#! /usr/bin/env python
# Author: Wojtek Jamrozy (www.wojtekrj.net)
 
def modexp(a, n, m):
	bits = []
	while n:
		bits.append(n%2)
		n /= 2
	solution = 1
	bits.reverse()
	for x in bits:
		solution = (solution*solution)%m
		if x:
			solution = (solution*a)%m
	return solution
 
print modexp(2,34, pow(10,10))
print modexp(2,pow(10,20), pow(10,10))

output:

7179869184
1787109376

Function pow(a,b) is Python’s function, which computes a^b
Function modexp(a, n, m) computes aⁿ mod m. Here is example, which shows running of the function a=2, n=34, m=10¹⁰
Firstly, we want to have bits of binary notation of n in list bits.
Before reversion, bits have following content: [0, 1, 0, 0, 0, 1] (2¹ + 2⁵ = 33)
After reversion of list: bits= [1, 0, 0, 0, 1, 0]
Bits of binary notation of n are in order from “bigger” powers to “lower” ones.
I will use (b) for binary notation of number: 33 = 100001(b)
At the beggining, solution = 2⁰ mod 10¹⁰ = 1. Then, during execution of for loop, solution has following values: 2^1(b) mod 10¹⁰, 2^10(b) mod 10¹⁰, 2^100(b) mod 10¹⁰, 2^1000(b) mod 10¹⁰, 2^10001(b) mod 10¹⁰ and finally 2^100010(b) mod 10¹⁰.
Using (1) and (2) we can calcualte every solution’s value if we know previous one.
Loop for is executed for each element of bits. Bits size is log₂n, so the complexity of whole alghorithm is O(log n)

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