# Computer Science Recursion Yuting Zhang Allegheny College, 04/24/06.

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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 boxes Recursion step: num(n) = 10*num(n-1) + 1 My number =1 Base case: num(0) = 1 My number = 10* the number inside + 1

16 Recursion Method Recursion step: num(n) = 10*num(n-1) + 1 Base case: num(0) = 1 int Num(int n) { if (n == 0) return 1; else return 10*Num(n-1) +1; }

17 Complete Program class 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(3) num(3) = 10 * num(2) + 1 = 10 * num(2) + 1 = 10 * (10 * num(1) + 1) + 1 = 10 * (10 * num(1) + 1) + 1 = 10 * (10 * (10 * num(0) + 1) +1) + 1 = 10 * (10 * (10 * num(0) + 1) +1) + 1 = 10 * (10 * (10 * 1 + 1) + 1 ) + 1 = 10 * (10 * (10 * 1 + 1) + 1 ) + 1 = 10 * (10 * 11 + 1 ) + 1 = 10 * (10 * 11 + 1 ) + 1 = 10 * 111 + 1 = 10 * 111 + 1 = 1111 = 1111 Num parameter 3 parameter 0 Num parameter 1 Num parameter 2 return 1111 return 1 return 11 return 111

21 Recursion and Method Call Stack Num(3) Num parameter 3 parameter 0 Num parameter 1 Num parameter 2 Num(2) Num(1) Num(0) 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 parameter 3 parameter 0 Num parameter 1 Num parameter 2 return 1111 return 1 return 11 return 111 Num(2) Num(1) Num(0) 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! = 1 Recursion 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 return 1 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); } IterationRecursion RepetitionExplicit 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 OverheadLessMore

26 Towers of Hanoi 4 5 1 2 3 1 2 3 Problem: How to move disks from peg 1 to 3, subjected to: - Only one disk is moved at a time - A larger disk cant be placed above a smaller disk at any time - Peg 2 is used for temporarily holding disks

27 Towers of Hanoi1 2 3 Case 1: move 1 disk from peg 1 to 3, using peg 2 -- Move disk 1 from peg 1 to peg 3 directly 1 1

28 Towers of Hanoi1 2 3 1 1 2 2 2 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 4 5 2 3 Towers of Hanoi1 2 3 1 1 4 5 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 4 5 2 3 4 5 2 3 4 5 2 3

30 Towers of Hanoi1 2 3 1 4 5 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? 4 5 3 4 5 3 2 Same way with 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; } 2 n -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 --> 3 2 --> 1 2 --> 3 1 --> 3 1 --> 2 3 --> 2 3 --> 1 2 --> 1 3 --> 2 1 --> 3 1 --> 2 3 --> 2 1 --> 3 2 --> 1 2 --> 3 1 --> 3 2 --> 1 3 --> 2 3 --> 1 2 --> 1 2 --> 3 1 --> 3 1 --> 2 3 --> 2 1 --> 3 2 --> 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