a list L, and returns the list of all objects y from L such that rxy holds. The type declaration should run: raccess :: Rel’ a -> a -> [a] -> [a] The concept of partitioning a set is made precise in the following definition. Definition 5.83 A family A of subsets of a set A is called a partition of A if , = 2 A (cid:15) ; = A, A for all X; Y S (cid:15) (cid:15) : if X 2 A = Y then X Y = . ; \ The elements of a partition are called its components. This definition says that every element of A is in some member of element of A is in more than one member of . A and that no A 1; 2 (trivially) is a parti- . Z can be partitioned into the negative integers, the positive integers, and ; is a partition of 1; 2; 3; 4 3; 4 ; g . g ff gg f ; f Example 5.84 tion of . g f 0 Exercise 5.85 Show the following: if is a partition of B, then j J Ai f Bj j i j 2 (i; j) I 2 g I is a partition of A and Bj f is a partition of A (cid:2) (cid:2) J g j B. Ai f (cid:2) 2 g Definition 5.86 (Quotients) Assume that R is an equivalence on the set A. The , is called the collection of equivalence classes of R, A=R = quotient of A modulo R. j j fj A 2 a a g The definition of partition was engineered in order to ensure the following: Theorem 5.87 Every quotient (of a set, modulo an equivalence) is a partition (of that set). Proof. This is nothing but a reformulation of Lemma 5.81. 6 5.6. EQUIVALENCECLASSESANDPARTITIONS 195 Example 5.88 (continued from examples 5.62 and 5.76) a g j a A . g 2 A=(cid:1)A = A=A2 = f f A . g f Example 5.89 (continued from examples 5.63 and 5.77) The partition R3= (cid:24) of R3 has components 0 f g and all half-lines. Example 5.90 (continued from examples 5.67 and 5.78) The partition Z=mod(4) of Z induced by the equivalence mod(4) has four com- of 0 (that also is the equivalence ponents: (i) the equivalence class n of 1, (iii) the class 4. . . ), (ii) the class class of 4, 8, 12,. . . , j , the class of 3. The quotient 4n + 3 g Z 4n + 2 f f g Z=mod(n) is usually written as Zn. Z 2 g 4n + 1 f Z n j (cid:0) of 2, and (iv) 4n 2 2 2 Z n n f g j j Example 5.91*(continued from examples 5.68 and 5.79) The quotient of the collection of finite sets modulo the relation “same number of elements” is —according to Frege and Russell— the set of natural numbers. Exercise 5.92 Give the partition that the relation of example 5.59 induces on N. Example 5.93 (continued from Example 5.72) (cid:24) Let R be a pre-order on A. Then lence relation. Consider the relation R relation R is a partial order on A= given by x on A= y : (cid:17) given by : (cid:17) (cid:24) called the po-set reflection of R. xRy ^ R x j j y (cid:24)j (cid:24) (cid:24) yRx is an equiva- xRy. The (cid:24) The definition is independent of the representatives, because assume x y and yRy0 by transitivity of R, x0Ry0. x0 and y0Ry, and x0Ry0. From x0Rx, xRy y0 and xRy. Then xRx0 x0Rx, yRy0 (cid:24) (cid:24) ^ ^ (cid:24) j R (cid:24) R y j j transitive because R is. (cid:24)j is reflexive, because R is. R is anti-symmetric, for suppose . Then xRy and yRx, so by definition of x j , (cid:24) x j j = y j (cid:24) y R x j j j . Finally R j (cid:24)j and is (cid:24) Thus, equivalences induce partitions. But the process can be inverted. Theorem 5.94 Every partition (of a set) is a quotient (of that set, modulo a certain equivalence). Specifically: Suppose that by A is a partition of A. Then the relation R on A defined xRy : K (x; y K) 2 2 A (cid:17) 9 196 CHAPTER5. RELATIONS (x and y member of the same component of the collection of equivalence classes of R. ) is an equivalence on A, and A is A Proof. Exercise 5.121. According to Theorems 5.87 and 5.94, equivalences and partitions are two sides of the same coin. Example 5.95 Consider the following relation on 0; 1; 2; 3; 4 f : g (0; 0); (0; 1); (1; 0); (1; 1); (2; 2); (2; 4); (3; 3); (4; 2); (4; 4) f : g This is an equivalence, and the corresponding partition is 0; 1 ; g ff f 2; 4 ; g 3 : gg f Exercise 5.96 Is the following relation an equivalence on give the corresponding partition. 0; 1; 2; 3; 4 f ? If so, g (0; 0); (0; 3); (0; 4); (1; 1); (1; 2); f (2; 1); (2; 2); (3; 0); (3; 3); (3; 4); (4; 0); (4; 3); (4; 4) : g Example 5.97 The equivalence corresponding to the partition 0 ; g ff is: f 1; 2; 3 ; g 4 f gg of 0; 1; 2; 3; 4 f g (0; 0); (1; 1); (1; 2); (1; 3); (2; 1); (2; 2); (2; 3); (3; 1); (3; 2); (3; 3); (4; 4) f : g Exercise 5.98 What are the equivalences corresponding to the following parti- tions?: 1. 2. 3. ff ff ff 0; 3 1; 2; 4 , of 0; 1; 2; 3; 4 ; g f Z n n < 0 2 j even numbers f ; g 0 n Z f 2 f j odd numbers gg ; g ; g f , g n > 0 gg , of N. gg , of Z, 5.6. EQUIVALENCECLASSESANDPARTITIONS 197 Exercise 5.99 A = 1; 2; 3; 4; 5 , g R = f (1; 1); (1; 2); (1; 4); (2; 1); (2; 2); (2; 4); f (3; 3); (3; 5); (4; 1); (4; 2); (4; 4); (5; 3); (5; 5) : g 1. Is R an equivalence on A? If so, answer 2 and 3: 2. Determine R. 2 j j 3. Determine A=R. Example 5.100 If on }(N) is given by (cid:24) A (cid:24) is reflexive. If A then and (cid:24) is finite. ; B : (A B) (B (cid:0) (cid:17) (cid:0) N is arbitrary, then (A [ (cid:18) A) is finite; A) (A [ (cid:0) (cid:0) A) = ; [ ; = , ; Exercise 5.101* Define the relation on }(N) by: (cid:24) Thus, N , since (N 6(cid:24) ; ) (cid:0) ; B : A (cid:24) (A (cid:17) ( ; (cid:0) [ (B B) (cid:0) [ N) = N A) is finite. (cid:0) = N is infinite. [ ; Show that (cid:24) is an equivalence (reflexivity is shown in example 5.100). Exercise 5.102 Define the relation R on all people by: aRb : common ancestor. Is R transitive? (cid:17) a and b have a Same question for the relation S defined by: aSb : ancestor along the male line. (cid:17) a and b have a common Example 5.103 For counting the partitions of a set A of size n, the key is to count the number of ways of partitioning a set A of size n into k non-empty classes. Let us use for this number, and see if we can find a recurrence for it, i.e., an equation in terms of the function for smaller arguments. Distributing n objects over 1 set can be done in only one way, so = 1. Similarly, distributing n n objects over n non-empty sets can be done in only one way, so n = 1. n 1 n k (cid:9) (cid:8) (cid:8) (cid:9) To distribute n objects over k different sets, we can either (i) put the last object into an equivalence class of its own, or (ii) put it in one of the existing classes. (cid:9) (cid:8) 198 CHAPTER5. RELATIONS ways, for this is the number of ways to distribute n n k 1 1 (cid:0) (cid:0) (i) can be done in objects over k are k classes to choose from, and classes. Thus, the recurrence we need is: (cid:9) (cid:0) (cid:0) k (cid:8) (cid:9) (cid:8) 1 1 non-empty classes. (ii) can be done in k n (cid:1) ways to distribute n (cid:8) (cid:0) 1 ways, for there 1 objects over k (cid:9) n 1 (cid:0) k (cid:0) (cid:27) (cid:27) In terms of this, the number of partitions b(n) is given by: (cid:26) (cid:27) (cid:26) (cid:26) n k = k (cid:1) 1 + n (cid:0) k n k 1 1 (cid:0) (cid:0) : b(n) = n k=1 (cid:26) X : n k (cid:27) The numbers bers. n k (cid:8) (cid:9) are called Stirling set numbers. The b(n) are called Bell num- Exercise 5.104 Implement functions bell and stirling to count the number of different partitions of a set of n elements. Exercise 5.105 Use the result of Exercise 5.104 to fill out the last column in the table of Exercise 5.71 on p. 191. Exercise 5.106 Show: conditions 2 and 3 of Definition 5.83 taken together are K. equivalent with: for every a A there exists exactly one K such that a 2 2 A 2 Exercise 5.107 Is the intersection R \ again an equivalence? And the union R [ example. (See the table of Exercise 5.50.) S of two equivalences R and S on a set A S? Prove, or supply a simple counter- Exercise 5.108* Suppose that R and S are equivalences on A such that R Show that every S-equivalence class is a union of R-equivalence classes. (cid:18) S. Exercise 5.109 A list partition is the list counterpart of a partition: list partitions are of type Eq a => [[a]], and a list partition xss of xs has the following prop- erties: (cid:15) (cid:15) (cid:15) [] is not an element of xss, xs and concat xss have the same elements, if ys and zs are distinct elements of xss, then ys and zs have no elements in common. 5.6. EQUIVALENCECLASSESANDPARTITIONS 199 Implement a function listPartition :: Eq a => [a] -> [[a]] -> Bool that maps every list xs to a check whether a given object of type [[a]] is a list partition of xs. Exercise 5.110 Implement a function listpart2equiv :: Ord a => [a] -> [[a]] -> Rel a that generates an equivalence relation from a list partition (see Exercise 5.109). The first argument gives the domain, the second argument the list partition. Gen- erate an error if the second argument is not a list partition on the domain given by the first argument.q Exercise 5.111 R = (0; 3); (1; 5); (2; 0) f ; A = g f 0; 1; 2; 3; 4; 5 . g 1. What is the smallest (in the sense of: number of elements) equivalence S R on A? 2. Determine A=S. 3. How many equivalences exist on A that include R? 4. Give the corresponding partitions. (cid:19) Exercise 5.112 Implement a function equiv2listpart :: Ord a => Set a -> Rel a -> [[a]] that maps an equivalence relation to the corresponding list partition (see Exercise 5.109). Generate an error if the input relation is not an equivalence. Exercise 5.113 Use the function equiv2listpart to implement a function equiv2part :: Ord a => Set a -> Rel a -> Set (Set a) that maps an equivalence relation to the corresponding partition. 200 CHAPTER5. RELATIONS Exercise 5.114* R is a relation on A. One of the following is the smallest equiva- lence on A that includes R. Which? Prove. R(cid:0) 1)+. ((cid:1)A [ (R(cid:0) [ 1)+. 1. ((cid:1)A R)+ [ R+ [ [ 2. (cid:1)A 3. (cid:1)A (R [ [ R(cid:0) 1)+. Exercise 5.115 Let R be a relation on A. Show that S = R(cid:3) (cid:3) is an equivalence on A. Next, show that the relation T on the quotient A=S given by a j j aR(cid:3)b is a partial order. b j j ST S : (cid:17) \ 1 R(cid:0) Note that the reasoning of Exercise 5.115 only uses reflexivity and transitivity of R(cid:3). So, in general, if R is a reflexive transitive relation on A (i.e., R is a pre-order), 1 is an equivalence on A, and the relation T on the quotient A=S then S = R given by S : a j j aRb is a partial order. R(cid:0) b j j \ ST (cid:17) Example 5.116 Consider the pre-order lence relation = of Example 5.25. Note that the equiva- j 1 is nothing other than the relation of logical equivalence. = j \ j =(cid:0) (cid:17) Remark. In constructions with quotients, it often happens that a relation on the quotient is defined in terms of a relation on the underlying set. In such a case one should always check that the definition is proper in the sense that it is independent of the representatives that are mentioned in it. Take again the example of exercise 5.115, of a set A with a reflexive transitive 1, then the definition of relation T on the quotient relation R on it. If S = R A=S given by S : S S together imply that a0Rb0. and b0 R(cid:0) aRb is proper, because it holds that aRb, a0 a j j \ (cid:17) b j j a j 2 j ST b j 2 j To see that this is so, assume a0 transitivity of R, a0Rb. Suppose b0 a0Rb this gives, again by transitivity of R, that a0Rb0. S and aRb. Then a0Sa, so a0Ra, and by a j S. Then b0Sb, so bRb0. Together with b j 2 j 2 j Exercise 5.117 Define the relations p q : p partition(s). (cid:0) (cid:25) (cid:17) 2 q Z, Z. Are these equivalences? If so, describe the corresponding on R by p and q : (cid:2) (cid:24) (cid:25) (cid:24) (cid:17) 2 p q Exercise 5.118 Define the relation R on R 3u v. (cid:0) R by (x; y)R(u; v) iff 3x y = (cid:0) (cid:2) 5.6. EQUIVALENCECLASSESANDPARTITIONS 201 1. Show that R is an equivalence. 2. Describe the equivalence classes of (0; 0) and (1; 1). 3. Describe R (cid:2) R=R in geometrical terms. Exercise 5.119 Define an equivalence on R centric circles with (0; 0) as centre. (cid:2) R that partitions the plane in con- Exercise 5.120 Q = on 0; 1; 2; 3; 4; 5 f it is given, that Q (0; 0); (0; 1); (0; 5); (2; 4); (5; 0) . For an equivalence R g R. R and (0; 2) = 2 (cid:18) f g 1. Show that (1; 5) R and (4; 5) R. 62 2 2. Give the partition corresponding to the smallest (in the sense of number of elements) equivalence Q. (cid:19) 3. How many equivalences S are there on and (0; 2) 62 S? Give the corresponding partitions. 0; 1; 2; 3; 4; 5 such that Q S (cid:18) g f Exercise 5.121 Prove Theorem 5.94. Hint. Make sure that you use all properties of partitions. Exercise 5.122* On a certain planet there are 20 countries. Every two of them either are at war with each other or have a peace treaty. Every two countries with a common enemy have such a peace treaty. What is the least possible number of peace treaties on this planet? Hint: Note that the relations P and W , for being at peace and at war, respectively, exclude one another, that they are both irreflexive and symmetric, and that they satisfy the principle that there are no war-triangles. For if x is at war with both y and z, then y and z have x as common enemy, hence y and z have a peace treaty: xyz((xW y 8 xW z) ^ ) yP z): Find a recurrence (see page 212) for the maximum number of countries at war among 2n countries. Next, use this to derive the minimum number of peace treaties among 2n countries. 202 CHAPTER5. RELATIONS 5.7 Integer Partitions Integer partitions of n exactly n. For example, the four integer partitions of 4 are 2 N+ are lists of non-zero natural numbers that add up to [4]; [1; 3]; [2; 2]; [1; 1; 2]; [1; 1; 1; 1]: The integer partitions of n correspond to the sizes of the set partitions of a set A with = n. Here is an algorithm for generating integer partitions, in lexico- graphically decreasing order: A j j (cid:15) (cid:15) The first integer partition of n is [n]. Let B be the last integer partition generated. If B consists of only 1’s, then done. Otherwise, there is a smallest non-1 part m. To generate the next partition, subtract 1 from m and collect all the units so as to match the new smallest part m 1. (cid:0) Examples The partition after [1; 1; 3; 3] is [1; 2; 2; 3], for after subtracting 1 from 3, we should pack the three units that result in parcels of size 2. The partition after [1; 1; 1; 1; 1; 1; 5] is [3; 4; 4], for after subtracting 1 from 5, we should pack the seven units that result in parcels with maximum size 4, which gives three units and one parcel of size 4, which in turn gives one parcel of size 3 and one of size 4. The partition after [3; 3; 5] is [1; 2; 3; 5]. The partition after [1; 3; 3; 4] is [2; 2; 3; 4]. Implementation An integer partition is represented as a list of integers. For convenience we count the number of 1’s, and remove them from the partition. This gives a compressed representation (2; [3; 3]) of [1; 1; 3; 3]. These compressed partitions have type CmprPart. type Part = [Int] type CmprPart = (Int,Part) Expansion of a compressed partition (n; p) is done by generating n 1’s followed by p: 5.7. INTEGERPARTITIONS 203 expand :: CmprPart -> Part expand (0,p) = p expand (n,p) = 1:(expand ((n-1),p)) In generating the next partition from (k,xs) we may assume that xs is non-empty, and that its elements are listed in increasing order. The partition that follows (k,x:xs) is generated by packing (k+x,x:xs) for maximum size x 1. This assumes that x is the smallest element in x:xs. (cid:0) nextpartition :: CmprPart -> CmprPart nextpartition (k,(x:xs)) = pack (x-1) ((k+x),xs) To pack a partition (m,xs) for size 1, there is nothing to do. To pack a partition (m,xs) for maximum size k > 1 and k > m, decrease the parcel size to k 1. To pack a partition (m,xs) for size k > 1 and k 6 m, use k units to generate one parcel of size k, and go on with (m-k,k:xs), for the same parcel size. (cid:0) pack :: Int -> CmprPart -> CmprPart pack 1 (m,xs) = (m,xs) pack k (m,xs) = if k > m then pack (k-1) (m,xs) else pack k (m-k,k:xs) To generate all partitions starting from a given partition (n,[]), just list the par- tition consisting of n units, for this is the last partition. To generate all partitions starting from a given partition (n,x:xs), list this partition, and then generate from the successor of (n,x:xs). generatePs :: CmprPart -> [Part] generatePs p@(n,[]) generatePs p@(n,(x:xs)) = = [expand p] (expand p: generatePs(nextpartition p)) 204 CHAPTER5. RELATIONS Generate all partitions starting from [n]. The case where n = 1 is special, for it is the only case where the compressed form of the first partition has a non-zero number of units. part :: Int -> [Part] part n | n < 1 = error "part: argument <= 0" = [[1]] | n == 1 | otherwise = generatePs (0,[n]) Here is what we get out: REL> part 5 [[5],[1,4],[2,3],[1,1,3],[1,2,2],[1,1,1,2],[1,1,1,1,1]] REL> part 6 [[6],[1,5],[2,4],[1,1,4],[3,3],[1,2,3],[1,1,1,3],[2,2,2], [1,1,2,2],[1,1,1,1,2],[1,1,1,1,1,1]] REL> length (part 20) 627 Exercise 5.123 Write a program change :: Int -> [Int] that returns change in EURO coins for any positive integer, in the least number of coins. Measure the values of the EURO coins 0.01, 0.02, 0.05, 0.10, 0.20, 0.50, 1, 2 in EURO cents, as 1, 2, 5, 10, 20, 50, 100, 200. Use pack for inspiration. Exercise 5.124 Modify the integer partition algorithm so that it generates all the possible ways of giving coin change for amounts of money up to 10 EURO, using all available EURO coins (0.01, 0.02, 0.05, 0.10, 0.20, 0.50, 1, 2). Measure the values of the EURO coins in EURO cents, as 1, 2, 5, 10, 20, 50, 100, 200. Exercise 5.125 How many different ways are there to give change for one EURO? Use the program from the previous exercise. 5.8 Further Reading More on relations in the context of set theory in [DvDdS78]. Binary relations on finite sets are studied in graph theory. See Chapters 4 and 5 of [Bal91]. Relations in the context of database theory are the topic of [AHV95]. Chapter 6 Functions Preview In mathematics, the concept of a function is perhaps even more important than that of a set. Also, functions are crucial in computer programming, as the functional programming paradigm demonstrates. This chapter introduces basic notions and then moves on to special functions, operations on functions, defining equivalences by means of functions, and compatibility of equivalences with operations. Many concepts from the abstract theory of functions translate directly into compu- tational practice. Most of the example functions mentioned in this chapter can be implemented directly in Haskell by just keying in their definitions, with domains and co-domains specified as Haskell types. Still, we have to bear in mind that an implementation of a function as a com- puter program is a concrete incarnation of an abstract object. The same func- If you key tion may be computed by vastly different computation procedures. in sum [2*k | k <- [1 .. 100] ] at the hugs prompt you get the answer 10100, and if you key in 100 * 101 you get the same answer, but the compu- tation steps that are performed to get at the answers are different. We have already seen that there is no mechanical test for checking whether two procedures perform the same task (Section 4.2), although in particular cases such results can be proved by mathematical induction (see Chapter 7). E.g., in Example 7.4 in Chapter 7 it n is proved by mathematical induction that the computational recipes k=1 2k and n(n + 1) specify the same function. P 205 206 CHAPTER6. FUNCTIONS module FCT where import List 6.1 Basic Notions A function is something that transforms an object given to it into another one. The objects that can be given to a function are called its arguments, and the results of the transformation are called values. The set of arguments is called the domain of the function. We say that a function is defined on its domain. If f is a function and x one of its arguments, then the corresponding value is denoted by f (x). A function value y = f (x) is called the image of x under f . That f (x) = y can also be indicated by f : x y. The domain of f is denoted by dom (f ). Its range is ran(f ) = x dom (f ) f (x) f j 7(cid:0)! 2 . g Example 6.1 A function can be given as a rule or a prescription how to carry out the transformation. (cid:15) First square, next add one is the function that transforms a real x x2 + 1. 2 R into Letting f stand for this function, it is customary to describe it by the equation f (x) = x2 + 1: The Haskell implementation uses the same equation: f x = x^2 + 1 The function described by (cid:15) = x j j (cid:26) x x (cid:0) if x > 0 if x < 0 6.1. BASICNOTIONS 207 transforms a real into its absolute value. The Haskell implementation given in Prelude.hs follows this definition to the letter: absReal x | x >= 0 = x | otherwise = -x (cid:15) The identity function 1A defined on A does not “transform” at all in the usual sense of the word: given an argument x A, it outputs x itself. A polymorphic identity function is predefined in Haskell as follows: 2 id :: a -> a id x = x Set theory has the following simple definition of the concept of a function. Definition 6.2 A function is a relation f that satisfies the following condition. (x; y) f 2 ^ (x; z) f = ) 2 y = z: That is: for every x f . 2 dom (f ) there is exactly one y ran(f ) such that (x; y) 2 2 dom (f ), then f (x) is by definition the unique object y ran(f ) for which 2 If x 2 (x; y) f . 2 Note that we use dom here in the sense defined for relations, but that the relation and the function-sense of the notion coincide: the domain dom (f ) of the function ; exactly as in the relation-case (cf. Definition 5.1 p. 162). f is y((x; y) f ) x f j 9 2 g Similarly, the range of f coincides with the range of f as a relation. The set-theoretic and the computational view on functions are worlds apart, for computationally a function is an algorithm for computing values. However, in cases of functions with finite domains it is easy to switch back and forth between the two perspectives, as the following conversions demonstrate. 208 CHAPTER6. FUNCTIONS list2fct :: Eq a => [(a,b)] -> a -> b list2fct [] _ = error "function not total" list2fct ((u,v):uvs) x | x == u = v | otherwise = list2fct uvs x fct2list :: (a -> b) -> [a] -> [(a,b)] fct2list f xs = [ (x, f x) | x <- xs ] The range of a function, implemented as a list of pairs, is given by: ranPairs :: Eq b => [(a,b)] -> [b] ranPairs f = nub [ y | (_,y) <- f ] If a function is defined on an enumerable domain, we can list its (finite or infinite) range starting from a given element. listValues listValues f i = (f i) : listValues f (succ i) :: Enum a => (a -> b) -> a -> [b] If we also know that the domain is bounded, we can generate the whole range as a finite list. listRange :: (Bounded a, Enum a) => (a -> b) -> [b] listRange f = [ f i | i <- [minBound..maxBound] ] Example 6.3 (cid:15) The function x x R 2 ^ 7! y = x2 + 1 g x2 +1 defined on R is identified with the relation x = R (x; x2 + 1) f j . g 2 (x; y) f j 6.1. BASICNOTIONS (cid:15) (1; 4); (2; 4); (3; 6) f = 4; 6 f f . g is a function. dom (f ) = g (1; 4); (2; 4); (2; 6) (cid:15) f is not a function. g 209 , ran(f ) = 1; 2; 3 f g (cid:1)X , the identity-relation on X (Definition 5.7, p. 163), also is a function from X to X. When viewed as a function, the identity relation on X is usually written as 1X . If f is a function, we clearly have that f = ran(f ) = dom (f ) f (a) a (a; f (a)) f a j 2 dom (f ) g , and f The relation j 2 is a function. dom ( ; ) = ). = ran( ; ; ; . g (cid:15) (cid:15) (cid:15) As is the case for relations, functions are more often given in a context of two sets. Compare 5.3 (p. 162). Definition 6.4 (From . . . to, On, Co-domain) Suppose that X and Y are sets. A function f is from X to Y ; notation: f : X Y; (cid:0)! if dom (f ) = X and ran(f ) with 5.3!) (cid:18) Y . (Note the difference in terminology compared In this situation, Y is called the co-domain of f . A function f is said to be defined on X if dom (f ) = X. Note that the set-theoretic way of identifying the function f with the relation R = has no way of dealing with this situation: it is not possible to (x; f (x)) f recover the intended co-domain Y from the relation R. As far as R is concerned, the co-domain of f could be any set that extends ran(R). X 2 x g j More Than One Argument. Functions as introduced here are unary, i.e., they apply to only one argument. But of course, functions with more than one argument (binary, ternary. . . ) do exist. E.g., addition and multiplication on N are of this kind. However, such a binary (ternary. . . ) function can be viewed as a unary one that applies to ordered pairs (resp., triples. . . ). As we have seen, Haskell has predefined operations curry and uncurry to switch back and forth between functions of types (a,b) -> c and a -> b -> c. We 210 CHAPTER6. FUNCTIONS can extend this to cases of functions that take triples, quadruples, etc. as arguments. As an example, here is the case for currying functions that take triples to functions that take three arguments, and for uncurrying functions that take three arguments to functions that take triples. curry3 :: ((a,b,c) -> d) -> a -> b -> c -> d curry3 f x y z = f (x,y,z) uncurry3 :: (a -> b -> c -> d) -> (a,b,c) -> d uncurry3 f (x,y,z) = f x y z Fact 6.5 (Function Equality) If f and g are functions that share their domain (dom (f ) = dom (g)) and, on it, carry out the same transformation (i.e., 2 dom (f )(f (x) = g(x))), then — according to Extensionality 4.1, p. 115 — we have that f = g. x 8 2 X with f (x) Y are different we have to find an To establish that two functions f; g : X = g(x). As we have seen in Section 4.2, there is no generic x algorithm for checking function equality. Therefore, to establish that two functions Y are equal, we need a proof. The general form, spelled out in full, of f; g : X such a proof is: ! ! Given: f; g : X Y . To be proved: f = g. Proof: ! Let x be an arbitrary object in X. To be proved: f (x) = g(x). Proof: . . . Thus f (x) = g(x). Thus f = g. Example 6.6 If f and g are defined on R by f (x) = x2 + 2x + 1, resp., g(x) = (x + 1)2, then f = g. Thus, functions are not distinguished by the details of how they actually transform (in this respect, f and g differ), but only with respect to their output-behaviour. 6 6.1. BASICNOTIONS 211 If functions f : X Z are given in the domain- Warning. co-domain-context, and X(f (x) = g(x)), then f and g count as equal only if we also have that Y = Z. The co-domains are taken as an integral part of the functions. Y and g : X ! ! 2 x 8 Function Definitions. in terms of an element x writing: 2 If t(x) is an expression that describes an element of Y Y by X, then we can define a function f : X (cid:0)! Let the function f : X ! Y be defined by f (x) = t(x). For completely specifying a function f three things are sufficient: Specify dom (f ), Specify the co-domain of f , Give an instruction for how to construct f (x) from x. (cid:15) (cid:15) (cid:15) Examples of such instructions are x definitions of functions that we have encountered in this book. x + 1, x 7! 7! x2, and in fact all the Haskell A very convenient notation for function construction is by means of lambda ab- straction (page 58). In this notation, (cid:21)x:x+1 encodes the specification x x+1. The lambda operator is a variable binder, so (cid:21)x:x+1 and (cid:21)y:y +1 denote the same function. In fact, every time we specify a function foo in Haskell by means of 7! foo x y z = t we can think of this as a specification (cid:21)xyz:t. If the types of x; y; z; t are known, this also specifies a domain and a co-domain. For if x :: a, y :: b, z :: c, t :: d, then (cid:21)xyz:t has type a -> b -> c -> d. Example 6.7 1. t(x) is the expression 2x2 + 3: Let the function g : R 2. t(x) is the expression R be defined by g(x) = 2x2 + 3. (cid:0)! x 0 y sin(y)dy: Let the function h : R R (cid:0)! R be defined by h(x) = x 0 y sin(y) dy. R 212 CHAPTER6. FUNCTIONS Example 6.8 The Haskell way of defining functions is very close to standard mathematical practice, witness the following examples (note that all these equa- tions define the same function): f1 x = x^2 + 2 * x + 1 g1 x = (x + 1)^2 f1’ = \x -> x^2 + 2 * x + 1 g1’ = \x -> (x + 1)^2 Recurrences versus Closed Forms. A definition for a function f : N A in terms of algebraic operations is called a closed form definition. A function defini- tion for f in terms of the values of f for smaller arguments is called a recurrence for f . The advantage of a closed form definition over a recurrence is that it allows for more efficient computation, since (in general) the computation time of a closed form does not grow exponentially with the size of the argument. ! Example 6.9 Consider the following recurrence. g 0 = 0 g n = g (n-1) + n A closed form definition of the same function is: g’ n = ((n + 1) * n ) / 2 Exercise 6.10 Give a closed form implementation of the following function: 6.1. BASICNOTIONS 213 h 0 = 0 h n = h (n-1) + (2*n) Exercise 6.11 Give a closed form implementation of the following function: k 0 = 0 k n = k (n-1) + (2*n-1) To show that a particular closed form defines the same function as a given recur- rence, we need a proof by induction: see Chapter 7. It is not always possible to find useful definitions in closed form, and recurrences are in general much easier to find than closed forms. E.g., a closed form for the factorial function n! would be an expression that allows us to compute n! with at most a fixed number of ‘standard’ operations. The number of operations should be independent of n, so hides a number of n! = product operations that does depend on n. No closed form for n! is known, and n n. k=1 k performs essentially the same calculation as n! = (n n! = Thus, computationally, there is nothing to choose between the following two im- plementations of n!. n does not count, for n k=1 k = 1 (cid:2) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) 1)! (n Q Q 1) (cid:0) (cid:2) (cid:0) (cid:2) fac 0 = 1 fac n = fac (n-1) * n fac’ n = product [ k | k <- [1..n] ] Note that there is no need to add fac’ 0 = 1 to the second definition, because of the convention that product [] gives the value 1. A simple example of defining a function in terms of another function is the follow- ing (see Figure 6.1). 214 4 3 2 1 0 CHAPTER6. FUNCTIONS 4 3 2 1 0 -2 -1 0 1 2 0 1 2 Figure 6.1: Restricting the function (cid:21)x:x2. Definition 6.12 (Restrictions) Suppose that f : X restriction of f to A is the function h : A notation for this function is f(cid:22)A. ! X. The Y and A Y defined by h(a) = f (a). The (cid:0)! (cid:18) Here is the implementation, for functions implemented as type a -> b: restrict :: Eq a => (a -> b) -> [a] -> a -> b restrict f xs x | elem x xs = f x | otherwise = error "argument not in domain" And this is the implementation for functions implemented as lists of pairs: restrictPairs :: Eq a => [(a,b)] -> [a] -> [(a,b)] restrictPairs xys xs = [ (x,y) | (x,y) <- xys, elem x xs ] Definition 6.13 (Image, Co-image) Suppose that f : X B Y . (cid:18) Y , A (cid:18) (cid:0)! X and 1. f [A] = f (x) f x A g 2 j is called the image of A under f ; 6.1. BASICNOTIONS 215 2. f (cid:0) 1[B] = x f 2 X j f (x) B g 2 is called the co-image of B under f . From this definition we get: 1. f [X] = ran(f ), 2. f (cid:0) 1[Y ] = dom (f ), A(y = f (x)), 3. y 4. x 2 2 f [A] x , 9 f (cid:0) 1[B] , 2 f (x) From 3. it follows that x necessarily have that f (x) x = 0, A = 1 .) f g 2 2 B. 2 A f [A] ) ) f (x) x 2 2 A. (Example: f = f [A]. But note that we do not , g (0; 2); (1; 2) f Two Types of Brackets. Distinguish f (a) and f [A]. The notation f (a) pre- dom (f ). Then, f (a) is the f -value of a. The notation f [A] supposes that a presupposes A dom (f ). Then f [A] is the set of values f (x) where x A. 2 (cid:18) 2 Remark. Many texts do not employ f [ ] but use f ( ) throughout. In that case, you have to figure out from the context what is meant. 1[B], the part f (cid:0) 1 has no meaning when taken by itself. The In the expression f (cid:0) 1 will be used later on for the inverse function corresponding to f . notation f (cid:0) Such an inverse only exists if f happens to be a bijection. However, the notation f (cid:0) 1[B] is always meaningful and does not presuppose bijectivity. Here are the implementations of image and co-image: image :: Eq b => (a -> b) -> [a] -> [b] image f xs = nub [ f x | x <- xs ] coImage :: Eq b => (a -> b) -> [a] -> [b] -> [a] coImage f xs ys = [ x | x <- xs, elem (f x) ys ] This gives: FCT> image (*2) [1,2,3] [2,4,6] FCT> coImage (*2) [1,2,3] [2,3,4] [1,2] 216 CHAPTER6. FUNCTIONS And here are the versions for functions represented as lists of pairs: imagePairs :: (Eq a, Eq b) => [(a,b)] -> [a] -> [b] imagePairs f xs = image (list2fct f) xs coImagePairs :: (Eq a, Eq b) => [(a,b)] -> [a] -> [b] -> [a] coImagePairs f xs ys = coImage (list2fct f) xs ys Exercise 6.14 Consider the relation R = (0; 4); (1; 2); (1; 3) f . g 1. Is R a function? If so, determine dom (R) and ran(R). 2. Remember: R(cid:0) 1 = f (Definition 5.7, p. 163). (b; a) (a; b) R g 2 j is the inverse of the relation R. 1 a function? Is R(cid:0) If so, determine dom (R(cid:0) 1) and ran(R(cid:0) 1). Exercise 6.15 Suppose that f : X Y and A (cid:18) (cid:0)! X. Verify: 1. f [A] = ran(f(cid:22)A), 2. f [dom (f )] = ran(f ), 3. f (cid:0) 1[B] = dom (f (X \ (cid:2) B)), 4. f (cid:0) 1[ran(f )] = dom (f ), 5. f(cid:22)A = f (A \ (cid:2) Y ). Exercise 6.16 Let X = 0; 1; 2; 3 f Determine f (cid:22) , f [ 1; 2; 3 f g with the implementation. 0; 3 f , Y = 2; 3; 4; 5 f g 1[ ] and f (cid:0) 2; 4; 5 f g (0; 3); (1; 2); (2; 4); (3; 2) , f = g ]. Next, check your answers g f . g 6.1. BASICNOTIONS 217 Exercise 6.17 Suppose that f : A Y , g : B Y , and A B = \ ! ! . Show that ; f [ g : A B [ ! Y: What if A B = ? ; \ Exercise 6.18* Let fA : A A Y is given. Show, that be a partition of X. For every component A fA : X Y . a function 2 A ! A 2A S ! Example 6.19 Suppose that f : X f [B]. B] f [A] Y and A; B (cid:18) ! X. We show that f [A (cid:0) (cid:19) (cid:0) Assume that y obtain x 2 y = f (x) 2 f [A] f [B]. Then y f [A] and y 2 (cid:0) 2 A such that y = f (x). From the second we see that x f [B]). So, x B, and y = f (x) f [A B]. A 62 62 f [B]. From the first we B (otherwise, 2 (cid:0) 2 (cid:0) To see that the inclusion cannot be replaced by an equality, take X = Y = A = , B = 0; 1 0 f g and f [A] , and let f be given by f (x) = 0. Then f [A f g f [B] = B] = f [ ] = (cid:0) = g f f 0 0 1 0 g (cid:0) f g (cid:0) f Next, suppose that f : X f (cid:0) f (cid:0) 1[D]: 1[C] ! (cid:0) . ; g Y and C; D Y . We show that f (cid:0) 1[C D] = (cid:0) (cid:18) f (cid:0) 1[C x 2 D] (cid:0) () () () () f (x) C D 2 (cid:0) C) 2 1[C] f (cid:0) 1[C] f (cid:0) (f (x) x x 2 2 ^ ^ (f (x) 62 x f (cid:0) 62 1[D]: f (cid:0) D) 1[D] (cid:0) The required equality follows using Extensionality. Exercise 6.20 Suppose that f : X 6.19. Show: ! Y , A; B (cid:18) X, and C; D (cid:18) Y . See example 1. A (cid:18) C (cid:18) 2. f [A f [A 3. f (cid:0) f (cid:0) [ \ 1[C 1[C B] [ \ f [B]; B D ) ) f [A] (cid:18) 1[C] f (cid:0) B] = f [A] (cid:18) f [B]; [ f (cid:0) 1[D], f [A] f [B], \ (cid:18) 1[C] D] = f (cid:0) 1[C] D] = f (cid:0) f (cid:0) f (cid:0) 1[D]; 1[D], [ \ 6 218 CHAPTER6. FUNCTIONS 4. f [f (cid:0) 1[C]] 1[f [A]] f (cid:0) C; A. (cid:18) (cid:19) Give simple examples to show that the inclusions in 2 and 4 cannot be replaced by equalities. 6.2 Surjections, Injections, Bijections If X is the domain of a function f , then for each x with (x; y) Functions for which this does not happen warrant a special name. 2 f . However, there may be other elements z 2 2 X there is only one y f . X with (z; y) 2 If f is a function from X to Y , then there may be elements of Y that are not in f [X]. Again, functions for which this does not happen warrant a special name. Definition 6.21 (Surjections, Injections, Bijections) A function f : X Y is called (cid:0)! 1. surjective, or a surjection, or onto Y, if every element b function value of at least one a X, i.e., if f [X] = Y ; 2 Y occurs as a 2 2. injective, an injection, or one-to-one, if every b X; a 2 Y is value of at most one 2 3. bijective or a bijection if it is both injective and surjective. Example 6.22 Most functions are neither surjective, nor injective. For instance, (cid:15) (cid:15) (cid:15) sin : R (sin 0 = sin (cid:25)). ! R is not surjective (e.g., 2 R is not a value) and not injective 2 The identity function 1X : X ! X is a bijection, whatever the set X. Let A be a set. According to Theorems 5.87 and 5.94, the function that trans- forms an equivalence R on A into its quotient A=R is a bijection between the set of equivalences and the set of partitions on A. (cid:15) Consider f = tion (0; 1); (1; 0); (2; 1) f . Thus, dom (f ) = g f 0; 1; 2 . The func- g f : 0; 1; 2 f g (cid:0)! f 0; 1 g 6.2. SURJECTIONS,INJECTIONS,BIJECTIONS 219 is surjective, but f : 0; 1; 2 f g (cid:0)! f 0; 1; 2 g is not surjective. The concept of surjectivity presupposes that of a co- domain. Cf. 6.5 (p. 210). However, whatever this co-domain, f clearly is not injective, since 0 and 2 have the same image. If the domain of a function is represented as a list, the injectivity test can be im- plemented as follows: injective :: Eq b => (a -> b) -> [a] -> injective f [] = True injective f (x:xs) = Bool notElem (f x) (image f xs) && injective f xs Similarly, if the domain and co-domain of a function are represented as lists, the surjectivity test can be implemented as follows: surjective :: Eq b => (a -> b) -> [a] -> [b] -> Bool surjective f xs [] = True surjective f xs (y:ys) = elem y (image f xs) && surjective f xs ys Exercise 6.23 Implement a test for bijectivity. Exercise 6.24 Implement tests injectivePairs, surjectivePairs, bijectivePairs for functions represented as lists of pairs. Proving that a Function is Injective/Surjective. The following implication is a useful way of expressing that f is injective: The proof schema becomes: f (x) = f (y) = ) x = y: 220 CHAPTER6. FUNCTIONS To be proved: f is injective. Proof: Let x; y be arbitrary, and suppose f (x) = f (y). ... Thus x = y. The contraposition of f (x) = f (y) = ) x = y, i.e., ) of course says the same thing differently, so an equivalent proof schema is: x = y = f (x) = f (y); To be proved: f is injective. Proof: Let x; y be arbitrary, and suppose x = y. ... Thus f (x) = f (y). That f : X ! Y is surjective is expressed by: b 8 This gives the following pattern for a proof of surjectivity: X f (a) = b: 2 2 Y 9 a To be proved: f : X Proof: Let b be an arbitrary element of Y . ! Y is surjective. ... Thus there is an a X with f (a) = b. 2 Exercise 6.25 Are the following functions injective? surjective? 6 6 6 6 6.2. SURJECTIONS,INJECTIONS,BIJECTIONS 221 1. sin : R+ R (N.B.: R+ = R x f 2 j 0 < x ), g ! [ ! (cid:0) 1; +1] 2. sin : R 1; +1], 3. sin : [ (cid:0) 1; +1], [ (cid:0) ! 4. ex : R R, ! 5. tan : R R, ! 6. log : R+ R, ! 7. p : R+ R+. ! Remark. The functions of Exercise 6.25 are all predefined in Haskell: sin, exp, tan, log, sqrt. The base of the natural logarithm, Napier’s number e, is given by exp 1. Exercise 6.26 Give a formula for the number of injections from an n-element set A to a k-element set B. Exercise 6.27 Implement a function injs :: [Int] -> [Int] -> [[(Int,Int)]] that takes a finite domain and a finite codomain of type Int and produces the list of all injections from domain to codomain, given as lists of integer pairs. Exercise 6.28 The bijections on a finite set A correspond exactly to the permuta- tions of A. Implement a function perms :: [a] -> [[a]] that gives all permutations of a finite list. The call perms [1,2,3] should yield: [[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]] Hint: to get the permutations of (x:xs), take all the possible ways of inserting x in a permutation of xs. 222 CHAPTER6. FUNCTIONS 6.3 Function Composition Definition 6.29 (Composition) Suppose that f : X Z. Thus, the co-domain of f coincides with the domain of g. The composition of f and g is the function g Z defined by Y and g : Y f : X (cid:0)! (cid:0)! (cid:14) (cid:0)! f )(x) = g(f (x)): (g (cid:14) (“First, apply f , next, apply g” — thanks to the usual “prefix”-notation for func- tions, the f and the g are unfortunately in the reverse order in the notation g f . To keep this reverse order in mind it is good practice to refer to g f as “g after f ”.) (cid:14) (cid:14) N.B.: The notation g g are the same. Furthermore, g (cid:14) f presupposes that the co-domain of f and the domain of f has the domain of f and the co-domain of g. (cid:14) Function composition is predefined in Haskell, as follows: (.) (f . g) x :: (b -> c) -> (a -> b) -> (a -> c) = f (g x) Example 6.30 Haskell has a general procedure negate for negating a number. The effect of first taking the absolute value, then negating can now be got by means of (negate . abs): Prelude> (negate . abs) 5 -5 Prelude> (negate . abs) (-7) -7 Prelude> Example 6.31 Another example from Prelude.hs is: even n odd = n ‘rem‘ 2 == 0 = not . even 6.3. FUNCTIONCOMPOSITION 223 Exercise 6.32 Implement an operation comp for composition of functions repre- sented as lists of pairs. Example 6.33 In analysis, compositions abound. E.g., if f : R x2 + 1, and g : R+ (g px, then g f )(x) = g(f (x)) = g(x2 + 1) = px2 + 1. R is x f : R ! 7! ! (cid:14) R is x ! 7! R+ is x 7! px2 + 1: (cid:14) The identity function behaves as a unit element for composition (see Example 6.1 for 1X): Fact 6.34 If f : X Y , then f 1X = 1Y (cid:14) f = f . (cid:14) ! Y , g : Y U . There are now two ways Suppose that f : X f ). The next to define a function from X to U : (i) (h (cid:14) lemma says that these functions coincide. Thus, composition is associative; and f if we mean either of these. we can safely write h Z and h : Z g) f , and (ii) = h ! ! ! (g (cid:14) (cid:14) (cid:14) g (cid:14) (cid:14) Lemma 6.35 If f : X h f ). (g (cid:14) (cid:14) Y , g : Y ! ! Z and h : Z U , then (h g) (cid:14) (cid:14) f = ! g) X. Then: Proof. Suppose that x f ))(x). f )(x) = (h ((h (cid:14) (cid:14) The functions (h f ) have the same domain (X) and co-domain g) (cid:14) (U ) and, on this domain, show the same action. Thus, they are (6.5, p. 210) equal. 2 g)(f (x)) = h(g(f (x))) = h((g f and h f )(x)) = (h (g (g (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) Lemma 6.36 Suppose that f : X Y , g : Y (cid:0)! (cid:0)! Z. Then: f injective, g surjective, 1. g (cid:14) 2. g f injective = ) f surjective = (cid:14) ) 3. f and g injective = ) 4. f and g surjective = g (cid:14) g f injective, f surjective. (cid:14) ) Proof. We prove 1, 2 and 4. 1. Given: g f injective, i.e., (g f )(a1) = (g To be proved: f injective, i.e., f (a1) = f (a2) (cid:14) (cid:14) f )(a2) ) a1 = a2. (cid:14) ) a1 = a2. 224 CHAPTER6. FUNCTIONS Proof: Assume f (a1) = f (a2). Then of course, g(f (a1)) = g(f (a2)). (Applying g twice to the same argument must produce the same value twice.) But then, f )(a2). The given now shows that (g (cid:14) a1 = a2. f )(a1) = g(f (a1)) = g(f (a2)) = (g (cid:14) (cid:14) f surjective. 2. Given: g To be proved: g surjective. I.e., Proof: Assume c there is a is the element looked for. 2 X such that (g c 8 Z. Wanted: b 2 (cid:14) 2 2 b 9 Y (g(b) = c). Z Y such that g(b) = c. Since g 2 f )(a) = c. But, g(f (a)) = (g f is surjective, f )(a). I.e., b = f (a) (cid:14) (cid:14) f surjective. I.e., every c 4. Given: f and g surjective. To be proved: g (cid:14) Proof: Assume c Since f is surjective, a g(f (a)) = g(b) = c. So, a is the element looked for. Z. Since g is surjective, b 2 2 2 2 Z is a value of g f . (cid:14) X exists such that f (a) = b. It follows that (g Y exists such that g(b) = c. f )(a) = (cid:14) Note that there are statements in the spirit of Lemma 6.36 that do not hold. In f is injective, it does not follow that g is injective. particular, from the fact that g to Y , and g an arbitrary Consider the case where f is the (empty) function from non-injective function from Y to ! Z, which is surely injective. But g by assumption is not injective. f is the (empty) function from Z. Then g (cid:14) ; ; (cid:14) For an example in which g take f : N (cid:14) ! f is bijective, but g not injective and f not surjective, N given by f (n) = n + 1 and g : N N given by g(n) = n n 1 (cid:0) (cid:26) ! if n = 0 if n > 1: Clearly, g (cid:14) f is the identity on N, but f is not a surjection, and g is not an injection. Exercise 6.37 Can you think up an example with g f bijective, but g not injective and f not surjective, with the domain of f finite? If yes, give the example, if no, prove that this is impossible. (cid:14) Exercise 6.38 The function f : following table: f 0; 1; 2; 3; 4 0; 1; 2; 3; 4 is defined by the g ! f g 0 1 1 2 2 0 3 0 4 3 x f (x) (f f )(x) (f f f )(x) (f f f f )(x) 6.3. FUNCTIONCOMPOSITION 225 1. Determine the compositions f the table. f , f f (cid:14) (cid:14) (cid:14) f and f f f (cid:14) (cid:14) (cid:14) f by completing 2. How many elements has the set are functions!) f; f f f; f f (cid:14) (cid:14) (cid:14) f; : : : g ? (N.B.: the elements 3. Exhibit a function g : 0; 1; 2; 3; 4 0; 1; 2; 3; 4 g ! f g g g; : : : g (cid:14) has 6 elements. f such that g; g f (cid:14) g; g (cid:14) A is a bijection. Then Exercise 6.39* Suppose that A is a finite set and f : A f 1 = f , f 2 = f f ,. . . all are bijections : A f , f 3 = f ! f (cid:14) (cid:14) (cid:14) A. ! 1. Show that, somewhere in this sequence, there occurs the bijection 1A. I.e., a number n exists such that f n = 1A. 2. Suppose that A has k elements. Can you give an upper bound for n? Exercise 6.40 Suppose that h : X a bijection. ! X satisfies h h (cid:14) (cid:14) h = 1X. Show that h is Exhibit a simple example of a set X and a function h : X 1X , whereas h = 1X . ! X such that h h = h (cid:14) (cid:14) Exercise 6.41 Prove Lemma 6.36.3. Exercise 6.42 Suppose that f : X Y and g : Y bijective. Show that f is surjective iff g is injective. ! ! Z are such that g f is (cid:14) Exercise 6.43 Suppose that limi Show that limi ai = a, and that f : N af (i) = a. (Cf. also Exercise 8.22, p. 318.) !1 !1 N is injective. ! 6 226 CHAPTER6. FUNCTIONS 6.4 Inverse Function ! If we consider f : X Y as a relation, then we can consider its relational f . However, there is no guarantee that inverse: the set of all (y; x) with (x; y) 2 the relational inverse of a function is again a function. In case f is injective, we know that the relational inverse of f is a partial function (some elements in the domain may not have an image). If f is also surjective, we know that the relational inverse of f is a function. Thus, an inverse function of f has to satisfy some special requirements. Figure 6.2: Inverse of the function (cid:21)x:x2 (restricted to R+). Definition 6.44 (Inverse Function) Suppose that f : X f = 1X , and (ii) f Y X is an inverse of f if both (i) g ! (cid:14) Y . A function g : g = 1Y . ! (cid:14) The next theorem says all there is to know about inverses. Its proof describes how to find an inverse if there is one. Note that, by the first part of the theorem, we can safely talk about the inverse of a function (provided there is one). Theorem 6.45 1. A function has at most one inverse. 2. A function has an inverse iff it is bijective. Proof. (1) Suppose that g and h are both inverses of f : X 1X 1Y = h. g) = h g = (h g = h f ) (f (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) Y . Then g = ! (2) (Only if.) Assume that g is inverse of f . Then since g Lemma 6.36.1 f also is injective. And since f also is surjective. (cid:14) f = 1X is injective, by g = 1Y is surjective, by 6.36.2 f (cid:14) 6.4. INVERSEFUNCTION 227 (If.) Suppose that f is a bijection. I.e., for every y such that f (x) = y. Thus, we can define a function g : Y the unique x such that f (x) = y. 2 ! Y there is exactly one x X X by letting g(y) be 2 Then g is inverse of f : firstly, if x then f (g(y)) = y. 2 X, then g(f (x)) = x; secondly, if y Y , 2 Notation. If f : X ! Y is a bijection, then its unique inverse is denoted by f (cid:0) 1. If f : X Remark. function : Y of Theorem 6.45 it is clear that these are the same. 1 can denote either the inverse X, or the inverse of f considered as a relation. But from the proof Y is a bijection, then f (cid:0) ! ! Example 6.46 The real function f that is given by f (x) = 9 to convert degrees Celcius into degrees Fahrenheit. The inverse function f (cid:0) given by f (cid:0) Celsius. Here are integer approximations: 5 x + 32 allows us 1 is 32); it converts degrees Fahrenheit back into degrees 1(x) = 5 9 (x (cid:0) c2f, f2c :: Int -> Int c2f x = div (9 * x) 5 + 32 f2c x = div (5 * (x - 32)) 9 *Left and Right-inverse. Note that there are two requirements on inverse func- f = 1X only, then g is Y , g : Y tions. If f : X called left-inverse of f and f right-inverse of g. X, and we have that g ! ! (cid:14) Example 6.47 The class Enum is (pre-)defined in Haskell as follows: class Enum a where succ, pred toEnum fromEnum :: a -> a :: Int -> a :: a -> Int fromEnum should be a left-inverse of toEnum: 228 CHAPTER6. FUNCTIONS fromEnum (toEnum x) = x This requirement cannot be expressed in Haskell, so it is the responsibility of the programmer to make sure that it is satisfied. Examples of use of toEnum and fromEnum from Prelude.hs: ord ord chr chr :: Char -> Int = fromEnum :: Int -> Char = toEnum Exercise 6.48 Show: if f : X is a bijection and g = h = f (cid:0) ! 1. Y has left-inverse g and right-inverse h, then f Exercise 6.49 Suppose that f : X are equivalent. ! Y and g : Y ! X. Show that the following 1. g (cid:14) f = 1X, 2. (f (x); x) f x j 2 X g (cid:18) g. ! ! Exercise 6.50 X = f : X 0; 1 Y . How many functions g : Y , Y = g f f 2; 3; 4; 5 , f = g X have the property, that g (0; 3); (1; 4) . We have that f = 1X ? g f ! (cid:14) Exercise 6.51 Give an example of an injection f : X g : Y X such that g f = 1X . Y for which there is no ! (cid:14) Exercise 6.52* Show that if f : X g = 1Y . such that f (cid:14) ! Y is surjective, a function g : Y X exists ! 6.5. PARTIALFUNCTIONS 229 Exercise 6.53 How many right-inverses are there to the function (domain: 0; 1; 2; 3; 4 f g 5; 6 )? g f (0; 5); (1; 5); (2; 5); (3; 6); (4; 6) g f , co-domain: Exercise 6.54 1. The surjection f : R three different right-inverses for f . ! R+ is defined by f (x) = x2. Give 2. Same question for g : [0; (cid:25)] [0; 1] defined by g(x) = sin x. ! Exercise 6.55 Suppose that f : X that the following are equivalent. ! Y is a surjection and h : Y X. Show ! 1. h is right-inverse of f , 2. h (f (x); x) (cid:18) f x X . g 2 j Exercise 6.56* Show: 1. Every function that has a surjective right-inverse is a bijection. 2. Every function that has an injective left-inverse is a bijection. 6.5 Partial Functions A partial function from X to Y is a function with its domain included in X and its range included in Y . If f is a partial function from X to Y we write this as f : X , X and f (cid:22) dom (f ) : dom (f ) Y . It is immediate from this definition that f : X , Y . Y iff dom (f ) ! ! (cid:18) ! A way of defining a partial function (using f (x) = ? t (cid:26) for ‘undefined’): ? if ::: otherwise 230 CHAPTER6. FUNCTIONS The computational importance of partial functions is in the systematic perspective they provide on exception handling. In Haskell, the crude way to deal with excep- tions is by a call to the error abortion function error. The code below implements partial functions succ0 and succ1. succ0 is partial because the pattern (x+1) only matches positive integers. succ1 has an explicit call to error. The disad- vantage of these implementations is that they are called by another program, the execution of that other program may abort. succ0 :: Integer -> Integer succ0 (x+1) = x + 2 succ1 :: Integer -> Integer succ1 = \ x -> if x < 0 then error "argument out of range" else x+1 This uses the reserved keywords if, then and else, with the obvious meanings. A useful technique for implementing partial functions is to represent a partial func- tion from type a to type b as a function of type a -> [b]. In case of an exception, the empty list is returned. If a regular value is computed, the unit list with the computed value is returned. succ2 :: Integer -> [Integer] succ2 = \ x -> if x < 0 then [] else [x+1] Composition of partial functions implemented with unit lists can be defined as follows: pcomp :: (b -> [c]) -> (a -> [b]) -> a -> [c] pcomp g f = \ x -> concat [ g y | y <- f x ] As an alternative to this trick with unit lists Haskell has a special data type for im- plementing partial functions, the data type Maybe, which is predefined as follows. 6.5. PARTIALFUNCTIONS 231 data Maybe a = Nothing | Just a deriving (Eq, Ord, Read, Show) maybe maybe n f Nothing maybe n f (Just x) = f x = n :: b -> (a -> b) -> Maybe a -> b Here is a third implementation of the partial successor function: succ3 :: Integer -> Maybe Integer succ3 = \ x -> if x < 0 then Nothing else Just (x+1) The use of the predefined function maybe is demonstrated in the definition of com- position for functions of type a -> Maybe b. mcomp :: (b -> Maybe c) -> (a -> Maybe b) -> a -> Maybe c mcomp g f = (maybe Nothing g) . f Of course, the maybe function allows for all kinds of ways to deal with excep- tions. E.g., a function of type a -> Maybe b can be turned into a function of type a -> b by the following part2error conversion. part2error :: (a -> Maybe b) -> a -> b part2error f = (maybe (error "value undefined") id) . f Exercise 6.57 Define a partial function stringCompare :: String -> String -> Maybe Ordering 232 CHAPTER6. FUNCTIONS for ordering strings consisting of alphabetic characters in the usual list order. If a non-alphabetic symbol occurs, the ordering function should return Nothing. Use isAlpha for the property of being an alphabetic character. 6.6 Functions as Partitions In practice, equivalences are often defined by way of functions. Par abus de lan- guage functions sometimes are called partitions for that reason. Examples of such functions on the class of all people: “the gender of x” (partitions in males and fe- males), “the color of x” (partitions in races), “the age of x” (some hundred equiv- alence classes). The next exercise explains how this works and asks to show that every equivalence is obtained in this way. Exercise 6.58 Suppose that f : A by: aRb : ! f (a) = f (b). Thus, R = (cid:17) f I is a surjection. Define the relation R on A f (a) = f (b) (a; b) . Show: A2 2 j g 1. R is an equivalence on A, 2. A=R = f (cid:0) f 1[ ] i g f j i I , g 2 3. for every equivalence S on A there is a function g on A such that aSb g(a) = g(b). , Z with n Example 6.59 For any n be given by RMn(m) := r where 0 6 r < n and there is some a m = an + r. Then RMn induces the equivalence = 0, let the function RMn :: Z n on Z. 2 2 Z ! Z with (cid:17) Here is a Haskell implementation of a procedure that maps a function to the equiv- alence relation inducing the partition that corresponds with the function: fct2equiv :: Eq a => (b -> a) -> b -> b -> Bool fct2equiv f x y = (f x) == (f y) You can use this to test equality modulo n, as follows: 6 6.6. FUNCTIONSASPARTITIONS 233 FCT> fct2equiv (‘rem‘ 3) 2 14 True Exercise 6.60* Suppose that f : A B. ! 1. Show: If f is an injection, then for all sets C and for every g : A C such that g = h is a function h : B f . C there ! ! (cid:14) C there is a function h : B C ! 2. Show: For all sets C, if to every g : A ! f , then f is an injection. such that g = h (cid:14) Exercise 6.61* Suppose that R is an equivalence on the set A. Show: for every equivalence S A=S such that, for a R on A there exists a function g : A=R A: ! a a R). (cid:19) S = g( j j j 2 j Exercise 6.62* Suppose that binary function such that for all a; b; x; y (cid:24) is an equivalence on A, and that f : A2 A is a ! A: 2 a x b (cid:24) ^ (cid:24) y = ) f (a; b) (cid:24) f (x; y): Show that a unique function f (cid:24) : (A= . f (a; b) ; a f (cid:24)( j j j j ) = b j j )2 (cid:24) ! B exists such that, for a; b A: 2 Exercise 6.63* Suppose that (cid:24) relation such that for all a; b; x; y is an equivalence on A, and that R A2 is a (cid:18) a x (cid:24) ^ Show that a unique relation R(cid:24) aRb. R(cid:24) a j j b j , j y ^ (A= aRb ) xRy: )2 exists such that for all a; b (cid:24) A: 2 (cid:24) (cid:18) A: 2 b Exercise 6.64* A and B are sets, with B (x; y) a = x. (cid:17) = . Define ; on A (cid:2) (cid:24) B by: (a; b) (cid:24) 1. Show that (cid:24) is an equivalence on A B. (cid:2) 2. Exhibit a bijection : (A to A. (cid:24) B)= (cid:2) (cid:24) (cid:0)! A from the quotient of A B modulo (cid:2) 6 234 CHAPTER6. FUNCTIONS 3. Exhibit, for every equivalence class, a bijection between the class and B. Equivalence classes (restricted to a list) for an equivalence defined by a function are generated by the following Haskell function: block :: Eq b => (a -> b) -> a -> [a] -> [a] block f x list = [ y | y <- list, f x == f y ] This gives: FCT> block (‘rem‘ 3) 2 [1..20] [2,5,8,11,14,17,20] FCT> block (‘rem‘ 7) 4 [1..20] [4,11,18] Exercise 6.65 Functions can be used to generate equivalences, or equivalently, partitions. In an implementation we use list partitions; see Exercise 5.109 for a definition. Implement an operation fct2listpart that takes a function and a domain and produces the list partition that the function generates on the domain. Some example uses of the operation are: Main> fct2listpart even [1..20] [[1,3,5,7,9,11,13,15,17,19],[2,4,6,8,10,12,14,16,18,20]] Main> fct2listpart (\ n -> rem n 3) [1..20] [[1,4,7,10,13,16,19],[2,5,8,11,14,17,20],[3,6,9,12,15,18]] Exercise 6.66 Give an formula for the number of surjections from an n-element B induces a partition. set A to a k-element set B. (Hint: each surjection f : A These partitions can be counted with the technique from Example 5.103.) ! 6.7 Products Definition 6.67 (Product) Suppose that, for every element i Xi is given. The product I and such that for all i I a non-empty set I Xi is the set of all functions f for which dom (f ) = i 2 I: f (i) 2 Xi. 2 Q 2 6.7. PRODUCTS 235 When I = f 1 , this product is also written as X0 (cid:2) (cid:1) (cid:1) (cid:1) (cid:2) 0; : : : ; n g I) are the same, Xi = X, the product is written as X I. Thus, X I is Xn (cid:0) 1. (cid:0) If all Xi (i the set of all functions f : I 2 X. ! Exercise 6.68* There are now two ways to interpret the expression X0 (cid:2) as x j g there is no harm in this? X1: (i) . Can you explain why Xi, and (ii) as X0 ^ X1g (x; y) i 2f 0;1 2 2 f y Q In our implementation language, product types have the form (a,b), (a,b,c), etcetera. Exercise 6.69* Let A be any set. Exhibit two different bijections between }(A) and A. 0; 1 f g Exercise 6.70* Suppose that X and Y are sets. On the set of functions Y X = there are bijections f f j i : Y is defined by: f f = g ! g Y and j : X X such that i , the relation f : X (cid:25) (cid:25) (cid:17) j. Y g ! ! (cid:14) (cid:14) 1. Show that (cid:25) is an equivalence. 2. Show: if f; g : X ! 3. Suppose that Y = Y are injective, then f (cid:25) 0; 1; 2; 3 g. . g 0; 1; 2 and X = f g (0; 0); (1; 0); (2; 1) f (a) Show that g (cid:25) f (b) How many equivalence classes has f representative. (0; 1); (1; 3); (2; 3) . g ? For every class, produce one (cid:25) Exercise 6.71* Suppose that X, Y and Z are sets and that h : Y F : Y X Z X by F (g) := h g. Show: Z. Define ! ! (cid:14) 1. if h is injective, then F is injective, 2. if h is surjective, then F is surjective. Exercise 6.72* Suppose that X Define F : Z Y , Y , and Z are sets and that h : X = ; Z X by F (g) := g (cid:14) h. Show: ! Y . ! 6 236 CHAPTER6. FUNCTIONS 1. if h is injective, then F is surjective, 2. if h is surjective, then F is injective. 6.8 Congruences A function f : X n plication are binary operations on N (on Z, on Q, on R, on C). X is called an n-ary operation on X. Addition and multi- ! If one wants to define new structures from old, an important method is taking quotients for equivalences that are compatible with certain operations. Definition 6.73 (Congruence) If f be an n-ary operation on A, and R an equiv- alence on A, then R is a congruence for f (or: R is compatible with f ) if for all x1; : : : ; xn; y1; : : : ; yn 2 A : x1Ry1; : : : ; xnRyn imply that f (x1; : : : ; xn)Rf (y1; : : : ; yn): If R is a congruence for f , then the operation induced by f on A=R is the operation fR : (A=R)n A=R given by ! a1j fR( j R; : : : ; R) := an j j f (a1; : : : ; an) j j R: If (A; f ) is a set with an operation f on it and R is a congruence for f , then (A=R; fR) is the quotient structure defined by R. (cid:17) n k0. Then (Proposition 5.65) there are a; b Example 6.74 Consider the modulo n relation on Z. Suppose m n m0 and Z with m0 = m + an and k n m0 + k0. This k0 = k + bn. Thus m0 + k0 = m + k + (a + b)n, i.e., m + k shows that n is a congruence for subtraction. It follows that we can define [m]n + [k]n := [m + k]n and [m]n n is a congruence for addition. Similarly, it can be shown that [k]n := [m k]n. (cid:17) (cid:17) (cid:17) (cid:17) 2 (cid:0) (cid:0) Exercise 6.75 Show that Z with n = 0. (cid:17) n on Z is a congruence for multiplication, for any n 2 6 6.8. CONGRUENCES 237 Example 6.76 Is possible to define exponentiation of classes in Zn by means of: n on Z (n = 0) a congruence for exponentiation, i.e., is it (cid:17) ([k]n)([m]n) := [km]n; for k Z; m N? 2 2 No, for consider the following example: ([2]3)([1]3) = [21]3 = [2]3. Since 1 (cid:17)3 4 we also have: ([2]3)([1]3) = ([2]3)([4]3) = [24]3 = [16]3 = [1]3 6 = [2]3. What this shows is that the definition is not independent of the class representatives. n is not a congruence for exponentiation. Therefore, (cid:17) Example 6.77 The definition of the integers from the natural numbers, in Section 7.2 below, uses the fact that the relation R on N2 given by (m1; m2)R(n1; n2) : (cid:17) m1 + n2 = m2 + n1 is an equivalence relation, that the relation Integer sumOdds’ n = sum [ 2*k - 1 | k <- [1..n] ] sumOdds :: Integer -> Integer sumOdds n = n^2 Note that the method of proof by mathematical induction may obscure the process of finding the relation in the first place. To see why the sum of the first n odd numbers equals n2, it is instructive to look at the following picture of the sum 1 + 3 + 5 + 7 + 9. 242 CHAPTER7. INDUCTIONANDRECURSION q q q q q q q q q q q q q q q q q q q q q q q q q Example 7.4 The Sum of the First n Even Numbers. What about the sum of the first n even natural numbers? Again, a picture suggests the answer. Look at the following representation of 2 + 4 + 6 + 8 + 10: q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q Again, a picture is not (quite) a proof. Here is proof by mathematical induction of the fact that n k=1 2k = n(n + 1). P Basis Putting k = 1 gives 2 = 1 2, which is correct. (cid:1) Induction step Assume n k=1 2k + 2(n + 1). Using the induction hypothesis we see that this is equal to n(n + P 1) + 2(n + 1) = n2 + 3n + 2 = (n + 1)(n + 2), and we are done. n k=1 2k = n(n + 1). Then n+1 k=1 2k = P P Notice that we left the term for k = 0 out of the sum. We might as well have n included it, for it holds that k=0 2k = n(n + 1). By the convention about empty sums, the two versions are equivalent. From what we found for the sum of the first n even natural numbers the formula for the sum of the first n positive natural numbers follows immediately: P k = n(n + 1) 2 : n k=1 X Again, we get improved computation procedures from the closed forms that we found: 7.1. MATHEMATICALINDUCTION 243 sumEvens’ :: Integer -> Integer sumEvens’ n = sum [ 2*k | k <- [1..n] ] sumEvens :: Integer -> Integer sumEvens n = n * (n+1) sumInts :: Integer -> Integer sumInts n = (n * (n+1)) ‘div‘ 2 Example 7.5 Summing Squares. Consider the problem of finding a closed formula for the sum of the first n squares 1)+n2; n > (a closed formula, as opposed to a recurrence f (0) = 0; f (n) = f (n 0). By direct trial one might find the following: (cid:0) 12 + 22 = 5 = 5 : (cid:1) 2 (cid:1) 3 6 12 + 22 + 32 = 5 + 9 = 14 = 7 : (cid:1) 3 (cid:1) 4 6 12 + 22 + 32 + 42 = 14 + 16 = 30 = 5 6 12 + 22 + 32 + 42 + 52 = 30 + 25 = 55 = 4 (cid:1) 9 : (cid:1) 5 (cid:1) 6 (cid:1) 6 11 : This suggests a general rule: 12 + : : : + n2 = n(n + 1)(2n + 1) 6 : But the trial procedure that we used to guess the rule is different from the procedure that is needed to prove it. Note that the fact that one can use mathematical induction to prove a rule gives no indication about how the rule was found in the first place. We will return to the issue of guessing closed forms for polynomial sequences in Sections 9.1 and 9.2. In Haskell, you can compute sums of squares in a naive way or in a sophisticated way, as follows: 244 CHAPTER7. INDUCTIONANDRECURSION sumSquares’ :: Integer -> Integer sumSquares’ n = sum [ k^2 | k <- [1..n] ] sumSquares :: Integer -> Integer sumSquares n = (n*(n+1)*(2*n+1)) ‘div‘ 6 Again, the insight that the two computation procedures will always give the same result can be proved by mathematical induction: Exercise 7.6 Summing Squares. Prove by mathematical induction: k2 = n(n + 1)(2n + 1) 6 : n k=1 X Example 7.7 Summing Cubes. Let us move on to the problem of summing cubes. By direct trial one finds: 13 + 23 = 9 = (1 + 2)2: 13 + 23 + 33 = 9 + 27 = 36 = (1 + 2 + 3)2: 13 + 23 + 33 + 43 = 36 + 64 = 100 = (1 + 2 + 3 + 4)2: 13 + 23 + 33 + 43 + 53 = 100 + 125 = 225 = (1 + 2 + 3 + 4 + 5)2: This suggests a general rule: 13 + (cid:1) (cid:1) (cid:1) + n3 = (1 + + n)2: (cid:1) (cid:1) (cid:1) We saw in Example 7.4 that k = n(n + 1) 2 ; n k=1 X so the general rule reduces to: 13 + (cid:1) (cid:1) (cid:1) + n3 = n(n + 1) 2 2 : (cid:19) (cid:18) So much about finding a rule for the sum of the first n cubes. In Sections 9.1 and 9.2 we will give an algorithm for generating closed forms for polynomial sequences. 7.1. MATHEMATICALINDUCTION 245 What we found is that sumCubes defines the same function as the naive procedure sumCubes’ for summing cubes: sumCubes’ :: Integer -> Integer sumCubes’ n = sum [ k^3 | k <- [1..n] ] sumCubes :: Integer -> Integer sumCubes n = (n*(n+1) ‘div‘ 2)^2 Again, the relation we found suggests a more sophisticated computation proce- dure, and proving the general relationship between the two procedures is another exercise in mathematical induction. Exercise 7.8 Summing Cubes. Prove by mathematical induction: n k=1 X k3 = n(n + 1) 2 2 : (cid:19) (cid:18) Exercise 7.9 Prove that for all n N: 32n+3 + 2n is divisible by 7. 2 8 x n 2 N : P (n) with that for an ordinary universal statement Remark. If one compares the proof strategy needed to establish a principle of the form A : P (x), where A is some domain of discourse, then the difference is that in the former case we can make use of what we know about the structure of N. In case we know nothing about A, and we have to prove P for an arbitrary element from A, we have to take our cue from P . In case we have to prove something about an arbitrary element n from N we know a lot more: we know that either n = 0 or n can be reached from 0 in a finite number of steps. The key property of N that makes mathematical induction work is the fact that the relation < on N is well- founded: any sequence m0 > m1 > m2 > terminates. This guarantees the existence of a starting point for the induction process. (cid:1) (cid:1) (cid:1) 2 8 (cid:18) For any A that is well-founded by a relation X A. If the following principle holds. Let (cid:30) 2 then X = A. In Section 11.1 we will say more about the use of well-foundedness as an induction principle. ) (cid:30) 2 2 8 a A( b 8 a(b X) a X); 246 CHAPTER7. INDUCTIONANDRECURSION 7.2 Recursion over the Natural Numbers Why does induction over the natural numbers work? Because we can think of any natural number n as the the result of starting from 0 and applying the successor op- eration +1 a finite number of times. Let us use this fact about the natural numbers to give our own Haskell implementation, as follows: data Natural = Z | S Natural deriving (Eq, Show) Here Z is our representation of 0, while S n is a representation of n + 1. The number 4 looks in our representation like S (S (S (S Z))). The symbol | is used to specify alternatives for the data construction. With deriving (Eq, Show) one declares Natural as a type in the classes Eq and Show. This ensures that objects of the type can be compared for equality and displayed on the screen without further ado. We can define the operation of addition on the natural numbers recursively in terms of the successor operation +1 and addition for smaller numbers: m + 0 := m m + (n + 1) := (m + n) + 1 This definition of the operation of addition is called recursive because the opera- tion + that is being defined is used in the defining clause, but for a smaller second argument. Recursive definitions always have a base case (in the example: the first line of the definition, where the second argument of the addition operation equals 0) and a recursive case (in the example: the second line of the definition, where the second argument of the addition operation is greater than 0, and the operation that is being defined appears in the right hand side of the definition). In proving things about recursively defined objects the idea is to use mathematical induction with the basic step justified by the base case of the recursive definition, the induction step justified by the recursive case of the recursive definition. This we will now demonstrate for properties of the recursively defined operation of addition for natural numbers. Here is the Haskell version of the definition of +, in prefix notation: 7.2. RECURSIONOVERTHENATURALNUMBERS 247 plus m Z = m plus m (S n) = S (plus m n) If you prefer infix notation, just write m ‘plus‘ n instead of plus m n. The back quotes around plus transform the two placed prefix operator into an infix operator. This gives the following equivalent version of the definition: m ‘plus‘ Z = m m ‘plus‘ (S n) = S (m ‘plus‘ n) Now, with diligence, we can prove the following list of fundamental laws of addi- tion from the definition. m + 0 m + n = = m + (n + k) = (m + n) + k m n + m (0 is identity element for +) (commutativity of +) (associativity of +) The first fact follows immediately from the definition of +. In proving things about a recursively defined operator to refer to clauses in the recursive definition, as follows: , clause in the definition of (cid:8) :2 to the second clause, and so on. (cid:8) (cid:8) (cid:8) it is convenient to be able :1 refers to the first Here is a proof by mathematical induction of the associativity of +: Proposition 7.10 m; n; k 8 2 N : (m + n) + k = m + (n + k): Proof. Induction on k. Basis (m + n) + 0 +:1= m + n +:1= m + (n + 0). Induction step Assume (m+n)+k = m+(n+k). We show (m+n)+(k+1) = m + (n + (k + 1)): (m + n) + (k + 1) +:2= ((m + n) + k) + 1 i.h. = (m + (n + k)) + 1 +:2 = m + ((n + k) + 1) +:2= m + (n + (k + 1)): 248 CHAPTER7. INDUCTIONANDRECURSION The inductive proof of commutativity of + uses the associativity of + that we just established. Proposition 7.11 m; n 8 2 N : m + n = n + m. Proof. Induction on n. Basis Induction on m. Basis 0 + 0 = 0 + 0. Induction Step Assume m+0 = 0+m. We show (m+1)+0 = 0+(m+1): (m + 1) + 0 +1 = +1= ih = prop 7:10 = m + 1 (m + 0) + 1 (0 + m) + 1 0 + (m + 1): Induction step Assume m + n = n + m. We show m + (n + 1) = (n + 1) + m: m + (n + 1) +:2= ih = +:2= ih = prop 7:10 = (m + n) + 1 (n + m) + 1 n + (m + 1) n + (1 + m) (n + 1) + m: Once we have addition, we can define multiplication in terms of it, again following a recursive definition: m 0 := 0 (cid:1) (n + 1) m (cid:1) := (m n) + m (cid:1) the multiplication operator. It is common to use mn as shorthand for n, or, in other words, one usually does not write the multiplication operator. We call m (cid:1) (cid:1) 7.2. RECURSIONOVERTHENATURALNUMBERS 249 Here is a Haskell implementation for our chosen representation (this time we give just the infix version): m ‘mult‘ Z = Z m ‘mult‘ (S n) = (m ‘mult‘ n) ‘plus‘ m Let us try this out: IAR> (S (S Z)) ‘mult‘ (S (S (S Z))) S (S (S (S (S (S Z))))) The following laws hold for , and for the interaction of (cid:1) and +: (cid:1) 1 = m (n + k) = m k) = (n = m (cid:1) m (cid:1) m (cid:1) n m n + m n) (cid:1) m (cid:1) (m (cid:1) n k (cid:1) k (cid:1) (1 is identity element for over +) (distribution of (cid:1) (associativity of ) (cid:1) (commutativity of ) (cid:1) ) (cid:1) (cid:1) (cid:1) Exercise 7.12 Prove these laws from the recursive definitions of + and laws that were established about +. , plus the (cid:1) If we now wish to implement an operation expn for exponentiation on naturals, the only thing we have to do is find a recursive definition of exponentiation, and implement that. Here is the definition: m0 mn+1 := 1 := (mn) m (cid:1) This leads immediately to the following implementation: expn m Z = (S Z) expn m (S n) = (expn m n) ‘mult‘ m This gives: IAR> expn (S (S Z)) S (S (S (S (S (S (S (S Z))))))) (S (S (S Z))) 250 CHAPTER7. INDUCTIONANDRECURSION Exercise 7.13 Prove by mathematical induction that km+n = km kn. (cid:1) We can define the relation 6 on N as follows: m 6 n : (cid:17) there is a k 2 N : m + k = n Instead of m 6 n we also write n > m, with the same meaning. We use m < n for m 6 n and m = n. Instead of m < n we also write n > m. This shows that we can define < or 6 in terms of addition. On further reflection, successor (+1) is enough to define 6, witness the following recursive definition: 0 6 m; m + 1 6 n + 1 if m 6 n: This is translated into Haskell as follows: leq Z _ = True leq (S _) Z = False leq (S m) (S n) = leq m n Note the use of _ for an anonymous variable: leq Z _ = True means that leq, applied to the zero element Z and to anything else, gives the value True. The expression (S _) specifies a pattern: it matches any natural that is a successor natural. To fully grasp the Haskell definition of leq one should recall that the three equa- tions that make up the definition are read as a list of three mutually exclusive cases. The Haskell system tries the equations one by one, from top to bottom, un- til it finds one that applies. Thus, leq Z _ applies to any pair of naturals with the first member equal to Z. leq (S _) Z applies to any pair of naturals with the first one starting with an S, and the second one equal to Z. The third equation, finally, applies to those pairs of naturals where both the first and the second member of the pair start with an S We can now define geq, gt, lt, in terms of leq and negation: 6 7.3. THENATUREOFRECURSIVEDEFINITIONS 251 geq m n = leq n m gt m n lt m n = not (leq m n) = gt n m Exercise 7.14 Implement an operation for cut-off subtraction subtr on naturals: the call subtr (S (S (S Z))) (S (S (S (S Z)))) should yield Z. Exercise 7.15 Implement operations quotient and remainder on naturals. Di- viding a by b yields quotient q and remainder r with 0 6 r < b, according to the b + r. (Hint: you will need the procedure subtr from Exercise formula a = q 7.14.) (cid:1) 7.3 The Nature of Recursive Definitions Not any set of equations over natural numbers can serve as a definition of an oper- ation on natural numbers. Consider f (0) := 1 f (n + 1) := f (n + 2): This does not define unique values for f (1), f (2), f (3), . . . , for the equations only require that all these values should be the same, not what the value should be. The following format does guarantee a proper definition: f (0) f (n + 1) := c := h(f (n)): Here c is a description of a value (say of type A), and h is a function of type A. A definition in this format is said to be a definition by structural recursion A over the natural numbers. The function f defined by this will be of type N A. ! ! This format is a particular instance of a slightly more general one called primitive recursion over the natural numbers. Primitive recursion allows c to depend on a number of parameters, so the function f will also depend on these parameters. But we will leave those parameters aside for now. Definition by structural recursion of f from c and h works like this: take a natural number n, view it as 1 + 0; + 1+ (cid:1) (cid:1) (cid:1) n times | {z } 252 CHAPTER7. INDUCTIONANDRECURSION replace 0 by c, replace each successor step 1+ by h, and evaluate the result: h( (cid:1) (cid:1) (cid:1) n times (h (c)) ): (cid:1)(cid:1) This general procedure is easily implemented in an operation foldn, defined as } follows: {z | foldn :: (a -> a) -> a -> Natural -> a foldn h c Z = c foldn h c (S n) = h (foldn h c n) Here is a first example application; please make sure you understand how and why this works. exclaim :: Natural -> String exclaim = foldn (’!’:) [] Now a function ‘adding m’ can be defined by taking m for c, and successor for h. For this we should be able to refer to the successor function as an object in its own right. Well, it is easily defined by lambda abstraction as (\ n -> S n). This is used in our alternative definition of plus, in terms of foldn. Note that there is no need to mention the two arguments of plus in the definition. plus :: Natural -> Natural -> Natural plus = foldn (\ n -> S n) Similarly, we can define an alternative for mult in terms of foldn. The recipe for the definition of mult m (‘multiply by m’) is to take Z for c and plus m for h: mult :: Natural -> Natural -> Natural mult m = foldn (plus m) Z 7.3. THENATUREOFRECURSIVEDEFINITIONS 253 Finally, for exponentiation expn m (‘raise m to power . . . ’) we take (S Z) for c and mult m for h: expn :: Natural -> Natural -> Natural expn m = foldn (mult m) (S Z) Exercise 7.16 Implement the operation of cut-off subtraction ( subtr m for ‘sub- tract from m’; Exercise (7.14)) in terms of foldn and a function for predecessor, on the basis of the following definition: 0 := x x _ (cid:0) (n + 1) x _ (cid:0) := p(x _ (cid:0) n); where p is the function for predecessor given by p(0) := 0; p(n + 1) := n. Call the predecessor function pre. Exercise 7.17 The Fibonacci numbers are given by the following recursion: F0 = 0; F1 = 1; Fn+2 = Fn+1 + Fn for n > 0: This gives: 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; 233; 377; 610; 987; 1597; 2584; 4181; : : : Prove with induction that for all n > 1: Fn+1Fn 1 (cid:0) (cid:0) F 2 n = ( 1)n: (cid:0) Exercise 7.18 A bitlist is a list of zeros and ones. Consider the following code bittest for selecting the bitlists without consecutive zeros. bittest :: [Int] -> Bool = True bittest [] = True bittest [0] = bittest xs bittest (1:xs) bittest (0:1:xs) = bittest xs bittest _ = False 254 CHAPTER7. INDUCTIONANDRECURSION 1. How many bitlists of length 0 satisfy bittest? How many bitlists of length 1 satisfy bittest? How many bitlists of length 2 satisfy bittest? How many bitlists of length 3 satisfy bittest? 2. Let an be the number of bitlists of length n without consecutive zeros. Give an induction proof to show that for every n > 0 it holds that an = Fn+2, where Fn is the n-th Fibonacci number. Take care: you will need two base cases (n = 0; n = 1), and an induction hypothesis of the form: “assume that the formula holds for n and for n + 1.” A further hint: the code for bittest points the way to the solution. Exercise 7.19* Consider the following two definitions of the Fibonacci numbers (Exercise 7.17): fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2) fib’ n = fib2 0 1 n where fib2 a b 0 = a fib2 a b n = fib2 b (a+b) (n-1) Use an induction argument to establish that fib and fib’ define the same function. Hint: establish the more general claim that for all i,n it holds that fib2 (fib i) (fib (i+1)) n = fib (i+n) by induction on n. Exercise 7.20 The Catalan numbers are given by the following recursion: C0 = 1; Cn+1 = C0Cn + C1Cn 1 + (cid:0) (cid:1) (cid:1) (cid:1) + Cn 1C1 + CnC0: (cid:0) This gives: [1; 1; 2; 5; 14; 42; 132; 429; 1430; 4862; 16796; 58786; 208012; 742900; 2674440; : : : Use this recursion to give a Haskell implementation. Can you see why this is not a very efficient way to compute the Catalan numbers? 7.4. INDUCTIONANDRECURSIONOVERTREES 255 Exercise 7.21 Let x0; : : : ; xn be a sequence of n + 1 variables. Suppose their product is to be computed by doing n multiplications. The number of ways to do the multiplications corresponds to the number of bracketings for the sequence. For instance, if n = 3 there are four variables x0; x1; x2; x3, and five possible bracketings: (x1x2)(x3x4); ((x1x2)x3)x4; (x1(x2x3))x4; x1((x2x3)x4); x1(x2(x3x4)): Show that the number of bracketings for n + 1 variables is given by the Cata- lan number Cn. (Hint: you will need strong induction, so-called because of its strengthened induction hypothesis. Your induction hypothesis should run: “For any i with 0 6 i 6 n, for any sequence of i + 1 variables x0 (cid:1) (cid:1) (cid:1) xi it holds that Ci gives the number of bracketings for that sequence.” Example 7.22 Balanced sequences of parentheses of length 2n are defined recur- sively as follows: the empty sequence is balanced; if sequence w is balanced then (w) is balanced; if sequences w and v are balanced then wv is balanced. Thus, ())(() is not balanced. The balanced sequences involving 3 left and 3 right paren- theses are: ()()(); (())(); ()(()); (()()); ((())): There is a one-to-one mapping between bracketings for sequences of n + 1 vari- ables and balanced sequences of parentheses with 2n parentheses. Let a bracketing xn be given. This can be changed into a balanced sequence of parenthe- for x0 (cid:1) (cid:1) (cid:1) ses as follows: We illustrate with the example x0((x2x3)x4). 1. Put one extra pair of parentheses around the bracketing: (x0((x2x3)x4)). 2. Insert multiplication operators at the appropriate places: (x0 (cid:1) 3. Erase the variables and the left-brackets: )). ) (cid:1) (cid:1)(cid:1) ((x2 (cid:1) x3) (cid:1) x4)). 