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Kontraktbaseret Programmering 1 Induction and Recursion Jens Bennedsen Agenda Syllabus for this week • Recursive definitions of sets • Recursive definitions of algorithms • Mathematical Induction – First Principle • Mathematical Induction – Second Principle Slide 2 Ingeniørhøjskolen i Århus Syllabus for this week • Discrete Structures Web Course Material – Section 5A: Recursive Definitions – Section 5D: Proof by Induction – But these web pages also contain material about • • • • • Propositional logic Predicate logic Sets Relations and Functions – Including on-line interactive exercises – May be handy for repetition before exam Slide 3 Ingeniørhøjskolen i Århus Agenda Syllabus for this week Recursive definitions of sets • Recursive definitions of algorithms • Mathematical Induction – First Principle • Mathematical Induction – Second Principle Slide 4 Ingeniørhøjskolen i Århus Recursive Definition of a set • The basis clause: – What are the basic starting elements of the set. • The inductive clause: – In which ways can existing elements be combined to produce new elements. • The extremal clause: – Unless an object can be shown to be a member of the set by applying the first two clauses it is not a member of the set. Slide 5 Ingeniørhøjskolen i Århus Example: The natural numbers • The set of the natural numbers can be defined as: – Basis clause: 0 N. – Inductive clause: For any element x in N, x + 1 is in N. – Extremal clause: Nothing is in N unless it is obtained from either the basis or the inductive clauses. • The “x + 1” is also called “succ(x)” in the literature • This was formalized by the mathematician Peano Slide 6 Ingeniørhøjskolen i Århus How to define the even natural numbers? • The set of the even natural numbers (NE) can be defined as: – Basis clause: 0 NE. – Inductive clause: For any element x in NE, x + 2 is in NE. – Extremal clause: Nothing is in NE unless it is obtained from either the basis or the inductive clauses. Slide 7 Ingeniørhøjskolen i Århus How to define the even integers? • The set of the even integers (EI) can be defined as: – Basis clause: 0 EI. – Inductive clause: For any element x in EI, x + 2 and x -2 are in EI. – Extremal clause: Nothing is in EI unless it is obtained from either the basis or the inductive clauses. Slide 8 Ingeniørhøjskolen i Århus How to define propositional forms? • Let V = {p,q,r,…} be a set of propositional variables, where V does not contain any of the following symbols: (, ), , , , , , T and F. Then the set of the propositional forms can be defined as: – Basis clause: T and F are propositional forms and if x V then x is a propositional form. – Inductive clause: If E1 and E2 are propositional forms then (E1), (E1 E2), (E1 E2), (E1 E2), (E1 E2) are all propositional forms. – Extremal clause: Nothing is a propositional form unless it is obtained from either the basis or the inductive clauses. Slide 9 Ingeniørhøjskolen i Århus Agenda Syllabus for this week Recursive definitions of sets Recursive definitions of algorithms • Mathematical Induction – First Principle • Mathematical Induction – Second Principle Slide 10 Ingeniørhøjskolen i Århus Recursive Thinking • Recursion is a problem-solving approach that can be used to generate simple solutions to certain kinds of problems that would be difficult to solve in other ways • Recursion splits a problem into one or more simpler versions of itself Slide 11 Ingeniørhøjskolen i Århus Recursion in practice Slide 12 Ingeniørhøjskolen i Århus Recursion Recipe 1. Handle the “base case”, where a recursive call is not made. 2. For the other cases, make a recursive call (the recursive “leap of faith”), which must make progress towards a goal (towards the base case) 3. Typically no need for an extremal clause If the base case is not handled, or the recursive call does not progress towards the base case, then the method will call itself infinitely (it will “diverge”). Slide 13 Ingeniørhøjskolen i Århus Recursive Definitions of Algorithms • The factorial N!, for any positive integer N, is defined to be the product of all integers between 1 and N inclusive • This definition can be expressed recursively as: 1! N! = = 1 N * (N-1)! • A factorial is defined in terms of another factorial • Eventually, the base case of 1! is reached Slide 14 Ingeniørhøjskolen i Århus Recursive Definitions 5! 120 5 * 4! 24 4 * 3! 6 3 * 2! 2 2 * 1! 