June 2006 - Haskell-Cafe

Editors for Haskell
by Walter Potter 08 Jun '06

08 Jun '06

All, I hope that this is the right place for this question. I'm using Haskell (GHC and Hugs) on several different platforms. Windows, OS X and Linux systems. I'd like to have an IDE that works well for medium to large size projects. I know of Eclipse and hIDE. Vim works fine but I'd like more. hiDE seems to be "in process." What would you suggest? I'll be asking my students to use the same IDE. Thanks, Walt

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HAppS in production?
by Joel Reymont 08 Jun '06

08 Jun '06

Folks, Is anyone using HAppS in production right now? It seems to be the most advanced Haskell web development platform right now but I would like to hear about others as well. Production (heavy) use is what I'm looking for. Thanks, Joel -- http://wagerlabs.com/

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Eclipse and Haskell
by Walter Potter 08 Jun '06

08 Jun '06

All, I intend to start using Eclipse as my IDE. Please pass along any suggestions that I may find useful. Thanks, Walt

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Polymorphic Sudoku solver
by Chris Kuklewicz 08 Jun '06

08 Jun '06

A while back there was a long thread about Sudoku solvers (some of which ended up on http://haskell.org/haskellwiki/Sudoku ). I contributed my brute-force dancing links solver at the time, and mentioned that I had a by-logic solver that, while a bit slow, was as good as most of those being discussed. At the time the code for my solver was too ugly to post. Attached is a cleaned up version. I have gone back and rewritten it, and come to the conclusion: There are only two deduction algorithms: "subsets" and "blocks". These subsume the other types of propagation and deduction. So I made this version as a "minimalist" example instead of going for performance. The "subsets" algorithm can be applied to all 6 permutations of row column and value, as well as 1 special case of value and block indices. The "blocks" algorithm can be applied 4 ways (in two flavors and to either permutation of row/column or column/row). There are newtypes for row, column, value, block index, and sub-block index. The state is held in an array of type DiffArray (R,C,V) Bool The actual computation is a series of concat/map/filter/group/sort operations on the assocs's of the array. The choice of which permutation is handled by leaning on the type system to reify the type into appropriate view,shuffle, and unshuffle functions. It should solve exactly the same number of puzzles as my older version, where "I sent the 36628 line sudoku17 puzzle through it and it could solve 31322 of the puzzles, leaving 5306 resistant." -- Chris Kuklewicz module Main (main) where import Data.Ix(inRange,range) import Data.Char(intToDigit,digitToInt) import Deduce(deduce,lo,hi) loC = intToDigit lo hiC = intToDigit hi unsetC = pred loC main = do all <- getContents let puzzles = zip [1..] (map parseBoard (lines all)) act (i,p) = do p' <- deduce p return (i,length p,length p') mapM_ (\ip -> act ip >>= print) puzzles parseBoard :: String -> [(Int,Int,Int)] parseBoard s = map toHint justSet where rcs = [ (r,c) | r <- range (lo,hi), c <- range (lo,hi) ] isHint vC = inRange (loC,hiC) vC justSet = filter (isHint . snd) (zip rcs s) toHint ((r,c),vC) = (r,c,digitToInt vC) {- By Chris Kuklewicz <haskell(a)list.mightyreason.com> -} module Deduce (deduce,lo,hi) where {- The exposed function deduce takes a list of (row,column,value) tuples that are the known parts of the solutions and returns a (hopefully longer) list in the same format. The indices can be any enumerated type in the range (lo,hi). -} import Data.Array.Diff (assocs,(!),(//),ixmap,range,inRange,accumArray,DiffArray,Ix) import Data.List(sortBy,groupBy,transpose,(\\)) import Control.Monad(liftM,guard) default () -- Typesafe values for indices -- This machinery allows for more type safety than if R,C,V,B,D were all Int or Char type E = Int newtype R = R E deriving (Eq,Ord,Ix,Enum,Show) -- Row index newtype C = C E deriving (Eq,Ord,Ix,Enum,Show) -- Column index newtype V = V E deriving (Eq,Ord,Ix,Enum,Show) -- Value index newtype B = B E deriving (Eq,Ord,Ix,Enum,Show) -- 3x3 Block index newtype D = D E deriving (Eq,Ord,Ix,Enum,Show) -- Inside 3x3 Block index lo,hi :: (Enum a) => a lo = toEnum 1 hi = toEnum 9 fullRange :: (Enum a,Ix a) => [a] fullRange = range (lo,hi) rcToBD (R r) (C c) = let (rq,rm) = quotRem (r-lo) 3 (cq,cm) = quotRem (c-lo) 3 b = lo + ( 3*rq + cq ) d = lo + ( 3*rm + cm ) in (B b, D d) bdToRC (B b) (D d) = let (bq,bm) = quotRem (b-lo) 3 (dq,dm) = quotRem (d-lo) 3 r = lo + ( 3*bq + dq ) c = lo + ( 3*bm + dm ) in (R r, C c) -- Typeclasses and Data for "shuffle" and "unshuffle" class (Show x, Ix x, Enum x, Ord x) => IE x instance IE R; instance IE C; instance IE V; instance IE B; instance IE D data Perms a b c = Perms { shuffle' :: (R,C,V) -> (a,b,c) , unshuffle' :: (a,b,c) -> (R,C,V) } -- Reify the types "a b c" to a value of type Perms class (IE a, IE b, IE c) => Perm a b c where perm :: Perms a b c instance Perm R C V where perm = Perms id id instance Perm R V C where perm = Perms (\ (r,c,v) -> (r,v,c)) (\ (r,v,c) -> (r,c,v)) instance Perm C V R where perm = Perms (\ (r,c,v) -> (c,v,r)) (\ (c,v,r) -> (r,c,v)) instance Perm C R V where perm = Perms (\ (r,c,v) -> (c,r,v)) (\ (c,r,v) -> (r,c,v)) instance Perm V R C where perm = Perms (\ (r,c,v) -> (v,r,c)) (\ (v,r,c) -> (r,c,v)) instance Perm V C R where perm = Perms (\ (r,c,v) -> (v,c,r)) (\ (v,c,r) -> (r,c,v)) -- Special cases instance Perm B D V where perm = Perms (\ (r,c,v) -> let (b,d) = rcToBD r c in (b,d,v)) (\ (b,d,v) -> let (r,c) = bdToRC b d in (r,c,v)) instance Perm V B D where perm = Perms (\ (r,c,v) -> let (b,d) = rcToBD r c in (v,b,d)) (\ (v,b,d) -> let (r,c) = bdToRC b d in (r,c,v)) shuffle :: (Perm a b c) => (R,C,V) -> (a,b,c) shuffle = shuffle' perm unshuffle :: (Perm a b c) => (a,b,c) -> (R,C,V) unshuffle = unshuffle' perm -- Array types, values and functions type Index = (R,C,V) type Cell = Bool on,off :: Cell on = True -- Means this might be part of the solution off = False -- Means this cannot be part of the solution boundsCube :: (Perm a b c) => ((a,b,c),(a,b,c)) boundsCube = ((lo,lo,lo),(hi,hi,hi)) type Cube = DiffArray Index Cell emptyCube :: Cube emptyCube = accumArray const on boundsCube [] type View a b c = DiffArray (a,b,c) Cell {-# INLINE view #-} view :: (Perm a b c) => Cube -> View a b c view cube = ixmap boundsCube unshuffle cube type Hints = [(Index,Cell)] isOn :: (Perm a b c) => View a b c -> [(a,b,c)] isOn = map fst . filter snd . assocs -- The goal is to create functions that turn the current Cube into a -- list of Hints. These Hints will be purely subtractive: they all -- turn a Cell from 'on' to 'off'. type Rule = Cube -> Hints -- Small utility functions fst3 (x,_,_) = x; snd3 (_,x,_) = x; thd3 (_,_,x) = x fst4 (x,_,_,_) = x; snd4 (_,x,_,_) = x; thd4 (_,_,x,_) = x; fth4 (_,_,_,x) = x by un bi = (\ left right -> (un left) `bi` (un right)) sortWith un = sortBy (by un compare) groupWith un = groupBy (by un (==)) groupSort un = groupWith un . sortWith un atLeastOne = not . null atLeastTwo (_:_:_) = True; atLeastTwo _ = False exactlyOne [_] = True; exactlyOne _ = False oneOrTwo [_] = True; oneOrTwo [_,_] = True; oneOrTwo _ = False {- ruleBlock1 : When operating on Perm V R C: Given a value V, look along each row and see which blocks that value may occupy. Find a row R1 for which the value is allowed in exactly one block B1 (an no other blocks). This occupies [(B1,d)] in B1. Eliminate V from the other locations in B1. Given a value V, look along each row and see which blocks that value may occupy. Find two rows [R1,R2] for which the value only is allowed in exactly the same two blocks [B1,B2] (and no others). These occupy [(B1,d11s)] in R1, [(B2,d12s)] in R1, [(B1,d21s)] in R2, and [(B2,d22s)] in R2. Eliminate V from the other locations in B1 and the other locations in B2. Works for R and C reversed, of course. -} {- ruleBlock2 : When operating on Perm V R C: Given a value V, look inside each block and see which rows that value may occupy. Find a block B1 for which only one row R1 is occupied (and no other rows). This occupies [(R1,c11s)] in B1. Eliminate V from all the other c's in row R1. Given a value V, look inside each block and see which rows that value may occupy. Find two blocks [B1,B2] for which the value only is allowed in exactly the same two rows [R1,R2] (and no others). These occupy [(R1,c11s)] in B1, [(R1,c12s)] in B2, [(R2,c21s)] in B1, and [(R2,c22s)] in B2 for some C's. Eliminate V from the other rows in R1 and the other columns in R2. Works for R and C reversed, of course. -} {- ruleBlockP There is enough similarity between ruleBlock1 and ruleBlock2 to parameterize over expand and contract. The V index is special because it is "orthogonal" to the 3x3 blocks. The line sVsA2 = map (filter atLeastTwo) $ sVsA -- drop unique ones is used to prevent propagating solved constraints, as rule4P should already do this in its k==1 case. (assocs view) becomes useful hints through a chain of map, sort, group, filter, and concat operations. The assemble function also takes care to remove redundant hints by consulting the view. -} {-# INLINE ruleBlockP #-} ruleBlockP :: forall a b c x y . (IE a, IE b, IE c, Perm V x y) => ( (V,x,y) -> (V,a,b,c) ) -- "expand" -> (V -> b -> [c] -> [Index]) -- "contract" -> View V x y -> Hints ruleBlockP expand contract view = let allOn :: [(V,a,b,c)] allOn = map expand . isOn $ view sV :: [ [(V,a,b,c)] ] sV = groupWith fst4 $ allOn -- group by V sVsA :: [[ [(V,a,b,c)] ]] sVsA = map (groupSort snd4) $ sV -- group by a sVsAsB,sVsAsB2 :: [[[ [(V,a,b,c)] ]]] sVsAsB = map (map (groupSort thd4)) $ sVsA -- group by b -- These filters are to remove empty and redundantly full possibilities sVsAsB2 = filter atLeastOne . map (filter oneOrTwo) $ sVsAsB sVsAsBgB :: [[ ([b],[[(V,a,b,c)]]) ]] -- The filter is to remove solved parts of the puzzle (punt to ruleSubsetP) sVsAsBgB = map (map getAllB . filter (atLeastTwo . concat)) $ sVsAsB2 where getAllB :: [[(V,a,b,c)]] -> ([b],[[(V,a,b,c)]]) getAllB vabcss = (map (thd4 . head) vabcss, vabcss) -- length (concat vabcss) >= 2 useful :: [ [([b],[[(V,a,b,c)]])] ] useful = concatMap (filter exactlySame . groupSort fst) $ sVsAsBgB where exactlySame :: [([b],[[(V,a,b,c)]])] -> Bool exactlySame sas@((sbs,_):_) = length sas == length sbs assemble :: [ ([b],[[(V,a,b,c)]]) ] -> Hints assemble stuff = [ (rcv,off) | rcv <- ixs, (view ! shuffle rcv) /= off ] where byVB :: [[(V,a,b,c)]] byVB = map concat . transpose . map snd $ stuff -- Regroup by identical 'b' -- byVB = groupSort