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如何在Agda中实现Floyd的Hare和Tortoise算法?

[英]How to implement Floyd's Hare and Tortoise algorithm in Agda?

I want to translate the following Haskell code into Agda: 我想将以下Haskell代码转换为Agda:

import Control.Arrow (first)
import Control.Monad (join)

safeTail :: [a] -> [a]
safeTail []     = []
safeTail (_:xs) = xs

floyd :: [a] -> [a] -> ([a], [a])
floyd xs     []     = ([], xs)
floyd (x:xs) (_:ys) = first (x:) $ floyd xs (safeTail ys)

split :: [a] -> ([a], [a])
split = join floyd

This allows us to efficiently split a list into two: 这使我们能够有效地将列表拆分为两个:

split [1,2,3,4,5]   = ([1,2,3], [4,5])
split [1,2,3,4,5,6] = ([1,2,3], [4,5,6])

So, I tried to convert this code to Agda: 所以,我试图将此代码转换为Agda:

floyd : {A : Set} → List A → List A → List A × List A
floyd xs        []        = ([] , xs)
floyd (x :: xs) (_ :: ys) = first (x ::_) (floyd xs (safeTail ys))

The only problem is that Agda complains that I'm missing the case for floyd [] (y :: ys) . 唯一的问题是Agda抱怨我错过了floyd [] (y :: ys) However, this case should never arise. 但是,这种情况永远不应该出现。 How can I prove to Agda that this case should never arise? 我怎样才能向Agda证明这种情况永远不会出现?


Here's a visual example of how this algorithm works: 以下是此算法如何工作的可视示例:

+-------------+-------------+
|   Tortoise  |     Hare    |
+-------------+-------------+
| ^ 1 2 3 4 5 | ^ 1 2 3 4 5 |
| 1 ^ 2 3 4 5 | 1 2 ^ 3 4 5 |
| 1 2 ^ 3 4 5 | 1 2 3 4 ^ 5 |
| 1 2 3 ^ 4 5 | 1 2 3 4 5 ^ |
+-------------+-------------+

Here's another example: 这是另一个例子:

+---------------+---------------+
|    Tortoise   |      Hare     |
+---------------+---------------+
| ^ 1 2 3 4 5 6 | ^ 1 2 3 4 5 6 |
| 1 ^ 2 3 4 5 6 | 1 2 ^ 3 4 5 6 |
| 1 2 ^ 3 4 5 6 | 1 2 3 4 ^ 5 6 |
| 1 2 3 ^ 4 5 6 | 1 2 3 4 5 6 ^ |
+---------------+---------------+

When the hare (the second argument to floyd ) reaches the end of the list, the tortoise (the first argument to floyd ) reaches the middle of the list. 当野兔( floyd的第二个参数)到达列表的末尾时,乌龟( floyd的第一个参数)到达列表的中间位置。 Thus, by using two pointers (the second one moving twice as fast as the first) we can efficiently split a list into two. 因此,通过使用两个指针(第二个指针移动速度是第一个指针的两倍),我们可以有效地将列表拆分为两个。

The same thing as Twan van Laarhoven suggests in the comments but with Vec s. Twan van Laarhoven在评论中提出的建议相同,但是对于Vec His version is probably better. 他的版本可能更好。

open import Function
open import Data.Nat.Base
open import Data.Product renaming (map to pmap)
open import Data.List.Base
open import Data.Vec hiding (split)

≤-step : ∀ {m n} -> m ≤ n -> m ≤ suc n
≤-step  z≤n     = z≤n
≤-step (s≤s le) = s≤s (≤-step le)

≤-refl : ∀ {n} -> n ≤ n
≤-refl {0}     = z≤n
≤-refl {suc n} = s≤s ≤-refl

floyd : ∀ {A : Set} {n m} -> m ≤ n -> Vec A n -> Vec A m -> List A × List A
floyd  z≤n            xs       []          = [] , toList xs
floyd (s≤s  z≤n)     (x ∷ xs) (y ∷ [])     = x ∷ [] , toList xs
floyd (s≤s (s≤s le)) (x ∷ xs) (y ∷ z ∷ ys) = pmap (x ∷_) id (floyd (≤-step le) xs ys)

split : ∀ {A : Set} -> List A -> List A × List A
split xs = floyd ≤-refl (fromList xs) (fromList xs)

open import Relation.Binary.PropositionalEquality

test₁ : split (1 ∷ 2 ∷ 3 ∷ 4 ∷ 5 ∷ []) ≡ (1 ∷ 2 ∷ 3 ∷ [] , 4 ∷ 5 ∷ [])
test₁ = refl

test₂ : split (1 ∷ 2 ∷ 3 ∷ 4 ∷ 5 ∷ 6 ∷ []) ≡ (1 ∷ 2 ∷ 3 ∷ [] , 4 ∷ 5 ∷ 6 ∷ [])
test₂ = refl

Also, functions like ≤-refl and ≤-step are somewhere in the standard library, but I was lazy to search. 此外,像≤-refl≤-step这样的函数在标准库中的某处,但我懒得搜索。


Here is a silly thing I like to do: 这是我喜欢做的傻事:

open import Function
open import Data.Nat.Base
open import Data.Product renaming (map to pmap)
open import Data.List.Base
open import Data.Vec hiding (split)

floyd : ∀ {A : Set}
      -> (k : ℕ -> ℕ)
      -> (∀ {n} -> Vec A (k (suc n)) -> Vec A (suc (k n)))
      -> ∀ n
      -> Vec A (k n)
      -> List A × List A
floyd k u  0            xs = [] , toList xs
floyd k u  1            xs with u xs
... | x ∷ xs' = x ∷ [] , toList xs'
floyd k u (suc (suc n)) xs with u xs
... | x ∷ xs' = pmap (x ∷_) id (floyd (k ∘ suc) u n xs')

split : ∀ {A : Set} -> List A -> List A × List A
split xs = floyd id id (length xs) (fromList xs)

open import Relation.Binary.PropositionalEquality

test₁ : split (1 ∷ 2 ∷ 3 ∷ 4 ∷ 5 ∷ []) ≡ (1 ∷ 2 ∷ 3 ∷ [] , 4 ∷ 5 ∷ [])
test₁ = refl

test₂ : split (1 ∷ 2 ∷ 3 ∷ 4 ∷ 5 ∷ 6 ∷ []) ≡ (1 ∷ 2 ∷ 3 ∷ [] , 4 ∷ 5 ∷ 6 ∷ [])
test₂ = refl

This is partly based on the Benjamin Hodgson suggestion in the comment below. 这部分是基于下面评论中的Benjamin Hodgson建议。

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