Triangle-free graphs that do not contain an induced subdivision of $K_4$ are 3-colorable

TL;DR

This paper proves that triangle-free graphs lacking an induced subdivision of K4 are always 3-colorable, confirming a conjecture and advancing understanding of graph coloring constraints.

Contribution

It establishes that such graphs are 3-colorable, resolving a conjecture by Trotignon and Vuskovic.

Findings

Triangle-free graphs without induced K4 subdivisions are 3-colorable.

Confirmed the conjecture of Trotignon and Vuskovic.

Provides a structural characterization relevant to graph coloring.

Abstract

We show that triangle-free graphs that do not contain an induced subgraph isomorphic to a subdivision of K4 are 3-colorable. This proves a conjecture of Trotignon and Vuskovic.

Equations

$$W_1=(D_1,x)$$

$$v\in V(G)\setminus V(W_1)$$

$$v$$

$$P^{\ast }$$

$$P^{\prime }$$

$$V(C)$$

$$v\in V(G)\setminus V(W_1)$$

$$W_1$$

$$v\in V(G)\setminus (V(W_1)\cup X)$$

$$v$$

$$P^{\ast }$$

$$i=j$$

$$i\le 2$$

$$z\in V(C)$$

$$x\in K$$

$$z$$

$$1\le i<t$$

$$i=1,\dots ,k$$

$$|V(C_k)|>1$$

$$V(G^{\prime })\setminus \left\{z\right\}$$

$$G^{\prime }$$

$$K$$

$$G^{\prime }\setminus z$$

Full text

1 Introduction

All graphs in this paper are finite and simple. Let $G$ be a graph. For a vertex $v\in V(G)$, we denote its set of neighbors by $N(v)$, and we let $N[v]=\left\{v\right\}\cup N(v)$. For $X,Y\subseteq V(G)$, we say that $X$ is complete to $Y$ if every vertex in $X$ is adjacent to every vertex in $Y$; $X$ is anticomplete to $Y$ if every vertex in $X$ is non-adjacent to every vertex in $Y$. A vertex $v\in V(G)$ is complete (anticomplete) to $X\subseteq V(G)$ if $\left\{v\right\}$ is complete (anticomplete) to $X$. A set $X\subseteq V(G)$ is a cutset for $G$ if there is a partition $(X,Y,Z)$ of $V(G)$ with $Y,Z\neq\emptyset$ and $Y$ anticomplete to $Z$. The cutset $X$ is a clique cutset if $X$ is a (possibly empty) clique in $G$. For a graph $H$, we say that $G$ contains $H$ if $H$ is isomorphic to an induced subgraph of $G$, and otherwise, $G$ is $H$-free. For a family $\mathcal{F}$ of graphs, we say that $G$ is $\mathcal{F}$-free if $G$ is $F$-free for every graph $F\in\mathcal{F}$.

For a graph $G$ and $X\subseteq V(G)$, $G|X$ denotes the induced subgraph of $G$ with vertex set $X$. For $X\subseteq V(G)$, we let $G\setminus X=G|(V(G)\setminus X)$ and for $x\in V(G)$, we let $G\setminus x=G|(V(G)\setminus\left\{x\right\})$. By a path in a graph we mean an induced path. Let $C$ be a cycle in $G$. The length of $C$ is $|V(C)|$. The girth of $G$ is the length of a shortest cycle, and is defined to be $\infty$ if $G$ has no cycle. A hole in a graph is an induced cycle of length at least four. An $\textnormal{ISK}_{4}$ is a graph that is isomorphic to a subdivision of $K_{4}$.

In [5] two of us studied the structure of $\textnormal{ISK}_{4}$-free graphs, and proposed the following conjecture (and proved several special cases of it):

Conjecture 1**.**

If $G$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free, then $\chi(G)\leq 3$.

In [2], Conjecture 1 was proved with 3 replaced by 4.

The main result of the present paper is the proof of Conjecture 1. In fact, we prove a stronger statement, from which Conjecture 1 easily follows:

Theorem 2**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. Then either $G$ has a clique cutset, $G$ is complete bipartite, or $G$ has a vertex of degree at most two.

For an induced subgraph $H$ of $G$ we write $v\in H$ to mean $v\in V(H)$. We use the same convention if $H$ is a path or a hole. For a path $P=p_{1}-\ldots-p_{k}$ we call the set $V(P)\setminus\{p_{1},p_{k}\}$ the interior of $P$, and denote it by $P^{*}$.

A wheel in a graph is a pair $W=(C,x)$ where $C$ is a hole and $x$ has at least three neighbors in $V(C)$. We call $C$ the rim of the wheel, and $x$ the center. The neighbors of $x$ in $V(C)$ are called the spokes of $W$. Maximal paths of $C$ that do not contain any spokes in their interior are called the sectors of $W$. We write $V(W)$ to mean $V(C)\cup\{x\}$.

A graph is series-parallel if it does not contain a subdivision of $K_{4}$ as a (not necessarily induced) subgraph.

Theorem 3** ([1]).**

Let $G$ be a series-parallel graph. Then $G$ is $\textnormal{ISK}_{4}$-free, wheel-free, and $K_{3,3}$-free, and $G$ contains a vertex of degree at most two.

The following two facts were proved in [3]:

Theorem 4** ([3]).**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. Then either $G$ is series-parallel, or $G$ contains a $K_{3,3}$ subgraph, or $G$ contains a wheel. If $G$ contains a subdivision of $K_{3,3}$ as an induced subgraph, then $G$ contains a $K_{3,3}$.

Theorem 5** ([3]).**

If $G$ is an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph and $G$ contains $K_{3,3}$, then either $G$ is complete bipartite, or $G$ has a clique cutset.

Thus to prove Theorem 2 we need to analyze $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graphs that contain wheels. This approach was already explored in [5], but we were able to push it further, as follows.

A wheel $W=(C,x)$ is proper if for every $v\in V(G)\setminus V(W)$

• •

there is a sector $S$ of $W$ such that $N(v)\cap V(C)\subseteq V(S)$.

• •

If $v$ has at least three neighbors in $V(C)$, then $v$ is adjacent to $x$.

(Please note that this definition is different from the one in [5].) We prove:

Theorem 6**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, and let $x$ be the center of a proper wheel in $G$. If $W=(C,x)$ is a proper wheel with a minimum number of spokes subject to having center $x$, then

every component of $V(G)\setminus N(x)$ contains the interior of at most one sector of $W$, and 2. 2.

for every $u\in N(x)$, the component $D$ of $V(G)\setminus(N(x)\setminus\{u\})$ such that $u\in V(D)$ contains the interiors of at most two sectors of $W$, and if $S_{1},S_{2}$ are sectors with $S_{i}^{*}\subseteq V(D)$ for $i=1,2$, then $V(S_{1})\cap V(S_{2})\neq\emptyset$.

Using Theorem 6 we can prove a variant of a conjecture from [5] that we now explain. For a graph $G$ and $x,y\in V(G)$, we say that $(x,y)$ is a non-center pair for $G$ if neither $x$ nor $y$ is the center of a proper wheel in $G$, and $x=y$ or $xy\in E(G)$. We prove:

Theorem 7**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph which is not series-parallel, and let $(x,y)$ be a non-center pair for $G$. Then some $v\in V(G)\setminus(N[x]\cup N[y])$ has degree at most two.

Here is the outline of the proof; the full proof is given in Section 5. We assume that $G$ is a counterexample to Theorem 7 with $|V(G)|$ minimum. Since $G$ is not series-parallel, it follows from Theorem 4 that $G$ contains a wheel, and we show in Lemma 9 that $G$ contains a proper wheel. Let $s\in V(G)$ be the center of a proper wheel chosen as in Theorem 6, and let $C_{1},\dots,C_{k}$ be the components of $G\setminus N[s]$. By Theorem 6, it follows that $k>1$. For each $i$, let $N_{i}$ be the set of vertices of $N(s)$ with a neighbor in $V(C_{i})$, and let $G_{i}=G|(V(C_{i})\cup N_{i}\cup\{s\})$. We analyze the structure of the graphs $G_{i}$ using the minimality of $|V(G)|$. It turns out that at most one $G_{i}$ is not series-parallel, and that (by contracting $C_{i}$’s) there is at most one value of $i$ for which $|V(C_{i})|>1$. Also, if $|V(C_{i})|>1$, then $\left\{x,y\right\}\cap V(C_{i})\neq\emptyset$. We may assume that $|V(C_{i})|=1$ for all $i\in\{1,\dots,k-1\}$, and that $\left\{x,y\right\}\cap V(C_{k})\neq\emptyset$. Now consider the bipartite graph $G^{\prime}$, which (roughly speaking) is the graph obtained from $G\setminus\{s\}$ by contracting $V(C_{k})\cup N_{k}$ to a single vertex $z$ if $|V(C_{k})|>1$. It turns out that $G^{\prime}$ is $\left\{\textnormal{ISK}_{4},K_{3,3}\right\}$-free and has girth at least $6$, while cycles that do not contain $z$ must be even longer. Now either there is an easy win, or we find a cycle in $G^{\prime}$ that contains a long path $P$ of vertices all of degree two in $G^{\prime}$ and with $V(P)\subseteq V(G)\setminus(N[x]\cup N[y])$. Further analysis shows that at least one of these vertices has degree two in $G$, and Theorem 7 follows.

This paper is organized as follows. In Section 2 we prove Theorem 6. Section 3 contains technical tools that we need to deduce that the graph $G^{\prime}$ described above has various useful properties. In Section 4 we develop techniques to produce a cycle with a long path of vertices of degree two. In Section 5 we put all of our knowledge together to prove Theorems 2 and 7 and deduce Conjecture 1.

Let us finish this section with an easy fact about $\textnormal{ISK}_{4}$-free graphs. Given a hole $C$ and a vertex $v\not\in C$, $v$ is linked to $C$ if there are three paths $P_{1},P_{2},P_{3}$ such that

• •

$P_{1}^{*}\cup P_{2}^{*}\cup P_{3}^{*}\cup\{v\}$ is disjoint from $C$;

• •

each $P_{i}$ has one end $v$ and the other end in $C$, and there are no other edges between $P_{i}$ and $C$;

• •

for $i,j\in\left\{1,2,3\right\}$ with $i\neq j$, $V(P_{i})\cap V(P_{j})=\left\{v\right\}$;

• •

if $x\in P_{i}$ is adjacent to $y\in P_{j}$ then either $v\in\{x,y\}$ or $\left\{x,y\right\}\subseteq V(C)$; and

• •

if $v$ has a neighbor $c\in C$, then $c\in P_{i}$ for some $i$.

Lemma 8**.**

If $G$ is $\textnormal{ISK}_{4}$-free, then no vertex of $G$ can be linked to a hole.

2 Wheels

Lemma 9**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph that contains a wheel. Then there is a proper wheel in $G$.

Proof.

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. Let $W=(C,x)$ be a wheel in $G$ with $|V(C)|$ minimum. We claim that $W$ is a proper wheel. Suppose $v\in V(G)\setminus V(W)$ violates the definition of a proper wheel.

If $v$ has at least three neighbors in the hole $x-S-x$ for some sector $S$ of $W$, then $(x-S-x,v)$ is a wheel with shorter rim than $W$, a contradiction. So $v$ has at most two neighbors in every sector of $W$ (and at most one if $v$ is adjacent to $x$). Therefore there exist sectors $S_{1},S_{2}$ of $W$ such that $v$ has a neighbor in $V(S_{1})\setminus V(S_{2})$ and a neighbor in $V(S_{2})\setminus V(S_{1})$. Also by the minimality of $|V(C)|$, every path of $C$ whose ends are in $N(v)$ and with interior disjoint from $N(v)$ contains at most two spokes of $W$, and we can choose $S_{1},S_{2}$ and for $i=1,2$, label the ends of $S_{i}$ as $a_{i},b_{i}$ such that either $b_{1}=a_{2}$, or $b_{1},a_{2}$ are the ends of a third sector $S_{3}$ of $W$ and $v$ has no neighbor in $S_{3}^{*}$. If possible, we choose $S_{1},S_{2}$ such that $b_{1}=a_{2}$. If $v$ has two neighbors in $S_{1}$, denote them $s,t$ such that $a_{1},t,s,b_{1}$ are in order in $C$. If $v$ has a unique neighbor in $S_{1}$, denote it by $s$. Let $z$ be the neighbor of $v$ in $S_{2}$ closest to $a_{2}$.

Assume first that $v$ is non-adjacent to $x$. Suppose $b_{1}\neq a_{2}$. By Lemma 8, $x$ cannot be linked to the hole $z-S_{2}-a_{2}-S_{3}-b_{1}-S_{1}-s-v-z$, and it follows that $z\neq b_{2}$. If $v$ has two neighbors in $S_{1}$, then $v$ can be linked to $x-S_{3}-x$ via the paths $v-s-S_{1}-b_{1}$, $v-t-S_{1}-a_{1}-x$, $v-z-S_{2}-a_{2}$; and if $v$ has a unique neighbor in $S_{1}$, then $s$ can be linked to $x-S_{3}-x$ via the paths $s-S_{1}-b_{1}$, $s-S_{1}-a_{1}-x$, $s-v-z-S_{2}-a_{2}$ (note that by the choice of $S_{1},S_{2}$ and since $b_{1}\neq a_{2}$, it follows that $s\neq b_{1}$). In both cases, this is contrary to Lemma 8. This proves that $b_{1}=a_{2}$. Let $y$ be the neighbor of $v$ in $S_{2}$ closest to $b_{2}$. Now if $v$ has two neighbors in $S_{1}$, then $v$ can be linked to $x-S_{1}-x$ via the paths $v-s$, $v-t$, $v-y-S_{2}-b_{2}-x$, contrary to Lemma 8. So $v$ has a unique neighbor in $S_{1}$, and similarly a unique neighbor in $S_{2}$. It follows that $s,b_{1}$ and $z$ are all distinct. Now we can link $x$ to $s-S_{1}-b_{1}-S_{2}-z-v-s$ via the paths $x-b_{1}$, $x-a_{1}-S_{1}-s$, and $x-b_{2}-S_{2}-z$, contrary to Lemma 8.

This proves that $v$ is adjacent to $x$, and so $v$ has at most one neighbor in every sector of $W$. If $b_{1}\neq a_{2}$, then $v$ can be linked to $x-S_{3}-x$ via the paths $v-s-S_{1}-b_{1}$, $v-x$, $v-z-S_{2}-a_{2}$, and if $b_{1}=a_{2}$, then $s,b_{1}$ and $z$ are all distinct and hence $x$ can be linked to the hole $s-S_{1}-b_{1}-S_{2}-z-v-s$ via the paths $x-b_{1}$, $x-v$, and $x-b_{2}-S_{2}-z$; in both cases contrary to Lemma 8. This proves that every $v\in V(G)\setminus V(W)$ satisfies the condition in the definition, and so $W$ is a proper wheel in $G$. ∎

Let $W=(C,v)$ be a wheel. We call $x$ proper for $W$ if either $x\in V(C)\cup\left\{v\right\}$; or

• •

all neighbors of $x$ in $V(C)$ are in one sector of $W$; and

• •

if $x$ has more than two neighbors in $V(C)$, then $x$ is adjacent to $v$.

