Hamiltonian paths & social graph algorithm

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I have a random undirected social graph.

I want to find a Hamiltonian path if possible. Or if not possible (or not possible to know if possible in polynomial time) a series of paths. In this "series of paths" (where all N nodes are used exactly once), I want to minimize the number of paths and maximize the average length of the paths. (So no trivial solution of N paths of a single node).

I have generated an adjacency matrix for the nodes and edges already.

Any suggestions? Pointers in the right direction? I realize this will require heuristics because of the NP-complete (?) nature of the problem, and I am OK with a "good enough" answer. Also I would like to do this in Java.

Thanks!

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Use a genetic algorithm (without crossover), where each individual is a permutation of the nodes. This gives you "series of paths" at each generation, evolving to a minimal number of paths (1) and a maximal avg. length (N).

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If I'm interpreting your question correctly, what you're asking for is still NP-hard, since the best solution to the "multiple paths" problem would be a Hamiltonian path, and determining whether one exists is known to be NP-hard. Moreover, even if you're guaranteed that a Hamiltonian path doesn't exist, solving this problem could still be NP-hard, since I could give you a graph with a single disconnected node floating in space, for which the best solution is a trivial path containing that node and a Hamiltonian path in the remaining graph. As a result, unless P = NP, there isn't going to be a polynomial-time algorithm for your problem.

Hope this helps, and sorry for the negative result!

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As you have realized there is no exact solution in polynomial time. You can try some random search methods though. My recommendation, start with genetic algorithm and try out tabu search.

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Angluin and Valiant gave a near linear-time heuristic that works almost always in a sufficiently dense Erdos-Renyi random graph. It's described by Wilf, on page 121. Probably your random graph is not Erdos-Renyi, but the heuristic might work anyway (when it "fails", it still gives you a (hopefully) long path; greedily take this path and run A-V again).