ted ted -4 years ago 157
Javascript Question

Pow and mod function optimization

I need to create an optimized function to count Math.pow(a,b) % c; in Javascript;

There's no problems while counting small numbers like:

Math.pow(2345,123) % 1234567;

But if you try to count:
Math.pow(2345678910, 123456789) % 1234567;

you'll get incorrect result because of Math.pow() function result that cannot count up "big" numbers;

My solution was:

function powMod(base, pow, mod){
var i, result = 1;
for ( i = 0; i < pow; i++){
result *= base;
result %= mod;
return result;

Though it needs a lot of time to be counted;

Is it possible to optimized it somehow or find more rational way to count up Math.pow(a, b) % c; for "big" numbers? (I wrote "big" because they are not really bigIntegers);

Answer Source

Based on SICP.

function expmod( base, exp, mod ){
  if (exp == 0) return 1;
  if (exp % 2 == 0){
    return Math.pow( expmod( base, (exp / 2), mod), 2) % mod;
  else {
    return (base * expmod( base, (exp - 1), mod)) % mod;

This one should be quicker than first powering and then taking remainder, as it takes remainder every time you multiply, thus making actual numbers stay relatively small.

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