A vertex $x$ is non-offensive for a wheel $W=(C,v)$ if there exist two sectors $S_{1},S_{2}$ of $W$ such that

• •

$x$ is adjacent to $v$;

• •

$x$ has neighbors in $S_{1}$ and in $S_{2}$;

• •

$N(x)\cap V(C)\subseteq V(S_{1})\cup V(S_{2})$;

• •

$S_{1}$ and $S_{2}$ are consecutive; and

• •

if $u\in V(G)\setminus V(W)$ is adjacent to $x$, then $N(u)\cap V(C)\subseteq V(S_{1})\cup V(S_{2})$.

If $V(S_{1})\cap V(S_{2})=\{a\}$, we also say that $x$ is $a$-non-offensive.

Lemma 10**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. Let $W=(C,v)$ be a wheel in $G$. Let $S_{1},S_{2}$ be consecutive sectors of $W$, and let $x\in N(v)\setminus V(W)$ be a vertex such that $N(x)\cap V(C)\subseteq V(S_{1})\cup V(S_{2})$ and $N(x)\cap V(S_{1}),N(x)\cap V(S_{2})\neq\emptyset$. Then $N(x)\cap V(C)\subseteq S_{1}^{*}\cup S_{2}^{*}$ and for $\left\{i,j\right\}=\left\{1,2\right\}$, $|N(x)\cap(V(S_{i})\setminus V(S_{j}))|\geq 3$.

Proof.

Since $x$ is adjacent to $v$ and $G$ is triangle-free, it follows that $N(x)\cap V(C)\subseteq S_{1}^{*}\cup S_{2}^{*}$. Suppose for a contradiction that $|N(x)\cap(V(S_{1})\setminus V(S_{2}))|\leq 2$.If $|N(x)\cap(V(S_{1})\setminus V(S_{2}))|=2$, then $x$ has exactly three neighbors in the hole $v-S_{1}-v$, contrary to Lemma 8. It follows that $|N(x)\cap(V(S_{1})\setminus V(S_{2}))|=1$. Let $z$ denote the neighbor of $x$ in $V(S_{1})$. Let $\left\{w\right\}=V(S_{1})\cap V(S_{2})$, and let $y$ denote the neighbor of $x$ in $V(S_{2})$ closest to $w$ along $S_{2}$. Then $x$ can be linked to the hole $v-S_{1}-v$ via the three paths $x-v$, $x-z$, and $x-y-S_{2}-w$. This is a contradiction to Lemma 8, and the result follows. ∎

We say that wheel $W=(C,v)$ is $k$-almost proper if there are spokes $x_{1},\dots,x_{k}$ of $W$ and a set $X\subseteq V(G)\setminus V(W)$ such that

• •

no two spokes in $\left\{x_{1},\dots,x_{k}\right\}$ are consecutive;

• •

$W$ is proper in $G\setminus X$;

• •

for every $x$ in $X$ there exists $i$ such that $x$ is $x_{i}$-non-offensive.

Lemma 11**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph, and let $W=(C,v)$ be a 1-almost proper wheel in $G$. Let $x_{1}$ and $X$ be as in the definition of a 1-almost proper wheel, and let $S_{1}$ and $S_{2}$ be the sectors of $W$ containing $x_{1}$.

Then there exists a proper wheel $W^{\prime}$ in $G$ with center $v$ and the same number of spokes as $W$. Moreover, either $W=W^{\prime}$, or $V(W^{\prime})\setminus V(W)=\left\{x^{*}\right\}$ where $x^{*}$ is a non-offensive vertex for $W$, and $V(W)\setminus V(W^{\prime})\subseteq V(S_{1}^{*})\cup V(S_{2}^{*})\cup\left\{x_{1}\right\}$.

Proof.

We may assume that $X\neq\emptyset$, for otherwise $W$ is proper in $G$. For $x\in X$, let $P(x)$ denote the longest path in $G|(V(S_{1})\cup V(S_{2}))$ starting and ending in a neighbor of $x$. Let $x^{*}\in X$ be a vertex with $|V(P(x^{*}))|$ maximum among vertices in $X$, and let $Y$ denote interior of $P(x^{*})$. Let $C^{\prime}=G|((V(C)\cup\left\{x^{*}\right\})\setminus Y)$. It follows that $W^{\prime}=(C^{\prime},v)$ is a wheel. Moreover, $N(v)\cap V(C^{\prime})=((N(v)\cap V(C))\setminus\left\{x_{1}\right\})\cup\left\{x^{*}\right\}$, and therefore $W^{\prime}$ has the same number of spokes as $W$.

If $W^{\prime}$ is proper, the result follows. Therefore, we may assume that there is a vertex $y\in V(G)\setminus V(W^{\prime})$ that is not proper for $W^{\prime}$. Let $S_{1}^{\prime},S_{2}^{\prime}$ denote the sectors of $W^{\prime}$ containing $S_{1}\setminus Y$ and $S_{2}\setminus Y$, respectively.

Suppose first that $y\in V(W)$, and consequently $y\in Y$. Since $x^{*}$ has at least two neighbors in each of $S_{1}$ and $S_{2}$ by Lemma 10, it follows that $|V(P(x^{*}))|\geq 4$. Consequently, either $N(y)\cap V(C^{\prime})\subseteq S_{1}^{\prime}$ or $N(y)\cap V(C^{\prime})\subseteq S_{2}^{\prime}$. Moreover, $N(y)\cap V(C^{\prime})\subseteq N[x]$, and therefore $|N(y)\cap V(C^{\prime})|\leq 1$. This implies that $y$ is proper for $W$, a contradiction. This proves that $y\not\in V(W)$.

Next, we suppose that $y\in X$. It follows that $N(y)\cap V(C^{\prime})\subseteq V(S_{1}^{\prime})\cup V(S_{2}^{\prime})$. Since $y$ is not proper for $W^{\prime}$, but $y$ is adjacent to $v$, it follows that $N(y)\cap V(S_{1}^{\prime}),N(y)\cap V(S_{2}^{\prime})\neq\emptyset$. By Lemma 10, $y$ has at least two neighbors in $S_{1}^{\prime}$. But then $V(P(x^{*}))\subsetneq V(P(y))$, a contradiction to the choice of $x^{*}$. This proves that $y\in V(G)\setminus(X\cup V(W))$.

If $y\not\in N(x^{*})$, then $N(y)\cap V(C^{\prime})\subseteq N(y)\cap V(C)$, and since $y\not\in X$, it follows that $y$ is proper for $W$ and thus $y$ is proper for $W^{\prime}$. Consequently $y\in N(x^{*})$. Since $x^{*}$ is non-offensive for $W$, it follows that $N(y)\cap V(C)\subseteq V(S_{1})\cup V(S_{2})$. Since $y$ is proper for $W$, we may assume by symmetry that $N(y)\cap V(C)\subseteq V(S_{1})$. It follows that $N(y)\cap V(C^{\prime})\subseteq V(S_{1}^{\prime})$. Since $y$ is adjacent to $x^{*}$ and $G$ is triangle-free, it follows that $y$ is non-adjacent to $v$, and hence $|N(y)\cap V(C)|\leq 2$. This implies that $|N(y)\cap V(C^{\prime})|\leq 3$, and by Lemma 8, it is impossible for $y$ to have exactly three neighbors in $C^{\prime}$ since $G$ is $\textnormal{ISK}_{4}$-free. Therefore, $|N(y)\cap V(C^{\prime})|\leq 2$, and therefore $y$ is proper for $W^{\prime}$. This is a contradiction, and it follows that $W^{\prime}$ is proper in $G$, and hence $W^{\prime}$ is the desired wheel. ∎

Lemma 12**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph, and let $W=(C,v)$ be a 2-almost proper wheel in $G$. Then there exists a proper wheel in $G$ with center $v$ and at most the same number of spokes as $W$.

Proof.

Let $x_{1},x_{2}$ and $X$ be as in the definition of a 2-almost proper wheel, and let $S_{1},S_{2}$ be the sectors of $W$ containing $x_{1}$. Let $X_{1}$ denote the set $x_{1}$-non-offensive vertices in $X$, and let $X_{2}=X\setminus X_{1}$. We may assume that $X_{1},X_{2}$ are both non-empty, for otherwise the result follows from Lemma 11. It follows that $W$ is 1-almost proper, but not proper, in $G\setminus X_{2}$. Let $W^{\prime}$, $x^{*}$ be as in Lemma 11. So $W^{\prime}$ is a proper wheel in $G\setminus X_{2}$. If $W^{\prime}$ is 1-almost proper in $G$, then the result of the Lemma follows from Lemma 11. So we may assume that $W^{\prime}$ is not 1-almost proper in $G$. Since every vertex of $V(G)\setminus X_{2}$ is proper for $W^{\prime}$, we deduce that some vertex of $x\in X_{2}$ is not proper and not $x_{2}$-non-offensive for $W^{\prime}$.

By the definition of $X_{1}$, and since $W$ is 2-almost proper in $G$, it follows that $N(x)\cap V(C)$ is contained in the sectors $S_{3},S_{4}$ of $W$ containing $x_{2}$. Since $x_{1}$ and $x_{2}$ are not consecutive, $S_{3},S_{4}\not\in\left\{S_{1},S_{2}\right\}$, and so by Lemma 11, $V(S_{3})\cup V(S_{4})\subseteq V(W^{\prime})$. Consequently, $S_{3}$ and $S_{4}$ are sectors of $W^{\prime}$. Since $G$ is triangle-free and every vertex in $X$ is adjacent to $v$, it follows that $x$ is not adjacent to $x^{*}$. Therefore, $N(x)\cap V(C^{\prime})\subseteq V(S_{3})\cup V(S_{4})$, $x$ has both a neighbor in $S_{3}$ and a neighbor in $S_{4}$, and $S_{3},S_{4}$ are the sectors of $W^{\prime}$ containing $x_{2}$. Let $s_{3}$ denote the neighbor of $x$ in $S_{3}$ furthest from $x_{2}$, and let $s_{4}$ denote the neighbors of $x$ in $S_{4}$ furthest from $x_{2}$. We may assume that among all vertices of $X_{2}$ that are not $x_{2}$-non-offensive for $W^{\prime}$, $x$ is chosen so that the path of $C$ from $s_{3}$ to $s_{4}$ containing $x_{2}$ is maximal.

Since $x$ is not $x_{2}$-non-offensive for $W^{\prime}$, there exists a vertex $u\in N(x)\setminus V(W^{\prime})$ with a neighbor in $V(C^{\prime})\setminus(V(S_{3})\cup V(S_{4}))$. Since $x$ is $x_{2}$-non-offensive for $W$, it follows that $u$ has a neighbor in $V(C^{\prime})\setminus V(C)=\left\{x^{*}\right\}$, and so $u$ is adjacent to $x$ and $x^{*}$.

Since $x$ and $x^{*}$ are non-offensive for $W$, it follows that $N(u)\cap V(C)\subseteq(V(S_{1})\cup V(S_{2}))\cap(V(S_{3})\cup V(S_{4}))$. Since $G$ is triangle-free, $u$ is non-adjacent to $v$, and therefore $u\not\in X$. Consequently, $u$ is proper for $W$, and all the neighbors of $u$ in $C$ belong to one sector of $W$. It follows that $u$ has at most one neighbor in $V(C)$. Suppose that $u$ has exactly one neighbor in $V(C)$. Then $u$ has three neighbors in the cycle arising from $C^{\prime}$ by replacing $s_{3}-S_{3}-x_{2}-S_{4}-s_{4}$ by $s_{3}-x-s_{4}$, contrary to Lemma 8. It follows that $u$ has no neighbors in $V(C)$.

Let $P_{1}^{\prime}$ denote the path of $C^{\prime}$ from $s_{3}$ to $x^{*}$ not containing $x_{2}$, and let $P_{1}$ be $x-s_{3}-P_{1}^{\prime}-x^{*}$. Let $P_{2}^{\prime}$ denote the path of $C^{\prime}$ from $s_{4}$ to $x^{*}$ not containing $x_{2}$, and let $P_{2}$ be $x-s_{4}-P_{1}^{\prime}-x^{*}$. Let $D=G|(V(P_{1})\cup\left\{u\right\})$. Since $x_{1}$ and $x_{2}$ are not consecutive, each of $P_{1}^{*},P_{2}^{*}$ contains at least one neighbor of $v$, and so $W^{\prime\prime}=(D,v)$ is a wheel with fewer spokes than $W$. Let $S_{3}^{\prime}$ denote the sector of $W^{\prime\prime}$ containing $x$ but not containing $u$. If $W^{\prime\prime}$ is proper in $G$, then the result follows. Therefore, we may assume that there is a vertex $y\in V(G)\setminus V(W^{\prime\prime})$ that is not proper for $W^{\prime\prime}$.

Since every vertex in $V(W^{\prime})\setminus V(W^{\prime\prime})$ and every vertex in $V(W)\setminus V(W^{\prime\prime})$ has at most one neighbor in $V(W^{\prime\prime})$, it follows that $y\not\in V(W)\cup V(W^{\prime})$. Suppose that $y\not\in N(u)$. If $y\in X_{2}$, then $N(y)\cap V(D)\subseteq(V(S_{3})\cup\left\{x\right\})\cap V(D)$, and so $y$ is proper for $W^{\prime\prime}$, since $y$ is adjacent to $v$, a contradiction. Thus $y\not\in X_{2}$, and so $y$ is proper for $W^{\prime}$. If $y\not\in N(x)$, then $N(y)\cap V(D)\subseteq N(y)\cap V(C^{\prime})$, and again $y$ is proper for $W^{\prime\prime}$, a contradiction. Thus $y\in N(x)$, but since $x$ is non-offensive for $W$, $N(y)\cap V(C)\subseteq V(S_{3})\cup V(S_{4})$, and so $N(y)\cap V(D)\subseteq V(S_{3}^{\prime})$. Since $y$ is not proper for $W^{\prime\prime}$, $y$ is non-adjacent to $v$ and has at least three neighbors in $S_{3}^{\prime}$. But $y$ is proper for $W^{\prime}$, and so $y$ has at most two neighbors in $S_{3}$; thus $y$ has exactly three neighbors in $S_{3}^{\prime}$ and hence in $D$ contrary to Lemma 8. This contradiction implies that $y\in N(u)$.

Since $y$ is not proper for $W^{\prime\prime}$, it follows that $y$ has a neighbor in $P_{1}^{*}$, and since $G$ is triangle-free, it follows that $y$ is non-adjacent to $x,x^{*}$. We claim that $y$ has no neighbor in $P_{2}$. Suppose that it does. If $y\in X_{2}$, then, since $y$ is adjacent to $u$ and has a neighbor in $P_{1}^{*}$, we deduce that $y$ is not $x_{2}$-non-offensive for $W^{\prime}$, and the claim follows from the maximality of the path of $C$ from $s_{3}$ to $s_{4}$ containing $x_{2}$. Thus we may assume that $y\not\in X_{2}$. Consequently, $y$ is proper for $W^{\prime}$, a contradiction. This proves the claim.

Let $z_{1}$ be the neighbor of $y$ in $V(P_{1})$ closest to $x$ along $P_{1}$, and let $z_{2}$ be the neighbor of $y$ in $V(P_{1})$ closest to $x^{*}$ along $P_{1}$. Let $D^{\prime}$ be the hole $x^{*}-P_{2}-x-u-x^{*}$. If $z_{1}\neq z_{2}$, we can link $y$ to $D^{\prime}$ via the paths $y-z_{1}-P_{1}-x$, $y-z_{2}-P_{1}-x^{*}$ and $y-u$, and if $z_{1}=z_{2}$, then we can link $z_{1}$ to $D^{\prime}$ via the paths $z_{1}-P_{1}-x$, $z_{1}-P_{1}-x^{*}$ and and $z_{1}-y-u$, in both cases contrary to Lemma 8. This proves Lemma 12. ∎

Throughout the remainder of this section $G$ is an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, and $W=(C,x)$ is a proper wheel in $G$ with minimum number of spokes (subject to having center $x$).

