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Data Structures
資料結構
Chapter 2
Recursion
What’s Recursion?

Definition


An algorithmic technique where a function,
in order to accomplish a task, calls itself
with some part of the task.
Every recursive solution involves two
major parts or cases, the second part
having three components.

Base case(s): in which the problem is
simple enough to be solved directly.
2
What’s Recursion? (cont.)

Recursive case(s): A recursive case has
three components.

Divide: the problem into one or more
simpler or smaller parts of the problem.

Call: the function (recursively) on each part.

Combine: the solutions of the parts into a
solution for the problem.
3

「遞迴」（Recursive）是程式設計的一
個重要觀念。
「遞迴函數」（Recursive Functions）
可以讓函數的程式碼變的很簡潔，但是
設計這類函數需要很小心，不然很容易
就掉入類似無窮迴圈的陷阱。

遞迴的觀念主要是在建立遞迴函數，其
基本定義，如下所示：
一個問題的內涵是由本身所定義的話，
稱之為遞迴。

遞迴函數是由上而下分析方法的一種特殊
的情況，因為子問題本身和原來問題擁有
相同的特性，只是範圍改變，範圍逐漸縮
小到一個終止條件

遞迴函數的特性，如下所示：

遞迴函數在每次呼叫時，都可以使問題範圍逐
漸縮小。

函數需要擁有一個終止條件，以便結束遞迴函
數的執行，否則遞迴函數並不會結束，而是持
續的呼叫自已。
2-1
Factorial - A Case Study
0! 1
N! 1 2  3  ...  N





0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
factor(n)
prod ← 1
for i ← 1 to n
prod ← prod * I
return prod
Iterative
algorithm
6
Factorial Function (cont.)
0! 1
N! 1 2  3  ...  N





0! = 1
1! = 1 * 0!
2! = 2 * 1!
3! = 3 * 2!
4! = 4 * 3!
Recursiv
e
algorithm
factor(n)
IF (n = 0) THEN
return 1
ELSE
return n * factor(n-1)
7
Factorial Function (cont.)

1
2
3
4
5
6
Before we can evaluate n!, we must first evaluate
(n-1)!.
5!=5*4!
4!=4*3!
3!=3*2!
2!=2*1!
1!=1*0!
0!=1
8
2-1
Factorial - A Case Study
 請參考課本
9
2-2
Designing Recursive Algorithms

The Design Methodology

Limitation of Recusion

Design Implemenation
16

Properties of Recursive Definitions 補充
or Algorithms

One important requirement for a
recursive algorithm is that it can not
generate an infinite sequence of calls on
itself.

There must be a “way out” of the
sequence of recursive calls.
說明
17

Properties of Recursive Definitions 補充
or Algorithms (cont.)

Without such a nonrecursive exit, no
recursive function can ever be
computed.

A recursive definition must eventually
reduce to some manipulation of one or
more simple, nonrecursive cases.
說明
18
2-3
Recursive Examples

Multiplication of Natural Numbers

Greatest Common Divisor

Fiboncci Numbers

The Towers of Honoi
19

Multiplication of Natural Numbers

補充
說明
a*b=a+a+a+a+a+a+a+a………..+a
b times
multi(a, b)
prod ← 0
for i ← 1 to b
prod ← prod + a
return prod
Recursive
multi(a, b)
IF (b = 1) THEN
return a
ELSE
Iterative
return multi(a,(b-1))+a
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Fibonacci Sequence


Fibonacci sequence is the sequence
of integers:
0,1,1,2,3,5,8,13,21,34,…
Each number is the sum of the
preceding two numbers.
0+1=1, 1+1=2, 1+2=3, 2+3=5,…
21
Fibonacci Sequence (cont.)


This famous sequence was published in
1202 by Leonardo Pisano, who is
sometimes called Leonardo Fibonacci.
His book contains the following
exercise:

How many pairs of rabbits can be produced
from a single pair in a year’s time?
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Fibonacci Sequence (cont.)


