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Unavoidable patterns in locally balanced colourings

Published online by Cambridge University Press:  01 June 2023

Nina Kamčev*
Affiliation:
Department of Mathematics, Faculty of Science, University of Zagreb, Zagreb, Croatia
Alp Müyesser
Affiliation:
University College London, London, UK
*
Corresponding author: Nina Kamčev; Email: nina.kamcev@math.hr
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Abstract

Which patterns must a two-colouring of $K_n$ contain if each vertex has at least $\varepsilon n$ red and $\varepsilon n$ blue neighbours? We show that when $\varepsilon \gt 1/4$, $K_n$ must contain a complete subgraph on $\Omega (\log n)$ vertices where one of the colours forms a balanced complete bipartite graph.

When $\varepsilon \leq 1/4$, this statement is no longer true, as evidenced by the following colouring $\chi$ of $K_n$. Divide the vertex set into $4$ parts nearly equal in size as $V_1,V_2,V_3, V_4$, and let the blue colour class consist of the edges between $(V_1,V_2)$, $(V_2,V_3)$, $(V_3,V_4)$, and the edges contained inside $V_2$ and inside $V_3$. Surprisingly, we find that this obstruction is unique in the following sense. Any two-colouring of $K_n$ in which each vertex has at least $\varepsilon n$ red and $\varepsilon n$ blue neighbours (with $\varepsilon \gt 0$) contains a vertex set $S$ of order $\Omega _{\varepsilon }(\log n)$ on which one colour class forms a balanced complete bipartite graph, or which has the same colouring as $\chi$.

Information

Type
Paper
Copyright
© The Author(s), 2023. Published by Cambridge University Press
Figure 0

Figure 1. The four types of colourings in $\mathcal{F}_t$. The circles denote cliques of size $t$, whereas the lines connecting the circles denote complete bipartite graphs. In the first row, one colour class forms a $K_{t,t}$, whereas on the second row, one colour class forms a $K_t$.

Figure 1

Figure 2. An illustration of the relevant two-coloured and totally two-coloured graphs.

Figure 2

Figure 3. A three-colouring of $K_n$ in which any locally balanced subgraph has at least $6$ vertices, where $6$ is the number of parts. This colouring is locally $(1/6)$-balanced. Colourings of this type (where two colour classes form a blow-up of an alternating cycle) show that there is no analogue of Theorem 1.4 for more than two colours.