实时持久队列总数 [英] Total real-time persistent queues

问题描述

Okasaki 描述了可以在 Haskell 中使用类型实现的持久实时队列

Okasaki describes persistent real-time queues which can be realized in Haskell using the type

data Queue a = forall x . Queue
  { front :: [a]
  , rear :: [a]
  , schedule :: [x]
  }

增量旋转保持不变

length schedule = length front - length rear

更多详情

如果您熟悉所涉及的队列，则可以跳过此部分.

More details

If you're familiar with the queues involved, you can skip this section.

旋转函数的样子

rotate :: [a] -> [a] -> [a] -> [a]
rotate [] (y : _) a = y : a
rotate (x : xs) (y : ys) a =
  x : rotate xs ys (y : a)

它被智能构造函数调用

exec :: [a] -> [a] -> [x] -> Queue a
exec f r (_ : s) = Queue f r s
exec f r [] = Queue f' [] f' where
  f' = rotate f r []

在每个队列操作之后.当 length s = length f - length r + 1 时总是调用智能构造函数，确保 rotate 中的模式匹配成功.

after each queue operation. The smart constructor is always called when length s = length f - length r + 1, ensuring that the pattern match in rotate will succeed.

我讨厌偏函数！我很想找到一种方法来表达类型中的结构不变量.通常的依赖向量似乎是一个可能的选择:

I hate partial functions! I'd love to find a way to express the structural invariant in the types. The usual dependent vector seems a likely choice:

data Nat = Z | S Nat

data Vec n a where
  Nil :: Vec 'Z a
  Cons :: a -> Vec n a -> Vec ('S n) a

然后(也许)

data Queue a = forall x rl sl . Queue
  { front :: Vec (sl :+ rl) a
  , rear :: Vec rl a
  , schedule :: Vec sl x
  }

问题是我无法弄清楚如何处理这些类型.似乎极有可能需要 一些 数量的 unsafeCoerce 来提高效率.但是，我还没有想出一种甚至可以勉强管理的方法.是否有可能在 Haskell 中很好地做到这一点?

The trouble is that I haven't been able to figure out how to juggle the types. It seems extremely likely that some amount of unsafeCoerce will be needed to make this efficient. However, I haven't been able to come up with an approach that's even vaguely manageable. Is it possible to do this nicely in Haskell?

推荐答案

这是我得到的:

open import Function
open import Data.Nat.Base
open import Data.Vec

grotate : ∀ {n m} {A : Set}
        -> (B : ℕ -> Set)
        -> (∀ {n} -> A -> B n -> B (suc n))
        -> Vec A n
        -> Vec A (suc n + m)
        -> B m
        -> B (suc n + m)
grotate B cons  []      (y ∷ ys) a = cons y a
grotate B cons (x ∷ xs) (y ∷ ys) a = grotate (B ∘ suc) cons xs ys (cons y a)

rotate : ∀ {n m} {A : Set} -> Vec A n -> Vec A (suc n + m) -> Vec A m -> Vec A (suc n + m)
rotate = grotate (Vec _) _∷_

record Queue (A : Set) : Set₁ where
  constructor queue
  field
    {X}      : Set
    {n m}    : ℕ
    front    : Vec A (n + m)
    rear     : Vec A m
    schedule : Vec X n

open import Relation.Binary.PropositionalEquality
open import Data.Nat.Properties.Simple

exec : ∀ {m n A} -> Vec A (n + m) -> Vec A (suc m) -> Vec A n -> Queue A
exec {m} {suc n} f r (_ ∷ s) = queue (subst (Vec _) (sym (+-suc n m)) f) r s
exec {m}         f r  []     = queue (with-zero f') [] f' where
 with-zero    = subst (Vec _ ∘ suc) (sym (+-right-identity m))
 without-zero = subst (Vec _ ∘ suc) (+-right-identity m)

 f' = without-zero (rotate f (with-zero r) [])

rotate 是根据 grotate 定义的，原因与 reverse 是根据 foldl (或 enumerategenumerate):因为 Vec A (suc n + m) 不是定义上的 Vec A (n + suc m)，而(B ∘ suc) m 定义为 B (suc m).

rotate is defined in terms of grotate for the same reason reverse is defined in terms of foldl (or enumerate in terms of genumerate): because Vec A (suc n + m) is not definitionally Vec A (n + suc m), while (B ∘ suc) m is definitionally B (suc m).

exec 与您提供的实现相同(以那些 subst 为模)，但我不确定类型: r 是否可以 必须不为空?

exec has the same implementation as you provided (modulo those substs), but I'm not sure about the types: is it OK that r must be non-empty?

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