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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\title{\bf Forbidden triples generating a finite set \\ of $3$-connected graphs}

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\author{Yoshimi Egawa\\
\small Dept. of Mathematical Information Science\\[-0.8ex]
\small Tokyo University of Science\\[-0.8ex] 
\small 1-3 Kagurazaka, Shinjuku-ku\\[-0.8ex] 
\small Tokyo 162-8601, Japan\\
\and
Jun Fujisawa\thanks{Supported by JSPS Grant-in-Aid for Young Scientists (B) (22740068) and Keio Gijuku Academic Development Funds.}\\
\small Faculty of Business and Commerce\\[-0.8ex]
\small Keio University\\[-0.8ex]
\small Hiyoshi 4-1-1, Kohoku-ku, Yokohama\\[-0.8ex]
\small Kanagawa 223-8521, Japan\\
\small\tt fujisawa@fbc.keio.ac.jp\\
\and
Michitaka Furuya\thanks{Supported by JSPS Grant-in-Aid for Young Scientists (B) (26800086).}\\
\small Dept. of Mathematical Information Science\\[-0.8ex]
\small Tokyo University of Science\\[-0.8ex] 
\small 1-3 Kagurazaka, Shinjuku-ku\\[-0.8ex] 
\small Tokyo 162-8601, Japan\\
\small\tt michitaka.furuya@gmail.com\\
\and
Michael D. Plummer\\
\small Dept. of Mathematics\\[-0.8ex] 
\small Vanderbilt University\\[-0.8ex] 
\small Nashville, Tennessee 37240, U.S.A\\
\small\tt michael.d.plummer@vanderbilt.edu\\
\and
Akira Saito\thanks{Partially supposed by JSPS Grant-in-Aid for Scientific Research (C) (22500018).}\\
\small Dept. of Information Science\\[-0.8ex] 
\small Nihon University\\[-0.8ex] 
\small Sakurajousui 3-25-40, Setagaya-ku\\[-0.8ex] 
\small Tokyo 156-8550, Japan\\
\small\tt asaito@chs.nihon-u.ac.jp
}






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\date{\dateline{}{}\\
\small Mathematics Subject Classifications: 05C75}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
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\begin{abstract}
For a graph $G$ and a set $\F$ of connected graphs, $G$ is said be $\F$-free if $G$ does not contain any member of $\F$ as an induced subgraph.
We let $\G _{3}(\F)$ denote the set of all $3$-connected $\F$-free graphs.
This paper is concerned with sets $\F$ of connected graphs such that $|\F|=3$ and $\G _{3}(\F)$ is finite.
Among other results, we show that for an integer $m\geq 3$ and a connected graph $T$ of order greater than or equal to $4$, $\G _{3}(\{K_{4},K_{2,m},T\})$ is finite if and only if $T$ is a path of order $4$ or $5$.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:}
forbidden subgraph; forbidden triple; $3$-connected graph
\end{abstract}








%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}\label{sec1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this paper, we consider only finite undirected simple graphs.

Let $\G$ denote the set of connected graphs with order greater than or equal to three.
For a graph $G$ and for $F\in \G$, $G$ is said to be {\it $F$-free} if $G$ does not contain $F$ as an induced subgraph and, for $\F \subseteq \G$, $G$ is said to be {\it $\F$-free} if $G$ is $F$-free for every $F\in \F$.
For an integer $k\geq 1$ and $\F\subseteq \G $, we let $\G _{k}$ denote the set of all $k$-connected graphs, and let $\G _{k}(\F)$ denote the set of all $\F$-free graphs belonging to $\G _{k}$.
Thus
$$
\G _{k}(\F):=\{G\mid G \mbox{ is a }k\mbox{-connected }\F\mbox{-free graph} \}.
$$
This paper is concerned with subsets $\F$ of $\G$ such that $\G _{k}(\F)$ is a finite set.
In this context, members of $\F$ are often referred to as forbidden subgraphs.
For detailed historical background and related results, we refer the reader to \cite{FPS}.

The following result can be found in \cite{D}. 
(Here $K_{n}$ denotes the complete graph of order $n$, $P_{l}$ denotes the path of order $l$ and, in general, $K_{m_{1},m_{2}}$ denotes the complete bipartite graph with partite sets having cardinalities $m_{1}$ and $m_{2}$.)

\begin{Thm}[Diestel~\cite{D}; Chapter 9]%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{Diestel th}
For $\F \subseteq \G $, $\G _{1}(\F)$ is finite if and only if $K_{n},K_{1,m},P_{l}\in \F $ for some integers $n\geq 3$, $m\geq 2$ and $l\geq 3$.
\end{Thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

For $k\geq 2$, it is unlikely that a general result like Theorem~\ref{Diestel th} holds.
Thus we confine ourselves to the case where $|\F|$ is ``small''.
It is known that for any $k\geq 2$, there is no $\F \subseteq \G $ with $|\F|=1$ such that $\G _{k}(\F)$ is finite.
(See \cite{FPS}; Theorem 2.)
Further, those subsets $\F$ of $\G$ with $|\F|=2$ for which $\G_{k}(\F)$ is finite are determined for $k\leq 6$ in \cite{FPS}.
Here we are interested in the case where $|\F|=3$.
Note that a connected $K_{1,2}$-free graph is a complete graph.
Hence if $K_{1,2}\in \F$, then $\G _{k}(\F)$ is finite if and only if $K_{n}\in \F$ for some $n\geq 3$, and there is no point in forbidding two more graphs.
%Note that for an integer $k\geq 2$ and a subset $\F$ of $\G$ with $K_{1,2}\in \F$, $\G _{k}(\F)$ is finite if and only if $K_{n}\in \F $ for some $n\geq 3$.
Thus when we discuss $\G _{k}(\F)$ with $|\F|=3$, we usually assume $K_{1,2}\not\in \F$.
For $k=2$, the following theorem is proved in \cite{FPS}.

\begin{Thm}[Fujisawa, Plummer and Saito~\cite{FPS}]%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{tri 2-conn}
Let $\F$ be a subset of $\G$ with $|\F|=3$ and $K_{1,2}\not\in \F$.
Then $\G _{2}(\F)$ is finite if and only if one of the following holds:
\begin{enumerate}
\item[{\upshape(i)}]
$\F =\{K_{3},K_{2,m},P_{l}\}$ for some integers $m$ and $l$ with $m\geq 3$ and $4\leq l\leq 5$;
\item[{\upshape(ii)}]
$\F =\{K_{3},K_{2,2},P_{6}\}$; or
\item[{\upshape(iii)}]
$\F =\{K_{n},K_{1,m},P_{l}\}$ for some integers $n$, $m$ and $l$ with $n\geq 3$, $m\geq 3$ and $l\geq 4$.
\end{enumerate}
\end{Thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In the present paper, we investigate the case where $|\F|=3$ and $k=3$.
It is easy to see $\G _{3}(\{K_{3},K_{1,3}\})=\emptyset $. (See \cite{FPS}.)
%Note that $\G _{3}(\{K_{3},K_{1,3}\})=\emptyset $.
Thus when we consider $\G _{3}(\F)$, we assume that $\{K_{3},K_{1,3}\}\not\subseteq \F $, in addition to the condition that $K_{1,2}\not\in \F $.
Before stating our results, we make some more definitions.

Let $n$ be an integer with $n\geq 2$.
Let $P=x_{1}x_{2}\cdots x_{n}$ be the path of order $n$, and let $y_{1}$, $y_{2}$, $z_{1}$ and $z_{2}$ be four distinct vertices different from $x_{1},\ldots ,x_{n}$.
We let $Y_{n}$, $Y^{*}_{n}$, $Q_{n}$ and $Q^{*}_{n}$ denote the graphs defined by
$$
V(Y_{n})=V(Q_{n})=V(P)\cup \{y_{1},y_{2}\},~~~V(Y^{*}_{n})=V(Q^{*}_{n})=V(P)\cup \{y_{1},y_{2},z_{1},z_{2}\},
$$
$$
E(Y_{n})=E(P)\cup \{x_{1}y_{1},x_{1}y_{2}\},~~~E(Y^{*}_{n})=E(P)\cup \{x_{1}y_{1},x_{1}y_{2},x_{n}z_{1},x_{n}z_{2}\},
$$
$$
E(Q_{n})=E(Y_{n})\cup \{y_{1}y_{2}\} \mbox{ ~~and~~ } E(Q^{*}_{n})=E(Y^{*}_{n})\cup \{y_{1}y_{2},z_{1}z_{2}\}
$$
(see Figure~\ref{f1}).

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\caption{Graphs $Y_{3}$, $Y^{*}_{3}$, $Q_{3}$ and $Q^{*}_{3}$}
\label{f1}
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A {\it caterpillar} is a tree for which the removal of all endvertices leaves a path.
A complete bipartite graph of the form $K_{1,m}$ with $m\geq 1$ is called a {\it star}.
Let $\T _{0}$ be the set of trees in $\G -\{K_{1,2},K_{1,3}\}$ having maximum degree at most $3$.
Note that $\T _{0}$ does not contain a star.
Let $\T _{1}$ be the set of those caterpillars belonging to $\T _{0}$ in which no two vertices of degree $3$ are adjacent.
Let $\T _{2}=\{P_{l},Y_{m},Y^{*}_{n}\mid l\geq 4,m\geq 3,n\geq 3\}$.
We have $\T _{0}\supseteq \T _{1}\supseteq \T _{2}$.

Let $G$ be a connected graph.
A vertex $v$ of $G$ is called a {\it cutvertex} if $G-v$ is disconnected.
If $G$ has a cutvertex, $G$ is said to be {\it separable}; otherwise, it is said to be {\it nonseparable}.
Note that $K_{1}$ is a nonseparable graph.
A maximal nonseparable subgraph of $G$ is called a {\it block} of $G$.
When $G$ is separable, the {\it block-cutvertex graph} of $G$ is defined to be the bipartite graph $Z$ such that $Z$ has as its partite sets the set of all cutvertices of $G$ and the set of all blocks of $G$ and, for a cutvertex $v$ and a block $B$, $v$ and $B$ are adjacent in $Z$ if and only if $v$ is a vertex of $B$ in $G$.
It is a well-known fact that the block-cutvertex graph of a connected graph is a tree.
A {\it cactus} is a connected graph every block of which is a complete graph of order two or a cycle.
Let $\T ^{*}_{0}$ be the set of those cacti $T$ in $\G -\{K_{1,2},K_{3}\}$ such that all cycles of $T$ are triangles and in the block-cutvertex graph of $T$, the distance between any two vertices corresponding to triangles of $T$ is a multiple of $4$.
Let $\T ^{*}_{1}$ be the set of those members of $\T ^{*}_{0}$ whose block-cutvertex graph is a path.
Let $\T ^{*}_{2}=\{P_{l},Q_{m},Q^{*}_{2n}\mid l\geq 4,m\geq 2,n\geq 1\}$.
We have $\T ^{*}_{0}\supseteq \T ^{*}_{1}\supseteq \T ^{*}_{2}$.

Our main result is as follows.

\begin{thm}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{nec}
Let $\F $ be a subset of $\G $ with $|\F |=3$, $K_{1,2}\not\in \F $ and $\{K_{1,3},K_{3}\}\not\subseteq \F $, and suppose that $\G _{3}(\F )$ is finite.
Then one of the following holds:
\begin{enumerate}
\item[{\upshape(i)}]
$\F =\{K_{3},K_{3,m},T\}$ with $m\geq 3$, where $T\in \T _{2}$;
\item[{\upshape(ii)}]
$\F =\{K_{4},K_{2,m},T\}$ with $m\geq 2$, where $T$ is a path;
\item[{\upshape(iii)}]
$\F =\{K_{3},K_{2,m},T\}$ with $m\geq 2$, where $T\in \T _{0}$ in the case where $m=2$, $T\in \T _{1}$ in the case where $3\leq m\leq 4$, and $T\in \T _{2}$ in the case where $m\geq 5$;
\item[{\upshape(iv)}]
$\F =\{K_{n},K_{1,m},T\}$ with $n\geq 4$ and $m\geq 4$, where $T$ is a path;
\item[{\upshape(v)}]
$\F =\{K_{3},K_{1,m},T\}$ with $m\geq 4$, where $T\in \T _{1}$ in the case where $m=4$, and $T\in \T _{2}$ in the case where $m\geq 5$; or
\item[{\upshape(vi)}]
$\F =\{K_{n},K_{1,3},T\}$ with $n\geq 4$, where $T\in \T ^{*}_{1}$ in the case where $n=4$, and $T\in \T ^{*}_{2}$ in the case where $n\geq 5$.
\end{enumerate}
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The converse of Theorem~\ref{nec} does not hold.
However, if (iv) of Theorem~\ref{nec} holds, then $\G _{3}(\F)$ is finite by Theorem~\ref{Diestel th}.
Also, as we shall state below in Theorem~\ref{ns4}, if (v) holds with $m\geq 5$, then $\G _{3}(\F)$ is finite.
Further when $m$ is ``large'' in cases (i) through (iii) of Theorem~\ref{nec}, we can determine $T$ as follows.

\begin{thm}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns1}
Let $m$ be an integer with $m\geq 4$, and let $T\in \G-\{K_{1,2},K_{1,3}\}$.
Then $\G _{3}(\{K_{3},K_{3,m},T\})$ is finite if and only if $T$ is a path of order $4$ or $5$ or $T=Y_{3}$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns2}
Let $m$ be an integer with $m\geq 3$, and let $T\in \G-\{K_{1,2}\}$.
Then $\G _{3}(\{K_{4},K_{2,m},T\})$ is finite if and only if $T$ is a path of order $4$ or $5$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3}
Let $m$ be an integer with $m\geq 5$, and let $T\in \G-\{K_{1,2},K_{1,3}\}$.
Then $\G _{3}(\{K_{3},K_{2,m},T\})$ is finite if and only if $T$ is either a path of order at most $7$ or an induced subgraph of $Y^{*}_{3}$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thm}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4}
Let $m$ be an integer with $m\geq 5$, and let $T\in \G-\{K_{1,2},K_{1,3}\}$.
Then $\G _{3}(\{K_{3},K_{1,m},T\})$ is finite if and only if $T\in \T _{2}$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We prove Theorem~\ref{nec} in Section~\ref{sec2}.
We prove Theorem~\ref{ns1} in Section~\ref{sec3}, Theorem~\ref{ns2} in Section~\ref{sec4}, Theorem~\ref{ns4} in Section~\ref{sec5}, and Theorem~\ref{ns3} in Section~\ref{sec6}.
Our notation and terminology are standard, and mostly taken from \cite{D}.
Exceptions are as follows.
Let $G$ be a graph.
For $u,v\in V(G)$, $d(u,v)$ denotes the distance between $u$ and $v$.
When $G$ is connected, we define the {\it diameter} $\mbox{diam}(G)$ of $G$ by $\mbox{diam}(G)=\max\{d(u,v)\mid u,v\in V(G)\}$.
Let $u\in V(G)$.
For an integer $i\geq 1$, we let $N_{i}(u)=\{x\in V(G)\mid d(u,x)=i\}$.
We write $N(u)$ for $N_{1}(u)$.
We let $d(u)$ denote the degree of $u$; thus $d(u)=|N(u)|$.
When we need to specify that the underlying graph is $G$, we write $N_{G}(u)$ and $d_{G}(u)$ for $N(u)$ and $d(u)$, respectively.
We let $\Delta (G)=\max\{d(u)\mid u\in V(G)\}$.
For $Y\subseteq V(G)$, we let $N(Y)$ denote the union of $N(u)$ as $u$ ranges over $Y$.
For $X,Y\subseteq V(G)$ with $X\cap Y=\emptyset $, $E(X,Y)$ denotes the set of edges joining a vertex in $X$ and a vertex in $Y$.
When $G$ is connected, a block of $G$ containing at most one cutvertex of $G$ is called an {\it endblock} of $G$.
When $G$ is not necessarily connected, by a cutvertex of $G$, we mean a cutvertex of a component of $G$.
Similarly, by a block (resp. an endblock) of $G$, we mean a block (resp. an endblock) of a component of $G$.
Note that isolated vertices of $G$ are endblocks of $G$.
For an endblock $B$ of $G$, a vertex of $B$ which is not a cutvertex of $G$ is called an {\it internal vertex} of $B$.
For a graph $H$ and an integer $s\geq 2$, we let $sH$ denote the disjoint union of $s$ copies of $H$.
For two graphs $H_{1}$ and $H_{2}$, we let $H_{1}+H_{2}$ denote the join of $H_{1}$ and $H_{2}$.
Finally for $s\geq 4$, $C_{s}$ denotes the cycle of order $s$ and, for $t\geq 5$, we let $W_{t}=C_{t-1}+K_{1}$ denote the {\it wheel graph} of order $t$.

In subsequent arguments, when we prove the finiteness of $\G _{3}(\F)$ for a given family $\F$, we bound the diameter and the maximum degree of a graph $G$ in $\G _{3}(\F)$ from above, and then bound the order in terms of the diameter and the maximum degree.
For this purpose, we make one easy observation.

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{new diam lem}
Let $m\geq 2$ and $k\geq 3$, and let $G$ be a graph with $\Delta (G)\leq m$ and $\mbox{diam}(G)\leq k$.
Then $|V(G)|\leq m^{k}$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $w\in V(G)$.
Then $|N_{i}(w)|\leq m(m-1)^{i-1}$ for each $1\leq i\leq k$.
Hence $|V(G)|\leq 1+m+(\sum_{2\leq i\leq k-1}m(m-1)^{i-1})+m(m-1)^{k-1}\leq m^{k-1}+(\sum_{2\leq i\leq k-1}m^{k-i}(m-1)^{i-1})+m(m-1)^{k-1}=m^{k}$.
\qed

The bound $m^{k}$ in the above lemma is far from sharp, but we use it for the sake of brevity.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A necessary condition}\label{sec2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section, we prove Theorem~\ref{nec}.
We start with several lemmas.
The first two lemmas are proved in \cite{K} and \cite{FPS}, respectively.

\begin{lem}[Kochol~\cite{K}]%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{nec lem1}
For every integer $g\geq 3$, there exists a $3$-connected $3$-regular graph with girth $g$.
In particular, for every integer $g\geq 3$, there exist infinitely many $3$-connected $3$-regular graphs with girth at least $g$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{lem}[Fujisawa, Plummer and Saito~\cite{FPS}]%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{nec lem2}
For $\F \subseteq \G$, if $\G _{3}(\F )$ is finite, then $\{K_{n},K_{m_{1},m_{2}}\}\subseteq \F $ for some integers $n$, $m_{1}$ and $m_{2}$ with $n\geq 3$, $m_{2}\geq m_{1}\geq 1$ and $m_{1}\leq 3$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{nec lem3}
Let $\F $ be a finite subset of $\G -\{K_{1,2},K_{1,3}\}$, and suppose that $\G _{3}(\F)$ is finite.
Then $\F $ contains a member of $\T _{0}$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $t=\max \{|V(F)|\mid F\in \F \}$, and let $\H =\{G\in \G _{3}\mid G$ is a $3$-regular graph with girth at least $t+1\}$.
By Lemma~\ref{nec lem1}, $\H $ is an infinite set.
Since $\G _{3}(\F )$ is finite and $\H $ is infinite, there exists $G\in \H $ such that $G$ contains a graph $F$ in $\F $ as an induced subgraph.
Since the girth of $G$ is strictly greater than $|V(F)|$, $F$ is a tree.
Since $G$ is $3$-regular and since $F\not=K_{1,2},K_{1,3}$ by the assumption that $\F \subseteq \G -\{K_{1,2},K_{1,3}\}$, it follows that $F\in \T _{0}$.
\qed

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{nec lem4}
Let $m_{1}$ and $m_{2}$ be integers such that $m_{2}\geq m_{1}\geq 1$, $m_{1}\leq 3$ and $(m_{1},m_{2})\not=(1,2),(1,3),(2,2)$, and let $T\in \G -\{K_{1,2},K_{1,3}\}$.
Set $\F =\{K_{3},K_{m_{1},m_{2}},T\}$, and suppose that $\G _{3}(\F )$ is finite.
Then the following hold.
\begin{enumerate}
\item[{\upshape(i)}]
We have $T\in \T _{1}$.
\item[{\upshape(ii)}]
If in addition, $(m_{1},m_{2})\not\in \{(1,4),(2,3),(2,4)\}$, then $T\in \T _{2}$.
\end{enumerate}
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
By Lemma~\ref{nec lem3}, $T\in \T_{0}$.
\begin{enumerate}
\item[{\upshape(i)}]
For each $s\geq 5$, the Cartesian product $C_{s}\times K_{2}$ is $3$-connected and $\{K_{3},K_{m_{1},m_{2}}\}$-free.
%For each $s\geq 5$, $C_{s}\times K_{2}$ is $3$-connected and $\{K_{3},K_{m_{1},m_{2}}\}$-free, where $C_{s}\times K_{2}$ denotes the Cartesian product of $C_{s}$ and $K_{2}$.
Since $\G _{3}(\F )$ is finite, this implies that there exists $s\geq 5$ such that $C_{s}\times K_{2}$ contains $T$ as an induced subgraph.
Since every member of $\T _{0}$ contained in $C_{s}\times K_{2}$ as an induced subgraph belongs to $\T _{1}$, $T\in \T _{1}$.
\item[{\upshape(ii)}]
By (i), $T\in \T _{1}$.
For each $s\geq 5$, let $C'_{s}$ denote the so-called lexicographic product of $C_{s}$ and the null graph of order two; that is to say, $V(C'_{s})=\{x_{i,j}\mid 1\leq i\leq s,1\leq j\leq 2\}$ and $E(C'_{s})=\{x_{i,j}x_{i+1,h}\mid 1\leq i\leq s,1\leq j,h\leq 2\}$, where first indices of the letter $x$ are to be read modulo $s$.
Then $C'_{s}$ is $3$-connected and $\{K_{3},K_{m_{1},m_{2}}\}$-free.
Hence there exists $s\geq 5$ such that $C'_{s}$ contains $T$ as an induced subgraph.
Since every member of $\T _{1}$ contained in $C'_{s}$ as an induced subgraph belongs to $\T _{2}$, $T\in \T _{2}$.
\qed
\end{enumerate}

\medbreak\noindent\textit{Proof of Theorem~\ref{nec}.}\quad
Let $\F $ be as in Theorem~\ref{nec}.
By Lemma~\ref{nec lem2}, $\{K_{n},K_{m_{1},m_{2}}\}\subseteq \F $ for some integers $n$, $m_{1}$ and $m_{2}$ with $n\geq 3$, $m_{2}\geq m_{1}\geq 1$ and $m_{1}\leq 3$.
Write $\F =\{K_{n},K_{m_{1},m_{2}},T\}$.

\medskip
\noindent
\textbf{Case 1:} $\F $ contains no star

In this case, $2\leq m_{1}\leq 3$, and we have $T\in \T _{0}$ by Lemma~\ref{nec lem3}.

\medskip
\noindent
\textbf{Subcase 1.1:} $m_{1}=3$

For each $s\geq 3$, $P_{3}+sK_{1}$ is $3$-connected and $K_{3,m_{2}}$-free.
Since $T$ is not a star and every tree contained in $P_{3}+sK_{1}$ as an induced subgraph is a star, $P_{3}+sK_{1}$ is also $T$-free.
Since $\G _{3}(\F )$ is finite, this implies that there exists $s\geq 3$ such that $P_{3}+sK_{1}$ contains $K_{n}$ as an induced subgraph.
Since $P_{3}+sK_{1}$ is $K_{4}$-free, this forces $n=3$.
Now by Lemma~\ref{nec lem4}(ii), $T\in \T _{2}$, and hence (i) of Theorem~\ref{nec} holds.

\medskip
\noindent
\textbf{Subcase 1.2:} $m_{1}=2$ and $n\geq 4$

For each $s\geq 3$, $K_{3}+sK_{1}$ is $3$-connected and $K_{2,m_{2}}$-free.
Since every tree contained in $K_{3}+sK_{1}$ as an induced subgraph is a star, $K_{3}+sK_{1}$ is also $T$-free.
Hence there exists $s\geq 3$ such that $K_{3}+sK_{1}$ contains $K_{n}$ as an induced subgraph, which implies $n=4$.
For each $t\geq 6$, $W_{t}$ is $3$-connected and $\{K_{4},K_{2,m_{2}}\}$-free.
Hence there exists $t\geq 6$ such that $W_{t}$ contains $T$ as an induced subgraph.
Since $T\in \T _{0}$ and every member of $\T _{0}$ contained in $W_{t}$ as an induced subgraph is a path, $T$ is a path.
Consequently (ii) of Theorem~\ref{nec} holds.

