How to raise an integer to fractional power efficiently?

Question

I have an binary search implemented in python. Now I want to check if element math.floor(n ^ (1/p)) is in my binary search. But p is a very, very large number. I wrote using fractions module :

binary_search.search(list,int (n**fractions.Fraction('1'+'/'+str(p))))

But I have an error OverflowError: integer division result too large for a float How can I take to n to the power, which is a fraction and do it fast?

Answer 1

Unless your values of n are also incredibly large, floor(n^(1/p)) is going to tend toward 1 for "very, very large" values of p. Since you're only interested in the integer portion, you could get away with a simple loop to test if 1^P, 2^p, 3^p and so on are greater than n.

Don't waste time finding exact values if you don't need them.

Answer 2

n^(1/p)=exp(ln(n)/p) ~~ 1+ln(n)/p for big p values

So you can compare p with natural logarithm of n. If the ratio p/ln(n) >> 1 (much larger), then you can use approximation above (which tends to 1)