问题描述
我们可以像这样枚举列表的元素:
We can enumerate the elements of a list like this:
-- enumerate-ℕ = zip [0..]
enumerate-ℕ : ∀ {α} {A : Set α} -> List A -> List (ℕ × A)
enumerate-ℕ = go 0 where
go : ∀ {α} {A : Set α} -> ℕ -> List A -> List (ℕ × A)
go n [] = []
go n (x ∷ xs) = (n , x) ∷ go (ℕ.suc n) xs
例如enumerate-ℕ (1 ∷ 3 ∷ 2 ∷ 5 ∷ []) 等于 (0 , 1) ∷ (1 , 3) ∷ (2 , 2) ∷ (3 , 5) ∷ [].假设 Agda 中有共享，则函数是线性的.
E.g. enumerate-ℕ (1 ∷ 3 ∷ 2 ∷ 5 ∷ []) is equal to (0 , 1) ∷ (1 , 3) ∷ (2 , 2) ∷ (3 , 5) ∷ []. Assuming there is sharing in Agda, the function is linear.
然而，如果我们尝试通过 Fins 而不是 ℕs 来枚举列表的元素，函数就会变成二次的:
However if we try to enumerate the elements of a list by Fins rather than ℕs, the function becomes quadratic:
enumerate-Fin : ∀ {α} {A : Set α} -> (xs : List A) -> List (Fin (length xs) × A)
enumerate-Fin [] = []
enumerate-Fin (x ∷ xs) = (zero , x) ∷ map (pmap suc id) (enumerate-Fin xs)
是否可以在线性时间内通过Fins进行枚举?
Is it possible to enumerate by Fins in linear time?
推荐答案
将此视为第一次尝试:
go : ∀ {m α} {A : Set α} -> Fin m -> (xs : List A) -> List (Fin (m + length xs) × A)
go i [] = []
go i (x ∷ xs) = (inject+ _ i , x) ∷ {!go (suc i) xs!}
i 在每次递归调用时都会增长，但存在不匹配:
i grows at each recursive call as it should, but there is a mismatch:
目标类型为List(Fin(.m + suc(length xs)) × .A)
孔内的表达式类型为List (Fin (suc (.m + length xs)) × .A)
证明两种类型相等很容易，但也很脏.这是一个常见的问题:一个参数变大而另一个变小，所以我们需要定义可交换的 _+_ 来处理这两种情况，但没有办法定义它.解决方案是使用 CPS:
It's easy to prove that two types are equal, but it's also dirty. This is a common problem: one argument grows and the other lowers, so we need definitionally commutative _+_ to handle both the cases, but there is no way to define it. The solution is to use CPS:
go : ∀ {α} {A : Set α} -> (k : ℕ -> ℕ) -> (xs : List A) -> List (Fin (k (length xs)) × A)
go k [] = []
go k (x ∷ xs) = ({!!} , x) ∷ go (k ∘ suc) xs
(k ∘ suc) (length xs) 与 k (length (x ∷ xs)) 相同，因此不匹配是固定的，但是什么是i 现在?洞的类型是Fin(k(suc(length xs)))，在当前上下文中是无人居住的，所以介绍一些居民:
(k ∘ suc) (length xs) is the same thing as k (length (x ∷ xs)), hence the mismatch is fixed, but what is i now? The type of the hole is Fin (k (suc (length xs))) and it's uninhabited in the current context, so let's introduce some inhabitant:
go : ∀ {α} {A : Set α}
-> (k : ℕ -> ℕ)
-> (∀ {n} -> Fin (k (suc n)))
-> (xs : List A)
-> List (Fin (k (length xs)) × A)
go k i [] = []
go k i (x ∷ xs) = (i , x) ∷ go (k ∘ suc) {!!} xs
新孔的类型为{n:ℕ}→Fin(k(suc(suc n))).我们可以用 i 填充它，但是 i 必须在每次递归调用时增长.但是 suc 和 k 不通勤，所以 suc i 是 Fin (suc (k (suc (_n_65 kix xs))))).所以我们在k下添加了suc的另一个参数，最终定义为
The type of the new hole is {n : ℕ} → Fin (k (suc (suc n))). We can fill it with i, but i must grow at each recursive call. However suc and k doesn't commute, so suc i is Fin (suc (k (suc (_n_65 k i x xs)))). So we add another argument that sucs under k, and the final definition is
enumerate-Fin : ∀ {α} {A : Set α} -> (xs : List A) -> List (Fin (length xs) × A)
enumerate-Fin = go id suc zero where
go : ∀ {α} {A : Set α}
-> (k : ℕ -> ℕ)
-> (∀ {n} -> Fin (k n) -> Fin (k (suc n)))
-> (∀ {n} -> Fin (k (suc n)))
-> (xs : List A)
-> List (Fin (k (length xs)) × A)
go k s i [] = []
go k s i (x ∷ xs) = (i , x) ∷ go (k ∘ suc) s (s i) xs
有效，因为 s : {n : ℕ} → Fin (kn) → Fin (k (suc n)) 可以被视为 {n : ℕ} → Fin (k (suc n)) → Fin (k (suc (suc n))).
which works, because s : {n : ℕ} → Fin (k n) → Fin (k (suc n)) can be treated as {n : ℕ} → Fin (k (suc n)) → Fin (k (suc (suc n))).
一个测试:C-c C-n enumerate-Fin(1 ∷ 3 ∷ 2 ∷ 5 ∷ [])给出
(zero , 1) ∷
(suc zero , 3) ∷
(suc (suc zero) , 2) ∷ (suc (suc (suc zero)) , 5) ∷ []
现在请注意，在 enumerate-Fin 中，k 在类型中总是跟在 Fin 之后.因此，我们可以抽象 Fin ∘ k 并得到一个通用版本的函数，该函数适用于 ℕ 和 Fin:
Now note that in enumerate-Fin k always follows Fin in the types. Hence we can abstract Fin ∘ k and get a generic version of the function that works with both ℕ and Fin:
genumerate : ∀ {α β} {A : Set α}
-> (B : ℕ -> Set β)
-> (∀ {n} -> B n -> B (suc n))
-> (∀ {n} -> B (suc n))
-> (xs : List A)
-> List (B (length xs) × A)
genumerate B s i [] = []
genumerate B s i (x ∷ xs) = (i , x) ∷ genumerate (B ∘ suc) s (s i) xs
enumerate-ℕ : ∀ {α} {A : Set α} -> List A -> List (ℕ × A)
enumerate-ℕ = genumerate _ suc 0
enumerate-Fin : ∀ {α} {A : Set α} -> (xs : List A) -> List (Fin (length xs) × A)
enumerate-Fin = genumerate Fin suc zero
这篇关于如何在线性时间内通过`Fin`s枚举列表的元素?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！