关于ess（f）与cnf_size（f）之间的差距
On the gap between ess(f) and cnf_size(f)
摘要：
给定一个布尔函数f，ess（f）表示证伪f的最大赋值集合，其中没有两个证伪f的公共蕴涵。虽然ess（f）$显然是cnf_size（f）的下界（在CNF公式中f的子句的最小数目），但Cepek等人表明它一般不是紧下界。他们给出了函数f的例子，其中ess（f）和cnf_size（f）之间有一个小的差距。我们展示了明显更大的差距。我们证明了对于任意的布尔函数，间隙可以是n中的指数，对于Horn函数，间隙可以是Theta（sqrt{n}），其中n是f的变量数。我们还引入了ess（f）的一个自然推广，我们称之为ess_k（f），它是最大的赋值集，其中没有k个赋值能证伪f的一个公共蕴涵。
英文摘要：
Given a Boolean function f, the quantity ess(f) denotes the largest set of assignments that falsify f, no two of which falsify a common implicate of f. Although ess(f)$ is clearly a lower bound on cnf_size(f) (the minimum number of clauses in a CNF formula for f), Cepek et al. showed that it is not, in general, a tight lower bound. They gave examples of functions f for which there is a small gap between ess(f) and cnf_size(f). We demonstrate significantly larger gaps. We show that the gap can be exponential in n for arbitrary Boolean functions, and Theta(sqrt{n}) for Horn functions, where n is the number of variables of f. We also introduce a natural extension of the quantity ess(f), which we call ess_k(f), which is the largest set of assignments, no k of which falsify a common implicate of f.
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