Lemma 13**.**

Let $P=p_{1}-\ldots-p_{k}$ be a path such that $p_{1},p_{k}$ have neighbors in $V(C)$, $V(P)\subseteq V(G)\setminus V(W)$, and there are no edges between $P^{*}$ and $V(C)$. Assume that no sector of $W$ contains $(N(p_{1})\cup N(p_{k}))\cap V(C)$. For $i\in\{1,k\}$, if $x$ is non-adjacent to $p_{i}$, then $p_{i}$ has a unique neighbor in $C$.

Proof.

Let $S_{1},S_{2}$ be distinct sectors of $W$ such that $N(p_{1})\cap V(C)\subseteq V(S_{1})$, and $N(p_{k})\cap V(C)\subseteq V(S_{2})$. We may assume that $p_{1}$ is non-adjacent to $x$, and so $p_{1}$ has at most two neighbors in $C$. Since $p_{1}$ cannot be linked to the hole $C$ (or the hole obtained from $C$ by rerouting $S_{2}$ through $p_{k}$) via two one-edge paths and $P$, it follows that $p_{1}$ has a unique neighbor in $C$. ∎

Theorem 14**.**

Let $P=p_{1}-\ldots-p_{k}$ be a path with $V(P)\subseteq V(G)\setminus V(W)$ such that $x$ has at most one neighbor in $P$.

If $P$ contains no neighbor of $x$, then there is a sector $S$ of $W$ such that every edge from $P$ to $C$ has an end in $V(S)$. 2. 2.

If $P$ contains exactly one neighbor of $x$, then there are two sectors $S_{1},S_{2}$ of $W$ such that $V(S_{1})\cap V(S_{2})\neq\emptyset$, and every edge from $P$ to $C$ has an end in $V(S_{1})\cup V(S_{2})$ (where possibly $S_{1}=S_{2}$).

Proof.

Let $P$ be a path violating the assertions of the theorem and assume that $P$ is chosen with $k$ minimum. Since $W$ is proper, it follows that $k>1$. Our first goal is to show that $x$ has a neighbor in $V(P)$.

Suppose that $x$ is anticomplete to $V(P)$. Then, by the minimality of $k$, there exist two sectors $S_{1},S_{2}$ of $W$ such that every edge from $\{p_{1},\dots,p_{k-1}\}$ to $V(C)$ has an end in $V(S_{1})$, and every edge from $\{p_{2},\dots,p_{k}\}$ to $V(C)$ has an end in $V(S_{2})$. It follows that $S_{1}\neq S_{2}$. Then $p_{1}$ has a neighbor in $V(S_{1})\setminus V(S_{2})$, and $p_{k}$ has a neighbor in $V(S_{2})\setminus V(S_{1})$, and every edge from $P^{*}$ to $V(C)$ has an end in $V(S_{1})\cap V(S_{2})$. For $i=1,2$ let $a_{i},b_{i}$ be the ends of $S_{i}$. We may assume that $a_{1},b_{1},a_{2},b_{2}$ appear in $C$ in this order and that $a_{1}\neq b_{2}$. Let $Q_{1}$ be the path of $C$ from $b_{2}$ to $a_{1}$ not using $b_{1}$, and let $Q_{2}$ be the path of $C$ from $b_{1}$ to $a_{2}$ not using $a_{1}$. We can choose $S_{1}$, $S_{2}$ with $|V(Q_{2})|$ minimum (without changing $P$). Let $s$ be the neighbor of $p_{1}$ in $S_{1}$ closest to $a_{1}$, $t$ the neighbor of $p_{1}$ in $S_{1}$ closest to $b_{1}$, $y$ the neighbor of $p_{k}$ in $S_{2}$ closest to $a_{2}$ and $z$ the neighbor of $p_{k}$ in $S_{2}$ closest to $b_{2}$. Then $s\neq b_{1}$ and $z\neq a_{2}$. It follows that $V(Q_{2})\cap\{s,z\}=\emptyset$. Moreover, if $V(S_{1})\cap V(S_{2})\neq\emptyset$, then $b_{1}=a_{2}$ and $V(Q_{2})=\{b_{1}\}$, and in all cases $V(Q_{1})$ is anticomplete to $P^{*}$.

Now $D_{1}=s-p_{1}-P-p_{k}-z-S_{2}-b_{2}-Q_{1}-a_{1}-S_{1}-s$ is a hole.

[TABLE]

Since $V(Q_{2})\cap V(D_{1})=\emptyset$ and $V(Q_{2})$ contains a neighbor of $x$, it follows that $x$ has fewer neighbors in $D_{1}$ than it does in $C$. It now suffices to show that $x$ has at least three neighbors in $Q_{1}$. Since $a_{1},b_{2}\in V(Q_{1})$, we may assume that $x$ has no neighbor in $Q_{1}^{*}$, and $Q_{1}$ is a sector of $W$. Since not every edge between $V(P)$ and $V(C)$ has an end in $V(Q_{1})$, it follows that $t\neq a_{1}$ or $y\neq b_{2}$. By symmetry, we may assume that $t\neq a_{1}$. Since $x$ cannot be linked to $W$ by Lemma 8, it follows that $x$ has at least four neighbors in $V(C)$, and therefore $V(S_{1})\cap V(S_{2})=\emptyset$. Consequently, $P^{*}$ is anticomplete to $V(C)$. It follows from Lemma 13 that $s=t$. Now we can link $s$ to the hole $a_{1}-Q_{1}-b_{2}-x-a_{1}$ via the paths $s-S_{1}-a_{1}$, $s-p_{1}-P-p_{k}-z-S_{2}-b_{2}$ and $s-S_{1}-b_{1}-x$, contrary to Lemma 8. This proves that $W_{1}=(D_{1},x)$ is a wheel with fewer spokes than $W$. This proves (1).

By the choice of $W$, it follows from (1) that $W_{1}$ is not proper. Let $S_{0}$ be the sector $a_{1}-S_{1}-s-p_{1}-P-p_{k}-z-S_{2}-b_{2}$ of $(D_{1},x)$. Since $W$ is a proper wheel and $W_{1}$ is not a proper wheel, we have that:

[TABLE]

Let $v$ be as in (2). First we show that $v\not\in V(C)$. The only vertices of $C$ that may have more than one neighbor in $D_{1}$ are $b_{1}$ and $a_{2}$, and that only happens if $b_{1}=a_{2}$. But $N(b_{1})\cap V(D_{1})\subseteq V(S_{0})$ and $b_{1}$ is adjacent to $x$, so $b_{1}$ does not satisfy the conditions described in the bullets. Thus $v\not\in V(C)$.

[TABLE]

If the first case of (2) holds, then the statement of (3) follows immediately from the minimality of $k$ (since $v$ is non-adjacent to $x$), and so we may assume that the second case of (2) holds. Observe that no vertex of $V(Q_{1})$ is contained both in a sector with end $a_{1}$ and in a sector with end $b_{2}$, and therefore we may assume that $v$ has a neighbor in a sector that does not have end $b_{2}$. If $v$ is non-adjacent to $x$, we get a contradiction to the minimality of $k$. So we may assume that $v$ is adjacent to $x$, and therefore $v$ has a neighbor in $S_{3}^{*}$, and $b_{2}\not\in V(S_{3})$. Let $S_{4}$ be the sector of $D_{1}$ such that $b_{2}\in V(S_{4})$ and $V(S_{4})\subseteq V(Q_{1})$. Suppose that $y\neq b_{2}$ or $V(S_{3})\cap V(S_{4})=\emptyset$. Let $i\in\{1,\dots,k\}$ be maximum such that $v$ is adjacent to $p_{i}$. Now the path $v-p_{i}-P-p_{k}$ violates the assertions of the theorem, and so it follows from the minimality of $k$ that $N(v)\cap V(P)\subseteq\{p_{1},p_{2}\}$. Therefore, since $G$ is triangle-free, it follows that $v$ has a unique neighbor in $P$, and (3) holds. So we may assume that $y=b_{2}$ and there exists $a_{3}\in V(C)$ such that $V(S_{4})\cap V(S_{3})=\{a_{3}\}$. Let $R$ be the path from $v$ to $a_{3}$ with $R^{*}\subseteq S_{3}^{*}$. Now we can link $v$ to $x-S_{4}-x$ via the paths $v-x$, $v-R-a_{3}$ and $v-p_{i}-P-p_{k}-b_{2}$, where $i$ is maximum such that $v$ is adjacent to $p_{i}$, contrary to Lemma 8. This proves (3).

In view of (3) let $N(v)\cap V(P)=\{p_{j}\}$. In the case of the first bullet of (2), since $v$ cannot be linked to the hole $x-S_{0}-x$ by Lemma 8, it follows that $v$ has at least four neighbors in $S_{0}$, and therefore at least three neighbors in $V(S_{1})\cup V(S_{2})$, contrary to the fact that $W$ is proper. So the case of the second bullet of (2) holds. Since $W$ is proper $N(v)\cap(V(S_{0})\setminus V(S_{3}))\subseteq V(P)$, and $N(v)\cap V(D_{1})\subseteq V(S_{0})\cup V(S_{3})$.

[TABLE]

Suppose not. By Lemma 13, $s=t$ and $y=z$. We claim that in this case $b_{1}\neq a_{2}$, for if $b_{1}=a_{2}$, then $b_{1}$ can be linked to the hole $x-a_{1}-S_{1}-s-p_{1}-P-p_{k}-z-S_{2}-b_{2}-x$ via the paths $b_{1}-x$, $b_{1}-S_{1}-s$ and $b_{1}-S_{2}-z$, contrary to Lemma 8. If $v$ has a unique neighbor $r$ in $C$, then $p_{j}$ can be linked to $C$ via the paths $p_{j}-P-p_{1}-s$, $p_{j}-P-p_{k}-z$ and $p_{j}-v-r$, contrary to Lemma 8, so $v$ has at least two neighbors in $C$. Recall that $N(v)\cap V(C)\subseteq V(S_{3})$. Let $D$ be the hole obtained from $C$ by rerouting $S_{3}$ through $v$. Then $s,z\in V(D)$, and $p_{j}$ can be linked to $D$ via the paths $p_{j}-P-p_{1}-s$, $p_{j}-P-p_{k}-z$ and $p_{j}-v$, contrary to Lemma 8. This proves (4).

If follows from (4) that $b_{1}=a_{2}$ and $b_{1}$ has neighbors in $P^{*}$. Now, by considering the path from a neighbor of $b_{1}$ in $P^{*}$ to $v$ with interior in $P^{*}$ if $v$ has a neighbor in $P^{*}$, and the paths $v-p_{1}$ or $v-p_{k}$ if $v$ has no neighbor in $P^{*}$, the minimality of $k$ implies that $v$ is adjacent to $x$ and one of $a_{1},b_{2}$ belongs to $S_{3}$.

By symmetry we may assume $a_{1}\in V(S_{3})$. Let $R$ be the path from $v$ to $a_{1}$ with $R^{*}\subseteq V(S_{3})$. Now $x$ can be linked to the hole $v-R-a_{1}-S_{1}-s-p_{1}-P-p_{j}-v$ via the paths $x-v$, $x-a_{1}$ and $x-b_{2}-S_{2}-z-p_{k}-P-p_{j}$, contrary to Lemma 8.

In summary, we have now proved:

[TABLE]

By (5), $x$ has a neighbor in $V(P)$, say $x$ is adjacent to $p_{i}$. Then $p_{i}$ is the unique neighbor of $x$ in $V(P)$. By the minimality of $k$, there exist two distinct sectors $S_{1},S_{2}$ of $W$ such that $p_{1}$ has a neighbor in $V(S_{1})\setminus V(S_{2})$, and $p_{k}$ has a neighbor in $V(S_{2})\setminus V(S_{1})$. By (5), if $1<i<j$, then every edge from $\{p_{1},\dots,p_{i-1}\}$ to $V(C)$ has an end in $V(S_{1})$, and every edge from $\{p_{i+1},\dots,p_{k}\}$ to $V(C)$ has an end in $V(S_{2})$; if $i=1$ then every edge from $V(P)\setminus\{p_{1}\}$ to $V(C)$ has and end in $V(S_{2})$; and if $i=k$ then every edge from $V(P)\setminus\{p_{k}\}$ to $V(C)$ has and end in $V(S_{1})$.

For $j=1,2$, let $a_{j},b_{j}$ be the ends of $S_{j}$.

[TABLE]

Suppose (6) is false. It follows that there are edges between $P^{*}$ and $V(C)$. Since $G$ is triangle-free, $p_{i}$ is anticomplete to $N(x)\cap V(C)$. Suppose that there is sector $S_{3}$ of $W$ and an edge from $S_{3}^{*}$ to $P^{*}$. By the minimality of $k$ we deduce that $S_{3}\not\in\{S_{1},S_{2}\}$, $1<i<k$ and $p_{i}$ has a neighbor in $S_{3}^{*}$. Again by the minimality of $k$ it follows that there exist sectors $S_{1}^{\prime}$, $S_{2}^{\prime}$ such that $V(S_{j}^{\prime})\cap V(S_{3})\neq\emptyset$ for $j=1,2$ and every edge from $\{p_{1},\dots,p_{i-1}\}$ to $C$ has an end in $S_{1}^{\prime}$, and every edge from $\{p_{i+1},\dots,p_{k}\}$ to $C$ has an end in $S_{2}^{\prime}$. Now we can choose $S_{1}=S_{1}^{\prime}$ and $S_{2}=S_{2}^{\prime}$. We may assume that $a_{1},b_{1},a_{2},b_{2}$ appear in $C$ in this order, and so $b_{1}$ and $a_{2}$ are the ends of $S_{3}$. Since $p_{i}$ has a neighbor in $S_{3}^{*}$, the minimality of $k$ implies that $\{p_{2},\dots,p_{i}\}$ is anticomplete to $V(S_{1})\setminus\{b_{1}\}$, and $\{p_{i},\dots,p_{k-1}\}$ is anticomplete to $V(S_{2})\setminus\{a_{2}\}$, and the second bullet is satisfied. So $P^{*}$ is anticomplete to $V(C)\setminus N(x)$. Since there are edges between $P^{*}$ and $V(C)$, and since $p_{i}$ is anticomplete to $N(x)\cap V(C)$, by symmetry we may assume that there is an edge between $\{p_{2},\dots,p_{i-1}\}$ and $t\in N(x)\cap V(C)$. Then $t\in V(S_{1})$. Let $S_{3}$ be the other sector of $W$ incident with $t$. By the minimality of $k$ it follows that $S_{2}$ can be chosen so that $V(S_{3})\cap V(S_{2})\neq\emptyset$, and again the case of the second bullet holds. This proves (6).

If the second bullet of (6) holds, let $Q_{1}$ be the path of $C$ from $b_{2}$ to $a_{1}$ not using $b_{1}$, and let $Q_{2}=S_{3}$. To define $Q_{1}$ and $Q_{2}$, let us now assume that the case of the first bullet holds. We may assume that $a_{1},b_{1},a_{2},b_{2}$ appear in $C$ in this order. Also, $a_{1},b_{1},a_{2},b_{2}$ are all distinct, since $P$ violates the assertion of the theorem. Let $Q_{1}$ be the path of $C$ from $b_{2}$ to $a_{1}$ not using $b_{1}$, and let $Q_{2}$ be the path of $C$ from $b_{1}$ to $a_{2}$ not using $a_{1}$. We may assume that $S_{1}$, $S_{2}$ are chosen with $|V(Q_{2})|$ minimum (without changing $P$).

Since $W$ is proper, it follows that $N(p_{1})\cap V(C)\subseteq V(S_{1})$ and $N(p_{k})\cap V(C)\subseteq V(S_{2})$. Let $s$ be the neighbor of $p_{1}$ in $S_{1}$ closest to $a_{1}$, $t$ the neighbor of $p_{1}$ in $S_{1}$ closest to $b_{1}$, $y$ the neighbor of $p_{k}$ in $S_{2}$ closest to $a_{2}$ and $z$ the neighbor of $p_{k}$ in $S_{2}$ closest to $b_{2}$. Then $s\neq b_{1}$ and $z\neq a_{2}$.

Let $D_{1}$ be the hole $a_{1}-S_{1}-s-p_{1}-P-p_{k}-z-S_{2}-b_{2}-Q_{1}-a_{1}$. Then $W_{1}=(D_{1},x)$ is a wheel with fewer spokes than $W$. We may assume that (subject to the minimality of $k$) $P$ was chosen so that $V(Q_{1})$ is (inclusion-wise) minimal. By Lemma 8, $x$ has a neighbor in $V(D_{1})\setminus\{a_{1},b_{1},p_{i}\}$, and so $x$ has a neighbor in $Q_{1}^{*}$.

Let $S_{0}$ be the sector $a_{1}-S_{1}-s-p_{1}-P-p_{i}$, and let $T_{0}$ be the sector $p_{i}-P-p_{k}-z-b_{2}$ of $(D_{1},x)$.

[TABLE]

Suppose (7) is false, and let $v\in V(G)\setminus V(W_{1})$ be such that $v$ has a neighbor in $V(S_{0})\setminus V(T_{0})$ and a neighbor in $V(T_{0})\setminus V(S_{0})$.

First we claim that $v$ is adjacent to $x$. Suppose $v$ has a neighbor in $V(a_{1}-S_{1}-s)$. Since $W$ is proper and $a_{1},s\not\in V(S_{2})$ (because $P$ violates the statement of the theorem), it follows that $v$ has no neighbor in $V(z-S_{2}-b_{2})$. Consequently $v$ has a neighbor in $V(T_{0})\setminus(V(S_{2})\cup V(S_{0}))$. Let $j$ be maximum such that $v$ is adjacent to $p_{j}$, then $j>i$. Now applying (5) to the path $v-p_{j}-P-p_{k}$ we deduce that $v$ is adjacent to $x$, as required. Thus we may assume that $N(v)\cap(V(S_{0})\cup V(T_{0}))\subseteq V(P)$. Let $j$ be minimum and $l$ maximum such that $v$ is adjacent to $p_{j}$, $p_{l}$. Then $j<i$ and $l>i$. Applying (5) to the path $p_{1}-P-p_{j}-v-p_{l}-P-p_{k}$, we again deduce that $x$ is adjacent to $v$. This proves the claim.

In view of the claim, Lemma 10 implies that $v$ has at least two neighbors in $V(T_{0})\setminus V(S_{0})$ and at least two neighbors in $V(S_{0})\setminus V(T_{0})$. But now, rerouting $P$ through $v$ (as in the previous paragraph), we get a contradiction to the minimality of $k$. This proves (7).

[TABLE]

Let $v$ be a non-offensive vertex for $W_{1}$. Since $W$ is proper, it follows that $N(v)\cap V(C)$ is included in a unique sector of $W$. Consequently, $v$ is either $a_{1}$-non-offensive, or $b_{2}$-non-offensive, or $p_{i}$ non-offensive. However, (7) implies that $v$ is not $p_{i}$-non-offensive, and (8) follows.

Let $X$ be the set of all non-offensive vertices for $W_{1}$. It follows from Lemma 12 that $W_{1}$ is not proper in $V(G)\setminus X$.

[TABLE]

We may assume that the first three bullets of (9) do not hold. Since $W$ is proper and $W_{1}$ is not, (possibly switching the roles of $S_{0}$ and $T_{0}$) there exists $v\in V(G)\setminus V(W_{1})$ and a sector $S_{4}$ of $W$ with $V(S_{4})\subseteq V(Q_{1})$, such that $v$ has a neighbor in $V(S_{4})\setminus V(S_{0})$, $v$ has a neighbor in $V(S_{0})\setminus V(S_{4})$, and $N(v)\cap V(C)\subseteq V(S_{4})$. But now (7) implies that the last bullet of (9) holds. This proves (9).

Let $v\in V(G)$ be as in (9). Next we show that:

[TABLE]

Suppose that $v$ has at least two neighbors in $P$. In the first two cases of (9) we get a contradiction to the minimality of $k$. The third case is impossible by (7). Thus we may assume that the case of the fourth bullet of (9) holds. We may assume that $N(v)\cap V(P)\subseteq V(S_{0})$, and in particular $v$ has a neighbor in $\{p_{1},\dots,p_{i-1}\}$. Suppose first that $v$ is non-adjacent to $x$. Since $v$ has a neighbor in $V(S_{4})\setminus V(S_{0})$, the minimality of $k$ implies that $t=a_{1}$ and $a_{1}\in V(S_{4})$, and also that $b_{2}\in V(S_{4})$, contrary to the fact that $x$ has a neighbor in $Q_{1}^{*}$. So $v$ is adjacent to $x$, and therefore $v$ has a neighbor in $S_{0}^{*}$.

Since $W$ is proper, $N(v)\cap(V(S_{0})\setminus V(S_{4}))\subseteq V(P)$. Let $Q$ be the path from $v$ to $p_{1}$ with $Q^{*}\subseteq V(P)$. Suppose first that $a_{1}\not\in V(S_{4})$. Let $S_{5}$ be the sector of $W$ with end $a_{1}$ and such that $V(S_{5})\subseteq V(Q_{1})$, and let $b_{3}$ be the other end of $S_{5}$. Since $Q$ is shorter than $P$, it follows from the minimality of $k$ that $V(S_{4})\cap V(S_{5})=\{b_{3}\}$ and $t=a_{1}$. Let $R$ be the path from $v$ to $b_{3}$ with $R^{*}\subseteq S_{4}^{*}$. Then $x$ has exactly three neighbors in the hole $v-R-b_{3}-S_{5}-a_{1}-p_{1}-Q-v$, contrary to Lemma 8. This proves that $a_{1}\in V(S_{4})$.

Let $b_{3}$ be the other end of $S_{4}$, let $S_{5}$ be the second sector of $W$ incident with $b_{3}$, and let $a_{3}$ be the other end of $S_{5}$. Since $v\not\in X$, it follows that $v$ has a neighbor $u\in V(G)\setminus V(W_{1})$ such that $u$ has a neighbor in $V(D_{1})\setminus(V(S_{4})\cup V(S_{0}))$. Since $G$ is triangle-free, $u$ is non-adjacent to $x$.

Suppose first that $u$ has a neighbor in $V(Q_{1})\setminus V(S_{4})$. Since $G$ is triangle-free and $v$ has at least two neighbors in $V(P)$, it follows that $i\geq 4$, and therefore $k\geq 4$. Consequently, the path $u-v$ is shorter than $P$, and so it follows from the minimality of $k$ that $N(u)\cap V(C)\subseteq V(S_{5})$. Let $R$ be the path from $v$ to $b_{3}$ with $R^{*}\subseteq S_{4}^{*}$, and let $D_{2}$ be the hole $v-R-b_{3}-x-v$. Let $p$ be the neighbor of $u$ in $V(S_{5})$ closest to $b_{3}$, and let $q$ be the neighbor of $u$ in $V(S_{5})$ closest to $a_{3}$. If $p\neq q$, we can link $u$ to $D_{2}$ via the paths $u-p-S_{5}-b_{3}$, $u-q-S_{5}-a_{3}-x$ and $u-v$, and if $p=q$ we can link $p$ to $D_{2}$ via the paths $p-u-v$, $p-S_{5}-b_{3}$ and $p-S_{5}-a_{3}-x$, in both cases contrary to Lemma 8. This proves that $u$ has no neighbor in $V(Q_{1})\setminus V(S_{4})$, and therefore $u$ has a neighbor in $V(T_{0})\setminus V(S_{0})$.

Next we define a new path $Q$. If $u$ has a neighbor in $V(T_{0})\cap V(S_{2})$, let $Q$ be the path $u-v$. If $u$ is anticomplete to $V(T_{0})\cap V(S_{2})$, let $j$ be maximum such that $u$ is adjacent to $p_{j}$; then $j>i$; let $Q$ be the path $v-u-p_{j}-P-p_{k}$. Since $i>4$, in both cases $|V(Q)|<k$ and $x$ has a unique neighbor in $V(Q)$. It follows from the minimality of $k$ that $z=y=b_{2}=a_{3}$. Since $P$ violates the theorem, it follows that $p_{1}$ has a neighbor in $V(S_{1})\setminus\{a_{1}\}$.

Let $T$ be the path from $v$ to $a_{1}$ with $T^{*}\subseteq V(S_{4})$. Suppose that $s\neq t$. Let $D_{3}$ be the hole $x-a_{1}-S_{1}-s-p_{1}-t-S_{1}-b_{1}-x$. Now $v$ can be linked to $D_{3}$ via the paths $v-x$, $v-Q-p_{1}$ (short-cutting through a neighbor of $b_{1}$ if possible) and $v-T-a_{1}$, contrary to Lemma 8. Thus $s=t$, and therefore $s\neq a_{1}$. Bow we can link $v$ to $x-S_{1}-x$ via the paths $v-x$, $v-Q-p_{1}-s$ (short-cutting through a neighbor of $b_{1}$ if possible) and $v-T-a_{1}$, contrary to Lemma 8. This proves (10).

In view of (10) let $p_{j}$ be the unique neighbor of $v$ in $V(P)$.

[TABLE]

Suppose first that the case of the first bullet of (9) happens. Then by Lemma 8 $v$ has at least four neighbors in the hole $x-S_{0}-x$, and so, in view of (10), $x$ has at least three neighbors in the path $a_{1}-S_{1}-s$, contrary to the fact that $W$ is proper. By symmetry it follows that the cases of first two bullets of (9) do not happen. Suppose that the case of the third bullet of (9) happens. Since by (10) $v$ has a unique neighbor in $V(P)$, it follows that $v$ has a neighbor in $(V(S_{0})\cup V(T_{0}))\setminus V(P)$. By symmetry we may assume that $v$ has a neighbor in $z-S_{2}-b_{2}$, and, since $W$ is proper, $v$ is anticomplete to $V(S_{0})\setminus V(P)$. Consequently, $p_{j}\in V(S_{0})\setminus V(T_{0})$, and so $j<i$. By the minimality of $k$ (applied to the path $p_{1}-P-p_{j}-v$), it follows that $j=k-1$, and therefore $i=k$. Then $\{v,p_{k}\}$ is anticomplete to $V(C)\setminus V(S_{2})$, since $W$ is proper. By (5) $v$ is adjacent to $x$. But now we get a contradiction to Lemma 10 applied to $v$ and $W_{1}$. This proves (11).

In the next claim we further restrict the structure of $P$.

[TABLE]

Suppose that (12) is false. Assume first that $j\not\in\{1,k\}$. Then $p_{j}$ is anticomplete to $V(C)$, since by assumption, there are no edges between $P^{*}$ and $C$. If $s=t$, $y=z$ and $v$ has a unique neighbor $r$ in $S_{4}$, then $r\in V(S_{4})\setminus(V(S_{1})\cup V(S_{2}))$, and $p_{j}$ can be linked to $C$ via the paths $p_{j}-P-p_{k}-z$, $p_{j}-v-r$ and $p_{j}-P-p_{1}-s$, contrary to Lemma 8. If some of $p_{1},p_{k},v$ have several neighbors in $C$, then similar linkages work for the holes obtained from $C$ by rerouting $S_{1}$ through $p_{1}$, $S_{2}$ through $p_{k}$, and $S_{4}$ through $v$, respectively. This proves that $j\in\{1,k\}$, and by symmetry may assume that $j=1$. Then $S_{4}$ cannot be chosen so that $a_{1}\not\in V(S_{4})$, for otherwise (12) holds. By the minimality of $k$ and by (5), since $S_{4}$ cannot be chosen so that $a_{1}\in V(S_{4})$, it follows that $x$ is adjacent to one of $p_{1},v$ and $k=2$. Since $G$ is triangle-free, $x$ has exactly one neighbor in $\{p_{1},v\}$. Let $R$ be the path from $v$ to $a_{1}$ with $R^{*}\subseteq V(C)\setminus\{b_{1}\}$. Let $Q_{1}^{\prime}$ be the subpath of $R$ from an end of $S_{4}$ to $a_{1}$. Then $V(Q_{1}^{\prime})\subseteq V(Q_{1})$ and $b_{2}\not\in V(Q_{1}^{\prime})$, and so the path $p_{1}-v$ contradicts the choice of $P$. This proves (12).

The goal of the next two claims is to obtain more information about $i$ and $j$.

[TABLE]

Suppose not; by symmetry we may assume that $j<i$. Suppose first that $x$ is non-adjacent to $v$. By (5) and the minimality of $k$, it follows that the first assertion of the theorem holds for the path $p_{1}-P-p_{j}-v$; therefore $a_{1}=t$ and $S_{4}$ can be chosen so that $a_{1}\in V(S_{4})$. Since $W$ is proper it follows that $v$ has at most two neighbors in $S_{4}$. If $v$ has exactly two neighbors, then, in view of (6), $v$ can be linked to $x-S_{4}-x$ via two one-edge paths and the path $v-p_{j}-P-p_{i}-x$, contrary to Lemma 8. Therefore $v$ has a unique neighbor $r$ in $S_{4}$. Now, again in view of (6), $p_{j}$ can be linked to $x-S_{4}-x$ via the paths $p_{j}-v-r$, $p_{j}-P-p_{1}-a_{1}$ and $p_{j}-P-p_{i}-x$, again contrary to Lemma 8. This proves that $v$ is adjacent to $x$, and, since $G$ is triangle-free, $v$ has a neighbor in $S_{4}^{*}$. It follows that the choice of $S_{4}$ is unique. Let $R$ be the path from $v$ to $a_{1}$ with $R^{*}\subseteq V(C)\setminus\{b_{1}\}$. Suppose $a_{1}\in V(S_{4})$. Then $R^{*}\subseteq S_{4}^{*}$. In this case, because of (6) and since $b_{1}\neq s$, $p_{j}$ can be linked to the hole $v-R-a_{1}-x-v$ via the path $p_{j}-v$, $p_{j}-P-p_{1}-s-S_{1}-a_{1}$ and $p_{j}-P-p_{i}-x$, contrary to Lemma 8. Thus $a_{1}\not\in V(S_{4})$. Let $S_{5}$ be the sector of $W$ with end $a_{1}$ such that $V(S_{5})\subseteq V(Q_{1})$. If $t=a_{1}$ and $V(S_{4})\cap V(S_{5})\neq\emptyset$, then $x$ has exactly three neighbors in the hole $v-R-a_{1}-p_{1}-P-p_{j}-v$, contrary to Lemma 8. Therefore the path $p_{1}-P-p_{j}-v$ violates the assertion of the theorem, and so the minimality of $k$ implies that $j=k-1$ and consequently $i=k$. Then by (6) $a_{2}$ is anticomplete to $V(P)\setminus\{p_{k}\}$. Since $j\neq k$ and $a_{1}\not\in V(S_{4})$ (the choice of $S_{4}$ is now unique), it follows from (12) that there are edges between $P^{*}$ and $V(C)$. Now by (6) there is a sector $S_{3}$ of $W$ with ends $a_{2},b_{1}$, and $b_{1}$ has a neighbor in $P^{*}$. Then there is a path $T$ from $b_{1}$ to $p_{k}$ with $T^{*}\subseteq P^{*}$, $b_{1}-S_{3}-a_{2}-S_{2}-y-p_{k}-T-b_{1}$ is a hole and $x$ has exactly three neighbors in it, contrary to Lemma 8 (observe that $y\neq b_{2}$ because $G$ has no triangles). This proves (13).

