Integer square root in python

Question

Is there an integer square root somewhere in python, or in standard libraries? I want it to be exact (i.e. return an integer), and bark if there's no solution.

At the moment I rolled my own naive one:

def isqrt(n):     i = int(math.sqrt(n) + 0.5)     if i**2 == n:         return i     raise ValueError('input was not a perfect square') 

But it's ugly and I don't really trust it for large integers. I could iterate through the squares and give up if I've exceeded the value, but I assume it would be kinda slow to do something like that. Also I guess I'd probably be reinventing the wheel, something like this must surely exist in python already...

Answer 1

Newton's method works perfectly well on integers:

def isqrt(n):     x = n     y = (x + 1) // 2     while y < x:         x = y         y = (x + n // x) // 2     return x 

This returns the largest integer x for which x * x does not exceed n. If you want to check if the result is exactly the square root, simply perform the multiplication to check if n is a perfect square.

I discuss this algorithm, and three other algorithms for calculating square roots, at my blog.