To solve this problem, we are told to
assume that each pair produces a new
pair of offspring every month, ant that
each new pair becomes fertile at the age
of one month.
Furthermore, the rabbits never die.
23
Fibonacci Sequence (cont.)
fib(n)
IF (n=0 or n=1) THEN
return n
ELSE
return fib(n-1) + fib(n-2)
24
Fibonacci Sequence (cont.)
fib(4) = fib(3) + fib(2)
= fib(2) + fib(1) + fib(1) + fib(0)
= fib(1) + fib(0) + fib(1) + fib(1) + fib(0)
=1+0+1+1+0
=3
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(Continued)


求最大公因數GCD
補充
說明
1. 判斷兩數中哪一個比較大，大的當被除數，小的當除數
2. 把被除數除以除數，得到商及餘數
3. 若餘數等於零則
GCD = 小數
否則
令小數成為新的被除數，餘數成為新的除數，
重複步驟 2
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

常見的遞迴應用
 求最大公因數GCD
// 求GCD之遞迴函數
int gcd(int dividend, int divisor)
{
int remainder;
if (dividend < divisor){ // 找出兩數之大者當被除數
swap(dividend,divisor); // 小者當除數
}
if (divisor !=0) {
remainder = dividend % divisor;
return gcd(divisor,remainder);
}
else
return dividend;
}
補充
說明
// 以參考值傳遞引數，
將x,y兩數對調
void swap(int &x, int &y)
{
int temp = x;
x = y;
y = temp;
}
32


常見的遞迴應用
 求最大公因數GCD
3
2
gcd(123,36)
gcd(36,15)
gcd(15,6)
gcd(6,3)




123
108
15
12
3
被除數 = 除數
123 = 36
36
= 15
15
6
6
3
36
30
6
6
0
*
*
*
*
*
商
3
2
2
2
2
補充
說明
2
+ 餘數
+ 15
+
6
+
3
+
0
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(Continued)
Recursive Solution
Hanoi(numDisk, src, dest, mid)
IF (numDisk = 1) THEN
move a disk from src to dest
ELSE
Hanoi(numDisk – 1, src, mid, dest)
Hanoi(1, src, dest, mid)
Hanoi(numDisk – 1, mid, dest, src)
48
To move n disks from A to C, using B as auxiliary:
1. If n == 1, then move the single disk from A to C and stop.
2. Move the top n - 1 disks from A to B, using C as auxiliary.
3. Move the remaining disk from A to C.
4. Move the n - 1 disks from B to C, using A as auxiliary.
49
50
move disk 1 from peg A to peg B
move disk 2 from peg A to peg C
move disk 1 from peg B to peg C
move disk 3 from peg A to peg B
move disk 1 from peg C to peg A
move disk 2 from peg C to peg B
move disk 1 from peg A to peg B
move disk 4 from peg A to peg C
move disk 1 from peg B to peg C
move disk 2 from peg B to peg A
move disk 1 from peg C to peg A
move disk 3 from peg B to peg C
move disk 1 from peg A to peg B
move disk 2 from peg A to peg C
move disk 1 from peg B to peg C
51
Towers of Hanoi
4
3
2
1
A
B
C
64 gold disks to be moved from tower A to tower C
52
 each tower operates as a stack

Towers of Hanoi
3
2
1
A

B
C
3-disk Towers Of Hanoi
53
Towers of Hanoi
2
1
A

3
B
C
3-disk Towers Of Hanoi
54
Towers of Hanoi

1
2
3
A
B
C
3-disk Towers Of Hanoi
55
Towers of Hanoi

1
3
2
A
B
C
3-disk Towers Of Hanoi
56
Towers of Hanoi
A

3
2
1
B
C
3-disk Towers Of Hanoi
57
Towers of Hanoi

3
2
1
A
B
C
3-disk Towers Of Hanoi
58
Towers of Hanoi
2
1
3
A

B
C
3-disk Towers Of Hanoi
59
Towers of Hanoi
3
2
1
A
B
3-disk Towers Of Hanoi
• 7 disk moves
C

60
Recursive Solution
1
A
B
C
n > 0 gold disks to be moved from A to C using B
61
 move top n-1 disks from A to B using C

Recursive Solution
1
A

B
C
move top disk from A to C
62
Recursive Solution
1
A

B
C
move top n-1 disks from B to C using A
63
Recursive Solution
1
A
B
C
moves(n) = 0 when n = 0
64
 moves(n) = 2*moves(n-1) + 1 = 2n-1 when n > 0

第二章 重點回顧
遞迴的基本原理
 遞迴在程式設計上有何優缺點？
 設計遞迴程式的原則
 常見的遞迴問題的解法




河內塔
費式數列
Ackerman 函數
65