Chapter 8 Recursion

Chapter 8 Recursion §8.1 Introduction §8.2 Methodology of Recursion Design §8.3 Recursion vs. Iteration §8.4 Case Studies

Objectives • To know the concept of recursive function and the benefits of using recursion • To learn the method of solving problems using recursive functions • To learn the use of overloaded helper function to derive a recursive function • To understand the relationship and difference between recursion and iteration.

§8.1 Introduction • Computing Factorial • 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!

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);

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

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

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(n -1) + fib(n -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

Fibonacci Numbers

§8.2 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

Problem Solving Using Recursion • General – thinking recursively • Divide and conquer • Sub-problems resemble the original void nPrintln(char * msg, int times) { if (times >= 1) { cout << msg << endl; nPrintln(msg, times - 1); } } bool isPalindrome(const char * const 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)); }

Problem Solving Using Recursion • Specific – making use of 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); }

§8.3 Recursion vs. Iteration • 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

f(n)= n! int factorial(int n){ int result; inti; 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

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; }

§8.4 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

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

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.

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

按字典序生成1~n的全排列 • 思路 • 先输出所有以1开头的排列，再输出所有以2开头的排列，…，最后才是以n开头的序列。 • 以1开头的排列的特点是：第一位是1，后面是2~9的排列。2~9的排列也必须按照字典序排列。（递归！） • 辅助函数 void print_permutation(前缀序列A，集合S) { if (S为空) 输出序列A; else 按照从小到大的顺序依次考虑S的每个元素v { print_permutation(在A的末尾填加v后得到的新序列, S-{v}); } }

按字典序生成1~n的全排列 void print_permutation(int n, int a[], int cur) { inti, j; if (cur==n) //递归边界 { for(inti=0;i<n;i++) printf("%d", a[i]); printf("\n"); } else for(inti=0;i<n;i++) //尝试在a[cur]中填各种整数i { int ok = 1; for (j=0;j<cur;j++) if (a[j]==i) ok = 0; //如果i已经在a[0]~a[cur-1]中出现，则不能再选 if (ok) { a[cur] = i; print_permutation(n,a,cur+1); //递归调用 } } }

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

Chapter 8 Recursion