Discrete
Mathematics
Chapter 4
歸納(Induction)與遞迴(Recursion)
大葉大學 資訊工程系 黃鈴玲
4.1 數學歸納法
例：爬「無限長」的梯子
已知：
1. 我們能到達梯子的第一階。
2. 假設能到達梯子的某一階，就必能到達梯子的下一階。
問：我們是否能到達這個梯子的每一階？
Ch4-2
Ch4-3
P(n) : 一個命題式函數 (例如： n  2n)
使用數學歸納法來證明 “P(n) 對所有正整數n皆為真”
包含以下兩個步驟:
1. 基礎步驟(Basis):
證明P(1)為真
(即：n代入1時公式成立)
(若 n 從 0 開始則證 P(0)為真，要證明式子在n最小時會成立)
2. 歸納步驟(Inductive) :
證明當k為任意正整數時，若P(k)成立則P(k+1)也成立
(即：假設n代入k時公式成立，
利用此假設，證明假設n代入k+1時公式成立)
注意 : 數學歸納法只能用來證明某個公式的正確性，不能用來
找出公式.
Ch4-4
n(n  1)
Example 1 證明當n為正整數時， 1  2  ...  n 
.
2
Proof.
1(1  1)
(Basis) n=1時：1  2
成立
(Inductive) 假設n=k時上式成立，
即： 1  2  ...  k  k (k  1)
2
當n=k+1時，
左式= 1+2+…+k+(k+1)
= k (k  1)  (k  1)
2
=
(k  1)( k  2)
= 右式，故由歸納法得證
2
Ch4-5
Example 2. 使用歸納法證明首n個正奇數的和是n2.
注意：其實不用歸納法也可以得証：
(1  2n - 1)n
1  3  5  ...  2n - 1 
 n2
2
Pf :
Basis : n=1時： 1 = 12 成立
Inductive : 假設n=k時上式成立，即:
1+3+5+…+(2k-1) = k2
當n=k+1時，左式= 1+3+5+…+(2k-1)+(2k+1)
= k2 + 2k+1 = (k+1)2 =右式
故根據數學歸納法得證。
Ch4-6
Example 5. 利用數學歸納法證明：
“對所有正整數n，不等式 n < 2n 都成立”。
pf :
Basis: n=1時，1 < 21 成立.
Inductive :
假設n=k 時上式成立，即 k < 2k.
當 n=k+1時 :
左式= k + 1
< 2k + 1
 2k + 1 =右式
∴ k + 1 < 2k + 1 也成立
故根據數學歸納法得證。
Ch4-7
Exercise 4.1
3. 證明當n為任意正整數時，
n(n  1)( 2n  1)
1  2  ...  n 
6
2
2
2
7. 證明當n為 0 的整數時，
3  3  5  3  5  ...  3  5 
2
n
3(5
n 1
- 1)
4
Ch4-8
(跳過)
Example 7. The harmonic numbers Hk, k =1,2,3,…, are
1 1
1
defined by H k  1    ...  . Use MI to show that
2 3
k
n
H 2n  1 
whenever n is a nonnegative integer.
2
Pf : Let P(n) be the proposition that “ H 2n  1  n / 2 ”.
Basis step : P(0) is true, since H 0  H1  1  1  0 / 2.
2
Inductive step : Assume that P(k) is true for some k,
i.e., H 2k  1  k / 2
Consider P(k+1) :
Ch4-9
(跳過)
H 2k 1
1 1
1
1
1
1
 1   k  k
 k
   k 1
2 3
2
2 1 2  2
2
1
1
1
 H 2k  k
 k
   k 1
2 1 2  2
2
k
1
1
1
 (1  )  k
 k
   k 1
2 2 1 2  2
2
k
1
1
1
 (1  )  k
 k
 k
k
k
2 2 2
2 2
2  2k
k
2k
 (1  )  k
2 2  2k
k 1
 1
2
∴P(k+1) is true.
By MI, P(n) is true for all nZ+.
Exercise : 7, 13
Ch4-10
(跳過)
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.
Ch4-11
(跳過)
Example 2. 證明若 n為 >1的整數，則 n 可以寫成質數的乘積。
Pf : 令 P(n) 表示 “n 可以寫成質數的乘積” 這一命題.
