Discrete Mathematics

Discrete Mathematics Chapter 4 Induction and Recursion 大葉大學資訊工程系黃鈴玲(Lingling Huang)

4.1 Mathematical Induction(數學歸納法) Note : Mathematical induction can be used only to prove results obtained in some other way. It is not a tool for discovering formulae or theorems. (p.265) P(n) : a propositional function (e.g. n ≦ 2n) A proof by mathematical induction(MI) that P(n) is true for every nZ+ consists of two steps : 1. Basis step : The proposition P(1) is shown to be true.(若 n 從 0 開始則證 P(0)為真) 2. Inductive step : the implication P(k) → P(k+1) is shown to be true for every kZ+

Example 2. Use MI to prove that the sum of the first n odd positive integers is n2. Note. 不用MI就可以得証: Pf : Let P(n) denote the proposition that Basis step :P(1) is true , since 1=12 Inductive step : Suppose that P(k) is true for a positive integer k, i.e., 1+3+5+…+(2k-1)=k2 Note that 1+3+5+…+(2k-1)+(2k+1) = k2+2k+1= (k+1)2 ∴ P(k+1) is true By induction, P(n) is true for all nZ+

Example 5. Use MI to prove the inequality n < 2n for all nZ+ pf : Let P(n) be the proposition “ n < 2n”. Basis step :P(1) is true since 1 < 21. Inductive step : Assume that P(k) is true for a positive integer k, i.e., k < 2k. Consider P(k+1) : k + 1 < 2k+ 1  2k+ 2k =2k + 1 ∴ P(k+1) is true. By MI, P(n) is true for all nZ+.

Example 7. The harmonic numbersHk, k =1,2,3,…, are defined by . Use MI to show that whenever n is a nonnegative integer. Pf : Let P(n) be the proposition that “ ”. Basis step :P(0) is true, since . Inductive step : Assume that P(k) is true for some k, i.e., Consider P(k+1) :

∴P(k+1) is true. By MI, P(n) is true for all nZ+. Exercise : 7, 13

4.2 Strong Induction(強數學歸納法) • Basis step 相同 • Inductive step : Assume all the statements P(1), P(2), …, P(k) are true. Show that P(k+1) is also true.

Example 2. Show that if nZ and n >1, then n can be written as the product of primes. Pf : Let P(n) be the proposition that n can be written as the product of primes. Basis :P(2) is true, since 2 is a prime number Inductive : Assume P(2), P(3), …, P(k) are true. Consider P(k + 1) : Case 1 : k + 1 is prime  P(k+1) is true Case 2 : k + 1 is composite, i.e., k + 1 = ab where 2 ab＜ k+1 By the induction hypothesis, both a and b can be written as the product of primes. P(k+1) is true. By Strong MI, P(k) is true if kZ and k >1. Note: 此題無法僅用 MI 證

Example 4. Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. Pf : Let P(n) be the statement that the postage of n cents can formed using just 4-cent and 5-cent stamps. Basis :P(12) is true, since 12 = 4  3; P(13) is true, since 13 = 4  2 + 5  1; P(14) is true, since 14 = 4  1 + 5  2; P(15) is true, since 15 = 5  3; Inductive : Assume P(12), P(13), …, P(k) are true. Consider P(k+1) : Suppose k-3= 4  m + 5  n. Then k+1= 4  (m+1) + 5  n. P(k+1) is true. By Strong MI, P(n) is true if nZ and n 12. (k-3  12) Exercise : 7

4.3 Recursive Definitions. Def. The process of defining an object in terms of itself is called recursion(遞迴). e.g. We specify the terms of a sequence using (1) an explicit formula: an=2n, n=0,1,2,… (2) a recursive form: a0=1, an+1=2an , n=0,1,2,… Example 1. Suppose that f is defined recursively by f(0)=3 , f(n+1)=2f(n)+3 Find f(1), f(2), f(3), f(4).

Example 2. Give an inductive (recursive) definition of the factorial function F(n) = n!. Sol : initial value : F(0) = 1 recursive form : F(n+1) = (n+1)! = n!  (n+1) = F(n)  (n+1) Def1, Example 5. The Fibonacci numbersf0, f1, f2…,are defined by : f0= 0 , f1 = 1 , fn = fn-1 + fn-2 , for n = 2,3,4,… what is f4 ? Sol : f4 = f3 + f2 = (f2 + f1) + (f1 + f0) = f2 + 2 = (f1 + f0) + 2 = 3

Example 6. Show that fn > a n-2 , where Pf:( By Strong MI ) Let P(n) be the statement fn>a n-2 . Basis: f3 = 2 > a so that P(3) and P(4) are true. Inductive: Assume that P(3), P(4), …, P(n) are true. We must show that P(n+1) is true. fn+1 = fn + fn-1 > a n-2 + a n-3 = a n-3(a +1) ∵ a +1= a2 ∴ fn+1 > a n-3  a2 = a n-1 We get that P(n+1) is true. By Strong MI , P(n) is true for all n  3