Since $G$ is triangle-free, (13) implies that $x$ is non-adjacent to $v$.

[TABLE]

Suppose (14) is false. By symmetry we may assume that $k-i>1$. Consequently $k>2$. Suppose that $S_{4}$ can be chosen so that $a_{1}\in V(S_{4})$. If $v$ has a unique neighbor $r$ in $V(S_{4})$, then, since $s\neq b_{1}$, $p_{i}$ can be linked to $x-S_{4}-x$ via the paths $p_{i}-v-r$, $p_{i}-x$ and $p_{i}-P-p_{1}-s-S_{1}-a_{1}$, a contradiction. Thus $v$ has at least two neighbors in $V(S_{4})$. Now, again using the fact that $s\neq b_{1}$, $p_{i}$ can be linked to the hole obtained from $x-S_{4}-x$ by rerouting $S_{4}$ through $v$ via the paths $p_{i}-v$, $p_{i}-x$ and $p_{i}-P-p_{1}-s-S_{1}-a_{1}$, again contrary to Lemma 8. Thus $S_{4}$ cannot be chosen so that $a_{1}\in V(S_{4})$. Let $S_{5}$ be the sector of $W$ with end $a_{1}$ such that $V(S_{5})\subseteq V(Q_{1})$. Since $i\leq k-2$, the minimality of $k$ applied to the path $p_{1}-P-p_{i}-v$ implies that $t=a_{1}$ and $V(S_{4})\cap V(S_{5})\neq\emptyset$. In particular, $i\neq 1$. It follows from (12) that there are edges between $P^{*}$ and $V(C)$, and by (6) there is a sector $S_{3}$ of $W$ with ends $b_{1},a_{2}$ and every edge from $p_{i}$ to $V(C)$ have an end in $S_{3}^{*}$. Together with the minimality of $k$ (using the path $p_{i}-v$), this implies that $p_{i}$ is anticomplete to $V(C)$. If $v$ has a unique neighbor $r$ in $S_{4}$ (and therefore $r\neq b_{2}$) and $p_{k}$ has a unique neighbor in $S_{2}$, then $p_{i}$ can be linked to $C$ via the paths $p_{i}-v-r$, $p_{i}-P-p_{1}-a_{1}$ (short-cutting through neighbors of $b_{1}$ if possible), and $p_{i}-P-p_{k}-z$ (short-cutting through neighbors of $a_{2}$ if possible). If $v$ has at least two neighbors in $V(S_{4})$ or or $p_{k}$ has at least two neighbors in $V(S_{2})$, then the same linkage works rerouting $S_{4}$ through $v$, and $S_{2}$ through $p_{k}$, respectively. This proves (14).

It follows from (13) and (14) that either

• •

$k=3$ and $i=j=2$, or

• •

$k=2$.

If $k=3$ and $i=j=2$, then by (12) there are edges between $P^{*}$ and $V(C)$, and so by (6) there is a sector $S_{3}$ with ends $a_{2},b_{1}$, so that $p_{2}$ has neighbors in $S_{3}^{*}$; now the path $p_{2}-v$ contradicts the minimality of $k$. Thus $k=2$, and we may assume that $i=1$, by symmetry. Since $G$ is triangle-free, it follows that $p_{1}$ is non-adjacent to $a_{1},b_{1}$. Since now $P^{*}=\emptyset$ is anticomplete to $V(C)$, it follows from (12) that we can choose $S_{4}$ with $a_{1}\in V(S_{4})$. Since $v$ is non-adjacent to $x$ and $W$ is proper, it follows that $v$ has at most two neighbors in $S_{4}$. If $v$ has exactly two neighbors in $S_{4}$, then $v$ can be linked to the hole $x-S_{4}-x$ via two one-edge paths, and the path $v-p_{1}-x$, contrary to Lemma 8. Thus $v$ has a unique neighbor $r$ in $V(S_{4})$. Now $p_{1}$ can be linked to $x-S_{4}-x$ via the paths $p_{1}-v-r$, $p_{1}-x$ and $p_{1}-s-S_{1}-a_{1}$, again contrary to Lemma 8. This completes the proof of Theorem 14. ∎

We can now prove Theorem 6 which we restate:

Theorem 15**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, and let $x$ be the center of a proper wheel in $G$. If $W=(C,x)$ is a proper wheel with a minimum number of spokes subject to having center $x$, then

every component of $V(G)\setminus N(x)$ contains the interior of at most one sector of $W$, and 2. 2.

for every $u\in N(x)$, the component $D$ of $V(G)\setminus(N(x)\setminus\{u\})$ such that $u\in V(D)$ contains the interiors of at most two sectors of $W$, and if $S_{1},S_{2}$ are sectors with $S_{i}^{*}\subseteq V(D)$ for $i=1,2$, then $V(S_{1})\cap V(S_{2})\neq\emptyset$.

Proof.

To prove the first statement, we observe that if some component of $V(G)\setminus N(x)$ contains the interiors of two sectors of $W$, then this component contains a path violating the first assertions of Theorem 14.

For the second statement, suppose $D$ contains the interiors of two disjoint sectors $S_{1},S_{2}$ of $W$. Since $|D\cap N(x)|=1$, we get a path in $D$ violating the second assertion of Theorem 14. This proves Theorem 15. ∎

3 Proper Wheel Centers

In the proof of our main theorem, we perform manipulations on $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graphs; in this section, we show that this preserves being $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free, and that no vertex becomes the center of a proper wheel.

Lemma 16**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, $s\in V(G)$, $K$ a component of $G\setminus N[s]$, and $N$ the set of vertices in $N(s)$ with a neighbor in $K$. Let $H=G|(V(K)\cup N\cup\left\{s\right\})$. Then $s$ is not the center of a proper wheel in $H$, and for $v\in V(H)\setminus\left\{s\right\}$, if $v$ is the center of a proper wheel in $H$, then $v$ is the center of a proper wheel in $G$.

Proof.

Since $H\setminus N[s]$ is connected, it follows that $s$ is not the center of a proper wheel in $H$ by Theorem 15. Let $v\in V(H)\setminus\left\{s\right\}$ be the center of a proper wheel $W=(C,v)$ in $H$. For all $w\in V(G)\setminus V(H)$, $N(w)\cap V(C)\subseteq N[s]$, and since $G$ is triangle-free, it follows that every vertex $w\in V(G)\setminus V(H)$ either has at most one neighbor in $V(C)$, or $N(w)\cap V(C)\subseteq N(s)$.

Suppose that $W$ is not proper in $G$. Then there exists a vertex $w$ such that either $w$ has more than two neighbors in a sector of $W$, but $w$ is not adjacent to $v$, or $w$ has neighbors in at least two sectors of $W$. It follows that $w$ has more than one neighbor in $V(C)$, and thus in $N(s)$. Suppose that $w$ has three distinct neighbors $a,b,c$ in $V(C)\cap N(s)$. Let $P$ be a shortest path connecting two of $a,b,c$, say $a$ and $b$, with interior in $K$; then $s$ is anticomplete to $P^{*}$. If $c$ is anticomplete to $P$, then $G|(V(P)\cup\left\{w,s,a,b,c\right\})$ is an $\textnormal{ISK}_{4}$. Otherwise, by the minimality of $|V(P)|$, $P^{*}$ consists of a single vertex $x$, and $\left\{w,s,x,a,b,c\right\}$ induces a $K_{3,3}$ subgraph in $G$, a contradiction. So $w$ has exactly two neighbors $a$ and $b$ in $V(C)$, and thus $a$ and $b$ are in different sectors of $W$. Since $a,b\in N(s)$ and $W$ is proper in $H$, it follows that $s\in V(C)$ and $s$ is a spoke of $W$; let $S,S^{\prime}$ be the two sectors of $W$ containing $s$. But then $v$ can be linked to the cycle $s-a-w-b-s$ via $v-s$ and the two paths with interiors in $S\setminus s$ and $S^{\prime}\setminus s$. This is a contradiction by Lemma 8 and it follows that $W$ is proper in $G$. This concludes the proof. ∎

We use the following well-known lemma, which we prove for completeness.

Lemma 17**.**

Let $G$ be a connected graph, $a,b,c\in V(G)$ with $d(a)=d(b)=d(c)=1$, and let $H$ be a connected induced subgraph of $G$ containing $a,b,c$ with $V(H)$ minimal subject to inclusion. Then either $H$ is a subdivision of $K_{1,3}$ with $a,b,c$ as the vertices of degree one, or $H$ contains a triangle.

Proof.

Let $G,a,b,c,H$ be as in the statement of the theorem. Let $P$ be a shortest $a-b$-path in $H$, and let $Q$ be a shortest path from $c$ to a vertex $d$ with a neighbor in $V(P)$. By the minimality of $V(H)$, it follows that $V(H)=V(P)\cup V(Q)$. Moreover, $P$ and $Q$ are induced paths and no vertex of $Q\setminus d$ has a neighbor in $V(P)$. If $d$ has exactly one neighbor in $V(P)$, then the result follows. If $d$ has two consecutive neighbors in $V(P)$, then $H$ contains a triangle. Otherwise, let $w\in V(P)$ such that $d$ has a neighbor both on the subpath of $P$ from $w$ to $a$ and on the subpath of $P$ from $w$ to $b$. It follows that $w\not\in\left\{a,b,c\right\}$, and that $H\setminus w$ is connected and contains $a,b,c$. This contradicts the minimality of $V(H)$, and the result follows. ∎

Lemma 18**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, $s$ the center of a proper wheel in $G$, $K$ a component of $G\setminus N[s]$, and $N$ the set of vertices in $N(s)$ with a neighbor in $K$. Let $G^{\prime}$ arise from $G$ by contracting $V(K)$ to a new vertex $z$. If $G|(V(K)\cup N\cup\left\{s\right\})$ is series-parallel, then $G^{\prime}$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free.

Proof.

$G^{\prime}$ does not contain a triangle, because $N_{G^{\prime}}(z)\subseteq N_{G}(s)$ is stable, and hence $z$ is not in a triangle in $G^{\prime}$. Suppose that $H$ is an induced subgraph of $G^{\prime}$ which is a $K_{3,3}$ or a subdivision of $K_{4}$. Then $z\in V(H)$. If $z$ has degree two in $H$ (and so $H$ is an $\textnormal{ISK}_{4}$), let $a,b$ denote its neighbors; we can replace $a-z-b$ by an $a-b$-path $P$ with interior in $K$ and obtain a subdivision of $H$, which is an $\textnormal{ISK}_{4}$, as an induced subgraph of $G$, a contradiction. Thus $z$ has degree three in $H$; let $a,b,c$ denote the neighbors of $z$ in $H$. Let $P$ be a shortest $a-b$-path with interior in $K$. Then $c$ has at most one neighbor on $P$, for otherwise $G|(V(P)\cup\left\{a,b,c,s\right\})$ is a wheel, contrary to the fact that $G|(V(K)\cup N\cup\left\{s\right\})$ is series-parallel and does not contain a wheel by Theorem 3. Let $Q$ be a shortest path from $c$ to $V(P)\setminus\left\{a,b\right\}$ with interior in $K$; then each of $a,b,c$ has a unique neighbor in $V(Q)\cup V(P)$ by symmetry. Let $H^{\prime}$ be a minimal connected induced subgraph of $G|(V(P)\cup V(Q))$ containing $a,b,c$. Since $G|(V(K)\cup N\cup\left\{s\right\})$ is series-parallel, it follows that $H^{\prime}$ is a subdivision of $K_{1,3}$ with $a,b,c$ as the vertices of degree one by Lemma 17. Therefore, $G|(V(H\setminus z)\cup V(H^{\prime}))$ is a subdivision of $H$, and by Theorem 4, it contains an $\textnormal{ISK}_{4}$ or a $K_{3,3}$ subgraph in $G$. This is a contradiction, and the result is proved. ∎

Lemma 19**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, $s$ the center of a proper wheel in $G$, $K$ a component of $G\setminus N[s]$, and $N$ the set of vertices in $N(s)$ with a neighbor in $K$, and let $H=G|(V(K)\cup N\cup\left\{s\right\})$ be series-parallel. Let $G^{\prime}$ arise from $G$ by contracting $V(K)$ to a new vertex $z$. Then $z$ is not the center of a proper wheel in $G^{\prime}$, and for $v\in V(G^{\prime})\setminus\left\{s,z\right\}$, if $v$ is the center of a proper wheel in $G^{\prime}$, then $v$ is the center of a proper wheel in $G$.

Proof.

Since $N_{G^{\prime}}(z)\subseteq N_{G^{\prime}}(s)$, it follows that $z$ is not the center of a proper wheel in $G^{\prime}$, for otherwise $s$ would have a neighbor in every sector of such a wheel. This proves the first statement of the lemma.

Throughout the proof, let $v\in V(G^{\prime})\setminus\left\{s,z\right\}$ be the center of a proper wheel in $G^{\prime}$, and let $W=(C,v)$ be such a wheel with a minimum number of spokes. Since $G^{\prime}$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free by Lemma 18, it follows that $W$ satisfies the hypotheses of Theorem 6. Our goal is to show that $v$ is the center of a proper wheel in $G$.

[TABLE]

Suppose that $z\in V(C)$. Let $a,b$ denote the neighbors of $z$ in $V(C)$. Let $P$ be a shortest $a-b$-path with interior in $K$. Then every vertex in $V(K)$ has at most two neighbors in $V(P)$. Let $W^{\prime}=(C^{\prime},v)$ be the wheel in $G$ that arises from $W$ by replacing the subpath $a-z-b$ of $C$ by $a-P-b$ to obtain $C^{\prime}$.

It remains to show that $W^{\prime}$ is a proper wheel in $G$. Suppose that some vertex $x\in V(G)\setminus V(K)$ has two or more neighbors in $P^{*}$. Then $x\in N\subseteq V(H)$, and $H|(\left\{a,b,x,s\right\}\cup V(P))$ is a wheel in $H$ with center $x$, a contradiction since $H$ is series-parallel Theorem 3.

