chopper draw lion4 - 5 months ago 23

Python 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 like 125, that should return a cube root of 5 but instead Python returns 4.99999

Example:

`>>> 125 ** (1.0/3.0)`

4.999999999999999

This is my code:

`processing = True`

n = 12000

while processing:

if (n ** (1.0/3.0)).is_integer() == True:

print((n ** (1.0/3.0)), "is the cube root of ", n)

processing = False

else:

n -= 1

Answer

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
```