Mirko Rahn <[EMAIL PROTECTED]> writes: (snip) > Correct (and more natural): > > nOf 0 _ = [[]] > nOf n (x:xs) = map (x:) (nOf (n-1) xs) ++ nOf n xs > nOf _ [] = []
Thanks very much - in both claims you're indeed correct. -- Mark _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe