How did Python implement the built-in function pow()?

Question

I have to write a program to calculate a**b % c where b and c are both very large numbers. If I just use a**b % c, it's really slow. Then I found that the built-in function pow() can do this really fast by calling pow(a, b, c).
I'm curious to know how does Python implement this? Or where could I find the source code file that implement this function?

Solution 1

If a, b and c are integers, the implementation can be made more efficient by binary exponentiation and reducing modulo c in each step, including the first one (i.e. reducing a modulo c before you even start). This is what the implementation of long_pow() does indeed. The function has over two hundred lines of code, as it has to deal with reference counting, and it handles negative exponents and a whole bunch of special cases.

At its core, the idea of the algorithm is rather simple, though. Let's say we want to compute a ** b for positive integers a and b, and b has the binary digits b_i. Then we can write b as

b = b_0 + b1 * 2 + b2 * 2**2 + ... + b_k ** 2**k

ans a ** b as

a ** b = a**b0 * (a**2)**b1 * (a**2**2)**b2 * ... * (a**2**k)**b_k

Each factor in this product is of the form (a**2**i)**b_i. If b_i is zero, we can simply omit the factor. If b_i is 1, the factor is equal to a**2**i, and these powers can be computed for all i by repeatedly squaring a. Overall, we need to square and multiply k times, where k is the number of binary digits of b.

As mentioned above, for pow(a, b, c) we can reduce modulo c in each step, both after squaring and after multiplying.

Solution 2

You might consider the following two implementations for computing (x ** y) % z quickly.

In Python:

def pow_mod(x, y, z):
    "Calculate (x ** y) % z efficiently."
    number = 1
    while y:
        if y & 1:
            number = number * x % z
        y >>= 1
        x = x * x % z
    return number

In C:

#include <stdio.h>

unsigned long pow_mod(unsigned short x, unsigned long y, unsigned short z)
{
    unsigned long number = 1;
    while (y)
    {
        if (y & 1)
            number = number * x % z;
        y >>= 1;
        x = (unsigned long)x * x % z;
    }
    return number;
}

int main()
{
    printf("%d\n", pow_mod(63437, 3935969939, 20628));
    return 0;
}

Solution 3

I don't know about python, but if you need fast powers, you can use exponentiation by squaring:

http://en.wikipedia.org/wiki/Exponentiation_by_squaring

It's a simple recursive method that uses the commutative property of exponents.

Comments

[JimB]

math.pow() does't have the modulo argument, and isn't the same function as the builtin pow(). Also FYI, Psyco is getting pretty stale, and no 64-bit support. NumPy is great for serious math.

[stackuser]

@Noctis, I tried running your Python implementation and got this:TypeError: ufunc 'bitwise_and' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe'' ---- As I'm learning Python right now, I thought you might have an idea about this error (a search suggests it might be a bug but I'm thinking that there's a quick workaround)

[Noctis Skytower]

@stackuser: It appears to be working fine in the following demonstration: ideone.com/sYzqZN

[kilojoules]

Can anyone explain why this solution works? I am having trouble understanding the logic behind this algorithm.

[Ben Sandler]

Why can we reduce by modulo c in each step?

[Sven Marnach]

@BenSandler: Because a ≡ a' (mod c) and b ≡ b' (mod c) imply ab ≡ a'b' (mod c), or in other words, it doesn't matter whether you first reduce a and b modulo c and then multiply them, or multiply them first and then reduce modulo c. See the Wikipedia article on modular arithmetic.

[Fabiano]

@NoctisSkytower, what would be the benefit of this considering the native python pow() builtin function supports this as well and seems faster? >>> st_pow = 'pow(65537L, 767587L, 14971787L) >>> st_pow_mod = 'pow_mod(65537L, 767587L, 14971787L)' >>> timeit.timeit(st_pow) 4.510787010192871 >>> timeit.timeit(st_pow_mod, def_pow_mod) 10.135776996612549

[Noctis Skytower]

@Fabiano My function is not supposed to be used. It is simply an explanation of how Python works behinds the scenes without referring to its source in C. I was trying to answer wong2's question about how pow was implimented.

[JohanC]

Note that long_pow is now defined at another line in that file: github.com/python/cpython/blob/master/Objects/…

[Sven Marnach]

@JohanC I've updated the link to include the commit hash, so it doesn't get out of date anymore.