On Thu, Feb 12, 2009 at 1:20 AM, Jan Snajder
Hi!
I'm trying to write a list permutation function, and there is in fact a nice explanation of how to do it here: http://sneakymustard.com/2008/12/23/shuffling-in-haskell
But for the start I wanted to keep things simple and avoid monad transformers (since I'm not into this yet). Instead, I'd like to write a function of type:
permute :: [a] -> IO [a]
and so this is what I did:
permute xs = do let n = length xs - 1 arr0 <- newListArray (0, n) xs arr <- foldM swap arr0 [n..1] getElems arr where swap arr n = do x <- readArray arr n r <- randomRIO (0, n) y <- readArray arr r writeArray arr n y writeArray arr r x return arr
Unfortunately, what I get is:
permute :: (MArray a1 a IO) => [a] -> IO [a]
and so when I try to apply this function:
permute [1,2,3]
this is what I get:
<interactive>:1:0: No instance for (MArray a1 t IO) arising from a use of `permute' at <interactive>:1:0-14 Possible fix: add an instance declaration for (MArray a1 t IO) In the expression: permute [1, 2, 3] In the definition of `it': it = permute [1, 2, 3]
How can I fix this?
Thanx, jan
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The simplest way to do this is to use base v4, which I believe contains a "permutations" function in Data.List. Alex