Integer cube root

Question

I'm looking for fast code for 64-bit (unsigned) cube roots. (I'm using C and compiling with gcc, but I imagine most of the work required will be language- and compiler-agnostic.) I will denote by ulong a 64-bit unisgned integer.

Given an input n I require the (integral) return value r to be such that

r * r * r <= n && n < (r + 1) * (r + 1) * (r + 1)

That is, I want the cube root of n, rounded down. Basic code like

return (ulong)pow(n, 1.0/3);

is incorrect because of rounding toward the end of the range. Unsophisticated code like

ulong
cuberoot(ulong n)
{
    ulong ret = pow(n + 0.5, 1.0/3);
    if (n < 100000000000001ULL)
        return ret;
    if (n >= 18446724184312856125ULL)
        return 2642245ULL;
    if (ret * ret * ret > n) {
        ret--;
        while (ret * ret * ret > n)
            ret--;
        return ret;
    }
    while ((ret + 1) * (ret + 1) * (ret + 1) <= n)
        ret++;
    return ret;
}

gives the correct result, but is slower than it needs to be.

This code is for a math library and it will be called many times from various functions. Speed is important, but you can't count on a warm cache (so suggestions like a 2,642,245-entry binary search are right out).

For comparison, here is code that correctly calculates the integer square root.

ulong squareroot(ulong a) {
    ulong x = (ulong)sqrt((double)a);
    if (x > 0xFFFFFFFF || x*x > a)
        x--;
    return x;
}

Answer 1

The book "Hacker's Delight" has algorithms for this and many other problems. The code is online here. EDIT: That code doesn't work properly with 64-bit ints, and the instructions in the book on how to fix it for 64-bit are somewhat confusing. A proper 64-bit implementation (including test case) is online here.

I doubt that your squareroot function works "correctly" - it should be ulong a for the argument, not n :) (but the same approach would work using cbrt instead of sqrt, although not all C math libraries have cube root functions).