QBF Solving Using Best First Search

Abstract

Quantified Boolean Formula or QBF is similar to a two-player strategy game on a theoretical level. While most search-based QBF solvers are based on a depth-first search based QDPLL procedure, many game solvers use best-first search algorithms such as proof number search. In this paper, we examine how to design a best-first search based QBF solver that combines some well-known QBF solving techniques such as backjumping and clause learning. We evaluate the performance of the resulting algorithm on a set of standard QBF benchmarks.

A Correctness Discussion of the BFS Based CDCL Solver

There is a critical problem we have not explained in algorithm 1, which is the reusability of the resolvent. We have mentioned in the paper that during backpropagation when we want to query the resolvent corresponding to a node, we only need to consider two cases. In the appendix, we would explain 1) why this is necessary, and 2) why this is correct. Recall that, we let $\phi [t]$ represent the QBF expression at iteration t. According to the definition of a BFS node, at iteration t each node i has an assignment $\mu ^{i}_{t}$ associated with it. If we only execute backjumping (i.e. CBJ + SBJ), the formula does not change during the solving procedure, hence for any t and $t'$ when node i is visited we have $\mu ^{i}_{t}~=~\mu ^{i}_{t'}$. Note that with careful implementation, even if we activate pure literal elimination, this property would hold. However, when CDCL is enabled, this property might no longer hold. We have discussed that in the backpropagation phase, we would use the resolvent we store in the children to derive the resolvent of the current node. However, in BFS each node can be visited more than once, and more clauses might be added to the formula which could create more unit propagation. In other words, suppose node i is visited in iteration t and $t'$, it is possible that $\mu ^{i}_{t}~\ne ~\mu ^{i}_{t'}$. When we attempt to derive the resolvent associated with a node i with the resolvents we stored in its children, it is possible that the resolvent we stored in the children corresponds to a different assignment. For example, suppose the left child was solved in iteration $t'~<~t$ and the information associated with the node is $(i^{l}, h^{l}_{t'},v^{l},\mu ^{l}_{t'}, R^{l}_{t'})$. When we want to calculate the resolvent of node i with state $(i, h,v,\mu _{t}, R)$, we need to make a query of the resolvent of the left child $i^{l}$ which corresponds to the assignment $\mu ^{l}_{t}=\mu _{t};v;U(\phi [t]_{\mu _{t};v})$. However, because of the additional unit propagation caused by clause learning, it is possible that $\mu ^{l}_{t'}~\ne ~\mu _{t};v;U(\phi [t]_{\mu _{t};v})$, hence, we can no longer use $R^{l}_{t'}$ as a valid resolvent for the left child! We would update the resolvent we have previously stored in a node as follows:

• If $\phi [t]_{\mu ^{l}_{t}}$ contains a contradicted clause, we can recalculate $R^{l}_{t}$ based on the conflict in linear time [6].

• Otherwise, we can reuse $R^{l}_{t'}$ as a resolvent for node $i^{l}$.

The correctness of this update method does not always hold, however, in our implementation of the generalized Schloeter’s method proposition 1 holds. With this proposition, we should realize the following two lemmas related to the $\mu$-contradicted clause and $\mu$-entailed cube trivially holds.

Lemma 1

For a QBF expression $\phi$, if C is a $\mu$-contradicted clause, and $\mu '$ is an assignment such that $A[\mu ']=A[\mu ]$ and $S(\mu )~\subseteq ~ S(\mu ')$, then C is also a $\mu '$-contradicted clause.

Proof

By the definition of $\mu$-contradicted clause, we must show that for any literal $l~\in ~C$, $l~\notin ~\mu '$, and for each existential literal $l~\in ~C$, we have $\lnot {l}~\in ~\mu '$. We consider two cases: 1) $l~\in ~C$ and l is existential. Then, since C is $\mu$-contradicted, we have $\lnot {l}~\in ~\mu$. Because $S(\mu )~\subseteq ~S(\mu ')$, we have $\lnot {l}~\in ~\mu '$ as well. In addition, since $\mu '$ is an assignment, whenever $\lnot {l}~\in ~\mu '$, $l~\notin ~\mu '$.

2) Suppose $l~\in ~C$ and l is universal. Then, because $A[\mu ']=A[\mu ]$, we can easily show that $l~\notin ~\mu '$.

By definition 2, we conclude that C is also a $\mu '$-contradicted clause.

Lemma 2

For a QBF expression $\phi$, if T is a $\mu$-entailed cube, and $\mu '$ is an assignment such that $A[\mu ']=A[\mu ]$ and $S(\mu )~\subseteq ~ S(\mu ')$, then T is also a $\mu '$-entailed cube.

Proof

Analog to the proof of lemma 1, the lemma trivially holds by the definition of $\mu$-entailed cubes.

Note that in iteration t, if $\phi [t]_{\mu ^{l}_{t}}$ contains a contradicted clause, we can easily replace $R^{l}_{t}$ with the contradicted clause. Otherwise, because of the consequence of proposition 1, we have

$$\begin{aligned} S(\mu ^{l}_{t'})~\subseteq ~S(\mu ^{l}_{t}) \end{aligned}$$

(2)

and,

$$\begin{aligned} A(\mu ^{l}_{t'})~=~A(\mu ^{l}_{t}) \end{aligned}$$

(3)

Since we only execute conflict-driven clause learning and solution-driven backjumping, the resolvent we initialized for a node is $\mu$-contradicted or $\mu$-entailed at the time it is calculated. According to lemma 1 and lemma 2, the $\mu ^{l}_{t'}$-contradicted clause (resp. $\mu ^{l}_{t'}$-entailed cube) we have calculated previously is still a $\mu ^{l}_{t}$-contradicted clause (resp. $\mu ^{l}_{t}$-entailed cube). Therefore, as long as $\phi [t]_{\mu ^{l}_{t}}$ does not contain a contradicted clause, we can reuse the reason we have calculated in a previous iteration, and the method we have discussed just now is correct.

Note that this discussion also explains the exact difficulty of combining BFS with SDCL. When SDCL is activated, universal unit propagation comes in, proposition 1 does not hold anymore, because when a node is visited more than once, it is possible that $A(\mu _{t'})~\ne ~A(\mu _{t})$ due to universal unit propagation. If this condition is falsified, lemma 1 does not hold, and the resolvent we previously stored in a node might not be reused.