3
$\begingroup$

Notation: given a CNF formula A over variables X, we write $[A(X)]$ for the set of valuations $v: X \to \{0,1\}$ such that $A(X/v)$ is true, i.e. the set of valuations that makes formula A true.

I suspect that the following problem is NExpTime-complete:

-given two CNF formulas $A(X)$ and $B(Y,X)$, and an integer $K=2^k$, determine if there exist $K$ valuations $v_1,v_2,...,v_K : Y \to {0,1}$ such that $[A(X)] \subseteq \bigcup_{v=v_1,v_2,...,v_K} [B(Y/v,X)]$, i.e. are there K valuations of the variables in Y such that all the valuations that makes $A(X)$ true are covered by valuations that makes $B(Y/v,X)$ true.

So this can be seen as a succinct encoding of the set cover problem using CNF formulas. My guess is that this result should be a consequence of the NP completness of set cover and some general results about succinct encodings of problems using CNF formulas. Unfortunately, I was not able to locate such a result in the literature...

$\endgroup$

1 Answer 1

1
$\begingroup$

As I suspected in my question, they are useful results in the literature that can be exploited to characterize the complexity of the problem. There is a reduction from the dominant set problem for graphs succinctly defined using CNF formulas to the problem that I have described. The dominant set problem is known to be NExpTime complete [1] for graph succinctly defined using CNF formulas.

An instance of DominantSet^{2^n}{CNF} is defined by an CNF formula E(X,Y) over X={x_1,x_2,…,x_n} and Y={y1,y2,…,y_n}, two sets of n Boolean variables, and an integer K (encoded in binary) . The formula E(X,Y) succinctly defines the following graph: the set of vertices is the set of all Boolean valuations over n variables: 2^n, so any valuation v for the set of variables X, or for the set of variables Y, defines a vertex. Let v_1 and v_2 be two such valuations, then there is an edge={v_1,v_2} in the graph iff E(v_1,v_2) is true or E(v_2,v_1) is true. An instance is positive, if there exists a set of K valuations V={v_1,v_2,…,v_K} such that \cup{v \in V} { v’ | E(v,v’) \lor E(v’,v) }=2^n (i.e. all vertices are adjacent to a vertex in V).

We can reduce in polynomial time this problem to our succinct set cover problem as follows:

Define 𝐴(U) to be the empty CNF formula over the set of n variables U (which is equivalent to true) and so [A(U)]=2^n. Then define B(Z,U) to be the CNF equivalent to E(Z,U) \lor E(U,Z). This last formula has a size which is polynomial in the size of the CNF E that defines the graph. Keep K unchanged.

Then it is direct to see that the instance of succinct set cover is positive iff the DominantSet instance is positive. Indeed, there is a set V of K valuations for the variables in Z such that \bigcup_{v \in V} { v' | B(v,v') } covers 2^n iff V is a dominant set.

[1] Patrick Schafenacker. Problems on succinctly encoded graphs. PhD dissertation. Universität Ulm 2017.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.