使用邻接表BFS遍历图中所有路径

Question

I am currently trying to traverse all paths from source to destination in a graph which uses adjacency matrix. I have been trying to do it in BFS way.Thanks for the help. I am getting only one path. How do I get to print other paths as well ?

public class AllPossiblePaths {
    static int v;
    static ArrayList<Integer> adj[];

    public AllPossiblePaths(int v) {
        this.v = v;
        adj = new ArrayList[v];
        for (int i = 0; i < v; i++) {
            adj[i] = new ArrayList<>();
        }
    }

    // add edge from u to v
    public static void addEdge(int u, int v) {
        adj[u].add(v);
    }

    public static void findpaths(int source, int destination) {
        LinkedList<ArrayList<Integer>> q = new LinkedList<>();
        boolean visited[] = new boolean[v];

        LinkedList<Integer> queue = new LinkedList<Integer>();

        queue.add(source);
        visited[source] = true;
        ArrayList<Integer> localPath = new ArrayList<>();
        while (!queue.isEmpty()) {
            // Dequeue a vertex from queue and print it
            int src = queue.poll();
            if (!localPath.contains(src)) {
                localPath.add(src);
            }
            if (src == destination) {
                System.out.println(localPath);
                localPath.remove(localPath.size() - 1);
                visited[src] = false;
            }

            Iterator<Integer> i = adj[src].listIterator();
            while (i.hasNext()) {
                int n = i.next();
                if (!visited[n]) {
                    queue.add(n);
                }
            }
        }
    }
}

Answer 1

Using the following class you can run a BFS to find a single path ( findPath ) or find multiple paths ( findAllPaths ). See comments:

import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;

public class AllPossiblePaths {

    private  boolean[] visited;
    //keep track of nodes already included in a path
    private  boolean[] includedInPath;
    private LinkedList<Integer> queue;
    private int numberOfNodes;
    private List<Integer>[] adj;
    //to find a path you need to store the path that lead to it
    private List<Integer>[] pathToNode;

    public AllPossiblePaths(int numberOfNodes) {

        this.numberOfNodes = numberOfNodes;
        adj = new ArrayList[numberOfNodes];
        pathToNode = new ArrayList[numberOfNodes];

        for (int i = 0; i < numberOfNodes; i++) {
            adj[i] = new ArrayList<>();
        }
    }

    // add edge from u to v
    public AllPossiblePaths addEdge(int from, int to) {
        adj[from].add(to);
        //unless unidirectional: //if a is connected to b
        //than b should be connected to a
        adj[to].add(from);
        return this; //makes it convenient to add multiple edges
    }

    public void findPath(int source, int destination) {

        System.out.println("------------Single path search---------------");
        initializeSearch(source);

        while (!queue.isEmpty()) {
            // Dequeue a vertex from queue and print it
            int src = queue.poll();
            visited[src] = true;

            if (src == destination) {
                System.out.println("Path from "+source+" to "
                        + destination+ " :- "+ pathToNode[src]);
                break; //exit loop if target found
            }

            Iterator<Integer> i = adj[src].listIterator();
            while (i.hasNext()) {
                int n = i.next();
                if (! visited[n] && ! queue.contains(n)) {
                    queue.add(n);
                    pathToNode[n].addAll(pathToNode[src]);
                    pathToNode[n].add(src);
                }
            }
        }
    }

    public void findAllpaths(int source, int destination) {

        System.out.println("-----------Multiple path search--------------");
        includedInPath = new boolean[numberOfNodes];
        initializeSearch(source);
        int pathCounter = 0;

        while(! allVisited() && !queue.isEmpty()) {

            while (!queue.isEmpty()) {
                // Dequeue a vertex from queue and print it
                int src = queue.poll();
                visited[src] = true;

                if (src == destination) {

                    System.out.println("Path " + ++pathCounter + " from "+source+" to "
                            + destination+ " :- "+ pathToNode[src]);
                    //mark nodes that are included in the path, so they will not be included
                    //in any other path
                    for(int i=1; i < pathToNode[src].size(); i++) {
                        includedInPath[pathToNode[src].get(i)] = true;
                    }
                    initializeSearch(source); //initialize before restarting
                    break; //exit loop if target found
                }

                Iterator<Integer> i = adj[src].listIterator();
                while (i.hasNext()) {
                    int n = i.next();
                    if (! visited[n] && ! queue.contains(n)
                            && ! includedInPath[n] /*ignore nodes already in a path*/) {
                        queue.add(n);
                        pathToNode[n].addAll(pathToNode[src]);
                        pathToNode[n].add(src);
                    }
                }
            }
        }
    }

    private void initializeSearch(int source) {

        queue = new LinkedList<>();
        queue.add(source);
        visited = new boolean[numberOfNodes];
        for (int i = 0; i < numberOfNodes; i++) {
            pathToNode[i]= new ArrayList<>();
        }
    }

    private boolean allVisited() {

        for( boolean b : visited) {
            if(! b ) return false; 
        }
        return true;
    }
}

For testing it, consider this graph:

Run test:

public static void main(String[] args){

    AllPossiblePaths app = new AllPossiblePaths(6);
    app.addEdge(0, 4)
    .addEdge(0, 1)
    .addEdge(1, 2)
    .addEdge(1, 4)
    .addEdge(4, 3)
    .addEdge(2, 3)
    .addEdge(2, 5)
    .addEdge(3, 5);

    app.findPath(0,5);
    app.findPath(5,0);
    app.findAllpaths(0,5);
}

output:

Answer 2

Apparently it is impossible to retrieve all paths from a given source to a given terminal via Breadth-First search. Consider the following class of graphs.

For any nonnegative integer n , let

V := {v_1,...,v2_n}                             // inner vertices
     union
     {s, t},                                    // source and terminal
E := { {v_i,v+2,} : i < 2n-2 }                  // horizontal edges
     union
     { {v_i,v_i+3} : i < 2n-3, i is odd }       // cross edges from top to bottom
     union
     { {v_i,v_i+3} : i < 2n-3, i is even }      // cross edges from bottom to top
     union
     { {s,v_1}, {s,v_2}, {t,v_2n-1}, {t,v_2n} } // source and terminal

Informally, the graph consists out of two rows of vertices with n columns each, to the left there is a source node and to the right there is a terminal node. For each path from s to t , you can choose for each column to stay in the current row or to switch to the other row.

In total, there are 2^n different paths from s to t , as for each column there are two possibilities to chose the row.

On the other hand, Breadth-First search yields a runtime bound which is polynomial in the encoding length of the graph; this means that Breadth-first search, in general, cannot generate all possible paths from a given source to a given terminal. Furthermore, if the graph contains a cycle, the number of paths might be inifinite via repetition of the cycle.