Isolation of cycles

TL;DR

This paper introduces a new graph parameter measuring the minimum vertex set removal to eliminate all cycles, proves an upper bound for connected graphs, and confirms the bound's sharpness, solving an open problem.

Contribution

It establishes a tight upper bound of n/4 for the cycle-elimination parameter in connected graphs, addressing a previously open problem.

Findings

Bound of n/4 for connected non-triangle graphs

The bound is proven to be sharp

Solves an open problem by Caro and Hansberg

Abstract

For any graph $G$, let $\iota_{\rm c}(G)$ denote the size of a smallest set $D$ of vertices of $G$ such that the graph obtained from $G$ by deleting the closed neighbourhood of $D$ contains no cycle. We prove that if $G$ is a connected $n$-vertex graph that is not a triangle, then $\iota_{\rm c}(G) \leq n/4$. We also show that the bound is sharp. Consequently, we solve a problem of Caro and Hansberg.

Equations

$$B_{n,F}=\left(V(P_{b_{n,k}}^{\ast })\cup \bigcup _{i=1}^{a_{n,k}}V(F_i),E(P_{b_{n,k}}^{\ast })\cup \{iv:i\in [a_{n,k}],v\in V(F_i)\}\cup \bigcup _{i=1}^{a_{n,k}}E(F_i)\right).$$

$$\iota (n,\mathcal{F},F)=\max \left\{\iota (G,\mathcal{F}):G\text{mbox}isaconnectedgraph,V(G)=[n],G\not\simeq F\right\}.$$

$${\iota }_{\mathrm{c}}(G)\le \frac{n}{4}.$$

$${\iota }_{\mathrm{c}}(n)={\iota }_{\mathrm{c}}(B_{n,K_3})=\lfloor \frac{n}{4}\rfloor .$$

$$\iota (B_{n,F},F)=\lfloor \frac{n}{k+1}\rfloor .$$

$$\lfloor \frac{n}{k+1}\rfloor =\iota (B_{n,F},F)\le \iota (n,F)\le \iota (n,\mathcal{F},F).$$

$$\iota (n,K_3)=\iota (B_{n,K_3},K_3)=\lfloor \frac{n}{4}\rfloor .$$

$$\iota (G,\mathcal{F})\le |X|+\iota (G-Y,\mathcal{F}).$$

$$\iota (G,\mathcal{F})=\sum _{H\in \mathrm{C}(G)}\iota (H,\mathcal{F}).$$

$${\iota }_{\mathrm{c}}(H)\le \frac{|V(H)|}{4}\text{mbox}foreachH\in \mathcal{H}\backslash {\mathcal{H}}^{\prime }.$$

$${\iota }_{\mathrm{c}}(G)\le 1+{\iota }_{\mathrm{c}}(G^{\prime })=1+\sum _{H\in \mathcal{H}}{\iota }_{\mathrm{c}}(H)\le 1+\sum _{H\in \mathcal{H}}\frac{|V(H)|}{4}=\frac{4+n^{\prime }}{4}\le \frac{n}{4}.$$

$${\iota }_{\mathrm{c}}(G)\le 1+{\iota }_{\mathrm{c}}(G-U_x^{+})=1+\sum _{H\in \mathrm{C}(G-U_x^{+})}{\iota }_{\mathrm{c}}(H)\le \frac{|U_x^{+}|}{4}+\sum _{H\in \mathrm{C}(G-U_x^{+})}\frac{|V(H)|}{4}=\frac{n}{4}.$$

$${\iota }_{\mathrm{c}}(G)\le 1+{\iota }_{\mathrm{c}}(G-Y)=\frac{|Y|}{4}+\sum _{H\in \mathrm{C}(G-Y)}{\iota }_{\mathrm{c}}(H)\le \frac{|Y|}{4}+\sum _{H\in \mathrm{C}(G-Y)}\frac{|V(H)|}{4}=\frac{n}{4}.$$

