CODEWITHSUNDEEP
modular-arithmetic
AspiringMat
AspiringMat

Reputation: 2269

Fastest Method to calculate 2^n Mod m where n and m are Random integers

I was wandering if there is any possible efficient way of finding the Remainder when 2^n is divided by m where n and m are random integers. Is there any equation where I sub n and m into to give me the remainder or do we have to follow recursive methods? Do note that I am a beginner and I'm just starting out so might not understand stuff that is too complicated.

Thanks in advance.

Upvotes: 3

Views: 4624

Answers (4)

Prasant Nair
Prasant Nair

Reputation: 1

This javascript code handles very large values of n correctly.

function fastMod(n, m){
var pow2 = 2
var result = 1
while(n > 0){
    if(n&1){
        result = (result * pow2) % m 
    }
   n/=2
    pow2 = (pow2 * pow2) % m   
}
console.log(result)
}
fastMod(77, 100)

Upvotes: 0

James Parker
James Parker

Reputation: 477

fkdosilovic's answer is correct but not the fastest.

His algorithm runs in O(n) time, but it is possible to achieve O(log(n)).

Since all numbers 2^n can be represented as products from a set {2^1, 2^2, 2^4, 2^8 ..., 2^floor(lg(n))}, we only need to compute these values and multiply them. E.g. 2^13 = 2^1 * 2^4 * 2^8. Here is a python code.

def fast_powmod(n, m):
    pow2 = 2
    result = 1
    while n > 0:
        if n % 2 == 1:
            result = (result * pow2) % m
        pow2 = (pow2 * pow2) % m
        n >>= 1

    return result

Upvotes: 3

Pablo M
Pablo M

Reputation: 241

Fermat's little theorem can help you in the cases where m is a prime number:

If p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p.

For example, if a = 2 and p = 7, 2^7 = 128, and 128 − 2 = 7 × 18 is an integer multiple of 7.

Upvotes: -1

fkdosilovic
fkdosilovic

Reputation: 116

Modular arithmetic for multiplication works like this:

(a * b) % x = ( (a % x) * (b % x) ) % x

Here is C++ code for that:

#include <cstdio>
#include <cstdlib>

using namespace std;

int powmod(int n, int m) {
  int ret = 1;
  for(int i = 0; i < n; ++i)
    ret = ( (ret % m) * (2 % m) ) % m; // expression from above
  return ret; // returns 2 to the power n modulo m
}

int main() {

  int n, m; scanf("%d%d", &n, &m);
  printf("%d\n", powmod(n, m));

  return 0;
}

Upvotes: 1

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