QuickCheck的推广功能的一般情况是什么？ [英] What is the general case of QuickCheck's promote function?

问题描述

具有类似于QuickCheck的促进函数的结构的函子的一般术语是什么，即函数的形式如下：

  promote ::（a  - > fb） - > f（a  - > b）
  

（这是翻转的倒数$ fmap（flip（$））:: f（a - > b） - >（a - > fb））。除了（ - >）r 和 Id 以外，是否还有使用这种操作的函数？ （我确定必须有）。谷歌搜索'quickcheck promotion'只是提供了QuickCheck文档，该文档在任何更一般的上下文AFAICS中都没有给出 promote ;搜索SO for'quickcheck promote'不会产生任何结果。

解决方案

到目前为止，我发现了构建 f 与宣传态度：

• f = Identity


• 如果f和g都有 promote ht =（ft，gt）如果f和g都有 promote ，那么

• $ c>，那么如果f的 ht = f（gt）也是

• / code>属性，g是任何反函数函数，则函数 ht = gt - > ft 具有宣传属性



财产可以推广到profunctors g，但然后f将只是一个整数，所以它可能不是很有用，除非你只需要profunctors。

现在，使用这四个我们可以找到许多函数的例子 f ，其中 promote 存在：

pre $ code> ft =（t，t）

ft =（t，b - > t）

ft =（ t - > a） - > t

f t =（（t，t） - > b） - > （t，t，t）

f t =（（t，t，c-> t，（t-> b） - > t） - > a） - > t

另请注意，宣传属性意味着指出 f 。

  point :: t  - > ft 
 point x = fmap（const x）（promote id）
  

基本上是同一个问题：函子的这个属性是否比单子更强？

What is the general term for a functor with a structure resembling QuickCheck's promote function, i.e., a function of the form:

promote :: (a -> f b) -> f (a -> b)

(this is the inverse of flip $ fmap (flip ($)) :: f (a -> b) -> (a -> f b)). Are there even any functors with such an operation, other than (->) r and Id? (I'm sure there must be). Googling 'quickcheck promote' only turned up the QuickCheck documentation, which doesn't give promote in any more general context AFAICS; searching SO for 'quickcheck promote' produces no results.

解决方案

So far I found these ways of constructing an f with the promote morphism:

• f = Identity
• if f and g both have promote then the pair functor h t = (f t, g t) also does
• if f and g both have promote then the composition h t = f (g t) also does
• if f has the promote property and g is any contrafunctor then the functor h t = g t -> f t has the promote property

The last property can be generalized to profunctors g, but then f will be merely a profunctor, so it's probably not very useful, unless you only require profunctors.

Now, using these four constructions, we can find many examples of functors f for which promote exists:

f t = (t,t)

f t = (t, b -> t)

f t = (t -> a) -> t

f t = ((t,t) -> b) -> (t,t,t)

f t = ((t, t, c -> t, (t -> b) -> t) -> a) -> t

Also note that the promote property implies that f is pointed.

point :: t -> f t
point x = fmap (const x) (promote id)

Essentially the same question: Is this property of a functor stronger than a monad?

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