If we want to know the last ten digits of number *a ^{n}* – we have to evaluate expression

*a*. Using brute force approach, we have to do

^{n}mod 10^{10}*O(n)*( If you don’t understand big „O” notation, visit: Big O notation – it’s VERY important) multiplications and modulo divisions. When

*n*is greater than

*10*, it will take a lot of time. Fortunately, we can use more „mathematical” approach, which has

^{9}*O(log n)*complexity.

Firstly, we have to notice, that:

(1) a

^{2c}mod r = (a

^{c})

^{2}mod r

(2) a

^{2c+1}mod r = a*(a

^{c})

^{2}mod r

#! /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

*a*. Here is example, which shows running of the function a=2, n=34, m=10

^{n}mod m^{10}

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] (2

^{1}+ 2

^{5}= 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*= 2

^{0}mod 10

^{10}= 1. Then, during execution of

*for*loop,

*solution*has following values: 2

^{1(b)}mod 10

^{10}, 2

^{10(b)}mod 10

^{10}, 2

^{100(b)}mod 10

^{10}, 2

^{1000(b)}mod 10

^{10}, 2

^{10001(b)}mod 10

^{10}and finally 2

^{100010(b)}mod 10

^{10}.

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

*log*, so the complexity of whole alghorithm is

_{2}n*O(log n)*

You can download code here:

**expon.py** (369 bytes, 927 hits)

Hi there,

Not enought information

Thank you

Eremeeff

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Super post, Need to mark it on Digg

Thank you

AnnaHopn

Hello,

Super post, Need to mark it on Digg

Thank you

Elcoj