Doaitse, For generate 4 this gives a.o. three equivalent solutions: [1,1,2], [1,2,1] and [2,1,1]. I guess the ultimate idea would be to prune permutations. Regards Erik Zuurbier Onderwerp: Re: [Haskell] Can anyone help me with partition numbers? Or (since we started to do someone's homework anyway) generate 0 = [[]] generate n = [x:rest | x <- [1..n], rest <- generate (n-x)] Doaitse Swierstra On 2005 nov 25, at 10:29, Tomasz Zielonka wrote:

On Thu, Nov 24, 2005 at 05:52:23PM +0100, Jan van Eijck wrote:

...

Like so:

generatePs :: (Int,[Int]) -> [[Int]] generatePs (n,[]) = [take n (repeat 1)] generatePs (n,(x:xs)) = (take n (repeat 1) ++ (x:xs)) : generatePs (pack (x-1) ((n +x),xs)) where pack :: Int -> (Int,[Int]) ->(Int,[Int]) pack 1 (m,xs) = (m,xs) pack k (m,xs) = if k > m then pack (k-1) (m,xs) else pack k (m-k,k:xs)

parts :: Int -> [[Int]] parts n | n < 1 = error "part: argument <= 0" | n == 1 = [[1]] | otherwise = generatePs (0,[n])

How about a shorter version?

part :: Integer -> [[Integer]] part = gen 1 where gen m 0 = [[]] gen m n = [ x:xs | x <- [m..n], xs <- gen x (n - x) ]

Best regards Tomasz

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