違反適用的函子定律

Question

從在線課程中鍛煉。

假設，對於標准列表 Applicative functor， <*>運算符以標准方式定義，而pure更改為

pure x = [x,x]

會違反 Applicative typeclass 的哪些規律？

• 同態： pure g <*> pure x ≡ pure (gx)
• 身份： pure id <*> xs ≡ xs
• 交換： fs <*> pure x ≡ pure ($ x) <*> fs
• Applicative-Functor: g <$> xs ≡ pure g <*> xs
• 組成： (.) <$> us <*> vs <*> xs ≡ us <*> (vs <*> xs)

我創建了以下文件：

newtype MyList a = MyList {getMyList :: [a]}
  deriving Show

instance Functor MyList where
  fmap f (MyList xs) = MyList (map f xs)

instance Applicative MyList where
  pure x = MyList [x,x]
  MyList gs <*> MyList xs = MyList ([g x | g <- gs, x <- xs])

fs = MyList [\x -> 2*x, \x -> 3*x]
xs = MyList [1,2]
x = 1
g = (\x -> 2*x)
us = MyList [(\x -> 2*x)]
vs = MyList [(\x -> 3*x)]

然后我嘗試：

同態： pure g <*> pure x ≡ pure (gx)

*Main> pure g <*> pure x :: MyList Integer
MyList {getMyList = [2,2,2,2]}
*Main> pure (g x) :: MyList Integer
MyList {getMyList = [2,2]}

身份： pure id <*> xs ≡ xs

*Main> pure id <*> xs :: MyList Integer
MyList {getMyList = [1,2,1,2]}
*Main> xs :: MyList Integer
MyList {getMyList = [1,2]}

交換： fs <*> pure x ≡ pure ($ x) <*> fs

*Main> fs <*> pure x
[2,3]
*Main> pure ($ x) <*> fs
[2,3]

Applicative-Functor: g <$> xs ≡ pure g <*> xs

*Main> g <$> xs
MyList {getMyList = [2,4]}
*Main> pure g <*> xs
MyList {getMyList = [2,4,2,4]}

組成： (.) <$> us <*> vs <*> xs ≡ us <*> (vs <*> xs)

*Main> (.) <$> us <*> vs <*> xs
MyList {getMyList = [6,12]}
*Main> us <*> (vs <*> xs)
MyList {getMyList = [6,12]}

不應違反組合，因為這里不使用pure 。

似乎同態、恆等和 Applicative-Functor 不起作用。 但是當我在課程中選擇它們時，它說答案是錯誤的。 那么，誰是傻瓜：我還是課程的作者？

Answer 1

根據@DavidFletcher 的評論，使用您的代碼，我看到交換測試的不同輸出：

> fs <*> pure x
MyList {getMyList = [2,2,3,3]}
> pure ($ x) <*> fs
MyList {getMyList = [2,3,2,3]}

所以你可能想仔細檢查那個。