Full text

1 Introduction

Unless stated otherwise, we use small letters such as $x$ to denote non-negative integers or elements of sets, and capital letters such as $X$ to denote sets or graphs. The set of positive integers is denoted by $\mathbb{N}$. For $n\geq 1$, $[n]$ denotes the set $\{1,\dots,n\}$ (that is, $[n]=\{i\in\mathbb{N}\colon i\leq n\}$). We take $[0]$ to be the empty set $\emptyset$. Arbitrary sets are assumed to be finite. For a set $X$, ${X\choose 2}$ denotes the set of $2$-element subsets of $X$ (that is, ${X\choose 2}=\{\{x,y\}\colon x,y\in X,x\neq y\}$).

If $Y$ is a subset of ${X\choose 2}$ and $G$ is the pair $(X,Y)$, then $G$ is called a graph, $X$ is called the vertex set of $G$ and is denoted by $V(G)$, and $Y$ is called the edge set of $G$ and is denoted by $E(G)$. A vertex of $G$ is an element of $V(G)$, and an edge of $G$ is an element of $E(G)$. We call $G$ an $n$-vertex graph if $|V(G)|=n$. We may represent an edge $\{v,w\}$ by $vw$. If $vw\in E(G)$, then we say that $w$ is a neighbour of $v$ in $G$ (and vice-versa). For $v\in V(G)$, $N_{G}(v)$ denotes the set of neighbours of $v$ in $G$, $N_{G}[v]$ denotes $N_{G}(v)\cup\{v\}$, and $d_{G}(v)$ denotes $|N_{G}(v)|$ and is called the degree of $v$ in $G$. For $S\subseteq V(G)$, $N_{G}[S]$ denotes $\bigcup_{v\in S}N_{G}[v]$ (the closed neighbourhood of $S$), $G[S]$ denotes $(S,E(G)\cap{S\choose 2})$ (the subgraph of $G$ induced by $S$), and $G-S$ denotes $G[V(G)\backslash S]$ (the graph obtained by deleting $S$ from $G$). Where no confusion arises, the subscript $G$ may be omitted from the notation above that uses it; for example, $N_{G}(v)$ may be abbreviated to $N(v)$.

If $G$ and $H$ are graphs, $f:V(H)\rightarrow V(G)$ is a bijection, and $E(G)=\{f(v)f(w)\,$: $vw\in E(H)\}$, then we say that $G$ is a copy of $H$, and we write $G\simeq H$. Thus, a copy of $H$ is a graph obtained by relabeling the vertices of $H$.

For $n\geq 1$, the graphs $([n],{[n]\choose 2})$ and $([n],\{\{i,i+1\}\colon i\in[n-1]\})$ are denoted by $K_{n}$ and $P_{n}$, respectively. For $n\geq 3$, $C_{n}$ denotes the graph $([n],\{\{1,2\},\{2,3\},\dots,\{n-1,n\},\{n,1\}\})$ ($=([n],E(P_{n})\cup\{n,1\})$). A copy of $K_{n}$ is called a complete graph. A copy of $P_{n}$ is called an $n$-path or simply a path. A copy of $C_{n}$ is called an $n$-cycle or simply a cycle. We call a $3$-cycle a triangle. Note that $K_{3}$ is the triangle $C_{3}$.

If $G$ and $H$ are graphs such that $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$, then $H$ is called a subgraph of $G$, and we say that $G$ contains $H$.

