Is cube root integer?

Question

This seems to be simple but I cannot find a way to do it. I need to show whether the cube root of an integer is integer or not. I used is_integer() float method in Python 3.4 but that wasn't successful. As

x = (3**3)**(1/3.0) 
is_integer(x)    
True

but

x = (4**3)**(1/3.0) 
is_integer(x)    
False

I tried x%1 == 0,x == int(x) and isinstance(x,int) with no success.

I'd appreciate any comment.

Answer 1

For small numbers (<~10¹³ or so), you can use the following approach:

def is_perfect_cube(n):
    c = int(n**(1/3.))
    return (c**3 == n) or ((c+1)**3 == n)

This truncates the floating-point cuberoot, then tests the two nearest integers.

For larger numbers, one way to do it is to do a binary search for the true cube root using integers only to preserve precision:

def find_cube_root(n):
    lo = 0
    hi = 1 << ((n.bit_length() + 2) // 3)
    while lo < hi:
        mid = (lo+hi)//2
        if mid**3 < n:
            lo = mid+1
        else:
            hi = mid
    return lo

def is_perfect_cube(n):
    return find_cube_root(n)**3 == n

Answer 2

In SymPy there is also the integer_nthroot function which will quickly find the integer nth root of a number and tell you whether it was exact, too:

>>> integer_nthroot(primorial(12)+1,3)
(19505, False)

So your function could be

def is_perfect_cube(x): return integer_nthroot(x, 3)[1]

(And because SymPy is open source, you can look at the routine to see how integer_nthroot works.)

Answer 3

If your numbers aren't big, I would do:

def is_perfect_cube(number):
    return number in [x**3 for x in range(15)]

Of course, 15 could be replaced with something more appropriate.

If you do need to deal with big numbers, I would use the sympy library to get more accurate results.

from sympy import S, Rational

def is_perfect_cube(number):
    # change the number into a sympy object
    num = S(number)
    return (num**Rational(1,3)).is_Integer