Graphs with girth $2\ell+1$ and without longer odd holes that contain an odd $K_{4}$ -subdivision

Rong Chen, Yidong Zhou

Center for Discrete Mathematics, Fuzhou University
Fuzhou, P. R. China

Abstract

We say that a graph $G$ has an odd $K_{4}$ -subdivision if some subgraph of $G$ is isomorphic to a $K_{4}$ -subdivision which if embedded in the plane the boundary of each of its faces has odd length and is an induced cycle of $G$ . For a number $\ell\geq 2$ , let $\mathcal{G}_{\ell}$ denote the family of graphs which have girth $2\ell+1$ and have no odd hole with length greater than $2\ell+1$ . Wu, Xu and Xu conjectured that every graph in $\bigcup_{\ell\geq 2}\mathcal{G}_{\ell}$ is 3-colorable. Recently, Chudnovsky et al. and Wu et al., respectively, proved that every graph in $\mathcal{G}_{2}$ and $\mathcal{G}_{3}$ is 3-colorable. In this paper, we prove that no $4$ -vertex-critical graph in $\bigcup_{\ell\geq 5}\mathcal{G}_{\ell}$ has an odd $K_{4}$ -subdivision. Using this result, Chen proved that all graphs in $\bigcup_{\ell\geq 5}\mathcal{G}_{\ell}$ are 3-colorable.

Key Words: chromatic number; odd holes.

¹¹1Mathematics Subject Classification: 05C15, 05C17, 05C69. Emails: rongchen13@163.com (R. Chen), zoed98@126.com (Y. Zhou). This research was partially supported by grants from the National Natural Sciences Foundation of China (No. 11971111).

1 Introduction

All graphs considered in this paper are finite, simple, and undirected. A proper coloring of a graph $G$ is an assignment of colors to the vertices of $G$ such that no two adjacent vertices receive the same color. A graph is $k$ -colorable if it has a proper coloring using at most $k$ colors. The chromatic number of $G$ , denoted by $\chi(G)$ , is the minimum number $k$ such that $G$ is $k$ -colorable.

The girth of a graph $G$ , denoted by $g(G)$ , is the minimum length of cycles in $G$ . A hole in a graph is an induced cycle of length at least four. An odd hole means a hole of odd length. For any integer $\ell\geq 2$ , let $\mathcal{G}_{\ell}$ be the family of graphs that have girth $2\ell+1$ and have no odd holes of length at least $2\ell+3$ . Robertson conjectured in [4] that the Petersen graph is the only graph in $\mathcal{G}_{2}$ that is 3-connected and internally 4-connected. Plummer and Zha [5] disproved Robertson’s conjecture and proposed the conjecture that all 3-connected and internally 4-connected graphs in $\mathcal{G}_{2}$ have bounded chromatic numbers, and proposed the strong conjecture that such graphs are 3-colorable. The first was proved by Xu, Yu, and Zha [9], who proved that all graphs in $\mathcal{G}_{2}$ are 4-colorable. The strong conjecture proposed by Plummer and Zha in [5] was solved by Chudnovsky and Seymour [2]. Wu, Xu, and Xu [7] showed that graphs in $\bigcup_{\ell\geq 2}\mathcal{G}_{\ell}$ are 4-colorable and conjectured

Conjecture 1.1.

([7], Conjecture 6.1.) For each integer $\ell\geq 2$ , every graph in $\mathcal{G}_{\ell}$ is $3$ -colorable.

Wu, Xu and Xu [8] recently proved that Conjecture 1.1 holds for $\ell=3$ .

We say that a graph $G$ has an odd $K_{4}$ -subdivision if some subgraph of $G$ is isomorphic to a $K_{4}$ -subdivision which if embedded in the plane the boundary of each of its faces has odd length and is an induced cycle of $G$ . Note that an odd $K_{4}$ -subdivision of $G$ maybe not induced. However, when $G\in\mathcal{G}_{\ell}$ for each integer $\ell\geq 2$ , all odd $K_{4}$ -subdivisions of $G$ are induced by Lemma 2.6 (2). In this paper, we prove the following theorem.

Theorem 1.2.

No $4$ -vertex-critical graph in $\bigcup_{\ell\geq 5}\mathcal{G}_{\ell}$ has an odd $K_{4}$ -subdivision.

Using Theorem 1.2, Chen [1] proved that Conjecture 1.1 holds for all $\ell\geq 5$ . Recently, following idea in this paper and [1], Wang and Wu [6] further proved that Conjecture 1.1 holds for $\ell=4$ .

2 Preliminaries

A cycle is a connected $2$ -regular graph. Let $G$ be a graph. A vertex $v\in V(G)$ is called a degree- $k$ vertex if it has exactly k neighbours. For any $U\subseteq V(G)$ , let $G[U]$ be the subgraph of $G$ induced on $U$ . For subgraphs $H$ and $H^{\prime}$ of $G$ , set $|H|:=|E(H)|$ and $H\Delta H^{\prime}:=E(H)\Delta E(H^{\prime})$ . Let $H\cup H^{\prime}$ denote the subgraph of $G$ whose vertex set is $V(H)\cup V(H^{\prime})$ and edge set is $E(H)\cup E(H^{\prime})$ . Let $H\cap H^{\prime}$ denote the subgraph of $G$ with edge set $E(H)\cap E(H^{\prime})$ and without isolated vertex. Let $N(H)$ be the set of vertices in $V(G)-V(H)$ that have a neighbour in $H$ . Set $N[H]:=N(H)\cup V(H)$ .

Let $P$ be an $(x,y)$ -path and $Q$ be a $(y,z)$ -path. When $P$ and $Q$ are internally disjoint, let $PQ$ denote the $(x,z)$ -path $P\cup Q$ . Evidently, $PQ$ is a path when $x\neq z$ , and $PQ$ is a cycle when $x=z$ . Let $P^{*}$ denote the set of internal vertices of $P$ . When $u,v\in V(P)$ , let $P(u,v)$ denote the subpath of $P$ with ends $u,v$ . For simplicity, we will let $P^{*}(u,v)$ denote $(P(u,v))^{*}$ .

A graph is $k$ -vertex-critical if $\chi(G)=k$ but $\chi(G\setminus v)<k$ for each $v\in V(G)$ . Dirac in [3] proved that every $k$ -vertex-critical graph is $(k-1)$ -edge-connected. Hence, we have

Lemma 2.1.

For each integer $k\geq 4$ , each $k$ -vertex-critical graph $G$ has no $2$ -edge-cut.

A theta graph is a graph that consists of a pair of distinct vertices joined by three internally disjoint paths. Let $C$ be a hole of a graph $G$ . A path $P$ of $G$ is a chordal path of $C$ if $V(P^{*})\cap V(C)=\emptyset$ and $C\cup P$ is an induced theta-subgraph of $G$ . Lemma 2.2 will be frequently used.

Lemma 2.2.

Let $\ell\geq 2$ be an integer and $C$ be an odd hole of a graph $G\in\mathcal{G}_{\ell}$ . Let $P$ be a chordal path of $C$ , and $P_{1},P_{2}$ be the internally disjoint paths of $C$ that have the same ends as $P$ . Assume that $|P|$ and $|P_{1}|$ have the same parity. If $|P_{1}|\neq 1$ , then $|P_{1}|>|P_{2}|$ and all chordal paths of $C$ with the same ends as $P_{1}$ have length $|P_{1}|$ .

Proof.

Since $|C|=2\ell+1$ , $|P_{1}|\neq 1$ and $|P|$ and $|P_{1}|$ have the same parity, $P\cup P_{2}$ is an odd hole. Moreover, since $g(G)=2\ell+1$ and all odd holes in $G$ have length $2\ell+1$ , we have $\ell+1\leq|P_{1}|=|P|$ and $|P_{2}|\leq\ell$ , so $|P_{1}|>|P_{2}|$ and all chordal paths of $C$ with the same ends as $P_{1}$ have length $|P_{1}|$ . ∎

Let $P$ be a path with $i$ vertices. If $G-V(P)$ is disconnected, then we say that $P$ is a $P_{i}$ -cut. Usually, a $P_{2}$ -cut is also called a $K_{2}$ -cut. Evidently, every $k$ -vertex-critical graph has no $K_{2}$ -cut. Chudnovsky and Seymour in [2] proved that every $4$ -vertex-critical graph $G$ in $\mathcal{G}_{2}$ has no $P_{3}$ -cut. Using the same argument as [2], Wu et al. [8] extend this result to graphs in $\bigcup_{\ell\geq 2}\mathcal{G}_{\ell}$ . Since the paper [8] does not include a proof of Lemma 2.3, we give a proof here for completeness.

Lemma 2.3.

([8]) For any number $\ell\geq 2$ , every $4$ -vertex-critical graph in $\mathcal{G}_{\ell}$ has neither a $K_{2}$ -cut nor a $P_{3}$ -cut.

Proof.

It is well-known that every $k$ -vertex-critical graph has no clique as a cut. Hence, it suffice to show that every $4$ -vertex-critical graph in $\mathcal{G}_{\ell}$ has no $P_{3}$ -cut. Let $G\in\mathcal{G}_{\ell}$ be a $4$ -vertex-critical graph. Assume to the contrary that $P=v_{1}v_{2}v_{3}$ is a path such that $G\backslash\{v_{1},v_{2},v_{3}\}$ is disconnected. Since $G$ has no $K_{3}$ as its cut, $v_{1}v_{3}\notin E(G)$ . Let $A_{1}$ be the a component of $G\backslash\{v_{1},v_{2},v_{3}\}$ , and let $A_{2}$ be the union of all other components. Set $G_{i}:=G[A_{i}\cup\{v_{1},v_{2},v_{3}\}]$ for $i=1,2$ . Since $G$ is $4$ -vertex-critical, both $G_{1}$ and $G_{2}$ are 3-colorable. Let $\phi_{i}:V(G_{i})\rightarrow\{1,2,3\}$ be a 3-coloring for $i=1,2$ . By symmetry we may assume that $\phi_{i}(v_{1})=1$ and $\phi_{i}(v_{2})=2$ for $i=1,2$ . Thus $\phi_{1}(v_{3}),\phi_{2}(v_{3})\in\{1,3\}$ . If $\phi_{1}(v_{3})=\phi_{2}(v_{3})$ , then $G$ is 3-colorable, which is a contradiction. Thus by symmetry we may assume that $\phi_{1}(v_{3})=1$ and $\phi_{2}(v_{3})=3$ . Let $H_{1}$ be the subgraph of $G_{1}$ induced on the set of vertices $v\in V(G_{1})$ with $\phi_{1}(v)\in\{1,3\}$ . If $v_{1},v_{3}$ belong to different components of $H_{1}$ , then by exchanging colors in the component containing $v_{3}$ , we obtain another 3-coloring of $G_{1}$ that can be combined with $\phi_{2}$ to show that $G$ is 3-colorable. So $v_{1},v_{3}$ belong to the same component of $H_{1}$ . Then there is an induced $(v_{1},v_{3})$ -path $P_{1}$ in $H_{1}$ having even length as $\phi_{1}(v_{1})=1=\phi_{1}(v_{3})$ . Similarly, there is an induced $(v_{1},v_{3})$ -path $P_{2}$ in $G_{2}$ having odd length as $\phi_{2}(v_{1})=1$ and $\phi_{2}(v_{3})=3$ . Moreover, since $PP_{1},PP_{2}$ are cycles of $G$ and $g(G)=2\ell+1$ , we have $|P_{1}|\geq 2\ell-1$ and $|P_{2}|\geq 2\ell$ , so $P_{1}\cup P_{2}$ is an odd hole of $G$ of length at least $4\ell-1$ , which is a contradiction as $G\in\mathcal{G}_{\ell}$ . ∎