If $\mathcal{F}$ is a set of graphs and $F$ is a copy of a graph in $\mathcal{F}$, then we call $F$ an $\mathcal{F}$-graph. If $G$ is a graph and $D\subseteq V(G)$ such that $G-N[D]$ contains no $\mathcal{F}$-graph, then $D$ is called an $\mathcal{F}$-isolating set of $G$. Let $\iota(G,\mathcal{F})$ denote the size of a smallest $\mathcal{F}$-isolating set of $G$. We abbreviate $\iota(G,\{F\})$ to $\iota(G,F)$. The study of isolating sets was introduced by Caro and Hansberg [1]. It is an appealing and natural generalization of the classical domination problem [2, 3, 4, 5, 6, 7]. Indeed, $D$ is a $\{K_{1}\}$-isolating set of $G$ if and only if $D$ is a dominating set of $G$ (that is, $N[D]=V(G)$), so the $\{K_{1}\}$-isolation number is the domination number (the size of a smallest dominating set). Let $\mathcal{C}$ denote $\{C_{k}\colon k\geq 3\}$. In this paper, we obtain a sharp upper bound for $\iota(G,\mathcal{C})$, and consequently we solve a problem of Caro and Hansberg [1].

We call a subset $D$ of $V(G)$ a cycle isolating set of $G$ if $G-N[D]$ contains no cycle (that is, $G-N[D]$ is a forest). We denote the size of a smallest cycle isolating set of $G$ by $\iota_{\rm c}(G)$. Thus, $\iota_{\rm c}(G)=\iota(G,\mathcal{C})$.

If $G_{1},\dots,G_{t}$ are graphs such that $V(G_{i})\cap V(G_{j})=\emptyset$ for every $i,j\in[t]$ with $i\neq j$, then $G_{1},\dots,G_{t}$ are vertex-disjoint. A graph $G$ is connected if, for every $v,w\in V(G)$, $G$ contains a path $P$ with $v,w\in V(P)$. A connected subgraph $H$ of $G$ is a component of $G$ if, for each connected subgraph $K$ of $G$ with $K\neq H$, $H$ is not a subgraph of $K$. Clearly, any two distinct components of $G$ are vertex-disjoint.

For $n,k\in\mathbb{N}$, let $a_{n,k}=\left\lfloor\frac{n}{k+1}\right\rfloor$ and $b_{n,k}=n-ka_{n,k}$. Thus, $a_{n,k}\leq b_{n,k}\leq a_{n,k}+k$. If $F$ is a $k$-vertex graph and $n\leq k$, then let $B_{n,F}=P_{n}$. If $F$ is a $k$-vertex graph and $n\geq k+1$, then let $F_{1},\dots,F_{a_{n,k}}$ be copies of $F$ such that $P_{b_{n,k}},F_{1},\dots,F_{a_{n,k}}$ are vertex-disjoint, let $P_{b_{n,k}}^{*}=([b_{n,k}],E(P_{a_{n,k}})\cup\{a_{n,k}i\colon i\in[b_{n,k}]\backslash[a_{n,k}]\})$, and let $B_{n,F}$ be the connected $n$-vertex graph given by

[TABLE]

Thus, $B_{n,F}$ is the graph obtained by taking $P_{b_{n,k}}^{*},F_{1},\dots,F_{a_{n,k}}$ and joining $i$ (a vertex of $P_{b_{n,k}}^{*}$) to each vertex of $F_{i}$ for each $i\in[a_{n,k}]$.

For any $n\in\mathbb{N}$, any family $\mathcal{F}$ of graphs, and any $F\in\mathcal{F}$, let

[TABLE]

We abbreviate $\iota(n,\{F\},F)$ to $\iota(n,F)$. Let $\iota_{\rm c}(n)=\iota(n,\mathcal{C},K_{3})$. In Section 2, we prove the following result.

Theorem 1.1

If $G$ is a connected $n$-vertex graph that is not a triangle, then

[TABLE]

Consequently, for any $n\geq 1$,

[TABLE]

The equality $\iota_{\rm c}(B_{n,K_{3}})=\left\lfloor\frac{n}{4}\right\rfloor$ is generalized in the following result.

Lemma 1.2

*Let $n,k\in\mathbb{N}$ and let $F$ be a $k$-vertex graph.

