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Discrete Mathematics

Discrete Mathematics. Chapter 3 Mathematical Reasoning, Induction, and Recursion. 感謝 大葉大學 資訊工程系 黃鈴玲老師 提供. 3.2 Sequences and Summations. ※ Sequence ( 數列 ) Def 1. A sequence is a function f from A  Z +

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Discrete Mathematics

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  1. Discrete Mathematics Chapter 3 Mathematical Reasoning, Induction, and Recursion 感謝 大葉大學 資訊工程系 黃鈴玲老師 提供

  2. 3.2 Sequences and Summations ※Sequence (數列) Def 1. A sequence is a function f from A Z+ (or A  N) to a set S. We use anto denote f(n), and call ana term (項) of the sequence. Example 1.{an}, where an= 1/n , n  Z+  a1 =1, a2 =1/2 , a3 =1/3, … Example 2.{bn} , where bn= (-1)n, n N  b0 = 1, b1 = -1 , b2 = 1, … 3.2.1

  3. Special Integer Sequence A common problem in discrete mathematics isfinding a formula for constructing the term of a sequence.方法：找出 aiai+1的變化；加減某數或者乘除某數？ Example 6. What is a rule that can produce the term of a sequence if the first 10 terms are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4? Sol : 規則：數字 i 出現 i次 數字 i出現之前共有 1+2+3…….+(i-1) = i(i -1)/2項 ai(i-1)/2 +1= ai(i-1)/2 +2 =… = ai(i+1)/2 = i 3.2.2

  4. Example 7. How can we produce the terms of a sequence if the first 10 terms are 5, 11, 17, 23, 29, 35,41, 47, 53, 59? Sol : a1= 5 a2=11 = 5 + 6 a3=17 = 11 + 6 = 5 + 6  2 : :  an= 5 + 6  (n-1) = 6n-1 3.2.3

  5. Example 8.Conjecture a simple formula for anif the first 10 terms of the sequence {an} are 1, 7, 25, 79, 241, 727, 2185, 6559, 19681,59047? Sol： 顯然非等差數列 後項除以前項的值接近3  猜測數列為 3n  … 比較： {3n} : 3, 9, 27, 81, 243, 729, 2187,… {an} : 1, 7, 25, 79, 241, 727, 2185,…  an= 3n - 2 , n  1 3.2.4

  6.  Summations Here, the variable j is call the index of summation,m is the lower limit, and n is the upper limit. Example 10. Example 13. (Double summation) 3.2.5

  7. Example 14. Table 2. Some useful summation formulae 3.2.6

  8. Cardinality Def 4.The sets A and B have the samecardinality (size) if and only if there is a one-to-one correspondence (1-1,onto 的function) from A to B. Def 5.A set that is either finite or has the same cardinality as Z+ (or N) is called countable (可數). A set that is not countable is called uncountable. 3.2.7

  9. Pf: (Figure 1) Z+ : 1 2 3 4 5 6 7 8 … …… { 正奇數 } : 1 3 5 7 9 11 13 15 … f : Z+ {正奇數} f (n) = 2n – 1 is 1-1 & onto. Example 18. Show that the set of odd positive integers is a countable set. 3.2.8

  10. 1 1 Example 19. Show that the set of positive rational number (Q+) is countable. Pf: Q+ = { a / b | a, b Z+ } (Figure 2) ∴ Z+ : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 … Q+ : Exercise : 9,13,17,38 (注意，因 等於 ，故 不算) ※Note. R is uncountable. (Example 20) 3.2.9

  11. 3.3 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.239) 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+ 3.3.1

  12. Example 1. 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+ 3.3.2

  13. Example 2. 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+. 3.3.3

  14. Example 6. 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) : 3.3.4

  15. ∴P(k+1) is true. By MI, P(n) is true for all nZ+. 3.3.5

  16. ※The 2nd principle of mathematical induction: ( 又稱為強數學歸納法 strong induction) • Basis step 相同 • Inductive step : Assume P(k) is true for all k  n Show that P(n+1) is also true. 3.3.6

  17. Example 14. 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(k) is true for all k  n. Consider P(n+1) : Case 1 : n+1 is prime  P(n+1) is true Case 2 : n+1 is composite, i.e., n+1=ab where 2 ab＜ n+1 By the induction hypothesis, both a and b can be written as the product of primes. P(n+1) is true. By 2nd MI, P(n) is true if nZ and n >1. Note: 此題無法用 1st MI 證 Exercise : 3,11,17 3.3.7

  18. 3.4 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). 3.4.1

  19. 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) 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 3.4.2

  20. Example 6. Show that fn > a n-2 , where Pf:( By 2nd 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(k) is true, 3 k  n, n  4. 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 2nd MI , P(n) is true for all n  3 3.4.3

  21. ※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 3.4.4

  22. Example 8. The set of strings over an alphabet is denoted by *. The empty string is denoted by 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 3.4.5

  23. 0l1 n個 n個 Exercise 3,13, 25, 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 3.4.6

  24. 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. 3.4.7

  25. 3.5 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. 3.5.1

  26. ∴ Algorithm 1. Procedurepower( a : nonzero real number, n : nonnegative integer ) ifn = 0 thenpower(a, n):=1 elsepower(a, n):= a*power(a, n-1). Example 1. 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. 3.5.2

  27. 從ai,ai+1,…aj中找 x Example 4. Find gcd(a,b) with 0a<b Algorithm 3. 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. 4 proceduresearch (i, j, x) if ai = xthenlocation := i elseif i = jthenlocation := 0 elsesearch(i+1, j, x) call search(1, n, x) 3.5.3

  28. 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. search x from ai, ai+1, …, aj Sol : Alg. 5 procedurebinary_search (x , i , j) 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) 3.5.4

  29. Example 7. Give the value of n!, nZ+ Sol : Note : n! = n (n-1)! Alg. 6 (Recursive Procedure) procedurefactorial (n: positive integer) ifn = 1 thenfactorial (n) := 1 elsefactorial (n) := nfactorial (n-1) Alg. 7 (Iterative Procedure) procedureiterative_factorial (n : positive integer) x := 1 for i := 1 ton x := i  x { x = n! } 3.5.5

  30. ※ 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. 8 (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) 3.5.6

  31. Alg.9 (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 : 5 , 27 3.5.7

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