4. Replace the ’s with left-brackets: (()()). (cid:1) This mapping gives a one-to-one correspondence between variable bracketings and balanced parentheses strings, so we get from the previous exercise that there are Cn different balanced parentheses strings of length 2n. 7.4 Induction and Recursion over Trees Here is a recursive definition of binary trees: 256 CHAPTER7. INDUCTIONANDRECURSION A single leaf node is a binary tree. (cid:15) If t1 and t2 are binary trees, then the result of joining t1 and t2 under a single node (called the root node) is a binary tree. A notation for this is: ( t1 t2) (cid:15) Nothing else is a binary tree. (cid:15) (cid:15) (cid:15) The depth of a binary tree is given by: is 0. The depth of (cid:15) The depth of ( (cid:15) (cid:15) (cid:15) t1 t2) is 1+ the maximum of the depths of t1 and t2. t1 t2), with both t1 and t2 balanced, and having the same depth. A binary tree is balanced if it either is a single leaf node ( (cid:15) We see the following: A balanced binary tree of depth 0 is just a single leaf node , so its number of nodes is 1. A balanced binary tree of depth 1 has one internal (cid:15) node and two leaves, so it has 3 nodes. A balanced binary tree of depth 2 has 3 internal nodes and 4 leaf nodes, so it has 7 nodes. A binary tree of depth 3 has 7 internal nodes plus 8 leaf nodes, so it has 15 nodes. , or it has the form (cid:15) Recursion and induction over binary trees are based on two cases t = ( (cid:15) t1 t2). (cid:15) and t = Example 7.23 Suppose we want to find a formula for the number of nodes in a balanced binary tree of depth d. q q q q q q q q q q q q q q q qq q qq q qq q qq Suppose we want to show in general that the number of nodes of a binary tree of depth n is 2n+1 1. Then a proof by mathematical induction is in order. We (cid:0) proceed as follows: Basis If n = 0, then 2n+1 1 = 21 of a binary tree of depth 0. (cid:0) 1 = 1. This is indeed the number of nodes (cid:0) Induction step Assume the number of nodes of a binary tree of depth n is 2n+1 (cid:0) 1. We have to show that the number of nodes of a binary tree of depth n + 1 equals 2n+2 1. (cid:0) 7.4. INDUCTIONANDRECURSIONOVERTREES 257 A binary tree of depth n + 1 can be viewed as a set of internal nodes con- stituting a binary tree of depth n, plus a set of leaf nodes, consisting of two new leaf nodes for every old leaf node from the tree of depth n. By induc- tion hypothesis, we know that a tree of depth n has 2n+1 1 nodes, so a tree of depth n + 1 has 2n+1 1 internal nodes. It is easy to see that a tree of depth n + 1 has 2n+1 leaf nodes. The total number of nodes of a tree of depth n + 2 is therefore 2n+1 1, and we have proved our induction step. 1 + 2n+1 = 2 1 = 2n+2 2n+1 (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:1) To illustrate trees and tree handling a bit further, here is a Haskell definition of binary trees, with a procedure for making balanced trees of an arbitrary depth n, and a procedure for counting their nodes in a naive way. We use L for a single leaf . The data declaration specifies that a BinTree either (cid:15) is an object L (a single leaf), or an object constructed by applying the constructor N to two BinTree objects (the result of constructing a new binary tree ( t1 t2) (cid:15) from two binary trees t1 and t2). The addition deriving Show ensures that data of this type can be displayed on the screen. data BinTree = L | N BinTree BinTree deriving Show makeBinTree :: Integer -> BinTree makeBinTree 0 = L makeBinTree (n + 1) = N (makeBinTree n) (makeBinTree n) count :: BinTree -> Integer count L = 1 count (N t1 t2) = 1 + count t1 + count t2 With this code you can produce binary trees as follows: IAR> makeBinTree 6 N (N (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N L L)) (N (N L L) (N L L)))) (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N L L)) (N (N L L) (N L L))))) (N (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N L L)) (N (N L L) (N L L)))) (N (N (N (N L L) (N L L)) (N (N L L) (N L L))) (N (N (N L L) (N 258 CHAPTER7. INDUCTIONANDRECURSION L L)) (N (N L L) (N L L))))) IAR> If you want to check that the depth of a the result of maketree 6 is indeed 6, or that maketree 6 is indeed balanced, here are some procedures. Note that the procedures follow the definitions of depth and balanced to the letter: depth :: BinTree -> Integer depth L = 0 depth (N t1 t2) = (max (depth t1) (depth t2)) + 1 balanced :: BinTree -> Bool balanced L = True balanced (N t1 t2) = (balanced t1) && (balanced t2) && depth t1 == depth t2 The programs allow us to check the relation between count (makeBinTree n) and 2^(n+1) - 1 for individual values of n: count (makeBinTree 6) == 2^7 - 1 IAR> True What the proof by mathematical induction provides is an insight that the relation holds in general. Mathematical induction does not give as clue as to how to find a formula for the number of nodes in a tree. It only serves as a method of proof once such a formula is found. So how does one find a formula for the number of nodes in a binary tree in the first place? By noticing how such trees grow. A binary tree of depth 0 has 1 node, and this node is a leaf node. This leaf grows two new nodes, so a binary tree of depth 1 has 1 + 2 = 3 nodes. In the next step the 2 leaf nodes grow two new nodes each, so we get 22 = 4 new leaf nodes, and the number of nodes of a binary tree of depth 2 equals 1 + 2 + 4 = 7. In general, a tree of depth n 1 is transformed into one of depth n by growing 2n new leaves, and the total number of leaves of + 2n. In other words, the number of nodes the new tree is given by 20 + 21 + (cid:0) (cid:1) (cid:1) (cid:1) 7.4. INDUCTIONANDRECURSIONOVERTREES 259 of a balanced binary tree of depth n is given by : here is a simple trick: n k=0 2k. To get a value for this, P 2k = 2 n k=0 X n n 2k (cid:0) (cid:1) k=0 X k=0 X 2k = (2 (cid:1) 2n + 2n + 2) (cid:0) (cid:1) (cid:1) (cid:1) (2n + (cid:1) (cid:1) (cid:1) + 1) = = 2 2n (cid:1) (cid:0) 1 = 2n+1 1: (cid:0) Example 7.24 Counting the Nodes of a Balanced Ternary Tree. Now suppose we want to find a formula for the number of nodes in a balanced ternary tree of depth n. The number of leaves of a balanced ternary tree of depth n is 3n, so the total number of nodes of a balanced ternary tree of depth n is given by n k=0 3k. We prove by induction that P n k=0 3k = 3 n+1 2 1 (cid:0) . Basis A ternary tree of depth n = 0 consists of just a single node. And indeed, P 3k = 1 = 31 1 : (cid:0) 2 0 k=0 X 3 n+1 Induction step Assume that the number of nodes of a ternary tree of depth n is . The number of leaf nodes is 3n, and each of these leaves grows 3 new leaves to produce the ternary tree of depth n + 1. Thus, the number of leaves of the tree of depth n + 1 is given by (cid:0) 2 1 n+1 n 3k = 3k + 3n+1: k=0 X k=0 X Using the induction hypothesis, we see that this is equal to 3n+1 2 1 (cid:0) + 3n+1 = 3n+1 2 1 2 (cid:1) + (cid:0) 3n+1 2 = 3n+2 2 1 : (cid:0) But how did we get at in the case of the binary trees: P n k=0 3k = 3 n+1 2 1 (cid:0) in the first place? In the same way as 2 (cid:1) n k=0 X n n 3k = 3 3k (cid:0) (cid:1) k=0 X k=0 X = 3 3k = (3 (cid:1) 3n + 3n + 3) (cid:0) (cid:1) (cid:1) (cid:1) (3n + (cid:1) (cid:1) (cid:1) + 1) = 3n (cid:1) (cid:0) 1 = 3n+1 1: (cid:0) 260 Therefore CHAPTER7. INDUCTIONANDRECURSION 3k = 3n+1 2 1 : (cid:0) n k=0 X Exercise 7.25 Write a Haskell definition of ternary trees, plus procedures for gen- erating balanced ternary trees and counting their node numbers. Example 7.26 Counting the Nodes of a Balanced m-ary Tree. The number of n k=0 mk = nodes of a balanced m-ary tree of depth n (with m > 1) is given by mn+1 m 1 . Here is a proof by mathematical induction. (cid:0) (cid:0) P 1 Basis An m-ary tree of depth 0 consists of just a single node. In fact, 1 = m m 1 1 . (cid:0) (cid:0) 0 k=0 mk = P mn+1 (cid:0) m 1 Induction step Assume that the number of nodes of an m-ary tree of depth n is . The number of leaf nodes is mn, and each of these leaves grows m new leaves to produce the ternary tree of depth n + 1. Thus, the number of leaves of the tree of depth n + 1 is given by (cid:0) 1 n+1 n mk = mk + mn+1: Using the induction hypothesis, we see that this is equal to k=0 X k=0 X mn+1 m (cid:0) 1 (cid:0) 1 + mn+1 = mn+1 m (cid:0) (cid:0) 1 1 + mn+2 m mn+1 1 = (cid:0) (cid:0) mn+2 m (cid:0) 1 : (cid:0) 1 Note that the proofs by mathematical induction do not tell you how to find the for- mulas for which the induction proof works in the first place. This is an illustration of the fact that mathematical induction is a method of verification, not a method of invention. Indeed, mathematical induction is no replacement for the use of creative intuition in the process of finding meaningful relationships in mathematics. Exercise 7.27 Geometric Progressions. Prove by mathematical induction (assuming q = 1; q = 0): n qk = qn+1 q (cid:0) 1 1 : k=0 X Note that this exercise is a further generalization of Example 7.26. (cid:0) 6 6 7.4. INDUCTIONANDRECURSIONOVERTREES 261 To get some further experience with tree processing, consider the following defi- nition of binary trees with integer numbers at the internal nodes: data Tree = Lf | Nd Int Tree Tree deriving Show We say that such a tree is ordered if it holds for each node N of the tree that the integer numbers in the left subtree below N are all smaller than the number at node N , and the number in the right subtree below N are all greater than the number at N . Exercise 7.28 Write a function that inserts a number n in an ordered tree in such a way that the tree remains ordered. Exercise 7.29 Write a function list2tree that converts a list of integers to an ordered tree, with the integers at the tree nodes. The type is [Int] -> Tree. Also, write a function tree2list for conversion in the other direction. Exercise 7.30 Write a function that checks whether a given integer i occurs in an ordered tree. Exercise 7.31 Write a function that merges two ordered trees into a new ordered tree containing all the numbers of the input trees. Exercise 7.32 Write a function that counts the number of steps that are needed to reach a number i in an ordered tree. The function should give 0 if i is at the top node, and 1 if i does not occur in the tree at all. (cid:0) A general data type for binary trees with information at the internal nodes is given by: data Tr a = Nil | T a (Tr a) (Tr a) deriving (Eq,Show) Exercise 7.33 Write a function mapT :: (a -> b) -> Tr a -> Tr b that does for binary trees what map does for lists. 262 CHAPTER7. INDUCTIONANDRECURSION Exercise 7.34 Write a function foldT :: (a -> b -> b -> b) -> b -> (Tr a) -> b that does for binary trees what foldn does for naturals. Exercise 7.35 Conversion of a tree into a list can be done in various ways, de- pending on when the node is visited: Preorder traversal of a tree is the result of first visiting the node, next visiting the left subtree, and finally visiting the right subtree. Inorder traversal of a tree is the result of first visiting the left subtree, next visit- ing the node, and finally visiting the right subtree. Postorder traversal of a tree is the result of first visiting the left subtree, next visiting the right subtree, and finally visiting the node. Define these three conversion functions from trees to lists in terms of the foldT function from Exercise 7.34. Exercise 7.36 An ordered tree is a tree with information at the nodes given in such manner that the item at a node must be bigger than all items in the left sub- tree and smaller than all items in the right subtree. A tree is ordered iff the list resulting from its inorder traversal is ordered and contains no duplicates. Give an implementation of this check. Exercise 7.37 An ordered tree (Exercise 7.36) can be used as a dictionary by putting items of of type (String,String) at the internal nodes, and defining the ordering as: (v; w) 6 (v0; w0) iff v 6 v0. Dictionaries get the following type: type Dict = Tr (String,String) Give code for looking up a word definition in a dictionary. The type declaration is: lookupD :: String -> Dict -> [String] 7.4. INDUCTIONANDRECURSIONOVERTREES 263 If (v; w) occurs in the dictionary, the call lookupD v d should yield [w], other- wise []. Use the order on the dictionary tree. Exercise 7.38 For efficient search in an ordered tree (Exercises 7.36 and 7.37) it is crucial that the tree is balanced: the left and right subtree should have (nearly) the same depth and should themselves be balanced. The following auxiliary function splits non-empty lists into parts of (roughly) equal lengths. split :: [a] -> ([a],a,[a]) split xs = (ys1,y,ys2) where ys1 = take n xs (y:ys2) = drop n xs n = length xs ‘div‘ 2 Use this to implement a function buildTree :: [a] -> Tr a for transforming an ordered list into an ordered and balanced binary tree. Here is a data type LeafTree for binary leaf trees (binary trees with information at the leaf nodes): data LeafTree a = Leaf a | Node (LeafTree a) (LeafTree a) deriving Show Here is an example leaf tree: ltree :: LeafTree String ltree = Node (Leaf "I") (Node (Leaf "love") (Leaf "you")) 264 CHAPTER7. INDUCTIONANDRECURSION Exercise 7.39 Repeat Exercise 7.33 for leaf trees. Call the new map function mapLT. Exercise 7.40 Give code for mirroring a leaf tree on its vertical axis. Call the func- tion reflect. In the mirroring process, the left- and right branches are swapped, and the same swap takes place recursively within the branches. The reflection of Node (Leaf 1) (Node (Leaf 2) (Leaf 3)) is Node (Node (Leaf 3) (Leaf 2)) (Leaf 1). Exercise 7.41 Let reflect be the function from Exercise 7.40. Prove with in- duction on tree structure that reflect (reflect t) == t. holds for every leaf tree t. A data type for trees with arbitrary numbers of branches (rose trees), with infor- mation of type a at the buds, is given by: data Rose a = Bud a | Br [Rose a] deriving (Eq,Show) Here is an example rose: rose = Br [Bud 1, Br [Bud 2, Bud 3, Br [Bud 4, Bud 5, Bud 6]]] Exercise 7.42 Write a function mapR :: (a -> b) -> Rose a -> Rose b that does for rose trees what map does for lists. For the example rose, we should get: IAR> rose Br [Bud 1, Br [Bud 2, Bud 3, Br [Bud 4, Bud 5, Bud 6]]] IAR> mapR succ rose Br [Bud 2, Br [Bud 3, Bud 4, Br [Bud 5, Bud 6, Bud 7]]] 7.5. INDUCTIONANDRECURSIONOVERLISTS 265 7.5 Induction and Recursion over Lists Induction and recursion over natural numbers are based on the two cases n = 0 and n = k + 1. Induction and recursion over lists are based on the two cases l = [] and l = x:xs, where x is an item and xs is the tail of the list. An example is the definition of a function len that gives the length of a list. In fact, Haskell has a predefined function length for this purpose; our definition of len is just for purposes of illustration. len [] len (x:xs) = 1 + len xs = 0 Similarly, Haskell has a predefined operation ++ for concatenation of lists. For purposes of illustration we repeat our version from Section (4.8). cat :: [a] -> [a] -> [a] cat [] cat (x:xs) ys = x : (cat xs ys) ys = ys As an example of an inductive proof over lists, we show that concatenation of lists is associative. Proposition 7.43 For all lists xs, ys and zs over the same type a: cat (cat xs ys) zs = cat xs (cat ys zs). Proof. Induction on xs. Basis cat (cat [] ys) zs cat:1= cat ys zs cat:1= cat [] (cat ys zs): 266 CHAPTER7. INDUCTIONANDRECURSION Induction step cat x:xs (cat ys zs) cat:2= x:(cat xs (cat ys zs) i.h. = x:(cat (cat xs ys) zs) cat:2= cat x:(cat xs ys) zs cat:2= cat (cat x:xs ys) zs: Exercise 7.44 Prove by induction that cat xs [] = cat [] xs. Exercise 7.45 Prove by induction: len (cat xs ys) = (len xs) + (len ys). A general scheme for structural recursion over lists (without extra parameters) is given by: f [] f (x : xs) := z := h x (f xs) For example, the function s that computes the sum of the elements in a list of numbers is defined by: s [] s(n : xs) := 0 := n + s xs Here 0 is taken for z, and + for h. As in the case for natural numbers, it is useful to implement an operation for this general procedure. In fact, this is predefined in Haskell, as follows (from the Haskell file Prelude.hs): foldr foldr f z [] foldr f z (x:xs) :: (a -> b -> b) -> b -> [a] -> b = z = f x (foldr f z xs) 7.5. INDUCTIONANDRECURSIONOVERLISTS 267 In general, z is the identity element of the operation f, i.e., the value you would start out with in the base case of a recursive definition of the operation. The identity element for addition is 0, for multiplication it is 1 (see Section 7.3). The base clause of the definition of foldr says that if you want to fold an empty list for operation f, the result is the identity element for f. The recursive clause says that to fold a non-empty list for f, you perform f to its first element and to the result of folding the remainder of the list for f. The following informal version of the definition of foldr may further clarify its meaning: foldr ( ) z [x1; x2; : : : ; xn] := x1 (cid:8) (cid:8) (x2 (cid:8) ( (cid:1) (cid:1) (cid:1) (xn (cid:8) z) (cid:1) (cid:1) (cid:1) )): The function add :: [Natural] -> Natural can now be defined as: add = foldr plus Z The function mlt :: [Natural] -> Natural is given by: mlt = foldr mult (S Z) And here is an alternative definition of list length, with values of type Natural. Note that the function (\ _ n -> S n) ignores its first argument (you don’t have to look at the items in a list in order to count them) and returns the successor of its second argument (for this second argument represents the number of items that were counted so far). ln :: [a] -> Natural ln = foldr (\ _ n -> S n) Z It is also possible to use foldr on the standard type for integers. Computing the result of adding or multiplying all elements of a list of integers can be done as follows: 268 CHAPTER7. INDUCTIONANDRECURSION Prelude> foldr (+) 0 [1..10] 55 Prelude> foldr (*) 1 [1..10] 3628800 Exercise 7.46 Use foldr to give a new implementation of generalized union and foldr1 to give a new implementation of generalized intersection for lists. (Look up the code for foldr1 in the Haskell prelude. Compare Exercise 4.53.) Consider the following definitions of generalized conjunction and disjunction: or :: [Bool] -> Bool or [] = False or (x:xs) = x || or xs and :: [Bool] -> Bool and [] = True and (x:xs) = x && and xs The function or takes a list of truth values and returns True if at least one member of the list equals True, while and takes a list of truth values and returns True if all members of the list equal True. (We have encountered and before, in Section 2.2.) In fact, Haskell has predefined these operations, in terms of foldr. To see how we can use foldr to implement generalized conjunction and disjunction, we only need to know what the appropriate identity elements of the operations are. Should the conjunction of all elements of [] count as true or false? As true, for it is indeed (trivially) the case that all elements of [] are true. So the identity element for conjunction is True. Should the disjunction of all elements of [] count as true or false? As false, for it is false that [] contains an element which is true. Therefore, the identity element for disjunction is False. This explains the following Haskell definition in Prelude.hs: and, or and or :: [Bool] -> Bool = foldr (&&) True = foldr (||) False 7.5. INDUCTIONANDRECURSIONOVERLISTS 269 Exercise 7.47 Define a function srt that sorts a list of items in class Ord a by folding the list with a function insrt. The operation foldr folds ‘from the right’. Folding ‘from the left’ can be done with its cousin foldl, predefined in Prelude.hs as follows: foldl foldl f z [] foldl f z (x:xs) :: (a -> b -> a) -> a -> [b] -> a = z = foldl f (f z x) xs An informal version may further clarify this: foldl ( ) z [x1; x2; : : : ; xn] := ( (cid:8) ((z x1) x2) (cid:8) (cid:8) (cid:8) (cid:1) (cid:1) (cid:1) xn: ) (cid:8) (cid:1) (cid:1) (cid:1) This can be used to flesh out the following recursion scheme: f z [] f z (x : xs) := z := f (h z x) xs This boils down to recursion over lists with an extra parameter, and in fact foldl can be used to speed up list processing. The case of reversing a list is an example: r zs [] r zs (x : xs) := zs := r (prefix zs x) xs; where prefix is given by prefix ys y = y:ys. Here is a definition in terms of foldl: rev = foldl (\ xs x -> x:xs) [] The list [1,2,3] is reversed as follows: rev [1,2,3] = foldl (\ xs x -> x:xs) [] [1,2,3] = foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [] 1) [2,3] = foldl (\ xs x -> x:xs) [1] [2,3] = foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [1] 2) [3] 270 CHAPTER7. INDUCTIONANDRECURSION = foldl (\ xs x -> x:xs) [2,1] [3] = foldl (\ xs x -> x:xs) ((\ xs x -> x:xs) [2,1] 3) [] = foldl (\ xs x -> x:xs) [3,2,1] [] = [3,2,1] Note that (\ xs x -> x:xs) has type [a] -> a -> [a]. An alternative defi- nition of rev, in terms of foldr, would need a swap function \ x xs -> xs ++ [x] :: a -> [a] -> [a] and would be much less efficient: rev’ = foldr (\ x xs -> xs ++ [x]) [] The inefficiency resides in the fact that ++ itself is defined recursively, as follows: [] (x:xs) ++ ys = x : (xs ++ ys) ++ ys = ys To see why rev’ is less efficient than rev, look at the following, where we write postfix for (\ x xs -> xs ++[x]). rev’ [1,2,3] = foldr postfix [] [1,2,3] = postfix 1 (foldr postfix [] [2,3]) = (foldr postfix [] [2,3]) ++ [1] = (postfix 2 (foldr postfix [] [3])) ++ [1] = (foldr postfix [] [3]) ++ [2] ++ [1] = (postfix 3 (foldr postfix [] [])) ++ [2] ++ [1] = (foldr postfix [] []) ++ [3] ++ [2] ++ [1] = [] ++ [3] ++ [2] ++ [1] = ([3] ++ [2]) ++ [1] = 3:([] ++ [2]) ++ [1] = [3,2] ++ [1] = 3:([2] ++ [1]) = 3:2:([] ++ [1]) = [3,2,1] 7.5. INDUCTIONANDRECURSIONOVERLISTS 271 If we compare the two list recursion schemes that are covered by foldr and foldl, then we see that the two folding operations h and h0 are close cousins: f [] f (x : xs) := z := h x (f xs) f 0 [] f 0(x : xs) := z := h0 (f 0 xs) x An obvious question to ask now is the following: what requirements should h and h0 satisfy in order to guarantee that f and f 0 define the same function? Exercise 7.48 and Example 7.49 invite you to further investigate this issue of the relation between the schemes. Exercise 7.48 Let h :: a -> b -> b and h’ :: b -> a -> b, and z :: b. Assume that for all x :: a it holds that h x (h’ y z) = h’ z (h x y). Show that for every x,y :: a and every finite xs :: [a] the following holds: h x (foldl h’ y xs) = foldl h’ (h x y) xs Use induction on xs. Example 7.49 Let h :: a -> b -> b and h’ :: b -> a -> b, and z :: b. Assume we have for all x,y :: a and all xs :: [a] the following: h x z = h’ z x h x (h’ y z) = h’ z (h x y) We show that we have for all finite xs :: [a] that foldr h z xs = foldl h’ z xs We use induction on the structure of xs. Basis Immediate from the definitions of foldr and foldl we have that: foldr h z [] = z = foldl h’ z [] Induction step Assume the induction hypothesis foldr h z xs = foldl h’ z xs. We have to show that foldr h z x:xs = foldl h’ z x:xs. Here is the reasoning: foldr h z x:xs foldr = h x (foldr h z xs) IH = h x (foldl h z xs) 7:48= foldl h0 (h x z) xs given = foldl h0 (h0 z x) xs foldl = foldl h0 z x:xs 272 CHAPTER7. INDUCTIONANDRECURSION For an application of Example (7.49), note that the functions postfix and prefix are related by: postfix x [] = [] ++ [x] = [x] == prefix [] x postfix x (prefix xs y) = (prefix xs y) ++ [x] = y:(xs ++ [x]) = y:(postfix x xs) = prefix (postfix x xs) y It follows from this and the result of the example that rev and rev’ indeed com- pute the same function. Exercise 7.50 Consider the following version of rev. = rev2 xs [] rev1 :: [a] -> [a] rev1 xs where rev2 [] ys = ys rev2 (x:xs) ys = rev2 xs (x:ys) Which version is more efficient, the original rev, the version rev’, or this version? Why? Exercise 7.51 Define an alternative version ln’ of ln using foldl instead of foldr. In Section 1.8 you got acquainted with the map and filter functions. The two operations map and filter can be combined: Prelude> filter (>4) (map (+1) [1..10]) [5,6,7,8,9,10,11] Prelude> map (+1) (filter (>4) [6,7,8,9,10,11] [1..10]) These outcomes are different. This is because the test (>4) yields a different result after all numbers are increased by 1. When we make sure that the test used in the filter takes this change into account, we get the same answers: 7.6. SOMEVARIATIONSONTHETOWEROFHANOI 273 Prelude> filter (>4) (map (+1) [1..10]) [5,6,7,8,9,10,11] Prelude> map (+1) (filter ((>4).(+1)) [5,6,7,8,9,10,11] [1..10]) Here (f . g) denotes the result of first applying g and next f. Note: ((>4).(+1)) defines the same property as (>3). Exercise 7.52 gives you an opportunity to show that these identical answers are no accident. Exercise 7.52 Let xs :: [a], let f :: a -> b, and let p :: b -> Bool be a total predicate. Show that the following holds: filter p (map f xs) = map f (filter (p f ) xs): (cid:1) Note: a predicate p :: b -> Bool is total if for every object x :: b, the appli- cation p x gives either true or false. In particular, for no x :: b does p x raise an error. 7.6 Some Variations on the Tower of Hanoi Figure 7.1: The Tower of Hanoi. The Tower of Hanoi is a tower of 8 disks of different sizes, stacked in order of decreasing size on a peg. Next to the tower, there are two more pegs. The task is to transfer the whole stack of disks to one of the other pegs (using the third peg as an auxiliary) while keeping to the following rules: (i) move only one disk at a time, (ii) never place a larger disk on top of a smaller one. Exercise 7.53 In this exercise, you are required to invent your own solution, and next prove it by mathematical induction. Make sure that the reasoning also es- tablishes that the formula you find for the number of moves is the best one can do. 274 CHAPTER7. INDUCTIONANDRECURSION 1. How many moves does it take to completely transfer a tower consisting of n disks? 2. Prove by mathematical induction that your answer to the previous question is correct. 3. How many moves does it take to completely transfer the tower of Hanoi? Exercise 7.54 Can you also find a formula for the number of moves of the disk of size k during the transfer of a tower with disks of sizes 1; : : : ; n, and 1 6 k 6 n? Again, you should prove by mathematical induction that your formula is correct. According to legend, there exists a much larger tower than the tower of Hanoi, the tower of Brahma, with 64 disks. Monks are engaged in transferring the disks of the Tower of Brahma, from the beginning of the universe to the present day. As soon as they will have completed their task the tower will crumble and the world will end. Exercise 7.55 How long will the universe last, given that the monks move one disk per day? For an implementation of the disk transfer procedure, an obvious way to represent the starting configuration of the tower of Hanoi is ([1,2,3,4,5,6,7,8],[],[]). For clarity, we give the three pegs names A, B and C. and we declare a type Tower: data Peg = A | B | C type Tower = ([Int], [Int], [Int]) There are six possible single moves from one peg to another: 7.6. SOMEVARIATIONSONTHETOWEROFHANOI 275 IAR> (display 88 . show . take 200 . hanoi) 8 [([1,2,3,4,5,6,7,8],[],[]),([2,3,4,5,6,7,8],[1],[]),([3,4,5,6,7,8],[1],[2]),([3,4,5,6,7, 8],[],[1,2]),([4,5,6,7,8],[3],[1,2]),([1,4,5,6,7,8],[3],[2]),([1,4,5,6,7,8],[2,3],[]),([ 4,5,6,7,8],[1,2,3],[]),([5,6,7,8],[1,2,3],[4]),([5,6,7,8],[2,3],[1,4]),([2,5,6,7,8],[3], [1,4]),([1,2,5,6,7,8],[3],[4]),([1,2,5,6,7,8],[],[3,4]),([2,5,6,7,8],[1],[3,4]),([5,6,7, 8],[1],[2,3,4]),([5,6,7,8],[],[1,2,3,4]),([6,7,8],[5],[1,2,3,4]),([1,6,7,8],[5],[2,3,4]) ,([1,6,7,8],[2,5],[3,4]),([6,7,8],[1,2,5],[3,4]),([3,6,7,8],[1,2,5],[4]),([3,6,7,8],[2,5 ],[1,4]),([2,3,6,7,8],[5],[1,4]),([1,2,3,6,7,8],[5],[4]),([1,2,3,6,7,8],[4,5],[]),([2,3, 6,7,8],[1,4,5],[]),([3,6,7,8],[1,4,5],[2]),([3,6,7,8],[4,5],[1,2]),([6,7,8],[3,4,5],[1,2 ]),([1,6,7,8],[3,4,5],[2]),([1,6,7,8],[2,3,4,5],[]),([6,7,8],[1,2,3,4,5],[]),([7,8],[1,2 ,3,4,5],[6]),([7,8],[2,3,4,5],[1,6]),([2,7,8],[3,4,5],[1,6]),([1,2,7,8],[3,4,5],[6]),([1 ,2,7,8],[4,5],[3,6]),([2,7,8],[1,4,5],[3,6]),([7,8],[1,4,5],[2,3,6]),([7,8],[4,5],[1,2,3 ,6]),([4,7,8],[5],[1,2,3,6]),([1,4,7,8],[5],[2,3,6]),([1,4,7,8],[2,5],[3,6]),([4,7,8],[1 ,2,5],[3,6]),([3,4,7,8],[1,2,5],[6]),([3,4,7,8],[2,5],[1,6]),([2,3,4,7,8],[5],[1,6]),([1 ,2,3,4,7,8],[5],[6]),([1,2,3,4,7,8],[],[5,6]),([2,3,4,7,8],[1],[5,6]),([3,4,7,8],[1],[2, 5,6]),([3,4,7,8],[],[1,2,5,6]),([4,7,8],[3],[1,2,5,6]),([1,4,7,8],[3],[2,5,6]),([1,4,7,8 ],[2,3],[5,6]),([4,7,8],[1,2,3],[5,6]),([7,8],[1,2,3],[4,5,6]),([7,8],[2,3],[1,4,5,6]),( [2,7,8],[3],[1,4,5,6]),([1,2,7,8],[3],[4,5,6]),([1,2,7,8],[],[3,4,5,6]),([2,7,8],[1],[3, 4,5,6]),([7,8],[1],[2,3,4,5,6]),([7,8],[],[1,2,3,4,5,6]),([8],[7],[1,2,3,4,5,6]),([1,8], [7],[2,3,4,5,6]),([1,8],[2,7],[3,4,5,6]),([8],[1,2,7],[3,4,5,6]),([3,8],[1,2,7],[4,5,6]) ,([3,8],[2,7],[1,4,5,6]),([2,3,8],[7],[1,4,5,6]),([1,2,3,8],[7],[4,5,6]),([1,2,3,8],[4,7 ],[5,6]),([2,3,8],[1,4,7],[5,6]),([3,8],[1,4,7],[2,5,6]),([3,8],[4,7],[1,2,5,6]),([8],[3 ,4,7],[1,2,5,6]),([1,8],[3,4,7],[2,5,6]),([1,8],[2,3,4,7],[5,6]),([8],[1,2,3,4,7],[5,6]) ,([5,8],[1,2,3,4,7],[6]),([5,8],[2,3,4,7],[1,6]),([2,5,8],[3,4,7],[1,6]),([1,2,5,8],[3,4 ,7],[6]),([1,2,5,8],[4,7],[3,6]),([2,5,8],[1,4,7],[3,6]),([5,8],[1,4,7],[2,3,6]),([5,8], [4,7],[1,2,3,6]),([4,5,8],[7],[1,2,3,6]),([1,4,5,8],[7],[2,3,6]),([1,4,5,8],[2,7],[3,6]) ,([4,5,8],[1,2,7],[3,6]),([3,4,5,8],[1,2,7],[6]),([3,4,5,8],[2,7],[1,6]),([2,3,4,5,8],[7 ],[1,6]),([1,2,3,4,5,8],[7],[6]),([1,2,3,4,5,8],[6,7],[]),([2,3,4,5,8],[1,6,7],[]),([3,4 ,5,8],[1,6,7],[2]),([3,4,5,8],[6,7],[1,2]),([4,5,8],[3,6,7],[1,2]),([1,4,5,8],[3,6,7],[2 ]),([1,4,5,8],[2,3,6,7],[]),([4,5,8],[1,2,3,6,7],[]),([5,8],[1,2,3,6,7],[4]),([5,8],[2,3 ,6,7],[1,4]),([2,5,8],[3,6,7],[1,4]),([1,2,5,8],[3,6,7],[4]),([1,2,5,8],[6,7],[3,4]),([2 ,5,8],[1,6,7],[3,4]),([5,8],[1,6,7],[2,3,4]),([5,8],[6,7],[1,2,3,4]),([8],[5,6,7],[1,2,3 ,4]),([1,8],[5,6,7],[2,3,4]),([1,8],[2,5,6,7],[3,4]),([8],[1,2,5,6,7],[3,4]),([3,8],[1,2 ,5,6,7],[4]),([3,8],[2,5,6,7],[1,4]),([2,3,8],[5,6,7],[1,4]),([1,2,3,8],[5,6,7],[4]),([1 ,2,3,8],[4,5,6,7],[]),([2,3,8],[1,4,5,6,7],[]),([3,8],[1,4,5,6,7],[2]),([3,8],[4,5,6,7], [1,2]),([8],[3,4,5,6,7],[1,2]),([1,8],[3,4,5,6,7],[2]),([1,8],[2,3,4,5,6,7],[]),([8],[1, 2,3,4,5,6,7],[]),([],[1,2,3,4,5,6,7],[8]),([],[2,3,4,5,6,7],[1,8]),([2],[3,4,5,6,7],[1,8 ]),([1,2],[3,4,5,6,7],[8]),([1,2],[4,5,6,7],[3,8]),([2],[1,4,5,6,7],[3,8]),([],[1,4,5,6, 7],[2,3,8]),([],[4,5,6,7],[1,2,3,8]),([4],[5,6,7],[1,2,3,8]),([1,4],[5,6,7],[2,3,8]),([1 ,4],[2,5,6,7],[3,8]),([4],[1,2,5,6,7],[3,8]),([3,4],[1,2,5,6,7],[8]),([3,4],[2,5,6,7],[1 ,8]),([2,3,4],[5,6,7],[1,8]),([1,2,3,4],[5,6,7],[8]),([1,2,3,4],[6,7],[5,8]),([2,3,4],[1 ,6,7],[5,8]),([3,4],[1,6,7],[2,5,8]),([3,4],[6,7],[1,2,5,8]),([4],[3,6,7],[1,2,5,8]),([1 ,4],[3,6,7],[2,5,8]),([1,4],[2,3,6,7],[5,8]),([4],[1,2,3,6,7],[5,8]),([],[1,2,3,6,7],[4, 5,8]),([],[2,3,6,7],[1,4,5,8]),([2],[3,6,7],[1,4,5,8]),([1,2],[3,6,7],[4,5,8]),([1,2],[6 ,7],[3,4,5,8]),([2],[1,6,7],[3,4,5,8]),([],[1,6,7],[2,3,4,5,8]),([],[6,7],[1,2,3,4,5,8]) ,([6],[7],[1,2,3,4,5,8]),([1,6],[7],[2,3,4,5,8]),([1,6],[2,7],[3,4,5,8]),([6],[1,2,7],[3 ,4,5,8]),([3,6],[1,2,7],[4,5,8]),([3,6],[2,7],[1,4,5,8]),([2,3,6],[7],[1,4,5,8]),([1,2,3 ,6],[7],[4,5,8]),([1,2,3,6],[4,7],[5,8]),([2,3,6],[1,4,7],[5,8]),([3,6],[1,4,7],[2,5,8]) ,([3,6],[4,7],[1,2,5,8]),([6],[3,4,7],[1,2,5,8]),([1,6],[3,4,7],[2,5,8]),([1,6],[2,3,4,7 ],[5,8]),([6],[1,2,3,4,7],[5,8]),([5,6],[1,2,3,4,7],[8]),([5,6],[2,3,4,7],[1,8]),([2,5,6 ],[3,4,7],[1,8]),([1,2,5,6],[3,4,7],[8]),([1,2,5,6],[4,7],[3,8]),([2,5,6],[1,4,7],[3,8]) ,([5,6],[1,4,7],[2,3,8]),([5,6],[4,7],[1,2,3,8]),([4,5,6],[7],[1,2,3,8]),([1,4,5,6],[7], [2,3,8]),([1,4,5,6],[2,7],[3,8]),([4,5,6],[1,2,7],[3,8]),([3,4,5,6],[1,2,7],[8]),([3,4,5 ,6],[2,7],[1,8]),([2,3,4,5,6],[7],[1,8]),([1,2,3,4,5,6],[7],[8]),([1,2,3,4,5,6],[],[7,8] ),([2,3,4,5,6],[1],[7,8]),([3,4,5,6],[1],[2,7,8]),([3,4,5,6],[],[1,2,7,8]),([4,5,6],[3], [1,2,7,8]),([1,4,5,6],[3],[2,7,8]),([1,4,5,6],[2,3],[7,8]),([4,5,6],[1,2,3],[7,8])] Figure 7.2: Tower of Hanoi (first 200 configurations) 276 CHAPTER7. INDUCTIONANDRECURSION move :: Peg move A B (x:xs,ys,zs) = move B A (xs,y:ys,zs) = move A C (x:xs,ys,zs) = move C A (xs,ys,z:zs) = move B C (xs,y:ys,zs) = move C B (xs,ys,z:zs) = -> Peg -> Tower -> Tower (xs,x:ys,zs) (y:xs,ys,zs) (xs,ys,x:zs) (z:xs,ys,zs) (xs,ys,y:zs) (xs,z:ys,zs) The procedure transfer takes three arguments for the pegs, an argument for the number of disks to move, and an argument for the tower configuration to move. The output is a list of tower configurations. The procedure hanoi, finally, takes a size argument and outputs the list of configurations to move a tower of that size. transfer :: Peg -> Peg -> Peg -> Int -> Tower -> [Tower] transfer _ _ _ 0 tower = [tower] transfer p q r n tower = transfer p r q (n-1) tower ++ transfer r q p (n-1) (move p q tower’) where tower’ = last (transfer p r q (n-1) tower) hanoi :: Int -> [Tower] hanoi n = transfer A C B n ([1..n],[],[]) The output for hanoi 8 is given in Figure 7.2. Here is the output for hanoi 4: IAR> hanoi 4 [([1,2,3,4],[],[]),([2,3,4],[1],[]),([3,4],[1],[2]),([3,4],[],[1,2]), ([4],[3],[1,2]),([1,4],[3],[2]),([1,4],[2,3],[]),([4],[1,2,3],[]), ([],[1,2,3],[4]),([],[2,3],[1,4]),([2],[3],[1,4]),([1,2],[3],[4]), ([1,2],[],[3,4]),([2],[1],[3,4]),([],[1],[2,3,4]),([],[],[1,2,3,4])] If you key in hanoi 64 and expect you can start meditating until the end of the world, you will discover that the program suffers from what functional program- mers call a space leak or black hole: as the execution progresses, the list of tower configurations that is kept in memory grows longer and longer, and execution will abort with an ‘out of memory’ error. All computation ceases before the world ends. 7.6. SOMEVARIATIONSONTHETOWEROFHANOI 277 Now consider the following. If there is a best way to transfer the tower of Hanoi (or the tower of Brahma), then in any given configuration it should be clear what the next move is. If they only have to look at the present configuration, the monks in charge of the tower of Brahma can go about their task in a fully enlightened manner, with complete focus on the here and now. Observe that there are 3n ways to stack n disks of decreasing sizes on 3 pegs in such a way that no disk is on top of a smaller disk. Since the number of moves to get from ([1..n],[],[]) to ([],[],[1..n]) is less than this, not all of these configurations are correct. How can we implement a check for correct configura- tions? Here is a checking procedure, with an argument for the size of the largest disk. check :: Int -> Tower -> Bool check 0 t = t == ([],[],[]) check n (xs,ys,zs) | xs /= [] && last xs == n = check (n-1) (init xs, zs, ys) | zs /= [] && last zs == n = check (n-1) (ys, xs, init zs) | otherwise = False Exercise 7.56 To see that the implementation of check is correct, we need an inductive proof. Give that proof. Here is a function for finding the largest disk in a configuration, and a function checkT for checking a configuration of any size. maxT :: Tower -> Int maxT (xs, ys, zs) = foldr max 0 (xs ++ ys ++ zs) checkT :: Tower -> Bool checkT t = check (maxT t) t The following exercise gives a more direct criterion. Exercise 7.57 Show that configuration (xs,ys,zs) with largest disk n is correct iff it holds that every disk m is either on top of a disk k with k m odd, or at the bottom of the source or destination peg, with n m even, or at the bottom of the auxiliary peg, with n k odd. (cid:0) (cid:0) (cid:0) 278 CHAPTER7. INDUCTIONANDRECURSION The previous exercise immediately gives a procedure for building correct config- urations: put each disk according to the rule, starting with the largest disk. This will give exactly two possibilities for every disk k: the largest disk can either go at source or at destination; if k + 1 is already in place, k can either go on top of k + 1 or to the only other place with the same parity as k + 1. For the implementation of a parity check, for the case where the largest disk has size n, we can take any configuration (xs,ys,zs), extend it to (xs ++ [n+1], ys ++ [n], zs ++ [n+1]) and then define the parities by means of par (x:xs) = x mod 2. Here is the implementation: parity :: Tower -> (Int,Int,Int) parity (xs,ys,zs) = par (xs ++ [n+1], ys ++ [n],zs ++ [n+1]) where n = maxT (xs, ys, zs) par (x:xs,y:ys,z:zs) = (mod x 2, mod y 2, mod z 2) Exercise 7.58 Show that if (xs,ys,zs) is a correct configuration, then parity (xs,ys,zs) (1; 1; 0); (1; 0; 1); (0; 1; 1) 2 f : g A little reflection shows that there are only two kinds of moves. Moves of the first kind move disk 1 (the smallest disk). Moves of the second kind move a disk other than 1. (cid:15) (cid:15) Moves of the first kind are fully determined, as can be seen by the fact that disk 1 should always be moved to the place with parity 0. Exercise 7.59 Give an argument for this. Moves of the second kind are also fully determined, for if there is one empty peg, then there are disks 1; k on top of the other pegs, and we can only move k to the 7.6. SOMEVARIATIONSONTHETOWEROFHANOI 279 empty peg, otherwise there are disks 1 < k < m on top of the pegs, and we can only move k on top of m. This gives a new algorithm for tower transfer, without space leak. We determine the target of a move of disk 1 by means of a parity check: target :: Tower -> Peg target t@(xs,ys,zs) | parity t == (0,1,1) = A | parity t == (1,0,1) = B | parity t == (1,1,0) = C The moves of disk 1 are now given by: move1 :: Tower -> Tower move1 t@(1:_,ys,zs) = move A (target t) t move1 t@(xs,1:_,zs) = move B (target t) t move1 t@(xs,ys,1:_) = move C (target t) t The moves of the middle disk are given by: move2 :: Tower -> Tower move2 t@(1:xs,[],zs) = move C B t move2 t@(1:xs,ys,[]) = move B C t move2 t@(1:xs,ys,zs) = if ys < zs then move B C t else move C B t move2 t@([],1:ys,zs) = move C A t move2 t@(xs,1:ys,[]) = move A C t move2 t@(xs,1:ys,zs) = if xs < zs then move A C t else move C A t move2 t@([],ys,1:zs) = move B A t move2 t@(xs,[],1:zs) = move A B t move2 t@(xs,ys,1:zs) = if xs < ys then move A B t else move B A t The check for completion is: 280 CHAPTER7. INDUCTIONANDRECURSION done :: Tower -> Bool done ([],[], _) = True done (xs,ys,zs) = False Transfer of a tower takes place by alternation between the two kinds of moves, until the tower is completely transferred. Since the last move has to be a move1, we only have to check for complete transfer right after such moves, i.e., in the function transfer2: transfer1, transfer2 :: Tower -> [Tower] transfer1 t = t : transfer2 (move1 t) transfer2 t = if done t then [t] else t : transfer1 (move2 t) And here is our new Hanoi procedure: hanoi’ :: Int -> [Tower] hanoi’ n = transfer1 ([1..n],[],[]) zazen :: [Tower] zazen = hanoi’ 64 By now we know enough about correct tower configurations to be able to order them. Exercise 7.60 Define and implement a total ordering on the list of all correct tower configurations. Exercise 7.60 makes clear that it should be possible to define a bijection between the natural numbers and the correct tower configurations, in their natural order. For this, we first define a function for finding the k-th correct configuration in the list of transitions for ([1..n],[],[]). 7.7. INDUCTIONANDRECURSIONOVEROTHERDATASTRUCTURES281 hanoiCount :: Int -> Integer -> Tower hanoiCount n k | k < 0 = error "argument negative" = error "argument not in range" | k > 2^n - 1 | k == 0 = ([1..n],[],[]) | k == 2^n - 1 = ([],[],[1..n]) | k < 2^(n-1) | k >= 2^(n-1) = (ys’, xs’, zs’ ++ [n]) = (xs ++ [n], zs, ys) where (xs,ys,zs) (xs’,ys’,zs’) = hanoiCount (n-1) (k - 2^(n-1)) = hanoiCount (n-1) k In terms of this we define the bijection. Note that for the definition we need the inverse of the function (cid:21)n:2n, i.e., the function (cid:21)n: log2 n. The predefined Haskell function logBase gives logarithms, and logarithms to base 2 are given by logBase 2. Since the results are in class Floating (the class of floating point numbers), we need conversion to get back to class Integral (the class consisting of Int and Integer). For this we use truncate. toTower :: Integer -> Tower toTower n = hanoiCount k m where n’ = fromInteger (n+1) k m = truncate (logBase 2 n’) = truncate (n’ - 2^k) Exercise 7.61 The function hanoiCount gives us yet another approach to the tower transfer problem. Implement this as hanoi’’ :: Int -> [Tower]. Exercise 7.62 Implement the function fromTower :: Tower -> Integer that is the inverse of toTower. 7.7 Induction and Recursion over Other Data Struc- tures A standard way to prove properties of logical formulas is by induction on their syntactic structure. Consider e.g. the following Haskell data type for propositional 282 formulas. CHAPTER7. INDUCTIONANDRECURSION data Form = P Int | Conj Form Form | Disj Form Form | Neg Form instance Show Form where show (P i) = ’P’:show i show (Conj f1 f2) = "(" ++ show f1 ++ " & " ++ show f2 ++ ")" show (Disj f1 f2) = "(" ++ show f1 ++ " v " ++ show f2 ++ ")" show (Neg f) = "~" ++ show f The instance Show Form ensures that the data type is in the class Show, and the function show indicates how the formulas are displayed. It is assumed that all proposition letters are from a list P0; P1; : : :. Then (P1 _ P2) is represented as Neg (Disj (P 1) (Neg (P 2))), and shown on the : : screen as ~(P1 v ~P2), and so on. We define the list of sub formulas of a formula as follows: subforms :: Form -> [Form] subforms (P n) = [(P n)] subforms (Conj f1 f2) = (Conj f1 f2):(subforms f1 ++ subforms f2) subforms (Disj f1 f2) = (Disj f1 f2):(subforms f1 ++ subforms f2) subforms (Neg f) = (Neg f):(subforms f) This gives, e.g.: IAR> subforms (Neg (Disj (P 1) (Neg (P 2)))) [~(P1 v ~P2),(P1 v ~P2),P1,~P2,P2] The following definitions count the number of connectives and the number of atomic formulas occurring in a given formula: 7.7. INDUCTIONANDRECURSIONOVEROTHERDATASTRUCTURES283 ccount :: Form -> Int ccount (P n) = 0 ccount (Conj f1 f2) = 1 + (ccount f1) + (ccount f2) ccount (Disj f1 f2) = 1 + (ccount f1) + (ccount f2) ccount (Neg f) = 1 + (ccount f) acount :: Form -> Int acount (P n) = 1 acount (Conj f1 f2) = (acount f1) + (acount f2) acount (Disj f1 f2) = (acount f1) + (acount f2) acount (Neg f) = acount f Now we can prove that the number of sub formulas of a formula equals the sum of its connectives and its atoms: Proposition 7.63 For every member f of Form: length (subforms f) = (ccount f) + (acount f). Proof. Basis If f is an atom, then subforms f = [f], so this list has length 1. Also, ccount f = 0 and acount f = 1. Induction step If f is a conjunction or a disjunction, we have: length (subforms f) = 1 + (subforms f1) + (subforms f2), ccount f = 1 + (ccount f1) + (ccount f2), acount f = (acount f1) + (acount f2), (cid:15) (cid:15) (cid:15) where f1 and f2 are the two conjuncts or disjuncts. By induction hypothe- sis: length (subforms f1) = (ccount f1) + (acount f1). length (subforms f2) = (ccount f2) + (acount f2). 284 CHAPTER7. INDUCTIONANDRECURSION The required equality follows immediately from this. If f is a negation, we have: length (subforms f) = 1 + (subforms f1), ccount f = 1 + (ccount f1), acount f = (acount f1), (cid:15) (cid:15) (cid:15) and again the required equality follows immediately from this and the in- duction hypothesis. If one proves a property of formulas by induction on the structure of the formula, then the fact is used that every formula can be mapped to a natural number that indicates its constructive complexity: 0 for the atomic formulas, the maximum of rank((cid:8)) and rank((cid:9)) plus 1 for a conjunction (cid:8) (cid:9), and so on. ^ 7.8 Further Reading Induction and recursion are at the core of discrete mathematics. See [Bal91]. A splendid textbook that teaches concrete (and very useful) mathematical skills in this area is Graham, Knuth and Patashnik [GKP89]. Recursion is also crucial for algorithm design. If you are specifically interested in the design and analysis of algorithms you should definitely read Harel [Har87]. Algorithm design in Haskell is the topic of [RL99]. Chapter 8 Working with Numbers Preview When reasoning about mathematical objects we make certain assumptions about the existence of things to reason about. In the course of this chapter we will take a look at integers, rational numbers, real numbers, and complex numbers. We will recall what they all look like, and we will demonstrate that reasoning about mathematical objects can be put to the practical test by an implementation of the definitions that we reason about. The implementations turn the definitions into procedures for handling representations of the mathematical objects. This chapter will also present some illuminating examples of the art of mathematical reasoning. module WWN where import List import Nats 285 286 CHAPTER8. WORKINGWITHNUMBERS 8.1 A Module for Natural Numbers The natural numbers are the conceptual basis of more involved number systems. The implementation of Natural from the previous chapter has the advantage that the unary representation makes the two cases for inductive proofs and recursive definitions, Z and S n, very clear. We repeat the code of our implementation of Natural, wrapped up this time in the form of a module, and integrated into the Haskell type system: see Figure 8.1. Note that an implementation of quotRem was added: quotRem n m returns a pair (q,r) consisting of a quotient and a remainder of the process of dividing n by m, i.e., q and r satisfy 0 6 r < m and q m + r = n. (cid:2) The integration of Natural in the system of Haskell types is achieved by means of instance declarations. E.g., the declaration instance Ord Natural makes the type Natural an instance of the class Ord. From an inspection of Prelude.hs we get that (i) a type in this class is assumed to also be in the class Eq, and (ii) a minimal complete definition of a type in this class assumes definitions of <= or compare. Condition (i) is fulfilled for Natural by the deriving Eq statement, while (ii) is taken care of by the code for compare in the module. The general code for class Ord provides methods for the relations <=, <, >=, >, and for the comparison operators max and min, so putting Natural in class Ord provides us with all of these. Similarly, the declaration instance Enum Natural ensures that Natural is in the class Enum. Definitions for the following functions for types in this class are provided in the module: succ :: Natural -> Natural pred :: Natural -> Natural toEnum :: Int -> Natural fromEnum :: Natural -> Int enumFrom :: Natural -> [Natural] The general code for the class provides, e.g., enumFromTo, so we get: Nats> [Z,S Z,S (S Z),S (S (S Z)),S (S (S (S Z))),S (S (S (S (S Z))))] enumFromTo Z (toEnum 5) Next, the declaration instance Num Natural ensures that Natural is in the class Num. This is where the functions for addition, multiplication, sign, abso- lute value belong, as well as a type conversion function fromInteger. Putting in the appropriate definitions allows the use of the standard names for these opera- tors. In terms of these, Prelude.hs defines subtraction, the type conversion function 8.1. AMODULEFORNATURALNUMBERS 287 module Nats where data Natural = Z | S Natural deriving (Eq, Show) instance Ord Natural where compare Z Z = EQ compare Z _ = LT compare _ Z = GT compare (S m) (S n) = compare m n instance Enum Natural where succ = \ n -> S n pred Z = Z pred (S n) = n toEnum fromEnum = toInt enumFrom n = map toEnum [(fromEnum n)..] = fromInt instance Num Natural where (+) = foldn succ (*) = \m -> foldn (+m) Z (-) = foldn pred abs = id signum Z = Z signum n = (S Z) fromInteger n | n < 0 = error "no negative naturals" = Z | n == 0 | otherwise = S (fromInteger (n-1)) foldn :: (a -> a) -> a -> Natural -> a foldn h c Z = c foldn h c (S n) = h (foldn h c n) instance Real Natural where toRational x = toInteger x % 1 instance Integral Natural where quotRem n d | d > n | otherwise = (S q, r) = (Z,n) where (q,r) = quotRem (n-d) d toInteger = foldn succ 0 Figure 8.1: A Module for Natural Numbers. 