How do I get a whole number as a result for a cube root?

Question

I am creating a problem which requires me to find the cube root of certain numbers, some of them have whole number roots, but a lot of them don\'t.

I have numbers li

Answer 1

The standard way to check for equality with floating point is to check for quality within a certain tolerance:

def floateq(a, b, tolerance=0.00000001):
    return abs(a-b) < tolerance

Now you can check if the rounded, converted-to-an-integer version of the cube root is equal to the cube root itself within a certain tolerance:

def has_integer_cube_root(n):
    floatroot = (n ** (1.0 / 3.0))
    introot = int(round(floatroot))
    return floateq(floatroot, introot)

Usage:

>>> has_integer_cube_root(125)
True
>>> has_integer_cube_root(126)
False

However, this is quite imprecise for your use case:

>>> has_integer_cube_root(40000**3)
True
>>> has_integer_cube_root(40000**3 + 1)
True

You can mess with the tolerance, but at some point, floating point numbers just won't be enough to have the accuracy you need.

EDIT: Yes, as the comment said, in this case you can check the result with integer arithmetic:

def has_integer_cube_root(n):
    floatroot = (n ** (1.0 / 3.0))
    introot = int(round(floatroot))
    return introot*introot*introot == n

>>> has_integer_cube_root(40000**3)
True
>>> has_integer_cube_root(40000**3 + 1)
False