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Chromatic number of randomly augmented graphs

Published 20 Jun 2024 in math.CO | (2406.14223v2)

Abstract: An extension of the Erd\H{o}s-Renyi random graph model Gn,pG_{n,p} is the model of perturbed graphs introduced by Bohman, Frieze and Martin (Bohman, Frieze, Martin 2003). This is a special case of the model of randomly augmented graphs studied in this paper. An augmented graph denoted by pertH,ppert_{H,p} is the union of a deterministic host graph and a random graph Gn,pG_{n,p}. Among the first problems in perturbed graphs has been the question how many random edges are needed to ensure Hamiltonicity of the graph. This question was answered in the paper by Bohman, Frieze and Martin. The host graph is often chosen to be a dense graph. In recent years several papers on combinatorial problems in perturbed graphs were published, e.g. on the emergence of powers of Hamiltonian cycles (Dudek, Reiher, Ruci\'nski, Schacht 2020), some positional games played on perturbed graphs (Clemens, Hamann, Mogge, Parczyk, 2020) and the behavior of multiple invariants e.g. fixed clique size (Bohman, Frieze, Krivelevich, Martin, 2004). In this paper we study the chromatic number of randomly augmented graphs. We concentrate on a host graph HH with chromatic number o(n)o(n), augmented by a Gn,pG_{n,p} with n<sup>13</sup>+δp(n)1δn<sup>{-\frac{1}{3}</sup> + \delta}\leq p(n) \leq 1-\delta for some δ(0,1)\delta \in (0,1). Our main result is an upper bound for the chromatic number: we show that asymptotically almost surely χ(pertH,p)(1+o(1))nlog(b)2(log(n)log(χ(H))\chi(pert_{H,p}) \leq (1+o(1)) \cdot \frac{n \log(b)}{2 (\log(n) - \log(\chi(H))} where b=(1p)<sup>1b = (1-p)<sup>{-1}. This result collapses to the famous theorem of Bollob\'as (1988), when HH is the empty host graph, thus our result can be regarded as a generalization of the latter. Our proof is not constructive. Further, we give a constructive coloring algorithm, when the chromatic number of the host graph is at most nlog(n)<sup>α,\frac{n}{\log(n)<sup>{\alpha}}, $\alpha&gt;\frac{1}{2}.$

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