Chapter 17 Recursion ( 递归 )

1. Chapter 17 Recursion (递归) §17.1 Introduction §17.2 Methodology of Recursion Design §17.3 Recursion vs. Iteration

2. §17.1 Introduction • Computing Factorial (阶乘) • Factorial in mathematics: f(n) = n! = 1*2*..*n 1 if n=0 = n*(n-1)! if n>0 • Factorial in C++ int factorial(int n){ int result; if (n==0) result =1; else result = n *factorial(n-1); return result; } Recursive Call!

3. Computing Factorial factorial(3) = = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) = 3 * ( 2 * ( 1 * 1))) = 3 * ( 2 * 1) = 3 * 2 = 6 factorial(0) = 1; factorial(n) = n*factorial(n-1);

4. Trace Recursive Factorial factorial(4) return 24 to caller Step 0: execute factorial(4) Step 9: return 24 Space Required for factorial(0) 4*factorial(3) Space Required Step 1: execute factorial(3) Step 8: return 6 for factorial(1) 3*factorial(2) Space Required Step 2: execute factorial(2) Step 7: return 2 for factorial(2) 2*factorial(1) Space Required Step 3: execute factorial(1) for factorial(3) Step 6: return 1 1*factorial(0) Space Required for factorial(4) Step 4: execute factorial(0) Step 5: return 1 return 1

5. Recursive Function • Recursive Function • A function with recursive call • Recursive call • A function calls itself, directly or indirectly f( ){ …… f( ); …… } f1( ){ f2( ){ …… …… f2( ); f1( ); …… …… } }

6. Fibonacci Numbers Finonacci series: 0 1 1 2 3 5 8 13 21 34 55 89 … indices: 0 1 2 3 4 5 6 7 8 9 10 11 … 0, if i = 0 fib(i) = 1, if i = 1 fib(i -1) + fib(i -2), if i >=2 fib(3) = fib(2) + fib(1) = (fib(1) + fib(0)) + fib(1) = (1 + 0) +fib(1) = 1 + fib(1) = 1 + 1 = 2 ComputeFibonacci

7. Fibonacci Numbers

8. §17.2 Methodology of Recursion Design • Characteristics of Recursion • Different cases using selection statement • One or more base cases (the simplest case) • To stop recursion • Every recursive call reduces the original problem • To bring it increasingly closer to and eventually to be the base case

9. Problem Solving Using Recursion • General – thinking recursively • Divide and conquer • Sub-problems resemble the original void nPrintln(string msg, int times) { if (times >= 1) { cout << msg << endl; nPrintln(msg, times - 1); } } bool isPalindrome(const string s) { if (strlen(s) <= 1) return true; else if (s[0] != s[strlen(s) - 1]) return false; else return isPalindrome(substring(s, 1, strlen(s) - 2)); }

10. Recursive Helper Function • Recursive helper function • A recursive function with additional parameters to reduce the problem • Especially useful for functions involving strings/arrays bool isPalindrome(const char * const s, int low, int high) { if (high <= low) return true; else if (s[low] != s[high]) return false; else return isPalindrome(s, low + 1, high - 1); } bool isPalindrome(const char * const s) { return isPalindrome(s, 0, strlen(s) - 1); }

11. Case Studies • Recursive Selection Sort • Find the largest number in the list and swaps it with the last number. • Ignore the last number and sort the remaining smaller list recursively. RecursiveSelectionSort

12. Recursive Binary Search • If the key is less than the middle element, recursively search the key in the first half of the array. • If the key is equal to the middle element, the search ends with a match. • If the key is greater than the middle element, recursively search the key in the second half of the array. RecursiveBinarySearch

13. Towers of Hanoi • There are n disks labeled 1, 2, 3, . . ., n, and three towers labeled A, B, and C. • No disk can be on top of a smaller disk at any time. • All the disks are initially placed on tower A. • Only one disk can be moved at a time, and it must be the top disk on the tower.

14. Solution to Towers of Hanoi Decompose the problem into three subproblems. n  n-1  n-2  … 1, move it directly!

15. 1 2 A B C A B C A B C 4 5 3 A B C A B C A B C 6 7 A B C A B C Towers of Hanoi TowersOfHanoi 1. 把n-1个盘从A搬到C，借助B 2. 把A上剩下的最大的一个盘搬到B 3. 再把n-1个盘从C搬到B，借助A 4. 当n==1时，直接从A搬到B即可

16. Eight Queens boolisValid(int row, int column) { for (int i = 1; i <= row; i++) if (queens[row - i] == column || queens[row - i] == column - i || queens[row - i] == column + i) return false; // There is a conflict return true; // No conflict } EightQueen

17. §17.3 Recursion vs. Iteration (迭代) • Negative aspects of recursion • High cost in both time and memory • Recursion  Iteration • Any recursive function can be converted to non-recursive (iterative) function • When to use recursion? Depending on the problem. • Recursion is suitable for “recursive” problems

18. f(n)= n! int factorial(int n){ int result; int i; for(i=1;i<=n;i++) result *= i; return result; } int factorial(int n){ int result; if(n==0) result =1; else result = n*factorial(n-1); return result; } Recursion vs. Iteration

19. Recursion vs. Iteration f(n) = 0 n=0 1 n=1 f(n-1)+f(n-2) n>1 int fib (int n) { int result, i, pre1, pre2 ; result = 0; i =2; pre2 = 1 ; while (i<=n) { pre1 = pre2 ; pre2 = result ; result = pre1 + pre2; i++; } return result; } int fib (int n){ int result; if (n==0) result =0; else if (n==1) result =1; else result = fib (n-1)+ fib (n-2); return result; }

20. Summary • Concept of recursion • Design of recursive functions • Recursion vs. iteration

21. Homework Questions 1. Please write a recursive function to solve the Sudoku problem. 2. How to find all possible solutions for the Eight Queens problem? • (Just for exercise by yourself. No need to submitting it.)