Ptharien の Flame の回答のおかげで (賛成票を投じてください!)、実行中の「Haskell を学ぶ」のモナドの例をうまく適応させることができました。私はいくつかの詳細をグーグルで調べなければならなかったので(Haskellの初心者であるため)、これが最後に行ったことです(「Learn ...」のflipThreeの例は、 [(True,9 % 40), (False,31 % 40 )]):
ファイル Control/Restricted.hs (短くするために、RApplicative、RMonadPlus などを削除しました):
{-# LANGUAGE ConstraintKinds, TypeFamilies, KindSignatures, FlexibleContexts, UndecidableInstances #-}
module Control.Restricted (RFunctor(..),
RMonad(..)) where
import Prelude hiding (Functor(..), Monad(..))
import Data.Foldable (Foldable(foldMap))
import Data.Monoid
import GHC.Exts (Constraint)
class RFunctor f where
type Restriction f a :: Constraint
fmap :: (Restriction f a, Restriction f b) => (a -> b) -> f a -> f b
class (RFunctor m) => RMonad m where
return :: (Restriction m a) => a -> m a
(>>=) :: (Restriction m a, Restriction m b) => m a -> (a -> m b) -> m b
(>>) :: (Restriction m a, Restriction m b) => m a -> m b -> m b
a >> b = a >>= \_ -> b
join :: (Restriction m a, Restriction m (m a)) => m (m a) -> m a
join a = a >>= id
fail :: (Restriction m a) => String -> m a
fail = error
ファイル Prob.hs:
{-# LANGUAGE ConstraintKinds, TypeFamilies, RebindableSyntax, FlexibleContexts #-}
import Data.Ratio
import Control.Restricted
import Prelude hiding (Functor(..), Monad(..))
import Control.Arrow (first, second)
import Data.List (all)
newtype Prob a = Prob { getProb :: [(a, Rational)] } deriving Show
instance RFunctor Prob where
type Restriction Prob a = (Eq a)
fmap f (Prob as) = Prob $ map (first f) as
flatten :: Prob (Prob a) -> Prob a
flatten (Prob xs) = Prob $ concat $ map multAll xs
where multAll (Prob innerxs, p) = map (\(x, r) -> (x, p*r)) innerxs
compress :: Eq a => Prob a -> Prob a
compress (Prob as) = Prob $ foldr f [] as
where f a [] = [a]
f (a, p) ((b, q):bs) | a == b = (a, p+q):bs
| otherwise = (b, q):f (a, p) bs
instance Eq a => Eq (Prob a) where
(==) (Prob as) (Prob bs) = all (`elem` bs) as
instance RMonad Prob where
return x = Prob [(x, 1%1)]
m >>= f = compress $ flatten (fmap f m)
fail _ = Prob []