Mark T.B. Carroll wrote:
nOf _ [] = [] nOf 1 xs = map return xs nOf n (x:xs) = map (x:) (nOf (n-1) xs) ++ nOf n xs
No! With this implementation we have nOf 0 _ == [] but it should be nOf 0 _ == [[]]: The list of all sublists of length 0 is not empty, it contains the empty list!
Correct (and more natural): nOf 0 _ = [[]] nOf n (x:xs) = map (x:) (nOf (n-1) xs) ++ nOf n xs nOf _ [] = [] BR, -- -- Mirko Rahn -- Tel +49-721 608 7504 -- --- http://liinwww.ira.uka.de/~rahn/ --- _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
