# Does faster exact algorithm for counting independent sets in comparability graphs than general graph exisits?

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?

• Please state your question precisely, but that sounds like a duplicate of this question. – Tsuyoshi Ito Aug 9 '11 at 16:56