\medskip
\noindent
\textbf{Subcase 1.3:} $m_{1}=2$ and $n=3$

Recall that $T\in \T _{0}$.
Also if $3\leq m_{2}\leq 4$, then $T\in \T _{1}$ by Lemma~\ref{nec lem4}(i); if $m_{2}\geq 5$, then $T\in \T _{2}$ by Lemma~\ref{nec lem4}(ii).
Thus (iii) of Theorem~\ref{nec} holds.

\medskip
\noindent
\textbf{Case 2:} $\F $ contains a star

Interchanging the roles of $K_{m_{1},m_{2}}$ and $T$ with each other if necessary, we may assume that $K_{m_{1},m_{2}}$ is the star of the smallest order contained in $\F $.
Then $m_{1}=1$, and $m_{2}\geq 3$ by the assumption that $K_{1,2}\not\in \F $, and $T\not=K_{1,2},K_{1,3}$ by the minimality of $m_{2}$.

\medskip
\noindent
\textbf{Subcase 2.1:} $m_{2}\geq 4$

By Lemma~\ref{nec lem3}, $T\in \T _{0}$.

\medskip
\noindent
\textbf{Subcase 2.1.1:} $n\geq 4$

For each $s\geq 7$, let $C_{s}^{2}$ denote the square of $C_{s}$; that is to say $V(C_{s}^{2})=\{x_{i}\mid 1\leq i\leq s\}$ and $E(C_{s}^{2})=\{x_{i}x_{i+1},x_{i}x_{i+2}\mid 1\leq i\leq s\}$, where indices of the letter $x$ are to be read modulo $s$.
Then $C_{s}^{2}$ is $3$-connected and $\{K_{n},K_{1,m_{2}}\}$-free.
Hence there exists $s\geq 7$ such that $C_{s}^{2}$ contains $T$ as an induced subgraph.
Since every tree contained in $C_{s}^{2}$ as an induced subgraph is a path, $T$ is a path.
Hence (iv) of Theorem~\ref{nec} holds.

\medskip
\noindent
\textbf{Subcase 2.1.2:} $n=3$

If $m_{2}=4$, then $T\in \T _{1}$ by Lemma~\ref{nec lem4}(i); if $m_{2}\geq 5$, then $T\in \T _{2}$ by Lemma~\ref{nec lem4}(ii).
Thus (v) of Theorem~\ref{nec} holds.

\medskip
\noindent
\textbf{Subcase 2.2:} $m_{2}=3$

By the assumption that $\{K_{1,3},K_{3}\}\not\subseteq \F $, we have $n\geq 4$ and $T\not=K_{3}$.
Thus $T\in \G -\{K_{1,2},K_{1,3},K_{3}\}$.

For a $3$-connected $3$-regular graph $G$, let $H_{G}$ be the graph obtained by expanding each vertex of $G$ to a triangle (see Figure~\ref{f5}); more precisely, we define $H_{G}$ by $V(H_{G})=\{x_{u,v}\mid (u,v)\in V(G)\times V(G), uv\in E(G)\}$ and $E(H_{G})=\{x_{u,v}x_{u,w}\mid u\in V(G),v,w\in N(u),v\not=w\}\cup \{x_{u,v}x_{v,u}\mid uv\in E(G)\}$.
Then $H_{G}$ is $3$-connected $3$-regular and $\{K_{n},K_{1,3}\}$-free and, for $u\in V(G)$, $B_{u}=x_{u,v_{1}}x_{u,v_{2}}x_{u,v_{3}}x_{u,v_{1}}$ $(\{v_{1},v_{2},v_{3}\}=N(u))$ is a triangle of $H_{G}$.
Also each cycle of $H_{G}$ which is not of the form $B_{u}$ with $u\in V(G)$ has length at least twice as large as the girth of $G$ and, for any $u,u'\in V(G)$ with $u\not=u'$, every induced path in $H_{G}$ joining $B_{u}$ and $B_{u'}$ has even order.
Hence every induced connected subgraph of $H_{G}$ having order greater than or equal to four and strictly less than twice the girth of $G$ belongs to $\T ^{*}_{0}$.
On the other hand, by Lemma~\ref{nec lem1}, the set $\{H_{G}\mid G$ is a $3$-connected $3$-regular graph with girth at least $(|V(T)|+1)/2\}$ is an infinite set.
Hence there exists a $3$-connected $3$-regular graph $G$ with girth at least $(|V(T)|+1)/2$ such that $H_{G}$ contains $T$ as an induced subgraph.
Note that $|V(T)|$ is less than twice the girth of $G$.
Consequently $T\in \T ^{*}_{0}$.

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Further for each $s\geq 7$, $C_{s}^{2}$ is $3$-connected and $\{K_{n},K_{1,3}\}$-free.
Hence there exists $s\geq 7$ such that $C_{s}^{2}$ contains $T$ as an induced subgraph.
Since every member of $\T ^{*}_{0}$ contained in $C_{s}^{2}$ as an induced subgraph belongs to $\T ^{*}_{1}$, $T\in \T ^{*}_{1}$.

Now assume $n\geq 5$.
For each $s\geq 4$, let $C^{*}_{s}$ denote the lexicographic product of $C_{s}$ and $K_{2}$; that is to say, $V(C^{*}_{s})=\{x_{i,j}\mid 1\leq i\leq s,1\leq j\leq 2\}$ and $E(C^{*}_{s})=\{x_{i,j}x_{i+1,h},x_{i,1}x_{i,2}\mid 1\leq i\leq s,1\leq j,h\leq 2\}$, where first indices of the letter $x$ are to be read modulo $s$.
Then $C^{*}_{s}$ is $3$-connected and $\{K_{n},K_{1,3}\}$-free.
Hence there exists $s\geq 4$ such that $C^{*}_{s}$ contains $T$ as an induced subgraph.
Since every member of $\T ^{*}_{1}$ contained in $C^{*}_{s}$ as an induced subgraph belongs to $\T ^{*}_{2}$, $T\in \T ^{*}_{2}$.

Thus we have $T\in \T ^{*}_{1}$ if $n=4$, and $T\in \T ^{*}_{2}$ if $n\geq 5$.
Hence (vi) of Theorem~\ref{nec} holds.

This completes the proof of Theorem~\ref{nec}.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$K_{3,m}$-free graphs}\label{sec3}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section, we prove Theorem~\ref{ns1}.
We first show that $\G _{3}(\{K_{3},K_{3,m},Y_{3}\})$ and $\G _{3}(\{K_{3},K_{3,m},P_{5}\})$ are finite.

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns1 prop1}
Let $m\geq 4$.
Then $\G _{3}(\{K_{3},K_{3,m},Y_{3}\})$ is finite.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $G\in \G _{3}(\{K_{3},K_{3,m},Y_{3}\})$.
We show that $|V(G)|\leq (m+1)^{3}$.

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns1 prop1 cl1}
$\mbox{diam}(G)\leq 3$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Suppose that $\mbox{diam}(G)\geq 4$.
Let $x,y\in V(G)$ be vertices with $d(x,y)=4$, and let $x_{0}x_{1}\cdots x_{4}$ be a shortest $x$-$y$ path in $G$.
Since $G$ is $3$-connected, $N(x_{2})-\{x_{1},x_{3}\}\not=\emptyset $.
Let $z\in N(x_{2})-\{x_{1},x_{3}\}$.
Since $G$ is $K_{3}$-free, $zx_{1},zx_{3}\not\in E(G)$.
Since $d(x_{0},x_{4})=4$, $z$ is adjacent to at most one of $x_{0}$ and $x_{4}$.
Hence $\{x_{0},x_{1},x_{2},x_{3},z\}$ or $\{x_{4},x_{3},x_{2},x_{1},z\}$ induces $Y_{3}$, which is a contradiction.
\qed

In view of Claim~\ref{ns1 prop1 cl1} and Lemma~\ref{new diam lem}, it suffices to show that $\Delta (G)\leq m+1$.
Suppose that $\Delta (G)\geq m+2$.
Let $w\in V(G)$ be a vertex such that $d(w)=\Delta (G)$, and let $x\in N(w)$.
Since $G$ is $K_{3}$-free, both $N(w)$ and $N(x)$ are independent.
Since $G$ is $3$-connected, $d(x)\geq 3$.
Take $y_{1},y_{2}\in N(x)-\{w\}$.
If $|N(w)-N(y_{i})|\geq 2$ for $i=1$ or $2$, say $a,b\in N(w)-N(y_{i})$, then $\{y_{i},x,w,a,b\}$ induces $Y_{3}$, a contradiction.
Thus $|N(y_{i})\cap N(w)|\geq d(w)-1$ for each $i=1,2$.
Consequently $|N(w)\cap N(y_{1})\cap N(y_{2})|\geq d(w)-2\geq m$, which implies that $G[\{w,y_{1},y_{2}\}\cup (N(w)\cap N(y_{1})\cap N(y_{2}))]$ contains $K_{3,m}$ as an induced subgraph, a contradiction.
\qed

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns1 prop2}
Let $m\geq 4$.
Then $\G _{3}(\{K_{3},K_{3,m},P_{5}\})$ is finite.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $G\in \G _{3}(\{K_{3},K_{3,m},P_{5}\})$.
We show that $|V(G)|\leq (4m-1)^{3}$.
Since $G$ is $P_{5}$-free, $\mbox{diam}(G)\leq 3$.
Thus in view of Lemma~\ref{new diam lem}, it suffices to show that $\Delta (G)\leq 4m-1$.
Suppose that $\Delta (G)\geq 4m$, and let $w\in V(G)$ be a vertex with $d(w)=\Delta (G)$.
Since $G$ is $K_{3}$-free, $N(w)$ is an independent set.

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns1 prop2 cl1}
Let $X\subseteq N(w)$, and let $Y$ be a minimal subset of $N_{2}(w)$ such that $N(Y)\supseteq X$.
Then $|Y|\leq 2$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Suppose that $|Y|\geq 3$.
Since $G$ is $K_{3}$-free, there exist vertices $y_{1},y_{2}\in Y$ such that $y_{1}y_{2}\not\in E(G)$.
By the minimality of $Y$, $(N(y_{i})\cap X)-N(y_{3-i})\not=\emptyset $ for each $i=1,2$.
For $i=1,2$, let $x_{i}\in (N(y_{i})\cap X)-N(y_{3-i})$.
Then $\{y_{1},x_{1},w,x_{2},y_{2}\}$ induces $P_{5}$, a contradiction.
\qed

Since $G$ is $3$-connected and $N(w)$ is independent, $N(N_{2}(w))\supseteq N(w)$.
Let $Y$ be a minimal subset of $N_{2}(w)$ such that $N(Y)\supseteq N(w)$.
Then $|Y|\leq 2$ by Claim~\ref{ns1 prop2 cl1}.
Hence there exists $y\in Y$ such that $|N(w)\cap N(y)|\geq d(w)/2\geq 2m$.
Since $G$ is $3$-connected and $N(w)$ is independent, $N(N_{2}(w)-\{y\})\supseteq N(w)\cap N(y)$.
Let $Y^{*}$ be a minimal subset of $N_{2}(w)-\{y\}$ such that $N(Y^{*})\supseteq N(w)\cap N(y)$ (see Figure~\ref{f2}).
Then $|Y^{*}|\leq 2$ by Claim~\ref{ns1 prop2 cl1}.
Hence there exists $y'\in Y^{*}$ such that $|N(w)\cap N(y)\cap N(y')|\geq m$.
Since $N(y)\cap N(y')\not=\emptyset $, we have $yy'\not\in E(G)$ by the assumption that $G$ is $K_{3}$-free.
Consequently $G[\{w,y,y'\}\cup (N(w)\cap N(y)\cap N(y'))]$ contains $K_{3,m}$ as an induced subgraph, a contradiction.
\qed

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\caption{Vertex $y$ and set $Y^{*}$}
\label{f2}
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\end{figure}

\medbreak\noindent\textit{Proof of Theorem~\ref{ns1}.}\quad
The `if' part follows from Propositions~\ref{ns1 prop1} and \ref{ns1 prop2}.
Thus it suffices to prove the `only if' part.
Suppose that $\G _{3}(\{K_{3},K_{3,m},T\})$ is finite.
From Theorem~\ref{nec}, it follows that $T\in \T _{2}$.
For each $s\geq 2$, let $H_{s}$ be the graph defined by $V(H_{s})=\{x_{i,j}\mid 1\leq i,j\leq 2\}\cup \{y_{i,j}\mid 1\leq i\leq 2,1\leq j\leq s\}$ and $E(H_{s})=\{x_{1,j}x_{2,h}\mid 1\leq j,h\leq 2\}\cup \{x_{i,j}y_{i,h}\mid 1\leq i,j\leq 2,1\leq h\leq s\}\cup \{y_{1,j}y_{2,j}\mid 1\leq j\leq s\}$.
Then $H_{s}$ is $3$-connected and $\{K_{3},K_{3,m}\}$-free since $m\geq 4$.
Since $\G _{3}(\{K_{3},K_{3,m},T\})$ is finite, there exists $s\geq 2$ such that $H_{s}$ contains $T$ as an induced subgraph.
By inspection, we see that if $F$ is a member of $\T _{2}$ contained in $H_{s}$ as an induced subgraph, then $F$ is a path of order $4$ or $5$ or $F=Y_{3}$.
Hence $T$ is a path of order $4$ or $5$ or $T=Y_{3}$.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$\{K_{4},K_{2,m}\}$-free graphs}\label{sec4}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section, we prove Theorem~\ref{ns2}.

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns2 prop1}
Let $m\geq 3$.
Then $\G _{3}(\{K_{4},K_{2,m},P_{5}\})$ is finite.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
By part (i) of Theorem~\ref{tri 2-conn}, there exists a positive integer $t=t(m)$ such that every $2$-connected $\{K_{3},K_{2,m},P_{5}\}$-free graph has order at most $t$.
Let $G\in \G _{3}(\{K_{4},K_{2,m},P_{5}\})$.
We show that $|V(G)|\leq (3(3m-1)t/2)^{3}$.
Note that $\mbox{diam}(G)\leq 3$.
Thus in view of Lemma~\ref{new diam lem}, it suffices to show that $\Delta (G)\leq 3(3m-1)t/2$.
Suppose that $\Delta (G)>3(3m-1)t/2$, and let $w\in V(G)$ be a vertex with $d(w)=\Delta (G)$.
Since $G$ is $\{K_{4},K_{2,m},P_{5}\}$-free, $G[N(w)]$ is $\{K_{3},K_{2,m},P_{5}\}$-free.

Let $F$ be a component of $G[N(w)]$.
If $F$ has two or more blocks which are not endblocks, then $F$ contains $P_{5}$ as an induced subgraph, a contradiction.
Thus $F$ has at most one block which is not an endblock, which implies that at least two thirds of the blocks of $F$ are endblocks.
Since $F$ is arbitrary, it follows that at least two thirds of the blocks of $G[N(w)]$ are endblocks.
Now suppose that $G[N(w)]$ has at most $3m-1$ endblocks.
Then $G[N(w)]$ has at most $3(3m-1)/2$ blocks.
Since $d(w)>3(3m-1)t/2$, it follows that there exists a block $B$ of $G[N(w)]$ with $|V(B)|>t$.
Since $G[N(w)]$ is $\{K_{3},K_{2,m},P_{5}\}$-free, this contradicts the definition of $t$.
Thus $G[N(w)]$ has at least $3m$ endblocks.
Let $B_{1},\ldots ,B_{3m}$ be endblocks of $G[N(w)]$.
Since $G$ is $3$-connected, we see that for each $1\leq i\leq 3m$, there exists an internal vertex $x_{i}$ of $B_{i}$ such that $N(x_{i})\cap (V(G)-(\{w\}\cup N(w)))\not=\emptyset $ (see Figure~\ref{f3}).
Set $X=\{x_{1},\ldots ,x_{3m}\}$.
Then $X$ is an independent set of $G$, and we have $N(N_{2}(w))\supseteq X$.
Let $Y$ be a minimal subset of $N_{2}(w)$ such that $N(Y)\supseteq X$.
If $|Y|\leq 3$, then there exists $y\in Y$ such that $|N(y)\cap X|\geq m$, and hence $G[\{w,y\}\cup (N(y)\cap X)]$ contains $K_{2,m}$ as an induced subgraph, a contradiction.
Thus $|Y|\geq 4$.
Since $G$ is $K_{4}$-free, there exist vertices $y_{1},y_{2}\in Y$ such that $y_{1}y_{2}\not\in E(G)$.
By the minimality of $Y$, $(N(y_{i})\cap X)-N(y_{3-i})\not=\emptyset $ for each $i=1,2$.
For $i=1,2$, let $x'_{i}\in (N(y_{i})\cap X)-N(y_{3-i})$.
Then $\{y_{1},x'_{1},w,x'_{2},y_{2}\}$ induces $P_{5}$, a contradiction.
\qed

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\label{f3}
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\medbreak\noindent\textit{Proof of Theorem~\ref{ns2}.}\quad
The `if' part follows from Proposition~\ref{ns2 prop1}.
Thus it suffices to prove the `only if' part.
Suppose that $\G _{3}(\{K_{4},K_{2,m},T\})$ is finite.
From Theorem~\ref{nec}, it follows that $T$ is a path.
For each $s\geq 2$, let $H_{s}$ be the graph defined by $V(H_{s})=\{x_{i},y_{i,j}\mid 1\leq i\leq 3,1\leq j\leq s\}$ and $E(H_{s})=\{x_{i}x_{j},y_{i,h}y_{j,h}\mid i\not=j,1\leq h\leq s\}\cup \{x_{i}y_{j,h}\mid i\not=j,1\leq h\leq s\}$ (see Figure~\ref{f6}).
Then $H_{s}$ is $3$-connected and $\{K_{4},K_{2,m}\}$-free.
Hence there exists $s\geq 2$ such that $H_{s}$ contains $T$ as an induced subgraph.
Since every induced path of $H_{s}$ has order at most $5$, $T$ has order at most $5$.
Since $T\not=K_{1,2}$ by assumption, it follows that $T$ is a path of order $4$ or $5$.
\qed

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% $y_{2,s}$
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% $y_{3,s}$
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\end{picture}%

\caption{Graph $H_{s}$}
\label{f6}
\end{center}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$K_{1,m}$-free graphs}\label{sec5}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section, we prove Theorem~\ref{ns4}.
We first prove a lemma, which we also use in Section~\ref{sec6}.

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4 lem1}
Let $n\geq 3$, and let $G$ be a $3$-connected $\{K_{3},Y^{*}_{n}\}$-free graph.
If $n=3$, let $t(n)=7$; if $n\geq 4$, let $t(n)=3n+10$.
Then $\mbox{diam}(G)\leq t(n)$.
In particular, $\mbox{diam}(G)\leq 3n+10$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $t=t(n)$, and suppose that $\mbox{diam}(G)\geq t+1$.
Let $x,y\in V(G)$ be vertices such that $d(x,y)=t+1$, and let $P=x_{0}x_{1}\cdots x_{t+1}$ be a shortest $x$-$y$ path in $G$.
Since $G$ is $3$-connected, $N(x_{i})-\{x_{i-1},x_{i+1}\}\not=\emptyset $ for each $1\leq i\leq t$.
Let $x'_{i}\in N(x_{i})-\{x_{i-1},x_{i+1}\}$.

\medskip
\noindent
\textbf{Case 1:} $n=3$

We first show that $N(x_{3})\cap N(x_{5})=\{x_{4}\}$.
By way of contradiction, suppose that $|N(x_{3})\cap N(x_{5})|\geq 2$.
Then we may assume we have chosen $x'_{3}$ and $x'_{5}$ so that $x'_{3}=x'_{5}\in (N(x_{3})\cap N(x_{5}))-\{x_{4}\}$.
Since $d(x_{0},x_{8})=8$ and $G$ is $K_{3}$-free, we get $N(x'_{3})\cap V(P)=\{x_{3},x_{5}\}$, $N(x'_{1})\cap V(P)\subseteq \{x_{1},x_{3}\}$, $N(x'_{7})\cap V(P)\subseteq \{x_{5},x_{7}\}$, and $x'_{1}x'_{3},x'_{3}x'_{7}\not\in E(G)$.
Since neither $\{x_{0},x'_{1},x_{1},x_{2},x_{3},x_{4},x'_{3}\}$ nor $\{x_{4},x'_{3},x_{5},x_{6},x_{7},x_{8},x'_{7}\}$ induces $Y^{*}_{3}$, this forces $N(x'_{1})\cap V(P)=\{x_{1},x_{3}\}$ and $N(x'_{7})\cap V(P)=\{x_{5},x_{7}\}$.
But then $\{x_{2},x'_{1},x_{3},x_{4},x_{5},x_{6},x'_{7}\}$ induces $Y^{*}_{3}$, a contradiction.
Thus $N(x_{3})\cap N(x_{5})=\{x_{4}\}$.

Choose $z\in N(x_{3})-\{x_{2},x_{4}\}$.
Since $\{x_{2},z,x_{3},x_{4},x_{5},x_{6},x'_{5}\}$ does not induce $Y^{*}_{3}$ and $N(x_{3})\cap N(x_{5})=\{x_{4}\}$, we get $zx'_{5}\in E(G)$.
Since $d(x_{1},x_{5})=4$, it follows that $x_{1}z\not\in E(G)$.
Since $z\in N(x_{3})-\{x_{2},x_{4}\}$ is arbitrary,  this implies $N(x_{1})\cap N(x_{3})=\{x_{2}\}$; in particular, $x'_{1}x_{3}\not\in E(G)$.
Letting $z=x'_{3}$, we also get $x'_{3}x'_{5}\in E(G)$ and $x_{1}x'_{3}\not\in E(G)$.
Since $\{x_{0},x'_{1},x_{1},x_{2},x_{3},x_{4},x'_{3}\}$ does not induce $Y^{*}_{3}$, we now obtain $x'_{1}x'_{3}\in E(G)$.
Similarly, $x'_{7}x'_{5}\in E(G)$.
But then $x_{0}x_{1}x'_{1}x'_{3}x'_{5}x'_{7}x_{7}x_{8}$ is an $x_{0}$-$x_{8}$ path of length $7$, which contradicts the fact that $d(x_{0},x_{8})=8$.

\medskip
\noindent
\textbf{Case 2:} $n\geq 4$

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4 lem1 cl1}
For $0\leq i\leq t-n-1$, if $|N(x_{i})\cap N(x_{i+2})|\geq 2$, then $x'_{i+n+1}x_{i+n-1}\in E(G)$.
For $n\leq i\leq t-1$, if $|N(x_{i})\cap N(x_{i+2})|\geq 2$, then $x'_{i-n+1}x_{i-n+3}\in E(G)$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $0\leq i\leq t-n-1$, and assume that $|N(x_{i})\cap N(x_{i+2})|\geq 2$.
Let $u\in (N(x_{i})\cap N(x_{i+2}))-\{x_{i+1}\}$.
Since $d(x_{0},x_{t+1})=t+1$ and $G$ is $K_{3}$-free, we have $N(u)\cap V(P)=\{x_{i},x_{i+2}\}$, $N(x'_{i+n+1})\cap \{x_{0},x_{1},\ldots ,x_{i+n+2}\}\subseteq \{x_{i+n-1},x_{i+n+1}\}$ and $ux'_{i+n+1}\not\in E(G)$.
Since $\{x_{i+1},u,x_{i+2},\ldots ,x_{i+n+1},x_{i+n+2},x'_{i+n+1}\}$ does not induce $Y^{*}_{n}$, this implies $x'_{i+n+1}x_{i+n-1}\in E(G)$.
Thus the first assertion is proved.
We can similarly verify the second assertion.
\qed

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4 lem1 cl2}
There is no integer $i$ with $n\leq i\leq t-n-2$ such that $|N(x_{i})\cap N(x_{i+2})|\geq 2$ and $|N(x_{i+1})\cap N(x_{i+3})|\geq 2$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Suppose that there exists an integer $i$ with $n\leq i\leq t-n-2$ such that $|N(x_{i})\cap N(x_{i+2})|\geq 2$ and $|N(x_{i+1})\cap N(x_{i+3})|\geq 2$.
Then by Claim~\ref{ns4 lem1 cl1}, $x'_{i-n+1}x_{i-n+3},x'_{i+n+2}x_{i+n}\in E(G)$.
This implies $|N(x_{i-n+1})\cap N(x_{i-n+3})|\geq 2$ and $|N(x_{i+n})\cap N(x_{i+n+2})|\geq 2$.
Let $\mathcal{Q}$ be the set of $x_{i}$-$x_{i+3}$ paths of order $4$, and let $X=(\bigcup_{Q\in \mathcal{Q}}V(Q))-\{x_{i},x_{i+3}\}$.
Since $G-\{x_{i},x_{i+3}\}$ is connected and $V(G)-(X\cup \{x_{i},x_{i+3}\})\not=\emptyset $, there exists $z\in X$ such that $N(z)-(X\cup \{x_{i},x_{i+3}\})\not=\emptyset $.
By the definition of $X$, $z$ is adjacent to $x_{i}$ or $x_{i+3}$.
By symmetry, we may assume $zx_{i}\in E(G)$.
Then there exists $z'\in X$ such that $x_{i}zz'x_{i+3}$ is an $x_{i}$-$x_{i+3}$ path.
Now take $w\in N(z)-(X\cup \{x_{i},x_{i+3}\})$.
Since $|N(x_{i+n})\cap N(x_{i+n+2})|\geq 2$, we obtain $wx_{i+3}\in E(G)$ by applying the second assertion of Claim~\ref{ns4 lem1 cl1} to the path $x_{0}x_{1}\cdots x_{i}zz'x_{i+3}x_{i+4}\cdots x_{t+1}$.
But then $x_{i}zwx_{i+3}\in \mathcal{Q}$, which contradicts the fact that $w\not\in X$.
\qed

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4 lem1 cl3}
For $n+2\leq i\leq t-2n$, we have $x'_{i-1}x'_{i+n-2}\in E(G)$ or $x'_{i}x'_{i+n-1}\in E(G)$ or $x'_{i+1}x'_{i+n}\in E(G)$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Suppose that $x'_{i}x'_{i+n-1}\not\in E(G)$.
Since $\{x_{i-1},x'_{i},x_{i},\ldots ,x_{i+n-1},x_{i+n},x'_{i+n-1}\}$ does not induce $Y^{*}_{n}$, $x'_{i}x_{i+2}\in E(G)$ or $x'_{i+n-1}x_{i+n-3}\in E(G)$.
First assume that $x'_{i}x_{i+2}\in E(G)$.
Since $0\leq i\leq t-n-1$, we have $x'_{i+n+1}x_{i+n-1}\in E(G)$ by Claim~\ref{ns4 lem1 cl1}.
Since $n\leq i\leq t-n-2$ and $n\leq i+n-2\leq t-n-2$, $x'_{i+1}x_{i+3},x'_{i+n}x_{i+n-2}\not\in E(G)$ by Claim~\ref{ns4 lem1 cl2}.
Since $\{x_{i},x'_{i+1},x_{i+1},\ldots ,x_{i+n},x_{i+n+1},x'_{i+n}\}$ does not induce $Y^{*}_{n}$ and $G$ is $K_{3}$-free, it follows that $x'_{i+1}x'_{i+n}\in E(G)$, as desired.
Next assume that $x'_{i+n-1}x_{i+n-3}\in E(G)$.
Arguing as above, we get $x'_{i-2}x_{i}\in E(G)$ by Claim~\ref{ns4 lem1 cl1}, and $x'_{i-1}x_{i+1},x'_{i+n-2}x_{i+n-4}\not\in E(G)$ by Claim~\ref{ns4 lem1 cl2}.
Since $\{x_{i-2},x'_{i-1},x_{i-1},\ldots ,x_{i+n-2},x_{i+n-1},x'_{i+n-2}\}$ does not induce $Y^{*}_{n}$ and $G$ is $K_{3}$-free, it follows that $x'_{i-1}x'_{i+n-2}\in E(G)$, as desired.
\qed

Let $j=n+2$.
Since $t\geq 3n+10$, we have $n+2\leq j\leq t-2n$.
Hence it follows from Claim~\ref{ns4 lem1 cl3} that $x'_{j-1}x'_{j+n-2}\in E(G)$ or $x'_{j}x'_{j+n-1}\in E(G)$ or $x'_{j+1}x'_{j+n}\in E(G)$.
Since $d(x_{j-1},x_{j+n-2})=d(x_{j},x_{j+n-1})=d(x_{j+1},x_{j+n})=n-1$, this forces $n=4$, and hence $t=22$ and $j=6$.
In particular, $x'_{5}x'_{8}\in E(G)$ or $x'_{6}x'_{9}\in E(G)$ or $x'_{7}x'_{10}\in E(G)$.
Let $s\in \{5,6,7\}$ be an integer such that $x'_{s}x'_{s+3}\in E(G)$.

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns4 lem1 cl4}
For $4\leq i\leq 11$, if $x'_{i}x'_{i+3}\in E(G)$, then $x'_{i+2}x'_{i+5}\in E(G)$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
If $x'_{i+3}x'_{i+6}\in E(G)$, then $x_{i}x'_{i}x'_{i+3}x'_{i+6}x_{i+6}$ is a path, which contradicts the fact that $d(x_{i},x_{i+6})=6$.
Thus $x'_{i+3}x'_{i+6}\not\in E(G)$.
Suppose that $x'_{i+2}x'_{i+5}\not\in E(G)$.
Then $x'_{i+2}x'_{i+5},x'_{i+3}x'_{i+6}\not\in E(G)$.
Since $n+2=6\leq i+2<i+3\leq 14=t-2n$, this together with Claim~\ref{ns4 lem1 cl3} implies $x'_{i+1}x'_{i+4}\in E(G)$ and $x'_{i+4}x'_{i+7}\in E(G)$.
It now follows that $x_{i+1}x'_{i+1}x'_{i+4}x'_{i+7}x_{i+7}$ is a path, which contradicts the fact that $d(x_{i+1},x_{i+7})=6$.
\qed

Recall that $x'_{s}x'_{s+3}\in E(G)$.
Since $4<s<s+2<s+4\leq 11$, by repetitively applying Claim~\ref{ns4 lem1 cl4}, we obtain $x'_{s+2}x'_{s+5},x'_{s+4}x'_{s+7},x'_{s+6}x'_{s+9}\in E(G)$.
Since $x'_{s}x'_{s+3},x'_{s+6}x'_{s+9}\in E(G)$ and $d(x_{s},x_{s+9})=9$, we get $N(x'_{s+3})\cap \{x_{s+5},x_{s+6},x_{s+7},x'_{s+6}\}=\emptyset $ and $N(x'_{s+6})\cap \{x_{s+2},x'_{s+3},x_{s+3},x_{s+4}\}=\emptyset $.
Since $G$ is $K_{3}$-free, it follows that $\{x_{s+2},x'_{s+3},x_{s+3},x_{s+4},x_{s+5},$ $x_{s+6},x_{s+7},x'_{s+6}\}$ induces $Y^{*}_{4}$, which is a contradiction.

This completes the proof of Lemma~\ref{ns4 lem1}.
\qed





\medbreak\noindent\textit{Proof of Theorem~\ref{ns4}.}\quad
Let $n\geq 3$, and let $G\in \G _{3}(\{K_{3},K_{1,m},Y^{*}_{n}\})$.
Then by Lemma~\ref{ns4 lem1}, $\mbox{diam}(G)\leq 3n+10$.
Since $G$ is $\{K_{3},K_{1,m}\}$-free, we also have $\Delta (G)\leq m-1$.
Hence $|V(G)|\leq (m-1)^{3n+10}$ by Lemma~\ref{new diam lem}.
Thus $\G _{3}(\{K_{3},K_{1,m},Y^{*}_{n}\})$ is finite.
Since $n\geq 3$ is arbitrary, this proves the `if' part.
The `only if' part follows from Theorem~\ref{nec}.
\qed


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$\{K_{3},K_{2,m}\}$-free graphs}\label{sec6}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section, we prove Theorem~\ref{ns3}.
We first prove several lemmas.

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem1}
Let $m\geq 2$, and let $G$ be a $\{K_{3},K_{2,m}\}$-free graph.
Let $H$ be a connected induced subgraph of $G$ with order $n\geq 2$ and let $x\in V(H)$, and suppose that $d_{G}(x)\geq (m-1)(n-2)+t+1$.
Then $G$ contains as an induced subgraph the graph obtained from $H$ by adding $t$ pendant edges to $x$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Since $G$ is $K_{3}$-free, $N_{G}(x)-V(H)$ is independent and no vertex in $N_{G}(x)-V(H)$ is adjacent to a vertex in $N_{G}(x)\cap V(H)$.
Since $G$ is $K_{2,m}$-free, we see that for each $x'\in V(H)-(\{x\}\cup N_{G}(x))$, $x'$ is adjacent to at most $m-1$ vertices in $N_{G}(x)-V(H)$.
Set $Y=N_{G}(x)\cap ((\bigcup _{x'\in V(H)-\{x\}}N_{G}(x'))\cup V(H))$.
It follows that $|Y|\leq (m-1)(n-1-d_{H}(x))+d_{H}(x)\leq (m-1)(n-2)+1$.
Hence $|N_{G}(x)-Y|\geq t$.
Now if we let $Z$ be a subset of $N_{G}(x)-Y$ with $|Z|=t$, then $V(H)\cup Z$ induces the desired graph.
\qed

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem2}
Let $m\geq 2$, and let $G$ be a $\{K_{3},K_{2,m}\}$-free graph.
Let $P$ be an induced path of $G$ with order $n\geq 2$, and suppose that both endvertices of $P$ have degree at least $(m-1)(n-1)+2$.
Then $G$ contains $P_{n+2}$ as an induced subgraph.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Applying Lemma~\ref{ns3 lem1} to one of the endvertices of $P$ with $H=P$, we get an induced path $P'$ of order $n+1$ and, applying Lemma~\ref{ns3 lem1} to the other endvertices of $P$ with $H=P'$, we obtain a path of the type desired.
\qed

Similarly, we obtain the following lemma.

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem3}
Let $m\geq 2$, and let $G$ be a $\{K_{3},K_{2,m}\}$-free graph, and let $P$ be an induced path of $G$ with order $n\geq 2$, and suppose that both endvertices of $P$ have degree at least $(m-1)n+3$.
Then $G$ contains $Y^{*}_{n}$ as an induced subgraph.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 prop1}
Let $m\geq 5$.
Then $\G _{3}(\{K_{3},K_{2,m},Y^{*}_{3}\})$ is finite.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $G\in \G _{3}(\{K_{3},K_{2,m},Y^{*}_{3}\})$.
We show that $|V(G)|\leq ((6m-3)(3m-2)(m-2)+1)^{7}$.
By Lemma~\ref{ns4 lem1}, $\mbox{diam}(G)\leq 7$.
Thus it suffices to show that $\Delta (G)\leq (6m-3)(3m-2)(m-2)+1$.
Suppose that $\Delta (G)\geq (6m-3)(3m-2)(m-2)+2$, and let $w\in V(G)$ be a vertex with $d(w)=\Delta (G)$.
Since $G$ is $K_{3}$-free, $N(w)$ is independent.
In particular, for any two vertices $x,x'$ in $N(w)$, $\{x,w,x'\}$ induces $P_{3}$.
Choose $a\in N(w)$ so that $d(a)\geq d(x)$ for all $x\in N(w)$.
If two vertices in $N(w)$ have degree at least $3m$, then by Lemma~\ref{ns3 lem3}, $G$ contains $Y^{*}_{3}$ as an induced subgraph, a contradiction.
Thus all vertex in $N(w)-\{a\}$ have degree at most $3m-1$.
Let $B\subseteq N(w)-\{a\}$ be a maximal set such that $N(b)\cap N(b')=\{w\}$ for any $b,b'\in B$ with $b\not=b'$.

Suppose that $|B|\leq 6m-3$.
Since $B\subseteq N(w)-\{a\}$, $|(\bigcup_{b\in B}N(b))-\{w\}|\leq (6m-3)(3m-2)$.
Since $G$ is $K_{2,m}$-free and $(\bigcup_{b\in B}N(b))-\{w\}\subseteq N_{2}(w)$, $|N(y)\cap N(w)|\leq m-1$ for all $y\in (\bigcup_{b\in B}N(b))-\{w\}$.
Hence $|\{x\in N(w)\mid N(x)\cap ((\bigcup _{b\in B}N(b))-\{w\})\not=\emptyset \}|\leq (6m-3)(3m-2)(m-1)$.
Since $|N(w)-\{a\}|\geq (6m-3)(3m-2)(m-1)+1$, there exists $b^{*}\in N(w)-\{a\}$ such that $(N(b^{*})\cap (\bigcup_{b\in B}N(b)))-\{w\}=\emptyset $.
Then $B\cup \{b^{*}\}$ satisfies the condition that $N(b)\cap N(b')=\{w\}$ for any $b,b'\in B\cup \{b^{*}\}$ with $b\not=b'$, which contradicts the maximality of $B$.
Thus $|B|\geq 6m-2$.

Since $G$ is $3$-connected, $|N(b)-\{w\}|\geq 2$ for every $b\in B$.
For each $b\in B$, let $v_{b},u_{b}\in N(b)-\{w\}$.
Fix a vertex $b_{0}\in B$.
For each $b\in B-\{b_{0}\}$, we have $E(\{v_{b_{0}},u_{b_{0}}\},\{v_{b},u_{b}\})\not=\emptyset $ because $\{v_{b_{0}},u_{b_{0}},b_{0},w,b,v_{b},u_{b}\}$ does not induce $Y^{*}_{3}$.
Since $|B-\{b_{0}\}|\geq 6m-3$, this implies that $v_{b_{0}}$ or $u_{b_{0}}$ is adjacent to at least $3m-1$ vertices in $\{v_{b},u_{b}\mid b\in B-\{b_{0}\}\}$.
We may assume $|N(v_{b_{0}})\cap \{v_{b},u_{b}\mid b\in B-\{b_{0}\}\}|\geq 3m-1$.
Then $d(v_{b_{0}})\geq 3m$.
Since $d(w)\geq 3m$ and $wb_{0}v_{b_{0}}$ is an induced path of $G$ of order $3$, $G$ contains $Y^{*}_{3}$ as an induced subgraph by Lemma~\ref{ns3 lem3}, a contradiction.

This completes the proof of Proposition~\ref{ns3 prop1}.
\qed

Before proving the finiteness of $\G _{3}(\{K_{3},K_{2,m},P_{7}\})$, we state an important part of the proof as a separate lemma.

\begin{lem}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem4}
Let $m\geq 5$, and let $G\in \G _{3}(\{K_{3},K_{2,m},P_{7}\})$.
Then there exist no vertices $a,b\in V(G)$ such that $d(a,b)=2$, $d(a)\geq 2(m-1)^{7}$ and $d(b)\geq 2(m-1)^{7}$.
\end{lem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Suppose that there exist two vertices $a,b\in V(G)$ such that $d(a,b)=2$, $d(a)\geq 2(m-1)^{7}$ and $d(b)\geq 2(m-1)^{7}$.
Since $d(a,b)=2$, $ab\not\in E(G)$.
Let $w\in N(a)\cap N(b)$, and let $A=N(a)-N(b)$ and $B=N(b)-N(a)$.
Since $G$ is $K_{3}$-free, $N(w)$, $N(a)$ and $N(b)$ are independent sets, and hence $N(a)-\{w\},N(b)-\{w\}\subseteq N_{2}(w)$.
Since $G$ is $K_{2,m}$-free, $|N(a)\cap N(b)|\leq m-1$, and hence $|A|\geq 2(m-1)^{7}-m+1$ and $|B|\geq 2(m-1)^{7}-m+1$.
Since $G$ is $3$-connected, $G-w$ is $2$-connected.
Let $k$ be the maximum order of an induced $a$-$b$ path in $G-w$.

We consider three cases according to the value of $k$.

\medskip
\noindent
\textbf{Case 1:} $k=3$

Note that it follows from the assumption of this case that $N(a)\cup N(b)$ is independent.
Since $G-w$ is $2$-connected, it also follows that $|(N(a)\cap N(b))-\{w\}|\geq 2$.
Let $c_{1},c_{2}\in (N(a)\cap N(b))-\{w\}$ with $c_{1}\not=c_{2}$.

Note that we have $N(x)\cap (N(a)\cup N(b))=\emptyset $ for each $x\in A\cup B$ because $N(a)\cup N(b)$ is independent.
Let $A_{0}=\{x\in A\mid N(x)\subseteq N(w)\}$ and $A_{1}=A-A_{0}$.
For each $x\in A$, we define $u_{x}\in N(x)$ as follows: if $x\in A_{0}$, let $u_{x}\in N(x)-\{a\}$; if $x\in A_{1}$, let $u_{x}\in N(x)-N(w)$.
For $A'\subseteq A$, let $U_{A'}=\{u_{x}\mid x\in A'\}$.
Let $B_{0}=\{y\in B\mid N(y)\subseteq N(w)\}$ and $B_{1}=B-B_{0}$.
For each $y\in B$, we define $v_{y}\in N(y)$ as follows: if $y\in B_{0}$, let $v_{y}\in N(y)-\{b\}$; if $y\in B_{1}$, let $v_{y}\in N(y)-N(w)$.
For $B'\subseteq B$, let $V_{B'}=\{v_{y}\mid y\in B'\}$.

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem4 cl1}
For $A'\subseteq A$, $|U_{A'}|\geq |A'|/(m-1)$.
For $B'\subseteq B$, $|V_{B'}|\geq |B'|/(m-1)$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $A'\subseteq A$.
Suppose that $|U_{A'}|<|A'|/(m-1)$.
Then there exists $u\in U_{A'}$ such that $u=u_{x}$ for some $m$ vertices $x$ in $A'$.
Hence $G[\{a,u\}\cup (N(a)\cap N(u))]$ contains $K_{2,m}$ as an induced subgraph, a contradiction.
Thus $|U_{A'}|\geq |A'|/(m-1)$.
Similarly, we get $|V_{B'}|\geq |B'|/(m-1)$ for $B'\subseteq B$.
\qed

We consider two subcases.

\medskip
\noindent
\textbf{Subcase 1.1:} $A_{1}=\emptyset $ or $B_{1}=\emptyset $

We may assume that $A_{1}=\emptyset $.
%Then $|A_{0}|=|A|\geq 2(m-1)^{7}-m+1\geq (m-1)^{2}$.
By Claim~\ref{ns3 lem4 cl1} and the fact that $|A_{0}|=|A|\geq 2(m-1)^{7}-m+1>(m-1)^{2}$, $|U_{A_{0}}|\geq m$.
Since $|B|\geq 2(m-1)^{7}-m+1>(m-1)^{2}$, $|V_{B}|\geq m$ by Claim~\ref{ns3 lem4 cl1}.
Since $G$ is $K_{3}$-free, it follows that if both $c_{1}$ and $c_{2}$ are adjacent to all vertices in $V_{B}$, then $G[\{c_{1},c_{2}\}\cup V_{B}]$ contains $K_{2,m}$ as an induced subgraph, a contradiction.
Thus $c_{h}v_{y}\not\in E(G)$ for some $h\in \{1,2\}$ and some $y\in B$.
Then since $v_{y}\not\in N(a)\cup N(b)$, $ac_{h}byv_{y}$ is an induced path.
Note that $U_{A_{0}}\subseteq N(w)$.
Since $G$ is $K_{2,m}$-free, $|N(c_{h})\cap U_{A_{0}}|\leq |N(c_{h})\cap N(w)|\leq m-1$.
Since $|U_{A_{0}}|\geq m$, there exists $x\in A_{0}$ such that $u_{x}c_{h}\not\in E(G)$.
Then $u_{x}xac_{h}b$ is an induced path.
Since $N(a)\cup N(b)$ is independent, $xy\not\in E(G)$.
Since $k=3$, we also have $u_{x}y,xv_{y},u_{x}v_{y}\not\in E(G)$.
Consequently the path $u_{x}xac_{h}byv_{y}$ is an induced path of order $7$, a contradiction.

\medskip
\noindent
\textbf{Subcase 1.2:} $A_{1}\not=\emptyset $ and $B_{1}\not=\emptyset $

Let $x\in A_{1}$ and $y\in B_{1}$.
Then since $u_{x},v_{y}\not\in N(a)\cup N(b)\cup N(w)$ and $k=3$, $u_{x}xawbyv_{y}$ is an induced path of order $7$, a contradiction.

\medskip
\noindent
\textbf{Case 2:} $k=4$

Let $H_{1},\ldots ,H_{p}$ be the components of $G[A\cup B]$.
Let $1\leq i\leq p$.
Suppose that there exists $x\in V(H_{i})$ such that $d_{H_{i}}(x)\geq m$.
We may assume that $x\in A$.
Then $N_{H_{i}}(x)\subseteq B$, which implies that $G[\{b\}\cup V(H_{i})]$ contains $K_{2,m}$ as an induced subgraph, a contradiction.
Thus $\Delta (H_{i})\leq m-1$.
Since $H_{i}$ is $P_{7}$-free, we also have $\mbox{diam}(H_{i})\leq 5$.
Consequently $|V(H_{i})|\leq (m-1)^{5}$ for each $1\leq i\leq p$.
Since $|A\cup B|\geq 2(2(m-1)^{7}-m+1)$, it follows that $p\geq 2(2(m-1)^{7}-m+1)/(m-1)^{5}>3(m-1)^{2}+2$.
Since $k=4$, $|V(H_{i})|\geq 2$ for some $i$.
We may assume that $|V(H_{1})|\geq 2$.
Let $a_{1}b_{1}\in E(H_{1})$ with $a_{1}\in A$ and $b_{1}\in B$.

Note that each vertex in $N(a)\cap N(b)$ is an isolated vertex in $G[N(a)\cup N(b)]$.
Since $G-\{a,b\}$ is connected, this implies that for each $2\leq i\leq p$, there exists $x_{i}\in V(H_{i})$ such that $N(x_{i})-(\{a,b\}\cup N(a)\cup N(b))\not=\emptyset $.
Let $X=\{x_{i}\mid 2\leq i\leq p\}$.
Let $A_{0}=\{x\in A\cap X\mid N(x)-(\{a,b\}\cup N(a)\cup N(b))\subseteq N(w)\}$ and $A_{1}=(A\cap X)-A_{0}$.
For each $x\in A\cap X$, we define $u_{x}\in N(x)$ as follows: if $x\in A_{0}$, let $u_{x}\in N(x)-(\{a,b\}\cup N(a)\cup N(b))$; if $x\in A_{1}$, let $u_{x}\in N(x)-(\{a,b\}\cup N(a)\cup N(b))-N(w)$.
For $A'\subseteq A\cap X$, let $U_{A'}=\{u_{x}\mid x\in A'\}$.
Let $B_{0}=\{y\in B\cap X\mid N(y)-(\{a,b\}\cup N(a)\cup N(b))\subseteq N(w)\}$ and $B_{1}=(B\cap X)-B_{0}$.
For each $y\in B\cap X$, we define $v_{y}\in N(y)$ as follows: if $y\in B_{0}$, let $v_{y}\in N(y)-(\{a,b\}\cup N(a)\cup N(b))$; if $y\in B_{1}$, let $v_{y}\in N(y)-(\{a,b\}\cup N(a)\cup N(b))-N(w)$.
For $B'\subseteq B\cap X$, let $V_{B'}=\{v_{y}\mid y\in B'\}$.
Note that $|A_{0}|+|A_{1}|+|B_{0}|+|B_{1}|=|X|=p-1$.
Arguing as in the proof of Claim~\ref{ns3 lem4 cl1}, we obtain the following claim.  

\begin{claim}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 lem4 cl2}
For $A'\subseteq A\cap X$, $|U_{A'}|\geq |A'|/(m-1)$.
For $B'\subseteq B\cap X$, $|V_{B'}|\geq |B'|/(m-1)$.
\end{claim}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Here we consider the following three subcases:\\
(1) $|A_{0}|>(m-1)^{2}$ or $|B_{0}|>(m-1)^{2}$;\\
(2) $|A_{0}|\leq (m-1)^{2}$ and $|B_{0}|\leq (m-1)^{2}$ and, moreover, we have $A_{1}\not=\emptyset $ and $B_{1}\not=\emptyset $;\\
(3) $|A_{0}|\leq (m-1)^{2}$ and $|B_{0}|\leq (m-1)^{2}$, but we have $A_{1}=\emptyset $ or $B_{1}=\emptyset $.

\medskip
\noindent
\textbf{Subcase 2.1:} $|A_{0}|>(m-1)^{2}$ or $|B_{0}|>(m-1)^{2}$

We may assume that $|A_{0}|>(m-1)^{2}$.
By Claim~\ref{ns3 lem4 cl2}, $|U_{A_{0}}|\geq m$.
Note that $U_{A_{0}}\subseteq N(w)$.
Since $G$ is $K_{2,m}$-free, $|N(a_{1})\cap U_{A_{0}}|\leq m-1$.
Since $|U_{A_{0}}|\geq m$, there exists $x\in A_{0}$ such that $u_{x}a_{1}\not\in E(G)$.
We may assume that $x=x_{2}$.
Since $x_{2}$ and $b_{1}$ belong to distinct components of $G[A\cup B]$, $x_{2}b_{1}\not\in E(G)$.
Since $|B|\geq 2(m-1)^{7}-m+1>2(m-1)^{5}\geq |V(H_{1})|+|V(H_{2})|$, $B-(V(H_{1})\cup V(H_{2}))\not=\emptyset $.
Let $y\in B-(V(H_{1})\cup V(H_{2}))$.
Then $ya_{1},yx_{2}\not\in E(G)$.
Since $k=4$, we also have $u_{x_{2}}b_{1},u_{x_{2}}y\not\in E(G)$.
Since $u_{x_{2}}\not\in N(a)\cup N(b)$, we now see that the path $u_{x_{2}}x_{2}aa_{1}b_{1}by$ is an induced path of order $7$, a contradiction.

\medskip
\noindent
\textbf{Subcase 2.2:} $|A_{0}|\leq (m-1)^{2}$ and $|B_{0}|\leq (m-1)^{2}$ and, moreover, we have $A_{1}\not=\emptyset $ and $B_{1}\not=\emptyset $

Let $x\in A_{1}$ and $y\in B_{1}$.
By the definition of $X$, $x$ and $y$ belong to distinct components of $G[A\cup B]$, and hence $xy\not\in E(G)$.
Since $u_{x},v_{y}\not\in N(a)\cup N(b)\cup N(w)$ and $k=4$, it follows that the path $u_{x}xawbyv_{y}$ is an induced path of order $7$, a contradiction.

\medskip
\noindent
\textbf{Subcase 2.3:} $|A_{0}|\leq (m-1)^{2}$ and $|B_{0}|\leq (m-1)^{2}$, but we have $A_{1}=\emptyset $ or $B_{1}=\emptyset $ 

We may assume that $A_{1}=\emptyset $.
Then $3(m-1)^{2}+1<p-1=|A_{0}|+|B_{0}|+|B_{1}|\leq |B_{1}|+2(m-1)^{2}$, and hence $|B_{1}|>(m-1)^{2}+1$.
On the other hand, since $|A|\geq 2(m-1)^{7}-m+1>(m-1)^{2}(m-1)^{5}\geq |A_{0}|(m-1)^{5}=|A_{0}\cup A_{1}|(m-1)^{5}=|A\cap X|(m-1)^{5}$ and since $|V(H_{i})|\leq (m-1)^{5}$ for each $2\leq i\leq p$, there exists $i$ with $2\leq i\leq p$ such that $V(H_{i})\cap A\not=\emptyset $ and $x_{i}\not\in A$.
We may assume that $V(H_{2})\cap A\not=\emptyset $ and $x_{2}\not\in A$.
Then $x_{2}\in B$, which implies $|V(H_{2})|\geq 2$.
Let $a_{2}b_{2}\in E(H_{2})$ with $a_{2}\in A$ and $b_{2}\in B$.
Now since $|B_{1}-\{x_{2}\}|>(m-1)^{2}$, $|V_{B_{1}-\{x_{2}\}}|\geq m$ by Claim~\ref{ns3 lem4 cl2}.
Since $G$ is $K_{3}$-free, it follows that if both $b_{1}$ and $b_{2}$ are adjacent to all vertices in $V_{B_{1}-\{x_{2}\}}$, then $G[\{b_{1},b_{2}\}\cup V_{B_{1}-\{x_{2}\}}]$ contains $K_{2,m}$ as an induced subgraph, a contradiction.
Thus $b_{h}v_{y}\not\in E(G)$ for some $h\in \{1,2\}$ and some $y\in B_{1}-\{x_{2}\}$.
We may assume that $y=x_{3}$.
Since $|A|\geq 2(m-1)^{7}-m+1>3(m-1)^{5}\geq |V(H_{1})|+|V(H_{2})|+|V(H_{3})|$, $A-(V(H_{1})\cup V(H_{2})\cup V(H_{3}))\not=\emptyset $.
Let $x\in A-(V(H_{1})\cup V(H_{2})\cup V(H_{3}))$.
Since $v_{x_{3}}\not\in N(a)\cup N(b)$ and $k=4$, the path $xaa_{h}b_{h}bx_{3}v_{x_{3}}$ is an induced path of order $7$, a contradiction.

\medskip
\noindent
\textbf{Case 3:} $k\geq 5$

If $k=5$, then by Lemma~\ref{ns3 lem2} and the assumption that $d(a)\geq 2(m-1)^{7}\geq 4(m-1)+2$ and $d(b)\geq 2(m-1)^{7}\geq 4(m-1)+2$, $G$ contains $P_{7}$ as an induced subgraph, a contradiction; if $k\geq 6$, then by Lemma~\ref{ns3 lem1} and the assumption that $d(a)\geq 2(m-1)^{7}\geq 4(m-1)+2$, $G$ contains $P_{7}$ as an induced subgraph, a contradiction.

This completes the proof of Lemma~\ref{ns3 lem4}.
\qed

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{ns3 prop2}
Let $m\geq 5$.
Then $\G _{3}(\{K_{3},K_{2,m},P_{7}\})$ is finite.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Let $G\in \G _{3}(\{K_{3},K_{2,m},P_{7}\})$.
We show that $|V(G)|\leq (4(m-1)^{7}(2(m-1)^{7}+m-2)^{5})^{5}$.
Since $\mbox{diam}(G)\leq 5$, it suffices to show that $\Delta (G)\leq 4(m-1)^{7}(2(m-1)^{7}+m-2)^{5}$.
Suppose that $\Delta (G)\geq 4(m-1)^{7}(2(m-1)^{7}+m-2)^{5}+1$, and let $w\in V(G)$ be a vertex with $d(w)=\Delta (G)$.
Since $G$ is $K_{3}$-free, $N(w)$ is independent.
Let $L=\{x\in V(G)\mid |N(x)\cap N_{2}(w)|\geq 2(m-1)^{7}\}$.
By Lemma~\ref{ns3 lem4}, $|L\cap N(w)|\leq 1$ and $L\cap N_{2}(w)=\emptyset $.

\medskip
\noindent
\textbf{Case 1:} $|L|\geq 2$ and $|L\cap N(w)|=1$

Let $a\in L\cap N(w)$ and $b\in L-N(w)$.
Since $b\not\in N(w)\cup N_{2}(w)$, we have $b\in N_{3}(w)$, and hence $ab\not\in E(G)$.
By Lemma~\ref{ns3 lem4}, $N(a)\cap N(b)=\emptyset $.
Let $y\in N(b)\cap N_{2}(w)$ and $v\in N(y)\cap N(w)$.
Since $ab\not\in E(G)$, $N(a)\cap N(b)=\emptyset $, and $N(w)$ is independent, it follows that the path $awvyb$ is an induced path of order $5$.
Hence by Lemma~\ref{ns3 lem2} and the assumption that $d(a)\geq 2(m-1)^{7}\geq 4(m-1)+2$ and $d(b)\geq 2(m-1)^{7}\geq 4(m-1)+2$, $G$ contains $P_{7}$ as an induced subgraph, a contradiction.

\medskip
\noindent
\textbf{Case 2:} $|L|\geq 2$ and $L\cap N(w)=\emptyset $

Let $a,b\in L$ with $a\not=b$.
Then $a,b\in N_{3}(w)$.
If $ab\in E(G)$, then $N(a)\cap N(b)=\emptyset $ by the assumption that $G$ is $K_{3}$-free and, if $ab\not\in E(G)$, then $N(a)\cap N(b)=\emptyset $ by Lemma~\ref{ns3 lem4}.
Hence in either case, we have $N(a)\cap N(b)=\emptyset $.
Let $A=N(a)\cap N_{2}(w)$ and $B=N(b)\cap N_{2}(w)$.
Let $y\in B$ and $v\in N(y)\cap N(w)$.
Since $G$ is $K_{2,m}$-free, $|N(y)\cap N(w)|\leq m-1$.
Since $G$ is $K_{2,m}$-free, it follows that $|((\bigcup_{u\in N(y)\cap N(w)}N(u))\cup N(y))\cap A|\leq (|N(y)\cap N(w)|+1)(m-1)\leq m(m-1)$.
Since $|A|\geq 2(m-1)^{7}\geq m(m-1)+1$, $A-((\bigcup_{u\in N(y)\cap N(w)}N(u))\cup N(y))\not=\emptyset $.
Take $x\in A-((\bigcup_{u\in N(y)\cap N(w)}N(u))\cup N(y))$.
Since $v\in N(y)\cap N(w)$, we have $xv,xy\not\in E(G)$.
Take $u\in N(x)\cap N(w)$.
By the choice of $x$, $u\not\in N(y)\cap N(w)$, and hence $uy\not\in E(G)$.
Consequently the path $xuwvyb$ is an induced path of order $6$.
Now by Lemma~\ref{ns3 lem1} and the assumption that $d(b)\geq 2(m-1)^{7}\geq 4(m-1)+2$, $G$ contains $P_{7}$ as an induced subgraph, a contradiction.

