Non-conformity colorings

Question

Motivation. Recently I attended a little party (adhering to physical distancing and in accordance to other COVID19-related laws). An attendee told me that he chose drink $Y$ since at least half of his acquaintances at the party had drink $X$. I was sober enough to put this into a graph-theoretic property - and that's what this question is about.

Formal version. If $G=(V,E)$ is a simple, undirected graph and $v\in V$, let $N(v) = \{w\in V:\{v,w\}\in E\}$. Given a cardinal $\kappa > 0$ we say a map $c:V\to \kappa$ is a non-conformity coloring if for all $v\in V$ with $N(v)\neq \varnothing$ we have $$|N(v)\cap c^{-1}(\{c(v)\})| < |N(v)\setminus c^{-1}(\{c(v)\})|.$$ (Note that this is the formal version of "$v$ has another drink than more than half of $v$'s neighbors" if we view $c$ as "drink assignment".) The non-conformity chromatic number $\chi_{nc}(G)$ is the smallest cardinal $\kappa$ such that there is a non-conformity coloring $c:V\to \kappa$.

If $K_3$ denotes the complete graph on $3$ vertices, it is easy to see that $\chi_{nc}(K_3) = 3$.

Question. Given a cardinal $\kappa > 3$, is there a graph $G$, finite or infinite, such that $\chi_{nc}(G) = \kappa$?

Answer 1

If $\kappa$ is an infinite cardinal, there is a graph $G$ of order $\kappa$ with $\chi_{nc}(G)=\kappa$. If $\kappa$ is regular, $G$ can be a complete graph; if $G$ is singular, $G$ can be a disjoint union of complete graphs.

If $G$ is a finite graph, then $\chi_{nc}(G)\le3$; just take a $3$-coloring of $G$ which minimizes the number of edges joining two vertices of the same color. The usual sort of compactness argument shows that $\chi_{nc}(G)\le3$ holds if $G$ is locally finite. Thus, if $\kappa\gt3$, there is no locally finite graph $G$ with $\chi_{nc}(G)=\kappa$.

For $n\lt\aleph_0$ the complete $n$-partite graph $G=K_{\underbrace{\aleph_0,\dots,\aleph_0}_n}$ is a countable graph with $\chi_{nc}(G)=n$.