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Tree versus tree of preorder induced by rainbow forbidden subgraphs

Published 9 Jan 2026 in math.CO | (2601.05497v1)

Abstract: A subgraph $H$ of an edge-colored graph $G$ is rainbow if all the edges of $H$ receive different colors. If $G$ does not contain a rainbow subgraph isomorphic to $H$, we say that $G$ is rainbow $H$-free. For connected graphs $H_1$ and $H_2$, if there exists an integer $t=t(H_1,H_2)$ such that every rainbow $H_1$-free edge-colored complete graph colored with $t$ or more colors is rainbow $H_2$-free, then we write $H_1\le H_2$. The binary relation $\le$ is reflexive and transitive, and hence it is a preorder. For graphs $H_1$ and $H_2$, we write $H_1 \equiv H_2$ if both $H_1 \le H_2$ and $H_2 \le H_1$ hold. Then $\equiv$ is an equivalence relation. If $H_1$ is a subgraph of $H_2$, then trivially $H_1\le H_2$ holds. On the other hand, there exists a pair $(H_1, H_2)$ such that $H_1$ is a proper supergraph of $H_2$ and $H_1\le H_2$ holds. Q.~Cui, Q.~Liu, C.~Magnant and A.~Saito [Discrete Math. {\bf 344} (2021) Article Number 112267] characterized these pairs. %On the other hand, there are few known results regarding the study of $\leq$ for the incomparable with respect to $\subseteq$. Cui et al. found pairs of graphs $H_1$ and $H_2$ such that $H_1 \leq H_2$ and $H_2 \leq H_1$, that is, non-singleton equivalence class with respect to $\le$. However, we have not found any other non-singleton equivalence class with respect to $\le$ except for those discovered by Cui et al. In this paper. we investigate the existence of non-singleton equivalence class with respect to $\le$ by focusing on trees.

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