Since $v\not\in V(K)$, it follows from the claim of the previous paragraph that $v$ has at most one neighbor in $P^{*}$, and no neighbor unless $v$ is adjacent to $z$, and therefore there are at most two sectors of $W^{\prime}$ intersecting $P^{*}$. We claim that if for a vertex $x$ we have $|N_{G}(x)\cap V(C^{\prime})|\geq 3$, then $x\not\in V(K)$ and $|N_{G^{\prime}}(x)\cap V(C)|\geq 3$. Suppose that $x$ is a vertex violating this claim. If $x\in K$, then $N_{G}(x)\cap V(C^{\prime})\subseteq V(P)$, and so $|N_{G}(x)\cap V(C^{\prime})|\leq 2$ by the minimality of $|V(P)|$, a contradiction; it follows that $x\not\in K$. Therefore, $|N_{G}(x)\cap P^{*}|\leq 1$, and thus $|N_{G}(x)\cap V(C^{\prime})|-|N_{G^{\prime}}(x)\cap V(C)|\leq 1$. But $|N_{G}(x)\cap V(C^{\prime})|>3$ by Lemma 8, and so $|N_{G^{\prime}}(x)\cap V(C)|\geq 3$, a contradiction. So the claim holds.

Now suppose that there is a vertex $x$ which is not proper for $W^{\prime}$. If $x$ has neighbors in at most one sector of $W^{\prime}$, then $|N_{G}(x)\cap V(C^{\prime})|\geq 3$, but we proved above that $|N_{G^{\prime}}(x)\cap V(C)|\geq 3$, and so, since $W$ is proper, $x$ is adjacent to $v$, a contradiction. It follows that $x$ has neighbors in more than one sector of $W^{\prime}$. Since $x$ is proper for $W$, it follows that $x$ has a neighbor in $P^{*}$ and thus, either $x\in V(K)$ or $x$ is adjacent to $z$. Since $x$ is proper for $W$, it follows that $N_{G}(x)\cap V(C^{\prime})$ is contained in the sectors of $W^{\prime}$ intersecting $P^{*}$. In particular, there are exactly two such sectors $S_{1}$ and $S_{2}$ of $W^{\prime}$, they are consecutive, and $v$ has a neighbor in $P^{*}$. Consequently, $v$ is adjacent to $z$ and $z$ is a spoke in $W$.

We claim that $x$ has at most two neighbors in $V(C^{\prime})$. If $x\in V(K)$ then $N_{G}(x)\cap V(C^{\prime})\subseteq V(P)$ and we have already shown that every vertex of $K$ has at most two neighbors in $P$. Thus we may assume that $x\not\in K$, and so $x$ is adjacent to $z$. Since $G^{\prime}$ is triangle-free by Lemma 18, it follows that $x$ is not adjacent to $v$. Since $x$ is proper for $W$, it follows that $x$ has at most two neighbors in $V(C)$, and hence in $V(C^{\prime})$, by our first claim. This proves our second claim. It follows that $x$ has exactly one neighbor $s_{1}$ in $S_{1}\setminus S_{2}$ and exactly one neighbor $s_{2}$ in $S_{2}\setminus S_{1}$. If $x$ is non-adjacent to $v$, then $G|(V(S_{1})\cup V(S_{2})\cup\left\{x,v\right\})$ is an $\textnormal{ISK}_{4}$ in $G$, a contradiction. Therefore, $x$ is adjacent to $v$ and can be linked to the cycle $G|(V(S_{1})\cup\left\{v\right\})$ via $x-v$, $x-s_{1}$, and a subpath of $x-s_{2}-S_{2}$. Therefore $W^{\prime}$ is a proper wheel in $G$. This proves (15).

By (15), we may assume that $z\not\in V(C)$. So $W$ is a wheel in $G$. Since $W$ is proper in $G^{\prime}$, there is a sector $S$ of $W$ containing all neighbors of $z$ in $C$. Then clearly the following holds.

[TABLE]

Next we claim the following.

[TABLE]

If $x\in G\setminus(V(C)\cup V(K))$, then $x$ is proper for $W$ is in $G$ as $x$ is proper for $W$ in $G^{\prime}$. Now consider a vertex $x\in V(K)$. Since $z$ is not adjacent to $v$, and $W$ is proper in $G^{\prime}$, it follows that $|N_{G^{\prime}}(z)\cap V(C)|\leq 2$. Then by (16), $|N_{G}(x)\cap V(C)|\leq 2$, and hence $x$ is proper for $W$ in $G$. This proves 17.

By (17), we may assume that $z$ is adjacent to $v$. Let $a$ and $b$ be the ends of $S$. We now define a sequence of wheels in $G$ with center $v$. Let $W_{1}=W$ and $S_{1}=S$. Assume that wheels $W_{1},\dots,W_{i}$ have been defined, and define $W_{i+1}$ as follows. If there is a vertex $x_{i}\in V(K)$ that is not adjacent to $v$ and has at least three neighbors in $S_{i}$, then let $S_{i+1}$ be the path from $a$ to $b$ in $G|(V(S_{i})\cup\left\{x_{i}\right\})$ that contains $x_{i}$, and (by (16)) let $W_{i+1}$ be the wheel obtained form $W_{i}$ by replacing $S_{i}$ by $S_{i+1}$. Since $S_{i+1}$ is strictly shorter than $S_{i}$, this sequence must stop at some point; say it stops with wheel $W_{t}$. For $1\leq i\leq t$, let $C_{i}$ be the rim of $W_{i}$ (so $V(C_{i})=(V(C)\setminus V(S))\cup V(S_{i})$). Then $W_{t}=(C_{t},v)$ is a wheel in $G$ such that every vertex of $K$ that has at least three neighbors in $S_{t}$ is adjacent to $v$. We will show that $W_{t}$ is a proper wheel in $G$, but first we show the following.

[TABLE]

Suppose that $y$ is proper for $W_{i}$ and not proper for $W_{i+1}$. Then $y$ is adjacent to $x_{i}$. Suppose first that $y$ is non-adjacent to $v$ and $|N_{G}(y)\cap V(C_{i+1})|\geq 3$. Since $y$ cannot have three neighbors in $C_{i+1}$ by Lemma 8, it follows that $|N_{G}(y)\cap V(C_{i+1})|>3$. Moreover, since $N_{G}(y)\cap V(C_{i+1})\subseteq\left\{x_{i}\right\}\cup(N_{G}(y)\cap V(C_{i}))$, it follows that $|N_{G}(y)\cap V(C_{i})|\geq 3$. But then $y$ is not proper for $W_{i}$, a contradiction. It follows that $y$ has a neighbor in $C_{i+1}\setminus S_{i+1}=C\setminus S$, and thus $y\not\in V(K)$. Therefore $y\in V(G^{\prime})$, and since $y$ is adjacent to $x_{i}$ in $G$, it follows that $y$ is adjacent to $z$ in $G^{\prime}$. Since $z$ is adjacent to $v$ in $G^{\prime}$ and $G^{\prime}$ is triangle-free by Lemma 18, it follows that $y$ is non-adjacent to $v$. Note that since $i+1>1$, it follows that a vertex of $K$ has a neighbor in $S^{*}$, and therefore $z$ has a neighbor in $S^{*}$. Since $W$ satisfies the hypotheses of Theorem 6, and since $y-z$ is a path containing exactly one neighbor of $v$, it follows that the neighbors of $y$ in $C$ are in a sector $S^{\prime}$ of $W$ consecutive with $S$. Since $y$ is non-adjacent to $v$, it follows that $y$ has at most two neighbors in $S^{\prime}$. Note that since $G^{\prime}$ is triangle-free, and $N_{G^{\prime}}(z)\cap V(C)\subseteq V(S)$, it follows that $z$ has no neighbors in $S^{\prime}$. If $y$ has exactly two neighbors in $S^{\prime}$, then $y$ can be linked in $G^{\prime}$ to the hole $v-S^{\prime}-v$ via two one-edge paths and the path $y-z-v$. So $y$ has exactly one neighbor $r$ in $S^{\prime}$ that is in $V(S^{\prime})\setminus V(S)$, and now $z$ can be linked to $v-S^{\prime}-v$ via the paths $z-v$, $z-y-r$, and a path with interior in $S$, contrary to Lemma 8. This concludes the proof of (18).

Every vertex in $G\setminus V(K)$ is proper for $W$ and hence it is proper for $W_{t}$ by (18). Suppose that there is a vertex $x\in V(K)$ that is not proper for $W_{t}$. By (16), $N_{G}(x)\cap V(C_{t})\subseteq V(S_{t})$. So $x$ is non-adjacent to $v$ and has at least three neighbors in $S_{t}$, contradicting the assumption that the wheel sequence terminates with $W_{t}$. Therefore, $W_{t}$ is a proper wheel in $G$ with center $v$.

∎

4 Tools

In this section we develop tools for our main theorem for finding a vertex of degree one, or a cycle with all but a few vertices of degree two.

Lemma 20**.**

Let $G$ be a graph, $x\in V(G)$, such that $G\setminus x$ is a forest. Then either $V(G)=N[x]$ and $G\setminus x$ is stable, or $V(G)\setminus N[x]$ contains a vertex of degree at most one in $G$, or $G$ contains an induced cycle $C$ containing $x$ such that every vertex of $V(C)\setminus\left\{x\right\}$ except for possibly one has degree two in $G$.

Proof.

If every component of $G\setminus x$ contains exactly one vertex, then either $V(G)=N[x]$ or $V(G)\setminus N[x]$ contains a vertex of degree zero. Hence, we may assume that there exists a component $T$ of $G\setminus x$ with at least two vertices, and $T$ is a tree. Let $A$ be the set of vertices of degree at least three in $T$. If $A$ is non-empty, then let $T^{\prime}$ be the subtree of $T$ that contains all vertices of $A$ and minimal with respect to this property, and let $a$ be a leaf of $T^{\prime}$. There is a path $P=v-\dots-v^{\prime}$ in $T$, whose ends are distinct leaves of $T$ and $P$ contains at most one vertex of degree three in $T$ (namely $a$). This is trivial is $A$ if empty, and follows from the definition of $a$ otherwise.

If $x$ is non-adjacent to $v$, then $v$ is a vertex in $V(G)\setminus N[x]$ of degree one in $G$, so we may assume that $x$ is adjacent to $v$, and similarly for $v^{\prime}$. Now, let $v^{\prime\prime}$ be the neighbor of $x$ in $P\setminus v$ closest to $v$ along $P$. We set $C=x-v-P-v^{\prime\prime}-x$ and observe that all vertices of $C$ except possibly $x$ and $a$ have degree two in $G$. ∎

Lemma 21**.**

Let $G$ be a series-parallel graph, and let $x,y\in V(G)$ with $x=y$ or $xy\in E(G)$. If $G\setminus\left\{x,y\right\}$ contains a cycle, then there is an induced cycle $C$ in $G$ such that $V(C)\cap\left\{x,y\right\}=\emptyset$ and all but at most two vertices of $C$ have degree two in $G$ (and are thus anticomplete to $\left\{x,y\right\}$), or $V(G)\setminus(N[x]\cup N[y])$ contains a vertex of degree at most one in $G$.

Proof.

By contracting the edge $xy$ and deleting any parallel edges that may arise, we may assume that $x=y$. We may further assume that every vertex except for possibly $x$ has degree at least two, because vertices of degree one in $N(x)$ can be deleted without affecting the hypotheses or the conclusions, and if there is a vertex of degree at most one in $V(G)\setminus N[x]$, then the conclusion holds.

Let $C$ be a cycle in $G\setminus x$. Since $G$ is series-parallel and by the definition of series-parallel graphs, it follows that there do not exist three paths from $x$ to $V(C)$ that are vertex disjoint except for $x$ in $G$. By Menger’s theorem [4], it follows that there is a partition $(X,Y,Z)$ of $V(G)$ with $X$ of size at most two, and $Y,Z\neq\emptyset$ such that $Y$ is anticomplete to $Z$ in $G$, $V(C)\subseteq Y\cup X$ and $x\in Z$.

We choose a partition $(X,Y,Z)$ with $|X|$ minimal, and subject to that, $|X\cup Y|$ minimal, such that $Y$ is anticomplete to $Z$ in $G$, $Y,Z\neq\emptyset$, $x\in Z$, and $G|(Y\cup X)$ contains a cycle. It follows that $|X|\leq 2$.

Suppose first that $X=\emptyset$. If $G|Y$ is an induced cycle, the result follows. Otherwise, since $G|Y$ contains a cycle, it follows that there is a vertex $x^{\prime}$ such that $G|(Y\setminus\left\{x^{\prime}\right\})$ contains a cycle. By induction applied to $G|Y$ and the vertex $x^{\prime}$, the result follows.

Next, suppose that $X=\left\{x^{\prime}\right\}$. If $G|Y$ is a forest, then $x^{\prime}\neq x$ and thus we obtain the desired result by applying Lemma 20 to $G|(X\cup Y)$. Otherwise, we apply induction to $G|(X\cup Y)$ and $x^{\prime}$, and again, the result follows.

It follows that $X=\left\{x^{\prime},y^{\prime}\right\}$, and therefore, the component of $G|(Z\cup X)$ containing $x$ contains $x^{\prime}$ and $y^{\prime}$, for otherwise $\left\{x^{\prime}\right\}$ or $\left\{y^{\prime}\right\}$ would be a better choice of $X$ for the partition. Suppose that $G|Y$ is connected. If there is a vertex $z$ such that every $x^{\prime}-y^{\prime}$-path with interior in $G|Y$ uses $z$, then $\left\{x^{\prime},z\right\}$ or $\left\{y^{\prime},z\right\}$ yields a better choice of $X$ and partition. Therefore, by Menger’s theorem [4], there are two disjoint paths $P_{1},P_{2}$ from $x^{\prime}$ to $y^{\prime}$ with interior in $Y$, and since $G|Y$ is connected, it follows that there is path $Q$ from $P_{1}$ to $P_{2}$ in $Y$. Moreover, there is a path $R$ from $x^{\prime}$ to $y^{\prime}$ with interior in $Z$ since the component of $G|(Z\cup X)$ containing $x$ also contains $x^{\prime}$ and $y^{\prime}$; but $P_{1}\cup P_{2}\cup Q\cup R$ is a (not necessarily induced) subdivision of $K_{4}$ in $G$, contrary to the fact that $G$ is series-parallel. Thus $G|Y$ is not connected. By the minimality of $X\cup Y$, for every component $K$ of $G|Y$, the graph $G|((X\setminus\left\{x\right\})\cup V(K))$ contains no cycle. However, $G|(X\cup Y)$ contains a cycle $C$ not using $x$, and so $C$ contains vertices from more than one component of $G|Y$. It follows that $x^{\prime},y^{\prime}\in V(C)$, and thus $x\not\in\left\{x^{\prime},y^{\prime}\right\}$. Therefore, for every component $K$ of $G|Y$, the graph $G|(X\cup V(K))$ is a tree. Since $K$ is connected, it follows that $x^{\prime},y^{\prime}$ are leaves. If $G|(X\cup V(K))$ contains a leaf other than $x^{\prime},y^{\prime}$, then the result follows. So each component is a path from $x^{\prime}$ to $y^{\prime}$, and no vertex of the path except for $x^{\prime},y^{\prime}$ has further neighbors in $G$. But then the union of two of those paths (there are at least two, since $G|Y$ is not connected) yields the desired cycle; the result follows. ∎

Theorem 22**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph, $x,y\in V(G)$ with $x=y$ or $xy\in E(G)$. Then either

• •

$V(G)=N[x]\cup N[y]$;

• •

there exists a vertex in $V(G)\setminus(N[x]\cup N[y])$ of degree at most one in $G$;

• •

there exists an induced cycle $C$ containing at least one of $x,y$ such that at most one vertex $v$ in $V(C)\setminus(N[x]\cup N[y])$ has $d(v)>2$; or

• •

there exists an induced cycle $C$ containing neither $x$ nor $y$ and a vertex $z\in V(C)$ such that at most one vertex $v$ in $V(C)\setminus N[z]$ has $d(v)>2$.