Basis : P(2) is true, 因為 2 是質數
Inductive : 假設 P(2), P(3), …, P(k) 都是true.
考慮 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
根據歸納法的假設, a 及 b都可以寫成質數的乘積.
 k + 1也是質數的乘積
故 P(k+1) is true.
根據強數學歸納法得證.
Note: 此題無法僅用普通的歸納法證明
Ch4-12
(跳過)
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.
(k-3  12)
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
Ch4-13
4.3 遞迴定義.
Def. 使用一個物件本身來定義它自己的過程稱為
recursion(遞迴).
例: 定義一個無窮數列 1, 2, 4, 8, 16, 32, 64, …時
(1) 使用明確的公式，可寫成:
an=2n, n=0,1,2,…
(2) 使用遞迴的形式，可寫成:
初始值： a0=1,
遞迴式： an+1=2an, n=0, 1, 2, …
Ch4-14
Example 1. 假設函數 f 以遞迴方式定義如下：
f(0)=3, f(n+1) = 2f(n)+3
求出 f(1), f(2), f(3), f(4).
Sol :
f(1) = 2f(0) + 3 =9
f(2) = 2f(1) + 3 =21
f(3) = 2f(2) + 3 =45
f(4) = 2f(3) + 3 =93
Ch4-15
Example 2. 寫出階乘函數 F(n) = n! 的遞迴定義。
Sol :
初始值 : F(0) = 1
遞迴式 : 因 F(n+1) = (n+1)!, F(n) =n!
故 F(n+1) = F(n)  (n+1)
Def1, Example 5. 費式數(Fibonacci numbers) f0, f1, f2, …
定義如下: 初始值： f0 = 0, f1 = 1,
遞迴式： fn = fn-1 + fn-2, for n = 2,3,4,…
What is f5 ?
Sol :
f2 = f1 + f0 =1
f3 = f2 + f1 =2
f4 = f3 + f2 =3
f5 = f4 + f3 =5
Ch4-16
Exercise 4.3
1. 根據 f(0)=1 與下列遞迴定義的 f(n) (n = 0, 1, 2, …)，
求出 f(1), f(2), f(3), f(4)。
(b) f(n+1) = 3f(n)
(d) f(n+1) = f(n)2 + f(n) + 1
3. 根據 f(0)= -1，f(1)= 2 與下列遞迴定義的 f(n)
(n = 0, 1, 2, …)，求出 f(2), f(3), f(4), f(5)。
(a) f(n+1) = f(n) + 3f(n-1)
8. 找出序列{an} (n = 1, 2, 3, …) 的遞迴定義，若
(a) an = 4n - 2
(c) an = n (n - 1)
Ch4-17
(跳過)
Example 6. Show that fn > a n-2 , where a 
1 5
,n 3
2
Pf: ( By Strong MI )
Let P(n) be the statement fn >a n-2 .
Basis: f3 = 2 > a
3 5
2
f4  3  a 
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.
fn+1 = fn + fn-1 > a n-2 + a n-3
= a n-3(a +1)
∵ a +1= a 2
∴ fn+1 > a n-3  a 2 = a n-1
We get that P(n+1) is true.
By Strong MI , P(n) is true for all n  3
Ch4-18
(跳過)
※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)
S=A
(1) 3  A ,
(2) if xA,yA, then 3|x and 3|y.
 3|(x+y)  x+yA
∴S  A
Ch4-19
(跳過)
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 }
la
lb
lc
* = { 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.
Ch4-20
(跳過)
Exercise 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, …}
0l1
∴當bit string a = 000…011…1 時
aA
n個
n個
Ch4-21
(跳過)

Ackermann’s function
A(m, n) =
2n
0
2
A(m-1, A(m, n-1))
if m = 0
if m  1 and n = 0
if m  1 and 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.
Ch4-22
4.4 遞迴演算法
※ 有時，我們能將有特定輸入集合的問題簡化，變成輸
入值較小的相同問題。
例. gcd(a, b) = gcd(b mod a, a)
(when a < b)
重複此法至b mod a = 0，就找到gcd了
Def 1. 藉由把問題簡化到更小輸入的相同問題來解決問
題的演算法，稱為遞迴演算法。
Ch4-23
Example 1. 求出計算 n!的遞迴演算法, 其中 n為
非負整數.