Lemma 2.4.

Let $\ell\geq 2$ be an integer and $x,y$ be non-adjacent vertices of a graph $G\in\mathcal{G}_{\ell}$ . Let $P$ be an induced $(x,y)$ -path of $G$ . If $|P|\leq\ell$ and all induced $(x,y)$ -paths have length $|P|$ , then no block of $G$ contains two non-adjacent vertices in $V(P)$ . In particular, each vertex in $P^{*}$ is a cut-vertex of $G$ .

Proof.

Assume not. Then there is a block $B$ of $G$ containing two consecutive edges of $P$ . Let $Q$ be an induced path in $B$ with ends in $V(P)$ and with $V(P)\cap V(Q^{*})=\emptyset$ . Since every pair of edges in a 2-connected graph is contained in a cycle, such a $Q$ exists. Without loss of generality we may further assume that $Q$ is chosen with $|Q|$ as small as possible. Let $C$ be the unique cycle in $P\cup Q$ . Then $C\Delta P$ is an $(x,y)$ -path. Since $Q$ is induced, the ends of $Q$ are not adjacent. Moreover, since $Q$ is chosen with $|Q|$ as small as possible, $C\Delta P$ is an induced $(x,y)$ -path, so $|C\Delta P|=|P|\leq\ell$ by the assumption of the lemma. Hence, $|C|\leq 2\ell$ , contrary to the fact $g(G)=2\ell+1$ . ∎

Lemma 2.5.

Let $\ell\geq 4$ be an integer and $x,y$ be non-adjacent vertices of a graph $G\in\mathcal{G}_{\ell}$ . Let $X$ be a vertex cut of $G$ with $\{x,y\}\subseteq X\subseteq N[\{x,y\}]$ , and $G_{1}$ be an induced subgraph of $G$ whose vertex set consists of $X$ and the vertex set of a component of $G-X$ . If all induced $(x,y)$ -paths in $G_{1}$ have length $k$ with $4\leq k\leq\ell$ , then $G$ has a degree- $2$ vertex, a $K_{1}$ -cut, or a $K_{2}$ -cut.

Proof.

Assume that $G$ has no degree-2 vertices. Let $P$ be an induced $(x,y)$ -path in $G_{1}$ . Let $uvw$ be a subpath of $P^{*}$ . Such $uvw$ exists as $k\geq 4$ . By the definition of $G_{1}$ , we have $v\notin X$ , so $N_{G}[v]=N_{G_{1}}[v]$ , which implies $d_{G_{1}}(v)\geq 3$ . By applying Lemma 2.4 to $G_{1}$ , there is a block $B$ of $G_{1}$ such that either $V(B)\cap V(P)=\{v\}$ , or $B$ is not isomorphic to $K_{2}$ and $V(B)\cap V(P)$ is $\{u,v\}$ or $\{u,v\}$ . When the first case happens, since $X\subseteq N[\{x,y\}]$ , $x,y\notin B$ and $v\notin X$ , we have $X\cap V(B)=\emptyset$ , for otherwise $P(x,v)$ or $P(v,y)$ is contained in a cycle of $P\cup B$ , so the vertex $v$ is a cut-vertex of $G$ as $X$ is a vertex cut of $G$ . When the latter case happens, by symmetry we may assume that $V(B)\cap V(P)=\{u,v\}$ . Since $B$ is a block of $G_{1}$ , $X\subseteq N[\{x,y\}]$ and $uvw$ is a subpath of $P^{*}$ , we have $V(B)\cap X=\{u\}\cap X$ , so $\{u,v\}$ is a $K_{2}$ -cut of $G_{1}$ and $G$ . ∎

Refer to caption — Figure 1: $u_{1},u_{2},u_{3},u_{4}$ are the degree-3 vertices of $H$ . All faces $C_{1},C_{2},C_{3},C_{4}$ of $H$ are odd holes. $\{P_{1},P_{2}\}$ , $\{Q_{1},Q_{2}\}$ , $\{L_{1},L_{2}\}$ are the pairs of vertex disjoint arrises of $H$ .

