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

2. Ch4-2 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 ≦ 2 n ) 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 +

3. Ch4-3 Example 2. Use MI to prove that the sum of the first n odd positive integers is n 2. 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=1 2 Inductive step : Suppose that P(k) is true for a positive integer k, i.e., 1+3+5+…+(2k  1)=k 2 Note that 1+3+5+…+(2k  1)+(2k+1) = k 2 +2k+1= (k+1) 2 ∴ P(k+1) is true By induction, P(n) is true for all n  Z +

4. Ch4-4 Example 5. Use MI to prove the inequality n < 2 n for all n  Z + pf : Let P(n) be the proposition “ n < 2 n ”. Basis step : P(1) is true since 1 < 2 1. Inductive step : Assume that P(k) is true for a positive integer k, i.e., k < 2 k. Consider P(k+1) : k + 1 < 2 k + 1  2 k + 2 k =2 k + 1 ∴ P(k+1) is true. By MI, P(n) is true for all n  Z +.

5. Ch4-5 Example 7. The harmonic numbers H k, k =1,2,3,…, are defined by. Use MI to show that 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) : whenever n is a nonnegative integer.

6. Ch4-6 ∴ P(k+1) is true. By MI, P(n) is true for all n  Z +. Exercise : 7, 13

7. Ch4-7 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.

8. Ch4-8 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

9. Ch4-9 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. Exercise : 7 (k  3  12)

10. Ch4-10 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: a n =2 n, n=0,1,2,… (2) a recursive form: a 0 =1, a n+1 =2a n, 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).

11. Ch4-11 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 numbers f 0, f 1, f 2 …, are defined by : f 0 = 0, f 1 = 1, f n = f n  1 + f n  2, for n = 2,3,4,… what is f 4 ? Sol : f 4 = f 3 + f 2 = (f 2 + f 1 ) + (f 1 + f 0 ) = f 2 + 2 = (f 1 + f 0 ) + 2 = 3

12. Ch4-12 Example 6. Show that f n >  n  2, where Pf: ( By Strong MI ) Let P(n) be the statement f n >  n  2. Basis: f 3 = 2 >  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. f n+1 = f n + f n  1 >  n  2 +  n  3 =  n  3 (  +1) ∵  +1=  2 ∴ f n+1 >  n  3   2 =  n  1 We get that P(n+1) is true. By Strong MI, P(n) is true for all n  3

13. Ch4-13 ※ 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

14. Ch4-14 Definition 2. The set of strings over an alphabet  is denoted by  *. The empty string is denoted by, , and wx  * whenever w  * and x . eg.  = { a, b, c }  * = {, 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( )=0 recursive def : l(wx)=l(w)+1 if w  *, x . a b c

15. Ch4-15 Exercise 3, 7, 13, 48, 49 Exercise 39. When does a string belong to the set A of bit strings defined recursively by  A 0x1  A if x  A. Sol : A={, 01, 0011, 000111, …} ∴當 bit string  = 000…011…1 時  A n個n個 n個n個 0 1

16. Ch4-16 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.

17. Ch4-17 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 recursive if it solves a problem by reducing it to an instance of the same problem with smaller input.

18. Ch4-18 Example 2. Give a recursive algorithm for computing a n, where a  R \ {0}, n  N. Sol : recursive definition of a n : initial value : a 0 =1 recursive def : a n = a  a n  1. Algorithm 2. Procedure power( a : nonzero real number, n : nonnegative integer ) if n = 0 then power(a, n):=1 else power(a, n):= a * power(a, n  1). ∴

19. Ch4-19 Example 4. Find gcd(a,b) with 0  a<b Sol : Algorithm 4. procedure gcd(a,b : nonnegative integers with a<b) if a=0 then gcd(a,b) := b else gcd(a,b) := gcd(b mod a, a). Example 5. Search x in a 1, a 2,…,a n by Linear Search Sol : Alg. 5 procedure search (i, j, x: integers) if a i = x then location := i else if i = j then location := 0 else search(i+1, j, x) 從 a i,a i+1,…a j 中找 x call search(1, n, x)

20. Ch4-20 Example 6. Search x from a 1,a 2,…,a n by binary search (recursive version). Sol : Alg. 5 procedure binary_search (x, i, j: integers) m :=  (i+j) / 2  if x = a m then location := m else if (x < a m and i < m) then binary_search(x, i, m  1) else if (x > a m and j > m) then binary_search(x, m+1, j) else location := 0 call binary_search(x, 1, n) search x from a i, a i+1, …, a j 表示左半邊 a i, a i+1, …, a m  1 至少還有一個元素

21. Ch4-21 Example 1. Give the value of n !, n  Z + Sol : Note : n! = n  (n  1)! Alg. 1 (Recursive Procedure) procedure factorial (n: positive integer) if n = 1 then factorial (n) := 1 else factorial (n) := n  factorial (n  1) Alg. (Iterative Procedure) procedure iterative_factorial (n : positive integer) x := 1 for i := 1 to n x := i  x { x = n! }

22. Ch4-22 ※ iterative alg. 的計算次數通常比 recursive alg. 少 ※ Find Fibonacci numbers (Note : f 0 =0, f 1 =1, f n =f n  1 +f n  2 for n  2 ) Alg. 7 (Recursive Fibonacci) procedure Fibonacci (n : nonnegative integer) if n = 0 then Fibonacci (0) := 0 else if n = 1 then Fibonacci (1) := 1 else Fibonacci (n) := Fibonacci (n  1)+Fibonacci (n  2)

23. Ch4-23 Alg.8 (Iterative Fibonacci) procedure iterative_fibonacci (n: nonnegative integer) if n = 0 then y := 0 // y = f 0 else begin x := 0 y := 1 // y = f 1 for i := 1 to n  1 begin z := x + y x := y y := z end {y is f n } Exercise : 11, 35 i = 1i = 2i = 3 zf2f2 f3f3 f4f4 xf1f1 f2f2 f3f3 yf2f2 f3f3 f4f4