Graph Tour with Uniform Cost Search in Java
- by user324817
Hi. I'm new to this site, so hopefully you guys don't mind helping a nub.
Anyway, I've been asked to write code to find the shortest cost of a graph tour on a particular graph, whose details are read in from file. The graph is shown below:
http://img339.imageshack.us/img339/8907/graphr.jpg
This is for an Artificial Intelligence class, so I'm expected to use a decent enough search method (brute force has been allowed, but not for full marks).
I've been reading, and I think that what I'm looking for is an A* search with constant heuristic value, which I believe is a uniform cost search. I'm having trouble wrapping my head around how to apply this in Java.
Basically, here's what I have:
Vertex class -
ArrayList<Edge> adjacencies;
String name;
int costToThis;
Edge class -
final Vertex target;
public final int weight;
Now at the moment, I'm struggling to work out how to apply the uniform cost notion to my desired goal path. Basically I have to start on a particular node, visit all other nodes, and end on that same node, with the lowest cost.
As I understand it, I could use a PriorityQueue to store all of my travelled paths, but I can't wrap my head around how I show the goal state as the starting node with all other nodes visited.
Here's what I have so far, which is pretty far off the mark:
public static void visitNode(Vertex vertex) {
ArrayList<Edge> firstEdges = vertex.getAdjacencies();
for(Edge e : firstEdges) {
e.target.costToThis = e.weight + vertex.costToThis;
queue.add(e.target);
}
Vertex next = queue.remove();
visitNode(next);
}
Initially this takes the starting node, then recursively visits the first node in the PriorityQueue (the path with the next lowest cost).
My problem is basically, how do I stop my program from following a path specified in the queue if that path is at the goal state? The queue currently stores Vertex objects, but in my mind this isn't going to work as I can't store whether other vertices have been visited inside a Vertex object.
Help is much appreciated!
Josh