thd4 . concat . concat . map snd $ stuff -- equivalent act :: [(V,a,b,c)] -> [Index] act allVB@((v,_,b,_):_) = contract v b (fullRange \\ map fth4 allVB) ixs :: [Index] ixs = concatMap act byVB in concatMap assemble useful {-# INLINE ruleBlock1 #-} ruleBlock1 :: (Perm V x y) => View V x y -> Hints ruleBlock1 = ruleBlockP expand contract where expand all@(v,x,y) = let (r,c,_) = unshuffle all (b,d) = rcToBD r c in (v,x,b,d) contract v b ds = map undo ds where undo d = let (r,c) = bdToRC b d in (r,c,v) {-# INLINE ruleBlock2 #-} ruleBlock2 :: (Perm V x y) => View V x y -> Hints ruleBlock2 = ruleBlockP expand contract where expand all@(v,x,y) = let (r,c,_) = unshuffle all (b,_) = rcToBD r c in (v,b,x,y) contract v x ys = map undo ys where undo y = unshuffle (v,x,y) ruleBlocks :: [Rule] ruleBlocks = [ (\ known -> ruleBlock1 (view known :: View V R C) ) , (\ known -> ruleBlock1 (view known :: View V C R) ) , (\ known -> ruleBlock2 (view known :: View V R C) ) , (\ known -> ruleBlock2 (view known :: View V C R) ) ] {- Given a list of locations, such as for the 9 columns of a row, look at the allowed values at each location. Find a subset of k columns for which the union of their allowed values [V..] has length k. Then eliminate [V..] from the (9-k) other columns. This clearly finds a list of N columns each with the same N values if such a thing exists, so it subsumes rule1P. This is fully symmetric in R C and V and depends on the constaints in R and V but not C. So the (View B D V) case also works. A useful property of this rule is that once there is only one way to place a value in a row or column or block then it will propagate that solution to the related contraints. This is the case when minK is 1. -} {-# INLINE ruleSubsetP #-} ruleSubsetP :: forall a b c.(Perm a b c) => View a b c -> Hints ruleSubsetP view = let allOn :: [(a,b,c)] allOn = isOn $ view sAsB :: [[ [(a,b,c)] ]] sAsB = map (groupWith snd3) . groupWith fst3 $ allOn sAsBgC :: [ [(Int,[c],[(a,b,c)])] ] sAsBgC = map (sortWith fst3 . map (\ abcs -> (length abcs ,map thd3 abcs ,abcs) ) ) $ sAsB makeChains :: [(Int,[c],[(a,b,c)])] -> Int -> [(a,[b],[c])] makeChains input k = filter check . map toChain . subsets k . upToK $ input where upToK :: [(Int,[c],[(a,b,c)])] -> [(Int,[c],[(a,b,c)])] upToK = takeWhile ((k>=).fst3) toChain :: [(Int,[c],[(a,b,c)])] -> (a,[b],[c]) toChain vals = ( fst3 . head . thd3 . head $ vals -- record "a" for easy retrieval later , map (snd3 . head . thd3) $ vals -- the bs, length [b] == length vals == k by property of subsets k , combine . map snd3 $ vals) -- union of the cs at each b in bs check :: (a,[b],[c]) -> Bool check (_,_,cs) = (k == length cs) -- check that length [c] == k as well getUseful :: [(Int,[c],[(a,b,c)])] -> [(a,[b],[c])] getUseful [] = [] getUseful input = concatMap (makeChains input) [minK .. maxK] -- assertion: (length input) == (length . combine . map snd3 $ input) where minK,maxK :: Int minK = fst3 . head $ input maxK = pred . length $ input useful :: [(a,[b],[c])] useful = concatMap getUseful sAsBgC {- The chains are (a,bs,cs) such that cs == nub . sort $ [ z | (x,y,z) <- allOn, x==a, y `elem` bs] and length bs == length cs. Thus all the (a,b in bs,c) in the final puzzle have distict c in cs. For location (a,b' not in bs,c') cannot have c' in cs, thus if c' is in cs then (a,b',c') should be turned off. -} assemble :: (a,[b],[c]) -> Hints assemble (a,inBs,inCs) = do -- List Monad b <- fullRange \\ inBs c <- inCs let abc :: (a,b,c) abc = (a,b,c) guard (view ! abc) return (unshuffle abc,off) in concatMap