递归方案允许递归调用之间的依赖关系(有序的变形?)
Recursion scheme allowing dependencies between recursive calls (an ordered catamorphism?)
我对编写递归代码的高阶方法(递归方案)感兴趣,其中递归调用之间可能存在依赖关系。
作为一个简化示例,考虑一个遍历整数树的函数,检查总和是否小于某个值。我们可以对整棵树求和并将其与最大值进行比较。或者,我们可以保持 运行 总和并在超过最大值后立即短路:
data Tree = Leaf Nat | Node Tree Tree
sumLT :: Nat -> Tree -> Bool
sumLT max t = sumLT' max t > 0
sumLT' :: Nat -> Tree -> Int
sumLT' max (Leaf n) = max - n
sumLT' max (Node l r) =
let max' = sumLT' max l
in if max' > 0 then sumLT' max' r else 0
有没有办法用递归方案来表达这个想法——本质上是有序遍历?我对尽可能一般地编写这样的有序遍历很感兴趣。
理想情况下,我想要某种方式来编写遍历,其中在数据结构上折叠(或展开)的函数决定遍历的顺序。无论我最终得到什么抽象,我都希望能够编写上面 sumLT'
遍历的逆序版本,我们从右到左进行遍历:
sumLT'' :: Nat -> Tree -> Int
sumLT'' max (Leaf n) = max - n
sumLT'' max (Node l r) =
let max' = sumLT'' max r
in if max' > 0 then sumLT'' max' l else 0
像往常一样,折叠成内函数给你一个处理顺序/状态传递的概念:
import Numeric.Natural
data Tree = Leaf Natural | Node Tree Tree
cata :: (Natural -> r) -> (r -> r -> r) -> Tree -> r
cata l n (Leaf a) = l a
cata l n (Node lt rt) = n (cata l n lt) (cata l n rt)
sumLT :: Natural -> Tree -> Bool
sumLT max t = cata (\ a max -> max - a) -- left-to-right
(\ l r max -> let max' = l max in
if max' > 0 then r max' else 0)
t max > 0
sumLT' :: Natural -> Tree -> Bool
sumLT' max t = cata (\ a max -> max - a) -- right-to-left
(\ l r max -> let max' = r max in
if max' > 0 then l max' else 0)
t max > 0
尝试一下:
> sumLT 11 (Node (Leaf 10) (Leaf 0))
True
> sumLT 11 (Node (Leaf 10) (Leaf 1))
False
> sumLT 11 (Node (Leaf 10) (Leaf undefined))
*** Exception: Prelude.undefined
> sumLT 11 (Node (Leaf 11) (Leaf undefined))
False
> sumLT 11 (Node (Leaf 10) (Node (Leaf 1) (Leaf undefined)))
False
> sumLT' 11 (Node (Leaf undefined) (Leaf 11))
False
和往常一样,Haskell的懒惰提供了短路/提前退出的能力。从示例中可以看出,如果 cata's
第二个参数(节点折叠函数)不需要其参数之一的值,则实际上根本不会访问相应的分支。
我会利用 Haskell 的懒惰来发挥我的优势。将树转为列表(这是一个变形),创建部分和,并找到第一个大于限制的。
{-# language DeriveFoldable #-}
module ShortSum where
import Data.Foldable
data Tree a = Leaf a | Node (Tree a) (Tree a)
deriving Foldable
type Nat = Int
type TreeN = Tree Nat
sumLt :: Nat -> TreeN -> Bool
sumLt mx = any (> mx) . scanl1 (+) . toList
我对编写递归代码的高阶方法(递归方案)感兴趣,其中递归调用之间可能存在依赖关系。
作为一个简化示例,考虑一个遍历整数树的函数,检查总和是否小于某个值。我们可以对整棵树求和并将其与最大值进行比较。或者,我们可以保持 运行 总和并在超过最大值后立即短路:
data Tree = Leaf Nat | Node Tree Tree
sumLT :: Nat -> Tree -> Bool
sumLT max t = sumLT' max t > 0
sumLT' :: Nat -> Tree -> Int
sumLT' max (Leaf n) = max - n
sumLT' max (Node l r) =
let max' = sumLT' max l
in if max' > 0 then sumLT' max' r else 0
有没有办法用递归方案来表达这个想法——本质上是有序遍历?我对尽可能一般地编写这样的有序遍历很感兴趣。
理想情况下,我想要某种方式来编写遍历,其中在数据结构上折叠(或展开)的函数决定遍历的顺序。无论我最终得到什么抽象,我都希望能够编写上面 sumLT'
遍历的逆序版本,我们从右到左进行遍历:
sumLT'' :: Nat -> Tree -> Int
sumLT'' max (Leaf n) = max - n
sumLT'' max (Node l r) =
let max' = sumLT'' max r
in if max' > 0 then sumLT'' max' l else 0
像往常一样,折叠成内函数给你一个处理顺序/状态传递的概念:
import Numeric.Natural
data Tree = Leaf Natural | Node Tree Tree
cata :: (Natural -> r) -> (r -> r -> r) -> Tree -> r
cata l n (Leaf a) = l a
cata l n (Node lt rt) = n (cata l n lt) (cata l n rt)
sumLT :: Natural -> Tree -> Bool
sumLT max t = cata (\ a max -> max - a) -- left-to-right
(\ l r max -> let max' = l max in
if max' > 0 then r max' else 0)
t max > 0
sumLT' :: Natural -> Tree -> Bool
sumLT' max t = cata (\ a max -> max - a) -- right-to-left
(\ l r max -> let max' = r max in
if max' > 0 then l max' else 0)
t max > 0
尝试一下:
> sumLT 11 (Node (Leaf 10) (Leaf 0))
True
> sumLT 11 (Node (Leaf 10) (Leaf 1))
False
> sumLT 11 (Node (Leaf 10) (Leaf undefined))
*** Exception: Prelude.undefined
> sumLT 11 (Node (Leaf 11) (Leaf undefined))
False
> sumLT 11 (Node (Leaf 10) (Node (Leaf 1) (Leaf undefined)))
False
> sumLT' 11 (Node (Leaf undefined) (Leaf 11))
False
和往常一样,Haskell的懒惰提供了短路/提前退出的能力。从示例中可以看出,如果 cata's
第二个参数(节点折叠函数)不需要其参数之一的值,则实际上根本不会访问相应的分支。
我会利用 Haskell 的懒惰来发挥我的优势。将树转为列表(这是一个变形),创建部分和,并找到第一个大于限制的。
{-# language DeriveFoldable #-}
module ShortSum where
import Data.Foldable
data Tree a = Leaf a | Node (Tree a) (Tree a)
deriving Foldable
type Nat = Int
type TreeN = Tree Nat
sumLt :: Nat -> TreeN -> Bool
sumLt mx = any (> mx) . scanl1 (+) . toList