Don Stewart wrote:
perfect :: [Int] perfect = [i | i<-[1..10000], i == sum (divisors i)]This should be a little faster , as sum will fuse, perfect :: [Int] perfect = [i | i<-[1..10000], i == sum' (divisors i)] where sum' = foldr (+) 0
sum' did not help. Times are about the same with Int type. Peter. _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
