How to determine if two nodes are connected?

Question

I'm concerned that this might be working on an NP-Complete problem. I'm hoping someone can give me an answer as to whether it is or not. And I'm looking for more of an answer than just yes or no. I'd like to know why. If you can say,"This is basically this problem 'x' which is/is not NP-Complete. (wikipedia link)"

(No this is not homework)

Is there a way to determine if two points are connected on an arbitrary non-directed graph. e.g., the following

Well
  |
  |
  A
  |
  +--B--+--C--+--D--+
  |     |     |     |
  |     |     |     |
  E     F     G     H
  |     |     |     |
  |     |     |     |
  +--J--+--K--+--L--+
                    |
                    |
                    M
                    |
                    |
                  House

Points A though M (no 'I') are control points (like a valve in a natural gas pipe) that can be either open or closed. The '+'s are nodes (like pipe T's), and I guess the Well and the House are also nodes as well.

I'd like to know if I shut an arbitrary control point (e.g., C) whether the Well and House are still connected (other control points may also be closed). E.g., if B, K and D are closed, we still have a path through A-E-J-F-C-G-L-M, and closing C will disconnect the Well and the House. Of course; if just D was closed, closing only C does not disconnect the House.

Another way of putting this, is C a bridge/cut edge/isthmus?

I could treat each control point as a weight on the graph (either 0 for open or 1 for closed); and then find the shortest path between Well and House (a result >= 1 would indicate that they were disconnected. There's various ways I can short circuit the algorithm for finding the shortest path too (e.g., discard a path once it reaches 1, stop searching once we have ANY path that connects the Well and the House, etc.). And of course, I can also put in some artificial limit on how many hops to check before giving up.

Someone must have classified this kind of problem before, I'm just missing the name.

Answer 1

You don't need Dijkstra's algorithm for this problem, as it uses a Heap which isn't needed and introduces a factor of log(N) to your complexity. This is just breadth first search - don't include the closed edges as edges.

Answer 2

See http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm , your one stop shop for all graph related problems. I believe your problem is in fact solvable in quadratic time.

Answer 3

Your description seems to indicate that you are just interested in whether two nodes are connected, not finding the shortest path.

Finding if two nodes are connected is relatively easy:

Create two sets of nodes:  toDoSet and doneSet
Add the source node to the toDoSet 
while (toDoSet is not empty) {
  Remove the first element from toDoList
  Add it to doneList
  foreach (node reachable from the removed node) {
    if (the node equals the destination node) {
       return success
    }
    if (the node is not in doneSet) {
       add it to toDoSet 
    }
  }
}

return failure.

If you use a hash table or something similar for toDoSet and doneSet, I believe this is a linear algorithm.

Note that this algorithm is basically the mark portion of mark-and-sweep garbage collection.

Answer 4

Any graph shortest path algorithm will be overkill if all you need is to find if a node is connected to another. A good Java library that accomplishes that is JGraphT. It's usage is quite simple, here's an example of an Integer graph:

public void loadGraph() {
    // first we create a new undirected graph of Integers
    UndirectedGraph<Integer, DefaultEdge> graph = new SimpleGraph<>(DefaultEdge.class);

    // then we add some nodes
    graph.addVertex(1);
    graph.addVertex(2);
    graph.addVertex(3);
    graph.addVertex(4);
    graph.addVertex(5);
    graph.addVertex(6);
    graph.addVertex(7);
    graph.addVertex(8);
    graph.addVertex(9);
    graph.addVertex(10);
    graph.addVertex(11);
    graph.addVertex(12);
    graph.addVertex(13);
    graph.addVertex(14);
    graph.addVertex(15);
    graph.addVertex(16);

    // then we connect the nodes
    graph.addEdge(1, 2);
    graph.addEdge(2, 3);
    graph.addEdge(3, 4);
    graph.addEdge(3, 5);
    graph.addEdge(5, 6);
    graph.addEdge(6, 7);
    graph.addEdge(7, 8);
    graph.addEdge(8, 9);
    graph.addEdge(9, 10);
    graph.addEdge(10, 11);
    graph.addEdge(11, 12);
    graph.addEdge(13, 14);
    graph.addEdge(14, 15);
    graph.addEdge(15, 16);

    // finally we use ConnectivityInspector to check nodes connectivity
    ConnectivityInspector<Integer, DefaultEdge> inspector = new ConnectivityInspector<>(graph);

    debug(inspector, 1, 2);
    debug(inspector, 1, 4);
    debug(inspector, 1, 3);
    debug(inspector, 1, 12);
    debug(inspector, 16, 5);
}

private void debug(ConnectivityInspector<Integer, DefaultEdge> inspector, Integer n1, Integer n2) {
    System.out.println(String.format("are [%s] and [%s] connected? [%s]", n1, n2, inspector.pathExists(n1, n2)));
}

This lib also offers all the shortest paths algorithms as well.

Answer 5

To me it seems like you are on to a solution, but it's possible I misunderstood the problem. If you do like you say, and give the closed edges 1 as weight, you can just apply Dijkstra's algorithm, http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm. This should solve your problem in O(E*lg(V))