Quadratic equation - Cuts

M(B) * M(D) = M(A)^2 * z^(nb/2 + 2) * y^(wb) * x^(hb) + M(C)^2

I'll prove the equation above with a bijection.

First consider the M(C)^2 term. We can represent that term by the matching polynomial of the following graph, since the snake C is symmetrical:



If one of the pairs of z-edges connecting to B are not being used in the matching then we can rearrange the parts of this graph to obtain a matching of the graph D U B (D union B).

But if both pairs of z-edges are being used, then the B parts of the graph have a common cut . We can cut the graph and rearrange the parts to get a matching of D U B:

Now, if we subtract all the matchings above from the equation, here's what is left:
  • On the left hand side, we have all the matchings of D U B that do not have a cut through the B part.
  • On the right hand side we have M(A)^2 * z^(nb/2 + 2) * y^(wb) * x^(hb)

    • But there is only one way to fill the B parts of D U B, such that there is not cut through the B parts. So, we want to prove that:



    The A parts are disconnected in D and the matchings of the B parts multiplied together gives us z^(nb/2 + 2) * y^(wb) * x^(hb).

        QED