Trung tâm cá nhân
sách gpt4 ai đã đi

haskell - 如何在Haskell中实现B+树？

In lại 作者：行者123 更新时间：2023-12-02 08:50:08 30 4
mua khóa gpt4 Nike

B+ 树的叶节点链接在一起。将 B+ 树的指针结构视为有向图，它不是循环的。但是忽略指针的方向并将其视为链接在一起的无向叶节点会在图中创建循环。

在 Haskell 中，如何将叶子构造为父内部节点的子节点，并同时构造来自相邻叶子节点的下一个链接。如何使用 Haskell 的代数数据类型做到这一点？看来 Haskell ADT 总体上使得循环类结构难以表达。

1 Câu trả lời

这是(不可变/“可变”重建/zipperable)ADT表示的一个想法(涉及不可变vectors):

module Data.BTree.Internal where

import Data.Vector

type Values v = Vector v

type Keys k = Vector k

data Leaf k v
  = Leaf
    { _leafKeys :: !(Keys k)
    , _leafValues :: !(Values v)
    , _leafNext :: !(Maybe (Leaf k v)) -- @Maybe@ is lazy in @Just@, so this strict mark
                                         -- is ok for tying-the-knot stuff.
    -- , _leafPrev :: !(Maybe (Leaf k v))
    -- ^ for doubly-linked lists of leaves
    }

type Childs k v = Vector (BTree k v)

data Node k v
  = Node
    { _nodeKeys :: !(Keys k)
    , _nodeChilds :: !(Childs k v)
    }

data BTree k v
  = BTreeNode !(Node k v)
  | BTreeLeaf !(Leaf k v)

newtype BTreeRoot k v
  = BTreeRoot (BTree k v)

  • 这应该是内部的，这样原始构造函数、访问器或模式匹配的不当使用就不会破坏树。

    • 可以添加

  • Keys,Values,Childs长度控制(通过运行时检查或可能通过GADT等) .

对于界面:

module Data.BTree ( {- appropriate exports -} ) where

import Data.Vector
import Data.BTree.Internal

-- * Building trees: "good" constructors.

keys :: [k] -> Keys k
keys = fromList

values :: [v] -> Values v
values = fromList

leaves :: [Leaf k v] -> Childs k v
leaves = fromList . fmap BTreeLeaf

leaf :: Keys k -> Values v -> Maybe (Leaf k v) -> Leaf k v
-- or
-- leaf :: Keys k -> Values v -> Maybe (Leaf k v) -> Maybe (Leaf k v) -> Leaf k v
-- for doubly-linked lists of leaves
leaf = Leaf

node :: Keys k -> Childs k v -> BTree k v
node ks = BTreeNode . Node ks

-- ...

-- * "Good" accessors.

-- ...

-- * Basic functions: insert, lookup, etc.

-- ...

然后是这种树:

B+tree example

可以构建为

test :: BTree Int ByteString
test = let
  root = node (keys [3, 5]) (leaves [leaf1, leaf2, leaf3])
  leaf1 = leaf (keys [1, 2]) (values ["d1", "d2"]) (Just leaf2)
  leaf2 = leaf (keys [3, 4]) (values ["d3", "d4"]) (Just leaf3)
  leaf3 = leaf (keys [5, 6, 7]) (values ["d5", "d6", "d7"]) Nothing
  in root

该技术称为 "tying the knot" 。叶子可以循环:

  leaf1 = leaf (keys [1, 2]) (values ["d1", "d2"]) (Just leaf2)
  leaf2 = leaf (keys [3, 4]) (values ["d3", "d4"]) (Just leaf3)
  leaf3 = leaf (keys [5, 6, 7]) (values ["d5", "d6", "d7"]) (Just leaf1)

或双向链接(假设_leafPrev和相应的leaf函数):

  leaf1 = leaf (keys [1, 2]) (values ["d1", "d2"]) (Just leaf2) (Just leaf3)
  leaf2 = leaf (keys [3, 4]) (values ["d3", "d4"]) (Just leaf3) (Just leaf1)
  leaf3 = leaf (keys [5, 6, 7]) (values ["d5", "d6", "d7"]) (Just leaf1) (Just leaf2)
<小时>

sử dụng mutable vectors 也可以实现完全可变的表示和 mutable references :

type Values v = IOVector v

type Keys k = IOVector k

type Childs k v = IOVector (BTree k v)

    , _leafNext :: !(IORef (Maybe (Leaf k v)))

等等，基本相同，但是使用IORefVàIOVector，在IO monad中工作。

关于haskell - 如何在Haskell中实现B+树？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20309501/

30 4 0
Bài viết được đề xuất: android - 指定 android.support.test.runner.AndroidJUnitRunner 时，“transformClassesWithMultidexlistForDebugAndroidTest”任务失败
Bài viết được đề xuất: spring - 如何创建一个使用特定配置文件执行 bootRun 的 gradle 任务？
Bài viết được đề xuất: vb6 - 如何引用 EXE 项目中的类？
Bài viết được đề xuất: Oracle 神秘的 Unicode 代码点
行者123
Hồ sơ cá nhân

Tôi là một lập trình viên xuất sắc, rất giỏi!

Nhận phiếu giảm giá Didi Taxi miễn phí
Mã giảm giá Didi Taxi
Các bài viết nóng hổi trên toàn bộ trang web
Giấy chứng nhận ICP Bắc Kinh số 000000
Hợp tác quảng cáo: 1813099741@qq.com 6ren.com