On 4 May 2011 13:13, Barbara Shirtcliff <ba...@gmx.com> wrote: > Hi, > > 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. > > [snip] > > 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
As a guess, I think it's from the usage of length on the right-hand size. Also, note that "lexOrder s@[_] = [s]" is nicer than "lexOrder s | length s == 1 = [s]". -- Ivan Lazar Miljenovic ivan.miljeno...@gmail.com IvanMiljenovic.wordpress.com _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