288 CHAPTER8. WORKINGWITHNUMBERS fromInt, and negate. Since negate is defined in terms of subtraction, and sub- traction for naturals is cut-off subtraction, we get: Nats> negate (S (S (S (S Z)))) Z The next instance declaration puts Natural in the class Real. The only code we have to provide is that for the type conversion function toRational. Putting Natural in this class is necessary for the next move: the instance declaration that puts Natural in the class Integral: types in this class are constrained by Prelude.hs to be in the classes Real and Enum. The benefits of making Natural an integral type are many: we now get implementations of even, odd, quot, rem, div, mod, toInt and many other functions and operations for free (see the contents of Prelude.hs for full details). The implicit type conversion also allows us to introduce natural numbers in short- hand notation: Nats> 12 :: Natural S (S (S (S (S (S (S (S (S (S (S (S Z))))))))))) The standard representation for natural numbers that we are accustomed to, deci- mal string notation, is built into the Haskell system for the display of integers. To display naturals in decimal string notation, all we have to do is use type conversion: Nats> toInt (S (S (S Z))) 3 Nats> (S (S Z)) ^ (S (S (S Z))) S (S (S (S (S (S (S (S Z))))))) Nats> toInt (S (S Z)) ^ (S (S (S Z))) 8 For other representations, i.e., binary string representation of an integral number n, divide by 2 repeatedly, and collect the remainders in a list. The reversal of this 1 + list gives the binary representation bk : : : b0 satisfying n = bk + b0. The nice thing is that we can provide this code for the whole class of 2k + bk 1 (cid:1) (cid:0) 2k (cid:0) (cid:1) (cid:1) (cid:1) (cid:1) Integral types: 8.2. GCDANDTHEFUNDAMENTALTHEOREMOFARITHMETIC 289 binary :: Integral a => a -> [Int] binary x = reverse (bits x) where bits 0 = [0] bits 1 = [1] bits n = toInt (rem n 2) : bits (quot n 2) To display this on the screen, we need intToDigit for converting integers into digital characters: showDigits :: [Int] -> String showDigits = map intToDigit bin :: bin = showDigits . binary Integral a => a -> String Exercise 8.1 Give a function hex for displaying numbers in type class Integral in hexadecimal form, i.e., in base 16 representation. The extra digits a, b, c, d, e, f for 10, 11, 12, 13, 14, 15 that you need are provided by intToDigit. The call hex 31 should yield "1f". (Hint: it is worthwhile to provide a more general function toBase for converting to a list of digits in any base in , and then define hex in terms of that.) 2; : : : ; 16 g f 8.2 GCD and the Fundamental Theorem of Arith- metic N with n > 1 has The fundamental theorem of arithmetic, stating that every n a unique prime factorization, was known in Antiquity. Let us reflect a bit to see why it is true. First note that the restriction on n is necessary, for m 0 = 0 for all N, so a factorization of 0 can never be unique, and since we have ruled out 1 m as a prime, 1 does not have a prime factorization at all. The fundamental theorem of arithmetic constitutes the reason, by the way, for ruling out 1 as a prime number. We have n = 1m N, so a factorization of n that admits 1m as a factor can never be unique. n for any n; m 2 2 2 (cid:1) (cid:1) 290 CHAPTER8. WORKINGWITHNUMBERS From the prime factorization algorithm (1.7) we know that prime factorizations of every natural number > 1 exist. To show uniqueness, we still have to establish that no number has more than one prime factorization. Euclid’s GCD algorithm The greatest common divisor of two natural numbers a; b, notation GCD(a; b), is the natural number d with the property that d divides both a and b, and for all natural numbers d0 that divide both a and b it holds that d0 divides d. For example, GCD(30; 84) = 6, for 6 divides 30 and 84, and every other common divisor of 30 and 84 divides 6 (these other common divisors being 1, 2 and 3). Clearly, if such d exists, it is unique, for suppose that d0 also qualifies. Then because d0 divides a and b, it follows from the definition of d that d0 divides d. Similarly, because d divides a and b, it follows from the definition of d0 that d divides d0. But if d divides d0 and d0 divides d then it follows that d = d0. The greatest common divisor of a and b can be found from the prime factorizations p(cid:12)k p(cid:11)k of a and b as follows. Let a = p(cid:11)1 k be prime factor- k izations of a and b, with p1; : : : ; pk distinct primes, and (cid:11)1; : : : ; (cid:11)k; (cid:12)1; : : : ; (cid:12)k natural numbers. Then the greatest common divisor of a and b equals the natural number p(cid:13)1 p(cid:13)k k , where each (cid:13)i is the minimum of (cid:11)i and (cid:12)i. For example, the greatest common divisor of 30 = 21 22 71 is given by 21 and b = p(cid:12)1 70 and 84 = 70 = 6. 1 (cid:1) (cid:1) (cid:1) 1 (cid:1) (cid:1) (cid:1) 1 (cid:1) (cid:1) (cid:1) 51 50 31 31 31 50 (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) But there is an easier way to find GCD(a; b). Here is Euclid’s famous algorithm (assume neither of a; b equals 0): WHILE a = b DO IF a > b THEN a := a b ELSE b := b a: (cid:0) (cid:0) Let us run this for the example case a = 30, b = 84. We get the following (conventions about variable names as in Section 1.7). a0 < b0 a1 < b1 a2 > b2 a3 < b3 a4 < b4 a5 < b5 a6 = b6 = 6 a0 = 30 a1 = 30 a2 = 30 a3 = 30 a4 = 6 a5 = 6 a6 = 6 (cid:0) b0 = 84 b1 = 84 b2 = 54 24 = 6 b3 = 24 b4 = 24 b5 = 18 b6 = 12 30 = 54 30 = 24 6 = 18 6 = 12 6 = 6 (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) Now why does this work? The key observation is the following. 6 8.2. GCDANDTHEFUNDAMENTALTHEOREMOFARITHMETIC 291 If d divides a and b and a > b then d divides a numbers m; n with m > n and a = md , b = nd, and therefore a d(m n)), and similarly, if d divides a and b and a < b then d divides b b (for then there are natural nd = (cid:0) a. b = md (cid:0) (cid:0) (cid:0) (cid:0) Conversely, if a > b and d divides a are natural numbers m; n with a d(m + n)), and similarly, if a > b and d divides b (cid:0) (cid:0) b and b, then d divides a (for then there b = md and b = nd, hence a = md + nd = a and a, then d divides a. (cid:0) Thus, if a > b then the set of common divisors of a and b equals the set of common b and b, and if b > a then the set of common divisors of a and b divisors of a a. equals the set of common divisors of a and b (cid:0) Since the sets of all common divisors are equal, the greatest common divisors must b; b), be equal as well. Therefore we have: if a > b then GCD(a; b) = GCD(a and if b > a then GCD(a; b) = GCD(a; b a). (cid:0) (cid:0) (cid:0) Using this we see that every iteration through the loop preserves the greatest com- mon divisor in the following sense: GCD(ai; bi) = GCD(ai+1; bi+1). Since we know that the algorithm terminates we have: there is some k with ak = bk. There- fore ak = GCD(ak; bk) = GCD(a; b). Haskell contains a standard function gcd for the greatest common divisor of a pair of objects of type Integral: gcd gcd 0 0 gcd x y :: Integral a => a -> a -> a = error "Prelude.gcd: gcd 0 0 is undefined" = gcd’ (abs x) (abs y) where gcd’ x 0 = x gcd’ x y = gcd’ y (x ‘rem‘ y) Exercise 8.2 If you compare the code for gcd to Euclid’s algorithm, you see that Euclid uses repeated subtraction where the Haskell code uses rem. Explain as precisely as you can how the Haskell version of the GCD algorithm is related to Euclid’s method for finding the GCD. We can use the GCD to define an interesting relation. Two natural numbers n and m are said to be co-prime or relatively prime if GCD(m; n) = 1. Here is an implementation: 292 CHAPTER8. WORKINGWITHNUMBERS coprime :: Integer -> Integer -> Bool coprime m n = (gcd m n) == 1 Exercise 8.3 Consider the following experiment: WWN> coprime 12 25 True WWN> 12 + 25 37 WWN> coprime 25 37 True WWN> 25 + 37 62 WWN> coprime 37 62 True WWN> 37 + 62 99 WWN> coprime 62 99 True WWN> 62 + 99 161 WWN> coprime 99 161 True WWN> This experiment suggests a general rule, for you to consider . . . Does it follow from the fact that a and b are co-prime with a < b that b and a+b are co-prime? Give a proof if your answer is ‘yes’ and a counterexample otherwise. Theorem 8.4 For all positive a; b GCD(a; b). 2 N there are integers m; n with ma + nb = Proof. Consider the pairs (a0; b0); (a1; b1); : : : ; (ak; bk) generated by Euclid’s algorithm. We know that (a0; b0) = (a; b) and that ak = bk = GCD(a; b). a0 satisfies a0 = ma + nb for m = 1; n = 0, b0 satisfies ma + nb = 1 for m = 0; n = 1. Suppose ai satisfies ai = m1a + n1b and bi satisfies bi = m2a + n2b. If ai > bi, n2)b and bi+1 satisfies bi+1 = then ai+1 satisfies ai+1 = (m1 (cid:0) m2)a + (n1 (cid:0) 8.3. INTEGERS 293 m2a + n2b. If ai < bi, then ai+1 satisfies ai+1 = m1a + n1b and bi+1 satisfies n1)b. Thus, every iteration through the loop of Euclid’s bi+1 = (m2(cid:0) algorithm preserves the fact that ai and bi are integral linear combinations ma+nb of a and b. m1)a+(n2(cid:0) This shows that there are integers m; n with ak = ma + nb, hence that ma + nb = GCD(a; b). Theorem 8.5 If p is a prime number that divides ab then p divides a or b. Proof. Suppose p divides ab and p does not divide a. Then GCD(a; p) = 1. By the previous theorem there are integers m; n with ma + np = 1. Multiplying both sides by b gives: mab + nbp = b: By the fact that p divides ab we know that p divides both mab and nbp. Hence p divides mab + nbp. Hence p divides b. Theorem 8.5 is the tool for proving the fundamental theorem of arithmetic. Theorem 8.6 (Fundamental Theorem of Arithmetic) Every natural number greater than 1 has a unique prime factorization. Proof. We established in 1.7 that every natural number greater than 1 has at least one prime factorization. To show that every natural number has at most one prime factorization, assume to the contrary that there is a natural number N > 1 with at least two different prime factorizations. Thus, N = p1 (cid:1) (cid:1) (cid:1) pr = q1 (cid:1) (cid:1) (cid:1) qs; with all of p1; : : : ; pr; q1; : : : ; qs prime. Divide out common factors if necessary. This gives a pi that is not among the q’s. But this is a contradiction with theorem qs but pi does not divide any of q1, . . . , qs, 8.5, because pi divides N = q1 (cid:1) (cid:1) (cid:1) since these are all prime numbers different from pi. 8.3 Integers Suppose n; m; k are natural numbers. When n = m + k we can view k as the difference of n and m, and it is tempting to write this as k = n m for the operation of subtraction. Addition and subtraction are called inverse operations, (cid:0) 294 CHAPTER8. WORKINGWITHNUMBERS for if the addition of n to m is followed by the subtraction of n, the end result is the original natural number m. In other words, we have: (m + n) n = m: (cid:0) But we have to be careful here. The operation of subtracting a natural number n from a natural number m will only result in a natural number if m > n. To make subtraction possible between any pair of numbers, we have to introduce negative numbers. This gives the domain of integers: Z = : : : ; f 3; (cid:0) 2; (cid:0) : 1; 0; 1; 2; 3; : : : g (cid:0) The symbol Z derives from Zahl, the German word for number. In fact, we need not consider the integers as given by God, but we can view them as constructed from the natural numbers. This can be done in several ways. The following Haskell code illustrates one possibility: data Sgn = P | N deriving (Eq,Show) type MyInt = (Sgn,Natural) myplus :: MyInt -> MyInt -> MyInt myplus (s1,m) (s2,n) | s1 == s2 = (s1,m+n) | s1 == P | s1 == P | otherwise && n <= m = (P,m-n) = (N,n-m) && n > m = myplus (s2,n) (s1,m) Another way is as follows. We represent every integer as a ‘difference pair’ of two 3 is natural numbers. For example, (0; 3) represents also represented by (1; 4) or (2; 5). N with m2 + k = In general, for all m1; m2 2 m1, and (m1; m2) represents k. If, on the other hand, m1 < m2 then there is a N2 be k defined as follows: N with m1 + k = m2, and (m1; m2) represents N, if m1 > m2 then there is a k 3, but the same number k. Thus, let R (cid:0) (cid:0) (cid:0) (cid:18) 2 2 (m1; m2)R(n1; n2) : (cid:17) m1 + n2 = m2 + n1: It is easy to see that R is an equivalence on N2. If m 2 represented by (0; m), but also by (k; k + m), for any k N, then the integer N. 2 m is (cid:0) Intuitively (m1; m2) and (n1; n2) are equivalent modulo R when their differences are the same. This is the case precisely when m1 + n2 = m2 + n1. In this case we 8.3. INTEGERS 295 say that (m1; m2) and (n1; n2) represent the same number. In Section 6.8 we saw that the equivalence relation R is a congruence for addition on N2, where addition on N2 is given by: (m1; m2) + (n1; n2) := (m1 + n1; m2 + n2); Denoting the equivalence class of (m1; m2) as [m1 (cid:0) m2], we get: [m1 (cid:0) m2] := (n1; n2) f 2 N2 : m1 + n2 = m2 + n1g j m2], for m1; m2 2 is not an operator. Call the equivalence classes We identify the integers with the equivalence classes [m1 (cid:0) N. Note that in this notation (cid:0) difference classes. If an integer m = [m1 (cid:0) m1 and m2, and put m = [m2 (cid:0) (cid:0) the original equivalence class. This reflects the familiar rule of sign m = m2], then we can swap the sign by swapping the order of m1]. Swapping the order twice gets us back to m). ( (cid:0) (cid:0) We define addition and multiplication on difference classes as follows: [m1 (cid:0) [m1 (cid:0) m2] + [n1 (cid:0) [n1 (cid:0) m2] (cid:1) n2] n2] := [(m1 + n1) := [(m1n1 + m2n2) (cid:0) (m2 + n2)] (m1n2 + n1m2)]: (cid:0) The purpose of this definition is to extend the fundamental laws of arithmetic to the new domain. To see whether we have succeeded in this we have to perform a verification, as a justification of the definition. In the verification that integers satisfy the law of commutativity for addition, we make use of the definition of addition for difference classes, and of laws of com- mutativity for addition on N: Proposition 8.7 For all m; n Z: m + n = n + m. 2 Proof. Representing m; n as difference classes we get: n2] = [definition of + for difference classes] (m2 + n2)] = [commutativity of + for N] (n2 + m2)] = [definition of + for difference classes] [m1 (cid:0) [(m1 + n1) m2] + [n1 (cid:0) (cid:0) [(n1 + m1) [(n1 (cid:0) (cid:0) n2] + [m1 (cid:0) m2)]: Note that the proof uses just the definition of + for difference classes and the com- mutativity of + on N. Note also that every equality statement is justified in the 296 CHAPTER8. WORKINGWITHNUMBERS proof by a reference to a definition or to a fundamental law of arithmetic, where these laws are in turn justified by inductive proofs based on the recursive defini- tions of the natural number operations. Thus, the fundamental laws of arithmetic and the definition of a new kind of object (difference classes) in terms of a familiar one (natural numbers) are our starting point for the investigation of that new kind of object. Exercise 8.8 Show (using the definition of integers as difference classes and the definition of addition for difference classes) that the associative law for addition holds for the domain of integers. In a similar way it can be shown from the definition of integers by means of dif- ference classes and the definition of multiplication for difference classes that the associative and commutative laws for multiplication continue to hold for the do- main of integers. As a further example of reasoning about this representation for the integers, we show that the distributive law continues to hold in the domain of integers (viewed as difference classes). In the verification we make use of the definition of addition and multiplication for difference classes, and of the fundamental laws for addition and multiplication on the natural numbers. Again, we justify every step by means of a reference to the definition of an operation on difference classes or to one of the laws of arithmetic for the natural numbers. Proposition 8.9 For all m; n; k Z : m(n + k) = mn + mk: 2 Proof. Representing m; n; k as difference classes we get: [m1 (cid:0) m2]([n1 (cid:0) n2] + [k1 (cid:0) k2]) = [definition of + for difference classes] [m1 (cid:0) = [definition of m2][(n1 + k1) (n2 + k2)] (cid:0) for difference classes] (cid:1) [(m1(n1 + k1) + m2(n2 + k2)) over + for N] = [distribution of (cid:1) [(m1n1 + m1k1 + m2n2 + m2k2) (m1(n2 + k2) + (n1 + k1)m2)] (cid:0) (m1n2 + m1k2 + m2n1 + m2k1)] (cid:0) 8.4. IMPLEMENTINGINTEGERARITHMETIC 297 = [commutativity of + for N] [(m1n1 + m2n2 + m1k1 + m2k2) (cid:0) = [definition of + for difference classes] (m1n2 + m2n1 + m1k2 + m2k1)] [(m1n1 + m2n2) (m1n2 + m2n1)] + [(m1k1 + m2k2) (m1k2 + m2k1)] (cid:0) (cid:0) = [definition of [m1 (cid:0) m2] for difference classes] n2] + [m1 (cid:0) (cid:1) [n1 (cid:0) m2] (cid:1) (cid:1) [k1 (cid:0) k2]: Once the integers and the operations of addition and multiplication on it are de- fined, and we have checked that the definitions are correct, we can forget of course about the representation by means of difference classes. It simply takes too much mental energy to keep such details of representation in mind. Mathematicians make friends with mathematical objects by forgetting about irrelevant details of their definition. The natural numbers have representations as difference classes too: a natural num- ber m is represented by [m 0]. In fact, the original numbers and their represen- tations in the new domain Z behave exactly the same, in a very precise sense: the 0] is one-to-one (it never maps different num- function that maps m bers to the same pair) and it preserves the structure of addition and multiplication, in the following sense (see also Section 6.8). If m + n = k then N to [m (cid:0) (cid:0) 2 [m and if mn = k then 0] + [n (cid:0) (cid:0) 0] = [(m + n) 0] = [k 0]; (cid:0) (cid:0) (cid:0) Again we forget about the differences in representation and we say: N (cid:0) (cid:0) (cid:0) (cid:1) [m 0] [n 0] = [mn 0] = [k 0]: Z. (cid:18) 8.4 Implementing Integer Arithmetic If we represent natural numbers as type Natural, then the moral of the above is that integers can be represented as pairs of naturals. E.g., minus five is represented by the pair (S Z,S(S(S(S(S(S Z)))))), or by the pair (Z,S(S(S(S(S Z))))), and so on. Here is an appropriate data type declaration: type NatPair = (Natural,Natural) 298 CHAPTER8. WORKINGWITHNUMBERS The gist of the previous section can be nicely illustrated by means of implementa- tions of the integer operations. For suppose we have the natural number operations plus for addition and times for multiplication available for the type Natural). Then addition for integer pairs is implemented in Haskell as: plus1 :: NatPair -> NatPair -> NatPair plus1 (m1, m2) (n1, n2) = (m1+n1, m2+n2) Subtraction is just addition, but with the sign of the second operand reversed. Sign reversal is done by swapping the elements of a pair, as we have seen. Thus, the implementation of subtraction can look like this: subtr1 :: NatPair -> NatPair -> NatPair subtr1 (m1, m2) (n1, n2) = plus1 (m1, m2) (n2, n1) Here is the implementation of multiplication: mult1 :: NatPair -> NatPair -> NatPair mult1 (m1, m2) (n1, n2) = (m1*n1 + m2*n2, m1*n2 + m2*n1) The implementation of equality for pairs of naturals is also straightforward: eq1 :: NatPair -> NatPair -> Bool eq1 (m1, m2) (n1, n2) = (m1+n2) == (m2+n1) Finally, it is useful to be able to reduce a naturals pair to its simplest form (such a simplest form is often called a canonical representation). The simplest form of a naturals pair is a pair which has either its first or its second member equal to Z. E.g., the simplest form of the pair (S(S Z),S(S(S(S Z)))) is (Z,S(S Z)). Reduction to simplest form can again be done by recursion. 8.5. RATIONALNUMBERS 299 reduce1 :: NatPair -> NatPair reduce1 (m1,Z) = (m1,Z) reduce1 (Z,m2) = (Z,m2) reduce1 (S m1, S m2) = reduce1 (m1, m2) Exercise 8.10 Define and implement relations leq1 for 6 and gt1 for > for dif- ference classes of naturals. 8.5 Rational Numbers A further assumption that we would like to make is that any integer can be divided by any non-zero integer to form a fraction, or rational number, and that any rational number m=n can be ‘canceled down’ to its lowest form by dividing m and n by the same number. E.g., 12=39 cancels down to 4=13. Z, n Again we can view the rational numbers as constructed by means of pairs, in this = 0. In this case the pairs are ‘ratio pairs’. case pairs (m; n) with m; n One and the same rational number can be represented in many different ways: (1; 2), (2; 4), (13; 26) all represent the rational number 1=2. Or in other words: the rational number m=n is nothing but the class of all (p; q) with p; q = 0, and the property that mq = np. In such a case we say that m=n and p=q represent the same number. Z, q 2 2 (Z As in the case of the representation of integers by means of difference classes we have an underlying notion of equivalence. Let S be the equivalence relation on Z mq = np. Note that (m; n) and (km; kn) ) given by (m; n)S(p; q) : g (cid:17) are equivalent modulo S (provided k = 0), which justifies the simplification of km=kn to m=n by means of canceling down. The set Q of rational numbers (or fractional numbers) can be defined as: (cid:0)f (cid:2) 0 Q := (Z (Z (cid:2) 0 ))=S: (cid:0) f g We write a class [(m=n)]S as [m=n]. Addition, multiplication and equality of ra- tional numbers are now defined in terms of addition and multiplication of integers, as follows: [m=n] + [p=q] := [(mq + pn)=nq] [m=n] [p=q] [m=n] = [p=q] (cid:1) := [mp=nq] mq = np : (cid:17) 6 6 6 300 CHAPTER8. WORKINGWITHNUMBERS Again, we have to check that these definitions make sense. It is easy to see that the sum, the difference, and the product of two ratio classes are again ratio classes. Z with For instance, if x = 0, and x = [m=n] and y = [p=q]. Then x + y = [m=n] + [p=q] = n [(mq + pn)=nq], and from mq + pn = 0 it follows that x + y Q, then there are integers m; n; p; q Z, and nq Q and y = 0, q Z, nq Q. 2 2 2 2 2 2 Proposition 8.11 The law of associativity for + holds for the rationals. Proof. If x; y; z z = [r=s], n 2 = 0, q Q then there are m; n; p; q; r; s = 0 and s = 0. Now: Z with x = [m=n], y = [p=q], 2 x + (y + z) = = = = = [definitions of x; y; z] n ] + ([ p [ m q ] + [ r s ]) ] qs [definition of + for Q] n ] + [ ps+rq [ m [definition of + for Q] [ mqs+(ps+rq)n nqs [distribution law for Z] [ mqs+psn+rqn nqs ] ] [assoc of for Z, dist law for Z] (cid:1) [ (mq+pn)s+rqn nqs ] ] + [ r s ] = [definition of + for Q] [ mq+pn nq = [definition of + for Q] n ] + [ p = [definitions of x; y; z] q ]) + [ r s ] ([ m (x + y) + z: In a similar way, it is not difficult (but admittedly a bit tedious) to check that the law of commutativity for + holds for the rationals, that the laws of associativity 6 6 6 6 6 6 8.5. RATIONALNUMBERS 301 and commutativity for for the rationals. (cid:1) hold for the rationals, and that the law of distribution holds A thing to note about Proposition 8.11 is that the statement about rational num- bers that it makes also tells us something about our ways of doing things in an implementation. It tells us that any program that uses the procedure add which implements addition for the rationals need not distinguish between (add (add (m,n) (p,q)) (r,s)) and (add (m,n) (add (p,q) (r,s))). Wherever one expression fits the other will fit. Again forgetting about irrelevant differences in representation, we see that Z for every integer n has a representation in the rationals, namely n=1. (cid:18) Q, 2 Q, x Q, then x + y = 0, If x; y Q. We say that the domain of rationals is closed under addition, we have x=y subtraction, and multiplication and ‘almost closed’ under division. If y = 0, then 1=y exists, which shows that every rational number y except 0 has an inverse: a rational number that yields 1 when multiplied with y. If, in addition y Q, xy Q. (cid:0) 2 2 2 2 y A domain of numbers that is closed under the four operations of addition, sub- traction, multiplication and division (with the extra condition on division that the = 0) is called a field. The domain of rational numbers is a field. divisor should be It is easy to see that each rational number m=n can be written in decimal form, by performing the process of long division. There are two possible outcomes: 1. the division process stops: m=n can be written as a finite decimal expansion. 2. the division process repeats: m=n can be written as a infinite decimal ex- pansion with a tail part consisting of a digit or group of digits which repeats infinitely often. Examples of the first outcome are 1=4 = 0:25, 0:25, 22=5 = 4:4. Examples of the second outcome are 1=3 = 0:3333 : : :, 1=6 = 0:166666 : : :, 29=7 = 4:142857142857142857 : : :, 23=14 = 1:6428571428571428571 : : :. A handy notation for this is 1=3 = 0:3, 1=6 = 0:6, 29=7 = 4:142857, 23=14 = 1:6428571. The part under the line is the part that repeats infinitely often. 1=4 = (cid:0) (cid:0) If the division process repeats for m=n then at each stage of the process there must be a non-zero remainder. This means that at each stage, the remainder must be in the range 1; : : : ; n 1. But then after at most n steps, some remainder k is bound to reoccur, and after the reappearance of k the process runs through exactly the same stages as after the first appearance of k. The same remainders turn up, in the same order. We are in a loop. (cid:0) 6 6 6 302 CHAPTER8. WORKINGWITHNUMBERS Conversely, it is not very difficult to see that every repeating decimal corresponds to a rational number. First an example. Suppose M = 0:133133133 : : : = 0:133. Then: 1000 0:133 = 133, so M = 133 999 . M = 133:133, and (1000 M = 133:133 M ) (cid:0) (cid:0) (cid:3) (cid:3) Now for the general case. Theorem 8.12 Every repeating decimal corresponds to a rational number. Proof. A repeating decimal can always be written in our over-line notation in the form 0:a1a2 (cid:1) (cid:1) (cid:1) where M is an integer and the ai and bj are decimal digits. For example, amb1b2 (cid:1) (cid:1) (cid:1) bn; M (cid:6) 2:6428571 = (cid:0) 2 (cid:0) (cid:0) 0:6428571. If we can prove that the part p = 0:a1a2 (cid:1) (cid:1) (cid:1) amb1b2 (cid:1) (cid:1) (cid:1) bn 37=14 = (cid:0) is rational, then we are done, for if p is rational and M is an integer then there are integers n; k with p = n=k, so M + p = M + n=k = kM +n , i.e., M + p is rational, and similarly for M p. Setting k (cid:0) A = 0:a1a2 (cid:1) (cid:1) (cid:1) am; B = 0:b1b2 (cid:1) (cid:1) (cid:1) bn; C = 0:b1b2 (cid:1) (cid:1) (cid:1) bn; we get amb1b2 (cid:1) (cid:1) (cid:1) If we can write this in terms of rational operations on A and B we are done. p = 0:a1a2 (cid:1) (cid:1) (cid:1) bn = A + 10(cid:0) mC: C = 0:b1b2 (cid:1) (cid:1) (cid:1) bn = 0:b1b2 (cid:1) (cid:1) (cid:1) bnb1b2 (cid:1) (cid:1) (cid:1) bn: Thus, C (cid:0) 10(cid:0) bnb1b2 (cid:1) (cid:1) (cid:1) bn = B: nC = 0:b1b2 (cid:1) (cid:1) (cid:1) = 0:b1b2 (cid:1) (cid:1) (cid:1) n , and we get: 10(cid:0) 1 (cid:0) Therefore, C = B bn 0:01 (cid:1) (cid:1) (cid:1) 0nb1b2 (cid:1) (cid:1) (cid:1) (cid:0) bn = p = A + so we have proved that p is rational. 10(cid:0) 1 mB 10(cid:0) n ; (cid:0) 8.5. RATIONALNUMBERS 303 r3 J r1 J J J J J J Jr 2 3 J J J J r 2 Figure 8.2: Constructing the fraction 2=3. There is a nice geometrical interpretation of the process of constructing rational numbers. Assume that we have been given the line of integers: 3 : : : (cid:0) (cid:15) 2 (cid:0) (cid:15) 1 (cid:0) (cid:15) 0 (cid:15) 1 (cid:15) 2 (cid:15) 3 (cid:15) : : : Place these on a (horizontal) x-axis and a (vertical) y-axis: x (cid:1) (cid:1) (cid:1) (cid:15) (cid:15) ... (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) (cid:1) (cid:1) (cid:1) (cid:15) (cid:15)... y Now construct further points on the x-axis by first drawing a straight line l through a point m on the x-axis and a point n on the y-axis, and next drawing a line l0 parallel to l through the point (0; 1). The intersection of l0 and the x-axis is the rational point m=n. (Use congruence reasoning for triangles to establish that the ratio between m and n equals the ratio between the intersection point m=n and 1.) Figure 8.2 gives the construction of the fraction 2=3 on the x-axis. Figures 8.3, 8.4, 8.5 and 8.6 give the geometrical interpretations of addition, nega- tion, multiplication and reciprocal. Subtraction is addition of a negated number, division is multiplication with a reciprocal, so these give all the rational operations. Note that these constructions can all be performed by a process of (i) connecting previously constructed points by a straight line, and (ii) drawing lines parallel to 304 CHAPTER8. WORKINGWITHNUMBERS a 1 0 1 a b a + b Figure 8.3: Geometrical Interpretation of Addition. a a (cid:0) 1 (cid:0) 0 1 a Figure 8.4: Geometrical Interpretation of Negation. 1 1 b 0 1 a b ab Figure 8.5: Geometrical Interpretation of Multiplication. 8.6. IMPLEMENTINGRATIONALARITHMETIC 305 a 1 1 a0 1 a Figure 8.6: Geometrical Interpretation of Reciprocal. previously constructed lines. These are the so-called linear constructions: the con- structions that can be performed by using a ruler but no compass. In particular, use of a compass to construct line segments of equal lengths is forbidden. 8.6 Implementing Rational Arithmetic Haskell has a standard implementation of the above, a type Rational, predefined as follows: data Integral a => Ratio a = a :% a deriving (Eq) type Rational = Ratio Integer To reduce a fraction to its simplest form, we first make sure that the denominator is non-negative. If (x,y) represents a fraction with numerator x and denominator y, then (x * signum y) (abs y) is an equivalent representation with positive denominator. Here abs gives the ab- solute value, and signum the sign (1 for positive integers, 0 for 0, 1 for negative integers). The reduction to canonical form is performed by: (cid:0) 306 CHAPTER8. WORKINGWITHNUMBERS (%) x % y :: Integral a => a -> a -> Ratio a = reduce (x * signum y) (abs y) reduce reduce x y | y == 0 | otherwise :: Integral a => a -> a -> Ratio a = error "Ratio.%: zero denominator" = (x ‘quot‘ d) :% (y ‘quot‘ d) where d = gcd x y Functions for extracting the numerator and the denominator are provided: numerator, denominator numerator (x :% y) denominator (x :% y) :: Integral a => Ratio a -> a = x = y Note that the numerator of (x % y) need not be equal to x and the denominator need not be equal to y: Prelude> numerator (2 % 4) 1 Prelude> denominator (2 % 10) 5 A total order on the rationals is implemented by: instance Integral a => Ord (Ratio a) where compare (x:%y) (x’:%y’) = compare (x*y’) (x’*y) The standard numerical operations from the class Num are implemented by: 8.6. IMPLEMENTINGRATIONALARITHMETIC 307 instance Integral a => Num (Ratio a) where (x:%y) + (x’:%y’) = reduce (x*y’ + x’*y) (y*y’) (x:%y) * (x’:%y’) = reduce (x*x’) (y*y’) negate (x :% y) abs (x :% y) signum (x :% y) = negate x :% y = abs x :% y = signum x :% 1 The rationals are also closed under the division operation (cid:21)xy: x operation (cid:21)x: 1 x . These are implemented as follows: y and the reciprocal instance Integral a => Fractional (Ratio a) where (x:%y) / (x’:%y’) recip (x:%y) = (x*y’) % (y*x’) = if x < 0 then (-y) :% (-x) else y :% x If you want to try out decimal expansions of fractions on a computer, here is a Haskell program that generates the decimal expansion of a fraction. decExpand :: Rational -> [Integer] decExpand x | x < 0 = error "negative argument" = [q] | r == 0 | otherwise = q : decExpand ((r*10) % d) where (q,r) = quotRem n d = numerator x n = denominator x d If the decimal expansion repeats, you will have to interrupt the process by typing control c: WWN> decExpand (1 % 7) [0, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5,{Interrupted!} 308 CHAPTER8. WORKINGWITHNUMBERS This problem can be remedied by checking every new quotient remainder pair against a list of quotient-remainder pairs, to spot a repetition. The main function decForm produces the integer part, and relegates the task of calculating the lists of non-repeating and repeating decimals to an auxiliary function decF. decForm :: Rational -> (Integer,[Int],[Int]) decForm x | x < 0 = error "negative argument" | otherwise = (q,ys,zs) where (q,r) n d (ys,zs) = decF (r*10) d [] = quotRem n d = numerator x = denominator x The function decF has a parameter for the list of quotient remainder pairs that have to be checked for repetition. The code for dForm uses elemIndex from the module List.hs to find the index of the first repeating digit, and splitAt to split a list at an index. decF :: Integer -> Integer -> [(Int,Integer)] -> ([Int],[Int]) decF n d xs | r == 0 = (reverse (q: (map fst xs)),[]) | elem (q,r) xs = (ys,zs) | otherwise = decF (r*10) d ((q,r):xs) where (q’,r) q xs’ Just k (ys,zs) = splitAt k (map fst xs’) = quotRem n d = toInt q’ = reverse xs = elemIndex (q,r) xs’ Here are a few examples: WWN> decForm (133 % 999) (0,[],[1,3,3]) WWN> decForm (1 % 7) (0,[],[1,4,2,8,5,7]) WWN> decForm (2 % 7) 8.7. IRRATIONALNUMBERS 309 (0,[],[2,8,5,7,1,4]) WWN> decForm (3 % 7) (0,[],[4,2,8,5,7,1]) WWN> decForm (4 % 7) (0,[],[5,7,1,4,2,8]) WWN> decForm (5 % 7) (0,[],[7,1,4,2,8,5]) WWN> decForm (6 % 7) (0,[],[8,5,7,1,4,2]) There is no upper limit to the length of the period. Here is an example with a period length of 99: WWN> decForm (468 % 199) (2,[],[3,5,1,7,5,8,7,9,3,9,6,9,8,4,9,2,4,6,2,3,1,1,5,5,7,7,8,8,9,4,4, 7,2,3,6,1,8,0,9,0,4,5,2,2,6,1,3,0,6,5,3,2,6,6,3,3,1,6,5,8,2,9,1,4,5, 7,2,8,6,4,3,2,1,6,0,8,0,4,0,2,0,1,0,0,5,0,2,5,1,2,5,6,2,8,1,4,0,7,0]) Exercise 8.13 Write a Haskell program to find the longest period that occurs in decimal expansions of fractions with numerator and denominator taken from the set 1; : : : ; 999 f . g 8.7 Irrational Numbers The Ancients discovered to their dismay that with just ruler and compass it is possible to construct line segments whose lengths do not form rational fractions. N In other words, it is possible to construct a length q such that for no m; n it holds that q = m n . Here is the famous theorem from Antiquity stating this disturbing fact, with its proof. 2 @ @ @ 1 @ @ x @ 1 @ @ @ 310 CHAPTER8. WORKINGWITHNUMBERS Theorem 8.14 There is no rational number x with x2 = 2. 2 Proof. Assume there is a number x n lowest form, i.e., there are no k; p; q Z, = 0 with (m=n)2 = 2. We can further assume that m=n is canceled down to its Z with k Q with x2 = 2. Then there are m; n = 1, m = kp and n = kq. 2 2 We have: 2 = (m=n)2 = m2=n2, and multiplying both sides by n2 we find 2n2 = m2. In other words, m2 is even, and since squares of odd numbers are always odd, m must be even, i.e., there is a p with m = 2p. Substitution in 2n2 = m2 gives 2n2 = (2p)2 = 4p2, and we find that n2 = 2p2, which leads to the conclusion that n is also even. But this means that there is a q with n = 2q, and we have a contradiction with the assumption that m=n was in lowest form. It Q with x2 = 2. The square root of 2 is not follows that there is no number x rational. 2 Of course, we all use p2 for the square root of 2. The theorem tells us that p2 = 2 Q. The collection of numbers that p2 does belong to is called the collection of real numbers, R. It is possible to give a formal construction of the reals from the rationals, but we will not do so here. Instead, we just mention that Q R, and we informally introduce the reals as the set of all signed (finite or infinite) decimal expansions. The domain of real numbers is closed under the four operations of addition, subtraction, multiplication and division, i.e., just like the rationals, the reals form a field. (cid:18) Exercise 8.15 Use the method from the proof of Theorem 8.14 to show that p3 is irrational. Exercise 8.16 Show that if p is prime, then pp is irrational. Exercise 8.17 Show that if n is a natural number with pn not a natural number, then pn is irrational. Using the fundamental theorem of arithmetic, we can give the following alternative proof of Theorem 8.14: If p2 = (p=q), then 2q2 = p2. In the representation of p2 as a product Proof. of prime factors, every prime factor has an even number of occurrences (for the square of p equals the product of the squares of p’s prime factors). In the represen- tation of 2q2 as a product of prime factors, the prime factor 2 has an odd number of occurrences. Contradiction with Theorem 8.6. 6 6 8.7. IRRATIONALNUMBERS 311 In Section 8.8 we will discuss an algorithm, the mechanic’s rule, for approaching the square root of any positive fraction p with arbitrary precision. The irrational numbers are defined here informally as the infinite non-periodic decimal expansions. This is not a very neat definition, for its relies on the use of decimals, and there is nothing special about using the number ten as a basis for doing calculations. Other bases would serve just as well. In any case, looking at Q and R in terms of expansions provides a neat perspective on the relation between rational and irrational numbers. What, e.g., are the odds against constructing a rational by means of a process of tossing coins to get an infinite binary expansion? Exercise 8.18 Does it follow from the fact that x + y is rational that x is rational or y is rational? If so, give a proof, if not, give a refutation. We have seen that writing out the decimal expansion of a real number like p2 does not give a finite representation. In fact, since there are uncountably many reals (see Chapter 11), no finite representation scheme for arbitrary reals is possible. In implementations it is customary to use floating point representation (or: scien- tific representation) of approximations of reals. E.g., the decimal fraction 1424213:56 106, and gets represented as 1:42421356E + 5. The is an approximation of p2 decimal fraction 0:00000142421356 is an approximation of p2 6, and gets represented as 1:42421356E 6. The general form is x:xxxEm, where x:xxx is a decimal fraction called the mantissa and m is an integer called the exponent. Here are the Hugs versions: 106 = p2 10(cid:0) (cid:2) (cid:2) (cid:0) Prelude> sqrt 2 * 10^6 1.41421e+06 Prelude> sqrt 2 / 10^6 1.41421e-06 Haskell has a predefined type Float for single precision floating point numbers, and a type Double for double precision floating point numbers. Together these two types form the class Floating: Prelude> :t sqrt 2 sqrt 2 :: Floating a => a Floating point numbers are stored as pairs (m; n), where m is the matrix and n the exponent of the base used for the encoding. If x is a floating point number, the base of its representation is given by floatRadix and its matrix and exponent, as a value of type (Integer,Int), by decodeFloat. Thus, if floatRadix x equals bn. floatDigits b and decodeFloat x equals (m; n), then x is the number m gives the number of digits of m in base b representation. In particular, if (m; n) (cid:1) 312 CHAPTER8. WORKINGWITHNUMBERS is the value of decodeFloat x, and d the value of floatDigits, then either m and n are both zero, or bd 1 6 m < bd. Here is an example: (cid:0) Prelude> floatRadix (sqrt 2) 2 Prelude> decodeFloat (sqrt 2) (11863283,-23) Prelude> 11863283 * 2^^(-23) 1.41421 Prelude> floatDigits (sqrt 2) 24 Prelude> 2^23 <= 11863283 && 11863283 < 2^24 True The inverse to the function decodeFloat is encodeFloat: Prelude> sqrt 2 1.41421 Prelude> encodeFloat 11863283 (-23) 1.41421 Scaling a floating point number is done by ‘moving the point’: Prelude> scaleFloat 4 (sqrt 2) 22.6274 Prelude> 2^4 * sqrt 2 22.6274 A floating point number can always be scaled in such a way that its matrix is in the interval ( 1; 1). The matrix in this representation is given by the function significand, the exponent by exponent: (cid:0) Prelude> significand (sqrt 2) 0.707107 Prelude> exponent (sqrt 2) 1 Prelude> 0.707107 1.41421 Prelude> scaleFloat 1 0.707107 1.41421 * 2^1 The definitions of exponent, significand and scaleFloat (from the Prelude): 8.8. THEMECHANIC’SRULE 313 exponent x = if m==0 then 0 else n + floatDigits x significand x = encodeFloat m (- floatDigits x) where (m,n) = decodeFloat x scaleFloat k x = encodeFloat m (n+k) where (m,_) = decodeFloat x where (m,n) = decodeFloat x 8.8 The Mechanic’s Rule Sequences of fractions can be used to find approximations to real numbers that themselves are not fractions (see Section 8.7). A well known algorithm for gen- erating such sequences is the so-called mechanic’s rule (also known as Newton’s method, a bit misleadingly, for the algorithm was already in use centuries before Newton): p > 0; a0 > 0; an+1 = 1 2 (an + p an ): In Exercise 8.19 you are asked to prove that this can be used to approximate the square root of any positive fraction p to any degree of accuracy. The Haskell im- plementation uses some fresh ingredients. The function recip takes the reciprocal of a fraction, the operation iterate iterates a function by applying it again to the result of the previous application, and takeWhile takes a property and a list and constructs the largest prefix of the list consisting of objects satisfying the property. In the present case, the property is (\ m -> m^2 <= p), having a square 6 p, and the list is the list of positive naturals. You should look up the implementations in Prelude.hs and make sure you understand. mechanicsRule :: Rational -> Rational -> Rational mechanicsRule p x = (1 % 2) * (x + (p * (recip x))) mechanics :: Rational -> Rational -> [Rational] mechanics p x = iterate (mechanicsRule p) x 314 CHAPTER8. WORKINGWITHNUMBERS sqrtM :: Rational -> [Rational] sqrtM p | p < 0 = error "negative argument" | otherwise = mechanics p s where s = if xs == [] then 1 else last xs xs = takeWhile (\ m -> m^2 <= p) [1..] As a demonstration, here are the first seven steps in the approximation to p2, the first seven steps in the approximation to p4, and the first, second, . . . , sixth step in the approximation to p50, respectively. This already gives greater accuracy than is needed in most applications; the algorithm converges very fast. WWN> take 5 (sqrtM 2) [1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832] WWN> take 7 (sqrtM 4) [2 % 1,2 % 1,2 % 1,2 % 1,2 % 1,2 % 1,2 % 1] WWN> sqrtM 50 !! 