Let $H$ be a graph that is isomorphic to a subdivision of $K_{4}$ , and let $P$ be a path of $H$ whose ends are degree-3 vertices in $H$ . If $P^{*}$ contains no degree-3 vertex of $H$ , then we say that $P$ is an arris of $H$ . Evidently, there are exactly six arrises of $H$ . See Figure 1.

Lemma 2.6.

For any integer $\ell\geq 2$ , if a graph $G\in\mathcal{G}_{\ell}$ has an an odd $K_{4}$ -subdivision $H$ , then the following statements hold.

(1)

Each pair of vertex disjoint arrises in $H$ have the same length and their lengths are at most $\ell$ .
(2)

$H$ is an induced subgraph of $G$ .
(3)

When $\ell\geq 3$ , every vertex in $V(G)-V(H)$ has at most one neighbour in $V(H)$ .

Proof.

Without loss of generality we may assume that $H$ is pictured as the graph in Figure 1. First, we prove that (1) is true. Assume that $|P_{1}|>|P_{2}|$ . Since $C_{1}$ and $C_{4}$ are odd holes, $|Q_{1}|<|Q_{2}|$ . Hence, $|P_{2}\cup Q_{1}\cup L_{2}|<|P_{1}\cup Q_{2}\cup L_{2}|$ , which is a contradiction to the fact that $C_{2}$ and $C_{3}$ are both odd holes. So $|P_{1}|=|P_{2}|$ . By symmetry each pair of vertex-disjoint arrises have the same length. Moreover, since $C_{1}\Delta C_{2}$ is an even cycle with length at least $2\ell+2$ , we have $|P_{1}|\leq\ell$ . By symmetry we have $|Q_{1}|,|L_{1}|\leq\ell$ . So (1) holds.

Secondly, we prove that (2) is true. Suppose not. Since odd holes have no chord, by symmetry we may assume that there is an edge $st$ in $G$ with $s\in V(P_{1}^{\ast})$ and $t\in V(P_{2}^{\ast})$ . On one hand, since $P_{1}(u_{1},s)stP_{2}(t,u_{4})Q_{2}$ and $P_{1}(u_{2},s)stP_{2}(t,u_{3})Q_{1}$ are cycles, by (1) we have

|P_{1}|+|P_{2}|+|Q_{1}|+|Q_{2}|+2=2(|P_{1}|+|Q_{1}|+1)\geq 2(2\ell+1).

On the other hand, since $|P_{1}|,|Q_{1}|\leq\ell$ by (1), we have $|P_{1}|=|Q_{1}|=\ell$ , implying that $|L_{1}|=|L_{2}|=1$ . Moreover, by the symmetry between $L_{1},L_{2}$ and $Q_{1},Q_{2}$ , we have $|Q_{1}|=|Q_{2}|=1$ , which is a contradiction as $|Q_{1}|=\ell\geq 2$ . So (2) holds.

Finally, we prove that (3) is true. Suppose to the contrary that some vertex $x\in V(G)-V(H)$ has at least two neighbours in $V(H)$ . Since a vertex not in an odd hole can not have two neighbours in the odd hole, $x$ has exactly two neighbours $x_{1},x_{2}$ in $V(H)$ . By symmetry we may further assume that $x_{1}\in V(P_{1}^{\ast})$ and $x_{2}\in V(P_{2}^{\ast})$ . Since $C^{\prime}_{1}=P_{1}(u_{1},x_{1})x_{1}xx_{2}P_{2}(x_{2},u_{3})L_{1}$ and $C^{\prime}_{2}=P_{1}(u_{1},x_{1})x_{1}xx_{2}P_{2}(x_{2},u_{4})Q_{2}$ are cycles whose lengths have different parity,

|C^{\prime}_{1}|+|C^{\prime}_{2}|=2\ell+1+2(2+|P_{1}(u_{1},x_{1})|)\geq 4\ell+3.

Hence, $|P_{1}(u_{1},x_{1})|=\ell-1$ and $x_{1}u_{2}\in E(H)$ as $|P_{1}|\leq\ell$ by (1). This implies that $u_{2}x_{1}xx_{2}$ is a chordal path of $C_{3}$ with length 3, which is a contradiction to Lemma 2.2 as $\ell\geq 3$ . ∎

By Lemma 2.6 (1), all odd $K_{4}$ -subdivisions of a graph $G\in\mathcal{G}_{\ell}$ have exactly $4\ell+2$ edges for each number $\ell\geq 2$ .

3 Proof of Theorem 1.2

Let $H_{1},H_{2}$ be vertex disjoint induced subgraphs of a graph $G$ . An induced $(v_{1},v_{2})$ -path $P$ is a direct connection linking $H_{1}$ and $H_{2}$ if $v_{i}$ is the only vertex in $V(P)$ having a neighbour in $V(H_{i})$ for each $i\in\{1,2\}$ . Evidently, $V(P)\cap V(H_{1}\cup H_{2})=\emptyset$ and the set of internal vertices of each shortest path joining $H_{1}$ and $H_{2}$ induces a direct connection linking $H_{1}$ and $H_{2}$ .

For convenience, Theorem 1.2 is restated here in another way.

Theorem 3.1.

Let $\ell\geq 5$ be an integer, and $G$ be a graph in $\mathcal{G}_{\ell}$ . If $G$ is $4$ -vertex-critical, then $G$ has no odd $K_{4}$ -subdivisions.

Proof.

Suppose not. Let $H$ be a subgraph of $G$ that is isomorphic to an odd $K_{4}$ -subdivision and pictured as the graph in Figure 1. By Lemma 2.6 (2), $H$ is an induced subgraph of $G$ . By Lemma 2.6 (1), we have

|P_{1}|=|P_{2}|\leq\ell,\ \ |Q_{1}|=|Q_{2}|\leq\ell,\ \ \text{and}\ \ |L_{1}|=|L_{2}|\leq\ell.

(3.1)

Without loss of generality we may assume that $P_{1},P_{2}$ are longest arrises in $H$ .

Let $e,f$ be the edges of $P_{2}$ incident with $u_{3},u_{4}$ , respectively. Since $G$ is 4-vertex-critical, $\{e,f\}$ is not an edge-cut of $G$ by Lemma 2.1, so there is a direct connection $P$ in $G\backslash\{e,f\}$ linking $P^{*}_{2}$ and $H-V(P^{*}_{2})$ . Let $v_{1},v_{2}$ be the ends of $P$ with $v_{2}$ having a neighbour in $P^{*}_{2}$ and $v_{1}$ having a neighbour in $H-V(P^{*}_{2})$ . By Lemma 2.6 (3), both $v_{1}$ and $v_{2}$ have a unique neighbour in $V(H)$ . Let $x,y$ be the neighbours of $v_{1}$ and $v_{2}$ in $V(H)$ , respectively. That is, $x\in V(H)-V(P^{*}_{2})$ and $y\in V(P^{*}_{2})$ . Set $P^{\prime}:=xv_{1}Pv_{2}y$ . Since $H$ is an induced subgraph of $G$ , so is $H\cup P^{\prime}$ .

3.1.1.

$x\notin\{u_{1},u_{2}\}$ .

Subproof..

Suppose not. By symmetry we may assume that $x=u_{1}$ . Set $C^{\prime}_{4}=L_{1}P^{\prime}P_{2}(y,u_{3})$ . Since $C_{4}$ is an odd hole, by symmetry we may assume that $C^{\prime}_{4}$ is an even hole and $C_{4}\Delta C^{\prime}_{4}$ is an odd hole. Since $P^{\prime}P_{2}(y,u_{3})$ is a chordal path of $C_{1}$ , by (3.1) and Lemma 2.2, we have $|L_{1}|=1$ . So $|P_{1}|=|Q_{1}|=\ell$ by (3.1) again. Since $P^{\prime}P_{2}(y,u_{4})$ is a chordal path of $C_{2}$ and $C_{4}\Delta C^{\prime}_{4}$ is an odd hole, $|P^{\prime}P_{2}(y,u_{4})|=|P_{1}L_{2}|=\ell+1$ by (3.1) and Lemma 2.2 again. Moreover, since $|P_{2}|=\ell$ and $|L_{1}|=1$ , we have $|C^{\prime}_{4}|\leq 2\ell$ , which is not possible. So $x\neq u_{1}$ . ∎

Set $d(H):=|P_{1}|-\mathrm{min}\{|Q_{1}|,|L_{1}|\}$ . We say that $d(H)$ is the difference of $H$ . Without loss of generality we may assume that among all odd $K_{4}$ -subdivisions, $H$ is chosen with difference as small as possible.

3.1.2.

$x\notin V(P_{1})$ .

Subproof..

Suppose to the contrary that $x\in V(P_{1})$ . Then $x\in V(P_{1}^{*})$ by 3.1.1. Without loss of generality we may assume that $|L_{1}|\geq|Q_{1}|$ . Set $C^{\prime}_{2}=Q_{2}P_{1}(u_{1},x)P^{\prime}P_{2}(y,u_{4})$ . Since $C_{4}$ is an odd hole, either $C^{\prime}_{2}$ or $C_{4}\Delta C^{\prime}_{2}$ is an odd hole. Suppose that $C_{4}\Delta C^{\prime}_{2}$ is an odd hole. Since $C_{1}\cup C_{3}\cup P^{\prime}$ is an odd $K_{4}$ -subdivision, by Lemma 2.6 (1) and (3.1), we have $|P^{\prime}|=|Q_{1}|$ , $|P_{1}(u_{1},x)|=|P_{2}(u_{4},y)|$ , and $|P_{1}(u_{2},x)|=|P_{2}(u_{3},y)|$ . So $C^{\prime}_{2}$ is an even hole of length $2(|Q_{2}|+|P_{1}(u_{1},x)|)$ by (3.1) again, implying $|L_{1}|+|P_{1}(u_{1},x)|\geq|Q_{2}|+|P_{1}(u_{1},x)|\geq\ell+1$ as $|L_{1}|\geq|Q_{1}|$ . Then $|C_{4}\Delta C^{\prime}_{2}\Delta C_{1}|=2|P_{1}(x,u_{2})Q_{1}|\leq 2\ell$ , contrary to the fact $g(G)=2\ell+1$ . So $C^{\prime}_{2}$ is an odd hole.

Since $C_{2}\cup C^{\prime}_{2}\cup C_{3}$ is an odd $K_{4}$ -subdivision, it follows from Lemma 2.6 (1) and (3.1) that

|P^{\prime}|=|L_{2}|,\ |P_{1}(u_{1},x)|=|P_{2}(u_{3},y)|,\ \text{and}\ |P_{1}(u_{2},x)|=|P_{2}(u_{4},y)|.

(3.2)

Then $|C_{2}\Delta C^{\prime}_{2}\Delta C_{1}|=2|L_{1}|+2\ell+1$ . Since $C_{2}\Delta C^{\prime}_{2}\Delta C_{1}$ is not an odd hole,

1\in\{|Q_{2}|,|P_{1}(u_{2},x)|,|P_{2}(u_{3},y)|\}.

(3.3)

When $|P_{1}(u_{2},x)|=1$ , since $|C_{2}\Delta C^{\prime}_{2}\Delta C_{1}|=2|L_{1}|+2\ell+1$ and $g(G)=2\ell+1$ , we have $|L_{1}|=|P^{\prime}|=\ell$ by (3.2), implying $|P_{1}|=\ell$ and $|Q_{1}|=1$ as $P_{1},P_{2}$ are longest arrises in $H$ . Hence, $d(H)=\ell-1$ . Then $G[V(C_{1}\cup C^{\prime}_{2}\cup P_{2})]$ is an odd $K_{4}$ -subdivision with difference $\ell-2$ , which is a contradiction to the choice of $H$ . So $|P_{1}(u_{2},x)|\geq 2$ . Assume that $|Q_{2}|=1$ . Then $|L_{1}|=|P_{1}|=\ell$ by (3.1). Since $|P_{1}(u_{2},x)|\geq 2$ , the graph $G[V(C^{\prime}_{2}\cup C_{2}\cup C_{3})]$ is an odd $K_{4}$ -subdivision whose difference is at most $\ell-2$ , which is a contradiction to the choice of $H$ as $d(H)=\ell-1$ . So $|Q_{2}|\geq 2$ . Then $yu_{3}\in E(H)$ by (3.3), implying $xu_{1}\in E(H)$ by (3.2). Hence, $|C_{4}\Delta C^{\prime}_{2}|=2+2|L_{1}|$ by (3.1) and (3.2), and so $|L_{1}|=\ell$ by (3.1) again. Since $|P_{1}|\geq|L_{1}|$ , we have $|P_{1}|=\ell$ and $|Q_{1}|=1$ by (3.1), which is a contradiction as $|Q_{2}|\geq 2$ . ∎

3.1.3.

When $x\in\{u_{3},u_{4}\}$ , the vertices $x$ and $y$ are adjacent, that is, $xy\in\{e,f\}$ .

Subproof..

By symmetry we may assume that $x=u_{3}$ . Assume to the contrary that $x,y$ are not adjacent. Set $C^{\prime}_{3}=P^{\prime}P_{2}(y,u_{3})$ . Since $P^{\prime}$ is a chordal path of $C_{3}$ , we have that $C_{3}^{\prime}$ is an odd hole by Lemma 2.2 and (3.1). Since $C_{3}^{\prime}\Delta C_{3}$ is an even hole, $|Q_{1}|=|L_{2}|=1$ by (3.1) and Lemma 2.2 again. Then $|P_{1}|=2\ell-1>\ell$ , which is a contradiction to (3.1). So $e=xy$ . ∎

3.1.4.

When $x\in V(L_{1}^{*})$ , we have that $|Q_{1}|=1$ , $|P_{1}|=|L_{1}|=\ell$ , $|P^{\prime}|=2\ell-1$ , $xu_{3},yu_{3}\in E(H)$ and $xu_{3}yP^{\prime}$ is an odd hole.

Subproof..

Set $C^{\prime}_{4}=L_{1}(x,u_{1})Q_{2}P_{2}(u_{4},y)P^{\prime}$ . We claim that $C_{4}\Delta C_{4}^{\prime}$ is an odd hole. Assume to the contrary that $C_{4}\Delta C_{4}^{\prime}$ is an even hole. Since $x\neq u_{3}$ , the subgraph $C_{1}\cup(C_{4}\Delta C_{4}^{\prime})$ is an induced theta subgraph. Hence, $xu_{3}\in E(H)$ by (3.1) and Lemma 2.2. Similarly, $yu_{3}\in E(H)$ . Since $P^{\prime}$ is a chordal path of $C_{4}$ , we get a contradiction to Lemma 2.2. So the claim holds, implying that $C^{\prime}_{4}$ is an even hole.

Since $x\neq u_{3}$ , the graph $C_{2}\cup C^{\prime}_{4}$ is an induced theta subgraph of $G$ . Moreover, since $C^{\prime}_{4}$ is an even hole, $|Q_{2}|=1$ by (3.1) and Lemma 2.2. Hence, $|P_{1}|=|L_{1}|=\ell$ by (3.1) again. Assume that $y,u_{3}$ are not adjacent. Since $C_{1}\cup C^{\prime}_{4}$ is an induced theta subgraph of $G$ , we have $xu_{1}\in E(H)$ , implying $|P(x,u_{3})|=\ell-1$ . Since $P^{\prime}$ is a chordal path of $C_{4}$ and $C_{4}\Delta C_{4}^{\prime}$ is an odd hole, $yu_{3}\in E(H)$ by Lemma 2.2, a contradiction. Hence, $yu_{3}\in E(H)$ . By symmetry we have $xu_{3}\in E(H)$ . This proves 3.1.4. ∎

3.1.5.

