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Sum-Of-Squares Bounds via Boolean Function Analysis

机译:通过布尔函数分析求平方和边界

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We introduce a method for proving bounds on the SoS rank based on Boolean Function Analysis and Approximation Theory. We apply our technique to improve upon existing results, thus making progress towards answering several open questions. We consider two questions by Laurent. First, finding what is the SoS rank of the linear representation of the set with no integral points. We prove that the SoS rank is between ceil[n/2] and ceil[~ n/2 +sqrt{n log{2n}} ~]. Second, proving the bounds on the SoS rank for the instance of the Min Knapsack problem. We show that the SoS rank is at least Omega(sqrt{n}) and at most ceil[{n+ 4 ceil[sqrt{n} ~]}/2]. Finally, we consider the question by Bienstock regarding the instance of the Set Cover problem. For this problem we prove the SoS rank lower bound of Omega(sqrt{n}).
机译:我们介绍了一种基于布尔函数分析和逼近理论的SoS等级边界证明方法。我们运用我们的技术来改善现有结果,从而在回答几个悬而未决的问题上取得进展。我们考虑洛朗的两个问题。首先,找到没有积分点的线性表示形式的SoS等级是多少。我们证明SoS等级在ceil [n / 2]和ceil [〜n / 2 + sqrt {n log {2n}}〜]之间。其次,证明最小背包问题的SoS等级界限。我们显示SoS等级至少为Omega(sqrt {n}),至多为ceil [{n + 4 ceil [sqrt {n}〜]} / 2]。最后,我们考虑Bienstock关于Set Cover问题实例的问题。对于此问题,我们证明了SoS等级为Omega(sqrt {n})的下限。

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