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Forbidden Subgraphs Generating Almost the Same Sets

Published online by Cambridge University Press:  11 July 2013

SHINYA FUJITA ,
MICHITAKA FURUYA  and
KENTA OZEKI
SHINYA FUJITA
Affiliation:
Department of Integrated Design Engineering, Maebashi Institute of Technology, 460-1 Kamisadori, Maebashi, Gunma 371-0816, Japan (e-mail: [email protected])
MICHITAKA FURUYA
Affiliation:
Department of Mathematical Information Science, Tokyo University of Science, 1-3 Kagurazaka, Shinjuku-ku, Tokyo 162-8601, Japan (e-mail: [email protected])
KENTA OZEKI
Affiliation:
National Institute of Informatics, 2-1-2 Hitotsubashi, Chiyoda-ku, Tokyo 101-8430, Japan JST, ERATO, Kawarabayashi Large Graph Project, Japan (e-mail: [email protected])
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Abstract

Let $\mathcal{H}$ be a set of connected graphs. A graph G is said to be $\mathcal{H}$-free if G does not contain any element of $\mathcal{H}$ as an induced subgraph. Let $\mathcal{F}_{k}(\mathcal{H})$ be the set of k-connected $\mathcal{H}$-free graphs. When we study the relationship between forbidden subgraphs and a certain graph property, we often allow a finite exceptional set of graphs. But if the symmetric difference of $\mathcal{F}_{k}(\mathcal{H}_{1})$ and $\mathcal{F}_{k}(\mathcal{H}_{2})$ is finite and we allow a finite number of exceptions, no graph property can distinguish them. Motivated by this observation, we study when we obtain a finite symmetric difference. In this paper, our main aim is the following. If $|\mathcal{H}|\leq 3$ and the symmetric difference of $\mathcal{F}_{1}(\{H\})$ and $\mathcal{F}_{1}(\mathcal{H})$ is finite, then either $H\in \mathcal{H}$ or $|\mathcal{H}|=3$ and H=C3. Furthermore, we prove that if the symmetric difference of $\mathcal{F}_{k}(\{H_{1}\})$ and $\mathcal{F}_{k}(\{H_{2}\})$ is finite, then H1=H2.

Keywords

05C75
Type
Paper
Information
Combinatorics, Probability and Computing , Volume 22 , Issue 5 , September 2013 , pp. 733 - 748
Copyright
Copyright © Cambridge University Press 2013 

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