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应用函子定律

applicative haskell

1

user4035 · 技术社区 · 6 年前

在线课程的练习。

假设,对于一个标准的列表应用函子 <*> 运算符是以标准方式定义的,而 pure 改为

pure x = [x,x]

应用程序类型类将违反哪些法律?

同态: pure g <*> pure x â¡ pure (g x)
身份: pure id <*> xs â¡ xs
交换: fs <*> pure x â¡ pure ($ x) <*> fs
应用函子: g <$> xs â¡ pure g <*> xs
组成: (.) <$> us <*> vs <*> xs â¡ 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)]

然后我试着:

同态: 纯g<*>纯x纯(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: 纯id<*>xs xs

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

交换: fs<*>纯x纯($x)<*>财政司司长

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

应用函子: g<$>纯g<*>xs

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

组成: ()<$>美国<*>vs<*>xs美国<*>(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 纯净的 这里不用。

看来同态、恒等式和应用函子都不起作用。但当我在课程中选择它们时,它表明答案是错误的。那么,谁是傻瓜:我还是这门课程的作者?

0 回复 | 直到 6 年前

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K. A. Buhr 6 年前

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]}

所以你可能想再检查一遍。