Question

Well, I have this bit of code that is slowing down the program hugely because it is linear complexity but called a lot of times making the program quadratic complexity. If possible I would like to reduce its computational complexity but otherwise I’ll just optimize it where I can. So far I have reduced down to:

def table(n):     a = 1     while 2*a <= n:       if (-a*a)%n == 1: return a        a += 1 

Anyone see anything I’ve missed? Thanks!

EDIT: I forgot to mention: n is always a prime number.

EDIT 2: Here is my new improved program (thank’s for all the contributions!):

def table(n):     if n == 2: return 1     if n%4 != 1: return      a1 = n-1     for a in range(1, n//2+1):         if (a*a)%n == a1: return a 

EDIT 3: And testing it out in its real context it is much faster! Well this question appears solved but there are many useful answers. I should also say that as well as those above optimizations, I have memoized the function using Python dictionaries…

Answer 1

Based off OP’s second edit:

def table(n):     if n == 2: return 1     if n%4 != 1: return      mod = 0     a1 = n - 1     for a in xrange(1, a1, 2):         mod += a          while mod >= n: mod -= n         if mod == a1: return a//2 + 1