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My graph contains no such edges which connect a vertex to itself. There is only one edge between two vertices. From Wikipedia i got to know about some algorithm which are used for calculating shortest path based on the given conditions. One of the most famous algorithm is Dijkstra's algorithm, which finds a shortest paths from source vertex to all other vertices in the graph.
But by using Dijkstra's algorithm, i am unnecessary exploring all the vertices, however my goal is just to find shortest path from single source to single destination. Which strategy should i use here? So that i need not to explore all other vertices.

One of my approach is to use bidirectional bfs. By bidirectional bfs i mean to apply two bfs one from source node, another one from destination node. As soon as for the first time when i find any same child in both tree,i can stop both bfs .Now the path from source to that child union path from child to destination would be my shortest path from source to destination.

But out of all the approaches described by Wikipedia and bidirectional bfs, which one suits best for my graph?

Tony Ennis
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ravi
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  • related: http://stackoverflow.com/questions/2655880/how-to-optimize-dijkstra-algorithm-for-a-single-shortest-path-between-2-nodes – K Mehta Apr 07 '12 at 11:09
  • @KshitijMehta: Should i apply [Dijkstra's algorithm](http://stackoverflow.com/questions/2655880/how-to-optimize-dijkstra-algorithm-for-a-single-shortest-path-between-2-nodes) here? Is this the best way for my graph? – ravi Apr 07 '12 at 11:11
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    You could work with A* search, but that involves using heuristic functions and is more of an "Artificial Intelligence" based approach. Other than that, Dijkstra's is your best option. – K Mehta Apr 07 '12 at 11:12
  • @KshitijMehta: I used `A* search` recently in another place. but `A* search` require two lists. I can maintain two lists but only when if that suits best for my graph. So what do you say? – ravi Apr 07 '12 at 11:23
  • If performance is not too big a concern, I'd suggest you go for Dijkstra's. A* requires a good heuristic to be effective. – K Mehta Apr 07 '12 at 11:36
  • i can't play with performance. i should explore heuristics in my graph. thank you very much. – ravi Apr 07 '12 at 11:38
  • In which case, you should make sure your heuristic is **consistent** to ensure that you get the optimal solution. – K Mehta Apr 07 '12 at 11:41

4 Answers4

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It depends if there's any apparent heuristic function that you could use or if you don't have any further usable information about your graph.

Your options are:

  • BFS: in general case or if you don't care about computation time that much.
  • Dijkstra (Dijkstra "is" BFS with priority queue): if your edges have weights/prices (non negative) and you care about CPU time.
  • A* (A* "is" Dijkstra with heuristic function - e.g. distance on a city map): if you want it to be really fast in average case, but you have to provide good heuristic function.

For some graph problems it may be possible to use Dynamic programming or other algorithmic tools. It depends on a situation. Further information can be found in tutorials from http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=alg_index ...

bubersson
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From Introduction to Algorithms (CLRS) second edition, page 581 :

Find a shortest path from u to v for a given vertices u and v. If we solve the single-source problem with source vertex u, we solve this problem also. Moreover, no algorithms for this problem are known that run asymptotically faster than the best single-source algorithms in the worst case.

So, stick to Dijkstra's algorithm :)

Priyank Bhatnagar
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The Wikipedia article spells out the answer for you:

If we are only interested in a shortest path between vertices source and target, we can terminate the search at line 13 if u = target.

Viktor Nordling
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You could use Dijkstra's algorithm and optimize it in the way that you stop exploring paths that are already longer than the shortest you found so far. Because those are not going to be shorter (provided that you don't have negative weighs on your edges).

DerMike
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