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