(i) If $n\neq k$ or $F\not\simeq P_{k}$, then*

[TABLE]

(ii) If $\mathcal{F}$ is a family of graphs, $F\in\mathcal{F}$, and $n\neq k=|V(F)|=\min\{|V(H)|\colon H\in\mathcal{F}\}$, then

[TABLE]

Proof. Let $B=B_{n,F}$. If either $n<k$ or $n=k$ and $F\not\simeq P_{k}$, then $\iota(B,F)=0$. Suppose $n\geq k+1$. Then, $\iota(B,F)\leq a_{n,k}$ as $[a_{n,k}]$ is a dominating set of $B$. Let $D$ be an $\{F\}$-isolating set of $B$ of size $\iota(B,F)$. For each $i\in[a_{n,k}]$, $D\cap(V(F_{i})\cup\{i\})\neq\emptyset$ as $B-N_{B}[D]$ does not contain the copy $F_{i}$ of $F$. Thus, $|D|\geq a_{n,k}$. Hence, (i) is proved.

Let $\mathcal{F}$ and $n$ be as in (ii). Since $n\neq|V(F)|$, $B\not\simeq F$. Since $|V(B)|=n$, we can choose a copy $B^{\prime}$ of $B$ with $V(B^{\prime})=[n]$. Since $B$ is connected, $B^{\prime}$ is connected. Thus, $\iota(B^{\prime},F)\leq\iota(n,F)$. Since $F\in\mathcal{F}$, the $\mathcal{F}$-isolating sets of a graph $G$ are $\{F\}$-isolating sets of $G$, so $\iota(G,F)\leq\iota(G,\mathcal{F})$. Thus, $\iota(n,F)\leq\iota(n,\mathcal{F},F)$. Now $\iota(n,B^{\prime})=\iota(n,B)=\left\lfloor\frac{n}{k+1}\right\rfloor$ by (i). Hence, (ii) is proved.  $\Box$

By the results above, $\left\lfloor\frac{n}{4}\right\rfloor$ is a sharp upper bound on $\iota(G,K_{3})$ for connected $n$-vertex graphs $G\not\simeq K_{3}$.

Theorem 1.3

For any $n\geq 1$,

[TABLE]

Proof. Let $G$ be a connected $n$-vertex graph that is not a copy of $K_{3}$. By Lemma 1.2 (ii), $\left\lfloor\frac{n}{4}\right\rfloor=\iota(B_{n,K_{3}},K_{3})\leq\iota(n,K_{3})\leq\iota(n,\mathcal{C},K_{3})$ as $K_{3}=C_{3}\in\mathcal{C}$. By Theorem 1.1, $\iota(n,\mathcal{C},K_{3})=\left\lfloor\frac{n}{4}\right\rfloor$. The result follows.  $\Box$

In [1], Caro and Hansberg showed that $\frac{1}{4}\leq\limsup_{n\rightarrow\infty}\frac{\iota_{\rm c}(n)}{n}\leq\frac{1}{3}$. In Problem 7.3 of the same paper, they asked for the value of $\limsup_{n\rightarrow\infty}\frac{\iota_{\rm c}(n)}{n}$. The answer is immediately given by Theorem 1.1.

Corollary 1.4

$\limsup_{n\rightarrow\infty}\frac{\iota_{\rm c}(n)}{n}=\frac{1}{4}$.

Proof. By Theorem 1.1, for any $n\in\mathbb{N}$, we have $\frac{1}{4}-\frac{3}{4n}=\frac{1}{n}\left(\frac{n-3}{4}\right)\leq\frac{\iota_{\rm c}(n)}{n}\leq\frac{1}{4}$, and, if $n$ is a multiple of $4$, then $\frac{\iota_{\rm c}(n)}{n}=\frac{1}{4}$. Thus, $\lim_{n\rightarrow\infty}\sup\left\{\frac{\iota_{\rm c}(k)}{k}\colon k\geq n\right\}=\lim_{n\rightarrow\infty}\frac{1}{4}=\frac{1}{4}$.  $\Box$

2 Proof of Theorem 1.1

In this section, we prove Theorem 1.1. We start with two lemmas that will be used repeatedly.

