On a colored Tur\'an problem of Diwan and Mubayi

TL;DR

This paper characterizes the maximum size of two red-blue graphs on the same vertex set avoiding a colored copy of a fixed graph H, generalizing Turán's theorem and extending to multiple colors and specific cycles.

Contribution

It provides an asymptotic formula for the extremal number x(n, H) based on the chromatic number and a new parameter al M(H), generalizing classical Tur1n results.

Findings

Asymptotic characterization of x(n, H) in terms of hi(H) and al M(H)

Extension of results to more than two colors with tight bounds

Analysis of x(n, H) for cycles with red-blue edge coloring, showing tight bounds

Abstract

Suppose that $R$ (red) and $B$ (blue) are two graphs on the same vertex set of size $n$, and $H$ is some graph with a red-blue coloring of its edges. How large can $R$ and $B$ be if $R\cup B$ does not contain a copy of $H$? Call the largest such integer $\mathrm{mex}(n, H)$. This problem was introduced by Diwan and Mubayi, who conjectured that (except for a few specific exceptions) when $H$ is a complete graph on $k+1$ vertices with any coloring of its edges $\mathrm{mex}(n,H)=\mathrm{ex}(n, K_{k+1})$. This conjecture generalizes Tur\'an's theorem. Diwan and Mubayi also asked for an analogue of Erd\H{o}s-Stone-Simonovits theorem in this context. We prove the following asymptotic characterization of the extremal threshold in terms of the chromatic number $\chi(H)$ and the \textit{reduced maximum matching number} $\mathcal{M}(H)$ of $H$. $$\mathrm{mex}(n, H)=\left(1- \frac{1}{2(\chi(H)-1)} - \Omega\left(\frac{\mathcal{M}(H)}{\chi(H)^2}\right)\right)\frac{n^2}{2}.$$ $\mathcal{M}(H)$ is, among the set of proper $\chi(H)$-colorings of $H$, the largest set of disjoint pairs of color classes where each pair is connected by edges of just a single color. The result is also proved for more than $2$ colors and is tight up to the implied constant factor. We also study $\mathrm{mex}(n, H)$ when $H$ is a cycle with a red-blue coloring of its edges, and we show that $\mathrm{mex}(n, H)\lesssim \frac{1}{2}\binom{n}{2}$, which is tight.