Almost-polynomial ratio ETH-hardness of approximating densest k-subgraph

Pasin Manurangsi
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引用次数: 143

Abstract

In the Densest k-Subgraph (DkS) problem, given an undirected graph G and an integer k, the goal is to find a subgraph of G on k vertices that contains maximum number of edges. Even though Bhaskara et al.'s state-of-the-art algorithm for the problem achieves only O(n1/4 + ϵ) approximation ratio, previous attempts at proving hardness of approximation, including those under average case assumptions, fail to achieve a polynomial ratio; the best ratios ruled out under any worst case assumption and any average case assumption are only any constant (Raghavendra and Steurer) and 2O(log2/3 n) (Alon et al.) respectively. In this work, we show, assuming the exponential time hypothesis (ETH), that there is no polynomial-time algorithm that approximates Densest k-Subgraph to within n1/(loglogn)c factor of the optimum, where c > 0 is a universal constant independent of n. In addition, our result has perfect completeness, meaning that we prove that it is ETH-hard to even distinguish between the case in which G contains a k-clique and the case in which every induced k-subgraph of G has density at most 1/n-1/(loglogn)c in polynomial time. Moreover, if we make a stronger assumption that there is some constant ε > 0 such that no subexponential-time algorithm can distinguish between a satisfiable 3SAT formula and one which is only (1 - ε)-satisfiable (also known as Gap-ETH), then the ratio above can be improved to nf(n) for any function f whose limit is zero as n goes to infinity (i.e. f ϵ o(1)).
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近似最密集k子图的几乎多项式比率eth -硬度
在密度k-子图(DkS)问题中,给定一个无向图G和一个整数k,目标是在k个顶点上找到包含最大边数的G的子图。尽管Bhaskara等人对该问题的最先进算法仅实现了O(n1/4 + λ)近似比,但之前证明近似硬度的尝试,包括在平均情况假设下的尝试,都未能实现多项式比;在任何最坏情况假设和任何平均情况假设下排除的最佳比率分别为任何常数(Raghavendra和Steurer)和2O(log2/3 n) (Alon等人)。在这项工作中,我们证明,假设指数时间假设(ETH),不存在多项式时间算法逼近最优的n1/(loglog)c因子,其中c > 0是一个独立于n的普遍常数。此外,我们的结果具有完美的完备性。这意味着我们证明了很难区分G包含k团的情况和G的每个诱导k子图在多项式时间内密度不超过1/n-1/(loglog)c的情况。此外,如果我们做出一个更强的假设,即存在一些常数ε> 0,使得没有子指数时间算法可以区分一个可满足的3SAT公式和一个只有(1 - ε)-可满足的(也称为Gap-ETH),那么对于任何函数f,其极限为零,当n趋于无穷(即f λ o(1)),上述比率可以改进为nf(n)。
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