Computer Science Recursion Yuting Zhang llegheny College, 04/24/06
Computer Science Recursion Yuting Zhang Allegheny College, 04/24/06.
Mar 29, 2015
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Computer Science
Recursion
Yuting Zhang
Allegheny College, 04/24/06
2
A Game
My number =10* the number inside
+ 1
3
A Game
My number =10* the number inside
+ 1
4
A Game
My number =10* the number inside
+ 1
5
A Game
My number =1
6
A Game
My number =10* 1 + 1 =
11
7
A Game
My number =10* 11 + 1
= 111
8
A Game
My number =10* 111 + 1
=1111
9
Examples of Recursion
10
Examples of Recursion
Divide the line into 16 segments:
11
Outline
The concept of recursion How to write and use recursive method How the recursive method are executed Recursion vs. iteration Examples: Tower of Hanoi Summary
12
Previous Game
My number =1
Simple case
Produce a same problem with smaller
size
13
Definition of Recursion
A computer programming technique involving the use of a procedure, subroutine, function, or algorithm that calls itself in a step having a termination condition so that successive repetitions are processed up to the critical step until the condition is met at which time the rest of each repetition is processed from the last one called to the first
14
Recursion Concepts Two parts to recursion:
Base case(s) – termination If the problem is easy, solve it
immediately.
Recursion step – call itself If the problem can't be solved
immediately, divide it into smaller problems, then
Solve the smaller problems by applying this procedure to each of them.
.
Divide
Conquer
Converge
15
Previous Game
For any n>0: num(n) -> num(n-1) -> num(n-2)….-> num(0)
n: # of boxesn: # of boxes
Recursion step: Recursion step: num(n) = 10*num(n-1) + 1num(n) = 10*num(n-1) + 1
My number =1
Base case: Base case: num(0) = 1num(0) = 1
My number =10* the number inside
+ 1
16
Recursion Method
Recursion step: num(n) = 10*num(n-1) + 1Base case: num(0) = 1
int Num(int n) {
if (n == 0) return 1;
else return 10*Num(n-1) +1;
}
17
Complete Programclass NumCalc{ int Num( int n ) { if ( n == 0 )
return 1; else
return 10*Num(n-1) +1; } } class NumCaclTester { public static void main ( String[] args) { NumCalc numcalc = new NumCalc(); int result = numcalc.Num( 3 ); System.out.println(“numcalc(3) is " + result ); } }
Result:
numcalc(3) is 1111
18
Recap
Skeleton for a recursive Java method
type solution(type para )
{
if ( base case ) {
return something easily computed
} else {
divide problem into pieces
return something calculated from the solution to each piece
}
}
19
Outline
The concept of recursion How to write and use recursive method How the recursive method are executed Recursion vs. iteration Examples: Tower of Hanoi Summary
20
Recursive Evaluation
num(num(33) ) = 10 * num(= 10 * num(22) + 1 ) + 1 = 10 * (10 * num(= 10 * (10 * num(11) + 1) + 1) + 1) + 1 = 10 * (10 * (10 * num(= 10 * (10 * (10 * num(00) + 1) +1) + 1) + 1) +1) + 1
= 10 * (10 * (10 * = 10 * (10 * (10 * 11 + 1) + 1 ) + 1 + 1) + 1 ) + 1 = 10 * (10 * = 10 * (10 * 1111 + 1 ) + 1 + 1 ) + 1 = 10 * = 10 * 111111 + 1 + 1 = = 11111111
Num
Num
parameter 3
parameter 0
Num
parameter 1
Num
parameter 2
return 1111
return1
return 11
return 111
21
Recursion and Method Call Stack
Num(3)
Num
Num
parameter 3
parameter 0
Num
parameter 1
Num
parameter 2Num(2)
Num(1)
Num(0)
top of stack
top of stack
top of stack
top of stack
top of stack
When a method is called,push the method and parameters into the stack
22
Recursion and Method Call Stack
Num(3)
Num
Num
parameter 3
parameter 0
Num
parameter 1
Num
parameter 2
return 1111
return1
return 11
return 111Num(2)
Num(1)
Num(0)
top of stack
top of stack
top of stack
top of stack
top of stack
When a method is returned,pop the method and parameters off the stack
23
Factorials
n! = n* (n-1) * … * 1, (n> 0)0! = 1Recursion step: n! = n*(n-1)!Base case: 0! = 1, 1! = 1
int factorial(int n) { if (n <= 1)
return 1; else
return n*factorial(n-1); }
1!