Proof.

Suppose first that $G$ is series-parallel. Define $H=G$ and $v=x$ if $x=y$, and define $H$ as the graph that arises from contracting the edge $xy$ to a new vertex $v$ if $x\neq y$. Then $H$ is series-parallel. Suppose that $H\setminus v$ is a forest, and apply Lemma 20. If the first outcome of Lemma 20 holds, then $V(H)=N_{H}(v)$, and so $V(G)=N_{G}(x)\cup N_{G}(y)$. If the second outcome of Lemma 20 holds, then $V(H)\setminus N_{H}[v]$ contains a vertex of degree at most one in $H$, and so $V(G)\setminus(N_{G}[x]\cup N_{G}[y])$ contains a vertex of degree at most one in $G$. Finally, if the third outcome of Lemma 20 holds, then $H$ contains an induced cycle $C$ containing $v$ such that every vertex of $V(C)\setminus\left\{v\right\}$ except for at most one has degree two in $H$, and so there is an induced cycle $C^{\prime}$ in $G$ containing at least one of $x,y$ such that every vertex of $V(C^{\prime})\setminus\left\{x,y\right\}$ except for possibly one has degree two in $G$. This proves the result in the case that $H\setminus v$ is a forest. So $H\setminus v$ contains a cycle, and thus $G\setminus\left\{x,y\right\}$ contains a cycle. By Lemma 21, either $V(G)\setminus(N[x]\cup N[y])$ contains a vertex of degree at most one in $G$, or $G$ contains a cycle $C$ with $V(C)\cap\left\{x,y\right\}=\emptyset$ and such that all but at most two vertices in $V(C)$ have degree two in $G$. In the former case, the second outcome of this theorem holds; in the latter case, the fourth outcome of this theorem holds by choosing $z\in V(C)$ with $d_{G}(z)$ maximum among vertices in $V(C)$.

Thus we may assume that $G$ contains a proper wheel by Lemma 9; let $z$ be the center of a proper wheel (where possibly $z\in\left\{x,y\right\}$). Let $W$ be such a wheel with minimum number of spokes. Let $Z=\left\{x,y\right\}\cap N(z)$. Since $x=y$ or $xy\in E(G)$, it follows that $x$ and $y$ are in the same component of $G\setminus(N[z]\setminus Z)$. Since $N(z)$ is stable, it follows that $|Z|\leq 1$. Therefore, by Theorem 6, the component of $G\setminus(N[z]\setminus Z)$ containing $\left\{x,y\right\}\setminus\left\{z\right\}$ includes the interiors of at most two sectors of $W$. Again by Theorem 6, the interior of every other sector of $W$ is contained in a separate component of $G\setminus(N[z]\setminus Z)$. Since $W$ has at least four sectors by Lemma 8, there is a component $K$ of $G\setminus(N[z]\setminus Z)$ that does not contain $x$ and $y$, and that contains no neighbor of $x,y$. Let $N$ be the set of neighbors of $z$ with a neighbor in $K$. Then, we apply induction to $H=G|(V(K)\cup N\cup\left\{z\right\})$ and $z$. By the choice of $H$ and $z$, the first outcome does not hold. If the second outcome holds for $H$ and $z$, then it holds for $G$ and $x,y$ as well, since $(N[x]\cup N[y])\cap V(H)\subseteq N[z]\cap V(H)$. If the third or fourth outcome holds for $H$ and $z$, then the third or fourth outcome holds for $G$ and $x,y$. ∎

5 Main Result

We say that $(G,x,y)$ has the property $\mathcal{P}$ if $V(G)\setminus(N[x]\cup N[y])$ contains a vertex of degree at most two in $G$.

We can now prove Theorem 7 which we restate:

Theorem 23**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free graph which is not series-parallel, and let $(x,y)$ be a non-center pair for $G$. Then $(G,x,y)$ has the property $\mathcal{P}$.

Proof.

Suppose for a contradiction that the theorem does not hold, and let $(G,x,y)$ be a counterexample with $|V(G)|$ minimum. Then every vertex in $V(G)\setminus(N[x]\cup N[y])$ has degree at least three in $G$. Since $G$ is not series-parallel, and $G$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free, it follows from Theorem 4 that $G$ contains a wheel and hence by Lemma 9, it follows that $G$ contains a proper wheel $W=(C,s)$. Let $C_{1},\dots,C_{k}$ denote the components of $V(G)\setminus N[s]$. For $i=1,\dots,k$ let $N_{i}$ denote the set of neighbors $v$ of $s$ such that $v$ has a neighbor in $C_{i}$, and let $G_{i}$ denote the induced subgraph of $G$ with vertex set $V(C_{i})\cup N_{i}\cup\left\{s\right\}$.

[TABLE]

Let $i\in\left\{1,\dots,k\right\}$ such that $G_{i}$ is series-parallel, and let $G_{i}\setminus(N[s]\cap\left\{x,y,s\right\})$ contain a cycle. Since $G$ is triangle-free, it follows that $1\leq|N[s]\cap\left\{x,y,s\right\}|\leq 2$. By Lemma 21 applied to $G_{i}$ and the vertices in $N[s]\cap\left\{x,y,s\right\}$, it follows that either there is a vertex in $V(G_{i})\setminus N[s]$ of degree at most one in $G_{i}$ anticomplete to $\left\{x,y\right\}\cap N(s)$, or $G_{i}\setminus(N[s]\cap\left\{x,y,s\right\})$ contains a cycle $C^{\prime}$ with at least two vertices of degree two in $G_{i}$. In both cases, there is a vertex $z$ in $V(G_{i})\setminus N[s]$ of degree at most two in $G_{i}$ and $z$ is anticomplete to $N[s]\cap\left\{x,y,s\right\}$, and hence its degree in $G$ is also at most two. Since $(G,x,y)$ does not satisfy property $\mathcal{P}$, it follows that $z\in N[x]\cup N[y]$, and thus $\left\{x,y\right\}\cap V(C_{i})\neq\emptyset$. This proves (19).

[TABLE]

Let $i\in\left\{1,\dots,k\right\}$ be such that $G_{i}$ is series-parallel, and suppose that $|V(C_{i})|>1$. Let $G^{\prime}$ be the graph that arises from $G$ by contracting $V(C_{i})$ to a new vertex $z$. We let $x^{\prime}=z$ if $x\in V(C_{i})$ and $x^{\prime}=x$ otherwise; and we let $y^{\prime}=z$ if $y\in V(C_{i})$ and $y^{\prime}=y$ otherwise. By Lemma 18, $G^{\prime}$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free. By Lemma 19, $(x^{\prime},y^{\prime})$ is a non-center pair for $G^{\prime}$. By the minimality of $|V(G)|$, it follows that $(G^{\prime},x^{\prime},y^{\prime})$ has the property $\mathcal{P}$. Let $v\in V(G^{\prime})\setminus(N[x^{\prime}]\cup N[y^{\prime}])$ be a vertex of degree at most two in $G^{\prime}$. From the definition of $x^{\prime}$ and $y^{\prime}$, it follows that $v\not\in N[x]\cup N[y]$. It follows that either $v=z$, or $v\neq z$ and $d_{G}(v)>2$, and so $v\in N[z]$.

Suppose first that $v=z$. Then $z\not\in N[x^{\prime}]\cup N[y^{\prime}]$, and so $V(G_{i})\cap\left\{x,y\right\}=\emptyset$. By (19), it follows that $G_{i}\setminus s$ is a tree. Since $v$ has degree at most two in $G^{\prime}$, it follows that $|N_{i}|\leq 2$, and since $G_{i}\setminus N[s]$ is connected, it follows that every vertex of $N_{i}$ is a leaf of $G_{i}\setminus s$. Thus, either $V(C_{i})$ contains a leaf of $G_{i}\setminus s$, or $G_{i}\setminus s$ is a path with ends in $N_{i}$, and so in both cases $V(C_{i})$ contains a vertex of degree at most two in $G$. This is a contradiction since $V(G_{i})\cap\left\{x,y\right\}=\emptyset$; it follows that $v\neq z$.

It follows that $v\in N(z)$. Since $d_{G}(v)>2$, it follows that $d_{G^{\prime}}(v)<d_{G}(v)$, and thus $v$ has more than one neighbor in $V(C_{i})$. Let $P$ be a path in $C_{i}$ between two neighbors of $v$, then $v-P-v$ is a cycle in $G_{i}\setminus(N[s]\cap\left\{x,y,s\right\})$. By (19), it follows that $V(C_{i})\cap\left\{x,y\right\}\neq\emptyset$. But then $z\in\left\{x^{\prime},y^{\prime}\right\}$, and so $v\in N[x^{\prime}]\cup N[y^{\prime}]$, a contradiction. This proves (20).

[TABLE]

Suppose not, and let $i\in\left\{1,\dots,k\right\}$ be such that $G_{i}$ contains a wheel and $V(C_{i})\cap\left\{x,y\right\}=\emptyset$. Since $N_{i}$ is a stable set, it follows that $|N_{i}\cap\left\{x,y\right\}|\leq 1$, and by symmetry, we may assume that $y\not\in N_{i}$. Let $y^{\prime}=s$, and let $x^{\prime}=x$ if $x\in N_{i}$ and $x^{\prime}=s$ otherwise. By Lemma 16, $(x^{\prime},y^{\prime})$ is a non-center pair for $G_{i}$. Since $G_{i}$ is an induced subgraph of $G$, it follows that $G_{i}$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free. Since $G$ is a minimum counterexample, it follows that $(G_{i},x^{\prime},y^{\prime})$ has the property $\mathcal{P}$. Let $v$ be a vertex of degree at most two in $G_{i}$ with $v\not\in N[x^{\prime}]\cup N[y^{\prime}]$. Since $v\not\in N[s]$, it follows that $d_{G}(v)=d_{G_{i}}(v)$. Therefore $v\in N[x]\cup N[y]$. Let $w\in\left\{x,y\right\}$ be such that $v\in N[w]$. By assumption, $w\not\in V(C_{i})$. If $w\in V(G_{i})$, then $w\in N_{i}$, and so $w=x=x^{\prime}$, and thus $v\not\in N[w]$, a contradiction. It follows that $w\not\in V(G_{i})$, and so $N[w]\cap V(G_{i})\subseteq N[s]$, and again $v\not\in N[w]$, a contradiction. This proves that $\left\{x,y\right\}\cap V(C_{i})\neq\emptyset$, and (21) follows.

It follows from Theorem 3 together with (21) and (20) that there is at most one $i\in\left\{1,\dots,k\right\}$ with $|V(C_{i})|>1$. We may assume $|V(C_{i})|=1$ for all $i\in\left\{1,\dots,k-1\right\}$.

[TABLE]

Observe that $G^{\prime}$ is bipartite with one side of the bipartition being $N(s)$.

Suppose that $C$ is a cycle of length four in $G^{\prime}$. Let $V(C)=\left\{a,b,c,d\right\}$ and $N(s)\cap V(C)=\left\{a,c\right\}$. If $d_{G}(b)\neq 2$, let $e$ be a neighbor of $b$ which is not $a,c$. Note that $e\in N(s)$. Then $\left\{a,b,c,d,e,s\right\}$ induces an $\textnormal{ISK}_{4}$ or a $K_{3,3}$, a contradiction. It follows that $d_{G}(b)=d_{G}(d)=2$, and moreover, $\left\{x,y\right\}\cap\left\{a,c\right\}\neq\emptyset$. So, by symmetry, say $x=a$, and we may assume that $d\neq y$.

Observe that $G\setminus b$ is not series-parallel by Theorem 3. By the minimality of $|V(G)|$, it follows that there exists $v\in V(G)\setminus(N[x]\cup N[y])$ with $d_{G\setminus\left\{b\right\}}(v)\leq 2$. Since $d_{G}(v^{\prime})=d_{G\setminus\left\{b\right\}}(v^{\prime})$ for all $v^{\prime}\in V(G)\setminus\left\{x,b,c\right\}$, it follows that $v=c$, and so $N_{G}(c)=\left\{b,d,s\right\}$, and so $\left\{s,a\right\}$ is a cutset in $G$. Let $G^{\prime\prime}=G\setminus\left\{b,c,d\right\}$, and if $y\in\left\{b,c,d\right\}$, let $y^{\prime}=x$, otherwise, $y^{\prime}=y$. Then $G^{\prime\prime}$ is not series-parallel. A proper wheel in $G^{\prime\prime}$ is proper in $G$, because each vertex in $\left\{b,c,d\right\}$ has at most one neighbor in the wheel, $s$ or $a$. Therefore, $(x,y^{\prime})$ is a non-center pair for $G^{\prime\prime}$. By the minimality of $|V(G)|$, it follows that $(G^{\prime\prime},x,y^{\prime})$ has the property $\mathcal{P}$. But this is a contradiction, since every vertex in $V(G^{\prime\prime})\setminus N[x]$ has the same degree in $G$ and $G^{\prime\prime}$. This proves that $G^{\prime}$ contains no 4-cycle.

Suppose $G^{\prime}$ contains a 6-cycle $C$. Then, since exactly three vertices in $V(C)$ are neighbors of $s$, it follows that $G|(V(C)\cup\left\{s\right\})$ is an $\textnormal{ISK}_{4}$, a contradiction. It follows that $G^{\prime}$ has girth at least eight, and so (22) is proved.

[TABLE]

Suppose not, and let $G^{\prime}=G\setminus s$. Then $G^{\prime}$ satisfies the hypotheses of Theorem 22. Since $s$ is the center of a proper wheel in $G$, it follows that there exists a vertex $z$ in $G^{\prime}$ that is not in $N[x]\cup N[y]\cup N[s]$, and so the first outcome of Theorem 22 does not hold. The second outcome does not hold, because every vertex in $V(G^{\prime})\setminus(N[x]\cup N[y])$ of degree one in $G^{\prime}$ has degree at most two in $G$, a contradiction.

Therefore, the third or fourth outcome of Theorem 22 holds, and hence there exists an induced cycle $C$ in $G^{\prime}$ with vertices $c_{1}-\ldots-c_{t}-c_{1}$, and $i,j\in\left\{1,\dots,t\right\}$, $l\in\left\{0,\dots,3\right\}$ such that all vertices of $C$ except for $c_{i},\dots,c_{i+l}$ (where $c_{t+1}=c_{1}$ and so on) and $c_{j}$ have degree two in $G^{\prime}$, do not coincide with $x,y$ and are non-neighbors of $x,y$. By (22), $t\geq 8$. Consequently, $G^{\prime}$ contains two adjacent vertices in $V(G^{\prime})\setminus(N[x]\cup N[y])$ of degree two in $G^{\prime}$. Since $G$ is triangle-free, it follows that one of them has degree two in $G$, a contradiction. Thus, $|V(C_{k})|>1$, and (23) is proved.

By (20), (21) and (23) we may assume that $x\in V(C_{k})$. Let $G^{\prime}$ arise from $G$ by contracting $V(C_{k})\cup N_{k}$ to a single vertex $z$, and by deleting $s$ and every vertex that is only adjacent to $z$. It follows that $G^{\prime}$ is bipartite. Our goal is to prove that $G^{\prime}\setminus z$ has girth at least 16, see (28). By (22), we know that $G^{\prime}\setminus z$ has girth at least eight.

