Haskell:hackage Control中适用仿函数定律的描述有缺陷.适用文章?:它说由Applicative确定Functor

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问题描述

我认为我在针对Control的黑客文章中发现了一个缺陷.适用 .作为适用函子定律的描述，它说:

I think I found a flaw in the hackage article for Control.Applicative.As a description of the applicative functor laws, it says:

class Functor f => Applicative f where

带有应用程序的函子，可为嵌入纯表达式( pure )，并进行序列计算并组合其结果(< *> ).

A functor with application, providing operations to embed pure expressions (pure), and sequence computations and combine their results (<*>).

最小的完整定义必须包括满足以下法律的这些功能的实现:

A minimal complete definition must include implementations of these functions satisfying the following laws:

身份

identity

pure id <*> v = v

组成

composition

pure (.) <*> u <*> v <*> w = u <*> (v <*> w)

同态

homomorphism

pure f <*> pure x = pure (f x)

交换

interchange

u <*> pure y = pure ($ y) <*> u

(请注意，这并没有说明fmap)，它指出这些法律决定了 Functor 实例:

(note that this does not say anything about fmap) and it states that these laws determine the Functor instance:

fmap f x = pure f <*> x

我认为起初这显然是错误的.也就是说，我猜想必须存在一个满足以下两个条件的类型构造器 t :

I thought at first this was obviously wrong. That is, I guessed there must exist a type constructor t satisfying the following two conditions:

t 是满足上述规则的 Applicative 的实例，并且
instance(Functor t)有两种不同的实现方式(即，有两个不同的函数 fmap1，fmap2 ::(a-> b)->(t a-> t b)，满足函子法则.

t is an instance of Applicative fulfilling the rules above, and
There are two different implementations of instance (Functor t)(i.e. there are two distinct functions fmap1, fmap2 :: (a->b) -> (t a->t b), satisfying the functor laws).

如果(且仅当)以上内容正确，则必须将以上语句重写为

If (and only if) the above is correct, the above statement must be rewritten to

fmap f x = pure f <*> x

作为这些法律的结果，这满足了 Functor 法律.

As a consequence of these laws, this satisfies the Functor laws.

这是正确的，无论我的猜测是否正确.

我的问题是 :我的猜测正确吗?有符合条件的 t 吗?

下面是对我自己回答这个问题时的想法的解释.

What follows is the explanation of what I thought during trying to answer this question by myself.

如果我们只是对实际的Haskell编程不感兴趣的数学家，我们可以轻松地肯定地回答这个问题.实际上，

If we were mere mathematicians uninterested in actual Haskell programming, we could easily answer this question affirmatively. In fact,

t = Identity
newtype Identity a = Identity {runIdentity :: a}

满足上述要求1和2(实际上，几乎所有操作都可以完成).实际上， Identity 只是 Functor 和 Applicative 的一个实例，如.(这满足 fmap f =(纯f< *>).)要定义 instance(Functor f)的另一个实现"，请为每种 a 类型采用两个函数

meets the requirements 1 and 2 above (in fact, almost anything will do). Indeed, Identity is trivially an instance of Functor and Applicative, as defined in Data.Functor.Identity. (This satisfies fmap f = (pure f <*>).)To define another "implementation" of instance (Functor f), Take, for each type a, two functions

transf_a, tinv_a :: a -> a

如此

tinv_a . transf_a = id

(这很容易进行理论上的设置).现在定义

(This is, set-theoretically, easy). Now define

instance (Functor Identity) where
 fmap (f :: a->b) = Identity . transf_b . f . tinv_a . runIdentity

这符合 Functor 法则，并且与琐碎的实现是不同的实现

This satisfies the Functor laws and is a different implementation from the trivial one if there are

x :: a
f :: a -> b

如此

f x /= transf_b $ f $ tinv_a x

但是我们是否可以在Haskell中做到这一点一点也不明显.是这样的:

But it is not at all obvious whether we can do it in Haskell. Is something like:

class (Isom a) where
 transf, tinv :: a -> a

instance (Isom a) where
 transf = id
 tinv = id

specialized instance (Isom Bool) where
 transf = not
 tinv = not

可能在Haskell吗?

possible in Haskell?

我忘了写非常重要的东西.我将 Control.Applicative 识别为GHC基本软件包的一部分，所以我也对我的问题的答案是否随任何GHC语言扩展(例如 FlexibleInstances 或 OverlappingInstances ，我还不了解.

I forgot to write something very important. I recognize Control.Applicative as part of GHC base package, so I am also interested in whether the answer to my question changes with any GHC language extensions, such as FlexibleInstances or OverlappingInstances, which I don't understand yet.

Functor

Haskell:hackage Control中适用仿函数定律的描述有缺陷.适用文章?:它说由Applicative确定Functor

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