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An in-depth analysis of recursive algorithms, focusing on the concept of recurrence relations and their application in determining the time complexity of various recursive methods, including recursive factorial, selection sort, binary search, and towers of hanoi. It covers the process of forming recurrence relations, solving them, and understanding the big-o complexity of each method.
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T(0) = c T(n) = b + T(n - 1) = b + b + T(n - 2) = b +b +b + T(n - 3) … = kb + T(n - k) When k = n, we have: T(n) = nb + T(n - n) = bn + T(0) = bn + c. Therefore method factorial is O(n). long factorial (int n) { if (n == 0) return 1; else return n * factorial (n – 1); }
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public int binarySearch (int target, int[] array, int low, int high) { if (low > high) return -1; else { int middle = (low + high)/2; if (array[middle] == target) return middle; else if(array[middle] < target) return binarySearch(target, array, middle + 1, high); else return binarySearch(target, array, low, middle - 1); } }
public static void hanoi(int n, char from, char to, char temp){ if (n == 1) System.out.println(from + " --------> " + to); else{ hanoi(n - 1, from, temp, to); System.out.println(from + " --------> " + to); hanoi(n - 1, temp, to, from); } }
Expanding: T(n) = 2T(n – 1) + b = 2[2T(n – 2) + b] + b = 22 T(n – 2) + 2b + b = 2^2 [2T(n – 3) + b] + 2b + b = 23 T(n – 3) + 2^2 b + 2b + b = 2^3 [2T(n – 4) + b] + 2^2 b + 2b + b = 24 T(n – 4) + 2^3 b + 2^2 b + 2^1 b + 2^0 b = …… = 2 k^ T(n – k) + b[2k- 1^ + 2k– 2^ +... 2^1 + 2^0 ] When k = n – 1, we have:
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