How to make perfect power algorithm more efficient?

Question

I have the following code:

def isPP(n):
  pos = [int(i) for i in range(n+1)]
  pos = pos[2:] ##to ignore the trivial n** 1 == n case
  y = []
  for i in pos:
            


        

Answer 1

I think a better way would be implementing this "hack":

import math

def isPP(n):
    range = math.log(n)/math.log(2)
    range = (int)(range)
    result = []
    for i in xrange(n):
        if(i<=1):
            continue
        exponent = (int)(math.log(n)/math.log(i))
        for j in [exponent-1, exponent, exponent+1]:
            if i ** j == n:
                result.append([i,j])
    return result

print isPP(10000)

Result:

[[10,4],[100,2]]

The hack uses the fact that:

if log(a)/log(b) = c,
    then power(b,c) = a

Since this calculation can be a bit off in floating points giving really approximate results, exponent is checked to the accuracy of +/- 1.

You can make necessary adjustments for handling corner cases like n=1, etc.