图中2个节点之间的所有路径

Question

I have to make an uninformed search (Breadth-first-Search) program which takes two nodes and return all the paths between them.

public void BFS(Nod start, Nod end) {
            Queue<Nod> queue = new Queue<Nod>();
            queue.Enqueue(start);
            while (queue.Count != 0)
            {
                Nod u = queue.Dequeue();
                if (u == end)  break;
                else
                {
                    u.data = "Visited";
                    foreach (Edge edge in u.getChildren())
                    {
                        if (edge.getEnd().data == "")
                        {
                            edge.getEnd().data = "Visited";
                            if (edge.getEnd() != end)
                            {
                                edge.getEnd().setParent(u);
                            }
                            else
                            {
                                edge.getEnd().setParent(u);
                                cost = 0; 
                                PrintPath(edge.getEnd(), true);
                                edge.getEnd().data = "";
                                //return;
                            }
                        }
                        queue.Enqueue(edge.getEnd());
                    }
                }
            }
        }

My problem is that i only get two paths instead of all and i don't know what to edit in my code to get them all. The input of my problem is based on this map :

Answer 1

Breadth first search is a strange way to generate all possible paths for the following reason: you'd need to keep track of whether each individual path in the BFS had traversed the node, not that it had been traversed at all.

Take a simple example

1----2
 \    \
  3--- 4----5

We want all paths from 1 to 5. We queue up 1, then 2 and 3, then 4, then 5. We've lost the fact that there are two paths through 4 to 5.

I would suggest trying to do this with DFS, though this may be fixable for BFS with some thinking. Each thing queued would be a path, not a single node, so one could see if that path had visited each node. This is wasteful memory wise, thoug

Answer 2

In the BFS algorithm you must not stop after you find a solution. One idea is to set data null for all the cities you visited except the first one and let the function run a little bit longer. I don't have time to write you a snippet but if ou don't get it i will write at least a pseudocode. If you didn't understood my idea post a comment with your question and i will try to explain better.

Answer 3

A path is a sequence of vertices where no vertex is repeated more than once. Given this definition, you could write a recursive algorithm which shall work as follows: Pass four parameters to the function, call it F(u, v, intermediate_list, no_of_vertices) , where u is the current source (which shall change as we recurse), v is the destination, intermediate_list is a list of vertices which shall be initially empty, and every time we use a vertex, we'll add it to the list to avoid using a vertex more than once in our path, and no_of_vertices is the length of the path that we would like to find, which shall be lower bounded by 2 , and upper bounded by V , the number of vertices. Essentially, the function shall return a list of paths whose source is u , destination is v , and whose length of each path is no_of_vertices . Create an initial empty list and make calls to F(u, v, {}, 2), F(u, v, {}, 3), ..., F(u, v, {}, V) , each time merging the output of F with the list where we intend to store all paths. Try to implement this, and if you still face trouble, I'll write the pseudo-code for you.

Edit: Solving the above problem using BFS: Breadth first search is an algorithm that could be used to explore all the states of a graph. You could explore the graph of all paths of the given graph, using BFS, and select the paths that you want. For each vertex v , add the following states to the queue: (v, {v}, {v}) , where each state is defined as: (current_vertex, list_of_vertices_already_visited, current_path) . Now, while the queue is not empty, pop off the top element of the queue, for each edge e of the current_vertex , if the tail vertex x doesn't already exist in the list_of_vertices_already_visited , push the new state (x, list_of_vertices_already_visited + {x}, current_path -> x) to the queue, and process each path as you pop it off the queue. This way you can search the entire graph of paths for a graph, whether directed, or undirected.

Answer 4

Sounds like homework. But the fun kind.

The following is pseudocode, is depth first instead of breath first (so should be converted to a queue type algorithm, and may contain bugs, but the general jist should be clear.

class Node{
  Vector[Link] connections;
  String name;
}

class Link{
  Node destination;
  int distance;
}

Vector[Vector[Node]] paths(Node source, Node end_dest, Vector[Vector[Node]] routes){
  for each route in routes{
    bool has_next = false;
    for each connection in source.connections{
      if !connection.destination in route {
        has_next = true;
        route.push(destination);
        if (!connection.destination == end_dest){
          paths(destination, end_dest, routes);
        }
      }
    }
    if !has_next {
      routes.remove(route) //watch out here, might mess up the iteration
    }
  }
  return routes;
}

Edit: Is this actually the answer to the question you are looking for? Or do you actually want to find the shortest path? If it's the latter, use Dijkstra's algorithm: http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm