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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
