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+
Ch4-2
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:
n
n
i 1
i 1
2
(
2
i
1
)

2
i
n

n
(
n

1
)
n

n


n
Pf : Let P(n) denote the proposition that
2
(
2
i
1
)

n

i 1
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+
Ch4-3
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+.
Ch4-4
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 2 n  1  n / 2 ”.
Basis step : P(0) is true, since H 20  H1  1  1  0 / 2.
Inductive step : Assume that P(k) is true for some k,
i.e., H 2 k  1  k / 2
Consider P(k+1) :
Ch4-5
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
Ch4-6
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.
Ch4-7
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
Ch4-8
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
Ch4-9
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).
Ch4-10
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 f0, 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
Ch4-11
Example 6. Show that fn > a
n-2
, where
1 5
a 
,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
Ch4-12
※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
Ch4-13
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.
Ch4-14
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, …}
0l1
∴當bit string a = 000…011…1 時
aA
n個
n個
Ch4-15

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.
Ch4-16
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.
Ch4-17
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.
∴
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).
Ch4-18
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 a1, a2,…,an by Linear Search
Sol : Alg. 5
從ai,ai+1,…aj 中找 x
procedure search (i, j, x: integers)
if ai = x then location := i
else if i = j then location := 0
else search(i+1, j, x)
call
search(1, n, x)
Ch4-19
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)
Ch4-20
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! }
Ch4-21
※ 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)
Ch4-22
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
Ch4-23