Agda中的Arity泛型编程 [英] Arity-generic programming in Agda

问题描述

如何在Agda中编写泛型函数?是否可以编写完全依赖的和Universe多态的Arity泛型函数?

How to write arity-generic functions in Agda? Is it possible to write fully dependent and universe polymorphic arity-generic functions?

推荐答案

我将以n元合成函数为例.

I'll take an n-ary composition function as an example.

open import Data.Vec.N-ary

comp : ∀ n {α β γ} {X : Set α} {Y : Set β} {Z : Set γ}
     -> (Y -> Z) -> N-ary n X Y -> N-ary n X Z
comp  0      g y = {!!}
comp (suc n) g f = {!!}

这是在Data.Vec.N-ary模块中定义N-ary的方式:

Here is how N-ary is defined in the Data.Vec.N-ary module:

N-ary : ∀ {ℓ₁ ℓ₂} (n : ℕ) → Set ℓ₁ → Set ℓ₂ → Set (N-ary-level ℓ₁ ℓ₂ n)
N-ary zero    A B = B
N-ary (suc n) A B = A → N-ary n A B

即comp接收一个数字n，一个函数g : Y -> Z和一个函数f，该函数具有Arity n和结果类型Y.

I.e. comp receives a number n, a function g : Y -> Z and a function f, which has the arity n and the resulting type Y.

在comp 0 g y = {!!}情况下

Goal : Z
y    : Y
g    : Y -> Z

因此g y可以轻松填充孔.

在comp (suc n) g f = {!!}情况下，N-ary (suc n) X Y减小为X -> N-ary n X Y，而N-ary (suc n) X Z减小为X -> N-ary n X Z.所以我们有

In the comp (suc n) g f = {!!} case, N-ary (suc n) X Y reduces to X -> N-ary n X Y and N-ary (suc n) X Z reduces to X -> N-ary n X Z. So we have

Goal : X -> N-ary n X Z
f    : X -> N-ary n X Y
g    : Y -> Z

C-c C-r将孔减小到λ x -> {!!}，现在是Goal : N-ary n X Z，可以用comp n g (f x)填充.所以整个定义是

C-c C-r reduces the hole to λ x -> {!!}, and now Goal : N-ary n X Z, which can be filled by comp n g (f x). So the whole definition is

comp : ∀ n {α β γ} {X : Set α} {Y : Set β} {Z : Set γ}
     -> (Y -> Z) -> N-ary n X Y -> N-ary n X Z
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

即comp接收类型为X的n自变量，将f应用于它们，然后将g应用于结果.

I.e. comp receives n arguments of type X, applies f to them and then applies g to the result.

g的类型为(y : Y) -> Z y

comp : ∀ n {α β γ} {X : Set α} {Y : Set β} {Z : Y -> Set γ}
     -> ((y : Y) -> Z y) -> (f : N-ary n X Y) -> {!!}
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

洞中应该有什么?我们不能像以前那样使用N-ary n X Z，因为Z现在是一个函数.如果Z是一个函数，我们需要将其应用于类型为Y的对象.但是获取Y类型的东西的唯一方法是将f应用于类型X的n自变量.就像我们的comp一样，但仅在类型级别上:

what should be there in the hole? We can't use N-ary n X Z as before, because Z is a function now. If Z is a function, we need to apply it to something, that has type Y. But the only way to get something of type Y is to apply f to n arguments of type X. Which is just like our comp but only at the type level:

Comp : ∀ n {α β γ} {X : Set α} {Y : Set β}
     -> (Y -> Set γ) -> N-ary n X Y -> Set (N-ary-level α γ n)
Comp  0      Z y = Z y
Comp (suc n) Z f = ∀ x -> Comp n Z (f x)

然后comp是

comp : ∀ n {α β γ} {X : Set α} {Y : Set β} {Z : Y -> Set γ}
     -> ((y : Y) -> Z y) -> (f : N-ary n X Y) -> Comp n Z f
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

带有不同类型参数的版本

有" Arity通用数据类型通用编程 "一文，其中描述了如何编写接收不同类型参数的泛型泛型函数.想法是将类型的向量作为参数传递，并以N-ary的样式将其折叠起来:

A version with arguments with different types

There is the "Arity-generic datatype-generic programming" paper, that describes, among other things, how to write arity-generic functions, that receive arguments of different types. The idea is to pass a vector of types as a parameter and fold it pretty much in the style of N-ary:

arrTy : {n : N} → Vec Set (suc n) → Set
arrTy {0}     (A :: []) = A
arrTy {suc n} (A :: As) = A → arrTy As

但是，即使我们提供了它的长度，Agda也无法推断该向量.因此，本文还提供了一种currying运算符，该运算符从一个函数中显式接收类型向量，一个函数，该函数接收n个隐式参数.