Assume that $P^{\prime}$ has the structure stated as in 3.1.4. Then no vertex in $V(G)-V(H\cup P^{\prime})$ has two neighbours in $H\cup P^{\prime}$ .

Subproof..

Assume to the contrary that some vertex $a\in V(G)-V(H\cup P^{\prime})$ has two neighbours $a_{1},a_{2}$ in $H\cup P^{\prime}$ . Since no vertex has two neighbours in an odd hole, it follows from Lemma 2.6 (3) and 3.1.4 that $a$ has exactly two neighbours in $H\cup P^{\prime}$ with $a_{1}\in V(H)-\{x,y,u_{3}\}$ and $a_{2}\in V(P)$ . When $xa_{2}\notin E(P^{\prime})$ , let $Q$ be the unique $(y,a_{1})$ -path in $G[V(P)\cup\{a_{1},y\}]$ . Since $Q^{*}$ is a direct connection in $G\backslash\{e,f\}$ linking $P^{*}_{2}$ and $H-V(P^{*}_{2})$ , by 3.1.2-3.1.4 and the symmetry between $P^{\prime}$ and $Q$ , we have $a_{1}\in\{x,u_{3}\}$ , contrary to the fact $a_{1}\in V(H)-\{x,y,u_{3}\}$ . So $xa_{2}\in E(P^{\prime})$ . Moreover, since $|P_{1}|=|L_{1}|=\ell\geq 5$ and $g(G)=2\ell+1$ , we have $a_{1}\notin V(P_{1})$ . Let $u^{\prime}_{1}$ be the neighbour of $u_{1}$ in $L_{1}$ . When $a_{1}\in V(C_{2})-V(P_{1})$ , since $aa_{2}$ is a direct connection in $G\backslash\{u_{1}u^{\prime}_{1},u_{3}x\}$ linking $L_{1}^{*}$ and $H-V(L_{1}^{*})$ , which is not possible by 3.1.4 and the symmetry between $P_{2}$ and $L_{1}$ . So $a\in V(L_{1}^{*})$ . Then $xa_{2}aa_{1}$ is a chordal path of $C_{1}$ with length 3, contrary to Lemma 2.2. ∎

3.1.6.

$x\in\{u_{3},u_{4}\}$ and $xy\in\{e,f\}$ .

Subproof..

By 3.1.1-3.1.3, it suffices to show that $x\notin V(L_{1}^{*}\cup L_{2}^{*}\cup Q_{1}^{*}\cup Q_{2}^{*})$ . Assume not. By symmetry we may assume that $x\in V(L_{1}^{*})$ . By 3.1.4, we have that

xu_{3}\in E(L_{1}),\ e=yu_{3},\ |P^{\prime}|=2\ell-1,\ |P_{1}|=|L_{1}|=\ell,\ \text{and}\ |Q_{1}|=1.

Since no 4-vertex-critical graph has a $P_{3}$ -cut by Lemma 2.3, to prove that 3.1.6 is true, it suffice to show that $\{x,y,u_{3}\}$ is a $P_{3}$ -cut of $G$ . Suppose not. Let $R$ be a shortest induced path in $G-\{x,y,u_{3}\}$ linking $P$ and $H-\{x,y,u_{3}\}$ . Let $s$ and $t$ be the ends of $R$ with $s\in V(P)$ . By 3.1.5, $|R|\geq 3$ and no vertex in $V(H\cup P^{\prime})-\{x,y,u_{3},s,t\}$ has a neighbour in $R^{*}$ .

We claim that $t\notin V(L_{1}\cup P_{2})-\{x,y,u_{3}\}$ . Suppose not. By symmetry we may assume that $t\in V(L_{1})-\{x,u_{3}\}$ . Let $R_{1}$ be the induced $(y,t)$ -path in $G[V(P^{\prime}\cup R)-\{x\}]$ . When $u_{3}$ has no neighbour in $R_{1}^{*}$ , set $R_{2}:=R_{1}$ and $C:=R_{2}L_{1}(t,u_{3})u_{3}y$ . When $u_{3}$ has a neighbour in $R_{1}^{*}$ , let $t^{\prime}\in V(R_{1}^{*})$ be a neighbour of $u_{3}$ closest to $t$ and set $R_{2}:=u_{3}t^{\prime}R_{1}(t^{\prime},t)$ and $C:=R_{2}L_{1}(t,u_{3})$ . Note that $C_{4}\Delta C$ is a hole, although $C$ maybe not a hole. Since $C\Delta C_{1}\Delta C_{2}$ is an odd hole with length at least $2\ell+3$ when $C$ is an odd cycle, to prove the claim, it suffices to show that $|C|$ is odd. When $x$ has a neighbour in $R_{2}^{*}$ , since $|R_{2}|\geq 2\ell$ by (3.1) and the fact that $g(G)=2\ell+1$ , the subgraph $C_{4}\Delta C$ is an even hole, which implies that $C$ is an odd cycle. So we may assume that $x$ has no neighbour in $R_{2}^{*}$ . When $u_{3}$ is an end of $R_{2}$ , since $R_{2}$ is a chordal path of $C_{1}$ , it follows from Lemma 2.2 and (3.1) that $C$ is an odd hole. When $y$ is an end of $R_{2}$ and $u_{3}yR_{2}$ is a chordal path of $C_{1}$ , for the similar reason, $C$ is an odd hole. Hence, we may assume that $y$ is an end of $R_{2}$ and $u_{3}yR_{2}$ is not a chordal path of $C_{1}$ , implying $xs\in E(P^{\prime})$ and $s\in V(R_{2})$ . Since $P\subset R_{2}$ , we have $|R_{2}|>2\ell$ , so $C_{4}\Delta C$ is an even hole, implying that $C$ is an odd cycle. Hence, this proves the claim.

