Re: [Boost-users] BGL A-star does not terminate

Aaron Windsor wrote:

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On Jan 15, 2008 2:13 AM, Lars-Peter Ellekilde wrote:

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Hi I have been trying to use the a* implementation in BGL to do a shortest path search in en undirected graph. In my application I can not guarantee that the graph only contains a single connected components, thus a solution may not always exist. Given a query where start and goal are in two different components the a* algorithm appears to loop forever.

Is this expected behavior or does it indicate that I might be using it wrong somehow? If it is expected, what will be the best way to make the algorithm terminate and indicate there is no solution (perhaps I use the visitor for this purpose).

Best regards and thanks - Lars-Peter

Hi Lars-Peter,

Can you give more information about the case were the implementation appears to loop forever? What platform are you using? Do you have a relatively small example that causes it to loop? There's a note in the code that mentions a case where the implementation can loop on x86 Linux.

The generally accepted way to break out of any BGL traversal early is to have the visitor throw an exception, so if you have an easy way of detecting that you won't be able to find the goal at all, you may want to consider this approach.

Regards, Aaron

On Jan 17, 2008 7:05 AM, Lars-Peter Ellekilde wrote: Hi Thanks for the reply. I haven't seen the comment about x86 Linux in the code, but guess it very well be my problem. The program I am writing constructs a graph with random data, thus it is difficult for me to reproduce the error in a small example.

Any good ideas on how best to detect if it enters an infinite loop? Would it be possible just to use the visitor and see if the vertex under examination is the same as the previous one?

The comment about looping seems to indicate that the A* visitor's "edge_not_relaxed" method will be repeatedly called. You may want to stick a std::cout in this method and see what's happening when the looping occurs - I don't know whether it will loop on the same edge or on a sequence of edges, but you may be able to short-circuit the search there. Regrds, Aaron