Lemma 2.1

If $G$ is a graph, $\mathcal{F}$ is a set of graphs, $X\subseteq V(G)$, and $Y\subseteq N[X]$, then

[TABLE]

Proof. Let $D$ be an $\mathcal{F}$-isolating set of $G-Y$ of size $\iota(G-Y,\mathcal{F})$. Clearly, $\emptyset\neq V(F)\cap Y\subseteq V(F)\cap N[X]$ for each $\mathcal{F}$-graph $F$ that is a subgraph of $G$ and not a subgraph of $G-Y$. Thus, $D\cup X$ is an $\mathcal{F}$-isolating set of $G$. The result follows.  $\Box$

For a graph $G$ and a set $\mathcal{F}$ of graphs, let ${\rm C}(G)$ denote the set of components of $G$, and let {\rm C}(G,\mathcal{F})=\{H\in{\rm C}(G)\colon H\mbox{ is an \mathcal{F}-graph}\}. We abbreviate ${\rm C}(G,\{K_{3}\})$ to ${\rm C}^{\prime}(G)$. Thus, ${\rm C}^{\prime}(G)=\{H\in{\rm C}(G)\colon H\mbox{ is a triangle}\}$.

Lemma 2.2

If $G$ is a graph and $\mathcal{F}$ is a set of graphs, then

[TABLE]

Proof. For each $H\in{\rm C}(G)$, let $D_{H}$ be a smallest $\mathcal{F}$-isolating set of $H$. Then, $\bigcup_{H\in{\rm C}(G)}D_{H}$ is an $\mathcal{F}$-isolating set of $G$, so $\iota(G,\mathcal{F})\leq\sum_{H\in{\rm C}(G)}|D_{H}|=\sum_{H\in{\rm C}(G)}\iota(H,\mathcal{F})$. Let $D$ be a smallest $\mathcal{F}$-isolating set of $G$. For each $H\in{\rm C}(G)$, $D\cap V(H)$ is an $\mathcal{F}$-isolating set of $H$. We have $\sum_{H\in{\rm C}(G)}\iota(H,\mathcal{F})\leq\sum_{H\in{\rm C}(G)}|D\cap V(H)|=|D|=\iota(G,\mathcal{F})$. The result follows.  $\Box$

Proof of Theorem 1.1. Let us first assume the bound in the first part of the theorem. Then, $\iota_{\rm c}(n)\leq\frac{n}{4}$. Since $\iota_{\rm c}(n)$ is an integer, $\iota_{\rm c}(n)\leq\left\lfloor\frac{n}{4}\right\rfloor$. Together with Lemma 1.2 (i), this gives us $\iota_{\rm c}(n)=\iota_{\rm c}(B_{n,K_{3}},K_{3})=\left\lfloor\frac{n}{4}\right\rfloor$.

We now prove the first part of the theorem. We use induction on $n$. Let $G$ be a connected $n$-vertex graph that is not a triangle. If $n\leq 3$, then, since $G$ is not a triangle, $G$ contains no cycle, and hence $\iota_{\rm c}(G)=0$. Suppose $n\geq 4$. Let $k=\max\{d(v)\colon v\in V(G)\}$. Since $G$ is connected, $k\geq 2$. Let $v\in V(G)$ such that $d(v)=k$. If $k=2$, then $G$ is a path or a cycle, $\{v\}$ is a cycle isolating set of $G$, and hence $\iota_{\rm c}(G)\leq 1\leq\frac{n}{4}$. Suppose $d(v)\geq 3$. Then, $|N[v]|\geq 4$. If $V(G)=N[v]$, then $\{v\}$ is a cycle isolating set of $G$, so $\iota(G)\leq 1\leq\frac{n}{4}$. Suppose $V(G)\neq N[v]$. Let $G^{\prime}=G-N[v]$ and $n^{\prime}=|V(G^{\prime})|$. Then, $n\geq n^{\prime}+4$ and $V(G^{\prime})\neq\emptyset$. Let $\mathcal{H}={\rm C}(G^{\prime})$ and $\mathcal{H}^{\prime}={\rm C}^{\prime}(G^{\prime})$. By the induction hypothesis,

