CS1101: Programming Methodology Recitation 9 – Recursion

1. CS1101: Programming MethodologyRecitation 9 – Recursion

2. Recursive Methods • Every recursive method has the following elements: • A test to stop or continue the recursion. • An end case (base case) that terminate the recursion. • A recursive call(s) that continues the recursion. • Example factorial method public int factorial (int N) { if (N == 1) { return 1; } else { return N * factoral (N-1); } } Test to stop or continue End case: recursion stops Recursive case: recursion continues with recursive call

3. Recursion for Mathematical Functions • Compute the sum of first N positive integers public int sum (int N) { if (N == 1) { return 1; } else { return N + sum (N-1); } } • Compute the exponential AN public double exponent (double A, int N) { if (N == 1) { return A; } else { return A * exponent (A, N-1); } } Pass two arguments: Value of A does not change in the calls, but the value of N is decremented after each recursive call

4. Recursion for Nonnumerical Applications • Compute the length of a string public int length (String str) { if (str.equals(“”)) { //str has no characters return 0; } else { return 1 + length (str.substring(1)); } } • Compute all the anagrams of a word. • An anagram is a word formed by reordering letters of another word • Example: anagrams of CAT are CTA, ATC, ACT, TCA, TAC Index of second position is 1

5. rotate left rotate left rotate left H L O A L A O H O H A L L H A O Anagram recursion Apply recursion to find all the anagrams of these three letters recursion recursion recursion

6. Recursive algorithm to find anagrams public void anagram (String word) { int numOfChars = word.length(); if (numOfChars == 1) { //end case – cannot recurse anymore } else { for (int i=1; i <= numOfChars; i++) { char firstLetter = word.charAt(0); suffix = word.substring (1, numOfChars); anagram (suffix); //recurse with remaining letters in word //rotate left word = suffix + firstLetter; } } } • What do we do when recursion stops ? • Print out the anagram found • But words passed in successive recursive calls are getting shorter • since we chop off the first letter • Pass two parameters – prefix and suffix

7. Recursive algorithm to find anagrams public void anagram (String prefix, String suffix) { int numOfChars = suffix.length(); if (numOfChars == 1) { //end case – print out one anagram System.out.println (prefix + suffix); } else { for (int i=1; i <= numOfChars; i++) { String newSuffix = suffix.substring (1, numOfChars); String newPrefix = prefix + suffix.charAt(0); anagram (newPrefix, newSuffix); //recursive call //rotate left to create a new rearranged suffix suffix = newSuffix + suffix.charAt(0); } } } • Call method initially with empty prefix and word as suffix • Anagram (“”, “HALO”); • What happens when user enter an empty string ? • Anagram (“”, “”);

8. Quicksort low high … Array number p partition number[i] < p number[i] > p p Quicksort Quicksort mid

9. Quicksort • Any element can be used as a pivot. • For simplicity, use number[low] as pivot p. • Scan array and move elements smaller than p to left half and • elements larger than p to upper half. • 4. Sort lower and upper halves recursively, using quicksort. • 5. Pivot p is placed at location mid. • 6. Recursion stops when low >= high.

10. public void quickSort (int[] number, int low, int high) { if (low < high) { int mid = partition (number, low, high); quickSort (number, low, mid-1); quickSort (number, mid+1, high ); } } private int partition (int[] number, int start, int end) { int pivot = number[start]; //start the pivot do { //look for a number smaller than pivot from the end while (start < end && number[end] >= pivot) { end--; } if (start < end) { //found a smaller number number[start] = number[end]; //now find a number larger than pivot from the start while (start < end && number[start] <= pivot) { start++; } if (start < end) { //found a larger number number[end] = number[start]; } } } while (start < end); number[start] = pivot; //done. Move pivot back to array return start; }

11. 23 12 12 12 12 23 12 17 17 17 17 17 17 17 5 5 5 5 5 5 5 90 90 23 90 90 90 90 12 12 90 12 90 12 90 48 48 48 48 48 48 48 37 37 37 37 37 37 37 83 83 83 83 83 83 83 77 77 77 77 77 77 77 mid start end pivot 0 1 2 3 4 5 6 7 8 23 pivot = number[start] end start 0 1 2 3 4 5 6 7 8 while (number[end] > pivot) end--; number[start] = number[end] end start 0 1 2 3 4 5 6 7 8 while (number[start] < pivot) start++; number[end] = number[start] start, end 0 1 2 3 4 5 6 7 8 while (number[end] > pivot) end--; number[start] = pivot