0 7 % 1 WWN> sqrtM 50 !! 1 99 % 14 WWN> sqrtM 50 !! 2 19601 % 2772 WWN> sqrtM 50 !! 3 768398401 % 108667944 WWN> sqrtM 50 !! 4 1180872205318713601 % 167000548819115088 Exercise 8.19 1. Prove that for every n, an+1 (cid:0) pp an+1 + pp = pp)2 (an (cid:0) (an + pp)2 : 2. From the first item it follows (by induction on n) that pp an+1 (cid:0) an+1 + pp = pp a0 (cid:0) a0 + pp (cid:19) (cid:18) n 2 : Derive from this that sqrtM p converges to pp for any positive rational number p. 8.9. REASONINGABOUTREALS 315 3. Show that the approximation is from above, i.e., show that n > 1 implies that an > pp. Exercise 8.20 1. Find a rule to estimate the number of correct decimal places try pp 6 t, with t an expression that in approximation an of pp. Use the result of Exercise 8.19. to find an inequality of the form an employs a0 and a1.) (Hint: (cid:0) 2. Use the previous item to give an estimate of the number of correct decimal places in the successive approximations to p2. 8.9 Reasoning about Reals Suppose one wants to describe the behaviour of moving objects by plotting their position or speed as a function of time. Moving objects do not suddenly disappear and reappear somewhere else (outside the Bermuda triangle, at least), so the path of a moving object does not have holes or gaps in it. This is where the notions of continuity and limit arise naturally. Analysis, with its emphasis on continuity and limit, is a rich source of examples where skill in quantifier reasoning is called for. The following Lemma illustrates that many common functions are continuous. Its proof is given as an example: it is a nice illustration of the use of the logical rules. In fact, logic is all there is to this proof: properties of real numbers are not needed at all. Lemma. The composition of two continuous functions is continuous. I.e., if f and g are continuous functions from reals to reals, then the function h defined by h(x) = g(f (x)) (cf. Definition 6.29 p. 222) is continuous as well. Proof. Given: f and g are continuous, i.e., (“"-(cid:14)-definition” p. 66; for clarity we use different variables for the arguments): x 8 " > 0 8 (cid:14) > 0 9 a 8 " > 0 8 (cid:14) > 0 9 8 x y ( j a b ( j y j b j (cid:0) (cid:0) 8 < (cid:14) = f (x) ) j < (cid:14) = g(a) ) j < "); f (y) j < "): g(b) j (cid:0) (cid:0) (8.1) (8.2) To be proved: x 8 " > 0 8 (cid:14) > 0 9 x y ( j 8 y j (cid:0) < (cid:14) = g(f (x)) ) j g(f (y)) < "). j (cid:0) 316 CHAPTER8. WORKINGWITHNUMBERS Proof (detailed version): Note that what is To be proved begins with two universal " > 0. Therefore, the proof has to start (recall the obligatory quantifiers 8 R opening that goes with -introduction!) with choosing two arbitrary values x and " > 0. We now have to show that x and 2 8 8 (cid:14) > 0 9 x y ( j 8 y j (cid:0) < (cid:14) = g(f (x)) ) j g(f (y)) j (cid:0) < "): (8.3) The next quantifier asks us to supply some example-(cid:14) that is > 0. There appears to be no immediate way to reduce the proof problem further, so we start looking at the givens in order to obtain such an example. It turns out that we have to use the second given (8.2) first. It can be used ( - 8 elimination) if we specify values for a and ". Later on it will show that a = f (x) and the earlier " will turn out to be useful. Thus, (8.2) delivers ( -elimination) some (cid:14)1 > 0 such that 9 f (x) b ( j 8 b j (cid:0) < (cid:14)1 = g(f (x)) ) j g(b) j (cid:0) < "): (8.4) Applying the given (8.1) to our x and " = (cid:14)1 ( elimination) (cid:14) > 0 such that -elimination) we obtain ( 8 - 9 x y ( j 8 y j (cid:0) < (cid:14) = f (x) ) j f (y) j (cid:0) < (cid:14)1): (8.5) And this is the example-(cid:14) we are looking for, i.e.: y 8 (cid:0) x y ( j Claim: < (cid:14) = (From this follows what we have to show using Proof: Suppose that (to prepare for ercise 3.18 p. 93) y is such that g(f (x)) ) j (cid:0) j 9 g(f (y)) j -introduction.) < "). -introduction and Deduction Rule — cf. Ex- 8 y (cid:0) < (cid:14). j x j From (8.5) (using this y — f (y) j < (cid:14)1. -elimination and Modus Ponens) you find, that 8 f (x) j (cid:0) Finally, from (8.4) (with b = f (y) — g(f (y)) j < ". The Claim follows. -elimination) you find that 8 g(f (x)) j (cid:0) Making up the score: the proof applies one rule for each of the fifteen (!) occur- rences of logical symbols in Given and To be proved. Make sure you understand every detail. Of course, the amount of detail is excessive and there is a more common concise version as well. The following version is the kind of argument you will find in 8.9. REASONINGABOUTREALS 317 analysis textbooks. As is usual, it leaves to the reader to fill in which rules have been applied, and where. Proof. Assume that x R and " > 0. 2 From (8.2), obtain (cid:14)1 > 0 such that f (x) b ( j 8 b j (cid:0) < (cid:14)1 = g(f (x)) ) j g(b) j (cid:0) < "): Applying (8.1) to x and (cid:14)1, obtain (cid:14) > 0 such that x y ( j 8 y j (cid:0) < (cid:14) = f (x) ) j f (y) j (cid:0) < (cid:14)1): (8.6) (8.7) x j Then if y (cid:0) follows that g(f (x)) j < (cid:14), by (8.7) we get that g(f (y)) j < ". (cid:0) j f (x) j f (y) j (cid:0) < (cid:14)1, and from (8.6) it This version of the proof will be considered very complete by every mathemati- cian. Nevertheless, the compression attained is approximately 4 : 1. The remaining examples of this section are about sequences of reals. Limits. Assume that a0; a1; a2; : : : is a sequence of reals and that a expression limi !1 quence”) by definition means that R. The ai = a (“the sequence converges to a”, “a is limit of the se- 2 " > 0 8 n 9 i > n ( a j 8 ai j (cid:0) < "): Generating concrete examples of converging sequences in Haskell is easy, as we have seen in Section 8.8. We have seen that the sequences produced by the me- chanic’s rule converge. We will prove some results about convergence. Theorem. Every sequence of reals has at most one limit. Proof. The situation should be analyzed as follows. !1 ai = b. ai = a, limi Given: limi To be proved: a = b. Proof: Proof by Contradiction is not a bad idea here, since the new given it pro- vides, nl., that a > 0. Thus, assume this. Proof by Contradiction now asks to look for something false. = b, is equivalent with the positive a j b j !1 (cid:0) In order to use the old given, you need to choose ( -elimination!) values for ". As 8 you’ll see later, it is useful to choose " = 1 . Note that " > 0. b a 2 j j (cid:0) 6 318 CHAPTER8. WORKINGWITHNUMBERS From the given limi !1 ai = a we obtain now, that n Thus ( -elimination) some n1 exists such that i > n ( a j ai < "). j 9 8 i > n1 ( a j 8 ai j (cid:0) (cid:0) < "). 9 ai = b we obtain, similarly, some n2 such that i > 8 From the given limi ai b n2( j < "). (cid:0) j !1 -elimination we now get from Define n = max(n1; n2). Since n > n1; n2, by these facts that both < " and 8 < ". an an a j (cid:0) j b j (cid:0) j Lastly, using the triangle-inequality it follows that (cid:0) — and this is the falsity looked for. a j a j b j (cid:0) = x + y j j an + an 6 (cid:0) ; j + x j j 6 b j y j a j an + b j j (cid:0) an j < 2" = a j (cid:0) b j (cid:0) Exercise 8.21 Write a concise version of the above proof. Exercise 8.22 Assume that limi !1 ai = a. 1. Show that limi !1 2. Assume that f : N Show that limi !1 a2i = a. N is a function such that ! af (i) = a. n m 8 9 8 i > m f (i) > n. Exercise 8.23 Assume that the sequences of reals limits a resp. b, and that a < b. Show that a number n exists such that n (am < bm). 1n=0 and an f bn f g g 1n=0 have m > 8 Exercise 8.24 Assume that limi Show that limi !1 (ai + bi) = a + b. !1 ai = a and that limi bi = b. !1 Exercise 8.25 Show that a function f : R f (a) whenever limi ai = a. !1 R is continuous iff limi !1 f (ai) = ! Cauchy. A sequence of reals " > 0 8 an f n 9 g i; j > n ( ai j 8 1n=0 is called Cauchy if aj j (cid:0) < "): 8.10. COMPLEXNUMBERS 319 Exercise 8.26 Assume that the sequence an f g 1n=0 is Cauchy. 1. Show that the sequence is bounded. I.e., that numbers b and c exist such that i (b < ai < c). 8 2. Assume that a R is such that < "). (The " > 0 existence of such an a follows from the sequence being bounded, but you are not asked to prove this.) Show that limi i > n ( a j ai = a. ai (cid:0) 2 n 8 9 8 j !1 It follows immediately from Exercise 8.19 that the sequences of rationals produced by the Mechanic’s rule are Cauchy. Thanks to that we can implement a program for calculating square roots as follows (for still greater precision, change the value of " in apprx): approximate :: Rational -> [Rational] -> Rational approximate eps (x:y:zs) | abs (y-x) < eps = y | otherwise = approximate eps (y:zs) apprx :: [Rational] -> Rational apprx = approximate (1/10^6) mySqrt :: Rational -> Rational mySqrt p = apprx (sqrtM p) Exercise 8.27 Just as we defined the integers from the naturals and the rationals from the integers by means of quotient sets generated from suitable equivalence classes, we can do so now, by defining the set of real numbers R as the set of all Cauchy sequences in Q modulo an appropriate equivalence relation. Define that equivalence relation and show that it is indeed an equivalence. 8.10 Complex Numbers In the domain of rational numbers we cannot solve the equation x2 in the domain of real numbers we can: its roots are x = p2 and x = 2 = 0, but p2. What (cid:0) (cid:0) 320 CHAPTER8. WORKINGWITHNUMBERS (cid:0) about solving x2 + 1 = 0? There are no real number solutions, for the square root of 1 does not exist in the realm of real numbers. The field of real numbers is not closed under the operation of taking square roots. To remedy this, we follow the by now familiar recipe of extending the number domain, by introducing numbers of a new kind. We extend the domain of reals to a domain of complex numbers, C, by introducing an entity i called ‘the imaginary unit’, and postulating i2 = 1. (cid:0) We do not want to lose closure under addition, subtraction, multiplication and division, so we should be able to make sense of p2 + i, 2i, i, and so on. In general, we want to be able to make sense of x + iy, where x and y are arbitrary real numbers, and we need rules for adding and multiplying complex numbers, in such a way that the laws of commutativity and associativity of + and , and the over + continue to hold. x is called the real part of x + iy, law of distribution of iy its imaginary part. Adding complex numbers boils down to adding real and imaginary parts separately: (cid:0) (cid:1) (cid:1) (x + iy) + (u + iw) = (x + u) + i(y + w): This also gives a recipe for subtraction: (x + iy) (u + iw) = (x + iy) + ( u + iw) = (x (cid:0) For multiplication, we use the fact that i2 = (cid:0) (cid:0) 1: (cid:0) u) + i(y w): (cid:0) (cid:0) (x + iy)(u + iw) = xu + iyu + ixw + i2yw = (xu yw) + i(yu + xw): (cid:0) Division uses the fact that (x + iy)(x iy) = x2 + y2. It is given by: x + iy u + iw = x + iy u + iw (cid:1) u u = xu + yw u2 + w2 + i yu xw u2 + w2 : (cid:0) (cid:0) iw iw (cid:0) (cid:0) We see that like the rationals and the reals, the complex numbers are closed under the four operations addition, subtraction, multiplication and division, so C is a field. Moreover, any real number a can be viewed as a complex number of the form a + 0i, so we have that R C. (cid:0) i. In general, solving the equation xn + an Solving the equation x2 + 1 = 0 in the domain C gives two roots, viz., x = i and 1 + : : : + a1x + a0 = 0, x = (cid:0) where a0; : : : ; an 1 may be either real or complex, gives n complex roots, for the fundamental theorem of algebra (which we will not prove here) states that every polynomial of degree n, 1xn (cid:0) (cid:0) (cid:18) 1xn f (x) = xn + an 1 + : : : + a1x + a0; (cid:0) (cid:0) can be factored into a product of exactly n factors, (x (cid:0) b1)(x b2) (x (cid:0) (cid:1) (cid:1) (cid:1) (cid:0) bn): 8.10. COMPLEXNUMBERS 321 Exercise 8.28 Check that the commutative and associative laws and the distribu- tive law hold for C. There is a standard Haskell module Complex.hs with an implementation of com- plex numbers. infix 6 :+ data (RealFloat a) => Complex a = !a :+ !a deriving (Eq,Read,Show) The exclamation marks in the typing !a :+ !a indicate that the real and imagi- nary parts of type RealFloat are evaluated in a strict way. The real part of a complex number x + iy is the real number x, the imaginary part the real number y. Notation for the real part of z: Re(z). Notation for the imaginary part of z: Im(z). The Haskell implementations are: realPart, imagPart :: (RealFloat a) => Complex a -> a realPart (x:+y) imagPart (x:+y) = x = y The complex number z = x + iy can be represented geometrically in either of two ways: 1. Associate with z the point with coordinates (x; y) in the plane R2. In this way, we view the plane R2 as the complex plane. 2. Associate with z the two-dimensional vector with components x and y. Think of this vector as a free vector, i.e., a vector that may be moved around freely as long as its direction remains unchanged. The two representations can be combined by attaching the vector z to the origin of the plane. We then get the picture of Figure 8.7. Call the horizontal axis through the origin the real axis and the vertical axis through the origin the imaginary axis. The conjugate of a complex number z = x + iy is the number (cid:22)z = x iy. Its geometrical representation is the reflection of z in the real axis (see again Figure 8.7). Its implementation is given by: (cid:0) 322 CHAPTER8. WORKINGWITHNUMBERS iy i i (cid:0) 1 (cid:0) (cid:18) 1 z x (cid:22)z Figure 8.7: Geometrical Representation of Complex Numbers. conjugate :: (RealFloat a) => Complex a -> Complex a = x :+ (-y) conjugate (x:+y) The magnitude or modulus or absolute value of a complex number z is the length r x2 + y2. of the z vector. Notation . The magnitude of z = x + iy is given by z j j Its Haskell implementation: p magnitude :: (RealFloat a) => Complex a -> a magnitude (x:+y) = scaleFloat k (sqrt ((scaleFloat mk x)^2 + (scaleFloat mk y)^2)) where k = max (exponent x) (exponent y) mk = - k The phase or argument of a complex number z is the angle (cid:18) of the vector z. Notation: arg(z). The phase (cid:18) of a vector of magnitude 1 is given by the vec- tor cos((cid:18)) + i sin((cid:18)) (see Figure 8.8), so the phase of z = x + iy is given by arctan y x + (cid:25) for x < 0. In case x = 0, the phase is x for x > 0, and arctan y 8.10. COMPLEXNUMBERS 323 iy i sin((cid:18)) (cid:18) cos((cid:18)) z x Figure 8.8: \(cid:18) = cos((cid:18)) + i sin((cid:18)). 324 CHAPTER8. WORKINGWITHNUMBERS 0. This computation is taken care of by the Haskell function atan2. Here is the implementation: phase phase (0:+0) phase (x:+y) :: (RealFloat a) => Complex a -> a = 0 = atan2 y x The polar representation of a complex number z = x + iy is r\(cid:18), where r = z j and (cid:18) = arg(z). The advantage of polar representation is that complex multiplica- tion looks much more natural: just multiply the magnitudes and add the phases. j (R\’)(r\(cid:18)) = Rr\(’ + (cid:18)): The polar representation of a complex number is given by: polar polar z :: (RealFloat a) => Complex a -> (a,a) = (magnitude z, phase z) Note that polar representations are not unique, for we have: : : : = r\((cid:18) (cid:0) 2(cid:25)) = r\(cid:18) = r\((cid:18) + 2(cid:25)) = r\((cid:18) + 4(cid:25)) = : : : To get from representation r\(cid:18) to the representation as a vector sum of real and imaginary parts, use r\(cid:18) = r(cos((cid:18)) + i sin((cid:18))) (see again Figure 8.8). Imple- mentation: mkPolar mkPolar r theta :: (RealFloat a) => a -> a -> Complex a = r * cos theta :+ r * sin theta Converting a phase (cid:18) to a vector in the unit circle is done by: cis cis theta :: (RealFloat a) => a -> Complex a = cos theta :+ sin theta 8.10. COMPLEXNUMBERS 325 The implementations of the arithmetical operations of addition, subtraction, mul- tiplication, negation, absolute value and signum are given by: instance (RealFloat a) => Num (Complex a) where (x:+y) + (x’:+y’) (x:+y) - (x’:+y’) (x:+y) * (x’:+y’) negate (x:+y) abs z signum 0 signum z@(x:+y) = (x+x’) :+ (y+y’) = (x-x’) :+ (y-y’) = (x*x’-y*y’) :+ (x*y’+y*x’) = negate x :+ negate y = magnitude z :+ 0 = 0 = x/r :+ y/r where r = magnitude z Note that the signum of a complex number z is in fact the vector representation sign for of arg(z). This suggests that arg(z) is a generalization of the + or real numbers. That this is indeed the case can be seen when we perform complex multiplication on real numbers: positive real numbers, represented in the complex plane, have phase 0, or in general, phase 2k(cid:25). Negative real number, represented in the complex plane, have phase (cid:25), or in general, (2k + 1)(cid:25). Multiplying two negative real numbers means multiplying their values and adding their phases, so we get phase 2(cid:25), which is the same as phase 0, modulo 2k(cid:25). (cid:0) Complex division is the inverse of multiplication, so it boils down to performing division on the magnitudes, and subtraction on the phases. The implementation in terms of the vector representations looks slightly more involved than this, however: instance (RealFloat a) => Fractional (Complex a) where (x:+y) / (x’:+y’) = (x*x’’+y*y’’) / d :+ (y*x’’-x*y’’) / d where x’’ = scaleFloat k x’ y’’ = scaleFloat k y’ k d = - max (exponent x’) (exponent y’) = x’*x’’ + y’*y’’ For the definition of further numerical operations on complex numbers we refer to the library file Complex.hs. Complex numbers lose their mystery when one gets well acquainted with their geometrical representations. Here are some examples, for getting the feel of them. The number 1 + i has magnitude p2 and phase (cid:25) 4 : see Figure 8.9. Squaring 326 CHAPTER8. WORKINGWITHNUMBERS i p2 (cid:25) 4 1 Figure 8.9: The number 1 + i. i (cid:25) 2 Figure 8.10: The number (1 + i)2. 8.10. COMPLEXNUMBERS 327 2p2 3(cid:25) 4 Figure 8.11: The number (1 + i)3. 328 CHAPTER8. WORKINGWITHNUMBERS this number involves squaring the magnitude and doubling the phase, so (1 + i)2 has magnitude 2 and phase (cid:25) 2 : see Figure 8.10. Raising 1 + i to the third power 4 and (cid:25) involves multiplying the magnitudes 2 and p2 and adding the phases (cid:25) 2 . This gives magnitude 2p2 and phase 3(cid:25) 4 . See Figure 8.11 for a picture of the number (1+i)3. Raising 1+i to the fourth power involves squaring (1+i)2, so the magnitude 2 is squared and the phase (cid:25) 2 is doubled, which gives magnitude 4 and phase (cid:25). Translating all of this back into vector sum notation, we get (1+i)2 = 2i, (1 + i)3 = 4. And sure enough, the Haskell library Complex.hs confirms these findings: 2 + 2i, and (1 + i)4 = (cid:0) (cid:0) Complex> (1 :+ 1)^2 0.0 :+ 2.0 Complex> (1 :+ 1)^3 (-2.0) :+ 2.0 Complex> (1 :+ 1)^4 (-4.0) :+ 0.0 Similarly, we see that multiplying i and 1 and adding the phases (cid:25) phase 2(cid:25) = 0, i.e., the number 1. Here is the Haskell confirmation: i involves multiplying the magnitudes 2 , with result the number with magnitude 1 and 2 and 3(cid:25) (cid:0) Complex> (0 :+ 1) * (0 :+ (-1)) 1.0 :+ 0.0 Exercise 8.29 You are encouraged to familiarize yourself further with complex numbers by checking the following by means of pictures: 1. Re(z) = 1 2 (z + (cid:22)z). (cid:22)z). 2i (z 2. Im(z) = 1 (cid:0) 3. tan(arg(z)) = Im(z) Re(z) . 4. R\’ r\(cid:18) = R (cid:18)). r \(’ ) = arg(z1) (cid:0) 5. arg( z1 z2 6. arg( 1 z ) = arg(z). (cid:0) arg(z2). (cid:0) Exercise 8.30 1. Use induction on n to prove De Moivre’s formula for n (cos(’) + i sin(’))n = cos(n’) + i sin(n’): N: 2 Draw a picture to see what is happening! 8.11. FURTHERREADING 329 2. Prove De Moivre’s formula for exponents in Z by using the previous item, plus: (cos(’) + i sin(’))(cid:0) m = 1 (cos(’) + i sin(’))m : 8.11 Further Reading A classic overview of the ideas and methods of mathematics, beautifully written, and a book everyone with an interest in mathematics should possess and read is Courant and Robbins [CR78, CrbIS96]. Here is praise for this book by Albert Einstein: A lucid representation of the fundamental concepts and methods of the whole field of mathematics. It is an easily understandable intro- duction for the layman and helps to give the mathematical student a general view of the basic principles and methods. Another beautiful book of numbers is [CG96]. For number theory and its history see [Ore88]. 330 CHAPTER8. WORKINGWITHNUMBERS Chapter 9 Polynomials Preview Polynomials or integral rational functions are functions that can be represented by a finite number of additions, subtractions, and multiplications with one indepen- dent variable. The closed forms that we found and proved by induction for the sums of evens, sums of odds, sums of squares, sums of cubes, and so on, in Chap- ter 7, are all polynomial functions. In this chapter we will first study the process of automating the search for polynomial functions that produce given sequences of integers. Next, we establish the connection with the binomial theorem, we im- plement a datatype for the polynomials themselves, and we use this datatype for the study of combinatorial problems. module POL where import Polynomials 331 332 CHAPTER9. POLYNOMIALS 9.1 Difference Analysis of Polynomial Sequences is a sequence of natural numbers, i.e., f = (cid:21)n:an is a function in N. The function f is a polynomial function of degree k if f can be presented an f g Suppose N in the form ! cknk + ck 1nk (cid:0) 1 + (cid:0) (cid:1) (cid:1) (cid:1) + c1n + c0; with ci 2 Q and ck = 0. Example 9.1 The sequence [1; 4; 11; 22; 37; 56; 79; 106; 137; 172; 211; 254; 301; 352; : : :] is given by the polynomial function f = (cid:21)n:(2n2 + n + 1). This is a function of the second degree. Here is the Haskell check: Prelude> take 15 (map (\ n -> 2*n^2 + n + 1) [0..]) [1,4,11,22,37,56,79,106,137,172,211,254,301,352,407] Consider the difference sequence given by the function The Haskell implementation looks like this: d(f ) = (cid:21)n:an+1 (cid:0) an: difs :: [Integer] -> [Integer] difs [] = [] difs [n] = [] difs (n:m:ks) = m-n : difs (m:ks) This gives: POL> difs [1,4,11,22,37,56,79,106,137,172,211,254,301] [3,7,11,15,19,23,27,31,35,39,43,47] 6 9.1. DIFFERENCEANALYSISOFPOLYNOMIALSEQUENCES 333 The difference function d(f ) of a polynomial function f is itself a polynomial function. E.g., if f = (cid:21)n:(2n2 + n + 1), then: d(f ) = (cid:21)n:(2(n + 1)2 + (n + 1) + 1 = (cid:21)n:4n + 3: (2n2 + n + 1) (cid:0) The Haskell check: POL> take 15 (map (\n -> 4*n + 3) [0..]) [3,7,11,15,19,23,27,31,35,39,43,47,51,55,59] POL> take 15 (difs (map (\ n -> 2*n^2 + n + 1) [0..])) [3,7,11,15,19,23,27,31,35,39,43,47,51,55,59] Proposition 9.2 If f is a polynomial function of degree k then d(f ) is a polyno- mial function of degree k 1. (cid:0) Proof. Suppose f (n) is given by cknk + ck 1nk (cid:0) 1 + (cid:0) (cid:1) (cid:1) (cid:1) + c1n + c0: Then d(f )(n) is given by + c1(n + 1) + c0 ck(n + 1)k+ck 1(n + 1)k (cid:0) (cid:0) (cknk + ck 1 + 1nk (cid:0) It is not hard to see that f (n + 1) has the form cknk + g(n), with g a polynomial 1. Since f (n) also is of the form cknk + h(n), with h a polynomial of degree k 1, d(f )(n) has the form g(n) of degree k h(n), so d(f ) is itself a polynomial 1. of degree k + c1n + c0): (cid:1) (cid:1) (cid:1) (cid:0) 1 + (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) It follows from Proposition 9.2 that if f is a polynomial function of degree k, then dk(f ) will be a constant function (a polynomial function of degree 0). Here is a concrete example of computing difference sequences until we hit at a constant sequence: -12 1 -11 16 6 22 45 28 39 6 67 6 17 6 112 34 101 6 213 40 141 6 354 46 541 187 We find that the sequence of third differences is constant, which means that the form of the original sequence is a polynomial of degree 3. To find the next number 334 CHAPTER9. POLYNOMIALS in the sequence, just take the sum of the last elements of the rows. This gives 6 + 46 + 187 + 541 = 780. Charles Babbage (1791–1871), one of the founding fathers of computer science, used these observations in the design of his difference engine. We will give a Haskell version of the machine. According to Proposition 9.2, if a given input list has a polynomial form of degree k, then after k steps of taking differences the list is reduced to a constant list: POL> difs [-12,-11,6,45,112,213,354,541,780,1077] [1,17,39,67,101,141,187,239,297] POL> difs [1,17,39,67,101,141,187,239,297] [16,22,28,34,40,46,52,58] POL> difs [16,22,28,34,40,46,52,58] [6,6,6,6,6,6,6] The following function keeps generating difference lists until the differences get constant: difLists :: [[Integer]]->[[Integer]] difLists [] = [] difLists lists@(xs:xss) = if constant xs then lists else difLists ((difs xs):lists) where constant (n:m:ms) = all (==n) (m:ms) constant = error "lack of data or not a polynomial fct" _ This gives the lists of all the difference lists that were generated from the initial sequence, with the constant list upfront. POL> difLists [[-12,-11,6,45,112,213,354,541,780,1077]] [[6,6,6,6,6,6,6], [16,22,28,34,40,46,52,58], [1,17,39,67,101,141,187,239,297], [-12,-11,6,45,112,213,354,541,780,1077]] The list of differences can be used to generate the next element of the original sequence: just add the last elements of all the difference lists to the last element of the original sequence. In our example case, to get the next element of the list 12; [ (cid:0) 11; 6; 45; 112; 213; 354; 541; 780; 1077] (cid:0) 9.1. DIFFERENCEANALYSISOFPOLYNOMIALSEQUENCES 335 add the list of last elements of the difference lists (including the original list): 6 + 58 + 297 + 1077 = 1438. To see that this is indeed the next element, note that the difference of 1438 and 1077 is 361, the difference of 361 and 297 is 64, and the difference of 64 and 58 is 6, so the number 1438 ‘fits’ the difference analysis. The following function gets the list of last elements that we need (in our example case, the list [6,58,297,1077]): genDifs :: [Integer] -> [Integer] genDifs xs = map last (difLists [xs]) A new list of last elements of difference lists is computed from the current one by keeping the constant element d1, and replacing each di+1 by di + di+1. nextD :: [Integer] -> [Integer] nextD [] = error "no data" nextD [n] = [n] nextD (n:m:ks) = n : nextD (n+m : ks) The next element of the original sequence is given by the last element of the new list of last elements of difference lists: next :: [Integer] -> Integer next = last . nextD . genDifs In our example case, this gives: POL> next [-12,-11,6,45,112,213,354,541,780,1077] 1438 All this can now be wrapped up in a function that continues any list of polynomial form, provided that enough initial elements are given as data: 336 CHAPTER9. POLYNOMIALS continue :: continue xs = map last (iterate nextD differences) [Integer] -> [Integer] where differences = nextD (genDifs xs) This uses the predefined Haskell function iterate, that is given by: iterate iterate f x :: (a -> a) -> a -> [a] = x : iterate f (f x) This is what we get: POL> take 20 (continue [-12,-11,6,45,112,213,354,541,780,1077]) [1438,1869,2376,2965,3642,4413,5284,6261,7350,8557,9888,11349, 12946,14685,16572,18613,20814,23181,25720,28437] If a given list is generated by a polynomial, then the degree of the polynomial can be computed by difference analysis, as follows: degree :: [Integer] -> Int degree xs = length (difLists [xs]) - 1 The difference engine is smart enough to be able to continue a list of sums of squares, or a list of sums of cubes: POL> take 10 (continue [1,5,14,30,55]) [91,140,204,285,385,506,650,819,1015,1240] POL> take 10 (continue [1,9,36,100,225,441]) [784,1296,2025,3025,4356,6084,8281,11025,14400,18496] Exercise 9.3 What continuation do you get for [3,7,17,39,79,143]? Can you reconstruct the polynomial function that was used to generate the sequence? 9.2. GAUSSIANELIMINATION 337 Difference analysis yields an algorithm for continuing any finite sequence with a polynomial form. Is it also possible to give an algorithm for finding the form? This would solve the problem of how to guess the closed forms for the functions that calculate sums of squares, sums of cubes, and so on. The answer is ‘yes’, and the method is Gaussian elimination. 9.2 Gaussian Elimination If we know that a sequence a0; a1; a2; a3; : : : has a polynomial form of degree 3, then we know that the form is a + bx + cx2 + dx3 (listing the coefficients in increasing order). This means that we can find the form of the polynomial by solving the following quadruple of linear equations in a; b; c; d: a = a0 a + b + c + d = a1 a + 2b + 4c + 8d = a2 a + 3b + 9c + 27d = a3 Since this is a set of four linear equations in four unknowns, where the equations are linearly independent (none of them can be written as a multiple of any of the others), this can be solved by eliminating the unknowns one by one. Example 9.4 Find the appropriate set of equations for the sequence and solve it. 7; [ (cid:0) 2; 15; 50; 109; 198; 323] (cid:0) Difference analysis yields that the sequence is generated by a polynomial of the third degree, so the sequence leads to the following set of equations: a = a + b + c + d = (cid:0) (cid:0) a + 2b + 4c + 8d = 15 a + 3b + 9c + 27d = 50 7 2 Eliminating a and rewriting gives: b + c + d = 5 2b + 4c + 8d = 22 3b + 9c + 27d = 57 338 CHAPTER9. POLYNOMIALS Next, eliminate the summand with factor d from the second and third equation. Elimination from the second equation is done by subtracting the second equation from the 8-fold product of the first equation. This gives 6b + 4c = 18, which can be simplified to 3b + 2c = 9. Elimination from the third equation is done by subtracting the third equation from the 27-fold product of the first, with result 24b + 18c = 78. We get the following pair of equations: 3b + 2c = 9 24b + 18c = 78 Elimination of c from this pair is done by subtracting the second equation from the 9-fold product of the first. This gives 3b = 3, whence b = 1. Together with 3b + 2c = 9 we get c = 3. Together with b + c + d = 5 we get d = 1. Thus, the polynomial we are looking for has the form (cid:21)n:(n3 + 3n2 + n 7). (cid:0) Exercise 9.5 Find the appropriate set of equations for the sequence [13; 21; 35; 55; 81; 113; 151] and solve it. Solving sets of linear equations can be viewed as manipulation of matrices of coefficients. E.g., the quadruple of linear equations in a; b; c; d for a polynomial of the third degree gives the following matrix: a0 1 0 0 a1 1 1 1 1 2 4 a2 1 3 9 27 a3 0 1 8 0 B B @ 1 C C A To solve this, we transform it to an equivalent matrix in so-called echelon form or left triangular form, i.e., a matrix of the form: a00 0 0 0 a01 a02 a03 a11 a12 a13 a22 a23 0 a33 0 0 b0 b1 b2 b3 0 B B @ 1 C C A From this form, compute the value of variable d from the last row, next eliminate this variable from the third row, and find the value of c. Then use the values of d and c to find the value of b from the second row, and finally, use the values of b; c; d to find the value of a from the first row. To handle matrices, the following type declarations are convenient. 9.2. GAUSSIANELIMINATION 339 type Matrix = [Row] type Row = [Integer] It is also convenient to be able to have functions for the numbers of rows and columns of a matrix. rows, cols :: Matrix -> Int rows m = length m cols m | m == [] = 0 | otherwise = length (head m) The function genMatrix produces the appropriate matrix for a list generated by a polynomial: genMatrix :: [Integer] -> Matrix genMatrix xs = zipWith (++) (genM d) [ [x] | x <- xs ] where d genM n = [ [ (toInteger x^m) | m <- [0..n] ] | x <- [0..n] ] = degree xs zipWith is predefined in the Haskell prelude as follows: zipWith zipWith z (a:as) (b:bs) zipWith _ _ _ :: (a->b->c) -> [a]->[b]->[c] = z a b : zipWith z as bs = [] In a picture: [(z x0 y0); (z x1 y1); : : : ; (z xn yn); [xn+1; xn+2; : : : * [yn+1; yn+2; : : : 340 CHAPTER9. POLYNOMIALS genMatrix gives, e.g.: POL> genMatrix [-7,-2,15,50,109,198,323] [[1,0,0,0,-7],[1,1,1,1,-2],[1,2,4,8,15],[1,3,9,27,50]] The process of transforming the matrix to echelon form is done by so-called for- ward elimination: use one row to eliminate the first coefficient from the other rows by means of the following process of adjustment (the first row is used to adjust the second one): adjustWith :: Row -> Row -> Row adjustWith (m:ms) (n:ns) = zipWith (-) (map (n*) ms) (map (m*) ns) To transform a matrix into echelon form, proceed as follows: 1. If the number of rows or the number of columns of the matrix is 0, then the matrix is already in echelon form. 2. If every row of rs begins with a 0 then the echelon form of rs can be found by putting 0’s in front of the echelon form of map tail rs. 3. If rs has rows that do not start with a 0, then take the first one of these, piv, and use it to eliminate the leading coefficients from the other rows. This gives a matrix of the form a00 0 0 0 ... 0 a03 a02 a01 a13 a12 a11 a23 a22 a21 a33 a32 a31 ... ... ... an1 an2 an3 b0 b1 b2 b3 ... bn (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) ... (cid:1) (cid:1) (cid:1) 0 B B B B B B B @ 1 C C C C C C C A where the first row is the pivot row. All that remains to be done in this case is to put the following sub matrix in echelon form: a13 a12 a11 a23 a22 a21 a33 a32 a31 ... ... ... an1 an2 an3 0 B B B B B @ (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) ... (cid:1) (cid:1) (cid:1) b1 b2 b3 ... bn 1 C C C C C A 9.2. GAUSSIANELIMINATION 341 The code for this can be found in the Haskell demo file Matrix.hs (part of the Hugs system): echelon echelon rs :: Matrix -> Matrix | null rs || null (head rs) = rs | null rs2 | otherwise where rs’ = map (0:) (echelon (map tail rs)) = piv : map (0:) (echelon rs’) = map (adjustWith piv) (rs1++rs3) = span leadZero rs (rs1,rs2) leadZero (n:_) = n==0 (piv:rs3) = rs2 Here is an example: POL> echelon [[1,0,0,0,-7],[1,1,1,1,-2],[1,2,4,8,15],[1,3,9,27,50]] [[1,0,0,0,-7],[0,-1,-1,-1,-5],[0,0,-2,-6,-12],[0,0,0,-12,-12]] Backward Gaussian elimination, or computing the values of the variables from a matrix in echelon form, is done by computing the value of the variable in the last row, eliminate that variable from the other rows to get a smaller matrix in echelon If form, and repeating that process until the values of all variables are found. = 0), and the coordinate of x is the we know that ax = c (we may assume a coordinate ain = b of ai1 j (cid:1) (cid:1) (cid:1) j d, then we can eliminate this ain 1 j (cid:0) coordinate by replacing ai1 j (cid:1) (cid:1) (cid:1) j ain 1 j (cid:0) d with the following: ain ain j j a ai1 j (cid:1) (cid:1) (cid:1) j (cid:1) a (cid:1) ain 1 j (cid:0) ad (cid:0) bc: It does make sense to first reduce x = c=a to its simplest form by dividing out common factors of c and a. The implementation of rational numbers does this for us if we express x as a number of type Rational (see Section 8.6 for further details). Note that an elimination step transforms a matrix in echelon form, minus its last row, into a smaller matrix in echelon form. Here is the implementation: 6 342 CHAPTER9. POLYNOMIALS eliminate :: Rational -> Matrix -> Matrix eliminate p rs = map (simplify c a) rs where c = numerator p a = denominator p simplify c a row = init (init row’) ++ [a*d - b*c] where d b row’ = map (*a) row = last row = last (init row) The implementation of backward substitution runs like this: backsubst :: Matrix -> [Rational] backsubst rs = backsubst’ rs [] where backsubst’ [] ps = ps backsubst’ rs ps = backsubst’ rs’ (p:ps) where a c p rs’ = (last rs) !! ((cols rs) - 2) = (last rs) !! ((cols rs) - 1) = c % a = eliminate p (init rs) We get: POL> backsubst [[1,0,0,0,-7],[0,-1,-1,-1,-5],[0,0,-2,-6,-12],[0,0,0,-12,-12]] [-7 % 1,1 % 1,3 % 1,1 % 1] To use all this to analyze a polynomial sequence, generate the appropriate matrix (appropriate for the degree of the polynomial that we get from difference analysis of the sequence), put it in echelon form, and compute the values of the unknowns by backward substitution. solveSeq :: [Integer] -> [Rational] solveSeq = backsubst . echelon . genMatrix 9.2. GAUSSIANELIMINATION 343 Recall that the sequence of sums of squares starts as follows: 0; 1; 5; 14; 30; : : :. Solving this sequence with solveSeq gives: POL> solveSeq [0,1,5,14,30] [0 % 1,1 % 6,1 % 2,1 % 3] This gives the form 2n3 + 3n2 + n 6 Here is a Haskell check (the use of the / operator creates a list of Fractionals): n(n + 1)(2n + 1) 6 n3 + n2 + n = 1 3 1 6 1 2 = : POL> map (\ n -> (1/3)*n^3 + (1/2)*n^2 + (1/6)*n) [0..4] [0.0,1.0,5.0,14.0,30.0] Similarly, 0; 1; 9; 36; 100; 225 is the start of the sequence of sums of cubes. Solving this with solveSeq gives: POL> solveSeq [0, 1, 9, 36, 100, 225] [0 % 1,0 % 1,1 % 4,1 % 2,1 % 4] This gives the form n4 + 1 4 n3 + 1 2 1 4 n2 = n4 + 2n3 + n2 4 = n2(n + 1)2 4 = n(n + 1) 2 2 : (cid:19) (cid:18) The running example from the previous section is solved as follows: POL> solveSeq [-12,-11,6,45,112,213,354,541,780,1077] [-12 % 1,-5 % 1,5 % 1,1 % 1] Thus, the sequence has the form n3 + 5n2 5n 12. (cid:0) (cid:0) Before we look at the confirmation, let us note that we are now in fact using rep- resentations of polynomial functions as lists of their coefficients, starting from the constant coefficient. It is easy to implement a conversion from these representa- tions to the polynomial functions that they represent: p2fct :: Num a => [a] -> a -> a p2fct [] x = 0 p2fct (a:as) x = a + (x * p2fct as x) We can use this in the confirmations, as follows: 344 CHAPTER9. POLYNOMIALS POL> [ n^3 + 5 * n^2 - 5 * n - 12 | n <- [0..9] ] [-12,-11,6,45,112,213,354,541,780,1077] POL> map (p2fct [-12,-5,5,1]) [0..9] [-12,-11,6,45,112,213,354,541,780,1077] Finally, here is the automated solution of Exercise 9.3: POL> solveSeq [3,7,17,39,79] [3 % 1,3 % 1,0 % 1,1 % 1] This represents the form n3 + 3n + 3. Exercise 9.6 Suppose you want to find a closed form for the number of pieces you can cut a pie into by making n straight cuts. After some experimentation it becomes clear that to obtain the maximum number of pieces no cut should be parallel to a previous cut, and no cut should pass through an intersection point of previous cuts. Under these conditions you find that the n-th cut can be made to 1 previous cuts, and can thereby made to split n of the old intersect all the n regions. This gives the recurrence C0 = 1 and Cn = Cn 1 + n, which yields the sequence 1; 2; 4; 7; 11; : : : Next, you use solveSeq to obtain: (cid:0) (cid:0) POL> solveSeq [1,2,4,7,11] [1 % 1,1 % 2,1 % 2] You conclude that n cuts can divide a pie into 1 Is there still need for an inductive proof to show that this answer is correct? 2 n + 1 = n(n+1) 2 n2 + 1 2 + 1 pieces. 9.3 Polynomials and the Binomial Theorem In this section we will establish a connection between polynomials and lists, namely lists of coefficients of a polynomial. Let f (x) be a polynomial of degree n, i.e., let f (x) be a function 1xn f (x) = anxn + an (cid:0) (cid:0) 1 + (cid:1) (cid:1) (cid:1) + a1x + a0; with ai constants, and an Substitution of y + c for x in f (x) gives a new polynomial of degree n in y, say = 0. Let c be a constant and consider the case x = y + c. 1yn f (x) = f (y + c) = bnyn + bn (cid:0) (cid:0) 1 + (cid:1) (cid:1) (cid:1) + b1y + b0: 6 9.3. POLYNOMIALSANDTHEBINOMIALTHEOREM 345 0 a 0 (cid:1) (a f (x)) 0 n (x ) g(x)) g(x)) (f (g(x))) (cid:1) 0 0 0 (f (x) (cid:6) (f (x) = 0; = a (cid:1) = nx 0 = f = f = g 0 0 0 f n(cid:0)1 (x); ; 0 g (cid:6) (x); (x) (x)g(x) + f (x)g (x) (g(x)): f 0 (cid:1) 0 (x)b; Figure 9.1: Differentiation Rules. To consider an example, take f (x) = 3x4 y + c for x we get: (cid:0) x + 2, and let c = 1. Substituting (cid:0) f (x) = f (y + c) = f (y 1)4 (y 12y3 + 18y2 12y3 + 18y2 = 3(y = (3y4 = 3y4 (cid:0) (cid:0) (cid:0) (cid:0) 1) 1) + 2 (cid:0) (cid:0) 12y + 3) 13y + 6: (cid:0) (cid:0) (y (cid:0) (cid:0) 1) + 2 We will see shortly that the coefficients bi can be computed in a very simple way. Substitution of x c for y in in f (y + c) gives (cid:0) f (x) = bn(x c)n + bn (cid:0) 1(x (cid:0) (cid:0) c)n 1 + (cid:0) + b1(x (cid:0) (cid:1) (cid:1) (cid:1) c) + b0: Calculation of f 0(x); f 00(x); : : : ; f (n)(x) (the first, second, . . . , n-th derivative of f ), is done with the familiar rules of Figure 9.1.1 In particular, this gives (b(x c)k)0 = kb(x 1, and we get: c)k (cid:0) (cid:0) (cid:0) c) + + nbn(x (cid:0) 2b3(x (cid:1) (cid:1) (cid:1) c) + (cid:0) (cid:1) (cid:1) (cid:1) (cid:0) + n(n c)n (cid:0) 1 1)bn(x c)n (cid:0) 2 (cid:0) (cid:0) f 0(x) = b1 + 2b2(x f 00(x) = 2b2 + 3 ... f (n)(x) = n(n 1)(n (cid:1) (cid:0) 2) 3 (cid:1) (cid:1) (cid:1) (cid:1) (cid:0) 2bn: 1If these rules are unfamiliar to you, or if you need to brush up your knowledge of analysis, you should consult a book like [Bry93]. 