oleg at okmij.org oleg at okmij.org
Fri Apr 12 12:49:43 CEST 2013

```> 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 #-}

import qualified Data.Set as S

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

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 []  = 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

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
{-
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14]
(0.09 secs, 31465384 bytes)
fromList [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
(0.18 secs, 62421208 bytes)
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)
-}

```