Why is pow(base, exponent) is more efficient than pow(base, exponent, mod) in Python? Got you.. It’s reverse let’s see why!

2 min readMay 24, 2021

Nothing fancy, just some part of pow(2, 10000) result.

I randomly chose to solve a question on Leetcode and I was given this question which is pretty easy to solve using python. You may want to look at the question in order to understand the next part of the blog.

So, I got Time Limit Exceeded(TLE) when I used pow(base, exp) % 1337for a scary and fascinating test case which apparently has the below-mentioned list as exponent! It’s HUGE!

[1,7,7,4,3,1,7,0,1,4,4,9,2,8,5,0,0,9,3,1,2,5,9,6,0,9,9,0,9,6,0,5,3,7,9,8,8,9,8,2,5,4,1,9,3,8,0,5,9,5,6,1,1,8,9,3,7,8,5,8,5,5,3,0,4,3,1,5,4,1,7,9,6,8,8,9,8,0,6,7,8,3,1,1,1,0,6,8,1,1,6,6,9,1,8,5,6,9,0,0,1,7,1,7,7,2,8,5,4,4,5,2,9,6,5,0,8,1,0,9,5,8,7,6,0,6,1,8,7,2,9,8,1,0,7,9,4,7,6,9,2,3,1,3,9,9,6,8,0,8,9,7,7,7,3,9,5,5,7,4,9,8,3,0,1,2,1,5,0,8,4,4,3,8,9,3,7,5,3,9,4,4,9,3,3,2,4,8,9,3,3,8,2,8,1,3,2,2,8,4,2,5,0,6,3,0,9,0,5,4,1,1,8,0,4,2,5,8,2,4,2,7,5,4,7,6,9,0,8,9,6,1,4,7,7,9,7,8,1,4,4,3,6,4,5,2,6,0,1,1,5,3,8,0,9,8,8,0,0,6,1,6,9,6,5,8,7,4,8,9,9,2,4,7,7,9,9,5,2,2,6,9,7,7,9,8,5,9,8,5,5,0,3,5,8,9,5,7,3,4,6,4,6,2,3,5,2,3,1,4,5,9,3,3,6,4,1,3,3,2,0,0,4,4,7,2,3,3,9,8,7,8,5,5,0,8,3,4,1,4,0,9,5,5,4,4,9,7,7,4,1,8,7,5,2,4,9,7,9,1,7,8,9,2,4,1,1,7,6,4,3,6,5,0,2,1,4,3,9,2,0,0,2,9,8,4,5,7,3,5,8,2,3,9,5,9,1,8,8,9,2,3,7,0,4,1,1,8,7,0,2,7,3,4,6,1,0,3,8,5,8,9,8,4,8,3,5,1,1,4,2,5,9,0,5,3,1,7,4,8,9,6,7,2,3,5,5,3,9,6,9,9,5,7,3,5,2,9,9,5,5,1,0,6,3,8,0,5,5,6,5,6,4,5,1,7,0,6,3,9,4,4,9,1,3,4,7,7,5,8,2,0,9,2,7,3,0,9,0,7,7,7,4,1,2,5,1,3,3,6,4,8,2,5,9,5,0,8,2,5,6,4,8,8,8,7,3,1,8,5,0,5,2,4,8,5,1,1,0,7,9,6,5,1,2,6,6,4,7,0,9,5,6,9,3,7,8,8,8,6,5,8,3,8,5,4,5,8,5,7,5,7,3,2,8,7,1,7,1,8,7,3,3,6,2,9,3,3,9,3,1,5,1,5,5,8,1,2,7,8,9,2,5,4,5,4,2,6,1,3,6,0,6,9,6,1,0,1,4,0,4,5,5,8,2,2,6,3,4,3,4,3,8,9,7,5,5,9,1,8,5,9,9,1,8,7,2,1,1,8,1,5,6,8,5,8,0,2,4,4,7,8,9,5,9,8,0,5,0,3,5,5,2,6,8,3,4,1,4,7,1,7,2,7,5,8,8,7,2,2,3,9,2,2,7,3,2,9,0,2,3,6,9,7,2,8,0,8,1,6,5,2,3,0,2,0,0,0,9,2,2,2,3,6,6,0,9,1,0,0,3,5,8,3,2,0,3,5,1,4,1,6,8,7,6,0,9,8,0,1,0,4,5,6,0,2,8,2,5,0,2,8,5,2,3,0,2,6,7,3,0,0,2,1,9,0,1,9,9,2,0,1,6,7,7,9,9,6,1,4,8,5,5,6,7,0,6,1,7,3,5,9,3,9,0,5,9,2,4,8,6,6,2,2,3,9,3,5,7,4,1,6,9,8,2,6,9,0,0,8,5,7,7,0,6,0,5,7,4,9,6,0,7,8,4,3,9,8,8,7,4,1,5,6,0,9,4,1,9,4,9,4,1,8,6,7,8,2,5,2,3,3,4,3,3,1,6,4,1,6,1,5,7,8,1,9,7,6,0,8,0,1,4,4,0,1,1,8,3,8,3,8,3,9,1,6,0,7,1,3,3,4,9,3,5,2,4,2,0,7,3,3,8,7,7,8,8,0,9,3,1,2,2,4,3,3,3,6,1,6,9,6,2,0,1,7,5,6,2,5,3,5,0,3,2,7,2,3,0,3,6,1,7,8,7,0,4,0,6,7,6,6,3,9,8,5,8,3,3,0,9,6,7,1,9,2,1,3,5,1,6,3,4,3,4,1,6,8,4,2,5]

Then I searched for a more efficient way and I found that pow(base, exp, mod) takes way less time than pow(base, exp)does. I wanted to know why and what makes it faster.

So, here’s what I learned…

Long story short, The two-argument form pow(base, exp) % mod is equivalent to using the power operator: (base ** exp) % mod.

Whereas, pow(base, exp, mod) is recursively solving in the below manner.

//Pseudo code:base = base % n;if (exp == 0)
    return 1;else if (exp == 1)
    return base;else if (exp % 2 == 0)
    return modPow(base * base % n, exp / 2, n);else
    return base * modPow(base, exp - 1, n) % n;

This all comes from the idea of breaking down exponents into parts and reuse them.

Like, b ^ e = (b ^ e/2) * (b ^ e/2)

Reusing the value of b ^ e/2 % naccrues a great benefit.

This a simple tiny tweak used to save lots and lots of time.

This might not be efficient or considerate for small numbers as shown in the figure but numbers having a large number of digits will definitely show great results in terms of Time.

I tried implementing the function and Implementation of pow(base, exp, mod) would be…

Code in C:

#include<stdio.h>int modPow(int base, int exp, int n) {base = base % n;if (exp == 0)
    return 1;else if (exp == 1)
    return base;else if (exp % 2 == 0)
    return modPow(base * base % n, exp / 2, n);else
    return base * modPow(base, exp - 1, n) % n;
}int main() {
    
    int base = 256437289;
    int exp = 10;
    int n = 1337;
    
    long int result = modPow(256437289, 10, 1337);
    printf("%ld", result);
    
    return 0;
}

Output:

This brings us to the end of the blog.

Happy learning 🎈

References:
1) A Document on Fast Modular Exponentiation.
2) Python documentation.

Why is pow(base, exponent) is more efficient than pow(base, exponent, mod) in Python? Got you.. It’s reverse let’s see why!

Written by Hima Bindu