Casey Hawthorne
For example, I have this: list1 = [a, b, c] list2 = [d, e, f] list3 = [g, h, i]
Think in abstract terms what you want to accomplish.
A bit more specifically, let's say the input is a list of lists, and you want to produce all combinations of drawing one element from each of the input lists¹: perms :: [[a]] -> [[a]] You need to consider two cases, when the input is empty, and when the input contains at least one list of elements: perms (l:ls) = ... perms [] = ... The second case shouldn't be so hard. Now, if you pretend that 'perms' is already implemented, then you can use it to generate all permutations for the tail of the input list. The first case boils down to combining the first input list with all permutations of the rest of the lists: perms (l:ls) = ... l ... perms ls Does this help? -k ¹ Using tuples is harder to generalize for length, but nicer typewise, since you'd get something like 'perms :: ([a],[b],..[x]) -> [(a,b,..,x)] -- If I haven't seen further, it is by standing in the footprints of giants