assemble useful -- All subsets of length 'k', order is stable subsets :: Int -> [a] -> [[a]] subsets 0 _ = [[]] subsets _ [] = [] subsets k (x:xs) = (fmap (x:) (subsets (pred k) xs)) ++ subsets k (xs) -- Hopefully efficient merge of (list of (sorted lists)), unique values only combine :: (Ord a) => [[a]] -> [a] combine [] = [] combine [x] = x combine xs = let (a,b) = split xs in merge (combine a) (combine b) where split [] = ([],[]) split [a] = ([a],[]) split (x:y:cs) = let (a,b) = split cs in (x:a,y:b) merge a [] = a merge [] b = b merge a@(x:a') b@(y:b') = case compare x y of EQ -> x : merge a' b' LT -> x : merge a' b GT -> y : merge a b' {-# INLINE eachPerm #-} eachPerm :: (forall x y z. (Perm x y z) => View x y z -> Hints) -> [ Rule ] eachPerm rule = [ (\ known -> rule ( view known :: View R C V ) ) , (\ known -> rule ( view known :: View R V C ) ) , (\ known -> rule ( view known :: View C V R ) ) , (\ known -> rule ( view known :: View C R V ) ) , (\ known -> rule ( view known :: View V R C ) ) , (\ known -> rule ( view known :: View V C R ) ) ] ruleSubsets :: [ Rule ] ruleSubsets = eachPerm ruleSubsetP ++ [ (\ known -> ruleSubsetP (view known :: View B D V) ) ] allRules :: [ Rule ] allRules = ruleSubsets ++ ruleBlocks -- Applying the rules {- The evolution strategy is simple: Apply each rule in turn, keeping track whether or not there were any changes. If all rules cause no change then it is done evolving. -} evolve :: Cube -> Cube evolve cube = let (cube',changed) = foldl step (cube,False) allRules in if changed then evolve cube' else cube' where step orig@(known,_) rule = case rule known of [] -> orig hints -> (known // hints,True) toCube :: (Monad m,Enum e) => [(e,e,e)] -> m Cube toCube locs = do hints <- liftM concat $ mapM setLoc locs return (emptyCube // hints) where setLoc:: (Enum e,Monad m) => (e,e,e) -> m Hints setLoc i@(re,ce,ve) = mapM checkM (rs ++ cs ++ vs) where r = toEnum $ fromEnum re c = toEnum $ fromEnum ce v = toEnum $ fromEnum ve rs = [((r',c,v),off) | r' <- fullRange, r' /= r] cs = [((r,c',v),off) | c' <- fullRange, c' /= c] vs = [((r,c,v'),off) | v' <- fullRange, v' /= v] checkM hint = if check hint then return hint else fail "Input location is out of range" check ((r,c,v),_) = and [ inRange (lo,hi) r , inRange (lo,hi) c , inRange (lo,hi) v ] fromCube :: (Enum e) => Cube -> [(e,e,e)] fromCube cube = map head . filter exactlyOne $ [ [ eee | v <- fullRange, cube ! (r,c,v) , let eee = (toEnum $ fromEnum r ,toEnum $ fromEnum c ,toEnum $ fromEnum v) ] | r <- fullRange, c <- fullRange ] consistent :: Cube -> Bool consistent known = and [ consistentView ( view known :: View R C V ) , consistentView ( view known :: View C V R ) , consistentView ( view known :: View R V C ) , consistentView ( view known :: View V B D ) ] where consistentView :: (Perm a b c) => View a b c -> Bool consistentView view = and [ atLeastOne [ () | c <- fullRange, view ! (a,b,c) ] | a <- fullRange, b <- fullRange ] checkCube :: (Monad m) => String -> Cube -> m Cube checkCube msg cube = if consistent cube then return cube else fail msg deduce :: (Monad m,Enum e) => [(e,e,e)] -> m [(e,e,e)] deduce locs = toCube locs >>= checkCube "Inconsistent locations passed in" >>= return . evolve >>= checkCube "Inconsistent cube deduced from input" >>= return . fromCube test :: [(E,E,E)] test = [(1,8,1),(1,9,2),(2,5,3),(2,6,5),(3,4,6),(3,8,7),(4,1,7),(4,7,3),(5,4,4),(5,7,8),(6,1,1),(7,4,1),(7,5,2),(8,2,8),(8,8,4),(9,2,5),(9,7,6)] testC :: IO Cube testC = toCube test check :: IO () check = do c <- testC print (consistent (evolve c))