346 CHAPTER9. POLYNOMIALS Substitution of x = c gives: f (c) = b0; f 0(c) = b1; f 00(c) = 2b2; : : : ; f (n)(c) = n!bn: This yields the following instruction for calculating the bk: b0 = f (c); b1 = f 0(c); b2 = f 00(c) 2 ; : : : ; bn = f (n)(c) n! : In general: : bk = f (k)(c) k! Applying this to the example f (x) = 3x4 b0 = f ( 1) = 3( (cid:0) b2 = f 00( 2 = 36( 1) (cid:0) 6 Another example is the expansion of (z + 1)n. Using the calculation method with derivatives, we get, for c = 0: 1)4 + 1 + 2 = 6, b1 = f 0( (cid:0) (cid:0) 1)2 72 6 = = (cid:0) (cid:0) 2 1, we see that 13, 1 = (cid:0) (cid:0) (cid:0) 12, b4 = f (4)( 1) 24 = 3. (cid:0) = 18, b3 = f (3)( x + 2, with c = 1) = 12( (cid:0) 1)3 (cid:0) (cid:0) 1) (cid:0) 1zn f (z) = (z + 1)n = bnzn + bn (cid:0) (cid:0) 1 + (cid:1) (cid:1) (cid:1) + b1z + b0; with the following derivatives: f 0(z) = n(z + 1)n 1; f 00(z) = n(n (cid:0) (cid:0) 1)(z + 1)n (cid:0) 2; : : : ; f (n)(z) = n!: Substituting z = 0 this gives: b0 = 1; b1 = n; b2 = n(n (cid:0) 2 1) ; (cid:1) (cid:1) (cid:1) ; bk = n(n (cid:0) 1) (n k+1) (cid:1)(cid:1)(cid:1) k! (cid:0) ; (cid:1) (cid:1) (cid:1) ; bn = n! n! = 1: The general coefficient bk has the form n(n bk = 1) (n (cid:1) (cid:1) (cid:1) k! (cid:0) k + 1) n(n 1) (cid:0) = k + 1) (n (cid:1) (cid:1) (cid:1) k! (cid:0) (n (n (cid:1) k)! k)! (cid:0) (cid:0) (cid:0) n! k! (n (cid:0) = Define: : k)! n k (cid:19) (cid:18) := n! (cid:0) : k)! k! (n Pronounce n k as ‘n choose k’ or ‘n over k’. We have derived: Theorem 9.7 (Newton’s binomial theorem) (cid:0) (cid:1) (z + 1)n = n k=0 (cid:18) X zk: n k (cid:19) 9.3. POLYNOMIALSANDTHEBINOMIALTHEOREM 347 = k. Thus, ways to pick a subset B from A with If A is a set of n objects, then there are n is also the number of k-sized subsets of an n-sized set. To B k (cid:0) j see why this is so, note that the number of k-sequences picked from A, without (cid:0) repetitions, equals (cid:1) (cid:1) j n k n (n 1) (n k + 1); (cid:0) for there are n ways to pick the first element in the sequence, n the second element in the sequence, . . . , and n element in the sequence. The number n 1 ways to pick 1) ways to pick the k-th k + 1) is equal to (n (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:0) 1) (n (k (cid:0) (cid:0) (cid:1) (cid:1) (cid:0) (cid:0) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:0) (n n! k)! . (cid:0) For picking k-sized subsets from A, order does not matter. There are k! ways of arranging sequences of size k without repetition. These are all equivalent. This k)! for the number of k-sized subsets of a set of size n. This connection gives (cid:0) explains the phrasing ‘n choose k’. n! k! (n Note the following: n 0 (cid:19) (cid:18) = 1; n 1 (cid:19) (cid:18) = n; n 2 (cid:19) (cid:18) = n(n 1) ; (cid:0) 2 n 3 (cid:19) (cid:18) n(n = (cid:0) 1)(n 6 2) : (cid:0) Here is a straightforward implementation of n k . (cid:0) (cid:1) choose n k = (product [(n-k+1)..n]) ‘div‘ (product [1..k]) The more general version of Newton’s binomial theorem runs: Theorem 9.8 (Newton’s binomial theorem, general version) (x + y)n = n k=0 (cid:18) X n k (cid:19) xkyn k: (cid:0) Note: a binomial is the sum of two terms, so (x + y) is a binomial. Proof: To get this from the special case (z + 1)n = set z = x xk yk , and multiply by yn: (cid:1) y to get ( x y + 1)n = n k=0 n k=0 n k n k (cid:0) P zk derived above, (x + y)n = x y n (cid:18) n + 1 P yn (cid:1) xk yk ! (cid:1) (cid:0) (cid:1) n yn = k=0 (cid:18) X xk yn yk = (cid:1) n k (cid:19) n k=0 (cid:18) X n k (cid:19) xkyn k: (cid:0) (cid:19) n k (cid:19) = k=0 (cid:18) X 348 CHAPTER9. POLYNOMIALS Because of their use in the binomial theorem, the numbers coefficients. What the binomial theorem gives us is: are called binomial n k (cid:0) (cid:1) (x + y)0 = 1x0y0 (x + y)1 = 1x1y0 + 1x0y1 (x + y)2 = 1x2y0 + 2x1y1 + 1x0y2 (x + y)3 = 1x3y0 + 3x2y1 + 3x1y2 + 1x0y3 (x + y)4 = 1x4y0 + 4x3y1 + 6x2y2 + 4x1y3 + 1x0y4 (x + y)5 = 1x5y0 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + 1x0y5 ... To see how this pattern arises, look at what happens when we raise x + y to the n-th power by performing the n multiplication steps to work out the product of (x + y)(x + y) (x + y) : (cid:1) (cid:1) (cid:1) n factors | {z } x + y x + y x2 + xy + xy + y2 x + y x3 + x2y + x2y + x2y + xy2 + xy2 + xy2 + y3 x + y (cid:2) (cid:2) (cid:2)... Every term in this expansion is itself the product of x-factors and y-factors, with k. Every a total number of factors always n, so that each term has the form xkyn k. binomial (x+y) in (x+n)n either contributes an x-factor or a y-factor to xkyn The number of ways to get at the term xkyn k equals the number of k-sized sub- sets from a set of size n (pick any subset of k binomials from the set of n binomial n factors). Thus, this term occurs exactly k times. (cid:0) (cid:0) (cid:0) We can arrange the binomial coefficients in the well-known triangle of Pascal. (cid:0) (cid:1) 1 0 3 1 (cid:1) (cid:0) (cid:0) (cid:1) 0 0 2 1 4 2 ... (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) (cid:0) 1 1 3 2 (cid:1) (cid:0) (cid:0) (cid:1) 2 2 4 3 (cid:1) (cid:0) (cid:0) (cid:1) 3 3 (cid:0) (cid:1) 4 4 (cid:0) (cid:1) 3 0 (cid:0) (cid:1) 4 0 (cid:0) (cid:1) 2 0 4 1 (cid:1) (cid:0) (cid:0) (cid:1) 9.3. POLYNOMIALSANDTHEBINOMIALTHEOREM 349 Working this out, we get: 1 3 1 4 1 1 1 3 1 4 1 1 1 2 6 ... Studying the pattern of Pascal’s triangle, we see that that it is built according to the following law: n k = n k 1 1 (cid:0) (cid:0) + n 1 (cid:0) k : (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) 2 This is called the addition law for binomial coefficients. To see that this law is correct, consider a set A of size n, and single out one of its objects, a. To count the number of ways of picking a k-sized subset B from A, consider the two cases B. The number of ways of picking a k-sized subset B from (i) a A with a 1-sized subset from . The number of ways of picking a k-sized subset B from A a B is equal to the number of ways of picking a k-sized subset from A with a = 2 , i.e., ways of picking a k-sized A a + subset from an n-sized set. B and (ii) a = 2 1 1 B is equal to the number of ways of picking a k (cid:1) . Thus, there are (cid:0) It is of course also possible to prove the addition law directly from the definition of . Assume k > 0. Then: 2 , i.e., (cid:0) f (cid:0) f (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) n (cid:0) n k n k 1 1 (cid:0) k (cid:0) k g g (cid:1) (cid:1) (cid:0) (cid:0) (cid:1) n 1 1 n k (cid:0) (cid:1) (cid:18) n k 1 1 (cid:0) (cid:0) + (cid:19) (cid:18) n (cid:0) k 1 (cid:19) = = = 1)! (n (cid:0) 1)! (n 1)! k k)! (cid:0) 1)! n k)! (k (cid:0) (n (cid:0) k! (n (n (cid:0) k! (n (cid:0) k)! (n (cid:0) + = + (n k! (n 1)! (n 1)! (cid:0) 1 (cid:0) k) (cid:0) k)! (cid:0) (cid:0) k! (n k)! (cid:0) n! k! (n k)! (cid:0) = n k : (cid:19) (cid:18) n k We can use the addition law for an implementation of . In accordance with as the number of k-sized subset of an n-sized set, we the interpretation of n = 1 (there is just one way to pick a 0-sized subset from any set) and will put 0 n = 0 for n < k (no ways to pick a k-sized subset from a set that has less k than k elements). A further look at Pascal’s triangle reveals the following law of (cid:0) symmetry: (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) (cid:1) (cid:0) n k n k (cid:19) (cid:18) = n (cid:18) n (cid:0) : k (cid:19) 350 CHAPTER9. POLYNOMIALS as the number of k-sized subsets This makes sense under the interpretation of of a set of size n, for the number of k-sized subsets equals the number of their complements. There is just one way to pick an n-sized subset from an n-sized set (pick the whole set), so = 1. This leads to the following implementation: (cid:1) (cid:0) n k n n (cid:0) (cid:1) choose’ n 0 = 1 choose’ n k | n < k = 0 = 1 | n == k | otherwise = choose’ (n-1) (k-1) + (choose’ (n-1) (k)) Exercise 9.9 Which implementation is more efficient, choose or choose’? Why? Exercise 9.10 Derive the symmetry law for binomial coefficients directly from the definition. We will now give an inductive proof of Newton’s binomial theorem. The proof uses the addition law for binomials, in the form n . + = n k n+1 k k 1 (cid:0) (cid:1) Theorem 9.11 (Newton’s binomial theorem again) (cid:0) (cid:0) (cid:1) (cid:0) (cid:1) (x + y)n = n k=0 (cid:18) X n k (cid:19) xkyn (cid:0) k: Proof. Induction on n. Basis: (x + y)0 = 1 = 0 0 (cid:18) (cid:19) 0 x0y0 = k=0 (cid:18) X 0 k (cid:19) Induction step: Assume (x + y)n = n k=0 (cid:18) X n k (cid:19) xkyn k: (cid:0) xky0 k: (cid:0) 9.3. POLYNOMIALSANDTHEBINOMIALTHEOREM 351 Then: (x + y)n+1 = (x + y)(x + y)n n ih = (x + y) n = x n k (cid:19) xkyn (cid:0) k n xkyn (cid:0) k + y k=0 (cid:18) X n k (cid:19) k=0 (cid:18) X n n k = xk+1yn (cid:0) k + k=0 (cid:18) X = xn+1 + (cid:19) n 1 (cid:0) k=0 (cid:18) X xk+1yn (cid:0) k + (cid:19) n n k xkyn (cid:0) k (cid:19) xky(n+1) (cid:0) k k=0 (cid:18) X n n k n k (cid:19) n k=0 (cid:18) X n k=1 (cid:18) X n k 1 (cid:19) (cid:0) n k=1 (cid:18)(cid:18) X n 1 (cid:19) k (cid:0) n + 1 k (cid:19) n k (cid:19) n k=1 (cid:18) X k + (cid:0) xky(n+1) xky(n+1) k + (cid:0) xky(n+1) (cid:0) k + yn+1 xky(n+1) (cid:0) k + yn+1 = = xn+1 + = xn+1 + add = xn+1 + k=1 (cid:18) X n + 1 k (cid:19) = n+1 k=1 (cid:18) X xky(n+1) (cid:0) k + yn+1 xky(n+1) (cid:0) k + yn+1 n k (cid:19) xky(n+1) (cid:0) k + yn+1 (cid:19) k=1 (cid:18) X n k (cid:18) (cid:19) n+1 k=0 (cid:18) X n + 1 k (cid:19) xky(n+1) (cid:0) k: Exercise 9.12 Show from the definition that if 0 < k 6 n then: n k (cid:19) (cid:18) = n k (cid:1) n k (cid:0) (cid:0) (cid:18) : 1 1 (cid:19) The law from exercise 9.12 is the so-called absorption law for binomial coeffi- cients. It allows for an alternative implementation of a function for binomial coef- ficients, for we have the following recursion: n 0 n k = 1; = 0 for n < k; (cid:18) (cid:19) (cid:18) (cid:19) Thus we get a more efficient function for (cid:18) n k (cid:1) n k 1 1 (cid:0) (cid:0) (cid:18) (cid:19) for 0 < k 6 n: n k (cid:19) n k = : (cid:0) (cid:1) 352 CHAPTER9. POLYNOMIALS binom n 0 = 1 binom n k | n < k = 0 | otherwise = (n * binom (n-1) (k-1)) ‘div‘ k Exercise 9.13 Use a combinatorial argument (an argument in terms of sizes the subsets of a set) to prove Newton’s law: n m (cid:18) (cid:1) (cid:18) (cid:19) m k n k = (cid:19) (cid:18) (cid:1) (cid:18) (cid:19) n m k k : (cid:19) (cid:0) (cid:0) Exercise 9.14 Prove: n n (cid:18) + (cid:19) (cid:18) n + 1 n + (cid:19) (cid:18) n + 2 n (cid:19) + (cid:1) (cid:1) (cid:1) + n + k n (cid:18) = (cid:19) (cid:18) n + k + 1 n + 1 (cid:19) : 9.4 Polynomials for Combinatorial Reasoning To implement the polynomial functions in a variable z, we will represent a poly- nomial f (z) = f0 + f1z + f2z2 + + fn 1 + fnzn 1zn (cid:0) (cid:0) (cid:1) (cid:1) (cid:1) as a list of its coefficients: [f0; f1; : : : ; fn]: As we have seen, the function p2fct maps such lists to the corresponding func- tions. The constant zero polynomial has the form f (z) = 0. In general we will avoid trailing zeros in the coefficient list, i.e., we will assume that if n > 0 then fn = 0. The constant function (cid:21)z:c will get represented as [c], so there is a map from in- tegers to polynomial representations, given by (cid:21)c:[c]. We will also allow rationals as coefficients, to there is also a map from rationals to polynomial representations, given by (cid:21)r:[r]. We need some conventions for switching back and forth between a polynomial and its list of coefficients. If f (z) is a polynomial, then we use f for its coefficient list. If this list of coefficients is non-empty then, as before, we will indicate the tail of f 6 9.4. POLYNOMIALSFORCOMBINATORIALREASONING 353 as f . Thus, if f = [f0; f1; : : : ; fn], then f = [f1; : : : ; fn], and we have the identity 1 + fnzn, then f = f0 : f . Moreover, if f (z) = f0 + f1z + f2z2 + (cid:0) (cid:1) (cid:1) (cid:1) 2 + fnzn 1. This convention yields the we use f (z) for f1 + f2z + (cid:0) (cid:1) (cid:1) (cid:1) following important equality: + fn + fn 1zn 1zn (cid:0) (cid:0) (cid:0) f (z) = f0 + zf(z): The identity function (cid:21)z:z will get represented as [0; 1], for this function is of the form (cid:21)z:f0 + f1z, with f0 = 0 and f1 = 1. This gives: z :: Num a => [a] z = [0,1] To negate a polynomial, simply negate each term in its term expansion. For if f (z) = f0 + f1z + f2z2 + . This gives: f (z) = f2z2 , then f1z f0 (cid:0) (cid:0) (cid:0) (cid:0) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:0) negate [] negate (f:fs) = [] = (negate f) : (negate fs) To add two polynomials f (z) and g(z), just add their coefficients, for clearly, if = gmzm, f (z) = f0 + f1z + f2z2 + then + fkzk and g(z) = b0 + b1z + g2z2 + (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) f (z) + g(z) = (f0 + g0) + (f1 + g1)z + (f2 + g2)z2 + (cid:1) (cid:1) (cid:1) This translates into Haskell as follows: + [] + gs fs [] (f:fs) + (g:gs) = f+g : fs+gs = fs = gs Note that this uses overloading of the + sign: in f+g we have addition of numbers, in fs+gs addition of polynomial coefficient sequences. 354 CHAPTER9. POLYNOMIALS The product of f (z) = f0 + f1z + f2z2 + + gmzm looks like this: g2z2 + + fkzk and g(z) = g0 + g1z + (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) g(z) = (f0 + f1z + f2z2 + f (z) (cid:1) (g0 + g1z + g2z2 + = f0g0 + (f0g1 + f1g0)z + (f0g2 + f1g1 + f2g0)z2 + = f0g0 + z(f0g(z) + g0f(z) + zf(z)g(z)) + fkzk) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) + gmzm) (cid:1) (cid:1) (cid:1) = f0g0 + z(f0g(z) + f (z)g(z)) 1, Here f(z) = f1 + f2z + i.e., f (z) and g(z) are the polynomials that get represented as [f1; : : : ; fk] and [g1; : : : ; gm], respectively. 1 and g(z) = g1 + g2z + + gmzm + fkzk (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) If f (z) and g(z) are polynomials of degree k, then for all n 6 k, in the list of coefficients for f (z)g(z), the n-th coefficient has the form f0gn + f1gn + 1g1 + fng0. The list of coefficients of the product is called the convolution of fn the lists of coefficients f and g. 1 + (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) Multiplying a polynomial by z boils down to shifting its sequence of coefficients one place to the right. This leads to the following Haskell implementation, where (.*) is an auxiliary multiplication operator for multiplying a polynomial by a nu- merical constant. Note that (*) is overloaded: f*g multiplies two numbers, but fs * (g:gs) multiplies two lists of coefficients. We cannot extend this overload- ing to multiplication of numbers with coefficient sequences, since Haskell insists on operands of the same type for (*). Hence the use of (.*). infixl 7 .* (.*) :: Num a => a -> [a] -> [a] = [] c .* [] c .* (f:fs) = c*f : c .* fs fs [] (f:fs) * (g:gs) = f*g : (f .* gs + fs * (g:gs)) = [] = [] * [] * gs Example 9.15 In Figure 9.2 the polynomials are declared as a data type in class Num. This entails that all Haskell operations for types in this class are available. We get: POL> (z + 1)^0 9.4. POLYNOMIALSFORCOMBINATORIALREASONING 355 [1] POL> (z + 1) [1,1] POL> (z + 1)^2 [1,2,1] POL> (z + 1)^3 [1,3,3,1] POL> (z + 1)^4 [1,4,6,4,1] POL> (z + 1)^5 [1,5,10,10,5,1] POL> (z + 1)^6 [1,6,15,20,15,6,1] This gives yet another way to get at the binomial coefficients. Now suppose we have a polynomial f (z). We are interested in the difference list f1; : : :]. It is easy to see that this difference list of its coefficients [f0; f1 (cid:0) is the list of coefficients of the polynomial (1 f0; f2 (cid:0) z)f (z): (cid:0) f (z) ; [ zf (z) ; [ (cid:0) z)f (z) ; [ (cid:0) (1 f0; 0; f0; f1; f0; f0; (cid:0) f1 (cid:0) f2; f1; f1; (cid:0) f2 (cid:0) f3; f2; f2; (cid:0) f3 (cid:0) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) ] ] ] This is implemented by the following function: delta :: Num a => [a] -> [a] delta = ([1,-1] *) This gives, for example: POL> delta [2,4,6] [2,2,2,-6] Note that the coefficient of z4 in [2; 4; 6] is 0, so this is correct. Note also that we are now looking at difference lists of coefficients, not at different lists of the result of mapping the polynomial function to [0::], as in Section 9.1. 356 CHAPTER9. POLYNOMIALS module Polynomials where infixl 7 .* (.*) :: Num a => a -> [a] -> [a] = [] c .* [] c .* (f:fs) = c*f : c .* fs z :: Num a => [a] z = [0,1] instance Num a => Num [a] where = [fromInteger c] = [] = (negate f) : (negate fs) = fs = gs fromInteger c negate [] negate (f:fs) fs [] (f:fs) + (g:gs) = f+g : fs+gs fs [] (f:fs) * (g:gs) = f*g : (f .* gs + fs * (g:gs)) = [] = [] + [] + gs * [] * gs delta :: Num a => [a] -> [a] delta = ([1,-1] *) shift :: [a] -> [a] shift = tail p2fct :: Num a => [a] -> a -> a p2fct [] x = 0 p2fct (a:as) x = a + (x * p2fct as x) comp :: Num a => [a] -> [a] -> [a] comp _ comp [] comp (f:fs) (0:gs) = f : gs * (comp fs (0:gs)) comp (f:fs) (g:gs) = ([f] + [g] * (comp fs (g:gs))) = error ".." = [] [] _ + (0 : gs * (comp fs (g:gs))) deriv :: Num a => [a] -> [a] deriv [] deriv (f:fs) = deriv1 fs 1 where = [] deriv1 [] deriv1 (g:gs) n = n*g : deriv1 gs (n+1) _ = [] Figure 9.2: A Module for Polynomials. 9.4. POLYNOMIALSFORCOMBINATORIALREASONING 357 The composition of two polynomials f (z) and g(z) is again a polynomial f (g(z)). It is given by: f (z) = f0 + f3z3 + f (g(z)) = f0 + f1g(z) + f2(g(z))2 + f3(g(z))3 + f2z2 + f1z + (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) We see from this: f (g(z)) = f0 + g(z) f (g(z)) (cid:1) This leads immediately to the following implementation (the module of Figure 9.2 has a slightly more involved implementation comp that gets explained in the next chapter, on page 389): comp1 :: Num a => [a] -> [a] -> [a] comp1 _ [] = error ".." comp1 [] _ = [] comp1 (f:fs) gs = [f] + (gs * comp1 fs gs) Example 9.16 We can use this to pick an arbitrary layer in Pascal’s triangle: POL> comp1 (z^2) (z+1) [1,2,1] POL> comp1 (z^3) (z+1) [1,3,3,1] POL> comp1 (z^12) (z+1) [1,12,66,220,495,792,924,792,495,220,66,12,1] We can also use it to generate Pascal’s triangle up to arbitrary depth: POL> comp1 [1,1,1,1,1,1] [[0],[1,1]] [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]] Note that this uses the composition of f (z) = 1 + z + z 2 + z3 + z4 + z5 + z6 with g(z) = (y + 1)z + 0. The result of this is f (g(z)) = 1 + (y + 1)z + (y + 1)2z2 + + (y + 1)6z6. (cid:1) (cid:1) (cid:1) If f (z) = f0 + f1z + f2z2 + 9.1): (cid:1) (cid:1) (cid:1) + fkzk, the derivative of f (z) is given by (Figure f 0(z) = f1 + 2f2z + + kfkzk 1: (cid:0) (cid:1) (cid:1) (cid:1) 358 CHAPTER9. POLYNOMIALS This has a straightforward implementation, as follows: deriv :: Num a => [a] -> [a] deriv [] deriv (f:fs) = deriv1 fs 1 where = [] deriv1 [] deriv1 (g:gs) n = n*g : deriv1 gs (n+1) _ = [] The close link between binomial coefficients and combinatorial notions makes polynomial reasoning a very useful tool for finding solutions to combinatorial problems. Example 9.17 How many ways are there of selecting ten red, blue or white mar- bles from a vase, in such a way that there are at least two of each color and at most five marbles have the same colour? The answer is given by the coefficient of z 10 in the following polynomial: This is easily encoded into a query in our implementation: (z2 + z3 + z4 + z5)3: POL> ([0,0,1,1,1,1]^3) !! 10 12 How many ways are there of selecting ten red, blue or white marbles from a vase, in such manner that there is even number of marbles of each colour: POL> ([1,0,1,0,1,0,1,0,1,0,1]^3) !! 10 21 We associate coefficient lists with combinatorial problems by saying that [f0; f1; f2; : : : ; fn] solves a combinatorial problem if fr gives the number of solutions for that prob- lem. Example 9.18 The polynomial (1+z)10 solves the problem of picking r elements from a set of 10. The finite list [1; 10; 45; 120; 210; 252; 210; 120; 45; 10; 1] solves the problem. It is implemented by: 9.5. FURTHERREADING 359 POL> (1 + z)^10 [1,10,45,120,210,252,210,120,45,10,1] Example 9.19 The list [1; 3; 6; 10; 15; 18; 19; 18; 15; 10; 6; 3; 1] is a solution for the problem of picking r marbles from a vase containing red, white or blue mar- bles, with a maximum of five of each colour. A polynomial for this problem is (1 + z + z2 + z3 + z4 + z5)3. In the implementation: POL> (1 + z + z^2 + z^3 + z^4 + z^5)^3 [1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1] Exercise 9.20 Use polynomials to find out how many ways there are of selecting ten red, blue or white marbles from a vase, in such manner that the number of marbles from each colour is prime. 9.5 Further Reading Charles Babbage’s difference engine is described in [Lar34] (reprinted in [Bab61]), and by Babbage himself in his memoirs [Bab94]. The memoirs are very amusing: Among the various questions which have been asked respecting the Difference Engine, I will mention a few of the most remarkable: one gentleman addressed me thus: ‘Pray, Mr Babbage, can you explain to me in two words what is the principle of your machine?’ Had the querist possessed a moderate acquaintance with mathematics I might in four words have conveyed to him the required information by an- swering, ‘The Method of Differences.’ [. . . ] On two occasions I have been asked - ‘Pray, Mr Babbage, if you put into the machine wrong figures, will the right answers come out?’ In one case a member of the Upper, and in the other a member of the Lower, House put this question. There are many good textbooks on calculus, but [Bry93] is particularly enlight- ening. An excellent book on discrete mathematics and combinatorial reasoning is [Bal91]. 360 CHAPTER9. POLYNOMIALS Chapter 10 Corecursion Preview In this chapter we will look the construction of infinite objects and at proof meth- ods suited to reasoning with infinite data structures. The most important kind of infinite data structures are streams (infinite lists), so the main topic of this chapter is the logic of stream processing. We will show how non-deterministic processes can be viewed as functions from random integer streams to streams. For the im- plementation of this we will use two functions from Random.hs, a module for random number generation and processing from the Haskell library. At the end of the chapter we will connect combinatorial reasoning with stream processing, via the study of power series and generating functions. Our Haskell treatment of power series is modeled after the beautiful [McI99, McI00]. module COR where import Random (mkStdGen,randomRs) import Polynomials import PowerSeries The default for the display of fractional numbers in Haskell is floating point nota- 361 362 CHAPTER10. CORECURSION tion. As we are going to develop streams of integers and streams of fractions in this chapter, it is convenient to have them displayed with unlimited precision in integer or rational notation. The default command takes care of that. default (Integer, Rational, Double) 10.1 Corecursive Definitions As we have seen, it is easy to generate infinite lists in Haskell. Infinite lists are often called streams. Here is the code again for generating an infinite list (or a stream) of ones: ones = 1 : ones This looks like a recursive definition, but there is no base case. Here is a definition of a function that generates an infinite list of all natural numbers: nats = 0 : map (+1) nats Again, the definition of nats looks like a recursive definition, but there is no base case. Definitions like this are called corecursive definitions. Corecursive def- initions always yield infinite objects. When you come to think of it, the funny explanation of the acronym GNU as GNU’s Not Unix is also an example. As we have seen in Section 3.7, generating the odd natural numbers can be done by corecursion. odds = 1 : map (+2) odds 10.1. CORECURSIVEDEFINITIONS 363 Exercise 10.1 Write a corecursive definition that generates the even natural num- bers. We can make the corecursive definitions more explicit with the use of iterate. The definition of iterate in the Haskell prelude is itself an example of corecur- sion: iterate iterate f x :: (a -> a) -> a -> [a] = x : iterate f (f x) Here are versions of the infinite lists above in terms of iterate: theOnes = iterate id 1 theNats = iterate (+1) 0 theOdds = iterate (+2) 1 Exercise 10.2 Use iterate to define the infinite stream of even natural numbers. The list [0..] can be defined corecursively from ones with zipWith. Suppose n is a natural number. Then its successor can be got by adding 1 to n. 0 is the first natural number. The second natural number, 1, is got by adding 1 to 0. The third natural number, 2, is got by adding 1 to the second natural number, and so on: theNats1 = 0 : zipWith (+) ones theNats1 The technique that produced theNats1 can be used for generating the Fibonacci numbers: theFibs = 0 : 1 : zipWith (+) theFibs (tail theFibs) 364 CHAPTER10. CORECURSION The process on Fibonacci numbers that was defined in Exercise 7.17 can be defined with corecursion, as follows: pr (x1:x2:x3:xs) = x1*x3 - x2*x2 : pr (x2:x3:xs) As we proved in Exercise 7.17, applying this process to theFibs gives the list (cid:21)n:( 1)n+1: (cid:0) COR> take 20 (pr theFibs) [-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1] The definition of the sieve of Eratosthenes (page 106) also uses corecursion: sieve :: [Integer] -> [Integer] sieve (0 : xs) = sieve xs sieve (n : xs) = n : sieve (mark xs 1 n) where mark (y:ys) k m | k == m = | otherwise = m) 0 : (mark ys y : (mark ys (k+1) m) 1 What these definitions have in common is that they generate infinite objects, and that they look like recursive definitions, except for the fact that there is no base case. Here is a faster way to implement the Sieve of Eratosthenes. This time, we actually remove multiples of x from the list on encountering x in the sieve. The counting procedure now has to be replaced by a calculation, for the removals affect the distances in the list. The property of not being a multiple of n is implemented by the function (\ m -> (rem m n) /= 0). Removing all numbers that do not have this property is done by filtering the list with the property. sieve’ :: [Integer] -> [Integer] sieve’ (n:xs) = n : sieve’ (filter (\ m -> (rem m n) /= 0) xs) primes’ :: [Integer] primes’ = sieve’ [2..] 10.2. PROCESSESANDLABELEDTRANSITIONSYSTEMS 365 How does one prove things about corecursive programs? E.g., how does one prove that sieve and sieve’ compute the same stream result for every stream argu- ment? Proof by induction does not work here, for there is no base case. Exercise 10.3* The Thue-Morse sequence is a stream of 0’s and 1’s that is pro- duced as follows. First produce 0. Next, at any stage, swap everything that was produced so far (by interchanging 0’s and 1’s) and append that. The first few stages of producing this sequence look like this: 0 01 0110 01101001 0110100110010110 Thus, if Ak denotes the first 2k symbols of the sequence, then Ak+1 equals Ak + +Bk, where Bk is obtained from Ak by interchanging 0’s and 1’s. Give a corecur- sive program for producing the Thue-Morse sequence as a stream. 10.2 Processes and Labeled Transition Systems The notion of a nondeterministic sequential process is so general that it is impos- sible to give a rigorous definition. Informally we can say that processes are inter- acting procedures. Typical examples are (models of) mechanical devices such as clocks, protocols for traffic control, vending machines, operating systems, client- server computer systems, and so on. A formal notion for modeling processes that has turned out to be extremely fruitful is the following. A labeled transition system (Q; A; T ) consists of a set of states Q, a set of action labels A, and a ternary relation T Q, the transition relation. q0. If (q; a; q0) T we write this as q Q (cid:18) (cid:2) (cid:2) A a (cid:0)! 2 Example 10.4 Perhaps the simplest example of a labeled transition system is the crack system given by the two states c and c0 and the two transitions c (cid:0)! c0 (Figure 10.1). This is a model of a clock that ticks until it gets unhinged. tick (cid:0)! c and c Note that the process of the ticking clock is nondeterministic. The clock keeps ticking, until at some point, for no reason, it gets stuck. To implement nondeterministic processes like the clock process from Example 10.4, we have to find a way of dealing with the nondeterminism. Nondeterministic 366 CHAPTER10. CORECURSION tick crack Figure 10.1: Ticking clock. behaviour is behaviour determined by random factors, so a simple way of modeling nondeterminism is by modeling a process as a map from a randomly generated list of integers to a stream of actions. The following function creates random streams of integers, within a specified bound [0; ::; b], and starting from a particular seed s. It uses randomRs and mkStdGen from the library module Random.hs. randomInts :: Int -> Int -> [Int] randomInts bound seed = tail (randomRs (0,bound) (mkStdGen seed)) Exercise 10.5 Note that randomInts 1 seed generates a random stream of 0’s and 1’s. In the long run, the proportion of 0’s and 1’s in such a stream will be 1 to 1. How would you implement a generator for streams of 0’s and 1’s with, in the long run, a proportion of 0’s and 1’s of 2 to 1? We define a process as a map from streams of integers to streams of action labels. To start a process, create an appropriate random integer stream and feed it to the process. type Process = [Int] -> [String] start :: Process -> Int -> Int -> [String] start process bound seed = process (randomInts bound seed) (cid:0) (cid:0) 10.2. PROCESSESANDLABELEDTRANSITIONSYSTEMS 367 The clock process can now be modeled by means of the following corecursion: clock :: Process clock (0:xs) = "tick" clock (1:xs) = "crack" : [] : clock xs This gives: COR> start clock 1 1 ["tick","crack"] COR> start clock 1 2 ["crack"] COR> start clock 1 25 ["tick","tick","tick","tick","crack"] The parameter for the integer bound in the start function (the second argument of start function) should be set to 1, to ensure that we start out from a list of 0’s and 1’s. Example 10.6 Consider a very simple vending machine that sells mineral water and beer. Water costs one euro, beer two euros. The machine has a coin slot and a button. It only accepts 1 euro coins. If a coin is inserted and the dispense button is pushed, it dispenses a can of mineral water. If instead of pushing the dispense button, another one euro coin is inserted and next the dispense button is pushed, it dispenses a can of beer. If, instead of pushing the button for beer, a third coin is inserted, the machine returns the inserted money (three 1 euro coins) and goes back to its initial state. This time we need four states, and the following transitions coin (cid:0)! moneyback (cid:0)! (Figure 10.2): q beer (cid:0)! coin (cid:0)! coin (cid:0)! water (cid:0)! q1, q1 q3, q3 q2, q2 q, q2 q, q1 q. Again, this is easily modeled. This time, the random stream is not needed for the transitions q for the first of these, and return of all the inserted money for the second. q, for insertion of a coin is the only possibility q1 and q3 moneyback (cid:0)! coin (cid:0)! 368 CHAPTER10. CORECURSION moneyback beer water coin coin coin Figure 10.2: A simple vending machine. (cid:0) (cid:0) (cid:0) (cid:0) 10.2. PROCESSESANDLABELEDTRANSITIONSYSTEMS 369 (0:xs) = "coin" (1:xs) = vending, vending1, vending2, vending3 :: Process : vending1 xs vending xs vending : vending2 xs vending1 (0:xs) = "coin" : vending xs vending1 (1:xs) = "water" : vending3 xs vending2 (0:xs) = "coin" xs vending2 (1:xs) = "beer" : vending xs vending3 (0:xs) = "moneyback": vending vending3 xs vending3 (1:xs) = vending This gives: COR> take 9 (start vending 1 1) ["coin","water","coin","water","coin","water","coin","coin","beer"] COR> take 8 (start vending 1 3) ["coin","water","coin","coin","coin","moneyback","coin","water"] COR> take 8 (start vending 1 22) ["coin","water","coin","water","coin","coin","coin","moneyback"] Example 10.7 A parking ticket dispenser works as follows. As long as pieces of 1 or 2 euro are inserted, the parking time is incremented by 20 minutes per euro. If the red button is pressed, all the inserted money is returned, and the machine returns to its initial state. If the green button is pressed, a parking ticket is printed indicating the amount of parking time, and the machine returns to its initial state. return(i) 2euro There are the following transitions: q(i) (cid:0)! (cid:0)! q(i + 2), q(0) q(0). Note that the number of states is infinite. time i*20 min (cid:0)! q(i + 1), q(i) no time (cid:0)! q(0), q(i) q(0), q(i) 1euro (cid:0)! Here is an implementation of the parking ticket dispenser: 370 CHAPTER10. CORECURSION ptd :: Process ptd = ptd0 0 ptd0 :: Int -> Process ptd0 0 (0:xs) = ptd0 0 xs ptd0 i (0:xs) = ("return " ++ show i ++ " euro") : ptd0 0 xs ptd0 i (1:xs) = "1 euro" : ptd0 (i+1) xs ptd0 i (2:xs) = "2 euro" : ptd0 (i+2) xs ptd0 0 (3:xs) = ptd0 0 xs ptd0 i (3:xs) = ("ticket " ++ show (i * 20) ++ " min") : ptd0 0 xs This yields: COR> take 6 (start ptd 3 457) ["1 euro","2 euro","2 euro","ticket 100 min","1 euro","ticket 20 min"] tick tick crack crack Figure 10.3: Another ticking clock. Example 10.8 Intuitively, the clock process of Example 10.4 is the same as the clock process of the following example (Figure 10.3): c1 c3, tick c4. It is clear that this is also a clock that ticks until it gets c2 (cid:0)! stuck. c2 and c2 crack (cid:0)! tick (cid:0)! crack (cid:0)! c2, c1 (cid:0) (cid:0) (cid:0) (cid:0) 10.2. PROCESSESANDLABELEDTRANSITIONSYSTEMS 371 moneyback beer coin coin coin coin water Figure 10.4: Another simple vending machine. (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 372 CHAPTER10. CORECURSION Exercise 10.9 Consider the vending machine given by the following transitions coin (cid:0)! q, q2, q2 q. Taking the machine from Example 10.6 and this machine to be black (Figure 10.4): q q4 boxes, how can a user find out which of the two machines she is operating? moneyback (cid:0)! coin (cid:0)! beer (cid:0)! water (cid:0)! coin (cid:0)! coin (cid:0)! q4, q1 q3, q3 q1, q q, q2 Exercise 10.10 Give a Haskell implementation of the vending machine from Ex- ercise 10.9. The key question about processes is the question of identity: How does one prove that two processes are the same? How does one prove that they are different? Proof methods for this will be developed in the course of this chapter. Before we end this brief introduction to processes we give a simple example of process interaction. Example 10.11 A user who continues to buy beer from the vending machine in Example 10.6 can be modeled by: u u. It is clear how this should be implemented. beer (cid:0)! coin (cid:0)! coin (cid:0)! u1, u1 u2, u2 How about a beer drinker who interacts with a vending machine? It turns out that we can model this very elegantly as follows. We let the user start with buying his (her?) first beer. Next, we feed the stream of actions produced by the vending machine as input to the user, and the stream of actions produced by the user as input to the vending machine, and they keep each other busy. This interaction can be modeled corecursively, as follows: actions responses = vending actions = user [0,0,1] responses user acts ~(r:s:p:resps) = acts ++ user (proc [r,s,p]) resps proc ["coin","coin","beer"] = [0,0,1] This gives: COR> take 8 actions [0,0,1,0,0,1,0,0] COR> take 8 responses ["coin","coin","beer","coin","coin","beer","coin","coin"] 10.3. PROOFBYAPPROXIMATION 373 The user starts by inserting two coins and pressing the button, the machine re- sponds with collecting the coins and issuing a can of beer, the user responds to this by inserting two more coins and pressing the button once more, and so on. One hairy detail: the pattern ~(r:s:p:resps) is a so-called lazy pattern. Lazy patterns always match, they are irrefutable. This allows the initial request to be submitted ‘before’ the list (r:s:p:resps) comes into existence by the response from the vending machine. 10.3 Proof by Approximation One of the proof methods that work for corecursive programs is proof by approx- imation. For this, we have to extend each data type to a so-called domain with a partial ordering (the approximation order). Every data type gets extended with an element . This is the lowest element in the approximation order. v Let (D; ) be a set D with a partial order on it, and let A D. An element x the least element of A if x for which such least and greatest elements do not exist. A is the greatest element of A if a a for all a v 2 2 v (cid:18) x for all a 2 A. Note that there are D with A A. x v 2 A is D (cid:18) ? v Exercise 10.12 Give an example of a set D, a partial order A of D such that A has no greatest and no least element. v on D, and a subset A. Use Au for the x 2 An element x set of all upper bounds of A, i.e., Au := D is an upper bound of A if a D 2 E.g., consider N with the usual order 6. Take n N x f f x for all a A a a v j 8 2 2; 4; 6 2 v g (cid:18) j 2 f An element x set of all upper bounds of A, i.e., Al := . But the set of even numbers has no upper bounds in N. 6 6 n g D is a lower bound of A if x D v j 8 D is the lub or least upper bound of A if x is the least element An element x of Au. The lub of A is also called the supremum of A. Notation A. Use Al for the a a for all a A x a 2 v . g A. 2 2 2 2 x f N. Then 2; 4; 6 u = g . g f E.g., consider R with the usual order 6. Take A = , Au = A = 4 ; : : : g 3 ; 3 2 ; 2 0; 1 f An element x element of Al. The glb of A is also called the infimum of A. Notation Caution: there may be A D is the glb or greatest lower bound of A if x is the greatest A. A do not exist. D for which r > 1 g n n+1 j f A = 1. R. Then A and , and g (cid:18) r f N R 2 2 t u 2 n j (cid:18) t u t 374 CHAPTER10. CORECURSION Exercise 10.13 Give an example of a set D, a partial order A of D such that A has no lub and no glb. v on D, and a subset A subset A of D is called a chain in D if the ordering on A is total, i.e., if for all is A either x x; y a chain in }(N) under x. E.g., the set k < n g j ff A = v v N N 2 2 2 n k g j y or y . (cid:18) A set D with a partial order A exists for every chain A in D. E.g., }(N) with v is called a domain if D has a least element and is a domain: the least element ? t is given by , ; tA is given by . A (cid:18) Exercise 10.14 Show that N, with the usual ordering 6, is not a domain. Can you extend N to a domain? S We will now define approximation orders that make each data type into a domain. For basic data types A the approximation order is given by: x y v : (cid:17) x = ? _ x = y: To see that the result is indeed a domain, observe that data type, and that the only chains in basic data types are Obviously, we have = a lub. tf?g = x, tf? , ? tf ; x ? x g g is the least element of the . x g g = x. Thus, every chain has , f?g and f? ; x f For pair data types A order is given by: (cid:2) B — represented in Haskell as (a,b) — the approximation (x; y) (u; v) (x; y) ? v v : (cid:17) x u y ^ v v v: Again, it is not difficult to see that every chain in a pair data type has a lub. For functional data types A ! B the approximation order is given by: f g v : (cid:17) 8 x 2 A (f x gx): v Here it is assumed that A and B are domains. If A is a basic data type, f; g can be viewed as partial functions, with f g indicating that g is defined wherever f is defined, and f and g agreeing on every x for which they are both defined. v Exercise 10.15 Show that functional data types A mation order form a domain. Assume that A and B are domains. ! B under the given approxi- 10.3. PROOFBYAPPROXIMATION 375 For list data types [A] the approximation order is given by: xs ? v [] xs v y : ys v : (cid:17) : (cid:17) x : xs xs = [] y x v xs ys v ^ Exercise 10.16 Show that list data types [a] under the given approximation order form a domain. Assume that a is a domain. ? (cid:1) (cid:1) (cid:1) Using x1 : ? that diverges. The value definition of the undefined object. one can create partial lists, where a partial list is a list of the form x0 : is a program execution error or a program : shows up in the Haskell prelude in the following weird . The Haskell guise of ? ? undefined undefined | False = undefined :: a A call to undefined always gives rise to an error due to case exhaustion. An example of a partial list would be ’1’:’2’:undefined. Partial lists can be used to approximate finite and infinite lists. It is easy to check that x0 : x0 : x1 : x0 : x1 : : xn : ? v ? v This finite set of approximations is a chain, and we have: ? v (cid:1) (cid:1) (cid:1) v (cid:1) (cid:1) (cid:1) ? v x0 : x1 : : xn : []: (cid:1) (cid:1) (cid:1) ; x0 : f? ; x0 : x1 : ? ; : : : ; x0 : x1 : ? G = x0 : x1 : : xn : ; x0 : x1 : ? : xn : [] g (cid:1) (cid:1) (cid:1) : xn : []: (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) Also, for infinite lists, we can form infinite sets of approximations: x0 : x0 : x1 : x0 : x1 : x2 : ? v ? v This infinite set of approximations is a chain. We will show (in Lemma 10.17) that for infinite lists xs = x0 : x1 : x2 : we have: ? v (cid:1) (cid:1) (cid:1) ? v (cid:1) (cid:1) (cid:1) ; x0 : f? ; x0 : x1 : ? ; x0 : x1 : x2 : ? G ; : : : = xs: ? g The function approx can be used to give partial approximations to any list: 376 CHAPTER10. CORECURSION approx :: Integer -> [a] -> [a] approx (n+1) [] approx (n+1) (x:xs) = x : approx n xs = [] Since n + 1 matches only positive integers, the call approx n xs, with n less than or equal to the length of the list, will cause an error (by case exhaustion) after generation of n elements of the list, i.e., it will generate x0 : x1 : . ? If n is greater than the length of xs, the call approx n xs will generate the whole list xs. We can now write : xn (cid:1) (cid:1) (cid:1) 1 : (cid:0) ; x0 : f? ; x0 : x1 : ? ; x0 : x1 : x2 : ? ; : : : g ? as G approx n xs : 1 n=0 G Lemma 10.17 (Approximation Lemma) For any list xs: approx n xs = xs: 1 n=0 G Proof. We have to show two things: 1. xs approx n xs n 2 f 2 2. xs is the least element of j N u. g approx n xs f n j 2 u. N g To show (1), we have to establish that for every n and every xs, approx n xs We prove this with induction on n, for arbitrary lists. v xs. Basis: approx 0 xs = xs for any list xs. ? v Induction step: Assume that for every xs, approx n xs that for any list xs, approx (n + 1) xs xs. xs. We have to show v v If xs = we assume xs = x:xs’. Then: ? or xs = [], then the result trivially holds by the definition of approx, so approx (n + 1) x : xs’ = def approx f x : approx n xs’ g v f induction hypothesis x : xs’: g 10.3. PROOFBYAPPROXIMATION 377 To show (2), we have to show that for any list ys, if ys then xs N n ys, i.e., we have to show that xs is the least element of v u. 2 f approx n xs N n g 2 approx n xs f j u, j 2 g approx n xs Assume ys j 2 f ys. We have to show that xs N u. This means that for all n N, approx n xs 2 v g 2 ys. n v Suppose xs 6v But then approx (k + 1) xs N u. g ys. Then there has to be a k with (approx (k + 1) xs) !! k ys , and contradiction with ys 6v 2 f approx n xs 6v ys !! k. n j 2 Theorem 10.18 (Approximation Theorem) xs = ys n (approx n xs = approx n ys): , 8 : Immediate from the fact that xs !! n = ys !! n, for every n ) N. 2 Proof. : ( n(approx n xs = approx n ys) property of lub = 8 ) f 1 (approx n xs) = (approx n ys) 1 n=0 G n=0 G Lemma 10.17 () f xs = ys: g g The approximation theorem is one of our tools for proving properties of streams. Example 10.19 Suppose we want to prove the following: map f (iterate f x) = iterate f (f x): This equality cannot be proved by list induction, as the lists on both sides of the equality sign are infinite. We can prove an equivalent property by induction on n, as follows. n(approx n (map f (iterate f x)) = approx n (iterate f (f x))): 8 Proof by induction on n. Basis. For n = 0, the property holds by the fact that approx 0 xs = xs. ? for all lists 378 CHAPTER10. CORECURSION Induction step. Assume (for arbitrary x): approx n (map f (iterate f x)) = approx n (iterate f (f x)): We have: = = = = = = approx (n + 1) (map f (iterate f x)) definition of iterate f g approx (n + 1) (map f (x : iterate f (f x))) definition of map f approx (n + 1) (f x : map f (iterate f (f x))) g definition of approx f f x : approx n (map f (iterate f (f x))) g induction hypothesis f f x : approx n (iterate f (f (f x))) g definition of approx f approx (n + 1) (f x : iterate f (f (f x))) g definition of iterate f approx (n + 1) (iterate f (f x)): g Exercise 10.20 In Exercise 7.52 you showed that for every finite list xs :: [a], every function f :: a -> b, every total predicate p :: b -> Bool the follow- ing holds: filter p (map f xs) = map f (filter (p f ) xs): (cid:1) Use proof by approximation to show that this also holds for infinite lists. Example 10.21 To reason about sieve, we would like to show that mark n k has the same effect on a list of integers as mapping with the following function: (cid:21)m:if rem m n = 0 then m else 0: This will only hold when mark n k is applied to sequences that derive from a sequence [q..] by replacement of certain numbers by zeros. Let us use [q..](cid:15) for the general form of such a sequence. Suppose xs equals [q..](cid:15) for some q, i.e., suppose that xs is the result of replacing some of the items in the infinite list [q; q + 1; q + 2; : : :] by zeros. We prove by approximation that if q = an + k, with 1 6 k 6 n, then: mark n k xs = map ((cid:21)m:if rem m n = 0 then m else 0) xs: 6 6 10.4. PROOFBYCOINDUCTION 379 Basis. For n0 = 0, the property holds by the fact that approx 0 xs = xs. ? for all lists Induction step. Assume (for arbitrary xs of the form [q..](cid:15) with q = an + k and 1 6 k 6 n): approx n0 (mark n k xs) = approx n0 (map ((cid:21)m:if rem m n = 0 then m else 0) xs): Two cases: (i) k = n, and (ii) k < n. q, i.e., rem x n = 0. We get: Case (i) is the case where n j approx (n0 + 1) (mark n k x:xs) = = = = = definition of mark f approx (n0 + 1) (0 : mark n 1 xs) g definition of approx f 0 : approx n0 mark n 1 xs induction hypothesis g f 0 : approx n0 ((cid:21)m:if rem m n g = 0 then m else 0) xs definition of approx f approx (n0 + 1) (0 : ((cid:21)m:if rem m n g = 0 then m else 0) xs) definition of map, plus the fact that rem x n = 0 f approx (n0 + 1) map ((cid:21)m:if rem m n g = 0 then m else 0) x:xs: Case (ii) is the case where n of the definition of mark. 6 j q. The reasoning for this case uses the other case A proof by approximation of the fact that sieve and sieve’ define the same function on streams can now use the result from Example 10.21 as a lemma. 