如何在 Haskell 中导出组合类型 [英] How to derive type of composition in Haskell

问题描述

我是 Haskell 的新手.我试图了解类型的组合是如何工作的.

(.) :: (b -> c) ->(a -> b) ->->Cfmap :: 函子 f =>(x -> y) ->f x ->f y地图.fmap :: (Functor f1, Functor f2) =>(x -> y) ->f1 (f2 x) ->f1 (f2 y)

我对上述类型信息的理解如下.

<预><代码>1. (x -> y) ->f1 x ->f1 y -- 第一个 fmap->(x' -> y') ->f2 x' ->f2 y' -- 第二个 fmap2.比较(1)和(.)类型签名，然后我们得到在 (b -> c)b = (x -> y)c = f1 x ->f1 y在 (a -> b)a = (x' -> y')b = f2 x' ->f2 y'3. 现在的结果是 ->c 但在 (b -> c) 中的 b 之前应该是 (a -> b) 中的 b(x -> y) === f2 x' ->f2 y'这意味着 x = f2 x' &y = f2 y'4. 结果是 ->C(x' -> y') ->f1 x ->f1 y将 (3) 结果代入此处(x' -> y') ->f1 (f2 x') ->f1 (f2 y')这是 Alpha 等价于 (x -> y) ->f1 (f2 x) ->f1 (f2 y)

为了测试我的理解，我尝试了各种数据类型，但在少数情况下取得了成功.以下是我无法弄清楚的类型签名.

<预><代码>f = 未定义 :: (x -> y -> w -> z -> a) ->g x ->y- f 的类型.f 序曲F .f :: (x -> (w1 -> z1 -> a1) -> w2 -> z2 -> a2) ->g (y -> x) ->y

我对上述方法的处理

<代码>1.(x -> y -> w -> z -> a) ->g x ->y->(x' -> y' -> w' -> z' -> a') ->g' x' ->g'y'2. 比较 (.) then在 (b -> c)b = (x -> y -> w -> z -> a)c = g x ->y在 (a -> b)a = (x' -> y' -> w' -> z' -> a')b = g' x' ->g'y'3. 比较 b 来自 (b -> c) &(a -> b)(x -> y -> w -> z -> a) === g' x' ->g'y'这意味着 x = g' x' &y ->w->z->a = g'y'4. 结果是 ->C(x' -> y' -> w' -> z' -> a') ->g x ->y现在没有直接的 y 我可以用 g y 代替

我看构图的方法不适合我.

解决方案

类型派生纯粹是机制 过程.

(.) :: (b -> c ) ->(a -> b ) ->(a -> c)fmap₂ :: 函子 f =>(t3 -> t4) ->(f t3 -> f t4)fmap₁ :: 函子 g =>((t1 -> t2) -> (g t1 -> g t2))-------------------------------------------------------------------------------------------a ~ (t1 -> t2) b ~ (g t1 -> g t2)b ~ (t3 -> t4) c ~ (f t3 -> f t4) b ~ (t3 -> t4)-------------------------------------------------------------------------------------------fmap₂ .fmap₁ :: (Functor f, Functor g)=>->C~ (t1 -> t2) ->(f t3 -> f t4 )~ (t1 -> t2) ->(f(g t1) -> f(g t2))

这对于 (.) 的表亲 (>>>) fmap₁ fmap₂ = fmap₂ 好得多.fmap₁:

(>>>) :: (a -> b ) ->(b->c)->(a -> c )fmap₁ :: 函子 g =>(t1 -> t2) ->(g t1 -> g t2)fmap₂ :: 函子 f =>(t3->t4)->(f t3 -> f t4)-------------------------------------------------------------------------------a ~ (t1 -> t2) b ~ (g t1 -> g t2)b ~ (t3 -> t4 ) c ~ (f t3 -> f t4)-------------------------------------------------------------------------------fmap₁>>>fmap₂ :: (Functor f, Functor g)=>->C~ (t1 -> t2) ->(f t3 -> f t4 )~ (t1 -> t2) ->(f(g t1) -> f(g t2))

<小时>

针对您的具体问题:

f = undefined :: (x -> y -> w -> z -> a) ->g x ->yf2 .f₁ = f₁ >>>f ::(x -> y -> w -> z -> a) ->((->) (g x) (g y ))((->) x2 ((->) y2 (w2 -> z2 -> a2))) ->(g2 x2 -> g2 y2)---------------------------------------------------------------------------------------x2 ~ g x g ~ ((->) y2) y ~ (w2 -> z2 -> a2)---------------------------------------------------------------------------------------(x -> y -> w -> z -> a) ->(g2 x2 -> g2 y2) ~(x -> y -> w -> z -> a) ->(g2 (g x) -> g2 y2) ~(x -> (w2 -> z2 -> a2) -> w -> z -> a) ->( g2 (y2 -> x) -> g2 y2)

这是你从 GHCi 得到的，直到重命名变量.

I am new to Haskell. I am trying to understand how composition of types work.