By symmetry we may therefore assume that $t\in V(P_{1})-\{u_{1}\}$ . Let $R_{1}$ be the induced $(y,t)$ -path in $G[V(P^{\prime}\cup R)-\{x\}]$ . By 3.1.1 and 3.1.2, either $xs\in E(P^{\prime})$ and $y$ has no neighbour in $R$ or some vertex in $\{x,u_{3}\}$ has a neighbour in $R_{1}^{*}$ . No matter which case happens, we have $|R_{1}|\geq 2\ell$ . That is, $R_{1}P_{2}(y,u_{4})$ is a chordal path of $C_{2}$ with length at least $3\ell-1$ , which is a contradiction to Lemma 2.2 as $t,u_{4}$ are non-adjacent. Hence, $\{x,y,u_{3}\}$ is a $P_{3}$ -cut of $G$ . ∎

By 3.1.6, there is a minimal vertex cut $X$ of $G$ with $\{u_{3},u_{4}\}\subseteq X\subset N[\{u_{3},u_{4}\}]$ and $\{u_{3},u_{4}\}=X\cap V(H)$ . Let $G_{1}$ be the induced subgraph of $G$ whose vertex set consists of $X$ and the vertex set of the component of $G-X$ containing $P_{2}^{\ast}$ . Since $\ell\geq 5$ , we have $|P_{2}|\geq 4$ by (3.1). If all induced $(u_{3},u_{4})$ -paths in $G_{1}$ have length $|P_{2}|$ , by Lemma 2.5, $G$ has a degree-2 vertex, a $K_{1}$ -cut or a $K_{2}$ -cut, which is not possible as $G$ is 4-vertex-critical. Hence, to finish the proof of Theorem 3.1, it suffices to show that all induced $(u_{3},u_{4})$ -paths in $G_{1}$ have length $|P_{2}|$ .

Let $Q$ be an arbitrary induced $(u_{3},u_{4})$ -path in $G_{1}$ . When $|L_{1}|\geq 2$ , since $QQ_{2}$ is a chordal path of $C_{1}$ by Lemma 2.6 (3) and the definition of $G_{1}$ , we have $|QQ_{2}|=|Q_{1}P_{1}|$ by Lemma 2.2, so $|Q|=|P_{1}|$ by (3.1). Hence, by (3.1) we may assume that $|L_{1}|=1$ and $|Q_{1}|=|P_{1}|=\ell$ . Since $Q_{1}L_{2}$ is an induced $(u_{3},u_{4})$ -path of length $\ell+1$ , either $|Q|=|P_{2}|=\ell$ or $|Q|\geq\ell+1$ and $|Q|$ has the same parity as $\ell+1$ . Assume that the latter case happens. Without loss of generality we may further assume that $Q$ is chosen with length at least $\ell+1$ and with $|P_{2}\cup Q|$ as small as possible. Since $|Q|$ and $|P_{2}|$ have different parity, $P_{2}\cup Q$ is not bipartite. Moreover, by the choice of $Q$ , the subgraph $P_{2}\cup Q$ contains a unique cycle $C$ and $|C|$ is odd. Since $Q=P_{2}\Delta C$ is an induced path of length at least $\ell+1$ , we have $|C\cap Q|>|C\cap P_{2}|\geq 2$ . So $C\Delta C_{3}\Delta C_{1}$ is an odd hole of length at least $2\ell+3$ , which is not possible. ∎

4 Acknowledgments

The authors thank the two referees for their careful reading of this manuscript and pointing out an error in our original version.

References

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Graphs with girth 2​ℓ+12\ell+1 and without longer odd holes that contain an odd K4K_{4}-subdivision

Abstract

1 Introduction

Conjecture 1.1.

Theorem 1.2.

2 Preliminaries

Lemma 2.1.

Lemma 2.2.

Proof.

Lemma 2.3.

Proof.

Lemma 2.4.

Proof.

Lemma 2.5.

Proof.

Lemma 2.6.

Proof.

3 Proof of Theorem 1.2

Theorem 3.1.

Proof.

3.1.1.

Subproof..

3.1.2.

Subproof..

3.1.3.

Subproof..

3.1.4.

Subproof..

3.1.5.

Subproof..

3.1.6.

Subproof..

4 Acknowledgments

References

Graphs with girth $2\ell+1$ and without longer odd holes that contain an odd $K_{4}$ -subdivision