※Recursively defined sets. Example 7. Let S be defined recursively by 3S x+yS if xS and yS. Show that S is the of positive integers divisible by 3 (i.e., S = { 3, 6, 9, 12, 15, 18, … } Pf: Let A be the set of all positive integers divisible by 3. We need to prove that A=S. (i) A  S : (By MI) Let P(n) be the statement that 3nS … (ii) S  A : (利用S的定義) (1) 3  A , (2) if xA,yA, then 3|x and 3|y.  3|(x+y)  x+yA ∴S A S = A

Definition 2. The set of strings over an alphabet is denoted by *. The empty string is denoted by l, l , and wx* whenever w* and x.eg.  = { a, b, c } * = { l, a , b , c , aa , ab , ac , ba , bb , bc, …abcabccba, …} Example 9. Give a recursive definition of l(w), the length of the string w* Sol : initial value : l(l)=0 recursive def : l(wx)=l(w)+1 if w*, x. lb la lc

0l1 n個 n個 Exercise 3, 7, 13, 48, 49 Exercise 39. When does a string belong to the set A of bit strings defined recursively by lA 0x1A if xA. Sol : A={l, 01 , 0011, 000111, …} ∴當bit string a = 000…011…1時 aA

Ackermann’s function A(m, n) = 2n if m = 0 0 if m  1 and n = 0 2 if m  1 and n = 1 A(m-1, A(m, n-1)) if m  1 and n  2 Exercise 49 Show that A(m,2)=4 whenever m  1 Pf : A(m,2) = A(m-1, A(m,1)) = A(m-1,2) whenever m  1. A(m,2) = A(m-1,2) = A(m-2,2) = … = A(0,2) = 4.

4.4 Recursive algorithms. ※ Sometimes we can reduce the solution to a problem with a particular set of input to the solution of the same problem with smaller input values. eg. gcd(a,b) = gcd(b mod a, a) (when a < b) Def 1. An algorithm is called recursiveif it solves a problem by reducing it to an instance of the same problem with smaller input.

∴ Algorithm 2. Procedurepower( a : nonzero real number, n : nonnegative integer ) ifn = 0 thenpower(a, n):=1 elsepower(a, n):= a*power(a, n-1). Example 2. Give a recursive algorithm for computing an, where aR \ {0}, nN. Sol : recursive definition of an : initial value : a0=1 recursive def : an = a  an-1.

從ai,ai+1,…aj中找 x Example 4. Find gcd(a,b) with 0a<b Algorithm 4. proceduregcd(a,b : nonnegative integers with a<b) ifa=0 thengcd(a,b) := b elsegcd(a,b) := gcd(b mod a, a). Sol : Example 5. Search x in a1, a2,…,an by Linear Search Sol : Alg. 5 proceduresearch (i, j, x: integers) if ai = xthenlocation := i elseif i = jthenlocation := 0 elsesearch(i+1, j, x) call search(1, n, x)

表示左半邊ai, ai+1, …, am-1 至少還有一個元素 Example 6. Search x from a1,a2,…,an by binary search (recursive version). search x from ai, ai+1, …, aj Sol : Alg. 5 procedurebinary_search (x , i , j: integers) m := (i+j) / 2 ifx = amthenlocation := m else if (x < am and i < m) then binary_search(x, i, m-1) else if (x > am and j > m) then binary_search(x, m+1, j) elselocation := 0 call binary_search(x, 1, n)

Example 1. Give the value of n!, nZ+ Sol : Note : n! = n (n-1)! Alg. 1 (Recursive Procedure) procedurefactorial (n: positive integer) ifn = 1 thenfactorial (n) := 1 elsefactorial (n) := nfactorial (n-1) Alg. (Iterative Procedure) procedureiterative_factorial (n : positive integer) x := 1 for i := 1 ton x := i  x { x = n! }

※ iterative alg. 的計算次數通常比 recursive alg.少 ※ Find Fibonacci numbers (Note : f0=0, f1=1, fn=fn-1+fn-2 for n2) Alg. 7 (Recursive Fibonacci) procedureFibonacci (n : nonnegative integer) if n = 0 thenFibonacci (0) := 0 elseifn = 1 then Fibonacci (1) := 1 else Fibonacci (n) := Fibonacci (n-1)+Fibonacci (n-2)

Alg.8 (Iterative Fibonacci) procedureiterative_fibonacci (n: nonnegative integer) if n = 0 theny := 0 // y = f0 else begin x := 0 y := 1 // y = f1 fori := 1 ton-1 begin z := x + y x := y y := z end end {y is fn} Exercise : 11, 35

Discrete Mathematics