On Fri, Jun 10, 2011 at 8:39 AM, Gregory Ewing <greg.ew...@canterbury.ac.nz> wrote: > Chris Angelico wrote: > >> Rather than find all prime numbers up to num, stop at sqrt(num) - it's >> not possible to have any prime factors larger than that. > > That's not quite true -- the prime factors of 26 are 2 and 13, > and 13 is clearly greater than sqrt(26).
Oops! My bad. I was thinking in terms of the "divide and conquer" algorithm, whereby the 13 would be the residuum after dividing by 2... > However, once you've divided out all the primes up to sqrt(n), > whatever is left, if greater than 1, must itself be prime, so > you can add it to your prime factors and stop. ... which is effectively the same as you describe here. It's a small algorithmic change but an extremely advantageous one. If you _don't_ look for the residuum, then you stop at n/2 instead of sqrt(n). Either way, though, you don't need to list the primes all the way up to n, which will improve performance significantly. Chris Angelico -- http://mail.python.org/mailman/listinfo/python-list
