Suppose we have two bipartite graphs $G_1$ and $G_2$ with perfect matching count $P_1$ and $P_2$ respectively then their disjoint union gives a bipartite graph with perfect matching $P_1P_2$.
Is there a graph construction in polynomial time to get a bipartite graph from $G_1$ and $G_2$ with perfect matching count $P_1+P_2$ without knowing $P_1$ and $P_2$?
There is a way to get a graph having $2^{\phi(n)}(P_1+P_2)$ perfect matching where $\phi(n)=O(n^2)$ and $n=\max(|V(G_1)|,|V(G_2)|)$ where $|V(H)|$ is number of vertices in graph $H$ through Valiant's construction of perfect matching and the fact deciding perfect matching is an $NP$ problem by essentially translating the graphs to $SAT$ instances and using additive closure there and converting back to balanced bipartite graph whose perfect matching count is $2^{\phi(n)}(P_1+P_2)$.
Another example conversion to $SAT$ is following. Let $G=(U+V, E)$ be the bipartite graph with $|U|=|V|$ and no isolated vertices. Create a Boolean variable $x_e$ for each edge $e \in E$. Intuitively, $x_e$ is set to true iff $e$ is selected in the matching. Then, for all (unordered) pairs of distinct edges $\{e, f\}$ such that $e$ and $f$ share an endpoint, add the clause $(\overline{x}_e \vee \overline{x}_f)$. These clauses ensure that the selected edges form a matching. Finally, for each vertex $u \in U$, create a new clause $\bigvee\limits_{ e=(u,v) \in E} x_e$. Intuitively, this clause is satisfied iff $u$ is matched. The $SAT$ formula $\varphi$ obtained as the conjunction of all the above clauses is satisfiable iff the balanced bipartite graph $G$ admits a perfect matching.
We can add number of solutions up to a scale as mentioned before in the special $SAT$ structure by creating a new $SAT$ instance from the two constructed $SAT$ instances so that number of solutions is added through closure of $NP$ but not directly. It is unlikely we can do something here. So it has to be a construction which avoids $SAT$. Is there an additive construction which is in deterministic polynomial time? I think the problem might be solvable in affirmative.
