Hi,

As described, your problem is NP-hard, which means that if you were to find a way to solve it in polynomial time in the number of vertices in the graph, you'd also be able to solve a lot of other hard problems that nobody's ever been able to solve in polynomial time. For example, given an algorithm that solves your problem, you could use that algorithm to find Hamiltonian paths in graphs: for any graph G with n vertices, just run the algorithm on that graph and the path (u, x_0, ..., x_{n-2}, v), where the x_i's are vertices not appearing in G (or, say, isolated vertices in G) and u and v are two vertices from G. If the algorithm outputs a path, it's a path of length n-1 and thus Hamiltonian.

In addition, there can easily be an exponential number of paths output by such an algorithm, so even if you could find an efficient algorithm, once you reach graphs with on the order of hundreds of nodes, there's no guarantee that you have enough space to store the results or time to read them.

Maybe there are some additional constraints on your actual problem or the graphs you're dealing with that make it simpler than what you've described above? If so, those would be good to know. If not, and you still want to write an algorithm to do this, I'd suggest that you don't try to do anything fancy with the BGL. Your best bet will just enumerating through all candidate paths of the appropriate length and checking if the candidate is an actual path in G.

-Aaron

On Thu, Feb 27, 2014 at 3:40 AM, Sensei <senseiwa@gmail.com> wrote:

Dear all,

I have a simple question about the most efficient way of finding shared nodes between a path and a graph. Let me be more explicit.

I have an oriented graph G, and a very simple graph P, which is actually a simple oriented path (n0 -> n1 -> ... -> nK). I need to find *all* the possible shared paths between P and G with the same length as P.

For instance, being

P = n1 -> n2 -> n3 -> n4 -> n5 -> n6

we may find that the G and P share only two sub-paths

n1 -> n2 -> n3

n5 -> n6

I'd like to find all sub-graphs that connect the two paths, for instance having

n1 -> n2 -> n3 -> n10 -> n5 -> n6

n1 -> n2 -> n3 -> n42 -> n5 -> n6

with the constraint that the solution must be a path with the same length as P, therefore a solution as these

n1 -> n2 -> n3 -> n9 -> n8 -> n5 -> n6

n1 -> n2 -> n3 -> n5 -> n6

would not be acceptable.

I know I can do it manually, but I think a BGL-aware solution is quite better that a manual search :)

Thanks & Cheers!

_______________________________________________

Boost-users mailing list

Boost-users@lists.boost.org

http://lists.boost.org/mailman/listinfo.cgi/boost-users