Hi, I'm back. I have some troubles understanding your code: Chaddaï Fouché-2 wrote:
findAllAns 0 0 = [[]] findAllAns 0 s = [] findAllAns n s = [ x:xs | x <- [0..s], xs <- findAllAns (n - 1) (s - x) ]
Let's try a test case (CMIIW) for findAllAns 2 1: findAllAns 2 1 = [ 0:(findAllAns 1 1) ] = [ 0:0:(findAllAns 0 1) ] = [ 0:0:[] ] // Not a solution = [ 0:(findAllAns 1 1) ] = [ 0:1:(findAllAns 0 0) ] = [ 0:1:[[]] ] // A solution = [ 1:(findAllAns 1 0) ] = [ 1:0:(findAllAns 0 0) ] = [ 1:0:[[]] ] // A solution ( 1:1:? is never reached. ) I don't understand why if the last element is [[]] then it's included in the result and not if it's []. -- View this message in context: http://www.nabble.com/Help-with-generalizing-function-tp18063291p18106999.ht... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.