[Haskell-cafe] Set monad

[ol...@okmij.org]

The question of Set monad comes up quite regularly, most recently at
http://www.ittc.ku.edu/csdlblog/?p=134

Indeed, we cannot make Data.Set.Set to be the instance of Monad type
class -- not immediately, that it. That does not mean that there is no
Set Monad, a non-determinism monad that returns the set of answers,
rather than a list. I mean genuine *monad*, rather than a restricted,
generalized, etc. monad.

It is surprising that the question of the Set monad still comes up
given how simple it is to implement it. Footnote 4 in the ICFP
2009 paper on ``Purely Functional Lazy Non-deterministic Programming''
described the idea, which is probably folklore. Just in case, here is
the elaboration, a Set monad transformer.

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}

module SetMonad where

import qualified Data.Set as S
import Control.Monad.Cont

-- Since ContT is a bona fide monad transformer, so is SetMonadT r.
type SetMonadT r = ContT (S.Set r)

-- These are the only two places the Ord constraint shows up

instance (Ord r, Monad m) => MonadPlus (SetMonadT r m) where
mzero = ContT $ \k -> return S.empty
mplus m1 m2 = ContT $ \k -> liftM2 S.union (runContT m1 k) (runContT m2 k)

runSet :: (Monad m, Ord r) => SetMonadT r m r -> m (S.Set r)
runSet m = runContT m (return . S.singleton)

choose :: MonadPlus m => [a] -> m a
choose = msum . map return

test1 = print =<< runSet (do
n1 <- choose [1..5]
n2 <- choose [1..5]
let n = n1 + n2
guard $ n < 7
return n)
-- fromList [2,3,4,5,6]

-- Values to choose from might be higher-order or actions
test1' = print =<< runSet (do
n1 <- choose . map return $ [1..5]
n2 <- choose . map return $ [1..5]
n <- liftM2 (+) n1 n2
guard $ n < 7
return n)
-- fromList [2,3,4,5,6]

test2 = print =<< runSet (do
i <- choose [1..10]
j <- choose [1..10]
k <- choose [1..10]
guard $ i*i + j*j == k * k
return (i,j,k))
-- fromList [(3,4,5),(4,3,5),(6,8,10),(8,6,10)]

test3 = print =<< runSet (do
i <- choose [1..10]
j <- choose [1..10]
k <- choose [1..10]
guard $ i*i + j*j == k * k
return k)
-- fromList [5,10]



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[Petr Pudlák]

One problem with such monad implementations is efficiency. Let's define

    step :: (MonadPlus m) => Int -> m Int

    step i = choose [i, i + 1]

    

    -- repeated application of step on 0:

    stepN :: (Monad m) => Int -> m (S.Set Int)

    stepN = runSet . f

      where

        f 0 = return 0

        f n = f (n-1) >>= step

Then `stepN`'s time complexity is exponential in its argument. This is because `ContT` reorders the chain of computations to right-associative, which is correct, but changes the time complexity in this unfortunate way. If we used Set directly, constructing a left-associative chain, it produces the result immediately:

    step' :: Int -> S.Set Int

    step' i = S.fromList [i, i + 1]

    

    stepN' :: Int -> S.Set Int

    stepN' 0 = S.singleton 0

    stepN' n = stepN' (n - 1) `setBind` step'

      where

        setBind k f = S.foldl' (\s -> S.union s . f) S.empty k

See also: Constructing efficient monad instances on `Set` (and other containers with constraints) using the continuation monad <http://stackoverflow.com/q/12183656/1333025>

Best regards,

Petr Pudlak

2013/4/11 <ol...@okmij.org>

[ol...@okmij.org]

> One problem with such monad implementations is efficiency. Let's define
>
> step :: (MonadPlus m) => Int -> m Int
> step i = choose [i, i + 1]
>
> -- repeated application of step on 0:
> stepN :: (Monad m) => Int -> m (S.Set Int)
> stepN = runSet . f
> where
> f 0 = return 0
> f n = f (n-1) >>= step
>
> Then `stepN`'s time complexity is exponential in its argument. This is
> because `ContT` reorders the chain of computations to right-associative,
> which is correct, but changes the time complexity in this unfortunate way.
> If we used Set directly, constructing a left-associative chain, it produces
> the result immediately:

The example is excellent. And yet, the efficient genuine Set monad is
possible.

