Introduction to the Theory of Computation AZADEH FARZAN WINTER 2010

Transcription

1 Introduction to the Theory of Computation AZADEH FARZAN WINTER 2010

2 PROOFS Proof by Contradiction Proof by Construction Jack sees Jill, who has just come in from outdoors Proof by Induction dry. Jack knows that it is not raining. For any n, there exist n consecutive composite integers. His proof: if it were raining (assuming the statement Proof: is false), here Jill are the woud n be consecutive wet (contradiction, composite integers or false for consequence). any n: Therefore, it must not be raining. (n+1)!+2, (n+1)!+3,..., (n+1)!+n+1

3 INDUCTION A method used to show that all elements of an infinite set have a specific property. Example: an arithmetic expression computes a desired quantity for every assignment to its variables. Example: program works correctly at all steps or for all input. Every proof induction consists of two parts: The basis. The induction step.

4 IDEA Let s take the infinite set to be N = {1, 2, 3,...}. Let s call the property P. The goal is to prove that P(k) is true for each natural number k. The basis: P(1) is true. The induction step: if P(i) is true, then P(i + 1) is true as well. P(1) P(2) P(3)...P(i)...

5 FORMALLY: MATHEMATICAL INDUCTION PRINCIPLE Let P () be a predicate defined on N. Then if (1) P (0) is true, and 1 (2) n N [P (n) P (n + 1)], then n N P (n) is true.

6 EXAMPLE Prove n i=1 i = n(n + 1)/2. Proof by induction: The basis (n=1): holds since 1 = 1 2/2. The induction step: if m m+1 i=1 i=1 i =(m + 1)(m + 2)/2. i = m(m + 1)/2, then m+1 i=1 m i = i +(m + 1) i=1 = m(m + 1)/2+(m + 1) = [m(m + 1) + 2(m + 1)] /2 =(m + 1)(m + 2)/2

7 LET S FORMALIZE!

8 WELL-ORDERING Principle of Well-ordering: Any non-empty subset of N has minimum element. A N, s.t. A =, [ a A s.t. ( b A, a b)]. Is the principle true for Z? Is the principle true for rational numbers in (0, 1)?

9 WHAT IS THE RELATION BETWEEN WELL-ORDERING AND INDUCTION?

10 Induction works as a proof method since: Principle of well-ordering implies principle of mathematical induction. proof on the board!

11 SIMPLE INDUCTION Principle of Simple Induction: Let A be a set that satisfies the following properties: (i) 0 is in A; (ii) for any i N, if i A, then i +1 A. Then N A. 0 A 1 A 2 A 3 A...

12 Problem: prove that every natural number greater than 1 can be written as a product of prime numbers. LET S LOOK AT A SLIGHTLY DIFFERENT KIND OF INDUCTION.

13 FORMALLY: STRONG INDUCTION PRINCIPLE If P (0) and then n P(n). n 0 [(P (0) P (1)... P (n)) P (n +1)] WAIT, WHY DOES THIS ONE WORK?

14 COMPLETE INDUCTION Principle of Complete Induction: Let A be a set that satisfies the following properties: (*) for any i N, if j < i, j A, then i A. Then N A. 0 A, 1 A, 2 A, 3 A...

15 THREE EQUIVALENT NOTIONS Theorem. The principles of (a) well-ordering, (b) simple induction, and (c) complete induction are equivalent.

16 ALL DEFINITIONS Principle of Well-ordering: Any non-empty subset of N has minimum element. Principle of Simple Induction: Let A be a set that satisfies the following properties: (i) 0 is in A; (ii) for any i N, if i A, then i +1 A. Then N A. Principle of Complete Induction: Let A be a set that satisfies the following properties: (*) for any i N, if j < i, j A, then i A. Then N A.

17 EXAMPLE For every natural number n, the power set of a set with n elements has 2 n elements. Basis: The power set of (with 0 elements) has 2 0 =1element ( itself). Induction step: if n element sets have 2 n subsets, then n +1 element sets have 2 n+1 subsets. Let X = n +1 and x X and let Y = X {x}. Then Y = n. X has two types of subsets: (i) those which contain x, and (ii) those which do not. Subsets of type (i) are subsets of Y as well, and there are 2 n of them by IH. Subsets of type (ii) are subsets of Y if we remove x from them, and there are 2 n of them by IH. Total number of subsets: 2 n +2 n =2 n+1.

18 NON-ZERO BASES For all n 6, we have 2 n > 10n. Basis: n =6, clearly 2 6 = 64 > 10.6 = 60. Induction Step: For an arbitrary i 6, 2 i > 10i 2 i+1 > 10(i + 1). 2 i+1 = 2.2 i > 2.10i = 10i + 10i > 10i + 10 = 10(i + 1)

19 A VARIATION Sometimes it is easier to have an explicit basis for complete induction. Basis: Prove that P(0) is true. Induction Step: Prove that, for each natural number i>0, if P(j) holds for all natural numbers j<i, then P(i) holds as well.

20 EXAMPLE P(n): If a full binary tree has n nodes, then n is an odd number. Prove P(n) is true for all n 1. Basis: n =1. A full binary tree with one node, and one is odd. Induction Hypothesis: Assume we have a binary tree with n>1 nodes. If T 1 is the left subtree and T 2 is the right subtree of our full binary tree T, then we have: nodes(t )=nodes(t 1 )+nodes(t 2 )+1

21 READING Chapter 0: to remind yourself of the basic notions and definitions. Chapters 1 for this week s discussions on induction.