On Wed, 2011-05-04 at 07:13 -0600, Barbara Shirtcliff wrote:
In the following solution to problem 24, why is nub ignored? I.e. if you do lexOrder of "0012," you get twice as many permutations as with "012," even though I have used nub.
lexOrder :: [Char] -> [[Char]] lexOrder s | length s == 1 = [s] | length s == 2 = z : [reverse z] | otherwise = concat $ map (\n -> h n) [0..((length s) - 1)] where z = sort $ nub s -- why is the nub ignored here? h :: Int -> [String] h n = map (z!!n :) $ lexOrder $ filter (\c -> lexI c z /= n) z
You are using (length s) in the otherwise case. If you want the results to be identical with duplicates, perhaps you meant to say (length z)? -- Chris