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Revision #1 to TR16-195 | 10th April 2017 11:27

Almost-Polynomial Ratio ETH-Hardness of Approximating Densest $k$-Subgraph

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Revision #1
Authors: Pasin Manurangsi
Accepted on: 10th April 2017 11:27
Downloads: 209
Keywords: 


Abstract:

In the Densest $k$-Subgraph 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 the state-of-the-art algorithm for the problem achieves only $O(n^{1/4 + \varepsilon})$ approximation ratio (Bhaskara et al., 2010), 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, 2010) and $2^{\Omega(\log^{2/3} n)}$ (Alon et al., 2011) 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 $n^{1/(\log \log n)^c}$ 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 $1/n^{-1/(\log \log n)^c}$ in polynomial time.
Moreover, if we make a stronger assumption that there is some constant $\varepsilon > 0$ such that no subexponential-time algorithm can distinguish between a satisfiable 3SAT formula and one which is only $(1 - \varepsilon)$-satisfiable (also known as Gap-ETH), then the ratio above can be improved to $n^{f(n)}$ for any function $f$ whose limit is zero as $n$ goes to infinity (i.e. $f \in o(1)$).



Changes to previous version:

A lemma proved in the previous version was in fact proved before by Alon. We add the correct reference and remove our proof, which is almost the same as Alon's proof, from the current version.


Paper:

TR16-195 | 19th November 2016 02:29

Almost-Polynomial Ratio ETH-Hardness of Approximating Densest $k$-Subgraph





TR16-195
Authors: Pasin Manurangsi
Publication: 5th December 2016 20:37
Downloads: 906
Keywords: 


Abstract:

In the Densest $k$-Subgraph 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 the state-of-the-art algorithm for the problem achieves only $O(n^{1/4 + \varepsilon})$ approximation ratio (Bhaskara et al., 2010), 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, 2010) and $2^{\Omega(\log^{2/3} n)}$ (Alon et al., 2011) 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 $n^{1/(\log \log n)^c}$ 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 $1/n^{-1/(\log \log n)^c}$ in polynomial time.
Moreover, if we make a stronger assumption that there is some constant $\varepsilon > 0$ such that no subexponential-time algorithm can distinguish between a satisfiable 3SAT formula and one which is only $(1 - \varepsilon)$-satisfiable (also known as Gap-ETH), then the ratio above can be improved to $n^{f(n)}$ for any function $f$ whose limit is zero as $n$ goes to infinity (i.e. $f \in o(1)$).



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