如何在Haskell中实现B +树?

 阳光-假日 发布于 2023-02-13 16:26
  • php
  • B +树将叶节点链接在一起.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无向的,链接在一起的叶节点在图中创建循环.

    在Haskell中,如何将叶子构造为父内部节点的子节点,同时构建相邻叶节点的下一个链接.怎样才能用Haskell的代数数据类型做到这一点?似乎Haskell ADT通常使得类似循环的结构难以表达.

    1 个回答
    • 这是(不可变/"可变" - 重构/可拉伸)ADT表示(涉及不可变向量)的想法:

      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,因此不正确使用原始构造函数,访问器或模式匹配不会破坏树.

      Keys,Values,Childs长度控制可以添加(与运行时检查或可能与GADTs并且这样的).

      对于一个界面:

      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.
      
      -- ...
      

      那么这种树:

      B +树的例子

      可以建成

      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
      

      这种技术被称为"打结".叶子可循环使用:

        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)
      

      或双重链接(假设_leafPrev和相应的leaf功能):

        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)
      

      使用可变载体和可变引用也可以完全可变的表示:

      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,但使用IORefIOVector,在IOmonad 工作.

      2023-02-13 16:29 回答
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