[Haskell] Can anyone help me with partition numbers?

E. Zuurbier (Erik) ezuurbier at adp.nl
Fri Nov 25 06:25:54 EST 2005

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    = []
>>         | 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
>
> _______________________________________________