omega - 3 months ago 14

Python Question

In python, how can you check if a number n is an exact power of base b?

Note: it needs to be generalized to any base which is given as a parameter.

Here is what I got:

Assume n and base are integers > 0.

`import math`

def is_power(n,base):

return math.log(n,base) == base**n

Answer

Assuming you have a specific logarithm operator, `log`

can be calculated as _{a}b`log`

and, if the log of _{x}b / log_{x}a`X`

in base `Y`

is an integer, then `X`

is a power of `Y`

.

But Python goes one better since it can work out the logarithm for an *arbitrary* base without that tricky equality above.

So I'd start with the following code, now with added edge-case detection:

```
# Don't even think about using this for negative powers :-)
def isPower (num, base):
if base == 1 and num != 1: return False
if base == 1 and num == 1: return True
if base == 0 and num != 1: return False
power = int (math.log (num, base) + 0.5)
return base ** power == num
```

See for example the following complete program which shows this in action:

```
import math
def isPower (num, base):
if base == 1 and num != 1: return False
if base == 1 and num == 1: return True
if base == 0 and num != 1: return False
power = int (math.log (num, base) + 0.5)
return base ** power == num
print isPower (127,2) # false
print isPower (128,2) # true
print isPower (129,2) # false
print
print isPower (26,3) # false
print isPower (27,3) # true
print isPower (28,3) # false
print isPower (3**10,3) # true
print isPower (3**129,3) # true
print
print isPower (5,5) # true
print isPower (1,1) # true
print isPower (10,1) # false
```

If you're the sort that's worried about floating point operations, you *can* do it with repeated multiplications but you should test the performance of such a solution since it's likely to be substantially slower in software than it is in hardware. That won't matter greatly for things like `isPower(128,2)`

but it may become a concern for `isPower(verybignum,2)`

.

For a non-floating point variant of the above code:

```
def isPower (num, base):
if base == 1 and num != 1: return False
if base == 1 and num == 1: return True
if base == 0 and num != 1: return False
testnum = base
while testnum < num:
testnum = testnum * base
return testnum == num
```

But make sure it's tested against your largest number and smallest base to ensure you don't get any performance shocks.

Source (Stackoverflow)

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