However Agda is unable to infer that vector, even if we provide its length. Hence the paper also provides an operator for currying, which makes from a function, that explicitly receives a vector of types, a function, that receives n implicit arguments.

此方法有效，但不能扩展到完全Universe多态函数.我们可以通过将Vec数据类型替换为_^_运算符来避免所有这些问题:

This approach works, but it doesn't scale to fully universe polymorphic functions. We can avoid all these problems by replacing the Vec datatype with the _^_ operator:

_^_ : ∀ {α} -> Set α -> ℕ -> Set α
A ^ 0     = Lift ⊤
A ^ suc n = A × A ^ n

A ^ n与Vec A n同构.然后我们的新N-ary是

A ^ n is isomorphic to Vec A n. Then our new N-ary is

_->ⁿ_ : ∀ {n} -> Set ^ n -> Set -> Set
_->ⁿ_ {0}      _      B = B
_->ⁿ_ {suc _} (A , R) B = A -> R ->ⁿ B

为简单起见，所有类型都位于Set中. comp现在是

All types lie in Set for simplicity. comp now is

comp : ∀ n {Xs : Set ^ n} {Y Z : Set}
     -> (Y -> Z) -> (Xs ->ⁿ Y) -> Xs ->ⁿ Z
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

以及具有从属项g的版本:

And a version with a dependent g:

Comp : ∀ n {Xs : Set ^ n} {Y : Set}
     -> (Y -> Set) -> (Xs ->ⁿ Y) -> Set
Comp  0      Z y = Z y
Comp (suc n) Z f = ∀ x -> Comp n Z (f x)

comp : ∀ n {Xs : Set ^ n} {Y : Set} {Z : Y -> Set}
     -> ((y : Y) -> Z y) -> (f : Xs ->ⁿ Y) -> Comp n Z f
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

完全依赖且全域多态的comp

关键思想是将类型的矢量表示为嵌套的依赖对:

Fully dependent and universe polymorphic comp

The key idea is to represent a vector of types as nested dependent pairs:

Sets : ∀ {n} -> (αs : Level ^ n) -> ∀ β -> Set (mono-^ (map lsuc) αs ⊔ⁿ lsuc β)
Sets {0}      _       β = Set β
Sets {suc _} (α , αs) β = Σ (Set α) λ X -> X -> Sets αs β

第二种情况的读法类似于存在一个X类型，使得所有其他类型都依赖于X的元素".我们新的N-ary很简单:

The second case reads like "there is a type X such that all other types depend on an element of X". Our new N-ary is trivial:

Fold : ∀ {n} {αs : Level ^ n} {β} -> Sets αs β -> Set (αs ⊔ⁿ β)
Fold {0}      Y      = Y
Fold {suc _} (X , F) = (x : X) -> Fold (F x)

一个例子:

postulate
  explicit-replicate : (A : Set) -> (n : ℕ) -> A -> Vec A n

test : Fold (Set , λ A -> ℕ , λ n -> A , λ _ -> Vec A n) 
test = explicit-replicate

但是现在Z和g的类型是什么?

But what are the types of Z and g now?

comp : ∀ n {β γ} {αs : Level ^ n} {Xs : Sets αs β} {Z : {!!}}
     -> {!!} -> (f : Fold Xs) -> Comp n Z f
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

回想一下f以前的类型是Xs ->ⁿ Y，但是Y现在隐藏在这些嵌套相关对的末尾，并且可以依赖Xs中任何X的元素. Z以前的类型为Y -> Set γ，因此现在我们需要将Set γ附加到Xs，使所有x都隐式:

Recall that f previously had type Xs ->ⁿ Y, but Y now is hidden in the end of these nested dependent pairs and can depend on an element of any X from Xs. Z previously had type Y -> Set γ, hence now we need to append Set γ to Xs, making all xs implicit:

_⋯>ⁿ_ : ∀ {n} {αs : Level ^ n} {β γ}
      -> Sets αs β -> Set γ -> Set (αs ⊔ⁿ β ⊔ γ)
_⋯>ⁿ_ {0}      Y      Z = Y -> Z
_⋯>ⁿ_ {suc _} (_ , F) Z = ∀ {x} -> F x ⋯>ⁿ Z

好，Z : Xs ⋯>ⁿ Set γ，g是什么类型?以前是(y : Y) -> Z y.再次，我们需要向嵌套的依赖对中添加一些内容，因为Y再次被隐藏，现在仅以依赖的方式进行了

OK, Z : Xs ⋯>ⁿ Set γ, what type has g? Previously it was (y : Y) -> Z y. Again we need to append something to nested dependent pairs, since Y is again hidden, only in a dependent way now:

Πⁿ : ∀ {n} {αs : Level ^ n} {β γ}
   -> (Xs : Sets αs β) -> (Xs ⋯>ⁿ Set γ) -> Set (αs ⊔ⁿ β ⊔ γ)
Πⁿ {0}      Y      Z = (y : Y) -> Z y
Πⁿ {suc _} (_ , F) Z = ∀ {x} -> Πⁿ (F x) Z

最后

Comp : ∀ n {αs : Level ^ n} {β γ} {Xs : Sets αs β}
     -> (Xs ⋯>ⁿ Set γ) -> Fold Xs -> Set (αs ⊔ⁿ γ)
Comp  0      Z y = Z y
Comp (suc n) Z f = ∀ x -> Comp n Z (f x)

comp : ∀ n {β γ} {αs : Level ^ n} {Xs : Sets αs β} {Z : Xs ⋯>ⁿ Set γ}
     -> Πⁿ Xs Z -> (f : Fold Xs) -> Comp n Z f
comp  0      g y = g y
comp (suc n) g f = λ x -> comp n g (f x)

一个测试:

length : ∀ {α} {A : Set α} {n} -> Vec A n -> ℕ
length {n = n} _ = n

explicit-replicate : (A : Set) -> (n : ℕ) -> A -> Vec A n
explicit-replicate _ _ x = replicate x

foo : (A : Set) -> ℕ -> A -> ℕ
foo = comp 3 length explicit-replicate

test : foo Bool 5 true ≡ 5
test = refl

请注意参数中的依赖项和explicit-replicate的结果类型.此外，Set位于Set₁中，而ℕ和A位于Set中–这说明了宇宙的多态性.

Note the dependency in the arguments and the resulting type of explicit-replicate. Besides, Set lies in Set₁, while ℕ and A lie in Set — this illustrates universe polymorphism.

AFAIK，对于隐式参数没有可理解的理论，所以我不知道当第二个函数(即f)接收到隐式参数时，所有这些东西将如何工作.这项测试:

AFAIK, there is no comprehensible theory for implicit arguments, so I don't know, how all this stuff will work, when the second function (i.e. f) receives implicit arguments. This test:

foo' : ∀ {α} {A : Set α} -> ℕ -> A -> ℕ
foo' = comp 2 length (λ n -> replicate {n = n})

test' : foo' 5 true ≡ 5
test' = refl

至少通过了.

comp无法处理函数.例如

explicit-replicate' : ∀ α -> (A : Set α) -> (n : ℕ) -> A -> Vec A n
explicit-replicate' _ _ _ x = replicate x

... because this would result in an invalid use of Setω ...
error : ∀ α -> (A : Set α) -> ℕ -> A -> ℕ
error = comp 4 length explicit-replicate'

但是，这对于阿格达来说很常见，例如您不能对其本身应用明确的id

But it's common for Agda, e.g. you can't apply explicit id to itself:

idₑ : ∀ α -> (A : Set α) -> A -> A
idₑ _ _ x = x

-- ... because this would result in an invalid use of Setω ...
error = idₑ _ _ idₑ

---

代码.

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