Is finding whether k different perfect matchings exist in a bipartite graph co-NP?

Question

Few definitions first. The co-NP problem is a decision problem where the answer "NO" can be verified in polynomial time. The perfect matching in a bipartite graph is a set of pairs of nodes (a pair is an edge in the graph) and where every node occurs in this set exactly once.

I am given an n x n bipartite graph, and I am trying to find out if the problem of finding whether k different perfect matchings exist in the graph, where k= polynomial(n), is a co-NP problem.

Work done so far

To initially simplify the problem, I believe that if k=2, then this is a co-NP problem. I think this is true, because the bipartite graph does not have 2 different perfect matchings, if there does not exist an exchange of neighbors between 2 nodes. I define the exchange of neighbors as the following. Let G1 be the first set in the graph, and G2 be the second set in the graph. The exchange occurs when we have a subset of G1, S1={A,B}, and a second subset of G2, S2={X,Y}, where {(A,X),(A,Y),(B,X),(B,Y)} belongs to the set of edges E. I call it exchange because if A was initially matched with X, and B with Y, then when A gets paired with Y, and B with X, A and B have exchanged their neighbors. I believe that the only way to have 2 different perfect matchings is to have at least one such exchange.

Now, we can verify that no such exchange exist in polynomial time. This is true since getting all the possible subsets S1 and S2 has O(n^4) time complexity. This because we need (n choose 2) from G1 multiplied by (n choose 2) from G2, and this gives us an upper bound of n^4.

Answer 1

I am not sure if this is a co-NP problem, but it is NP for certain. I think you have a little mixed up the definition of "verifying an answer". In complexity theory verify an answer means that you provide a certificate that proves that your answer is correct, and such certificate may be checked (verified) in polynomial time.

For example, in the case of your problem, if you have a set k different perfect matchings, that will be a good certificate, verifying it means checking that it is indeed a set of perfect matchings in your input graph. You can check this in polynomial time by checking that all edges are in you graph and in each matching no two edges share a vertex, and all of them are different. Since the number of edges in a matching is linear, then verifying each matching can be done in polynomial time, then, since k is polynomial, we verify that property for all matchings also in polynomial time. Finaly, checking that all are different can be done in k square times something polynomial on n, yielding a polynomial complexity. So yes, your problem can be verified in polynomial time, and thus it is in NP.

Now, if you can find such certificate in polynomial time that will be proof enough that you problem is in P, and all problems in P are in NP and in co-NP. So I see two possible ways to solve this, you may prove that your problem is in P, that will yield a yes answer to your question, or you may prove that your problem is NP-complete, that will prove that your answer is no, since all NP-complete problems are not in co-NP (unless P = NP).

Any other way of proving that your problem is or is not in co-NP, might be very difficult and confusing, in fact the work you have done so far was moving towards proving that you can decide negative cases in polynomial time which is a different thing as verifying them, that would prove that it is co-NP, but because you proved that it is in P.