Why is pow(a, d, n) so much faster than a**d % n?

lyallcooper picture lyallcooper · Jan 3, 2013 · Viewed 14.7k times · Source

I was trying to implement a Miller-Rabin primality test, and was puzzled why it was taking so long (> 20 seconds) for midsize numbers (~7 digits). I eventually found the following line of code to be the source of the problem:

x = a**d % n

(where a, d, and n are all similar, but unequal, midsize numbers, ** is the exponentiation operator, and % is the modulo operator)

I then I tried replacing it with the following:

x = pow(a, d, n)

and it by comparison it is almost instantaneous.

For context, here is the original function:

from random import randint

def primalityTest(n, k):
    if n < 2:
        return False
    if n % 2 == 0:
        return False
    s = 0
    d = n - 1
    while d % 2 == 0:
        s += 1
        d >>= 1
    for i in range(k):
        rand = randint(2, n - 2)
        x = rand**d % n         # offending line
        if x == 1 or x == n - 1:
            continue
        for r in range(s):
            toReturn = True
            x = pow(x, 2, n)
            if x == 1:
                return False
            if x == n - 1:
                toReturn = False
                break
        if toReturn:
            return False
    return True

print(primalityTest(2700643,1))

An example timed calculation:

from timeit import timeit

a = 2505626
d = 1520321
n = 2700643

def testA():
    print(a**d % n)

def testB():
    print(pow(a, d, n))

print("time: %(time)fs" % {"time":timeit("testA()", setup="from __main__ import testA", number=1)})
print("time: %(time)fs" % {"time":timeit("testB()", setup="from __main__ import testB", number=1)})

Output (run with PyPy 1.9.0):

2642565
time: 23.785543s
2642565
time: 0.000030s

Output (run with Python 3.3.0, 2.7.2 returns very similar times):

2642565
time: 14.426975s
2642565
time: 0.000021s

And a related question, why is this calculation almost twice as fast when run with Python 2 or 3 than with PyPy, when usually PyPy is much faster?

Answer

BrenBarn picture BrenBarn · Jan 3, 2013

See the Wikipedia article on modular exponentiation. Basically, when you do a**d % n, you actually have to calculate a**d, which could be quite large. But there are ways of computing a**d % n without having to compute a**d itself, and that is what pow does. The ** operator can't do this because it can't "see into the future" to know that you are going to immediately take the modulus.