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Large independent sets in recursive Markov random graphs

Published 10 Jul 2022 in math.CO, cs.DM, math.OC, and math.PR | (2207.04514v4)

Abstract: Computing the maximum size of an independent set in a graph is a famously hard combinatorial problem that has been well-studied for various classes of graphs. When it comes to random graphs, only the classical Erd\H{o}s-R\'enyi-Gilbert random graph Gn,pG_{n,p} has been analysed and shown to have largest independent sets of size Θ(log⁡n)\Theta(\log{n}) w.h.p. This classical model does not capture any dependency structure between edges that can appear in real-world networks. We initiate study in this direction by defining random graphs Gn,prG^{r}_{n,p} whose existence of edges is determined by a Markov process that is also governed by a decay parameter r∈(0,1]r\in(0,1]. We prove that w.h.p. Gn,prG^{r}_{n,p} has independent sets of size (1−r2+ϵ)nlog⁡n(\frac{1-r}{2+\epsilon}) \frac{n}{\log{n}} for arbitrary ϵ>0\epsilon > 0, which implies an asymptotic lower bound of Ω(π(n))\Omega(\pi(n)) where π(n)\pi(n) is the prime-counting function. This is derived using bounds on the terms of a harmonic series, Tur\'an bound on stability number, and a concentration analysis for a certain sequence of dependent Bernoulli variables that may also be of independent interest. Since Gn,prG^{r}_{n,p} collapses to Gn,pG_{n,p} when there is no decay, it follows that having even the slightest bit of dependency (any r<1r < 1) in the random graph construction leads to the presence of large independent sets and thus our random model has a phase transition at its boundary value of r=1r=1. For the maximal independent set output by a greedy algorithm, we deduce that it has a performance ratio of at most 1+log⁡n(1−r)1 + \frac{\log{n}}{(1-r)} w.h.p. when the lowest degree vertex is picked at each iteration, and also show that under any other permutation of vertices the algorithm outputs a set of size Ω(n1/1+τ)\Omega(n^{1/1+\tau}), where τ=1/(1−r)\tau=1/(1-r), and hence has a performance ratio of O(n12−r)O(n^{\frac{1}{2-r}}).

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