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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