违反适用函子法

Question

来自在线课程的练习题。

假设，对于标准列表 Applicative functor，<*> 运算符以标准方式定义，而 pure 更改为

pure x = [x,x]

Applicative typeclass 会违反哪些规则？

同态：pure g <*> pure x ≡ pure (g x)
身份：pure id <*> xs ≡ xs
交汇处：fs <*> pure x ≡ pure ($ x) <*> fs
应用函子：g <$> xs ≡ pure g <*> xs
组成：(.) <$> us <*> vs <*> xs ≡ us <*> (vs <*> xs)

我创建了以下文件：

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)]

然后我尝试了：

同态：pure g <*> pure x ≡ pure (g x)

*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]}

交汇处：fs <*> pure x ≡ pure ($ x) <*> fs

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

应用函子：g <$> xs ≡ pure g <*> xs

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

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

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

作文不应该被违反，因为这里没有使用pure。

同态、恒等式和应用函子好像不行。但是当我在课程中select他们时，它说答案是错误的。那么，谁是傻瓜：我还是课程作者？

Answer 1

根据@DavidFletcher 的评论，使用您的代码，我看到交换测试的不同输出：

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

所以你可能想仔细检查那个。

违反适用函子法

Applicative functor laws violation

haskell

applicative