[TABLE]

Suppose first that there is a vertex $v\in V(G^{\prime})\setminus\left\{z\right\}$ with at least two neighbors $a,b\in N_{k}$ in $G$. Since $v\in V(G^{\prime})$ and in $G^{\prime}$ there are no vertices of degree one adjacent to $z$, it follows that $v$ has another neighbor $c\in N(s)\setminus N_{k}$. Let $P$ be a path connecting $a$ and $b$ with interior in $V(C_{k})$. Such a path exists, since $a,b\in N_{k}$. It follows that $G|(V(P)\cup\left\{a,b,c,v,s\right\})$ is an $\textnormal{ISK}_{4}$ in $G$, a contradiction. This implies the first statement of (24).

Suppose that $z$ is contained in a 4-cycle with vertex set $\left\{a,b,c,z\right\}$ in $G^{\prime}$ such that $a,c\in N_{G^{\prime}}(z)$. Note that $a,c\not\in N(s)$ and $b\in N(s)\setminus N_{k}$. By (22), $G\setminus(\left\{s\right\}\cup V(C_{k}))$ contains no 4-cycle, and thus $a$ and $c$ have no common neighbor in $N_{k}$. Let $a^{\prime},c^{\prime}$ be a neighbor of $a$ and $c$ in $N_{k}$, respectively; $a^{\prime}$ and $c^{\prime}$ exists since $a,c\in N_{G^{\prime}}(z)$. Let $P$ be a shortest path between $a^{\prime}$ and $c^{\prime}$ with interior in $C_{k}$. Since $b\not\in N_{k}$, it follows that $b$ is anticomplete to $V(P)$. Therefore, $G|(\left\{a,b,c,s\right\}\cup V(P))$ is an $\textnormal{ISK}_{4}$ in $G$, a contradiction. This proves (24).

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Since $G^{\prime}$ is bipartite, it follows that $G^{\prime}$ is triangle-free. Suppose that $G^{\prime}$ contains an induced subgraph $H$ which is either a $K_{3,3}$ or an $\textnormal{ISK}_{4}$. Since $G$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free, it follows that $z\in V(H)$. Suppose that $z$ has degree two in $H$. By (24), the neighbors of $z$ in $V(H)$ do not have a common neighbor in $N_{k}$. Let $P$ be a path in $G$ connecting the neighbors of $z$ in $V(H)$ with interior in $V(C_{k})\cup N_{k}$ containing exactly two vertices in $N_{k}$. Then $G|((V(H)\setminus\left\{z\right\})\cup V(P))$ is an induced subdivision of $H$ in $G$. By Theorem 4, it follows that $G$ is not $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free, a contradiction.

It follows that $z$ has degree three in $H$. Let $a,b,c$ be the neighbors of $z$ in $H$. By (24), each of $a,b,c$ has a unique neighbor in $N_{k}$. Let $a^{\prime},b^{\prime},c^{\prime}$ be neighbors of $a,b,c$ in $N_{k}$. Let $H^{\prime}$ be a minimal induced subgraph of $G|(V(C_{k})\cup\left\{a,b,c,a^{\prime},b^{\prime},c^{\prime}\right\})$ which is connected and contains $\left\{a,b,c\right\}$. It follows that each of $a,b,c$ has a unique neighbor (namely $a^{\prime},b^{\prime},c^{\prime}$, respectively), in $H^{\prime}$. By Lemma 17, $H^{\prime}$ is a subdivision of $K_{1,3}$ in which $a,b,c$ are the vertices of degree one. Consequently, $G|(V(H\setminus z)\cup V(H^{\prime}))$ is an induced subgraph of $G$ which is a subdivision of $H$. But then $G$ is not $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free by Theorem 4. Hence $G^{\prime}$ is $\left\{\textnormal{ISK}_{4},\textnormal{triangle},K_{3,3}\right\}$-free. This proves (25).

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Suppose $v\neq z$ is the center of a proper wheel $G^{\prime}$. By Theorem 6, there is a component $C$ of $G^{\prime}\setminus N[v]$ that is disjoint from $N[z]$. Let $N$ denote the set of vertices in $N(v)$ with a neighbor in $C$.

Then $H=G^{\prime}|(N\cup V(C)\cup\left\{v\right\})$ satisfies the hypotheses of Theorem 22. Since $V(C)\neq\emptyset$, it follows that the first outcome of Theorem 22 does not hold. Moreover, every vertex in $V(H)\setminus N[v]$ of degree one in $H$ has degree at most two in $G$, since such a vertex belongs to $C$ and $C$ is disjoint from $N[z]$, and the only additional neighbor that such a vertex may have in $G$ is $s$. Furthermore, such a vertex is in $V(G^{\prime})\setminus N[z]$, and hence in $V(G)\setminus(N[x]\cup N[y])$ as $x\in V(C_{k})$. It follows that the second outcome of Theorem 22 does not hold.

Therefore, the third or fourth outcome of Theorem 22 holds, and hence there exists an induced cycle $C^{\prime}$ in $H$ with vertices $c_{1}-\ldots-c_{t}-c_{1}$, and $i,j\in\left\{1,\dots,t\right\}$, $l\in\left\{0,\dots,3\right\}$ such that all vertices of $C^{\prime}$ except for $c_{i},\dots,c_{i+l}$ (where $c_{t+1}=c_{1}$ and so on) and $c_{j}$ have degree two in $H$, do not coincide with $v$ and are non-neighbors of $v$. By (22), $t\geq 8$, since $z\not\in V(H)$. Consequently, $G^{\prime}$ contains two adjacent vertices in $V(G^{\prime})\setminus N[z]$ of degree two in $G^{\prime}$. Since $G$ is triangle-free, it follows that one of them is non-adjacent to $s$ and thus has degree two in $G$, a contradiction. Hence (26) is proved.

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Suppose not, and let $K$ be a component of $G^{\prime}\setminus N[z]$ such that $G^{\prime}|(V(K)\cup N(z))$ is not a forest. Suppose first that $H=G^{\prime}|(V(K)\cup N[z])$ is not series-parallel. Then $H$ contains a proper wheel by Lemma 9. Let $v$ be the center of a proper wheel in $H$. Since $H\setminus N[z]$ is connected, it follows from Theorem 6 that $v\neq z$. By Lemma 16, it follows that $v$ is the center of a proper wheel in $G^{\prime}$, contrary to (26).

It follows that $H$ is series-parallel, and by our assumption, $H\setminus z$ contains a cycle. By applying Lemma 21 to $H$ and $z$, it follows that there is either a vertex in $V(H)\setminus N[z]$ of degree one, or a cycle $C$ not containing $z$, with all but at most two vertices of degree two in $H$. In the latter case, since $G^{\prime}\setminus z$ has girth at least eight, $C$ contains two adjacent vertices in $V(H)\setminus N[z]$ of degree two in $H$, and thus of degree two in $G^{\prime}$. Since $G$ is triangle-free, it follows that in both cases $G$ contains a vertex of degree at most two not in $N[z]$, and thus not in $N[x]\cup N[y]$. This is a contradiction, and (27) is proved.

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Suppose that this is false. Let $C$ be an induced cycle in $G^{\prime}\setminus z$ of length less than 16. Since by (27), for every component $K$ of $G^{\prime}\setminus N[z]$, we have that $G^{\prime}|(V(K)\cup N(z))$ is a forest, it follows that $C\setminus N[z]$ has at least two components. Since $z$ is not contained in a 4-cycle in $G^{\prime}$ by (24), and $G^{\prime}$ is bipartite, it follows that each component of $C\setminus N[z]$ has at least three vertices. If $C\setminus N[z]$ has at least four components, it follows that $C$ has length at least 16. If $C\setminus N[z]$ has exactly three components, then $G|(V(C)\cup\left\{z\right\})$ is an $\textnormal{ISK}_{4}$, a contradiction. So $C\setminus N[z]$ has exactly two components. For every component $K$ of $G^{\prime}\setminus N[z]$, by (27) we have that $G^{\prime}|(V(K)\cup N(z))$ is a forest. Therefore, the two components of $C\setminus N[z]$ are contained in two different components of $G^{\prime}\setminus N[z]$; say $A$ and $B$. Let $N_{A},N_{B}$ denote the set vertices in $N(z)$ with a neighbor in $A$, $B$, respectively. Suppose that $|N_{A}|\geq 3$. Since $|V(C)\cap N(z)|=2$, it follows that there is a path $P$ from a vertex $c$ in $N_{A}\setminus V(C)$ to $V(C)$ with interior in $V(A)$. Since $G^{\prime}|(V(A)\cup N_{A})$ and $G^{\prime}|(V(B)\cup N_{B})$ are trees, it follows that $c$ has at most one neighbor in each component $K$ of $C\setminus N[z]$. Therefore, $G^{\prime}|(V(P)\cup V(C)\cup\left\{z\right\})$ contains an induced subgraph of $G^{\prime}$ which is either a subdivision of $K_{4}$ or of $K_{3,3}$, a contradiction by Theorem 4 and (25). So $|N_{A}|=2$. Since $G^{\prime}|(V(A)\cup N_{A})$ is a tree, it follows that either $A$ contains a vertex of degree one in $G^{\prime}$, non-adjacent to $z$, or $G^{\prime}|(V(A)\cup N_{A})$ is a path containing at least five vertices, and hence $A$ contains two adjacent vertices of degree two in $G^{\prime}$, non-adjacent to $z$. Since $G$ is triangle-free, it follows that in either case $G$ contains a vertex of degree at most two not in $N[z]$, and thus not in $N[x]\cup N[y]$. This is a contradiction, and (28) is proved.

Recall that $\left\{x,y\right\}\cap V(C_{k})\neq\emptyset$, and we may assume that $x\in V(C_{k})$, and thus $y\in V(C_{k})\cup N_{k}$. Let $G^{\prime\prime}$ be the graph that arises from $G$ by deleting $\left\{s\right\}\cup(V(C_{k})\setminus\left\{x\right\})\cup(N_{k}\setminus\left\{y\right\})$, and every vertex other than $x$ with neighbors only in $N_{k}$ (this last operation does not change the degree of any vertex in $V(G^{\prime\prime})$ except for possibly $y$). Then $N_{G^{\prime\prime}}(x)\subseteq\left\{y\right\}$. It follows from (28) that $G^{\prime\prime}\setminus\left\{y\right\}$ has girth at least 16, and from (22) that $G^{\prime\prime}$ has girth at least eight. If $y\in V(G^{\prime\prime})$, let $y^{\prime}=y$; otherwise, let $y^{\prime}=x$. It follows that if $y^{\prime}=y$, then $y\in N_{k}$.

Since $G^{\prime\prime}$ is an induced subgraph of $G$, it follows that $G^{\prime\prime}$ and $x,y^{\prime}$ satisfy the hypotheses of Theorem 22.

Since $s$ is the center of a proper wheel, it follows from Theorem 6 that there are at least two components of $G^{\prime\prime}\setminus N[s]$ in which $y^{\prime}$ has no neighbors. Consequently, $V(G^{\prime\prime})\neq N[x]\cup N[y^{\prime}]$, and thus the first outcome of Theorem 22 does not hold.

The second outcome of Theorem 22 does not hold, because if $G^{\prime\prime}$ contains a vertex $v$ of degree one non-adjacent to $y^{\prime}$, then $v$ has degree at most two in $G$, and $v\not\in N[x]\cup N[y]$, a contradiction.

Suppose that the third outcome holds, and so $G^{\prime\prime}$ contains an induced cycle $C$ containing $y^{\prime}$ (since $d_{G^{\prime\prime}}(x)\leq 1$) such that at most one vertex in $V(C)\setminus N[y]$ has degree more than two. Since $G^{\prime\prime}$ has girth at least eight, $|V(C)|\geq 8$, and in particular $C$ contains a vertex $v$ of distance three from $y$ in $C$ and degree two in $G^{\prime\prime}$. Let $y-a-b-v$ be the three-edge path from $y$ to $v$ in $C$. Then $v$ is not adjacent to $s$ in $G$, because $G|(V(G^{\prime\prime})\cup\left\{s\right\})$ is bipartite and $ys\in E(G)$. Moreover, $v$ anticomplete to $N_{k}$, because otherwise $z-a-b-v-z$ is a 4-cycle in $G^{\prime}$ using $z$, contradicting (24). So $v$ has degree two in $G$ and is not in $N[x]\cup N[y]$, a contradiction.

Thus, the fourth outcome holds, and so $G^{\prime\prime}$ contains an induced cycle $C$ not containing $x,y^{\prime}$ and containing a vertex $z^{\prime}$ such that at most one vertex in $V(C)\setminus N[z^{\prime}]$ has degree more than two in $G^{\prime\prime}$. Since $|V(C)|\geq 16$, it follows that $C$ contains a path $P=p_{1}-\ldots-p_{6}$ of six vertices, all of degree two in $G^{\prime\prime}$ and non-adjacent to $x,y^{\prime}$. We may assume that $N(s)\cap V(P)\subseteq\left\{p_{1},p_{3},p_{5}\right\}$ by symmetry. Since $z$ is not in a 4-cycle in $G^{\prime}$ by (24), not both $p_{2}$ and $p_{4}$ have a neighbor in $N_{k}$. It follows that either $p_{2}$ or $p_{4}$ has degree two in $G$, a contradiction. This completes the proof of Theorem 23. ∎

We can now prove Theorem 2 which we restate:

Theorem 24**.**

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. Then either $G$ has a clique cutset, $G$ is complete bipartite, or $G$ has a vertex of degree at most two.

Proof.

Let $G$ be an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph. If $G$ is series-parallel, then $G$ contains a vertex of degree at most two Theorem 3. If $G$ contains $K_{3,3}$ as a subgraph, then by Theorem 5, either $G$ is complete bipartite or $G$ has a clique cutset. If $G$ is not series-parallel and $K_{3,3}$-free, then $G$ contains a vertex of degree at most two by Theorem 7 applied to the graph obtaining from $G$ by adding an isolated vertex $x$ with the non-center pair $(x,x)$. This implies the result. ∎

Note that the outcome of a clique cutset in Theorem 24 cannot be avoided, as the following example shows. Let $G$ be any $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph (e. g. a $C_{5}$, or a wheel), and let $H$ arise from $G$ by adding $|V(G)|$ disjoint copies of $K_{3,3}$ to $G$ and identifying each vertex of $G$ with a vertex of a different copy of $K_{3,3}$. The resulting graph is $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free, not series-parallel, and not bipartite if $G$ is not bipartite, and it contains no vertex of degree at most two.

Finally we are ready to prove the following.

Theorem 25**.**

If $G$ is an $\left\{\textnormal{ISK}_{4},\textnormal{triangle}\right\}$-free graph, then $G$ is 3-colorable.

Proof.

The proof is by induction on $|V(G)|$ using Theorem 2. If $G$ is complete bipartite, then $G$ is 2-colorable. If $G$ has a vertex $v$ of degree at most two, then, by induction, $G\setminus v$ is 3-colorable, and hence $G$ is 3-colorable. If $G$ has a clique cutset $C$ such that $(A,B,C)$ is a partition of $V(G)$ with $A$ anticomplete to $B$ and $C$ a clique, then $\chi(G)=\max\left\{\chi(G|(A\cup C)),\chi(G|(B\cup C))\right\}$, and again by induction, $G$ is 3-colorable. ∎

Acknowledgments

We are thankful to Paul Seymour and Mingxian Zhong for many helpful discussions.