1 Slide 15 Ingeniørhøjskolen i Århus A recursive C++ algorithm for factorial // Precondition n >= 1 // Postcondition result is n! int recursiveFac(int n) Base case { if(n==1) return 1; else return n*recursiveFac(n-1); } Recursive case Slide 16 Ingeniørhøjskolen i Århus Fibonacci numbers • Developed by Leonardo Pisano in 1202. – Investigating how fast rabbits could breed under idealized circumstances. – Assumptions • A pair of male and female rabbits always breed and produce another pair of male and female rabbits. • A rabbit becomes sexually mature after one month, and that the gestation period is also one month. – Pisano wanted to know the answer to the question how many rabbits would there be after one year? Slide 17 Ingeniørhøjskolen i Århus Fibonacci Numbers • The Fibonacci Numbers are defined by – The first two numbers are 1 and 1. – Each subsequent number is the sum of the preceding 2 numbers. • 1,1,2,3,5,8,13,21,34,55,etc. • Recursively it is defined as: • • • • Fibonachi(n) = 1 if n = 0 Fibonachi(n) = 1 if n = 1 Fibonachi(n) = Fibonachi(n-2) + Fibonachi(n-1) otherwise Non-recursively: n n 1 5 1 5 F ( n) 5 2n Slide 18 Ingeniørhøjskolen i Århus The fibonacci Method // Precondition: n >= 0 // Postcondition: The corresponding fibonachi number is // returned long fibonacci(long n){ int fib; if (n <= 1) return 1; Base case else return fibonacci(n-1) + fibonacci(n-2); } Recursive case Slide 19 Ingeniørhøjskolen i Århus Execution Trace (decomposition) fibonacci(3) fibonacci(1) fibonacci(2) Slide 20 Ingeniørhøjskolen i Århus Execution Trace (decomposition) fibonacci(3) fibonacci(1) fibonacci(2) fibonacci(1) fibonacci(0) Slide 21 Ingeniørhøjskolen i Århus Execution Trace (composition) fibonacci(3) + fibonacci(1) fibonacci(2) + fibonacci(1)->1 fibonacci(0)->1 Slide 22 Ingeniørhøjskolen i Århus Execution Trace (composition) fibonacci(3) + fibonacci(2)->2 Slide 23 fibonacci(1)->1 Ingeniørhøjskolen i Århus Execution Trace (composition) fibonacci(3)->3 Slide 24 Ingeniørhøjskolen i Århus Towers of Hanoi In the great temple of Brahma in Benares, on a brass plate under the dome that marks the center of the world, there are 64 disks of pure gold that the priests carry one at a time between these diamond needles according to Brahma's immutable law: No disk may be placed on a smaller disk. In the beginning of the world all 64 disks formed the Tower of Brahma on one needle. Now, however, the process of transfer of the tower from one needle to another is in mid course. When the last disk is finally in place, once again forming the Tower of Brahma but on a different needle, then will come the end of the world and all will turn to dust. R. Douglas Hofstadter. Metamagical themas. Scientific American, 248(2):16-22, March 1983. Slide 25 Ingeniørhøjskolen i Århus Towers of Hanoi Problem with Three Disks Slide 26 Ingeniørhøjskolen i Århus Towers of Hanoi: Three Disk Solution Slide 27 Ingeniørhøjskolen i Århus Towers of Hanoi: Three Disk Solution Slide 28 Ingeniørhøjskolen i Århus Towers of Hanoi: Recursive Function The general algorithm to solve this problem is 1. Move the top n-1 disks from needle 1 to needle 2 using needle 3 as the intermediate needle 2. Move disk number n from needle 1 to needle 3 3. Move the top n-1 disks from needle 2 to needle 3 using needle 1 as the intermediate needle Slide 29 Ingeniørhøjskolen i Århus Recursive Towers of Hanoi in C++ // Precondition: count >= 0 // Postcondition: count disks are moved from from_needle // to to_needle using via_needle for help void moveDisks(int count, int from_needle, int to_needle, int via_needle) { if(count > 0) Recursive call { moveDisks(count - 1, from_needle, via_needle, to_needle); cout<<"Move disk "<<count<<“ from "<<from_needle <<“ to "<<to_needle<<"."<<endl; moveDisks(count - 1, via_needle, to_needle, from_needle); } } Recursive call Slide 30 Ingeniørhøjskolen i Århus General problem solving strategy • Divide and conquer solve(p: Problem) { if <p is a simple problem> solution = <solve it directly> else <split p in two disjoint problems p1 and p2> res1 = solve(p1) res2 = solve(p2) solution = join(p1, p2) Slide 31 Ingeniørhøjskolen i Århus Example: search private bool search(int m, List<int> p) { if (p.Count == 0) return false; else if (p.Count == 1) return p[0] == m; else { List<int> p1 = new List<int>(); List<int> p2 = new List<int>(); for (int i = 0; i < p.Count / 2; i++) p1[i] = p[i]; for (int j = (p.Count / 2) + 1; j < p.Count; j++) p2[j - ((p.Count / 2) + 1)] = p[j]; return search(m, p1) || search(m, p2); } } Slide 32 Ingeniørhøjskolen i Århus Agenda Syllabus for this week Recursive definitions of sets Recursive definitions of algorithms Mathematical Induction – First Principle • Mathematical Induction – Second Principle Slide 33 Ingeniørhøjskolen i Århus Mathematical Induction • In the mathematical induction we have the law already formulated. We must prove that it holds generally • The basis for mathematical induction is the property of the well-ordering principle for the natural numbers Slide 34 Ingeniørhøjskolen i Århus 1. principle of Mathematical Induction • • • Also called weak induction Suppose P(n) is a statement involving an integer n Then to prove that P(n) is true for every n ≥ n0 it is sufficient to show that these two things hold: 1. P(n0) is true 2. For any k ≥ n0, if P(k) is true, then P(k+1) is true – In general we have a basis step – Followed by an inductive step. Slide 35 Ingeniørhøjskolen i Århus Example Induction Proof • Prove that for any natural number n it holds that: 0+1+…+n = n(n + 1)/2 – Basis step: If n = 0 then LHS = 0 and RHS = 0(0+1)/2=0 – Induction step: Let the induction hypothesis be: For an arbitrary k it holds that 0+1+…+k = k(k + 1)/2 Let us try to express LHS for (k+1). Using the induction hypothesis the LHS = k(k+1)/2 + (k+1) If we factor out (k+1) we can rewrite this to: (k+1)(k+2)/2 This is identical to the RHS for (k+1) QED. Slide 36 Ingeniørhøjskolen i Århus Example Induction Proof • Prove that for any natural number n it holds that: 1+3+…+(2n+1) = (n + 1)2 – Check for n = 0: 2*0 + 1 = (0 + 1)2 – Assume that: For an arbitrary k it holds that 1+3+…+(2k+1) = (k + 1)2 Then for k+1 we need to prove: 1+3+…+(2k+1)+(2(k+1)+1) = ((k+1)+1)2 Using the induction hypothesis we get: 1+3+…+(2k+1)+(2(k+1)+1) = (k+1)2 + (2(k+1) +1) Since (k+1)2 + 2(k+1)+1 = (k+1+1)2 we have proved our objective. QED. Slide 37 Ingeniørhøjskolen i Århus More on Inductive Proofs General strategy: Clarify on which variable you are going to do the induction Calculate some small cases n=0,1,2,3,… (Come up with your conjecture) Make clear what the induction step nn+1 is Prove it, and say that you proved it. Slide 38 Ingeniørhøjskolen i Århus Agenda Syllabus for this week Recursive definitions of sets Recursive definitions of algorithms Mathematical Induction – First Principle Mathematical Induction – Second Principle Slide 39 Ingeniørhøjskolen i Århus 2. principle of Mathematical Induction • • • Also called strong induction Suppose P(n) is a statement involving an integer n Then to prove that P(n) is true for every n ≥ n0 it is sufficient to show that this holds: 1. P(n0) is true 2. For any k ≥ n0, if P(x) is true for all x smaller than k, then P(k) is true Slide 40 Ingeniørhøjskolen i Århus Strong induction • Weak mathematical induction assumes P(k) is true, and uses that (and only that!) to show P(k+1) is true • Strong mathematical induction assumes P(1), P(2), …, P(k) are all true, and uses that to show that P(k+1) is true. P(1) P(2) P(3) ... P(k ) P(k 1) Slide 41 Ingeniørhøjskolen i Århus Strong induction example • Show that any number > 1 can be written as the product of primes • Base case: P(2) – 2 is the product of 2 (remember that 1 is not prime!) • Inductive hypothesis: P(1), P(2), P(3), …, P(k) are all true • Inductive step: Show that P(k+1) is true Slide 42 Ingeniørhøjskolen i Århus Strong induction example • Inductive step: Show that P(k+1) is true • There are two cases: – k+1 is prime • It can then be written as the product of k+1 – k+1 is composite • It can be written as the product of two composites, a and b, where 2 ≤ a ≤ b < k+1 • By the inductive hypothesis, both P(a) and P(b) are true – QED Slide 43 Ingeniørhøjskolen i Århus Induction in Computer Science Inductive proofs play an important role in computer science because of their similarity with recursive algorithms. Analyzing recursive algorithms often require the use of recurrent equations, which require inductive proofs. Also, recursive algorithms are constructive such that we often get a solution “for free”. Slide 44 Ingeniørhøjskolen i Århus Summary • The notion of Recursion and Induction – Recursive definitions of sets – Recursive definitions of algorithms – Mathematical Induction • Read Section 5A: Recursive Definitions and Section 5D: Proof by Induction from the web pages on slide 3 • Carry out as many exercises as possible from the web pages Slide 45 Ingeniørhøjskolen i Århus