Any Better logic for this problem..

[Chris Angelico]

On Fri, Jun 10, 2011 at 8:39 AM, Gregory Ewing
<greg.ewing at 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