Dijkstra's algorithm finding all possible shortest paths

Question

I'm working on Dijkstra's algorithm,and I need to find all possible shortest paths. Dijkstra's algorithm returns only one short path, if another path has the same cost I would like to print it. I'm out of ideas, please help me.

Thank you.

Here's my algorithm:

public class Dijkstra {
  private static final Graph.Edge[] GRAPH = {
    new Graph.Edge("a", "b", 7),
    new Graph.Edge("a", "c", 9),
    new Graph.Edge("a", "f", 14),
    new Graph.Edge("b", "c", 10),
    new Graph.Edge("b", "d", 13),
    new Graph.Edge("c", "d", 11),
    new Graph.Edge("c", "f", 2),
    new Graph.Edge("d", "e", 6),
    new Graph.Edge("e", "f", 9),
  };
  private static final String START = "a";
  private static final String END = "e";

  public static void main(String[] args) {
    Graph g = new Graph(GRAPH);
    g.dijkstra(START);
    g.printPath(END);
    //g.printAllPaths();
  }
}

import java.io.*;
import java.util.*;

class Graph {
  private final Map<String, Vertex>
      graph; // mapping of vertex names to Vertex objects, built from a set of Edges

  /** One edge of the graph (only used by Graph constructor) */
  public static class Edge {
    public final String v1, v2;
    public final int dist;

    public Edge(String v1, String v2, int dist) {
      this.v1 = v1;
      this.v2 = v2;
      this.dist = dist;
    }
  }

  /** One vertex of the graph, complete with mappings to neighbouring vertices */
  public static class Vertex implements Comparable<Vertex> {
    public final String name;
    public int dist = Integer.MAX_VALUE; // MAX_VALUE assumed to be infinity
    public Vertex previous = null;
    public final Map<Vertex, Integer> neighbours = new HashMap<>();

    public Vertex(String name) {
      this.name = name;
    }

    private void printPath() {
      if (this == this.previous) {
        System.out.printf("%s", this.name);
      } else if (this.previous == null) {
        System.out.printf("%s(unreached)", this.name);
      } else {
        this.previous.printPath();
        System.out.printf(" -> %s(%d)", this.name, this.dist);
      }
    }

    public int compareTo(Vertex other) {
      return Integer.compare(dist, other.dist);
    }
  }

  /** Builds a graph from a set of edges */
  public Graph(Edge[] edges) {
    graph = new HashMap<>(edges.length);

    //one pass to find all vertices
    for (Edge e : edges) {
      if (!graph.containsKey(e.v1)) graph.put(e.v1, new Vertex(e.v1));
      if (!graph.containsKey(e.v2)) graph.put(e.v2, new Vertex(e.v2));
    }

    //another pass to set neighbouring vertices
    for (Edge e : edges) {
      graph.get(e.v1).neighbours.put(graph.get(e.v2), e.dist);
      //graph.get(e.v2).neighbours.put(graph.get(e.v1), e.dist); // also do this for an undirected graph
    }
  }

  /** Runs dijkstra using a specified source vertex */
  public void dijkstra(String startName) {
    if (!graph.containsKey(startName)) {
      System.err.printf("Graph doesn't contain start vertex \"%s\"\n", startName);
      return;
    }
    final Vertex source = graph.get(startName);
    NavigableSet<Vertex> q = new TreeSet<>();

    // set-up vertices
    for (Vertex v : graph.values()) {
      v.previous = v == source ? source : null;
      v.dist = v == source ? 0 : Integer.MAX_VALUE;
      q.add(v);
    }

    dijkstra(q);
  }

  /** Implementation of dijkstra's algorithm using a binary heap. */
  private void dijkstra(final NavigableSet<Vertex> q) {
    Vertex u, v;
    while (!q.isEmpty()) {

      u = q.pollFirst(); // vertex with shortest distance (first iteration will return source)
      if (u.dist == Integer.MAX_VALUE)
        break; // we can ignore u (and any other remaining vertices) since they are unreachable

      //look at distances to each neighbour
      for (Map.Entry<Vertex, Integer> a : u.neighbours.entrySet()) {
        v = a.getKey(); //the neighbour in this iteration

        final int alternateDist = u.dist + a.getValue();
        if (alternateDist < v.dist) { // shorter path to neighbour found
          q.remove(v);
          v.dist = alternateDist;
          v.previous = u;
          q.add(v);
        } else if (alternateDist == v.dist) {
          // Here I Would do something
        }
      }
    }
  }

  /** Prints a path from the source to the specified vertex */
  public void printPath(String endName) {
    if (!graph.containsKey(endName)) {
      System.err.printf("Graph doesn't contain end vertex \"%s\"\n", endName);
      return;
    }

    graph.get(endName).printPath();
    System.out.println();
  }
  /** Prints the path from the source to every vertex (output order is not guaranteed) */
  public void printAllPaths() {
    for (Vertex v : graph.values()) {
      v.printPath();
      System.out.println();
    }
  }

  public void printAllPaths2() {
    graph.get("e").printPath();
    System.out.println();
  }
}

If there are multiple paths to a node, then a given node may have multiple previous nodes, each of which has the same dist value. So, for one thing, you'd need to make Vertex.previous into a collection of some kind. — Andy Turner
Note that finding all shortest pathes may not be able to be done in polynomial time. For example, there will be too many shortest pathes when there are many individual choice from two loads with same length like this. — MikeCAT

BKE BKE · Accepted Answer · 2016-06-01T21:14:55

Have a look into so called k-shortest path algorithms. These solve the problem of enumerating the first, second, ..., kth shortest path in a graph. There are several algorithms in the literature, see for example this paper, or Yen's algorithm.

Note, that most algorithms do not require that you specify k upfront, ie. you can use them to enumerate shortest paths in an increasing order, and stop when the length has strictly increased.

Dijkstra's algorithm finding all possible shortest paths

1 Answers