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Recursion

Recursion. Yuting Zhang. Allegheny College, 04/24/06. A Game. My number = 10* the number inside + 1. A Game. My number = 10* the number inside + 1. A Game. My number = 10* the number inside + 1. A Game. My number =1. A Game. My number = 10* 1 + 1 = 11. A Game.

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Recursion

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  1. 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 Produce a same problem with smaller size My number =1 Simple case

  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. . Conquer Converge Divide

  15. Previous Game My number =1 My number = 10* the number inside + 1 n: # of boxes Recursion step: num(n) = 10*num(n-1) + 1 Base case: num(0) = 1 For any n>0: num(n) -> num(n-1) -> num(n-2)….-> num(0)

  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 parameter 1 Num return 111 parameter 2 return 1111 Num num(3) = 10 * num(2) + 1 = 10 * (10 * num(1) + 1) + 1 = 10 * (10 * (10 * num(0) + 1) +1) + 1 = 10 * (10 * (10 * 1 + 1) + 1 ) + 1 = 10 * (10 * 11 + 1 ) + 1 = 10 * 111 + 1 = 1111 return1 parameter 0 return 11 Num parameter 3

  21. Recursion and Method Call Stack Num parameter 1 Num parameter 2 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 Num parameter 0 Num(0) Num(1) Num(2) Num Num(3) parameter 3

  22. Recursion and Method Call Stack Num parameter 1 Num return 111 parameter 2 top of stack top of stack top of stack top of stack top of stack return 1111 When a method is returned, pop the method and parameters off the stack Num return1 parameter 0 Num(0) return 11 Num(1) Num(2) Num Num(3) parameter 3

  23. Factorials 1! return1 parameter 1 2!=2*1! parameter 2 return 2 3! =3*2! return 6 parameter 3 4! =4*3! return 24 parameter 4 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); }

  24. Iteration n! = n* (n-1) * … * 1, 0! = 1 int factorial(int n) { int result = 1; for (int i = n; i>=1; i--); result *= i; return result; }

  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); }

  26. Towers of Hanoi • 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 3 1 2 5 4 3 2 1

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

  28. Towers of Hanoi 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 3 1 2 2 2 1 1 2

  29. Towers of Hanoi 5 5 5 5 5 4 4 4 4 4 3 3 3 3 3 2 2 2 2 2 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 3 1 2 1 1

  30. Towers of Hanoi How? 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 3 1 2 Same way with less disks 5 4 5 5 3 4 4 2 3 3 1 2

  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? 2n -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; }

  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

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