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Counting the number of bipartite matchings is #P-hard. Thus every #SAT problem can be reduced to counting the number of bipartite matchings. If a SAT problem is unsatisfiable however, it will have 0 solutions and thus the corresponding bipartite matching problem will have 0 solutions which is detectable in polynomial time (since a single bipartite matching can be found in polynomial time).

This would enable us to solve SAT in poly-time. What's going wrong with this reasoning?

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What, precisely, are the requirements on the reductions you're working with here? Are you sure that "$>0$ satisfactions" has to reduce to a problem instance of the form "$>0$ bipartite matchings"? –  Henning Makholm Nov 9 '11 at 2:17
The reduction I'm working of is given here secure.wikimedia.org/wikipedia/en/wiki/… –  PartialD Nov 9 '11 at 2:18
From the description of the reduction, it seems that the number of #SAT solutions is equal to the number of bipartite matchings multiplied by a constant - so the answer to your question is yes. –  PartialD Nov 9 '11 at 6:15
It's a different notion of reduction. It is not true that the reduction maps unsatisfable SAT instances to 01 matrices with permanent 0. See cstheory.stackexchange.com/questions/8749/… –  sdcvvc Nov 9 '11 at 18:46

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