A Short Implicant of a CNF Formula with Many Satisfying Assignments

Abstract

Consider any Boolean function $F(X_1,\ldots ,X_N)$ that has more than $2^{-N^{\delta }}\cdot 2^N$ satisfying assignments for some $\delta$, $0<\delta <1$, and that can be expressed by a CNF formula with at most $N^d$ clauses for some $d>0$. Then how many variables do we need to fix in order to satisfy $F$? We show that one can always find some “short” partial assignment on which $F$ evaluates to $1$ by fixing at most $\alpha N$ variables for some constant $\alpha <1$; that is, $F$ has an implicant of size $\le \alpha N$. A lower bound for such $\alpha$ is also shown in terms of $\delta$ and $d$. We also discuss an algorithm for obtaining a short partial assignment. For any $\delta$ and $\varepsilon$ such that $0<\delta +\varepsilon <1$, we show a deterministic algorithm that finds a short partial assignment in ${\widetilde{O}}(2^{N^{\beta }})$-time for some $\beta <1$ for any CNF formula with at most $N^{1+\varepsilon }$ clauses having more than $2^{-N^{\delta }}\cdot 2^N$ satisfying assignments. (This is an extended abstract, and some detailed explanations are omitted; see [6] for the details.)