10.4 Proof by Coinduction To compare two streams xs and ys, intuitively it is enough to compare their obser- vational behaviour. The key observation on a stream is to inspect its head. If two streams have the same head, and their tails have the same observational behaviour, then they are equal. Similarly, to compare two infinite binary trees it is enough to observe that they have the same information at their root nodes, that their left daughters have the same observational behaviour, and that their right daughters have the same observational behaviour. And so on, for other infinite data struc- tures. 6 6 6 6 380 CHAPTER10. CORECURSION To make this talk about observational behaviour rigorous, it is fruitful to consider the infinite data structures we are interested in as labeled transition systems. The observations are the action labels that mark a transition. A bisimulation on a labeled transition system (Q; A; T ) is a binary relation R on Q with the following properties. If qRp and a A then: 2 1. If q 2. If p a (cid:0)! a (cid:0)! q0 then there is a p0 2 p0 then there is an q0 2 Q with p p0 and q0Rp0. Q with p p0 and q0Rp0. a (cid:0)! a (cid:0)! Example 10.22 Take decimal representations of rational numbers with over-line notation to indicate repeating digits (Section 8.5). The following representations all denote the same number 1 7 : 0:142857; 0:1428571; 0:14285714; 0:142857142857142: Why? Because if you check them against each other decimal by decimal, you will never hit at a difference. The relation ‘having the same infinite expansion’ is a bisimulation on decimal representations, and the representations for 1 7 are all bisimilar. Exercise 10.23 Show that the relation of equality on a set A is a bisimulation on A. Example 10.24 Collect the transition for the two clock processes of Examples 10.4 and 10.8 together in one transition system. Then it is easy to check that the relation R given by cRc1; cRc2; c0Rc3; c0Rc4 is a bisimulation on this transition system (Figure 10.5). The bisimulation connects the states c and c1. The two clock processes are indeed indistinguishable. Exercise 10.25 Show that there is no bisimulation that connects the starting states of the two vending machines of Examples 10.6 and 10.9. Exercise 10.26 Show that the union of two bisimulations is again a bisimulation. Exercise 10.27 Show that there is always a greatest bisimulation on a set A under the inclusion ordering. (Hint: show that the union of all bisimulations on A is itself a bisimulation.) Exercise 10.28 The bisimulation relation given in Example 10.24 is not the great- est bisimulation on the transition system. Find the greatest bisimulation. 10.4. PROOFBYCOINDUCTION 381 tick tick tick crack crack crack Figure 10.5: Bisimulation between ticking clocks. (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 382 CHAPTER10. CORECURSION (cid:24) Use for the greatest bisimulation on a given set A (Exercise 10.27). Call two elements of A bisimilar when they are related by a bisimulation on A. Being bisimilar then coincides with being related by the greatest bisimulation: a b (cid:24) R(R is a bisimulation, and aRb): , 9 To show that two infinite objects x and y are equal, we show that they exhibit the same behaviour, i.e. we show that x y. Such a proof is called a proof by coinduction. (cid:24) In the particular case of comparing streams we can make the notion of a bisimula- tion more specific. There are only two kinds of observations on a stream: observ- ing its head and observing its tail. Comparison of the heads should exhibit objects that are the same. Comparison of the tails should exhibit bisimilar streams. When talking about streams f = [f0; f1; f2; f3; : : :] it is convenient to refer to the tail of the stream as f. Thus, a stream f = [f0; f1; f2; f3; : : :] always has the form f = f0 : f , where f0 is the head and f is the tail. f f0 f Figure 10.6: A stream viewed as a transition system. f f0 f (cid:1) g g0 g Figure 10.7: Bisimulation between streams. Viewing a stream f as a transition system, we get transitions f f , and so on (Figure 10.6). A bisimulation between streams f and g that connects f and f0 (cid:0)! (cid:0) (cid:0) (cid:1) (cid:2) (cid:2) (cid:2) (cid:2) 10.4. PROOFBYCOINDUCTION 383 g is given in Figure 10.7. It is immediate from the picture that the notion of a bisimulation between streams boils down to the following. A stream bisimulation on a set A is a relation R on [A] with the following prop- erty. If f; g [A] and f Rg then both f0 = g0 and f Rg. 2 The general pattern of a proof by coinduction using stream bisimulations of f where f; g a stream bisimulation, with f Rg. g, [A], is simply this. Define a relation R on [A]. Next, show that R is (cid:24) 2 Example 10.29 As an example, we will show: map f (iterate f x) iterate f (f x): (cid:24) Let R be the following relation on [A]. (map f (iterate f x); iterate f (f x)) f j f :: a ! a; x :: a : g Suppose: (map f (iterate f x)) R (iterate f (f x)): We show that R is a bisimulation. We have: map f (iterate f x) iterate = map f x : (iterate f (f x)) map = (f x) : map f (iterate f (f x)) iterate f (f x) iterate = (f x) : iterate f (f (f x)): This shows: map f (iterate f x) map f (iterate f x) iterate f (f x) iterate f (f x) head (cid:0)! tail (cid:0)! head (cid:0)! tail (cid:0)! (f x) map f x : (iterate f (f x)) (f x) iterate f (f (f x)) (1) (2) (3) (4) Clearly, map f (iterate f x) and iterate f (f x) have the same heads, and also, their tails are R-related. This shows that R is a bisimimulation that connects map f (iterate f x) and iterate f (f x). Hence (map f (iterate f x)) (iterate f (f x)): (cid:24) 384 CHAPTER10. CORECURSION Given: A labeled transition system (Q; A; T ) . . . To be proved: q Proof: (cid:24) p Let R be given by . . . and suppose qRp. To be proved: R is a bisimulation. a q0 Suppose q (cid:0)! To be proved: There is a p0 Proof: . . . Suppose p (cid:0)! To be proved: There is a q0 Proof: . . . p0 a Q with p Q with q 2 2 a (cid:0)! a (cid:0)! p0 and q0Rp0. q0 and q0Rp0. Thus R is a bisimulation with qRp. Thus q p. (cid:24) Given: . . . To be proved: f Proof: g (cid:24) Let R be given by . . . and suppose f Rg. To be proved: R is a stream bisimulation. To be proved: f0 = g0. Proof: . . . To be proved: f Rg. Proof: . . . Thus R is a stream bisimulation with f Rg. Thus f g. (cid:24) Figure 10.8: Proof Recipes for Proofs by Coinduction. 10.5. POWERSERIESANDGENERATINGFUNCTIONS 385 Exercise 10.30 Show by means of a coinduction argument that for every infinite list xs :: [a], every f :: a -> b, and every total predicate p :: b -> Bool the following holds: filter p (map f xs) = map f (filter (p:f ) xs): 10.5 Power Series and Generating Functions module PowerSeries where import Polynomials instance Fractional a => Fractional [a] where / [] / gs = [fromRational c] = error "division by 0 attempted" = [] fromRational c fs [] (0:fs) / (0:gs) = fs / gs (_:fs) / (0:gs) = error "division by 0 attempted" (f:fs) / (g:gs) = let q = f/g in q : (fs - q.*gs) / (g:gs) int :: Fractional a => [a] -> [a] int fs = 0 : int1 fs 1 where int1 [] int1 (g:gs) n = g/n : (int1 gs (n+1)) _ = [] expz = 1 + (int expz) Figure 10.9: A Module for Power Series. In chapter 9 we have seen how polynomials can get represented as finite lists of their coefficients, and how operations on polynomials can be given directly on these list representations. We will now generalize this to infinite sequences. A 386 CHAPTER10. CORECURSION possible motivation for this (one of many) is the wish to define a division operation on polynomials. Suppose we want to divide f (z) = f0 + f1z + f2z2 + + fkzk (cid:1) (cid:1) (cid:1) by The outcome h(z) = h0 + h1z + h2z2 + (cid:1) (cid:1) (cid:1) g(z) = g0 + g1z + g2z2 + + gmzm: has to satisfy: (cid:1) (cid:1) (cid:1) This gives: f (z) = h(z) g(z): (cid:1) f0 + zf(z) = (h0 + zh(z))g(z) = h0g(z) + zh(z)g(z) = h0(g0 + zg(z)) + zh(z)g(z) = h0g0 + z(h0g(z) + h(z)g(z)): Thus, f0 = h0g0, hence h0 = f0 g0 h(z) = f (z) h0g(z) (cid:0) g(z) process of long division: , and f (z) = h0g(z) + h(z)g(z), so from this . We see from this that computing fractions can be done by a f (z) g(z) = f0 g0 f(z) + z (cid:0) (f0=g0)g(z) g(z) : (div) An example case is 1 (cid:0) z 1 1 z . Long division gives: (cid:0) 1 z z = 1 + z + z 1 1 1 1 1 = (cid:0) (cid:0) = 1 + z(1 + z z 1 (cid:0) 1 z (cid:0) ) = 1 + z(1 + z(1 + z 1 )) = 1 z (cid:1) (cid:1) (cid:1) (cid:0) 1], so in terms of computation on sequences we (cid:0) (cid:0) z 1 z is [1; The representation of 1 get: (cid:0) [1] [1; 1] (cid:0) = = = = 1 1 1 1 1 1 1 1 : : : : = : : : 1] [1] (cid:0) : [1; 1 1 1 1 1 1 : : [1] (cid:0) : [1; 1 1 1 1 : 1] [1] [1; 1 1 (cid:0) : 1] [1] [1; 1] (cid:0) 10.5. POWERSERIESANDGENERATINGFUNCTIONS 387 This shows that 1 1 z does not have finite degree. It is not a polynomial. (cid:0) To get at a class of functions closed under division we define power series to be functions of the form f (z) = f0 + f1z + f2z2 + = (cid:1) (cid:1) (cid:1) 1 fkzk: k=0 X A power series is represented by an infinite stream [f0; f1; f2; : : :]. Power series in a variable z can in fact be viewed as approximations to complex numbers (Section 8.10), but we will not pursue that connection here. The implementation of division follows the formula (div). Since trailing zeros are suppressed, we have to take the fact into account that [] represents an infinite stream of zeros. / [] / gs = error "division by 0 attempted" = [] fs [] (0:fs) / (0:gs) = fs / gs (_:fs) / (0:gs) = error "division by 0 attempted" (f:fs) / (g:gs) = let q = f/g in q : (fs - q.*gs)/(g:gs) Here are some Haskell computations involving division: COR> take 10 (1/(1-z)) [1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1] COR> take 10 (1/(1+z)) [1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1,1 % 1,-1 % 1] Example 10.31 To get a feel for ‘long division with sequences’, we compute 3 z , 3 (cid:0) as a calculation on sequences. The sequence representation of 3 1], so we get: z is [3; (cid:0) (cid:0) [3] [3; 1] (cid:0) = = = = 3 3 3 3 3 3 3 3 : : : : = : : : [1] (cid:0) : [3; 1 3 1 3 1 3 : : 1] [1=3] 1] [3; [1=9] 1 [3; 9 1] 1 1 27 9 (cid:0) : (cid:0) : : [1=27] 1] [3; (cid:0) 388 CHAPTER10. CORECURSION This is indeed what Haskell calculates for us: COR> take 9 (3 /(3-z)) [1 % 1,1 % 3,1 % 9,1 % 27,1 % 81,1 % 243,1 % 729,1 % 2187,1 % 6561] Integration involves division, for z 0 f (t)dt is given by 1 1 2 3 To give an example, if g(z) = 1 z , then 1 (cid:0) 1 2 R 0 + f0z + (cid:1) (cid:1) (cid:1) z 0 g(t)dt equals: f1z2 + f2z3 + 0 + z + R z2 + z3 + 1 3 (cid:1) (cid:1) (cid:1) Integration is implemented by: int :: Fractional a => [a] -> [a] int [] = [] int fs = 0 : int1 fs 1 where int1 [] int1 (g:gs) n = g/n : (int1 gs (n+1)) _ = [] We get: COR> take 10 (int (1/(1-z))) [0 % 1,1 % 1,1 % 2,1 % 3,1 % 4,1 % 5,1 % 6,1 % 7,1 % 8,1 % 9] To extend composition to power series, we have to be careful, for the equation f(g(z)) has a snag: the first coefficient depends on all of f , f (g(z)) = f0 + g(z) which is a problem if f is a power series. Here is an example: (cid:1) COR> comp1 ones [0,2] where ones = 1 : ones ERROR - Garbage collection fails to reclaim sufficient space To solve this, we must develop the equation for composition a bit further: f (g(z)) = f0 + g(z) (cid:1) f(g(z)) = f0 + (g0 + zg(z)) = (f0 + g0 (cid:1) f(g(z)) (cid:1) f (g(z))) + zg(z) f (g(z)) (cid:1) 10.5. POWERSERIESANDGENERATINGFUNCTIONS 389 In the special case where g0 = 0 we can simplify this further: f (zg(z)) = f0 + zg(z) f(zg(z)): (cid:1) This leads immediately to the following implementation (part of the module for Polynomials): comp :: Num a => [a] -> [a] -> [a] comp _ comp [] comp (f:fs) (0:gs) = f : gs * (comp fs (0:gs)) comp (f:fs) (g:gs) = ([f] + [g] * (comp fs (g:gs))) = error ".." = [] [] _ + (0 : gs * (comp fs (g:gs))) This gives: COR> take 15 (comp ones [0,2]) where ones = 1 : ones [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384] Figure 10.9 gives a module for power series. We will now show how this is can be used as a tool for combinatorial calculations. We are interested in finite lists [f0; : : : ; fn] or infinite lists [f0; f1; f2; : : :] that can be viewed as solutions to com- binatorial problems. We associate sequences with combinatorial problems by saying that [f0; f1; f2; : : :] solves a combinatorial problem if fr gives the number of solutions for that prob- lem (cf. the use of polynomials for solving combinatorial problems in section 9.4, which is now generalized to power series). We call a power series f (z) a generating function for a combinatorial problem if the list of coefficients of f (z) solves that problem. If f (z) has power series expansion f0 + f1z + f2z2 + f3z3 + , then f (z) is a generating function for [f0; f1; f2; f3; : : :]. (cid:1) (cid:1) (cid:1) Example 10.32 Find a generating function for [1; 1; 1; 1; : : :]. Solution: as we saw above . The list of coefficients is [1; 1; 1; 1; : : :]. z has power series expansion 1 + z + z2 + z3 + z4 + (cid:0) (cid:1) (cid:1) (cid:1) 1 1 The generating function 1 z is an important building block for constructing other 1 (cid:0) generating functions, so we should give it a name. In fact, we already did: ones names the list [1,1,1,1,..]. As long as we don’t use division, this is OK. 390 CHAPTER10. CORECURSION Example 10.33 Find a generating function for the list of natural numbers. Solu- tion: recall the corecursive definition of the natural numbers in terms of zipWith (+). This can now be expressed succinctly in terms of addition of power series: COR> take 20 nats where nats = 0 : (nats + ones) [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19] This gives the following specification for the generating function: 1 (cid:0) 1 ) z g(z) = z(g(z) + g(z) (cid:0) zg(z) = g(z) = z 1 z (cid:0) z z)2 (1 (cid:0) Here is the check: COR> take 20 (z * ones^2) [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19] Another way of looking at this: differentiation of 1 + z + z 2 + z3 + z4 + 1 + 2z + 3z2 + 4z3 + function for the naturals. gives 0 = z z)2 is a generating (cid:0) . This means that z (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (1 (cid:0) 1 1 z (cid:16) (cid:17) Example 10.34 Find the generating function for [0; 0; 0; 1; 2; 3; 4; : : :]. Solution: multiplication by z has the effect of shifting the sequence of coefficients one place to the right and inserting a 0 at the front. Thus the generating function can be got from that of the previous example through multiplying by z 2. The generating function is: z3 z)2 . Here is the check: (cid:0) (1 COR> take 20 (z^3 * ones^2) [0,0,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17] Example 10.35 Find the generating function for the sequence of powers of two (the sequence (cid:21)n:2n). Solution: start out from a corecursive program for the pow- ers of two: COR> take 15 powers2 where powers2 = 1 : 2 * powers2 [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384] 10.5. POWERSERIESANDGENERATINGFUNCTIONS 391 This immediately leads to the specification g(z) = 1 + 2zg(z) (the factor z shifts one place to the right, the summand 1 puts 1 in first position). It follows that g(z) = 1 2z . Here is the confirmation: (cid:0) 1 COR> take 10 (1/(1-2*z)) [1 % 1,2 % 1,4 % 1,8 % 1,16 % 1,32 % 1,64 % 1,128 % 1,256 % 1,512 % 1] z , for multiplication by 1 Example 10.36 If g(z) is the generating function for (cid:21)n:gn, then (cid:21)n:gn+1 is gen- erated by g(z) z shifts the coefficients of the power series one place to the left. Thus, the generating function for the sequence (cid:21)n:n + 1 is 1 z)2 . This sequence can also be got by adding the sequence of naturals and the (cid:0) z . This identification (cid:0) (1 sequence of ones, i.e., it is also generated by makes algebraic sense: z)2 + 1 (1 (cid:0) 1 z z (cid:0) (1 z)2 + 1 1 (cid:0) = z (1 z (cid:0) z)2 + 1 (1 z z)2 = (cid:0) (cid:0) 1 (cid:0) (1 z)2 : Exercise 10.37 Find the generating functions for for and for [0; 0; 0; 1; 1; 1; : : :]; [1; 1; 1; 0; 0; 0; : : :]; [1; 1 2 ; 1 4 ; 1 8 ; : : :]: Check your answers by computer. Example 10.38 Find the generating function for [0; 1; 0; 1; 0; 1; : : :]. Solution: the sequence is generated by the following corecursive program. COR> take 20 things where things = 0 : 1 : things [0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1] This leads to the specification g(z) = 0 + z(1 + zg(z)), which reduces to g(z) = z + z2g(z). From this, g(z) = z z2 . In a similar way we can derive the generat- (cid:0) ing function for [1; 0; 1; 0; 1; 0; : : :], which turns out to be g(z) = 1 z2 . Alterna- (cid:0) tively, observe that [1; 0; 1; 0; 1; 0; : : :] is the result of shifting [0; 1; 0; 1; 0; 1; : : :] one place to the left, so division by z does the trick. 1 1 392 CHAPTER10. CORECURSION Theorem 10.39 summarizes a number of recipes for finding generating functions. Theorem 10.39 Suppose f (z) is a generating function for (cid:21)n:fn and g(z) is the generating function for (cid:21)n:gn. Then: c 1 2. 1. z is the generating function for (cid:21)n:c. (cid:0) z z)2 is the generating function for (cid:21)n:n. (cid:0) 3. f (z) z is the generating function for (cid:21)n:fn+1. (1 4. cf (z) + dg(z) is the generating function for (cid:21)n:cfn + dgn. 5. (1 z)f (z) is the generating function for the difference sequence (cid:0) (cid:21)n:fn fn 1. (cid:0) (cid:0) 6. 1 z z f (z) is the generating function for the difference sequence (cid:0) (cid:21)n:fn+1 (cid:0) fn. 7. 1 1 z f (z) is the generating function for (cid:21)n:f0 + f1 + (cid:0) + fn. (cid:1) (cid:1) (cid:1) 8. f (z)g(z) is the generating function for (cid:21)n:f0gn + f1gn 1 + + fn 1g1 + (cid:0) (cid:1) (cid:1) (cid:1) (cid:0) fng0 (the convolution of f and g). 9. zf 0(z) is the generating function for (cid:21)n:nfn. 10. 1 z z 0 f (t)dt is the generating function for (cid:21)n: fn n+1 . R Exercise 10.40 Prove the statements of theorem 10.39 that were not proved in the examples. Example 10.41 Find the generating function for the sequence (cid:21)n:2n + 1. Solu- tion: the odd natural numbers are generated by the following corecursive program: COR> take 20 (ones + 2 * nats) [1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39] This immediately gives g(z) = 1 1 (cid:0) z + 2 (1 z)2 . (cid:0) Example 10.42 Find the generating function for the sequence (cid:21)n:(n + 1)2. So- lution: the generating function is the sum of the generating functions for (cid:21)n:n2, (cid:21)n:2n and (cid:21)n:1. One way to proceed would be to find these generating functions and take their sum. But there is an easier method. Recall that (n + 1)2 is the sum of the first n odd numbers, and use the recipe from Theorem 10.39 for forming 10.5. POWERSERIESANDGENERATINGFUNCTIONS 393 the sequence (cid:21)n:a0 + Example 10.41 by 1 + an, by means of dividing the generating function from 1 (cid:1) (cid:1) (cid:1) z. This immediately gives z)2 + 2z (1 z)3 . (cid:0) (1 (cid:0) (cid:0) Alternatively, Theorem 10.39 tells us that (cid:21)n:(n + 1)2 can also be defined from the naturals with the help of differentiation: COR> take 20 (deriv nats) [1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400] It follows that the generating function g(z), for g(z) should satisfy: Working this out we find that g(z) = 1 (cid:0) (1 g(z) = (cid:18) z 0 : (1 z)2 (cid:19) (cid:0) z)2 + 2z (1 z)3 . (cid:0) Finally, here is computational confirmation: COR> take 20 (ones^2 + 2 * z * ones^3) [1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400] Example 10.43 Find the generating function for (cid:21)n:n2. Solution: shift the solu- tion for the previous example to the right by multiplication by z. This gives: COR> take 20 (z * ones^2 + 2 * z^2 * ones^3) [0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361] So the generating function is g(z) = z (cid:0) get the desired sequence from the naturals by shift and differentiation: z)3 . Sure enough, we can also (cid:0) z)2 + 2z2 (1 (1 COR> take 20 (z * deriv nats) [0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361] Example 10.44 Find the generating function for the sequence of Fibonacci num- bers. Solution: consider again the corecursive definition of the Fibonacci numbers. We can express this succinctly in terms of addition of power series, as follows: COR> take 20 fibs where fibs = 0 : 1 : (fibs + (tail fibs)) [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181] 394 CHAPTER10. CORECURSION This is easily translated into an instruction for a generating function: g(z) = z2 g(z) + (cid:18) g(z) z (cid:19) + z: Explanation: multiplying by z2 inserts two 0’s in front of the fibs sequence, and adding z changes the second of these to 1 (the meaning is: the second coefficient is obtained by add 1 to the previous coefficient). g(z) gives the tail of the fibs z sequence. From this we get: g(z) (cid:0) zg(z) (cid:0) z2g(z) = z g(z) = z z z2 (cid:0) 1 (cid:0) And lo and behold: COR> take 10 (z/(1-z-z^2)) [0 % 1,1 % 1,1 % 1,2 % 1,3 % 1,5 % 1,8 % 1,13 % 1,21 % 1,34 % 1] Exercise 10.45 The Lucas numbers are given by the following recursion: L0 = 2; L1 = 1; Ln+2 = Ln + Ln+1: This gives: [2; 1; 3; 4; 7; 11; 18; 29; 47; 76; 123; 199; 322; 521; 843; 1364; 2207; 3571; : : :] The recipe is that for the Fibonacci numbers, but the initial element of the list is 2 instead of 0. Find a generating function for the Lucas numbers, starting out from an appropriate corecursive definition. Example 10.46 Recall the definition of the Catalan numbers from Exercise 7.20: C0 = 1; Cn+1 = C0Cn + C1Cn 1 + (cid:0) (cid:1) (cid:1) (cid:1) + Cn 1C1 + CnC0: (cid:0) Clearly, Cn+1 is obtained from [C0; : : : ; Cn] by convolution of that list with itself. This leads to the following corecursive definition of the Catalan numbers: COR> take 15 cats where cats = 1 : cats * cats [1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440] 10.5. POWERSERIESANDGENERATINGFUNCTIONS 395 From this, we get at the following specification for the generating function: g(z) = 1 + z z(g(z))2 (cid:0) g(z) + 1 = 0 (g(z))2 (cid:1) Considering g(z) as the unknown, we can solve this as a quadratic equation ax2 + pb2 b bx + c = 0, using the formula x = (cid:0) . We get the following generating (cid:6) 2a functions: 4ac (cid:0) g(z) = 1 (cid:6) 4z : (cid:0) p1 2z Consider the following power series. g(z) = 1 + z + z2 2! + z3 3! + (cid:1) (cid:1) (cid:1) = zn n! 1 n=0 X Note that by the definition of the derivative operation for power series, g(z)0 = g(z). The function with this property is the function ez, where e is the base of the natural logarithm, Napier’s number e. The number e is defined by the following infinite sum: 1 + 1 1! + 1 2! + 1 3! + (cid:1) (cid:1) (cid:1) = 1 n! 1 n=0 X Therefore we call g(z) the exponential function ez. Note that by the rules for integration for power series, etdt = 0 + z + z2 2! + z3 3! + (cid:1) (cid:1) (cid:1) = ez 1; (cid:0) z 0 Z and therefore, ez = 1 + sion, as follows: R z 0 etdt. This gives us a means of defining ez by corecur- expz = 1 + (int expz) We get: COR> take 9 expz [1 % 1,1 % 1,1 % 2,1 % 6,1 % 24,1 % 120,1 % 720,1 % 5040,1 % 40320] 396 CHAPTER10. CORECURSION Since we know (Theorem 10.39) that z f (z) gives incremental sums of the se- (cid:0) quence generated by f (z), we now have an easy means to approximate Napier’s number (the number e): 1 1 COR> take 20 (1/(1-z) * expz) [1 % 1,2 % 1,5 % 2,8 % 3,65 % 24,163 % 60,1957 % 720,685 % 252, 109601 % 40320,98641 % 36288,9864101 % 3628800,13563139 % 4989600, 260412269 % 95800320,8463398743 % 3113510400, 47395032961 % 17435658240,888656868019 % 326918592000, 56874039553217 % 20922789888000,7437374403113 % 2736057139200, 17403456103284421 % 6402373705728000,82666416490601 % 30411275102208] 10.6 Exponential Generating Functions Up until now, the generating functions were used for solving combinatorial prob- lems in which order was irrelevant. These generating functions are sometimes called ordinary generating functions. To tackle combinatorial problems where order of selection plays a role, it is convenient to use generating functions in a slightly different way. The exponential generating function for a sequence (cid:21)n:fn is the function f (z) = f0 + f1z + f2z2 2! + f3z3 3! + (cid:1) (cid:1) (cid:1) = fnzn n! 1 n=0 X Example 10.47 The (ordinary) generating function of [1; 1; 1; : : :] is z , the ex- ponential generating function of [1; 1; 1; : : :] is ez. If ez is taken as an ordinary generating function, it generates the sequence [1; 1 (cid:0) 1! ; 1 2! ; 1 3! ; 1 4! ; : : :]. 1 1 Example 10.48 (1 + z)r is the ordinary generating function for the problem of picking an n-sized subset from an r-sized set, and it is the exponential generating function for the problem of picking a sequence of n distinct objects from an r-sized set. We see from Examples 10.47 and 10.48 that the same function g(z) can be used for generating different sequences, depending on whether g(z) is used as an ordinary or an exponential generating function. It makes sense, therefore, to define an operation that maps ordinarily generated sequences to exponentially generated sequences. 10.6. EXPONENTIALGENERATINGFUNCTIONS 397 o2e :: Num a => [a] -> [a] o2e [] = [] o2e (f:fs) = f : o2e (deriv (f:fs)) With this we get: COR> take 10 (o2e expz) [1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1] A function for converting from exponentially generated sequences to ordinarily generated sequences is the converse of this, so it uses integration: Fractional a => [a] -> [a] e2o :: e2o [] e2o (f:fs) = [f] + (int (e2o (fs))) = [] This gives: COR> take 9 (e2o (1/(1-z))) [1 % 1,1 % 1,1 % 2,1 % 6,1 % 24,1 % 120,1 % 720,1 % 5040,1 % 40320] Example 10.49 Here is how (z + 1)10 is used to solve the problem of finding the number of ways of picking subsets from a set of 10 elements: COR> (1+z)^10 [1,10,45,120,210,252,210,120,45,10,1] And here is how the same function is used to solve the problem of finding the number of ways to pick sequences from a set of 10 elements: COR> o2e ((1+z)^10) [1,10,90,720,5040,30240,151200,604800,1814400,3628800,3628800] Example 10.50 Suppose a vase contains red, white and blue marbles, at least four of each kind. How many ways are there of arranging sequences of marbles from 398 CHAPTER10. CORECURSION the vase, with at most four marbles of the same colour? Solution: the exponential generating function is (1 + z + z2 6 + z4 24 )3. The idea: if you pick n marbles of the same colour, then n! of the marble orderings become indistinguishable. Here is the Haskell computation of the solution. 2 + z3 COR> o2e ([1,1,1/2,1/6,1/24]^3) [1 % 1,3 % 1,9 % 1,27 % 1,81 % 1,240 % 1,690 % 1,1890 % 1,4830 % 1, 11130 % 1,22050 % 1,34650 % 1,34650 % 1] Example 10.51 Suppose a vase contains red, white and blue marbles, an unlim- ited supply of each colour. How many ways are there of arranging sequences of marbles from the vase, assume that at least two marbles are of the same colour? Solution: a suitable exponential generating function is (ez 6 + z4 24 + 1)3 = ( z2 2 + z3 )3: (cid:0) (cid:0) z (cid:1) (cid:1) (cid:1) COR> take 10 (o2e ((expz - z - 1)^3)) [0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,90 % 1,630 % 1,2940 % 1,11508 % 1] This gives the number of solutions for up to 9 marbles. For up to 5 marbles there are no solutions. There are 90 = 6! 23 sequences containing two marbles of each colour. And so on. Exercise 10.52 Suppose a vase contains red, white and blue marbles, at least four of each kind. How many ways are there of arranging sequences of marbles from the vase, with at least two and at most four marbles of the same colour? 10.7 Further Reading Domain theory and proof by approximation receive fuller treatment in [DP02]. Generating functions are a main focus in [GKP89]. The connection between gen- erating functions and lazy semantics is the topic of [Kar97, McI99, McI00]. A coinductive calculus of streams and power series is presented in [Rut00]. Core- cursion is intimately connected with circularity in definitions [BM96] and with the theory of processes and process communication [Fok00, Mil99]. Chapter 11 Finite and Infinite Sets Preview Some sets are bigger than others. For instance, finite sets such as , g are smaller than infinite ones such as N and R. But there are varieties of infinity: the infinite set R is bigger than the infinite set N, in a sense to be made precise in this chapter. 0; 1; 2 and f ; This final chapter starts with a further discussion of the Principle of Mathematical Induction as the main instrument to fathom the infinite with, and explains why some infinite sets are incomparably larger than some others. module FAIS where 11.1 More on Mathematical Induction The natural numbers form the set N = . The Principle of Mathemati- f cal Induction allows you to prove universal statements about the natural numbers, N E(n). that is, statements of the form 0; 1; 2; : : : g n 8 2 399 400 CHAPTER11. FINITEANDINFINITESETS Fact 11.1 (Mathematical Induction) For every set X N, we have that: (cid:18) if 0 2 X and N(n n 8 2 2 X ) n + 1 2 X), then X = N. (cid:18) X and N satisfies both 0 This looks not impressive at all, since it is so obviously true. For, suppose that X). Then X X (and, hence, that X = N) as follows. First, we have as a we see that N (cid:18) -elimination, n = 0) yields first Given, that 0 X. Applying the second Given again that 0 ( X. And so 8 on, for all natural numbers. 2 1 -elimination, n = 1) we obtain 1 X. Therefore, 1 X X. Then the second Given ( X; therefore, 2 n + 1 N(n ) ) ) X X 2 2 2 2 2 2 2 2 2 2 n 8 8 2 Nevertheless, despite its being so overly true, it is difficult to overestimate the importance of induction. It is essentially the only means by with we acquire infor- mation about the members of the infinite set N. Sets versus Properties. With every property E of natural numbers there corre- sponds a set of natural numbers. Thus, induction can also be formulated using properties instead of sets. It then looks as follows: E(n) N 2 n g f j If E is a property of natural numbers such that (i) E(0), (ii) n 2 8 N E(n). then n 8 2 N [E(n) E(n + 1)], ) Induction: Basis, Step, Hypothesis. According to Induction, in order that E(n) is true for all n N, it suffices that 2 (i) this holds for n = 0, i.e., E(0) is true, and (ii) this holds for a number n + 1, provided it holds for n. As we have seen, the proof of (i) is called basis of a proof by induction, the proof of (ii) is called the induction step. By the introduction rules for , in order to carry out the induction step, you are granted the Given E(n) for an arbitrary number n, whereas you have to show that E(n + 1). The assumption that E(n) is called the induction hypothesis. The nice thing about induction is that this induc- tion hypothesis comes entirely free. and ) 8 Fact 11.2 (Scheme for Inductions) Viewed as a rule of proof, induction can be schematized as follows. 11.1. MOREONMATHEMATICALINDUCTION 401 E(n) ... E(0) E(n + 1) nE(n) 8 A proof using the Principle of Induction of a statement of the form n E(n) is called a proof with induction with respect to n. It can be useful to mention the parameter n when other natural numbers surface in the property E. 8 According to the above schema, the conclusion n E(n) can be drawn on the basis of (i) the premiss that E(0), and (ii) a proof deriving E(n + 1) on the basis of the induction hypothesis E(n). 8 The following strong variant of induction is able to perform tasks that are beyond the reach of the ordinary version. This time, we formulate the principle in terms of properties. Fact 11.3 (Strong Induction) For every property E of natural numbers: if n 8 2 N [ ( m < n E(m)) 8 E(n) ], then N E(n). n 8 2 ) Using strong induction, the condition esis from which you have to derive that E(n). 8 m < n E(m) acts as the induction hypoth- This is (at least, for n > 1) stronger than in the ordinary form. Here, to establish that E(n), you are given that E(m) holds for all m < n. In ordinary induction, you are required to show that E(n + 1) on the basis of the Given that it holds for n only. Note: using ordinary induction you have to accomplish two things: basis, and induction step. Using strong induction, just one thing suffices. Strong induction has no basis. E(n) ]. Proof. (of 11.3.) Suppose that n N [ ( m < n E(m)) f 8 n ) j 8 8 2 Define the set X by X = . Assumption (1) says that every m < n E(m) element of X has property E. Therefore, it suffices showing that every natural number is in X. For this, we employ ordinary induction. Basis. 0 I.e.: since there are no natural numbers < 0. The implication m < 0 trivially satisfied. Induction step. E(m)]. This holds trivially E(m) is m < 0 E(m); written differently: m[m < 0 X. ) ) 2 8 8 g (1) 402 CHAPTER11. FINITEANDINFINITESETS Assume that (induction hypothesis) n m < n E(m). As- 2 sumption (1) implies, that E(n) is true. Combined with the induction hypothesis, we obtain that X. The induction step is completed. Induction now implies, that X = N. According to the remark above, we have that m < n + 1 E(m). This means, that n + 1 X. That means: 2 8 8 N E(n). n 8 2 It is useful to compare the proof schemes for induction and strong induction. Here they are, side by side: E(n) ... E(0) E(n + 1) nE(n) 8 8 m < n E(m) ... E(n) nE(n) 8 An alternative to strong induction that is often used is the minimality principle. Fact 11.4 (Minimality Principle) Every non-empty set of natural numbers has a least element. Proof. Assume that A is an arbitrary set of natural numbers. It clearly suffices to show that, for each n N, 2 A n 2 ) A has a least element. For, if also is given that A , then some n ; applied to such an element yields the required conclusion. = 2 A must exist. Thus the implication Here follows a proof of the statement using strong induction w.r.t. n. Thus, assume that n is an arbitrary number for which: Induction hypothesis: for every m < n: m A 2 A has a least element. ) To be proved: n A 2 ) A has a least element. A. There are two cases. Proof: Assume that n (i) n is (by coincidence) the least element of A. Thus, A has a least element. (ii) n is not least element of A. Then some m can apply the induction hypothesis and again find that A has a least element. A exists such that m < n. So we 2 2 Remark. The more usual proof follows the logic from Chapter 2. This is much more complicated, but has the advantage of showing the Minimality Principle to be equivalent with Strong Induction. 6 11.1. MOREONMATHEMATICALINDUCTION 403 Strong Induction is the following schema: n[ m < n E(m) 8 8 E(n)] ) n E(n): ) 8 Since E can be any property, it can also be a negative one. Thus: n[ m < n E(m) 8 8 : E(n)] ) : n E(n): ) 8 : The contrapositive of this is (Theorem 2.10 p. 45): n E(n) :8 Applying Theorem 2.40 (p. 65): : ) :8 n[ m < n E(m) 8 : E(n)]: ) : n E(n) 9 Again Theorem 2.10: ) 9 n [ 8 : m < n E(m) : n E(n) n[ m < n E(m) : Some final transformations eventually yield: ) 9 8 9 E(n)]: ) : E(n)]: ^ :: n E(n) 9 n[E(n) ) 9 ^ :9 m < n E(m)] — which is Minimality formulated using properties. Exercise 11.5 Prove Induction starting at m, that is: for every X and N(m 6 n n > m(n X), then n + 1 X n n N, if m (cid:18) X). X 2 8 2 ) 2 ) 2 Hint. Apply induction to the set Y = 2 j N is such that 1 X). Show that every odd number is in X. Exercise 11.6 Suppose that X n + 2 (cid:18) f n 2 X . g X and 8 2 m + n n 8 2 N(n X 2 ) Definition 11.7 A relation on A is called well-founded if no infinite sequence a0 exists in A. Formulated differently: every sequence a0 (cid:31) (cid:30) a1 (cid:30) a2 (cid:30) a2 (cid:31) (cid:1) (cid:1) (cid:1) in A eventually terminates. (cid:1) (cid:1) (cid:1) (cid:30) a1 (cid:31) Example 11.8 According to Exercise 4.34 (p. 136), the relation by: a b iff }(a) = b, is well-founded. on sets defined (cid:30) 2 (cid:30) An easier example is in the next exercise. Exercise 11.9 Show that < is well-founded on N. That is: there is no infinite sequence n0 > n1 > n2 > . (cid:1) (cid:1) (cid:1) Hint. Show, using strong induction w.r.t. n, that for all n E(n) signifies that no infinite sequence n0 = n > n1 > n2 > Alternatively, use Exercise 11.10. 2 N, E(n); where starts at n. (cid:1) (cid:1) (cid:1) 404 CHAPTER11. FINITEANDINFINITESETS Exercise 11.10* Well-foundedness as an induction principle. Let tion on a set A. (cid:30) be a rela- 1. Suppose that (cid:30) is well-founded. Assume that X A satisfies (cid:18) a 8 2 A( b 8 (cid:30) a(b X) a 2 ) 2 X): Show that X = A. Hint. Show that any a0 2 in A. a1 (cid:31) a0 (cid:31) 2. Conversely: a2 (cid:31) (cid:1) (cid:1) (cid:1) A (cid:0) X can be used to start an infinite sequence (cid:18) Suppose that every X coincides with A. Show that Hint. Suppose that a0 (cid:31) . a0; a1; a2; : : : g f (cid:30) a1 (cid:31) A satisfying 2 b A( a 8 8 is well-founded. a2 (cid:31) (cid:1) (cid:1) (cid:1) a(b X) a 2 ) 2 X) (cid:30) , and consider the set X = A (cid:0) Exercise 11.11* Let R be a relation on a set A. Recall that R(cid:3) denotes the reflexive transitive closure of R, cf. Exercise 5.46 p. 173. 1. Assume that for all a; b1; b2 2 that b1Rc and b2Rc. Show that R is confluent, that is: for all a; b1; b2 2 A exists such that b1R(cid:3)c and b2R(cid:3)c. then c A, if aRb1 and aRb2, then c 2 A exists such 2 A, if aR(cid:3)b1 and aR(cid:3)b2, 2. Assume that R is weakly confluent, that is: for all a; b1; b2 2 2 A exists such that b1R(cid:3)c and b2R(cid:3)c. aRb2, then c A counter-example to confluence is called bad. That is: a is bad iff there A we have that are b1; b2 2 b1R(cid:3)c and b2R(cid:3)c. Show: if a is bad, then a bad b exists such that aRb. A such that aR(cid:3)b1 and aR(cid:3)b2, and for no c 2 A, if aRb1 and 3. (In abstract term rewriting theory, the following result is known as New- man’s Lemma.) Assume that R is weakly confluent. Furthermore, assume that R(cid:0) 1 is well-founded. Show that R is confluent. 11.1. MOREONMATHEMATICALINDUCTION 405 Hint. Use part 2; alternatively, use Exercise 11.10. Remark. That R(cid:0) For example, R = confluent. 1 is well-founded is necessary here. (1; 0); (1; 2); (2; 1); (2; 3) f g is weakly confluent but not Exercise 11.12 Suppose that exists such that for all n element m X such that for all n 2 = X N X, n 6 m. Show that X has a maximum, that is: an N, and that X is bounded, i.e.: that m ; 6 (cid:18) 2 2 2 Induction w.r.t. m. Thus, E(m) is: every non-empty X X, n 6 m. Hint. n X (n 6 m), has a maximum. 2 8 Exercise 11.13 Suppose that f : N Show by means of an induction argument that for all n ! N is such that n < m N such that (cid:18) f (n) < f (m). ) N: n 6 f (n). 2 Exercise 11.14 Suppose that a0; a1; a2; : : : is an infinite sequence of natural num- bers. Prove that i; j N exist such that both i < j and ai 6 aj. 2 Exercise 11.15 The function g : N+ N+ (cid:2) 1. If n < m, then g(n; m) = g(n; m (cid:0) ! n). 2. g(n; m) = g(m; n). 3. g(n; n) = n. N+ has the following properties. Show that g(n; m) is the gcd (greatest common divisor) of n and m. Exercise 11.16* You play Smullyan’s ball game on your own. Before you is a box containing finitely many balls. Every ball carries a natural number. Next to the box, you have a supply of as many numbered balls as you possibly need. A move in the game consists in replacing one of the balls in the box by arbitrarily (possibly zero, but finitely) many new balls that carry a number smaller than that on the one you replace. (Thus, 999 balls numbered 7 can replace one ball numbered 8, but a ball numbered 0 can only be taken out of the box since there are no natural numbers smaller than 0.) Repeat this move. Show that, no matter how you play the game, you’ll end up with an empty box eventually. (Thus, your last moves necessarily consist in throwing away balls num- bered 0.) 