手动推导zipWith的类型。 uncurry`

Question

我试图派生zipWith . uncurry的类型zipWith . uncurry zipWith . uncurry

zipWith . uncurry = (.) zipWith uncurry zipWith . uncurry = (.) zipWith uncurry连接为函数

(.) :: (b1 -> c1) -> (a1 -> b1) -> a1 -> c1
zipWith :: (a2 -> b2 -> c2) -> [a2] -> [b2] -> [c2]
uncurry :: (a3 -> b3 -> c3) -> (a3, b3) -> c3

b1 ~ (a2 -> b2 -> c2)
c1 ~ [a2] -> [b2] -> [c2]
a1 ~ (a3 -> b3 -> c3)
b1 ~ (a3, b3) -> c3

用(a3 -> b3 -> c3)替换a1并用[a2] -> [b2] -> [c2]替换c1我得到：

(.) zipWith uncurry :: a1 -> c1 ~ (a3 -> b3 -> c3) -> [a2] -> [b2] -> [c2]

如果我向GHCi查询:t zipWith . uncurry :t zipWith . uncurry我得到： (a -> b1 -> b -> c) -> [(a, b1)] -> [b] -> [c] :t zipWith . uncurry (a -> b1 -> b -> c) -> [(a, b1)] -> [b] -> [c] :t zipWith . uncurry (a -> b1 -> b -> c) -> [(a, b1)] -> [b] -> [c]

我认为关键是在b1 ~ (a2 -> b2 -> c2) ~ (a3, b3) -> c3但是我不知道如何使结果与预期相符。

有什么帮助吗？

谢谢，
塞巴斯蒂安。

Answer 1

如果我们从

(.) :: (b -> c) -> (a -> b) -> (a -> c)
zipWith :: (x -> y -> z) -> [x] -> [y] -> [z]
uncurry :: (u -> v -> w) -> (u, v) -> w

然后我们将zipWith的类型与(b -> c)匹配， zipWith的类型与(b -> c) uncurry (a -> b)匹配：

b             -> c
(x -> y -> z) -> ([x] -> [y] -> [z])

和

a             -> b
(u -> v -> w) -> ((u, v) -> w)

然后我们必须将b项等价

b ~ x      -> (y -> z)
b ~ (u, v) -> w

所以w ~ (y -> z)和x ~ (u, v) 。 到目前为止我们已经确定的表格：

a ~ u -> v -> w
b ~ x -> y -> z
c ~ [x] -> [y] -> [z]
x ~ (u, v)
w ~ y -> z

因此我们可以用x和w代替：

a ~ u -> v -> (y -> z)
b ~ (u, v) -> y -> z
c -> [(u, v)] -> [y] -> [z]

现在我们可以完成zipWith . uncurry的类型zipWith . uncurry zipWith . uncurry ：

(u -> v -> y -> z) -> ([(u, v)] -> [y] -> [z])

哪个也与GHCi的答案一致。

这些技巧只是将箭头对齐，然后再代入，直到您不再缩小它为止。