It's well known that the powerset of a list:
{1,2,3,4} is {{},{1},{2},{1,2},{3},{1,3},{2,3},{1,2,3},{4},{1,4},{2,4},{1,2,4},{3,4},{1,3,4},{2,3,4},{1,2,3,4}}
the haskell code I got for that problem is:
potencia [] = [[]]
potencia (a:bs) = potencia bs ++ map (a:) (potencia bs)
Now, how would I get a list of sublists of the same length?, for example, the list above would generate the next list of sublists of length 3 = {{1,2,3},{1,2,4},{1,3,4}}
I'm a student sorry for my english, thanks in advance... XD
How about
sublists _ 0 = [[]]
sublists [] _ = []
sublists (x:xs) n = sublists xs n ++ map (x:) (sublists xs $ n - 1)
Which is very similar to the code you had but just has two decreasing parameters, the length and the list.
Also, for more advanced Haskellers
powerset = flip runCont id . foldM step [[]]
where step xs x = cont $ \c -> c xs ++ c (map (x:) xs)
is a powerset implementation without recursion using continuations. Doing the same with the sublists function is an interesting challenge.