我正在尝试将Relation a b定义为Category实例。在我看来,作曲家的定义明确,并遵守关联法。当涉及到ID时,我找不到正确的定义。我在做什么错了?
import Data.Map as M
import Data.Set as S
import Control.Category as Cat
newtype Relation a b = R (Map a (Set b)) deriving (Show, Eq)
-- instance Cat.Category Relation where
-- id =
-- (.) = (°)
-- GHC.Base.id r1
-- > R (fromList [(10,fromList "abef"),(30,fromList "GRTXa")])
r1 = R $ M.fromList [(10,S.fromList "abfe"),(30,S.fromList "aXGRT")]
r2 = R $ M.fromList [('a',S.fromList [Just "Apple",Just "Ask"]),('b',S.fromList [Just "book",Just "brother"]),('T',S.fromList [Just "Table"]),('?',S.fromList [Just "Apple",Just "brother"])]
-- ex. r1 ° r2 = R (fromList [(10,fromList [Just "Apple",Just "Ask",Just "book",Just "brother"]),(30,fromList [Just "Apple",Just "Ask",Just "Table"])])
(°) :: (Ord a, Ord k, Ord b) => Relation a k -> Relation k b -> Relation a b
R mp1 ° R mp2
| M.null mp1 || M.null mp2 = R M.empty
| otherwise = R $ M.foldrWithKey (\k s acc -> M.insert k (S.foldr (\x acc2 -> case M.lookup x mp2 of
Nothing -> acc2
Just s2 -> S.union s2 acc2
) S.empty s) acc
) M.empty mp1
答案 0 :(得分:7)
Relation
不能是Category
的实例:
Relation
仅表示有限关系(当视为对对时),但是无限集合上的恒等关系是无限的;
组合不是在所有类型上都定义,仅在Ord
的实例上定义。
这是解决这些问题并使Relation
成为Category
实例的一种方法:
Option
从一个半群中形成一个半模像元的想法相同)Ord
个实例。这可以使用GADT完成。
{-# LANGUAGE GADTs #-}
data Relation a b where
Id :: Relation a a
R :: (Ord a, Ord b) => Map a (Set b) -> Relation a b
instance Category Relation where
id = Id
Id . r = r
r . Id = r
R r1 . R r2 = ...