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Q:
Optimized Graph traversal algorithm
I am creating a graph with edges drawn using the following code, and I am having trouble to find the best algorithm to traverse the graph. Basically what I am trying to do is finding the shortest path between two nodes in the graph. Please see this image,
As you can see in the figure I have a binary tree, with two leaves at each node. When user enters the node (say node B) as the start point, my traversal algorithm should first check if there is any path between the start node and the end node, if there is a path then it should return the nodes that are on the path. If there is no path then it should check the children of the node. This should be done recursively until the node is reached or until a dead end is found. The traversal algorithm should return the nodes that are on the path.
The problem is there can be multiple paths between two nodes, in the example in the picture there are three paths. Now the shortest path, path1, has a length of 3. But it should return both nodes A and B. If the algorithm finds the first node on the path (which is A in this case), then it should return node A and B since node A has a path to node B. If it finds the second node on the path (which is B in this case), it should return node A and B since node B also has a path to node A. So the traversal algorithm should return the shortest path between nodes A and B.
I tried Breadth-first Search but it's a waste of time and resources to traverse the graph for every possible combination of nodes. I think some form of DFS is the best option, but I am not really good at DFS. Please help, and I'd be really thankful if you also explain the concepts behind the algorithm.
A:
This is a typical application of Dijkstra's algorithm. Starting with the node that you have entered and following edges until you get to a node with no edges in the direction you're going.
You can implement this using a queue or similar. Starting from the start node, for each other node, add it to the queue. Once the queue is full, you be359ba680
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