CSE 311 Foundations of Computing I Lecture 15 Strong Induction and Recursive Definitions Spring 2013 1.

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Presentation transcript:

CSE 311 Foundations of Computing I Lecture 15 Strong Induction and Recursive Definitions Spring

Announcements Reading assignments – Today: th Edition th Edition Midterm Friday, May 10, MGH 389 – Closed book, closed notes – Tables of inference rules and equivalences will be included on test – Sample questions from old midterms are now posted 2

Highlights from last lecture Mathematical Induction Induction proof layout: 1.By induction we will show that P(n) is true for every n≥0 2.Base Case: Prove P(0) 3.Inductive Hypothesis: Assume that P(k) is true for some arbitrary integer k ≥ 0 4.Inductive Step: Prove that P(k+1) is true using Inductive Hypothesis that P(k) is true 5.Conclusion: Result follows by induction P(0)  k≥0 (P(k)  P(k+1))   n≥0 P(n) 3

Strong Induction P(0)  k ((P(0)  P(1)  P(2)  …  P(k))  P(k+1))   n P(n) Follows from ordinary induction applied to Q(n) = P(0)  P(1)  P(2)  …  P(n) 4

Strong Induction English Proofs 1.By induction we will show that P(n) is true for every n≥0 2.Base Case: Prove P(0) 3.Inductive Hypothesis: Assume that for some arbitrary integer k ≥ 0, P(j) is true for every j from 0 to k 4.Inductive Step: Prove that P(k+1) is true using the Inductive Hypothesis (that P(j) is true for all values  k) 5.Conclusion: Result follows by induction 5

Every integer ≥ 2 is the product of primes 6

Recursive Definitions of Functions F(0) = 0; F(n + 1) = F(n) + 1 for all n  0 G(0) = 1; G(n + 1) = 2  G(n) for all n  0 0! = 1; (n+1)! = (n+1)  n! for all n  0 H(0) = 1; H(n + 1) = 2 H(n) for all n  0 7

Fibonacci Numbers f 0 = 0; f 1 = 1; f n = f n-1 + f n-2 for all n  2 8

Bounding the Fibonacci Numbers Theorem: 2 n/2-1  f n  2 n for all n  2 9

Fibonacci numbers and the running time of Euclid’s algorithm 10 each quotient q i ≥ 1 r 1 ≥ 1=f 1 “r 0 ”=0=f 0

Recursive Definitions of Sets Recursive definition – Basis step: 0  S – Recursive step: if x  S, then x + 2  S – Exclusion rule: Every element in S follows from basis steps and a finite number of recursive steps 11

Recursive definitions of sets Basis: 6  S; 15  S; Recursive: if x, y  S, then x + y  S; Basis: [1, 1, 0]  S, [0, 1, 1]  S; Recursive: if [x, y, z]  S,  in R, then [  x,  y,  z]  S if [x 1, y 1, z 1 ], [x 2, y 2, z 2 ]  S then [x 1 + x 2, y 1 + y 2, z 1 + z 2 ]  S Powers of 3 12

Recursive Definitions of Sets: General Form Recursive definition – Basis step: Some specific elements are in S – Recursive step: Given some existing named elements in S some new objects constructed from these named elements are also in S. – Exclusion rule: Every element in S follows from basis steps and a finite number of recursive steps 13

Strings An alphabet  is any finite set of characters. The set  * of strings over the alphabet  is defined by – Basis:   * ( is the empty string) – Recursive: if w   *, x  , then wx   * 14

Palindromes Palindromes are strings that are the same backwards and forwards Basis: is a palindrome and any a ∈  is a palindrome Recursive step: If p is a palindrome then apa is a palindrome for every a ∈  15

All binary strings with no 1’s before 0’s 16