"Richard O'Keefe" <o...@cs.otago.ac.nz> writes: > factors n = [m | m <- [1..n], mod n m == 0]
-- saves about 10% time, seems to give the same result: factors n = [m | m <- [1..n `div` 2], mod n m == 0]++[n] (But checking against primes is even faster, it seems) -k -- If I haven't seen further, it is by standing in the footprints of giants _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
