Now i wrote a crude parts function to do this
nparts 0 = [] nparts n = [n] : [x:xs | x <- [1..n`div`2], xs <- nparts(n - x)]
which is a bad bad example of recursion coz it’s slow!
It's not the recursion that makes it slow, but its asymptotic runtime, which is O(2^n). This one might run slightly faster: intParts :: Integer -> [[Integer]] intParts n = do x <- [1..n] let r = n - x if r > 0 then map (x :) (intParts (n - x)) else [[x]] Verify that this function takes a lot of time for an argument as small as 20, because it finds 2^19 partitionings. You need to eliminate candidates early. For example if you know that you only need 2-partitionings, then there is no need at all for the recursion and you can get away with O(n). Alternatively you can eliminate certain partitions by changing the third line. x <- [2..n] This finds partitions of at least 2. Or introduce a stop-early condition. Greets, Ertugrul