Sorry for not-precise question. :-(

There are several papers concerning exact counting (maximum) independent sets in general graphs. Actually, they concerns counting of solutions of 2SAT. The best of them is $O(1.23^n)$. But the algorithms do not use the specific information of comparability graphs.

So I wonder whether there exists more powerful(faster) exact algorithm for counting independent sets in comparability graphs?

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    $\begingroup$ Please state your question precisely, but that sounds like a duplicate of this question. $\endgroup$ Aug 9, 2011 at 16:56

1 Answer 1


Bipartite graphs are comparability graphs, and it's known (since Provan and Ball) that counting the number of independent sets in bipartite graphs is #P-complete.

J. S. Provan and M. O. Ball. The complexity of Counting Cuts and of Computing the Probability that a Graph is Connected. SIAM J. Comput. 12 (1983) 777-788.

If I haven't mistaken, this should answer your question.

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    $\begingroup$ I'm afraid this wasn't an intended answer. Probably, I need a clarification on what the "reduction" in the question exactly means. $\endgroup$ Aug 9, 2011 at 15:43
  • $\begingroup$ I am sorry, I have just edited my question. the older one is not preise. $\endgroup$
    – Peng Zhang
    Aug 9, 2011 at 17:22

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