Re: [Haskell-cafe] Review request for my permutations implementation

Thanks everyone, Thanks Daniel for this really detailed explanation - thank you very much. Regards, Kashyap

From: Daniel Fischer To: haskell-cafe@haskell.org Cc: CK Kashyap Sent: Thu, January 7, 2010 4:16:33 PM Subject: Re: [Haskell-cafe] Review request for my permutations implementation

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Hi All,

I've written this piece of code to do permutations -

perms :: String -> [String] Nothing in the algorithm needs the list elements to be Chars, there's no type class involved, so it should be

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perms []= [] This should actually be

Am Donnerstag 07 Januar 2010 09:37:42 schrieb CK Kashyap: perms :: [a] -> [[a]] perms [] = [[]]

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perms (x:[])= [[x]] That is then superfluous. perms (x:xs)= concat (f [x] (perms xs))

'f' is a good name for a function parameter, not for a top level binding. Why not perms (x:xs) = concat (map (spread [x]) (perms xs)) whcih you can reformulate as perms (x:xs) = concatMap (spread [x]) (perms xs) or, if you like Monads, since concatMap is just the bind operator of the []-monad, perms (x:xs) = perms xs >>= spread [x] Which can be written as a simple do-block: perms (x:xs) = do prm <- perms xs spread [x] prm or a list-comprehension perms (x:xs) = [permutation | tailPerm <- perms xs, permutation <- spread [x] tailPerm]

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spread :: String -> String -> [String] -- interpolate first string at various positions of second string spread str1 str2 = _spread str1 str2 (length str2) where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1))

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f xs = map (spread xs)

The number of outcomes seem to indicate that correctness of the algo .. Apart from the case of empty input, it is correct. however, I'd be very obliged if I could get some feedback on the Haskellness etc of this ... also any performance pointers ... Re performance: I think the repeated (take k) and (drop k) in your spread are likely to be slower than using inits and tails, but it would need measuring the performance to be sure. I don't see anything that would automatically give bad performance. But there's the question of repeated elements.

import Data.List spread short long = zipWith (\a b -> a ++ short ++ b) (inits long) (tails long) If you only use spread for perms, you never interpolate anything but single element lists, so you might consider spread' :: a -> [a] -> [[a]] spread' x xs = zipWith (\a b -> a ++ x:b) (inits xs) (tails xs) But if you import Data.List, you could also say perms = permutations and be done with it :) (except if you 1. need the permutations in a particular order, which is different from the one Data.List.permutations generates, or 2. you need it to be as fast as possible - Data.List.permutations was written to also cope with infinite lists, so a few things that could speed up generation of permutations for short lists couldn't be used). perms "aaaaabbbbb" spills out 3628800 permutations, but there are only 252 distinct permutations, each of them appearing 120^2 = 14400 times. If your input may contain repeated elements and you're 1. only interested in the distinct permutations (and 2.) or 2. don't care about the order in which the permutations are generated, distinctPerms :: Ord a => [a] -> [[a]] distinctPerms = foldr inserts [[]] . group . sort inserts :: [a] -> [[a]] -> [[a]] inserts xs yss = yss >>= (mingle xs) mingle :: [a] -> [a] -> [[a]] mingle xs [] = [xs] mingle [] ys = [ys] mingle xxs@(x:xs) yys@(y:ys) = [x:zs | zs <- mingle xs yys] ++ [y:zs | zs <- mingle xxs ys] generates the distinct permutations much faster if there are many repeated elements; if you want each distinct permutation repeated the appropriate number of times, the modification is easy.

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Regards, Kashyap