""" * Binary Exponentiation with Multiplication * This is a method to find a*b in a time complexity of O(log b) * This is one of the most commonly used methods of finding result of multiplication. * Also useful in cases where solution to (a*b)%c is required, * where a,b,c can be numbers over the computers calculation limits. * Done using iteration, can also be done using recursion * @author chinmoy159 * @version 1.0 dated 10/08/2017 """ def b_expo(a, b): res = 0 while b > 0: if b & 1: res += a a += a b >>= 1 return res def b_expo_mod(a, b, c): res = 0 while b > 0: if b & 1: res = ((res % c) + (a % c)) % c a += a b >>= 1 return res """ * Wondering how this method works ! * It's pretty simple. * Let's say you need to calculate a ^ b * RULE 1 : a * b = (a+a) * (b/2) ---- example : 4 * 4 = (4+4) * (4/2) = 8 * 2 * RULE 2 : IF b is ODD, then ---- a * b = a + (a * (b - 1)) :: where (b - 1) is even. * Once b is even, repeat the process to get a * b * Repeat the process till b = 1 OR b = 0, because a*1 = a AND a*0 = 0 * * As far as the modulo is concerned, * the fact : (a+b) % c = ((a%c) + (b%c)) % c * Now apply RULE 1 OR 2, whichever is required. """