2!=2*1!
3!=3*2!
4!=4*3!
parameter 4
parameter 1
parameter 2
parameter 3
return 24
return1
return 2
return 6
24
Iteration
int factorial(int n) { int result = 1; for (int i = n; i>=1; i--); result *= i; return result;}
n! = n* (n-1) * … * 1, 0! = 1
25
Recursion vs. Iteration
int factorial(int n) { int result = 1; for (int i = n; i>=1; i--); result *= i; return result;}
int factorial(int n) { if (n <= 1)
return 1; else
return n*factorial(n-1); }
Iteration Recursion
Repetition Explicit statement e.g. for, while
Repeat method calls
Termination
test
Loop-continuation condition fails
Base case is reached
Approach termination
Modify a counter toward fail condition
Produce simpler version of the original problem
Overhead Less More
26
Towers of Hanoi
45
123
11 22 33
Problem: How to move disks from peg 1 to 3, subjected to:- Only one disk is moved at a time- A larger disk can’t be placed above a smaller disk at any time- Peg 2 is used for temporarily holding disks
27
Towers of Hanoi
11 22 33
Case 1: move 1 disk from peg 1 to 3, using peg 2
-- Move disk 1 from peg 1 to peg 3 directly
11 11
28
Towers of Hanoi
11 22 33
11 1122
2222
Case 2: move 2 disks from peg 1 to 3, using peg 2
- Move disk 2 from peg 1 to 2- Move disk 1 from peg 1 to 3 - Move disk 2 from peg 2 to 3
29
45
23
Towers of Hanoi
11 22 33
111
45
23
Case 3: move 5 disks from peg 1 to 3
- Move 4 disks(2-5) from peg 1 to 2, using peg 3- Move disk 1 from peg 1 to 3 - Move 4 disks(2-5) from peg 2 to 3, using peg 1
45
23
45
2345
23
30
Towers of Hanoi
11 22 33
1
45
2
3
Case 3: move 5 disks from peg 1 to 3
- Move 4 disks(2-5) from peg 1 to 2, using peg 3- Move disk 1 from peg 1 to 3 - Move 4 disks(2-5) from peg 2 to 3, using peg 1
How?How?
45
3
45
32
Same way Same way with less diskswith less disks
31
Towers of Hanoi
General Case : move n disks from peg 1 to 3, using peg 2
- Move n-1 disks from peg 1 to 2, using peg 3- Move last disk from peg 1 to 3 - Move n-1 disks from peg 2 to 3, using peg 1
void solveTowers (int disks, int srcpeg, int destpeg, int temppeg){
if (disks == 1) { System.out.printf(“\n %d -> %d”, srcpge,destpeg);
} else {solveTowers(disks -1, srcpeg, temppeg, destpeg);System.out.printf(“\n%d -> %d”, srcpeg,destpeg);solveTowers(disks -1, temppeg, destpeg, srcpeg);
}return;
}
32
Towers of Hanoi
Question:
How many moves are need to move n disks from peg 1 to 3?
void solveTowers (int disks, int srcpeg, int destpeg, int temppeg){
if (disks == 1) { System.out.printf(“\n %d -> %d”, srcpge,destpeg);
} else {solveTowers(disks -1, srcpeg, temppeg, destpeg);System.out.printf(“\n%d -> %d”, srcpeg,destpeg);solveTowers(disks -1, temppeg, destpeg, srcpeg);
}return;
}
2n -1
33
Towers of Hanoi
Case : move 5 disks from peg 1 to 3, using peg 2
Result:
1 --> 3 1 --> 2 3 --> 2 1 --> 32 --> 1 2 --> 3 1 --> 3 1 --> 23 --> 2 3 --> 1 2 --> 1 3 --> 21 --> 3 1 --> 2 3 --> 2 1 --> 32 --> 1 2 --> 3 1 --> 3 2 --> 13 --> 2 3 --> 1 2 --> 1 2 --> 31 --> 3 1 --> 2 3 --> 2 1 --> 32 --> 1 2 --> 3 1 --> 3
34
Other Examples
Fibonacci Series String Permutations Fractals Binary Search …
35
Summary
The concept of recursion How to write and use recursive method How the recursive method are executed Recursion vs. iteration Examples: Tower of hanoi
In order to understand recursion you must first understand recursion.
— Unknown
http://cs-people.bu.edu/danazh/cs111-recursion/index.html
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