406 CHAPTER11. FINITEANDINFINITESETS Hint. Proof by Contradiction. Suppose that you can play ad infinitum, and that Bk is how the box looks after your k-th move. Derive a contradiction applying strong induction w.r.t. the greatest number n present on one of the balls in the box B0 you start from. If m is the number of balls in B0 carrying n, apply a second strong induction, now w.r.t. m. Exercise 11.17 Implement a simplified version of Smullyan’s ball game from the previous exercise, where (i) the box of balls is represented as a list of integers, (ii) it is always the first integer n > 1 that gets replaced, (iii) an integer n > 1 gets replaced by two copies of n 1, (iv) the game terminates with a list consisting of just ones. The type declaration should run: (cid:0) ballgame :: [Integer] -> [[Integer]]. How long will it take before ballgame [50] terminates? Minutes? Hours? Days? Years? Exercise 11.18 The following theorem implies that, e.g., all inhabitants of Ams- terdam have the same number of hairs. Theorem: in every set of n people, everyone has the same number of hairs. Proof. Induction w.r.t. n. Basis. n = 0 (or n = 1): trivial. Induction step. Induction hypothesis: the statement holds for sets of n people. Now assume that A is an (n + 1)-element set of humans. Arbitrarily choose dif- ferent p and q in A; we show that p and q have the same number of hairs. A p g have n elements, thus the induction hypothesis applies to these sets. and A q (cid:0) f Choose r . Then r and q have the same number of hairs (they are both p; q g 2 (cid:0) f in the set A ); and r and p have the same number of hairs (similar reason). p g Thus, p and q have the same number of hairs as well. (cid:0) f (cid:0) f g A Explain this apparent contradiction with common sense observation. 11.2 Equipollence In order to check whether two finite sets have the same number of elements, it is not necessary at all to count them. For, these numbers are the same iff there is a bijection between the two sets. Sometimes, it is much easier to construct a bijection than to count elements. Imag- ine a large room full of people and chairs, and you want to know whether there are as many people as there are chairs. In order to answer this question, it suffices to ask everyone to sit down, and have a look at the resulting situation. 11.2. EQUIPOLLENCE 407 This observation motivates the following definition. Definition 11.19 (Equipollence) Two sets A and B are called equipollent if there is a bijection from A to B. Notation: A B. (cid:24) Example 11.20 (Trivial but Important) The set N is equipollent with its proper subset N+ = N n+1. We can generate the graph of this function in Haskell by means of: . For, this is witnessed by the successor function n g 7(cid:0)! (cid:0)f 0 succs = [ (n, succ n) | n <- [0..] ]. Of course, that a set is equipollent with one of its proper subsets can only happen in the case of infinite sets. The example shows that the notion of equipollence can have surprising properties when applied to infinite sets. The following theorem shows that is an equivalence. (cid:24) Theorem 11.21 For all sets A; B; C: 1. A 2. A 3. A (cid:24) (cid:24) (cid:24) A B = ) B B B ^ (cid:24) A (cid:24) C = ) A C (cid:24) (reflexivity), (symmetry), (transitivity). Proof. 1. 1A is a bijection from A to itself. 2. If f : A 3. If f : A as well. (Cf. Lemma 6.36, p. 223) B is a bijection, then f (cid:0) B and g : B ! ! ! 1 is a bijection : B C are bijections, then g A. ! f : A (cid:14) C is a bijection ! The standard example of an n-element set is serves as well.) This motivates part 1 of the following definition. 0; : : : ; n (cid:0) g f 1 . (Of course, 1; : : : ; n f g Definition 11.22 (Finite, Infinite) 1. A set has n elements if it is equipollent with 0; : : : ; n f 1 . g (cid:0) 2. It is finite if n 2 N exists such that it has n elements. 408 CHAPTER11. FINITEANDINFINITESETS 3. It is infinite if it is not finite. Example 11.23 1. The set 2. If A has n elements and x is arbitrarily chosen, then A elements (depending on whether x has 0 elements, hence is finite. ; ; A). Thus, if A is finite then so is A [ f g x has n or n + 1 x . g [ f 2 The proof of the following unsurprising theorem illustrates the definitions and the use of induction. Theorem 11.24 N is infinite. Proof. We show, using induction w.r.t. n, that for all n N, 2 The induction step applies Exercise 11.25. N 0; : : : ; n 6(cid:24) f 1 : g (cid:0) Basis. If n = 0, then equipollent with Induction step. 0; : : : ; n f . ; 1 g (cid:0) = . And a non-empty set like N cannot be ; 1 (cid:0) . g 6(cid:24) f 0; : : : ; n 6(cid:24) f 0; : : : ; n Induction hypothesis: N To be proved: N . g Proof: Assume that, nevertheless, a bijection from N to 0; : : : ; n g exists. According to Exercise 11.25 we may assume there is a bi- jection f : N Its restriction f (cid:22) (N 0 g to ) (This is the Triv- g ial but Important Example 11.20). Conclusion (Theorem 11.21.3): N ) (Definition 6.12 p. 214) then is a bijection from N 0 g (cid:0)f 0; : : : ; n . But this contradicts the induction hypothesis. . We have that N g such that f (0) = n. 0; : : : ; n 0; : : : ; n ! f (cid:0) f (cid:0)f (N (cid:0) (cid:24) f f g 1 1 0 (cid:24) f (cid:0) g Exercise 11.25 Suppose that A f : A B exists such that f (a) = b. (cid:24) B, a A and b 2 2 B. Show that a bijection ! (cid:24) B, we have a bijection g : A B. If, by coincidence, g(a) = b, Hint. By A we let f be g. Thus, assume g(a) = b0 A exists such that g(a0) = b. Make a picture of the situation and look whether you can find f by suitably modifying g. = b. Since g is surjective, a0 ! 2 6 11.2. EQUIPOLLENCE 409 Exercise 11.26* Suppose that A bijection, using f (cid:3) : }(A) witnesses A B. ! (cid:24) B. Show that }(A) }(B). Write out the }(B), with f (cid:3)(X) = f [X], for f a bijection that (cid:24) (cid:24) Exercise 11.27 Show that, for every set A: }(A) 0; 1 A. (cid:24) f g Hint. Associate with X (cid:18) defined by: (cid:31)X (a) = 1 iff a A sets to functions) is the bijection you are looking for. A its characteristic function, that is, the function (cid:31)X : (cid:31)X (that sends X. The function X ! f 0; 1 7! 2 g Exercise 11.28 Suppose that A B. Show: (cid:24) 1. if A has n elements,then so has B, 2. if A is finite, then so is B, 3. if A is infinite, then so is B. Exercise 11.29 f is a function. Show that f dom (f ). (cid:24) Exercise 11.30* Suppose that R is an equivalence on A. V = A=R is the corre- sponding quotient. Show that the set of all partitions of V is equipollent with the set of all equivalences Q on A for which R Q. (cid:18) Exercise 11.31* Suppose that n; m 0; : : : ; m 1 (cid:0) f Hint. Use induction to show that f Employ the method of proof of Theorem 11.24. n < m 8 . g N and n < m. Show that 0; : : : ; n f 1 (cid:0) g 6(cid:24) 2 0; : : : ; n 1 0; : : : ; m (cid:0) g 6(cid:24) f . 1 g (cid:0) Exercise 11.32 Suppose that X m). Show that X is finite. (cid:18) N and m 2 N are give such that n 8 2 X(n < The following exercise explains how to apply induction to finite sets. Exercise 11.33* Prove the following induction principle for finite sets. If E is a property of sets such that ), 1. E( ; 2. for every set A and every object x A: if E(A), then also E(A x ), g [ f 62 410 CHAPTER11. FINITEANDINFINITESETS then E holds for every finite set. Hint. Apply induction to the property E 0 of numbers, defined by E 0(n) has n elements E(A)]. ) A[A (cid:17) 8 Exercise 11.34* Show that a subset of a finite set is finite. Exercise 11.35* Show that the union of two finite sets is finite. Exercise 11.36* Suppose that h is a finite injection with dom (h) rng (h) (cid:18) such that f B. Show that a bijection f : A h. (And what, if h is infinite?) B. Suppose that A (cid:24) ! A and B exists (cid:18) (cid:19) Hint. Induction on the number n of elements of h. (The case n = 1 is Exer- cise 11.25.) Exercise 11.37* Show that a proper subset of a finite set never is equipollent with that set. Exercise 11.38 Suppose that A and B are finite sets and f : A Show: ! B a bijection. 1. (B A) (cid:0) (cid:24) (A (cid:0) B), 2. there exists a bijection g : A B A [ ! [ B such that f g. (cid:18) Exercise 11.39* Show: a set A is finite iff the following condition is satisfied for every collection E E), then E and A }(A): if A (B ; 2 g 2 [ f E. (cid:18) B E 2 2 8 8 a a 2 11.3 Infinite Sets One of the great discoveries of Cantor is that for any set A, the set }(A) is ‘larger’ (in a sense to made precise below) than A. This shows the existence of ‘Cantor’s paradise’ of an abundance of sets with higher and higher grades of infinity. 11.3. INFINITESETS 411 At Most Equipollent To Recall Definition 11.19: sets are equipollent means that there is a bijection be- tween the two. Definition 11.40 (At Most Equipollent To) The set A is at most equipollent to B if an injection exists from A into B. Notation: A B. (cid:22) Example 11.41 Z R+. (cid:22) Theorem 11.42 For every infinite set A: N A. (cid:22) Proof. Suppose that A is infinite. An injection h : N produced in the following way. ! A as required can be (since is finite). Thus, it i s possible to choose an element h(0) A. h(0) is finite). Thus, another element h(1) A 2 g (cid:0)f f 2 h(0) g Note: A = = Now, A exist. f ; h(0) ; (for, g Again, A = h(0); h(1) , etc. f g Going on, this argument produces different elements h(0); h(1); h(2); : : : in A; thus, the corresponding function h : N A is an injection. ! Thus, N is the “simplest” infinite set. Exercise 11.43 Show: 1. A 2. A 3. A 4. A (cid:22) (cid:24) (cid:22) (cid:18) A, B = ) B, A (cid:22) B ^ B = ) C = ) C, A (cid:22) B, B A (cid:22) (cid:22) Exercise 11.44* Show the reverse of Theorem 11.42: if N A, then A is infinite. (cid:22) Hint. Cf. the proof of Theorem 11.24, p. 408. A slight modification (that uses a modification of Exercise 11.25) shows that for all n, N ; this implies what is required. 0; : : : ; n 6(cid:22) f (cid:0) 1 g 6 6 6 412 CHAPTER11. FINITEANDINFINITESETS 2 A (cid:0) rng (h). Define f : N Exercise 11.45 Suppose that h : A b f (3) = h(f (2)) = h(h(f (1))) = h(h(h(f (0)))) = h(h(h(b))). Show that f (n) is different from f (0); : : : ; f (n Conclusion: f is a injection, and N 1) (n A. ! (cid:0) 2 A is an injection that is not surjective. Say, ! A by: f (0) = b, f (n + 1) = h(f (n)). E.g., N). (Induction w.r.t. n.) (cid:22) Exercise 11.46 Show: if N exists. (cid:22) A, then a non-surjective injection h : A A ! Exercise 11.47 Show: a set is infinite iff it is equipollent with one of its proper subsets. Hint. Use Theorem 11.42 and Exercises 11.44, 11.45 and 11.46. Exercise 11.48 Suppose that A is finite and that f : A iff f is injective. ! A. Show: f is surjective Hint. ( : use Exercises 11.44 and 11.45. Countably Infinite The prototype of a countably infinite set is N. Definition 11.49 (Countable) A set A is countably infinite (or: denumerable) if N A. (cid:24) Exercise 11.50* Show: a subset of a countably infinite set is countably infinite or finite. Exercise 11.51 Show: 1. Z (the set of integers) is countably infinite, 2. a union of two countably infinite sets is countably infinite. Theorem 11.52 N2 = (n; m) f n; m N g 2 j is countably infinite. Proof. Define S(p) = The sets S(p) are pairwise disjoint, and N2 = S(0) S(p) has exactly p+1 elements: (0; p);(1; p n + m = p (n; m) . g f j S(1) : : : Moreover, S(2) 1);: : : ;(p; 0). Verify that the function [ [ [ (cid:0) 11.3. INFINITESETS 413 j : N2 the pairs of the subsequent sets S(p). N that is defined by j(n; m) = 1 ! 2 (n + m)(n + m + 1) + n enumerates Note that the function j enumerates N2 as follows: (0; 0); (0; 1); (1; 0); (0; 2); (1; 1); (2; 0); (0; 3); (1; 2); (2; 1); (3; 0); : : : Look at all pairs (n; m) as the corresponding points in two-dimensional space. Visualize j as a walk along these points. Of course, there are many other “walks” along these points that prove this theorem. E.g., (visualize!) (0; 0); (1; 0); (1; 1); (0; 1); (0; 2); (1; 2); (2; 2); (2; 1); (2; 0); (3; 0); (3; 1); : : : Theorem 11.53 The set of positive rationals Q+ is countably infinite. N2 for which q = Proof. Identify a positive rational q n m and for which n and m are co-prime. Use Theorem 11.52 and Exercise 11.50. Q+ with the pair (n; m) 2 2 Exercise 11.54 Show that Q is countably infinite. Exercise 11.55 Show that N(cid:3) is countably infinite. Exercise 11.56 Show that a union of countably infinitely many countably infinite sets is countably infinite. Uncountable Definition 11.57 (Less Power Than) A set A has power less than B if both A B and A B. (cid:22) 6(cid:24) Notation: A B. (cid:30) Thus: A B (cid:30) () A B A 6(cid:24) ^ B: (cid:22) Example 11.58 N R (Theorem 11.60). 0; : : : ; n f (cid:30) 1 g (cid:30) (cid:0) N (Theorem 11.24); Warning. That A surjective injection from A into B. (cid:30) B implies but is not equivalent with: there exists a non- 414 CHAPTER11. FINITEANDINFINITESETS B means by definition: A That A injection f : A A and B; in particular, the injection f cannot be surjective. B exists. If, moreover, A B and A ! 6(cid:24) (cid:30) (cid:22) B. That A B means that an (cid:22) B, then no bijection exists between 6(cid:24) Counter-examples to the converse: (Example 11.20) the successor-function n 7(cid:0)! n + 1 is a non-surjective injection : N N; (cid:30) ! the identity function 1N is a non-surjective injection from N into Q, but we do not have (Exercise 11.54), that N N, but of course it is false that N Q. (cid:30) Definition 11.59 (Uncountable) A set A is uncountable in case N A. (cid:30) The following is Cantor’s discovery. Theorem 11.60 R is uncountable. Proof. (i) That N that no bijection h : N That is: (cid:22) ! R is clear (Exercise 11.43.3, N R). (ii) It must be shown, (cid:18) R exists. In fact, there is no surjection from N to R. Claim. To every function h : N Proof. Suppose that h : N real h(n): h(n) = pn + 0:rn decimals rn 0; 1; 2; : : : ; 9 i 2 f ! 0 rn R there exists a real r such hat r ran(h). ! R. Write down the decimal expansion for every 1 rn Z, pn 6 h(n) < pn + 1, and 2 (cid:1) (cid:1) (cid:1) . (E.g., , where pn p2 = 2 + 0; 15 62 2 ) (cid:0) (cid:0) 0; 1; 2; : : : ; 9 n. The real g then differs from h(n) in its n-th decimal digit (n = 0; 1; 2; : : :). g For every n, choose a digit rn r = 0; r0r1r2 (cid:1) (cid:1) (cid:1) = h(n). For, a real can have two However, even if pn = 0, this does not imply r decimal notations. E.g., 0:5000 . But, a tail of zeros vs. a tail of = 0:4999 nines is the only case for which this happens. So, this problem can be avoided if we require rn to be different from 0 and 9. such that rn = rn 2 f (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) Recall, that }(A) = is the collection of all subsets of A (Defini- tion 4.22, p. 132). The powerset operation produces in a direct way sets of greater power. X X (cid:18) A g f j In particular, not every uncountable set is equipollent with R. Theorem 11.61 A }(A). (cid:30) Proof. (i) The injection a (ii) To show that A h : A 6(cid:24) }(A), an element D 7! f a g ! }(A) (cid:0) 2 rng (h). from A into }(A) shows that A }(A). }(A), we exhibit, as in the previous proof, for every function (cid:22) 6 6 11.3. INFINITESETS 415 A a f 2 j A. 2 Such an element can be simply described in this context: we take D = a h(a) 62 . g rng (h), then D would be the value of some d D, or d D. 62 2 If we would have that D Now there are two cases: either d 2 2 If d diction. And if d hence d 62 2 Conclusion: D rng (h). 62 D; again a contradiction. D, then, by definition of D, we would have that d D, then, since D = h(d), we would have that d 62 h(d) = D: contra- h(d), and 62 Corollary 11.62 1. N }(N) }(}(N)) (cid:30) (cid:1) (cid:1) (cid:1) (cid:30) 2. for every set A there exists a set B such that A (cid:30) , B. (cid:30) Exercise 11.63 Show: 1. A 2. A A, B 6(cid:30) (cid:22) 3. A B () B A B (cid:30) C = A (cid:24) _ B, A C. (cid:30) ) 4. What is wrong with the following “proof ” for 2 (cid:24) (cid:30) ^ Given is that A B. Suppose that f : A (a) f is (by accident) surjective. Then A (b) f is not surjective. Then A (cid:22) ) ?: B is an injection. B. B, hence A ! (cid:24) B. 6(cid:24) (cid:30) Exercise 11.64 Show: if A is finite, then A (The converse of this is true as well) (cid:30) N. Exercise 11.65 Show: the real interval (0; 2 able. 9 ] = R r f 2 j 0 < r 6 2 9 g is uncount- Exercise 11.66 A is a set. Define h : A . a g h(a) A 2 62 a f j }(A) by h(a) = ! . Determine a f g Exercise 11.67 Show that N Exercise 11.68* Show that N 0; 1 N. g (cid:30) f Hint. Produce, for every function ’ : N (cid:30) NN. (NN is the set of all functions : N NN, a function f NN N.) ! rng (’). ! 2 (cid:0) 416 CHAPTER11. FINITEANDINFINITESETS Exercise 11.69 Suppose that h : N Q is surjective and that we apply the ! procedure from the proof of Theorem 11.60 to h to produce a real r. Is r a rational or not? *Cantor-Bernstein Theorem The following result is of fundamental importance in the theory of equipollence. Theorem 11.70 (Cantor-Bernstein) A B (cid:22) B (cid:22) ^ A = ) B. A (cid:24) The proof of this theorem is delegated to the exercises. Note that we have two examples of uncountable sets: R (Theorem 11.60) and }(N) (Theorem 11.61). It turns out that these sets are of the same magnitude. Theorem 11.71 R }(N). (cid:24) Proof. We show that R Bernstein Theorem produces the required conclusion. }(Q) }(N) (cid:24) f 0; 1 (cid:24) (cid:22) g (cid:22) N R. From this, the Cantor- 1. R }(Q). (cid:22) }(Q). The function r q 7(cid:0)! f 2 Q j q < r g is an injection from R into }(N). Choose a bijection h : Q N (Exercise 11.54, p. 413). Now h[X] is a bijection between }(Q) and }(N) (Exercise 11.26, p. 409). ! (cid:24) 2. }(Q) X 7(cid:0)! 3. }(N) 0; 1 N. Cf. Exercise 11.27. (cid:24) f N g 0; 1 4. has decimal expansion 0; h(0)h(1)h(2) R. Associate with h : N (cid:22) g f 0; 1 . g ! f (cid:1) (cid:1) (cid:1) the real (in the interval [0; 1 9 ]) that Continuum Problem and Hypothesis. Since N (If so, such an A exists for which A (cid:18) Problem. (The continuum is the set R.) (cid:30) R, it is tempting to ask whether sets A exist such that N R. R.) This question is Cantor’s Continuum (cid:30) (cid:30) A Cantor’s Continuum Hypothesis asserts that such a set does not exist. The usual set-theoretic axioms cannot answer the question. G¨odel proved in 1938 that the axioms cannot show that the Continuum Hypothesis is false. Cohen proved in 1963 that the axioms cannot show that the Continuum Hypothesis is true. Indeed, as far as the axioms are concerned, the power of R can be unimag- R such that inably big and there can be arbitrarily many sets A; B; C; : : : N (cid:18) R. B A C (cid:30) (cid:30) (cid:30) (cid:30) (cid:1) (cid:1) (cid:1) (cid:30) 11.3. INFINITESETS 417 Example 11.72 We show that [0; 1] [0; 1). Although these intervals differ in only one element, establishing a bijection between them is far from trivial. (The reader who doubts this should give it a try before reading on.) Let f : [0; 1] [0; 1) be an arbitrary injection; say, f (x) = 1 h : [0; 1] other arguments r ! 2 x. Consider the following function 4 to f ( 1 8 etc.; on 2 to f ( 1 n in [0; 1], you let h(r) = r. [0; 1): h sends 1 to f (1) = 1 4 ) = 1 2 ) = 1 4 , 1 2 , 1 ! (cid:24) = 2(cid:0) 2 = s i and s = 2(cid:0) n, then by definition of h we have h(r) = r Verify that h is bijective. To check injectivity, let r is of the form 2(cid:0) n, say r = 2(cid:0) are of the form 2(cid:0) j i 1 of h, h(r) = 2(cid:0) = 2(cid:0) (cid:0) other is not, then one of h(r); h(s) is of the form 2(cid:0) h(r) not of the form 2(cid:0) of the form 2(cid:0) is an s = h(s). This shows that h is injective. For surjectivity, let r n then h(r) = r, so there is an s 2 i+1) = 2(cid:0) i with i > 0, then h(2(cid:0) [0; 1]. If neither of r; s = s = h(s). If both = j, then, by definition n and the n and the other is not, so again [0; 1). If r is 2 [0; 1] with h(s) = r. If r is i = r. Again, there j with i 1 = h(s). If one of r; s is of the form 2(cid:0) n, say r = 2(cid:0) [0; 1] with h(s) = r. (cid:0) 2 Lemma 11.73 If A B (cid:22) (cid:18) A, then A B. (cid:24) Exercise 11.74* Prove the Lemma. Hint. Generalize the solution for Example 11.72. Exercise 11.75* Prove Theorem 11.70. Hint. Apply Lemma 11.73 to the composition of the two functions that are given. Exercise 11.76 Show the following variations on the fact that [0; 1] [0; 1): (cid:24) 1. [0; 1] [0; 2 3 ), (cid:24) (x; y) j j (x; y) f f 2. 3. x2 + y2 6 1 x2 + y2 6 1 (x; y) (x; y) j x2 + y2 < 1 , g . 2 g ; x j < 1 y j j j j g (cid:24) f g (cid:24) f Exercise 11.77* Suppose that A R A is uncountable. Can you show that (R (cid:18) (cid:0) A) (cid:0) (cid:24) R? R is finite or countably infinite. Show that Exercise 11.78 Show that (R Q) (cid:0) (cid:24) R. 6 6 6 6 6 6 418 CHAPTER11. FINITEANDINFINITESETS 11.4 Cantor’s World Implemented The following program illustrates that N2 is denumerable: natpairs = [(x,z-x) | z <- [0..], x <- [0..z]] This gives: FAIS> natpairs [(0,0),(0,1),(1,0),(0,2),(1,1),(2,0),(0,3),(1,2),(2,1), (3,0),(0,4),(1,3),(2,2),(3,1),(4,0),(0,5),(1,4),(2,3), (3,2),(4,1),(5,0){Interrupted!} Exercise 11.79 Implement the function pair :: (Int,Int) -> Int that is the inverse of natpairs. It should hold for all natural numbers n that pair (natpairs !! n) = n. Exercise 11.80 Implement a function natstar :: [[Int]] to enumerate N(cid:3) (cf. Exercise 11.55). The following code illustrates that Q is denumerable: rationals = [ (n,m) | (n,m) <- natpairs, m /= 0, gcd n m == 1 ] This gives: FAIS> rationals [(0,1),(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(2,3),(3,2),(4,1),(1,5), (5,1),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(1,7),(3,5),(5,3),(7,1), (1,8),(2,7),(4,5),(5,4),(7,2),(8,1),(1,9),(3,7),(7,3),(9,1),(1,10), (2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2),(10,1),(1,11),(5,7), (7,5),(11,1),(1,12),(2,11),(3,10),(4,9),(5,8),(6,7),(7,6),(8,5),(9,4), (10,3),(11,2),(12,1),(1,13),(3,11),(5,9),(9,5),(11,3),(13,1),(1,14), 11.4. CANTOR’SWORLDIMPLEMENTED 419 (2,13),(4,11),(7,8),(8,7),(11,4),(13,2),(14,1),(1,15),(3,13),(5,11), (7,9),(9,7),(11,5),(13,3),(15,1),(1,16),(2,15),(3,14),(4,13),(5,12), (6,11),(7,10),(8,9),(9,8),(10,7),(11,6),(12,5),(13,4),(14,3),(15,2), (16,1),(1,17),(5,13),(7,11),(11,7),(13,5),(17,1),(1,18),(2,17),(3,16), (4,15),(5,14),(6,13),(7,12),(8,11),(9,10),(10,9),(11,8),(12,7),(13,6), (14,5),(15,4),(16,3),(17,2),(18,1),(1,19),(3,17),(7,13),(9,11),(11,9), (13,7),(17,3),(19,1),(1,20),(2,19),(4,17),(5,16),(8,13),(10,11), (11,10),(13,8),(16,5){Interrupted!} The following code illustrates that True; False g f N is not denumerable: diagonal :: (Integer -> [Bool]) -> Integer -> Bool diagonal f n = not ((f n)!!(fromInteger n)) f :: Integer -> [Bool] f 0 = cycle [False] f (n+1) = True : f n Now [ f n | n <- [0..] ] is a list of streams of booleans, all different, and diagonal f is a new stream of booleans, different from all members of [ f n | n <- [0..] ]. Here is an illustration for initial segments of the lists: FAIS> [ take 11 (f n) | n <- [0..10] ] [[False,False,False,False,False,False,False,False,False,False,False], [True,False,False,False,False,False,False,False,False,False,False], [True,True,False,False,False,False,False,False,False,False,False], [True,True,True,False,False,False,False,False,False,False,False], [True,True,True,True,False,False,False,False,False,False,False], [True,True,True,True,True,False,False,False,False,False,False], [True,True,True,True,True,True,False,False,False,False,False], [True,True,True,True,True,True,True,False,False,False,False], [True,True,True,True,True,True,True,True,False,False,False], [True,True,True,True,True,True,True,True,True,False,False], [True,True,True,True,True,True,True,True,True,True,False]] FAIS> [ diagonal f n | n <- [0..10] ] [True,True,True,True,True,True,True,True,True,True,True] 420 CHAPTER11. FINITEANDINFINITESETS 11.5 *Cardinal Numbers By Theorem 11.21, equipollence is an equivalence on the collection of all sets. Definition 11.81 (Cardinal Number) A cardinal number is an equivalence class w.r.t. equipollence. A j j denotes the cardinal number of the set A modulo . (cid:24) The following is immediate (cf. Lemma 5.80, p. 193): Lemma 11.82 A j j = B j j () B. A (cid:24) Usually, the cardinal number with the natural number n. 0; : : : ; n jf 1 gj (cid:0) of the n-element sets is identified Aleph-zero. @0 = .1 N j j The concept of a cardinal number can be considered as a generalisation of that of a natural number. It is possible to generalize the definitions of addition, multiplication and exponen- tiation to cardinal numbers, and to prove natural laws that generalize those for the natural numbers. j [ B = A j A j B j (provided that A The operations are defined as follows. B = + j : only if A and B are disjoint sets of n resp. m elements does it follow that their ; union has n + m elements), j Thus, by Theorem 11.71, we have that j (cid:2) = 2@0 . @0 is the starting point of an infinite series of cardinal numbers called The cardinal ! is the smallest cardinal bigger ; ( alephs: @1 < @0 < @ n). than every @ @2 < AB j R j j ! < j (cid:2) j A j A j A j and j = (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) B j < = B B \ @ j j j . Using cardinals sometimes makes for amazingly compact proofs. An example is the following theorem. Theorem 11.83 R line. (cid:2) R (cid:24) R, i.e.: there are as many points in the plane as on a 1@ (aleph) is the first letter of the Hebrew alphabet. 11.5. CARDINALNUMBERS 421 Proof. R j R j (cid:2) R j 2@0 R = j j (cid:2) j = 2@0 (cid:2) = 2@0+ @0 = 2@0 R : = j j The third equality uses the rule np Exercise 11.51, p. 412. (cid:2) nq = np+q, the fourth that @0 + @0 = @0, cf. *Further Exercises B2 = B2. Show: A2 and B1 (cid:24) B1 (cid:24) Exercise 11.84 Suppose that A1 (cid:24) B1 = A2 \ A2 (cid:2) B1 (cid:24) AB2 2 1. if A1 \ 2. A1 (cid:2) B2j = 3.* AB1 1 (cid:24) B2 , for we don’t have a rule of exponentiation for cardinal numbers as A2 j j yet. Instead, show how to establish a bijection between AB1 1 A2j j and AB2 2 .) (Hint: it does not do to say , then A1 [ ; A2 [ B1j = j A1j j A1 j B2, B2, = B1 j j B2. Show: A2 and B1 (cid:22) B2, A2 [ B1 (cid:22) Exercise 11.85 Suppose that A1 (cid:22) B2 = 1. if A2 \ 2. A1 (cid:2) 3. }(A1) , then A1 [ ; B2, A2 (cid:2) }(A2), B1 (cid:22) (cid:22) 4.* if A2 6 = can pick a (Hint: Use the fact that, since A2 6 A2 for the definition of the injection that you need.) 1 (cid:22) AB2 2 , then AB1 ; 2 = , you ; Exercise 11.86 Give counter-examples to the following implications: 1. A1 (cid:30) A2 ) A1 [ B A2 [ B (A1 \ B = A2 \ (cid:30) B = ), ; 422 CHAPTER11. FINITEANDINFINITESETS 2. A1 (cid:30) 3. A1 (cid:30) 4. A1 (cid:30) A2 ) A2 ) A2 ) A2 (cid:2) B, A1 (cid:2) AB 1 (cid:30) B (cid:30) AB 2 , BA1 (cid:30) BA2. Exercise 11.87 Show: , then AB ; C [ (cid:24) AB (cid:2) AC, 1. if B 2. (A (cid:2) C = \ B)C 3.* (AB)C (cid:24) (cid:24) AB AC BC (cid:2) C. (Hint: use the currying operation.) (cid:2) Exercise 11.88 Show (n > 1): 1. 2. 0; 1 0; 1 f f N R g g (cid:24) f (cid:24) f 0; : : : ; n 0; : : : ; n NN NR N R (cid:24) (cid:24) (cid:24) (cid:24) RN RR (cid:24) (cid:24) R, (}(R))R (RR)R. (cid:24) g g Exercise 11.89 Show, for all n RN R). (cid:24) N+: Nn 2 (cid:24) N (n.b.: NN N) and Rn R (n.b.: (cid:24) 6(cid:24) Exercise 11.90 Show: 1 < y 6 2 . g (x; y) f R2 j 2 x2 + y2 6 1 ^ y > 0 (x; y) g (cid:24) f R2 j 2 Exercise 11.91 Show: if A is infinite and B finite, then (A B) (B [ (cid:0) A) (cid:24) (cid:0) A. The Greek Alphabet Mathematicians are in constant need of symbols, and most of them are very fond of Greek letters. Since this book might be your first encounter with this new set of symbols, we list the Greek alphabet below. upper case (cid:0) (cid:1) (cid:2) (cid:3) (cid:4) (cid:5) (cid:6) (cid:7) (cid:8) (cid:9) (cid:10) name alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu nu xi pi rho sigma tau upsilon phi chi psi omega lower case (cid:11) (cid:12) (cid:13) (cid:14) " (cid:16) (cid:17) (cid:18) (cid:19) (cid:20) (cid:21) (cid:22) (cid:23) (cid:24) (cid:25) (cid:26) (cid:27) (cid:28) (cid:29) ’ (cid:31) ! 423 424 Bibliography [AHV95] S Abiteboul, R. Hull, and V. Vianu. Foundations of Databases. Ad- dison Wesley, 1995. [Bab61] [Bab94] C. Babbage. On the Principles and Development of the Calculator. Dover, 1961. Edited and with an introduction by P. Morrison and E. Morrison. C. Babbage. Passages from the Life of a Philosopher. Rutgers Uni- versity Press and IEEE-Press, New Brunswick, New Jersey and Pis- cataway, New Jersey, 1994. Edited with a new introduction by Martin Campbell-Kelly. Originally published 1864. [Bal91] V. K. Balakrishnan. Introductory Discrete Mathematics. Dover, 1991. [Bar84] [Bir98] [BM96] [Bry93] [Bur98] H. Barendregt. The Lambda Calculus: Its Syntax and Semantics (2nd ed.). North-Holland, Amsterdam, 1984. R. Bird. Prentice Hall, 1998. Introduction to Functional Programming Using Haskell. J. Barwise and L. Moss. Vicious Circles: On the Mathematics of Non-wellfounded Phenomena. CSLI Publications, 1996. V. Bryant. Yet another introduction to analysis. Cambridge University Press, 1993. Stanley N. Burris. Logic for Mathematics and Computer Science. Prentice Hall, 1998. [CG96] J.H. Conway and R.K. Guy. The Book of Numbers. Springer, 1996. [CR78] R. Courant and H. Robbins. What is Mathematics? An Elementary Approach to Ideas and Methods. Oxford University Press, Oxford, 1978. 425 426 BIBLIOGRAPHY [CrbIS96] R. Courant and H. Robbins (revised by I. Stewart). What is Math- ematics? An Elementary Approach to Ideas and Methods (Second Edition). Oxford University Press, Oxford, 1996. [Doe96] [DP02] H.C. Doets. Wijzer in Wiskunde. CWI, Amsterdam, 1996. Lecture notes in Dutch. B.A. Davey and H.A. Priestley. Introduction to Lattices and Order (Second Edition). Cambridge University Press, Cambridge, 2002. First edition: 1990. [DvDdS78] K. Doets, D. van Dalen, and H. de Swart. Sets: Naive, Axiomatic, and Applied. Pergamon Press, Oxford, 1978. [Ecc97] P. J. Eccles. An Introduction to Mathematical Reasoning. Cambridge University Press, 1997. [EFT94] H.-D. Ebbinghaus, J. Flum, and W. Thomas. Mathematical Logic. Springer-Verlag, Berlin, 1994. Second Edition. [Euc56] Euclid. The Thirteen Books of the Elements, with Introduction and Commentary by Sir Thomas L. Heath. Dover, 1956. [Fok00] W. Fokkink. Introduction to Process Algebra. Springer, 2000. [GKP89] R.L. Graham, D.E. Knuth, and O. Patashnik. Concrete Mathematics. Addison Wesley, Reading, Mass, 1989. 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The Hugs98 user manual. http: //www.haskell.org/hugs/. J. Karczmarczuk. Generating power of lazy semantics. Theoretical Computer Science, 187, 1997. [Knu92] D.E. Knuth. Literate Programming. CSLI Lecture Notes, no. 27. CSLI, Stanford, 1992. [Lar34] D. Lardner. Babbage’s calculating engine. Edinburgh Review, 1834. [McI99] M.D. McIlroy. Power series, power serious. Journal of Functional Programming, 9:323–335, 1999. [McI00] M.D. McIlroy. The music of streams. Elsevier Preprints, 2000. [Mil99] R. Milner. Communicating and Mobile Systems: the (cid:25) Calculus. Cambridge University Press, 1999. [Ore88] O. Ore. Number Theory and its History. Dover, 1988. [Pol57] [RL99] [Rus67] [Rut00] [SKS01] [Tho99] G. Polya. How to Solve It. A New Aspect of Mathematical Method. Princeton University Press, Princeton, 1957. F. Rabhi and G. Lapalme. Algorithms: a Functional Programming Approach. Addison-Wesley, 1999. B. Russell. Letter to Frege. 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Index (a; b), 136 :=, 32 , 61 : (cid:17) A=R, 194 Ac, 131 1, 163 R(cid:0) (cid:1)A, 163 , 127 , 127 \ [ dom (R), 162 , 126 ; , 29, 31 ^ , 29, 32 _ , 29, 35 , , 29, 30 : :n, 247 (cid:8) }(X), 132 , 29, 33 ) ran(R), 162 , 125 a g P A x j g 2 , 118 x P j g , 192 a j n k=1 ak, 54, 241 f f f j , 118 &&, 17, 31, 141 P ||, 32 (.), 69 (.*), 354 (op x), 21 (x op), 21 (x1,x2), 139 +, 6 --, 122 ->, 9, 142 ., 69, 222 /, 15 /=, 5, 124 ::, 9 :l, 2 :r, 5 :t, 9 <, 124 <+>, 36 <=, 124 <=, 13 <=>, 36 =, 5 ==, 5, 124 ==>, 33 <, 8 >=, 8 @, 175 [a], 18 [n..m], 120 Zn, 195 \\, 150 \, 46 _, 141 n , 346 k , 46, 84 (cid:1) (cid:0) ? dom (f ), 206 n, 189 (cid:17) (cid:21)x:t, 58 (mod n), 189 429 430 INDEX ran(f ), 206 , 197 n k , 46 (cid:9) (cid:8) > ^, 6, 119 n, 56 d j e, 221 n!, 213 |, 6, 246 , 407 (cid:24) ‘abc’-formula, 59 absReal, 207 add, 267 addElem, 151 adjustWith, 340 Alexander the Great, v algebra of sets, 127 all, 68 and, 42, 268 antisymmetric relation, 167 any, 68 approx, 375 approximate, 319 apprx, 319 Apt, K.R., ix arbitrary object, 92 assignment statement, 24 asymmetric relation, 167 average, 15 axiomatics, 114 Babbage, C., 334 backsubst, 342 ballgame, 406 bell, 198 Bell numbers, 198 Benthem, J. van, ix Bergstra, J., ix bijection, 218 bijective, 219 bijectivePairs, 219 binary, 288 binding power of operators, 33 binomial theorem, 346, 347 BinTree, 257 bisimulation, 380 black hole, 276 Bool, 8, 30 Boolean function, 39 brackets, 215 Brouwer, L.E.J, 29 Brunekreef, J., ix Cantor’s continuum hypothesis, 416 Cantor’s theorem, 414 Cantor, G., 114 Cantor-Bernstein theorem, 416 cardinal number, 420 case, 142 cat, 265 Catalan numbers, 254, 394 chain, 169 characteristic function, 182 chr, 228 Church, A., 1, 64 class, 124 class, 227 clock, 367 closed form, 212 closures of a relation, 170 co-domain function, 209 co-image, 214 co-prime, 291 coprime, 291 Cohen, P.J., 416 coImage, 215 coImagePairs, 216 coinduction, 382 cols, 339 comp, 223, 389 compare, 141 INDEX comparison property, 169 completeness, 168 complex numbers, 320 complR, 178 composition function –, 69, 222 confluence, 404 congruence, 236 conjunction, 31 constructor, 12 constructor identifier, 12 continue, 335 continuity, 66, 315 continuum problem, 416 contradiction, 48 contraposition, 35 converse, 35 conversion, 15 convolution, 354 Coquand, T., ix corecursion, 362 corecursive definition, 362 countable, 412 curry, 183 Curry, H.B., 1, 183 curry3, 210 CWI, ix data, 139, 257 data, 30, 145 database query, 145 decExpand, 307 decForm, 308 decodeFloat, 311 deduction rule, 78 default, 362 delete, 149 deleteSet, 156 delta, 355 denumerable, 412 deriv, 358 431 deriving, 140 destructive assignment, 24 diagonal, 419 difLists, 334 difs, 332 disjunction, 32 display, 158 div, 20 divergence, 122 divides, 5, 183 domain, 162 Double, 17, 311 e2o, 397 echelon, 341 echelon matrix form, 338 elem, 123 elem, 151 elem’, 150 elemIndex, 308 eliminate, 341 else, 230 emptySet, 156 encodeFloat, 312 Enum, 227 enum_2, 107 EQ, 141 Eq, 124, 246 eq, 184 eq1, 298 equalSize, 190 equation guarding, 6 equational reasoning, 24 equipollent, 406 equiv2listpart, 199 equiv2part, 199 equivalence, 35 equivalence, 188 equivalence’, 188 Eratosthenes sieve of —, 105 432 INDEX error, 8, 230 Erven, T. van, ix Euclid, 103, 290 Euclid’s GCD algorithm, 290 Euler, L., 104 even, 81, 222 evens, 81, 362 evens1, 118 evens2, 119 every, 69 exception handling, 230 exclaim, 252 exp, 221 expn, 253 exponent, 312 fac, 213 fac’, 213 False, 4, 30 fasterprimes, 107 fct2equiv, 232 fct2list, 207 fct2listpart, 234 Fermat, P. de, 104 fib, 254 fib’, 254 Fibonacci numbers, 253, 254, 363 field, 301 filter, 22 fixity declaration, 33 flip, 184 Float, 15, 311 floatDigits, 312 Floating, 311 floatRadix, 311 Fokkink, W., ix foldl, 269 foldn, 252 foldr, 266 foldr1, 268 foldT, 262 forall, 12 Fraenkel, A., 114 fromEnum, 227 fromInt, 15 fromTower, 281 fst, 139 function, 205 domain, 206 function composition, 69, 222 fundamental theorem of algebra, 320 of arithmetic, 293 G¨odel, K., 416 Gaussian elimination, 337 gcd, 291 genDifs, 335 genMatrix, 339 GIMPS, 108 GNU’s Not Unix, 362 Goldreyer, D., 28 Goris, E., ix greatest common divisor definition, 290 Euclid’s algorithm for —, 290 properties, 405 GT, 141 gt1, 299 guard, 7 guarded equation, 6 Haan, R. de, ix halting problem, 121 hanoi, 276 hanoi’, 280 hanoiCount, 280 Haskell, 1 head, 142 Heman, S., ix hex, 289 Hoogland, E., ix INDEX id, 207 idR, 175 Iemhoff, R., ix if then else, 230 iff, 30 ILLC, ix image, 214 image, 215 imagePairs, 216 implication, 33 import, 72 in, 15 induction, 239, 240, 400 strong, 401 infinity of primes, 103 infix, 4 infix, 33 infix notation, 21 infixl, 156 infixr, 156 init, 143 injection, 218 injective, 219 injectivePairs, 219 injs, 221 inorder tree traversal, 262 inR, 178 insertSet, 156 inSet, 155 instance of a type class, 124 instance, 140 Int, 11 int, 388 Integer, 9, 11 integers, 294 Integral, 11 integral rational functions, 331 intersect, 150 intransitive relation, 168 433 intToDigit, 288, 289 intuitionism, 29 inverse, 184 invR, 178 irrational numbers, 311 irrationality of p2, 309 irreflexive relation, 166 irreflR, 179 isAlpha, 232 isEmpty, 156 iterate, 313, 336, 363 Jongh, D. de, ix Just, 231 Kaldewaij, A., ix labeled transition system, 365 lambda abstraction, 46, 58, 145, 211 lambda term, 58 last, 143 law associativity, 46, 247, 249 commutativity, 46, 247, 249 contraposition, 45 DeMorgan, 46, 132 distribution, 46 distributivity, 249 dominance, 48 double negation, 45 excluded middle, 48 idempotence, 45 identity, 48 lazy list, 23, 105 lazy pattern, 373 LD, 4 ldp, 23 ldpf, 23 leaf tree, 263 LeafTree, 263 left triangular matrix form, 338 434 INDEX len, 265 length, 15 leq, 250 leq1, 299 lessEq, 184 let, 15 lexicographical order, 141 limit, 315 linear relation, 169 list, 16, 139 list comprehension, 42, 54, 118 list2fct, 207 list2set, 156 listpart2equiv, 199 listPartition, 199 listRange, 208 lists, 139 listValues, 208 ln, 267 ln’, 272 load Haskell file, 2 logBase, 281 LT, 141 Lucas numbers, 394 Main>, 5 mantissa, 311 map, 21, 151, 272 mapLT, 264 mapR, 264 mapT, 261 Matrix, 338 matrix, 338 maxBound, 208 Maybe, 231 maybe, 231 mechanic’s rule, 313 mechanics, 313 mechanicsRule, 313 Menaechmus, v Mersenne, M., 104 min, 13 minBound, 208 mkStdGen, 366 mlt, 267 mnmInt, 13 mod, modulo, 190 module, 2 modulo, 190 Modus Ponens, 80 mult, 252 mult1, 298 mySqrt, 319 n-tuples, 139 Napier’s number, 221, 395 natpairs, 418 natstar, 418 Natural, 246 natural logarithm, 221 natural number, 246 naturals, 118, 362 necessary condition, 35 negate, 222 negation, 30 Newman’s Lemma, 404 Newman, B., 28 Newton’s method, 313 Newton, I., 346 newtype, 153 next, 335 nextD, 335 nondeterminism, 365 not, 30 notElem, 151 Nothing, 231 Nuall´ain, B. ´O, ix nub, 144 null, 143 Num, 124 o2e, 396 odd, 222 INDEX odds, 362 odds1, 119 oddsFrom3, 107 of, 142 one-to-one, 218 ones, 125 ones, 362 onto, 218 Oostrom, V. van, ix open problems, 109, 111 operation, 236 operator precedence, 42 or, 268 Ord, 124 Ord, 141 ord, 228 order partial —, 168 strict partial —, 168 total —, 169 Ordering, 141 otherwise, 7 overloading, 125 p2fct, 343 pair, 418 pairs, 136 paradox halting —, 121 Russell —, 120 part2error, 231 partial functions, 229 partial order, 168 Pascal’s triangle, 348 Pascal, B., 348 pattern matching, 13, 143, 250 perms, 221 Platonism, 28 plus, 247 plus, 252 plus1, 298 435 po-set reflection, 195 polynomial, 320 polynomials, 331 polynomials and coefficient lists, 344 Ponse, A., ix postorder tree traversal, 262 power series, 387 powerList, 151 powerSet, 156 pre, 253 pre-order, 168 pred, 227 prefix, 4 prefix, 17 prefix notation, 21 Prelude>, 2 Prelude, 26 preorder tree traversal, 262 primCompAux, 142 prime, 23 prime factorization algorithm, 19 prime’’, 165 prime0, 8 primes definition, 60 Mersenne —, 105 primes, 106 primes’, 364 primes0, 22 primes1, 23 principle comprehension —, 114 minimality —, 402 Process, 366 product of sets, 136 product, 213 propositional function, 39 ptd, 369 quasi-order, 168 436 quotient, 194 quotient, 251 quotRem, 286 raccess, 194 random numbers, 366 random streams, 366 Random.hs, 361 randomInts, 366 randomRs, 366 ranPairs, 208 Rational, 341 rationals, 299 countability of —, 413 rationals, 418 rclosR, 182 reals, 310 uncountability of —, 414 recip, 313 recurrence, 212 recursion, 246 recursive definition, 7, 246 reduce1, 298 reflect, 264 reflexive closure, 171 reflexive relation, 166 reflexive transitive closure, 171 reflR, 178 relation antisymmetric —, 167 asymmetric —, 167 between, 162 domain, 162 from—to, 162 intransitive —, 168 irreflexive —, 166 linear —, 169 range, 162 reflexive —, 166 symmetric —, 166 transitive —, 167 INDEX relatively prime, 291 reload Haskell file, 5 rem, 5 remainder, 251 removeFst, 14 reserved keywords, 12 restrict, 214 restrictPairs, 214 restrictR, 182 rev, 269 rev’, 270 rev1, 272 Rodenburg, P., ix rose tree, 264 rows, 339 royal road to mathematics, v Russell, B., 57, 120 Rutten, J., ix sclosR, 182 sections, 21 Set a, 153 Show, 246 showDigits, 289 showSet, 155 sieve, 106, 364 sieve of Eratosthenes, 105 sieve’, 364 significand, 312 sin, 221 singleton, 125 small_squares1, 119 small_squares2, 120 Smullyan’s ball game, 405 Smullyan, R., 405 snd, 139 solveQdr, 59 solveSeq, 342 solving quadratic equations, 59 some, 69 soundness, 168 INDEX 437 space leak, 276 split, 263 splitAt, 308 splitList, 144 sqrt, 59, 221 sqrtM, 313 srtInts, 14 start, 366 stirling, 198 Stirling set numbers, 198 stream, 362 stream bisimulation, 383 strict partial order, 168 String, 16 String, 16 stringCompare, 232 sub domain, 53 substitution principle, 48 subtr, 251 subtr, 253 subtr1, 298 succ, 227 successor, 246 sufficient condition, 35 sum, 241 sum, 15, 54, 241 sumCubes, 245 sumCubes’, 245 sumEvens, 242 sumEvens’, 242 sumInts, 242 sumOdds, 241 sumOdds’, 241 sumSquares, 243 sumSquares’, 243 surjection, 218 surjective, 219 surjectivePairs, 219 Swaen, M., ix symmetric closure, 171 symmetric difference, 132 symmetric relation, 166 symR, 179 tail, 142 take, 107 takeWhile, 313 tan, 221 tclosR, 182 Terlouw, J., ix theFibs, 363 then, 230 theNats, 363 theNats1, 363 theOdds, 363 theOnes, 363 toBase, 289 toEnum, 227 total order, 169 totalR, 178 toTower, 281 tower of Hanoi, 273 transClosure’, 187 transition system, 365 transitive closure, 171 transitive relation, 167 transR, 179 tree leaf –, 263 rose —, 264 tree traversal, 262 trivially true statements, 34 Tromp, J., ix True, 4, 30 truncate, 281 truth function, 39 Turing, A., 64 type, 8, 17, 53, 121, 140 type, 145 type conversion, 15 type declaration, 9 type judgment, 10 INDEX 438 type variables, 18 uncountable, 413 uncurry, 183 uncurry3, 210 undefined, 122 undefined, 375 union, 150 Van Benthem, J., ix variable identifier, 12 vending, 367 Venema, Y., ix Visser, A., ix Vries, F.J. de, ix weak confluence, 404 Wehner, S., ix well-founded, 245, 403 where, 2, 14 Who is Afraid of Red, Yellow and Blue, 28 wild card, 12 wildcard, 141 z, 353 Zermelo, E., 114 zipWith, 339, 363