BTW, a simpler example to see the problem with the original CPS monad is to
repeat
choose [1,1] >> choose [1,1] >>choose [1,1] >> return 1

and observe exponential behavior. But your example is much more
subtle.

Enclosed is the efficient genuine Set monad. I wrote it in direct
style (it seems to be faster, anyway). The key is to use the optimized
choose function when we can.

{-# LANGUAGE GADTs, TypeSynonymInstances, FlexibleInstances #-}

module SetMonadOpt where

import qualified Data.Set as S
import Control.Monad

data SetMonad a where
SMOrd :: Ord a => S.Set a -> SetMonad a
SMAny :: [a] -> SetMonad a

instance Monad SetMonad where
return x = SMAny [x]

m >>= f = collect . map f $ toList m

toList :: SetMonad a -> [a]
toList (SMOrd x) = S.toList x
toList (SMAny x) = x

collect :: [SetMonad a] -> SetMonad a
collect [] = SMAny []
collect [x] = x
collect ((SMOrd x):t) = case collect t of
SMOrd y -> SMOrd (S.union x y)
SMAny y -> SMOrd (S.union x (S.fromList y))
collect ((SMAny x):t) = case collect t of
SMOrd y -> SMOrd (S.union y (S.fromList x))
SMAny y -> SMAny (x ++ y)

runSet :: Ord a => SetMonad a -> S.Set a
runSet (SMOrd x) = x
runSet (SMAny x) = S.fromList x

instance MonadPlus SetMonad where
mzero = SMAny []
mplus (SMAny x) (SMAny y) = SMAny (x ++ y)
mplus (SMAny x) (SMOrd y) = SMOrd (S.union y (S.fromList x))
mplus (SMOrd x) (SMAny y) = SMOrd (S.union x (S.fromList y))
mplus (SMOrd x) (SMOrd y) = SMOrd (S.union x y)

choose :: MonadPlus m => [a] -> m a
choose = msum . map return

test1 = runSet (do

n1 <- choose [1..5]
n2 <- choose [1..5]
let n = n1 + n2
guard $ n < 7
return n)
-- fromList [2,3,4,5,6]

-- Values to choose from might be higher-order or actions

test1' = runSet (do

n1 <- choose . map return $ [1..5]
n2 <- choose . map return $ [1..5]
n <- liftM2 (+) n1 n2
guard $ n < 7
return n)
-- fromList [2,3,4,5,6]

test2 = runSet (do

i <- choose [1..10]
j <- choose [1..10]
k <- choose [1..10]
guard $ i*i + j*j == k * k
return (i,j,k))
-- fromList [(3,4,5),(4,3,5),(6,8,10),(8,6,10)]

test3 = runSet (do

i <- choose [1..10]
j <- choose [1..10]
k <- choose [1..10]
guard $ i*i + j*j == k * k
return k)
-- fromList [5,10]

-- Test by Petr Pudlak

-- First, general, unoptimal case

step :: (MonadPlus m) => Int -> m Int
step i = choose [i, i + 1]

-- repeated application of step on 0:

stepN :: Int -> S.Set Int

stepN = runSet . f
where
f 0 = return 0
f n = f (n-1) >>= step

-- it works, but clearly exponential
{-
*SetMonad> stepN 14
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14]
(0.09 secs, 31465384 bytes)
*SetMonad> stepN 15
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
(0.18 secs, 62421208 bytes)
*SetMonad> stepN 16
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
(0.35 secs, 124876704 bytes)
-}

-- And now the optimization
chooseOrd :: Ord a => [a] -> SetMonad a
chooseOrd x = SMOrd (S.fromList x)

stepOpt :: Int -> SetMonad Int
stepOpt i = chooseOrd [i, i + 1]

-- repeated application of step on 0:

stepNOpt :: Int -> S.Set Int
stepNOpt = runSet . f

where
f 0 = return 0

f n = f (n-1) >>= stepOpt

{-
stepNOpt 14
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14]
(0.00 secs, 515792 bytes)
stepNOpt 15
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
(0.00 secs, 515680 bytes)
stepNOpt 16
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
(0.00 secs, 515656 bytes)

stepNOpt 30
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]
(0.00 secs, 1068856 bytes)
-}

[Petr Pudlák]

Oleg, thanks a lot for this example, and sorry for my late reply. I
really like the idea and I'm hoping to a similar concept soon for a
monad representing probability computations.

With best regards,
Petr