有没有办法遍历可变树以获得随机节点? [英] Is there a way to iterate over a mutable tree to get a random node?

问题描述

我正在尝试更新树结构的节点.随机选择要更新的节点.要使用 Reservoir Sampling 算法对树中的节点进行采样，我必须遍历这些节点，因此我尝试为我的 Node 枚举创建一个 Iterator.>

问题是，一方面，我必须在堆栈或队列中存储子节点的引用，但另一方面，我必须返回父节点的可变引用.Rust 不允许为一个值创建多个可变引用，也不允许将不可变引用转换为可变引用.

有没有办法遍历可变树?或者是否有另一种方法可以随机获取对树中节点的可变引用?

这是我的代码.

#![feature(box_syntax, box_patterns)]外部板条箱兰特；//简单的二叉树结构#[派生(调试)]枚举节点{叶(u8),分支(框<节点>，框<节点>)，}实现节点{fn iter_mut(&mut self) ->迭代器{迭代器{堆栈: vec![self],}}fn pick_random_node_mut<'a>(&'a mut self) ->&'一个mut节点{//Revervoir采样让 rng = &mut rand::thread_rng();rand::seq::sample_iter(rng, self.iter_mut(), 1).ok().and_then(|mut v| v.pop()).unwrap()}}//`Node` 的迭代器struct IterMut<'a>{堆栈:Vec<&'a mut Node>，}impl <'a>IterMut<'a>的迭代器{type Item = &'a mut Node;fn next(&mut self) ->选项<&'a mut Node>{让节点 = self.stack.pop()?;//我被困在这里:不能一次多次借用 `*node` 作为可变的if let &mut Node::Branch(box ref mut a, box ref mut b) = node {self.stack.push(b);self.stack.push(a);}一些(节点)}}fn 主(){使用节点::*;let mut tree: Node = Branch(box Leaf(1), box Leaf(2));println!("{:?}", 树);{让节点:&mut Node = tree.pick_random_node_mut();*节点=叶(3)；}println!("{:?}", 树);}

解决方案

不，编写对树节点的可变引用的迭代器是不安全的.

假设我们有这个树结构:

 +-++----+ +----+|+-+ |||||+--v-+ +--v--+|50 ||100 |+----+ +-----+

如果存在这样的迭代器，我们可以这样称呼它:

let mut all_nodes: Vec<&mut Node>= tree.iter_mut().collect();

假设父节点在索引 0 中结束，左节点在索引 1 中，右节点在索引 2 中.

let (head, tail) = all_nodes.split_at_mut(1);让 x = 匹配 &mut head[0] {Branch(ref mut l, _) =>升，叶(_) =>无法访问！(),};让 y = &mut tail[1];

现在 x 和 y 是彼此可变的别名.我们在完全安全的代码中违反了基本的 Rust 要求.这就是为什么这样的迭代器是不可能的.

<小时>

您可以实现对树中值的可变引用的迭代器:

impl<'a>IterMut<'a>的迭代器{type Item = &'a mut u8;fn next(&mut self) ->选项<Self::Item>{环形 {让节点 = self.stack.pop()?;匹配节点{节点::分支(a, b) =>{self.stack.push(b);self.stack.push(a);}节点::叶(l) =>返回一些(l)，}}}}

这是安全的，因为没有办法从一个可变引用到另一个值.然后，您可以在此基础上构建您的随机选择:

<代码>{让rando = 匹配rand::seq::sample_iter(&mut rand::thread_rng(), tree.iter_mut(), 1) {Ok(mut v) =>v.pop().unwrap(),错误(_)=>恐慌！(没有足够的元素")，};*随机+= 1;}

I am trying to update a node of a tree structure. A node which is to be updated is selected randomly. To sample a node in the tree using the Reservoir Sampling algorithm, I have to iterate over the nodes, so I have tried to make an Iterator for my Node enum.

The problem is that, on the one hand, I have to store references for child nodes in a stack or queue, however on the other hand, I have to return a mutable reference for a parent node. Rust does not allow to make multiple mutable references for one value, neither to convert an immutable reference into a mutable reference.

Is there a way to iterate over a mutable tree? Or is there another approach to randomly get a mutable reference to a node in a tree?

Here is my code.

#![feature(box_syntax, box_patterns)]
extern crate rand;

// Simple binary tree structure
#[derive(Debug)]
enum Node {
    Leaf(u8),
    Branch(Box<Node>, Box<Node>),
}

impl Node {
    fn iter_mut(&mut self) -> IterMut {
        IterMut {
            stack: vec![self],
        }
    }

    fn pick_random_node_mut<'a>(&'a mut self) -> &'a mut Node {
        // Revervoir sampling
        let rng = &mut rand::thread_rng();
        rand::seq::sample_iter(rng, self.iter_mut(), 1)
            .ok().and_then(|mut v| v.pop()).unwrap()
    }
}

// An iterator for `Node`
struct IterMut<'a> {
    stack: Vec<&'a mut Node>,
}

impl <'a> Iterator for IterMut<'a> {
    type Item = &'a mut Node;

    fn next(&mut self) -> Option<&'a mut Node> {
        let node = self.stack.pop()?;

        // I am stucking here: cannot borrow `*node` as mutable more than once at a time
        if let &mut Node::Branch(box ref mut a, box ref mut b) = node {
            self.stack.push(b);
            self.stack.push(a);
        }
        Some(node)
    }
}

fn main() {
    use Node::*;

    let mut tree: Node = Branch(box Leaf(1), box Leaf(2));
    println!("{:?}", tree);

    {
        let node: &mut Node = tree.pick_random_node_mut();
        *node = Leaf(3);
    }
    println!("{:?}", tree);

}

解决方案

No, it is not safe to write an iterator of the mutable references to the nodes of a tree.

Assume we have this tree structure:

         +-+
    +----+ +----+
    |    +-+    |
    |           |
    |           |
 +--v-+      +--v--+
 | 50 |      | 100 |
 +----+      +-----+

If such an iterator existed, we could call it like this:

let mut all_nodes: Vec<&mut Node> = tree.iter_mut().collect();

Assume that the parent node ends up in index 0, the left node in index 1, and the right node in index 2.

let (head, tail) = all_nodes.split_at_mut(1);

let x = match &mut head[0] {
    Branch(ref mut l, _) => l,
    Leaf(_) => unreachable!(),
};

let y = &mut tail[1];

Now x and y are mutable aliases to each other. We have violated a fundamental Rust requirement in completely safe code. That's why such an iterator is not possible.

---

You could implement an iterator of mutable references to the values in the tree:

impl<'a> Iterator for IterMut<'a> {
    type Item = &'a mut u8;

    fn next(&mut self) -> Option<Self::Item> {
        loop {
            let node = self.stack.pop()?;

            match node {
                Node::Branch(a, b) => {
                    self.stack.push(b);
                    self.stack.push(a);
                }
                Node::Leaf(l) => return Some(l),
            }
        }
    }
}

This is safe because there's no way to go from one mutable reference to a value to another one. You can then build your random selection on top of that:

{
    let rando = match rand::seq::sample_iter(&mut rand::thread_rng(), tree.iter_mut(), 1) {
        Ok(mut v) => v.pop().unwrap(),
        Err(_) => panic!("Not enough elements"),
    };

    *rando += 1;
}

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