Home Reading Searching Subscribe Sponsors Statistics Posting Contact Spam Lists Links About Hosting Filtering Features Download Marketing Archives FAQ Blog From: Roberto Zunino di.unipi.it> Subject: Monad Set via GADT Newsgroups: gmane.comp.lang.haskell.cafe Date: Wednesday 3rd January 2007 23:38:59 UTC (over 11 years ago) To improve my understanding of GADT, I tried to define a Set datatype, with the usual operations, so that it can be made a member of the standard Monad class. Here I report on my experiments. First, I recap the problem. Data.Set.Set can not be made a Monad because of the Ord constraint on its parameter, while e.g. (return :: a -> m a) allows any type inside the monad m. This problem can be solved using an alternative monad class (restricted monad) so that the Ord context is actually provided for the monad operations. Rather, I aimed for the standard Monad typeclass. To avoid reinventing Data.Set.Set, my datatype is based on that. Basically, when we know of an Ord context, we use a Set. Otherwise, we use a simple list representation. Using a list is just enough to allow the implementation of return (i.e. (:[])) and (>>=) (i.e. map and (++)), so while not being very efficient, it is simple. Other non-monadic operators require an Ord context, so that we can turn lists into Set. My first shot at this was: data SetM a where L :: [a] -> SetM a SM :: Ord a => Set.Set a -> SetM a However, this was not enough to convince the type checker to use the Ord context stored in SM. Specifically, performing union with union (SM m1) (SM m2) = SM (m1 Set.union m2) causes GHC to report "No instance for (Ord a)". After some experiments, I found the following, using a type equality witness: data Teq a b where Teq :: Teq a a data SetM a where L :: [a] -> SetM a SM :: Ord w => Teq a w -> Set.Set w -> SetM a Now if I use union (SM p1 m1) (SM p2 m2) = case (p1,p2) of (Teq,Teq) -> SM Teq (m1 Set.union m2) it typechecks! This rises some questions I cannot answer: 1) Why the first version did not typececk? 2) Why the second one does? 3) If I replace (Teq a w) with (Teq w a), as in SM :: Ord w => Teq w a -> Set.Set w -> SetM a then union above does not typecheck! Why? I guess the type variable unification deriving from matching Teq is not symmetric as I expect it to be... Below, I attach the working version. Monad and MonadPlus instances are provided for SetM. Conversions from/to Set are also provided, requiring an Ord context. "Efficient" return and mzero are included, forcing the Set representation to be used, and requiring Ord (these could also be derived from fromSet/toSet, however). Comments are very welcome, of course, as well as non-GADT related alternative approaches. Regards, Roberto Zunino. ============================================================ \begin{code} {-# OPTIONS_GHC -Wall -fglasgow-exts #-} module SetMonad ( SetM() , toSet, fromSet , union, unions , return', mzero' ) where import qualified Data.Set as S import Data.List hiding (union) import Control.Monad -- Type equality witness data Teq a b where Teq :: Teq a a -- Either a list or a real Set data SetM a where L :: [a] -> SetM a SM :: Ord w => Teq a w -> S.Set w -> SetM a instance Monad SetM where return = L . (:[]) m >>= f = case m of L l -> unions (map f l) SM Teq s -> unions (map f (S.toList s)) instance MonadPlus SetM where mzero = L [] mplus = union -- Efficient variants for Ord types return' :: Ord a => a -> SetM a return' = SM Teq . S.singleton mzero' :: Ord a => SetM a mzero' = SM Teq S.empty -- Set union: use the best representation union :: SetM a -> SetM a -> SetM a union (L l1) (L l2) = L (l1 ++ l2) union (SM p1 m1) (SM p2 m2) = case (p1,p2) of (Teq,Teq) -> SM Teq (m1 S.union m2) union (L l1) (SM p m2) = case p of Teq -> SM Teq (m2 S.union S.fromList l1) union s1 s2 = union s2 s1 -- Try to put a SM first before folding, to improve performance unions :: [SetM a] -> SetM a unions = let isSM (SM _ _) = True isSM _ = False in foldl' union (L []) . uncurry (++) . break isSM -- Conversion from/to Set requires Ord toSet :: Ord a => SetM a -> S.Set a toSet (L l) = S.fromList l toSet (SM p m) = case p of Teq -> m fromSet :: Ord a => S.Set a -> SetM a fromSet = SM Teq -- Tests test :: IO () test = do let l = [1..3] :: [Int] s = fromSet (S.fromList l) g x = return' x mplus return' (x+100) print $S.toList$ toSet $do x <- s y <- s return' (x+y) -- [2,3,4,5,6] print$ S.toList $toSet$ do x <- s g x -- [1,2,3,101,102,103] print $S.toList$ toSet $do x <- s y <- g x return' y -- [1,2,3,101,102,103] print$ S.toList $toSet$ do x <- s y <- g x g y -- [1,2,3,101,102,103,201,202,203] print $S.toList$ toSet \$ do x <- s y <- return (const x) -- no Ord! return' (y ()) -- [1,2,3] \end{code}
CD: 3ms