Sol : n! 的遞迴定義:
初始值: 0! =1
遞迴式: n! = n  (n-1)!.
要求出4! 時:
3! = 32!,
4! = 43!,
0! = 1, 代回得到 1! = 1,
∴
2! = 21!,
2! = 2,
1! = 10!,
3! = 6,
4! = 24
Algorithm 1.
Procedure factorial(n : nonnegative integer)
if n = 0 then factorial(n):=1
else factorial(n):= n  factorial(n-1)
Ch4-24
Algorithm 1.
Procedure factorial(n : nonnegative integer)
if n = 0 then factorial(n):=1
else factorial(n):= n  factorial(n-1)
遞迴呼叫自己
也可改寫為:
Algorithm 1.
Procedure factorial(n : nonnegative integer)
if n = 0 then return(1)
else return(n  factorial(n-1))
Ch4-25
用Algorithm 1求出4!的過程：
factorial(4) = 4  factorial(3) = 4  6 = 24
= 3  factorial(2) = 3  2 =6
= 2  factorial(1) = 2  1 =2
= 1  factorial(0) = 1  1=1
Algorithm 1.
Procedure factorial(n : nonnegative integer)
if n = 0 then factorial(n):=1
else factorial(n):= n  factorial(n-1)
Ch4-26
Example 2. 求出計算 an,的遞迴演算法, 其中
a為實數，n為非負整數。
Sol :
an的遞迴定義 :
初始值: a0=1
遞迴式: an = a  an-1.
∴
Algorithm 2.
Procedure power(a : number, n : nonnegative integer)
if n = 0 then power(a, n):=1
else power(a, n):= a  power(a, n-1).
Ch4-27
Exercise 4.4
補充: 根據演算法2的步驟，以a=2, n=4為輸入，
詳細寫出求出24的值之過程。
Algorithm 2.
Procedure power(a : number, n : nonnegative integer)
if n = 0 then power(a, n):=1
else power(a, n):= a  power(a, n-1).
Ch4-28
Example 4. 寫出求 gcd(a,b) 的遞迴演算法，其中 a<b.
Sol :
遞迴式:gcd(a,b) := gcd(b mod a, a)
初始值:gcd(0,b) := b
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).
Ch4-29
Exercise 4.4
6. 根據演算法4的步驟，求出gcd(8, 13)的值。
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).
29. 設計一個遞迴演算法來找出數列的第n項，其中
a0=1，a1=2，而an= an-1 an-2，n=2, 3, …。
Ch4-30
(跳過)
Example 5. Search x in a1, a2,…,an by Linear Search
Sol :
從ai,ai+1,…aj 中找 x
Alg. 5
procedure search (i, j, x: integers)
if ai = x then location := i
else if i = j then location := 0
call
else search(i+1, j, x)
search(1, n, x)
Ch4-31
(跳過)
Example 6. Search x from a1,a2,…,an by binary
search (recursive version).
search x from ai, ai+1, …, aj
Sol : Alg. 5
procedure binary_search (x , i , j: integers)
m := (i+j) / 2
表示左半邊
if x = am then location := m
ai, ai+1, …, am-1
else if (x < am and i < m) then
至少還有一個元素
binary_search(x, i, m-1)
else if (x > am and j > m) then
binary_search(x, m+1, j)
else location := 0
call binary_search(x, 1, n)
Ch4-32
(跳過)
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! }
Ch4-33
(跳過)
※ 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)
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)
Ch4-34
(跳過)
Alg.8 (Iterative Fibonacci)
procedure iterative_fibonacci (n: nonnegative integer)
if n = 0 then y := 0 // y = f0
else begin
x := 0
y := 1 // y = f1
for i := 1 to n-1
begin
i=1 i=2 i=3
z := x + y
z
f2
f3
f4
x := y
y := z
x
f1
f2
f3
end
y
f2
f3
f4
end
{y is fn }
Exercise : 11, 35
Ch4-35