Well it looks I was beat to the answer by Mariano, but here's my solution anyway. It's a little more expository, hope that helps. Note that we both decided to use a right fold in our solutions. {- Define a binary operation. Note the type signature. When used in the fold operation below, the 2nd argument denotes the set product of the first couple of lists (counting from the right-hand side) that were in your initial list. -} f :: [a] -> [[a]] -> [[a]] f [] _ = [] f (x : xs) yss = map (x :) yss ++ f xs yss {- Now for the product, we use a right fold operation to get the product of the first two lists (counting from the right), then distribute the elements of the next list over the result, then the elements of the next list, and so on. -} setProd :: [[a]] -> [[a]] setProd [] = [] setProd xss = foldr f (map (: []) (last xss)) (init xss) test = [[1,2,3],[4,5],[6,7]] On 4/8/14, Christian Maeder wrote:

Some time ago I discovered that the monadic "sequence" function operates quite interestingly on lists. I don't know why you call it multiplication. I was looking for something like combinations:

Prelude> sequence [[1,2,3], [4,5], [6,7]]

HTH Christian

Am 08.04.2014 07:00, schrieb Nishant:

...

hi,

I am trying to implement a set multiplication program in haskell.

Input : [1,2,3] [4,5] [6,7] Ouput : [ [1,4,6] , [1,4,7] , [1,5,6] , [1,5,7] ...]

I implemented it for constant number of inputs like

setMul xs ys zs = [ [x] ++ [y] ++ [z] | x <- xs , y<-ys ,z <- zs]

I am not able to generalize this for any number of lists.

type signature would be :

setMulMany :: [[a]] -> [[a]]

Example :

Input : [ [1,2,3] , [4,5] , [6,7]] Ouput : [ [1,4,6] , [1,4,7] , [1,5,6] , [1,5,7] ...]

Regards. Nishant

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