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Profiling Gtk2HS applications
by Dmitry V'yal 08 Jun '06

08 Jun '06

Hello. How can one profile a program which uses gtk2hs? I get this: $ ghc --make MainGui.hs -prof Chasing modules from: MainGui.hs Could not find module `Graphics.UI.Gtk.Mogul': use -v to see a list of the files searched for (imported from ./TreeViewHelpers.hs) I tried to specify gtk2hs source path using -i option with no success.

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Undecideable instances for one instance?
by Christophe Poucet 07 Jun '06

07 Jun '06

Dear, Recently I was wrapping something in a typeclass (you may argue the reason for the typeclass, I will admit that's no longer needed, yet the compile error still begs questioning). Basically if you have only one global instance of a typeclass (so there is no overlapping), GHC will still complain about overlapping. Is there any reason this is so or is this a remnant of more complicated overlapping situations? The sample code is: {-# OPTIONS_GHC -fglasgow-exts #-} module Main where data Located a = L a class Locatable a where value :: Located a -> a wrap :: a -> Located a instance Locatable a where value (L a) = a wrap a = L a main :: IO () main = do print . value . wrap $ 1 -- Illegal instance declaration for `Locatable a' -- (There must be at least one non-type-variable in the instance head -- Use -fallow-undecidable-instances to permit this) -- In the instance declaration for `Locatable a' Christophe(vincenz) -- Christophe Poucet Ph.D. Student Phone:+32 16 28 87 20 E-mail: Christophe (dot) Poucet (at) imec (dot) be Website: http://notvincenz.com/ IMEC vzw – Register of Legal Entities Leuven VAT BE 0425.260.668 – Kapeldreef 75, B-3001 Leuven, Belgium – www.imec.be *****DISCLAIMER***** This e-mail and/or its attachments may contain confidential information. It is intended solely for the intended addressee(s). Any use of the information contained herein by other persons is prohibited. IMEC vzw does not accept any liability for the contents of this e-mail and/or its attachments. **********

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Fwd: [Haskell-cafe] Undecideable instances for one instance?
by Christophe Poucet 07 Jun '06

07 Jun '06

Hello Bulat, You are indeed correct. However I fail to see how there is any undecideability. If instead one specified it as follows, it would be fine: class Locatable a b | a -> b, b -> a where value :: a -> b wrap :: b -> a instance Locatatable (Located a) a where value (L a) = a wrap a = L a It compiles fine, yet this is semantically the same. On 6/7/06, Bulat Ziganshin <bulat.ziganshin(a)gmail.com> wrote: > > Hello Christophe, > > Wednesday, June 7, 2006, 12:27:22 PM, you wrote: > > > global instance of a typeclass (so there is no overlapping), GHC will > > still complain about overlapping. Is there any reason this is so or is > > GHC compains here about undecidability, not overlapping :) > > > -- Illegal instance declaration for `Locatable a' > > -- (There must be at least one non-type-variable in the instance head > > -- Use -fallow-undecidable-instances to permit this) > > -- In the instance declaration for `Locatable a' > > > Christophe(vincenz) > > > > > -- > Best regards, > Bulat mailto:Bulat.Ziganshin@gmail.com > >

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FunDeps conflict
by Joel Björnson 07 Jun '06

07 Jun '06

Hi. I have a question regarding type classes and FunDeps. Consider the following code : > class Class2 a b | a -> b > class IsFoo a > data Bar a = Bar a > instance IsFoo a => Class2 a a > instance IsFoo a => Class2 (Bar a) a The last two instantiations will yield a 'Functional dependencies conflict error'. From what I understand, this is because (Bar a) *MAY* instantiate the type class isFoo, which would obviously lead to a conflict between the two instantiations. Is there anyway to get around this, knowing that (Bar a) won`t instantiate the isFoo class, or is the right answer simply to get rid of the FunDep ? Thanks /Joel

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Problem in code
by Jenny678 06 Jun '06

06 Jun '06

hello, I can not find the problem, I use WinHugs and shows me(Syntax error in input (unexpected keyword "where"). here is my code: f :: (Float,Float) -> Float f (x,y) = (a(x,y) + 4) * (b(x,y) + 3) where a :: (Float,Float) -> Float a(x,y) = (x + y)*2 b :: (Float,Float) -> Float b(x,y) = (y - x)*5 How can I define this code with let in? Thanks for any help. -- View this message in context: http://www.nabble.com/Problem-in-code-t1742233.html#a4734684 Sent from the Haskell - Haskell-Cafe forum at Nabble.com.

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Combinations
by developer＠imaginando.net 06 Jun '06

06 Jun '06

Hi, I need a functions which takes as argument a list of lists like this one: [[1,2],[3],[4]] and gives me a list of list with all the possible combinations like this one: [[1,3,4],[2,3,4]] In this case there are only 2 combinations but if there was more than one element in the other lists the results were much more. I'm a little stuck in this simple problem. First i was thinking in cartesian product but then i realized that i may not be what i really need. Any tip? Many thx, Nuno Santos

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