On 11/27/07, Sebastian Sylvan
That is indeed a nice and clear version that's pretty fast. It's basically the same as the C version except "backwards" (i.e. examine a number and work backwards through its divisors, rather than filling in a "map" of all multiples of a known prime). Let me suggest the following slight modification (primeFactors in your version doesn't actually return prime factors - it returns prime factors *or* a list of just the number itself),
primes :: [Integer] primes = 2 : filter (null . primeFactors) [3,5..]
primeFactors :: Integer-> [Integer] primeFactors n = factor n primes where factor m (p:ps) | p*p > m = [] | m `mod` p == 0 = p : factor (m `div` p) (p:ps) | otherwise = factor m ps
This is roughly 35% faster on my machine with GHC 6.7.20070730 too, but the point wasn't to make it faster, but clearer. -- Sebastian Sylvan +44(0)7857-300802 UIN: 44640862
Great remark, it's even simpler like this. By the way I just found the article I stole this algorithm from: http://www.haskell.org/haskellwiki/99_questions/31_to_41 last one in problem 35. Cheers, Olivier.