(.) :: (b -> c) -> (a -> b) -> a -> c

fmap :: Functor f => (x -> y) -> f x -> f y
fmap . fmap :: (Functor f1, Functor f2) => (x -> y) -> f1 (f2 x) -> f1 (f2 y)

How I understood the above type information is as below.

1. (x -> y) -> f1 x -> f1 y  -- First fmap
      -> (x' -> y') -> f2 x' -> f2 y'  -- Second fmap
2. Compare the (1) with (.) type signature then we get
     In (b -> c)
        b = (x -> y)
        c = f1 x -> f1 y
     In (a -> b)
        a = (x' -> y')
        b = f2 x' -> f2 y'
3. Now the result is a -> c but before that b in (b -> c) should be b in (a -> b)
     (x -> y) === f2 x' -> f2 y'
     That means x = f2 x' & y = f2 y'
4. Result is a -> c
     (x' -> y') -> f1 x -> f1 y
     Substituting (3) results here
     (x' -> y') -> f1 (f2 x') -> f1 (f2 y')
     This is Alpha equivalent to  (x -> y) -> f1 (f2 x) -> f1 (f2 y)

To test my understanding I tried with various data types and I am success in few. Below is the type signature I can't figure out.

 f = undefined :: (x -> y -> w -> z -> a) -> g x -> g y
 -- Type of f . f in prelude
 f . f :: (x -> (w1 -> z1 -> a1) -> w2 -> z2 -> a2) -> g (y -> x) -> g y

My approach to the above one

1. (x -> y -> w -> z -> a) -> g x -> g y
     -> (x' -> y' -> w' -> z' -> a') -> g' x' -> g' y'
2. Comparing with (.) then 
     In (b -> c)
        b = (x -> y -> w -> z -> a)
        c = g x -> g y
     In (a -> b)
        a = (x' -> y' -> w' -> z' -> a')
        b = g' x' -> g' y'
3. Comparing b from (b -> c) & (a -> b)
     (x -> y -> w -> z -> a) === g' x' -> g' y'
     That means x = g' x'  &  y -> w -> z -> a = g' y'
4. Result is a -> c
     (x' -> y' -> w' -> z' -> a') -> g x -> g y
     Now there is no direct y I can substitute in g y

My approach of seeing the composition is not working out for me.

解决方案

Type derivation is a purely mechanical process.

(.)  ::              (b          -> c             ) -> (a          -> b             ) -> 
                                                       (a                             -> c)
fmap₂ :: Functor f => (t3 -> t4) -> (f t3 -> f t4)
fmap₁ :: Functor g =>                                  ((t1 -> t2) -> (g t1 -> g t2))
-------------------------------------------------------------------------------------------
                                                   a ~ (t1 -> t2)  b ~ (g t1 -> g t2)
                 b ~ (t3 -> t4)  c ~ (f t3 -> f t4)                b ~ (t3   -> t4  )
-------------------------------------------------------------------------------------------
fmap₂ . fmap₁ :: (Functor f, Functor g) 
              =>  a          ->  c
           ~      (t1 -> t2) ->  (f t3     -> f t4    )
           ~      (t1 -> t2) ->  (f (g t1) -> f (g t2))

This is much nicer with (.)'s cousin, (>>>) fmap₁ fmap₂ = fmap₂ . fmap₁:

(>>>)  ::            (a           -> b                               ) -> 
                     (               b              -> c             ) -> 
                     (a                             -> c             )
fmap₁ :: Functor g =>  (t1 -> t2) -> (g t1 -> g t2)
fmap₂ :: Functor f =>                (t3   -> t4  ) -> (f t3 -> f t4) 
------------------------------------------------------------------------------
                  a ~ (t1 -> t2)   b ~ (g t1 -> g t2)
                                   b ~ (t3   -> t4  )  c ~ (f t3 -> f t4)
------------------------------------------------------------------------------
fmap₁ >>> fmap₂ :: (Functor f, Functor g) 
              =>    a             ->                   c
           ~          (t1 -> t2)  ->                   (f t3     -> f t4    )
           ~          (t1 -> t2)  ->                   (f (g t1) -> f (g t2))

---

To your specific problem:

f = undefined :: (x -> y -> w -> z -> a) -> g x -> g y

f₂ . f₁ = f₁ >>> f₂ :: 
 (x -> y -> w -> z -> a) -> ((->) (g x) (g       y               ))
                            ((->) x2    ((->) y2 (w2 -> z2 -> a2))) -> (g2 x2 -> g2 y2)
 ---------------------------------------------------------------------------------------
                       x2 ~ g x     g ~ ((->) y2)    y ~ (w2 -> z2 -> a2)
 ---------------------------------------------------------------------------------------
 (x -> y                -> w -> z -> a)                    ->  ( g2 x2        -> g2 y2)  ~
 (x -> y                -> w -> z -> a)                    ->  ( g2 (g     x) -> g2 y2)  ~
 (x -> (w2 -> z2 -> a2) -> w -> z -> a)                    ->  ( g2 (y2 -> x) -> g2 y2)

which is what you got from GHCi, up to the renaming of variables.

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