Transcript Chapter 2
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.
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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.
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「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
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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)
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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
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2-1
Factorial - A Case Study
請參考課本
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2-2
Designing Recursive Algorithms
The Design Methodology
Limitation of Recusion
Design Implemenation
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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
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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
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2-3
Recursive Examples
Multiplication of Natural Numbers
Greatest Common Divisor
Fiboncci Numbers
The Towers of Honoi
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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,…
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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.
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Fibonacci Sequence (cont.)
fib(n)
IF (n=0 or n=1) THEN
return n
ELSE
return fib(n-1) + fib(n-2)
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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;
}
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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)
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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.
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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
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Towers of Hanoi
4
3
2
1
A
B
C
64 gold disks to be moved from tower A to tower C
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each tower operates as a stack
Towers of Hanoi
3
2
1
A
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
2
1
A
3
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
1
2
3
A
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
1
3
2
A
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
A
3
2
1
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
3
2
1
A
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
2
1
3
A
B
C
3-disk Towers Of Hanoi
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Towers of Hanoi
3
2
1
A
B
3-disk Towers Of Hanoi
• 7 disk moves
C
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Recursive Solution
1
A
B
C
n > 0 gold disks to be moved from A to C using B
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move top n-1 disks from A to B using C
Recursive Solution
1
A
B
C
move top disk from A to C
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Recursive Solution
1
A
B
C
move top n-1 disks from B to C using A
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Recursive Solution
1
A
B
C
moves(n) = 0 when n = 0
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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
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