On Mon, 21 May 2007, Greg Meredith wrote:
HC-er's,
Find below some simple-minded code from a naive Haskeller for generating all partitions of a multiset about which i have two questions.
mSplit :: [a] -> [([a], [a])] mSplit [x] = [([x],[])]
What about [] ? See http://www.haskell.org/haskellwiki/Base_cases_and_identities
mSplit (x:xs) = (zip (map (x:) lxs) rxs) ++ (zip lxs (map (x:) rxs)) where (lxs,rxs) = unzip (mSplit xs)
1. Is there a clever way to reduce the horrid complexity of the naive approach? 2. How lazy is this code? Is there a lazier way?
The code looks good. Maybe instead of doing zip ... ++ zip ... you should interleave the generated lists. This will probably reduce the need of constructing elements if only a prefix of (mSplit xs) is requested. mSplitLazy [] = [([],[])] mSplitLazy (x:xs) = let (lxs,rxs) = unzip (mSplitLazy xs) lys = zip (map (x:) lxs) rxs rys = zip lxs (map (x:) rxs) in concat (zipWith (\a b -> [a,b]) lys rys) If you are also after elegance - how about the List monad? mSplitMonad [] = [([],[])] mSplitMonad (x:xs) = do (lxs,rxs) <- mSplitMonad xs [(x:lxs,rxs), (lxs,x:rxs)] Or more condensed: mSplitFoldr = foldr (\x -> concatMap (\(lxs,rxs) -> [(x:lxs,rxs), (lxs,x:rxs)])) [([],[])]