\medskip
\noindent
\textbf{Case 3:} $|L|\leq 1$

Let $H_{1},\ldots ,H_{p}$ be the components of $G[(N(w)\cup N_{2}(w))-L]$.
Let $1\leq i\leq p$.
Take $x\in V(H_{i})$.
Since $G$ is $\{K_{3},K_{2,m}\}$-free, $|N(x)\cap N(w)|\leq m-1$.
Since $x\not\in L$, we also have $|N(x)\cap N_{2}(w)|\leq 2(m-1)^{7}-1$.
Hence $d_{H_{i}}(x)\leq |N(x)\cap N(w)|+|N(x)\cap N_{2}(w)|\leq 2(m-1)^{7}+m-2$.
Since $x\in V(H_{i})$ is arbitrary, $\Delta (H_{i})\leq 2(m-1)^{7}+m-2$.
Since $H_{i}$ is $P_{7}$-free, we also have $\mbox{diam}(H_{i})\leq 5$.
Thus $|V(H_{i})|\leq (2(m-1)^{7}+m-2)^{5}$ for each $1\leq i\leq p$.
We have $V(H_{i})\cap N_{2}(w)\not=\emptyset $ for each $1\leq i\leq p$ because $N(w)$ is independent.
Let $q=|\{i\mid V(H_{i})\cap N(w)\not=\emptyset \}|$.
(Note that in the case where $\emptyset \not=L\subseteq N(w)$, it is possible that $V(H_{i})\cap N(w)=\emptyset $ for some $i$.)
Without loss of generality, we may assume that $V(H_{i})\cap N(w)\not=\emptyset $ for each $1\leq i\leq q$ and $V(H_{i})\cap N(w)=\emptyset $ for each $q+1\leq i\leq p$.
Since $|N(w)-L|\geq d(w)-1\geq 4(m-1)^{7}(2(m-1)^{7}+m-2)^{5}$, we have $q\geq |N(w)-L|/(2(m-1)^{7}+m-2)^{5}\geq 4(m-1)^{7}$.
Since $G$ is $3$-connected, $G-(\{w\}\cup L)$ is connected.
Let $P$ be a shortest $V(H_{1})$-$((N(w)\cup N_{2}(w))-L-V(H_{1}))$ path in $G-(\{w\}\cup L)$.
Then $|V(P)|\geq 3$.
Let $y_{1}$, $z_{1}$ and $z_{2}$ be the first three vertices of $P$.
Then $y_{1}\in V(H_{1})\cap N_{2}(w)$ and $z_{1}\in N_{3}(w)$.
Let $wx_{1}y_{1}$ be a shortest $w$-$y_{1}$ path in $G[V(H_{1})\cup \{w\}]$.
Then $wx_{1}y_{1}z_{1}z_{2}$ is an induced $w$-$z_{2}$ path of order $5$.
Since $z_{1}\not\in L$, $|N(z_{1})\cap N_{2}(w)|\leq 2(m-1)^{7}-1$.
In particular, $z_{1}$ is adjacent to at most $2(m-1)^{7}-1$ of the $H_{i}$, $1\leq i\leq q$.
If $z_{2}\not\in N_{2}(w)$, then similarly, $z_{2}$ is adjacent to at most $2(m-1)^{7}-1$ of the $H_{i}$, $1\leq i\leq q$; if $z_{2}\in N_{2}(w)$, then clearly $z_{2}$ is adjacent to at most one of the $H_{i}$, $1\leq i\leq q$ (note that it is possible that $z_{2}$ belongs to $H_{i}$ for some $i$ with $q+1\leq i\leq p$).
Thus $z_{2}$ is adjacent to at most $2(m-1)^{7}-1$ of the $H_{i}$.
Since $q\geq 4(m-1)^{7}$, there exists $j$ with $2\leq j\leq q$ such that neither $z_{1}$ nor $z_{2}$ is adjacent to $H_{j}$.
Since $V(H_{j})\cap N(w)\not=\emptyset $ and $V(H_{j})\cap N_{2}(w)\not=\emptyset $, there exist $x_{2}\in V(H_{j})\cap N(w)$ and $y_{2}\in V(H_{j})\cap N_{2}(w)$ such that $x_{2}y_{2}\in E(G)$.
Then $y_{2}x_{2}wx_{1}y_{1}z_{1}z_{2}$ is an induced path of order $7$, a contradiction.

This completes the proof of Proposition~\ref{ns3 prop2}.
\qed

\medbreak\noindent\textit{Proof of Theorem~\ref{ns3}.}\quad
The `if ' part follows from Propositions~\ref{ns3 prop1} and \ref{ns3 prop2}.
Thus it suffices to prove the  `only if' part.
Suppose that $\G _{3}(\{K_{3},K_{2,m},T\})$ is finite.
From Theorem~\ref{nec}, it follows that $T\in \T _{2}$.
For each $s\geq 2$, let $H^{(1)}_{s}$ be the graph defined by $V(H^{(1)}_{s})=\{x_{1},x_{2},x_{3}\}\cup \{y_{i,h},z_{i,h}\mid 1\leq i\leq 2,1\leq h\leq s\}$ and $E(H^{(1)}_{s})=\{x_{1}x_{2}\}\cup \{x_{i}y_{i,h},x_{3}z_{i,h}\mid 1\leq i\leq 2,1\leq h\leq s\}\cup \{y_{i,h}z_{j,h}\mid 1\leq i,j\leq 2,1\leq h\leq s\}$.
Then $H^{(1)}_{s}$ is $3$-connected and $\{K_{3},K_{2,m}\}$-free.
Since $\G _{3}(\{K_{3},K_{2,m},T\})$ is finite, there exists $s\geq 2$ such that $H^{(1)}_{s}$ contains $T$ as an induced subgraph.
Since every tree contained in $H^{(1)}_{s}$ as an induced subgraph has diameter at most $6$, $\mbox{diam}(T)\leq 6$.
Now for each $t\geq 3$, let $P=x_{1}x_{2}$ be a path of order $2$ and $C=y_{1}y_{2}\cdots y_{2t}y_{1}$ be a cycle of order $2t$, and let $H^{(2)}_{t}$ be the graph defined by $V(H^{(2)}_{t})=V(C)\cup V(P)$ and $E(H^{(2)}_{t})=E(C)\cup E(P)\cup \{x_{1}y_{2i-1},x_{2}y_{2i}\mid 1\leq i\leq t\}$.
Then $H^{(2)}_{t}$ is $3$-connected and $\{K_{3},K_{2,m}\}$-free.
Hence there exists $t\geq 3$ such that $H^{(2)}_{t}$ contains $T$ as an induced subgraph.
Observe that if $F$ is a member of $\T _{2}$ contained in $H^{(2)}_{t}$ as an induced subgraph, then $F$ is a path or an induced subgraph of $Y^{*}_{3}$.
Therefore $T$ is either a path of order at most $7$ or an induced subgraph of $Y^{*}_{3}$.
\qed

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% 4 3600 1500 3600 1600 5 0 0 0
% $y_{2,s}$
\put(36.0000,-16.0000){\makebox(0,0){$y_{2,s}$}}%
% STR 2 0 3 0 Black White
% 4 3600 2100 3600 2200 5 0 0 0
% $z_{2,s}$
\put(36.0000,-22.0000){\makebox(0,0){$z_{2,s}$}}%
% STR 2 0 3 0 Black White
% 4 3000 1500 3000 1600 5 0 0 0
% $y_{1,s}$
\put(30.0000,-16.0000){\makebox(0,0){$y_{1,s}$}}%
% STR 2 0 3 0 Black White
% 4 3000 2100 3000 2200 5 0 0 0
% $z_{1,s}$
\put(30.0000,-22.0000){\makebox(0,0){$z_{1,s}$}}%
% STR 2 0 3 0 Black White
% 4 2200 2900 2200 3000 5 0 0 0
% $x_{3}$
\put(22.0000,-30.0000){\makebox(0,0){$x_{3}$}}%
% STR 2 0 3 0 Black White
% 4 4195 1300 4195 1400 5 0 0 0
% $x_{1}$
\put(41.9500,-14.0000){\makebox(0,0){$x_{1}$}}%
% STR 2 0 3 0 Black White
% 4 4195 2100 4195 2200 5 0 0 0
% $x_{2}$
\put(41.9500,-22.0000){\makebox(0,0){$x_{2}$}}%
% STR 2 0 3 0 Black White
% 4 5595 1100 5595 1200 2 0 0 0
% 
\put(55.9500,-12.0000){\makebox(0,0)[lb]{}}%
% STR 2 0 3 0 Black White
% 4 5595 1070 5595 1170 5 0 0 0
% $y_{2t}$
\put(55.9500,-11.7000){\makebox(0,0){$y_{2t}$}}%
% STR 2 0 3 0 Black White
% 4 5165 1190 5165 1290 5 0 0 0
% $y_{1}$
\put(51.6500,-12.9000){\makebox(0,0){$y_{1}$}}%
% STR 2 0 3 0 Black White
% 4 5025 1410 5025 1510 5 0 0 0
% $y_{2}$
\put(50.2500,-15.1000){\makebox(0,0){$y_{2}$}}%
% STR 2 0 3 0 Black White
% 4 4995 1700 4995 1800 5 0 0 0
% $y_{3}$
\put(49.9500,-18.0000){\makebox(0,0){$y_{3}$}}%
% STR 2 0 3 0 Black White
% 4 5005 1980 5005 2080 5 0 0 0
% $y_{4}$
\put(50.0500,-20.8000){\makebox(0,0){$y_{4}$}}%
% STR 2 0 3 0 Black White
% 4 5215 2240 5215 2340 5 0 0 0
% $y_{5}$
\put(52.1500,-23.4000){\makebox(0,0){$y_{5}$}}%
% STR 2 0 3 0 Black White
% 4 5595 2340 5595 2440 5 0 0 0
% $y_{6}$
\put(55.9500,-24.4000){\makebox(0,0){$y_{6}$}}%
% STR 2 0 3 0 Black White
% 4 890 760 890 860 5 0 0 0
% $H^{(1)}_{s}$
\put(8.9000,-8.6000){\makebox(0,0){$H^{(1)}_{s}$}}%
% STR 2 0 3 0 Black White
% 4 4145 760 4145 860 5 0 0 0
% $H^{(2)}_{t}$
\put(41.4500,-8.6000){\makebox(0,0){$H^{(2)}_{t}$}}%
\end{picture}%
\caption{Graphs $H^{(1)}_{s}$ and $H^{(2)}_{t}$}
\label{f4}
\end{center}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Concluding remarks}\label{sec7}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this paper, we have considered three forbidden subgraphs which generate a finite set in the class of $3$-connected graphs.
As we have seen in Theorem~\ref{nec}, there are six types.
For many of them, we have given a characterization.
The cases which remain uncharacterized are the following:\\
(a) $\F=\{K_{3},K_{3,3},T\}$, where $T\in \T _{2}$;\\
(b) $\F=\{K_{4},K_{2,2},T\}$, where $T$ is a path of order at least $4$;\\
(c) $\F=\{K_{3},K_{2,4},T\}$ and $\{K_{3},K_{2,3},T\}$, where $T\in \T _{1}$;\\
(d) $\F=\{K_{3},K_{2,2},T\}$, where $T\in \T _{0}$;\\
(e) $\F=\{K_{3},K_{1,4},T\}$, where $T\in \T _{1}$;\\
(f) $\F=\{K_{n},K_{1,3},T\}$, where $n\geq 5$ and $T\in \T ^{*}_{2}$;\\
(g) $\F=\{K_{4},K_{1,3},T\}$, where $T\in \T ^{*}_{1}$.\\
In cases (a)--(d), $\F$ does not contain a star.
In these cases, we can give a bound on the diameter of $T$.

\begin{prop}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\label{diam finite prop}
Let $\F$ be a subset of $\G$ with $|\F|=3$.
Suppose $\F$ does not contain a star and $\G _{3}(\F)$ is finite.
Then $\F$ contains a tree of diameter at most $8$.
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\proof
Since $\F$ does not contain a star, it follows from Theorem~\ref{nec} that $\F$ can be written in the form $\F =\{K_{n},K_{m_{1},m_{2}},T\}$, where $n\in \{3,4\}$, $m_{1}\in \{2,3\}$, $m_{2}\geq m_{1}$ and $T\in \T _{0}$.

Let $s\geq 2$.
Let $C^{(i)}=a^{(i)}_{1}b^{(i)}_{1}c^{(i)}_{1}a^{(i)}_{2}b^{(i)}_{2}c^{(i)}_{2}a^{(i)}_{1}$ be a cycle of order $6$ for each $1\leq i\leq s$, and define a graph $H_{s}$ by $V(H_{s})=\{a_{0},b_{0},c_{0}\}\cup (\bigcup _{1\leq i\leq s}V(C^{(i)}))$ and $E(H_{s})=\{a_{0}a^{(i)}_{j},b_{0}b^{(i)}_{j},c_{0}c^{(i)}_{j}\mid 1\leq i\leq s,1\leq j\leq 2\}\cup (\bigcup _{1\leq i\leq s}E(C^{(i)}))$.
Then $H_{s}$ is $3$-connected and $\{K_{n},K_{m_{1},m_{2}}\}$-free.
Hence there exists $s\geq 2$ such that $H_{s}$ contains $T$ as an induced subgraph.
However, $H_{s}$ does not contain $P_{10}$ as an induced subgraph.
Therefore the diameter of $T$ is at most $8$.
\qed

In (a)--(d), we have $T\in \T _{0}$ and hence $\Delta (T)\leq 3$.
By combining this fact and Proposition~\ref{diam finite prop}, we see that the order of $T$ is bounded.
Thus the number of triples in these cases is finite.
On the other hand, in cases (e)--(g), where the triple contains a star, Proposition~\ref{diam finite prop} gives no further information about $\F$.
In fact, Theorem~\ref{Diestel th} shows that in these cases, there exist infinitely many $\F$ such that $\G _{3}(\F)$ is finite.

We add that for (a) and (b), and for the case where $\F=\{K_{3},K_{2,4},T\}$ in (c), the determination of $T$ has recently been completed in \cite{EF}.

\section*{Acknowledgment}
The authors are grateful to the referees for helpful comments.
In particular, one of them has pointed out errors in the first version of the paper.


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% then copy and past the contents of your .bbl file into your .tex file

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