A Monotone-2CNF formula is a CNF formula where each clause is composed by exactly 2 positive literals.Now, I have a Monotone-2CNF formula $F$. Let $S$ be the set of $F$'s satisfying assignments. I also have an oracle $O$ which is able to give the following information:The cardinality of the set $S$ (i.e. the number of solutions of $F$).Given a variable $x$:The number of solutions in $S$ containing the positive literal $x$.The number of solutions in $S$ containing the negative literal $\lnot x$.Given 2 variables $x_1$ and $x_2$:The number of solutions in $S$ containing $x_1 \land x_2$.The number of solutions in $S$ containing $x_1 \land \lnot x_2$.The number of solutions in $S$ containing $\lnot x_1 \land x_2$.The number of solutions in $S$ containing $\lnot x_1 \land \lnot x_2$.Note that the oracle $O$ is limited: it works only on $F$, it can't be used on a formula $F' \neq F$. Question:Given 3 variables $x_1$, $x_2$, $x_3$  is it possible to determine the number  of solutions in $S$ containing $\lnot  x_1 \land \lnot x_2 \land \lnot x_3$ in  polynomial time, using $F$ and  the information provided by $O$?Note:You can replace $\lnot  x_1 \land \lnot x_2 \land \lnot x_3$ in the  question with whatever else of the 8 possible  combinations of $x_1$, $x_2$, $x_3$. The problem  would remain the same.Empirical fact:I came across the following empirical fact one  week ago. Let $S_{\lnot x_1 \land \lnot x_2} \subset S$  be the set of those solutions containing $\lnot x_1 \land \lnot x_2$,  and let $S_{\lnot x_1 \land \lnot x_2 \land x_3} \subset S$ be the set  of those solutions containing $\lnot x_1 \land \lnot x_2 \land x_3$. Now, it  seems to be the case that, if condition $C$ holds, this relationship also holds:  $\frac{|S_{\lnot x_1 \land \lnot x_2}|}{|S_{\lnot x_1 \land \lnot x_2 \land x_3}|} \simeq \phi$  where $\phi = 1.618033...$ is the golden ratio. Condition $C$ seems to be the following: $x_1$, $x_2$, $x_3$ are mentioned in $F$ almost the same number of times.
Counting solutions of Monotone-2CNF formulas
cc.complexity theory;ds.algorithms;sat;counting complexity
To use that empirical fact you really want to know whether approximate numbers can give others approximate numbers. But for the exact case, I think there may be a straightforward way to show this is hard. Here's a sketch.First note that satisfying assignments correspond to independent sets in a graph. I'll use the phrase S-projections of I(G) to describe the function mapping $T \subset S$ to the number of independent sets I with $I \cap S = T$. The k-projections are the S-projections for all subsets S of V with $|S|=k$.Proof outline:If 2-projections give 3-projections, they also give k-projections in polytime for each k.If 2-projections give 4-projections, then the number of independent sets of a graph is in FP, so FP=#P.(1)Let $k\geq 3$ such that (k-1)-projections give k-projections. Given a graph, its k-projections, and $x_1,...,x_k,v \in G$, we will compute the projections onto ${x_1,...,x_k,v}$.Define the graph $G'$ by attaching a fresh vertex to v. This can be seen as weighting v. The (k-1)-projections of $G'$ can be computed because we know the k-projections of G. So then we have the k-projections of $G'$. And this gives ${x_1,...,x_k,v}$-projections of G.(2)Given a graph, order the edges ${e_1,...,e_m}$ and define $G_k$ to have edges ${e_1,...,e_k}$. The 2-projections of $G_{k+1}$ can be computed from the 4-projections of $G_k$. The number of independent sets in $G_0$ is $2^{|G|}$. Iteratively the 4-projections of G can be computed in polynomial time.