Question

在我的代码中，我在图表中找到哈密尔顿路径来解决另一个问题我现在正在测试我的代码，我想采用没有边的一般图形并在其中构造哈密顿路径。在那个图上（现在有了形成哈密尔顿路径的边）我将根据Erdős–Rényi model添加随机边。通过这种方式，我可以看到我的代码处理具有不同边数的图形的速度有多快我能处理的有效图表如何：

对于矩阵中的每个单元格，我添加一个顶点。
顶点u和v 可以连接，如果它们在矩阵中相邻。
我的目标是使用哈密顿路径生成随机有效图。

问题在于我无法找到一种有效的方法来构建一个哈密尔顿路径，而不会重复出现所有可能的路径并找到一个遍历所有顶点的路径。

例如：

The matrix: Possible path: Not possible: ------------- | 1 | 2 | 3 | 1 - 2 - 3 1 - 2 - 3 _ ------------- | | | 4 | 5 | 6 | 4 - 5 - 6 4 - 5 - 6 | ------------- | | | | 7 | 8 | 9 | 7 - 8 - 9 7 - 8 - 9_/ -------------

第二条路径是不可能的，因为3和9在矩阵中不相邻有没有办法在给定矩阵的线性时间内构造哈密顿路径？

Answer 1

package hamiltonian_path;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;



public class Main {

    int[] solution;
    HashMap<Integer, List<Integer>> candidates;

    public static void main(String args[]) {
        Main main = new Main();
        main.solution = new int[10];//stores the solution; index 0 is not used, i will use indexes from 1 to 9
        main.candidates = new HashMap<Integer, List<Integer>>();//for each position (1 to 9) in the solution, stores a list of candidate elements for that position

        List<Integer> oneToNine = new LinkedList<Integer>(Arrays.asList(1,2,3,4,5,6,7,8,9));
        /*
         * because no solution can start from matrix elements 2,4,6 or 8, 
         * for the sake of optimization, the above list can be written as
         * Arrays.asList(1,3,5,7,9)
         * the way it is right now is useful to follow the way the program 
         * does the backtracking, when it accidentally starts with either 2,4,6 or 8
         */
        Collections.shuffle(oneToNine);//a random permutation of the list
        main.candidates.put(1, oneToNine);
        main.buildSol(1);

    }

    //backtracking
    public void buildSol(int k)
    {
        if(k==10)
        {
            System.out.print("the solution is ");
            printSolution();
            return;
        }

        List<Integer> candList = candidates.get(k);
        if(candList.isEmpty())
        {
            cleanSolution(k);
            buildSol(k-1); //if no candidates for current step, go one step back
        }
        else
        {
            int firstCandidate = candList.get(0);
            candList.remove(0);
            candidates.put(k, candList);
            solution[k] = firstCandidate;//for the position k in the solution, pick the first element in the candidates list

            List<Integer> neighbors = getNeighbors(solution[k]);
            List<Integer> prevElems = getPreviousElementsInSolution(k);
            candidates.put(k+1, generateCandidates(neighbors, prevElems));//while being at step k, generate candidate elements for step k+1
            //these candidates are the neighbors (in the matrix) of the current element (solution[k]), 
            //which are not already part of the solution at an earlier position

            System.out.println("step "+k);
            System.out.print("partial solution: ");
            printSolution();
            System.out.println();


            buildSol(k+1);//go to next step
        }
    }



    //candidates are those elements which are neighbors, and have not been visited before
    public List<Integer> generateCandidates(List<Integer> neighbors, List<Integer> previousElements) 
    {
        List<Integer> cnd = new ArrayList<Integer>();
        for(int i=0;i<neighbors.size();i++)
            if(!previousElements.contains(neighbors.get(i)))
                cnd.add(neighbors.get(i));

        return cnd;
    }

    //get the set of previous elements in the solution, up to  solution[k]
    public List<Integer> getPreviousElementsInSolution(int step)
    {
        List<Integer> previousElements = new ArrayList<Integer>();
        for(int i=1; i<=step-1;i++)
            previousElements.add(solution[i]);

        return previousElements;
    }

    //get neighbors of the matrix element which corresponds to solution[k]

    public  List<Integer> getNeighbors(int element) {

        List<Integer> neighboursList = new ArrayList<Integer>();

        switch (element) {

            case 1: neighboursList = Arrays.asList(2, 4);
                    break;

            case 2: neighboursList = Arrays.asList(1, 3, 5);
                    break;

            case 3: neighboursList = Arrays.asList(2, 6);
                    break;

            case 4: neighboursList = Arrays.asList(1, 5, 7);
                    break;

            case 5: neighboursList = Arrays.asList(2, 4, 6, 8);
                    break;

            case 6: neighboursList = Arrays.asList(3, 5, 9);
                    break;

            case 7: neighboursList = Arrays.asList(4, 8);
                    break;

            case 8: neighboursList = Arrays.asList(5, 7, 9);
                    break;

            case 9: neighboursList = Arrays.asList(6, 8);
                    break;

            default: neighboursList = new ArrayList<Integer>();
                    break;
        }

        return neighboursList;
    }



    public void printSolution()
    {
        for(int i=1;i<solution.length;i++)
            System.out.print(solution[i]+" ");
    }

    public void cleanSolution(int k)
    {
        for(int i=k;i<solution.length;i++)
            solution[i]=0;
    }
}

在图中构建随机哈密顿路径

1 个答案: