关于随机公式的最大可满足性
On the Maximum Satisfiability of Random Formulas
摘要：
最大可满足性是一个典型的NP-难优化问题，在随机情况下经验上很难。假设一个由$k$ -子句组成的合取范式（CNF）公式是$p$ -可满足的，如果存在一个真值赋值满足所有子句的$1-2^{-k}+p 2^{-k}$（注意每个$k$-CNF都是0-可满足的）.同样，让$F_k(n,m)$表示通过从所有可能的$k$ -子句中均匀且独立地选择$m$而形成的$n$变量上的随机$k$-CNF.很容易证明，对于每个$k>1$和$(0,1]$中的每个$p$，存在$R_k(p)$，使得如果$r >R_k(p)$，则当$n$趋于无穷大时，$F_k(n,rn)$是$p$ -可满足的概率趋于0。我们证明了存在序列$δ_k \to 0$使得如果$r <(1-δ_k) R_k(p)$，则当$n$趋于无穷大时，$F_k(n,rn)$是$p$ -可满足的概率趋于1。序列$δ_k$在$k$中以指数速度趋向于0。
英文摘要：
Maximum satisfiability is a canonical NP-hard optimization problem that appears empirically hard for random instances. Let us say that a Conjunctive normal form (CNF) formula consisting of $k$-clauses is $p$-satisfiable if there exists a truth assignment satisfying $1-2^{-k}+p 2^{-k}$ of all clauses (observe that every $k$-CNF is 0-satisfiable). Also, let $F_k(n,m)$ denote a random $k$-CNF on $n$ variables formed by selecting uniformly and independently $m$ out of all possible $k$-clauses. It is easy to prove that for every $k>1$ and every $p$ in $(0,1]$, there is $R_k(p)$ such that if $r >R_k(p)$, then the probability that $F_k(n,rn)$ is $p$-satisfiable tends to 0 as $n$ tends to infinity. We prove that there exists a sequence $δ_k \to 0$ such that if $r <(1-δ_k) R_k(p)$ then the probability that $F_k(n,rn)$is $p$-satisfiable tends to 1 as $n$ tends to infinity. The sequence $δ_k$ tends to 0 exponentially fast in $k$.
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