On 7/19/07, Alexteslin <[EMAIL PROTECTED]> wrote:
What wrong with my original solution?allEqual2 :: [Int] -> Bool allEqual2 xs = length xs == length (filter isEqual xs) where isEqual n = (head xs) == n It looks simpler to me
I believe it's correct, but the use of "length" and "filter" seems forced. For example, for a very long list that has the second element mismatch, the versions with explicit recursion (or fold variants) will terminate with False as soon as a mismatch is found; the "length" version will not terminate until both have been fully traversed. If you're allowed to use everything, perhaps the clearest would be: allEqual3 [] = True allEqual3 (x:xs) = all (== x) xs (with "all" built on top of a fold.) Colin _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
