有趣的问题。这是我的看法 - 让我们看看我是否没有在任何地方搞砸！
首先，我将用（稍微不那么伪的）Haskell 拼写你的签名：
return :: a -> PSet (r -> a)
(>>=) :: PSet (r -> a) -> (a -> PSet (r -> b)) -> PSet (r -> b))
在继续之前，值得一提的是两个实际问题。首先，正如您已经观察到的，由于Eq 和/或Ord 约束，给集合Functor 或Monad 实例是不平凡的；无论如何，there are ways around it。其次，更令人担忧的是，对于您为 (>>=) 建议的类型，有必要从 PSet (r -> a) 中提取 as 而没有任何明显的 rs 供应 - 或者，在换句话说，您的(>>=) 需要遍历函数函子(->) r。当然，这在一般情况下是不可能的，即使在可能的情况下也往往是不切实际的——至少就 Haskell 而言。无论如何，为了我们的推测目的，假设我们可以通过将函数应用于所有可能的r 值来遍历(->) r。我将通过一个手工波浪形的universe :: PSet r 集合来表明这一点，以向this package 致敬。我还将使用universe :: PSet (r -> b)，并假设我们可以判断两个r -> b 函数是否同意某个r，即使不需要Eq 约束。 （伪 Haskell 确实越来越假了！）
初步评论，这是我的伪 Haskell 版本的方法：
return :: a -> PSet (r -> a)
return x = singleton (const x)
(>>=) :: PSet (r -> a) -> (a -> PSet (r -> b)) -> PSet (r -> b))
m >>= f = unionMap (\x ->
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (f (x r)))
(universe :: PSet (r -> b)))
(universe :: PSet r)) m
where
unionMap f = unions . map f
intersectionMap f = intersections . map f
接下来，单子定律：
m >>= return = m
return y >>= f = f y
m >>= f >>= g = m >>= \y -> f y >>= g
（顺便说一句，在做这种事情时，最好记住我们正在使用的课程的其他演示文稿 - 在这种情况下，我们有 join 和 (>=>) 作为 @ 的替代品987654347@ -- 因为切换演示文稿可能会使您选择的实例更愉快。这里我将坚持使用Monad 的(>>=) 演示文稿。）
继续第一定律...
m >>= return = m
m >>= return -- LHS
unionMap (\x ->
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (singleton (const (x r))))
(universe :: PSet (r -> b)))
(universe :: PSet r)) m
unionMap (\x ->
intersectionMap (\r ->
filter (\rb ->
const (x r) r == rb r)
(universe :: PSet (r -> b)))
(universe :: PSet r)) m
unionMap (\x ->
intersectionMap (\r ->
filter (\rb ->
x r == rb r)
(universe :: PSet (r -> b)))
(universe :: PSet r)) m
-- In other words, rb has to agree with x for all r.
unionMap (\x -> singleton x) m
m -- RHS
一个下来，两个去。
return y >>= f = f y
return y -- LHS
unionMap (\x ->
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (f (x r)))
(universe :: PSet (r -> b)))
(universe :: PSet r)) (singleton (const y))
(\x ->
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (f (x r)))
(universe :: PSet (r -> b)))
(universe :: PSet r)) (const y)
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (f (const y r)))
(universe :: PSet (r -> b)))
(universe :: PSet r)
intersectionMap (\r ->
filter (\rb ->
any (\rb' -> rb' r == rb r) (f y)))
(universe :: PSet (r -> b)))
(universe :: PSet r)
-- This set includes all functions that agree with at least one function
-- from (f y) at each r.
因此，return y >>= f 可能是一个比f y 大得多的集合。我们违反了第二定律；因此，我们没有 monad —— 至少在此处提出的实例中没有。
附录：这是一个实际的、可运行的函数实现，它至少可以用于小类型。它利用了前面提到的universe 包。
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE ScopedTypeVariables #-}
module FunSet where
import Data.Universe
import Data.Map (Map)
import qualified Data.Map as M
import Data.Set (Set)
import qualified Data.Set as S
import Data.Int
import Data.Bool
-- FunSet and its would-be monad instance
newtype FunSet r a = FunSet { runFunSet :: Set (Fun r a) }
deriving (Eq, Ord, Show)
fsreturn :: (Finite a, Finite r, Ord r) => a -> FunSet r a
fsreturn x = FunSet (S.singleton (toFun (const x)))
-- Perhaps we should think of a better name for this...
fsbind :: forall r a b.
(Ord r, Finite r, Ord a, Ord b, Finite b, Eq b)
=> FunSet r a -> (a -> FunSet r b) -> FunSet r b
fsbind (FunSet s) f = FunSet $
unionMap (\x ->
intersectionMap (\r ->
S.filter (\rb ->
any (\rb' -> funApply rb' r == funApply rb r)
((runFunSet . f) (funApply x r)))
(universeF' :: Set (Fun r b)))
(universeF' :: Set r)) s
toFunSet :: (Finite r, Finite a, Ord r, Ord a) => [r -> a] -> FunSet r a
toFunSet = FunSet . S.fromList . fmap toFun
-- Materialised functions
newtype Fun r a = Fun { unFun :: Map r a }
deriving (Eq, Ord, Show, Functor)
instance (Finite r, Ord r, Universe a) => Universe (Fun r a) where
universe = fmap (Fun . (\f ->
foldr (\x m ->
M.insert x (f x) m) M.empty universe))
universe
instance (Finite r, Ord r, Finite a) => Finite (Fun r a) where
universeF = universe
funApply :: Ord r => Fun r a -> r -> a
funApply f r = maybe
(error "funApply: Partial functions are not fun")
id (M.lookup r (unFun f))
toFun :: (Finite r, Finite a, Ord r) => (r -> a) -> Fun r a
toFun f = Fun (M.fromList (fmap ((,) <$> id <*> f) universeF))
-- Set utilities
unionMap :: (Ord a, Ord b) => (a -> Set b) -> (Set a -> Set b)
unionMap f = S.foldl S.union S.empty . S.map f
-- Note that this is partial. Since for our immediate purposes the only
-- consequence is that r in FunSet r a cannot be Void, I didn't bother
-- with making it cleaner.
intersectionMap :: (Ord a, Ord b) => (a -> Set b) -> (Set a -> Set b)
intersectionMap f s = case ss of
[] -> error "intersectionMap: Intersection of empty set of sets"
_ -> foldl1 S.intersection ss
where
ss = S.toList (S.map f s)
universeF' :: (Finite a, Ord a) => Set a
universeF' = S.fromList universeF
-- Demo
main :: IO ()
main = do
let andor = toFunSet [uncurry (&&), uncurry (||)]
print andor -- Two truth tables
print $ funApply (toFun (2+)) (3 :: Int8) -- 5
print $ (S.map (flip funApply (7 :: Int8)) . runFunSet)
(fsreturn (Just True)) -- fromList [Just True]
-- First monad law demo
print $ fsbind andor fsreturn == andor -- True
-- Second monad law demo
let twoToFour = [ bool (Left False) (Left True)
, bool (Left False) (Right False)]
decider b = toFunSet
(fmap (. bool (uncurry (&&)) (uncurry (||)) b) twoToFour)
print $ fsbind (fsreturn True) decider == decider True -- False (!)