Question

我一直在努力重现Christiansen，Danilenko和Dylus在All Sorts of Permutations (Functional Pearl)中提及的一篇文章，即2016年ICFP即将发表的论文。第8节（“最终评论”）声称通过选择一个特定的非确定性谓词，monadic合并排序可以按字典顺序产生序列的所有排列。

我们只考虑非确定性谓词 coinCmp ，而有其他非确定性谓词可用于影响枚举的顺序。例如，以下函数将谓词 cmp 提升为非确定性上下文。
liftCmp :: MonadPlus μ
        ⇒ (α → α → Bool) → Cmp α μ
liftCmp p x y = return (p x y) ⊕ return (not (p x y))
当我们使用这个函数解除比较函数并将其传递给monadic版本的合并排序时，我们得到一种特殊的排列函数：它按字典顺序枚举排列。

我很确定我在这里写的是合并排序，但是在运行时订购并不像宣传的那样。

import Control.Applicative (Alternative((<|>)))
import Control.Monad (MonadPlus, join)
import Data.Functor.Identity (Identity)

-- Comparison in a context
type Comparison a m = a -> a -> m Bool

-- Ordering lifted into the Boring Monad
boringCmp :: (a -> a -> Bool) -> Comparison a Identity
boringCmp p x y = return (p x y)

-- Arbitrary ordering in a non-deterministic context
cmp :: MonadPlus m => Comparison a m
cmp _ _ = return True <|> return False

-- Ordering lifted into a non-deterministic context
liftCmp :: MonadPlus m => (a -> a -> Bool) -> Comparison a m
liftCmp p x y = let b = p x y in return b <|> return (not b)

mergeM :: Monad m => Comparison a m -> [a] -> [a] -> m [a]
mergeM _ ls         []         = return ls
mergeM _ []         rs         = return rs
mergeM p lls@(l:ls) rrs@(r:rs) = do
    b <- p l r
    if b
    then (l:) <$> mergeM p ls rrs
    else (r:) <$> mergeM p lls rs

mergeSortM :: Monad m => Comparison a m -> [a] -> m [a]
mergeSortM _ []  = return []
mergeSortM _ [x] = return [x]
mergeSortM p xs  = do
    let (ls, rs) = deinterleave xs
    join $ mergeM p <$> mergeSortM p ls <*> mergeSortM p rs
  where
    deinterleave :: [a] -> ([a], [a])
    deinterleave [] = ([], [])
    deinterleave [l] = ([l], [])
    deinterleave (l:r:xs) = case deinterleave xs of (ls, rs) -> (l:ls, r:rs)

λ mergeSortM (boringCmp (<=)) [2,1,3] :: Identity [Int]
Identity [1,2,3]

λ mergeSortM cmp [2,1,3] :: [[Int]]
[[2,3,1],[2,1,3],[1,2,3],[3,2,1],[3,1,2],[1,3,2]]

λ mergeSortM (liftCmp (<=)) [2,1,3] :: [[Int]]
[[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]]

实际的词典排序供参考 -

λ sort it
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Answer 1

让我们尝试一下deinterleave的变体，它会分割列表的前半部分和后半部分，而不是像发布的代码那样拆分偶数和奇数索引的元素：

deinterleave :: [a] -> ([a], [a])
deinterleave ys = splitAt (length ys `div` 2) ys

结果：

> mergeSortM (liftCmp (<=)) [2,1,3] :: [[Int]]
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

不幸的是，正如我最初希望的那样，这并没有解决问题，正如Rowan Blush在下面指出的那样。： - /

非确定性合并排序不按字典顺序排列排列

1 个答案: