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The UGC Hardness Threshold of the L^sub p^ Grothendieck Problem

机译:L ^ sub p ^ Grothendieck问题的UGC硬度阈值

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For p ≥ 2, the authors consider the problem of, given an n x n matrix A = (aij) whose diagonal entries vanish, approximating in polynomial time the number. When p = 2 this is simply the problem of computing the maximum eigenvalue of A, whereas for p = ∞ it is the Grothendieck problem on the complete graph, which was shown to have a O log n approximation algorithm in Nemirovski et al. The authors design a polynomial time algorithm which, given p ≥ 2 and an n x n matrix A = (aij) with zeros on the diagonal, computes Opt p (A) up to a factor p/e+30 log p. On the other hand, assuming the unique games conjecture (UGC) they show that it is NP-hard to approximate Opt p (A) up to a factor smaller than p/e +1/4. Hence as p [arrow right] ∞ the UGC-hardness threshold for computing Opt p (A) is exactly (p/e) 1+o(1)). Show less
机译:对于p≥2,作者考虑的问题是,给定一个n x n矩阵A =(aij),其对角线项消失,在多项式时间内近似该数。当p = 2时,这只是计算A的最大特征值的问题,而对于p =∞而言,这是完整图中的Grothendieck问题,在Nemirovski等人的研究中,该算法具有O log n近似算法。作者设计了多项式时间算法,该算法在p≥2且对角线上为零的n x n矩阵A =(aij)时,计算出Opt p(A)直到系数p / e + 30 log p。另一方面,假设唯一游戏猜想(UGC),他们表明对Opt p(A)进行近似直到小于p / e +1/4的因数都是NP难的。因此,当p [向右箭头]∞时,用于计算Opt p(A)的UGC硬度阈值正好是(p / e)1 + o(1))。显示较少

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