是所有可区分类型Monads [英] Are all differentiable types Monads

问题描述

给定可区分类型，我们知道 Zipper 是一个 Comonad 。对此，丹·伯顿问道：如果派生出一个共同点，那么这是否意味着整合成了一个单子？或者说这是无稽之谈？。我想给这个问题一个具体的含义。 如果一个类型是可区​​分的，它是否必然是一个monad？这个问题的一个表述是询问，给定以下定义

  data Zipper ta = Zipper {diff :: D ta，here :: a} 
 
导出实例Diff t => Functor（Zipper t）
 
 class（Functor t，Functor（D t））=> Diff t其中
 type D t :: *  - > * 
 up ::拉链t a  - > t a 
 down：t a  - >我们可以写出类似于 
的签名的函数吗？我们可以写一些类似于>的签名的函数吗？ 
 
  return ::（Diff t）=> a  - > t a 
（>> =）::（Diff t）=> t a  - > （a  - > t b） - > tb 
  
服从 Monad laws 。
 
 
 
链接的问题，有两种成功的方法来解决为 Zipper 派生 Comonad 实例的类似问题。第一种方法是展开 Diff 类以包含>> = 并使用偏分化。第二种方法是要求该类型是两次或无限可区分。
解决方案
没有。 void functor  data V a 是可区分的，但是 return 无法实现。
 
Given a differentiable type, we know that its Zipper is a Comonad. In response to this, Dan Burton asked, "If derivation makes a comonad, does that mean that integration makes a monad? Or is that nonsense?". I'd like to give this question a specific meaning. If a type is differentiable, is it necessarily a monad? One formulation of the question would be to ask, given the following definitions
data Zipper t a = Zipper { diff :: D t a, here :: a }

deriving instance Diff t => Functor (Zipper t)

class (Functor t, Functor (D t)) => Diff t where
    type D t :: * -> *
    up :: Zipper t a -> t a
    down :: t a -> t (Zipper t a)
can we write functions with signatures similar to 
return :: (Diff t) => a -> t a
(>>=) :: (Diff t) => t a -> (a -> t b) -> t b
obeying the Monad laws.

In the answers to the linked questions, there were two successful approaches to a similar problem of deriving Comonad instances for the Zipper. The first approach was to expand the Diff class to include the dual of >>= and use partial differentiation. The second approach was to require that the type be twice or infinitely differentiable.
解决方案
No. The void functor data V a is differentiable, but return cannot be implemented for it.

                        
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