哈密​​顿路径

Question

我要实现的是使用backtracking algorithm 的Hamiltonian path finder。

注意： 虽然我设法找到的大多数数学论文和 SO 主题要么与回答“是否哈密顿 路径/循环'存在或致力于寻找哈密顿循环，我的问题 略有不同 - 我需要找出顶点的顺序， 哈密​​顿量路径经过

对于示例图（未使用的边为黑色，顶点索引为从 0 开始）：

我已经建立了以下表示的邻接矩阵（顶点索引从 1 开始）：

const vertexes = [ { vertex: 1, peers: [ 3, 8, 15 ] },
  { vertex: 2, peers: [ 7, 14, 23 ] },
  { vertex: 3, peers: [ 1, 6, 13, 22 ] },
  { vertex: 4, peers: [ 5, 12, 21 ] },
  { vertex: 5, peers: [ 4, 11, 20 ] },
  { vertex: 6, peers: [ 3, 10, 19 ] },
  { vertex: 7, peers: [ 2, 9, 18 ] },
  { vertex: 8, peers: [ 1, 17 ] },
  { vertex: 9, peers: [ 7, 16 ] },
  { vertex: 10, peers: [ 6, 15 ] },
  { vertex: 11, peers: [ 5, 14 ] },
  { vertex: 12, peers: [ 4, 13 ] },
  { vertex: 13, peers: [ 3, 12, 23 ] },
  { vertex: 14, peers: [ 2, 11, 22 ] },
  { vertex: 15, peers: [ 1, 10, 21 ] },
  { vertex: 16, peers: [ 9, 20 ] },
  { vertex: 17, peers: [ 8, 19 ] },
  { vertex: 18, peers: [ 7 ] },
  { vertex: 19, peers: [ 6, 17 ] },
  { vertex: 20, peers: [ 5, 16 ] },
  { vertex: 21, peers: [ 4, 15 ] },
  { vertex: 22, peers: [ 3, 14 ] },
  { vertex: 23, peers: [ 2, 13 ] } ]

接下来，从作为根的顶点 18 和包含单个路径 [[18]] 的 paths 数组开始，我尝试改变该数组，将其替换为包含原始路由（如果不是死路）的临时副本向前走，直到没有找到长度为n（顶点总数）的潜在路线或路径：

while(paths.length>0){
    let tempPath = [];
    for(path of paths){
     const nextSteps = vertexes.find(({vertex}) => vertex == path[path.length-1]).peers.filter(v => !path.includes(v));
     if(!nextSteps.length) continue;
     else if(path.length == n-1) return [...path, nextSteps[0]];
     else nextSteps.forEach(step => tempPath.push([...path,step]));
    }
    paths = tempPath;
}

所以，问题是上面的代码永远不会退出循环，也没有返回所需的输出 ([18, 7, 9, 16, 20, 5, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 12, 13, 23, 2])。

您对上述代码为何失败以及如何修复它以返回预期输出的想法将不胜感激。

编辑：感谢@DavidSampson 和@trincot 指出我的拼写错误，我的代码现在可以工作了：

while(paths.length>0){
    let tempPath = [];
    for(let path of paths){
     const nextSteps = vertexes.find(({vertex}) => vertex == path[path.length-1]).peers.filter(v => !path.includes(v));
     if(!nextSteps.length) continue;
     else if(path.length == n-1) return [...path, nextSteps[0]];
     else tempPath.push(...nextSteps.map(v => [...path,v]));
    }
    paths = tempPath;
}

现在性能是我最关心的问题，有什么建议可以提高吗？

Answer 1

主要问题是拼写错误，如果您使用严格模式，则可以避免：

变化：

path = tempPath;

到：

paths = tempPath;

同时声明你的变量。喜欢这里：

for (let path of paths) {
//   ^^^^

然后它对我有用：

"use strict";

function hamiltonian(vertexes, start) {
    let n = vertexes.length;
    let paths = [[start]];  
    while(paths.length>0) {
        let tempPath = [];
        for(let path of paths){
            const nextSteps = vertexes.find(({vertex}) => vertex == path[path.length-1]).peers.filter(v => !path.includes(v));
            if(!nextSteps.length) continue;
            else if(path.length == n-1) return [...path, nextSteps[0]];
            else nextSteps.forEach(step => tempPath.push([...path,step]));
        }
        paths = tempPath;
    }
}

const vertexes = [ { vertex: 1, peers: [ 3, 8, 15 ] },{ vertex: 2, peers: [ 7, 14, 23 ] },{ vertex: 3, peers: [ 1, 6, 13, 22 ] },{ vertex: 4, peers: [ 5, 12, 21 ] },{ vertex: 5, peers: [ 4, 11, 20 ] },{ vertex: 6, peers: [ 3, 10, 19 ] },{ vertex: 7, peers: [ 2, 9, 18 ] },{ vertex: 8, peers: [ 1, 17 ] },{ vertex: 9, peers: [ 7, 16 ] },{ vertex: 10, peers: [ 6, 15 ] },{ vertex: 11, peers: [ 5, 14 ] },{ vertex: 12, peers: [ 4, 13 ] },{ vertex: 13, peers: [ 3, 12, 23 ] }, { vertex: 14, peers: [ 2, 11, 22 ] },{ vertex: 15, peers: [ 1, 10, 21 ] },{ vertex: 16, peers: [ 9, 20 ] },{ vertex: 17, peers: [ 8, 19 ] },{ vertex: 18, peers: [ 7 ] },{ vertex: 19, peers: [ 6, 17 ] },{ vertex: 20, peers: [ 5, 16 ] },{ vertex: 21, peers: [ 4, 15 ] },{ vertex: 22, peers: [ 3, 14 ] },{ vertex: 23, peers: [ 2, 13 ] } ];

let result = hamiltonian(vertexes, 18);
console.log('result', result);

请注意，使用动态编程可以提高运行时间。查看Hamiltonian path problem 上的不同方法。