Joost Behrends wrote:
since about three weeks i am learning Haskell now. One of my first exercises is
to decompose an Integer into its primefactors.
How about separating the candidate prime numbers from the recursion
factorize :: Integer -> [Integer]
factorize = f primes'
where
primes' = 2:[3,5..]
f (p:ps) n
| r == 0 = p : f (p:ps) q
| p*p > n = [n]
| otherwise = f ps n
where
(q,r) = n `divMod` p
For a faster factorization, just plug in a better primes' .
Regards,
apfelmus
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