[TABLE]

If $\mathcal{H}^{\prime}=\emptyset$, then, by Lemma 2.1 (with $X=\{v\}$ and $Y=N[v]$) and Lemma 2.2,

[TABLE]

Suppose $\mathcal{H}^{\prime}\neq\emptyset$. For any $H\in\mathcal{H}$ and any $x\in N(v)$ such that $xy\in E(G)$ for some $y\in V(H)$, we say that $H$ is linked to $x$ and that $x$ is linked to $H$. Since $G$ is connected, each member of $\mathcal{H}$ is linked to at least one member of $N(v)$. Let $L=\{x\in N(v)\colon x\mbox{ is linked to some member of }\mathcal{H}^{\prime}\}$. Since $\mathcal{H}^{\prime}\neq\emptyset$, $L\neq\emptyset$. Let $x\in L$. Let $\mathcal{H}^{\prime}_{x}=\{H\in\mathcal{H}^{\prime}\colon H\mbox{ is linked to }x\}$ and \mathcal{H}_{x}^{*}=\{H\in\mathcal{H}\backslash\mathcal{H}^{\prime}\colon H\mbox{ is linked to x only}\}. Let $U=\bigcup_{H\in\mathcal{H}^{\prime}}V(H)$. Let $U_{x}=N(x)\cap U$ and $U_{x}^{+}=\{x\}\cup U_{x}$. Note that if a component $A$ of $G-U_{x}^{+}$ is a triangle, then $V(A)=N[v]\backslash\{x\}$.

Suppose $|U_{x}|\geq 3$. If no component of $G-U_{x}^{+}$ is a triangle, then, by Lemma 2.1 (with $X=\{x\}$ and $Y=U_{x}^{+}$), Lemma 2.2, and the induction hypothesis, we have

[TABLE]

Suppose that a component $A$ of $G-U_{x}^{+}$ is a triangle. Then, $V(A)=N[v]\backslash\{x\}$. Let $Y=U_{x}^{+}\cup V(A)$. Since $G-Y$ contains no triangle, $\iota_{\rm c}(G-Y)\leq\frac{n-|Y|}{4}$ by Lemma 2.2 and the induction hypothesis. Let $D_{G-Y}$ be a cycle isolating set of $G-Y$ of size $\iota_{\rm c}(G-Y)$. Since $v\in N(x)\cap V(A)$ and $U_{x}^{+}\subset N[x]$, $\{x\}\cup D_{G-Y}$ is a cycle isolating set of $G$. Thus, $\iota_{\rm c}(G)\leq 1+\frac{n-|Y|}{4}$. Since $|Y|\geq 7$, $\iota_{\rm c}(G)<\frac{n}{4}$.

Now suppose $|U_{x}|\leq 2$. Then, $1\leq|\mathcal{H}_{x}^{\prime}|\leq 2$.

Case 1: $|\mathcal{H}_{x}^{\prime}|=1$. Let $T$ be the member of $\mathcal{H}_{x}^{\prime}$. We have $xy\in E(G)$ for some $y\in V(T)$. Let $Y=\{x\}\cup V(T)$. Then, $Y\subseteq N[y]$. Also, $G-Y$ has a component $T^{\prime}$ with $N[v]\backslash\{x\}\subseteq V(T^{\prime})$. Since $T$ is the only member of $\mathcal{H}^{\prime}$ that is linked to $x$, the components of $G-Y$ are $T^{\prime}$ and the members of $\mathcal{H}_{x}^{*}$. Recall that no member of $\mathcal{H}_{x}^{*}$ is a triangle.

If $T^{\prime}$ is not a triangle, then, by Lemma 2.1 (with $X=\{y\}$), Lemma 2.2, and the induction hypothesis, we have $\iota_{\rm c}(G)\leq 1+\iota_{\rm c}(G-Y)\leq 1+\frac{n-|Y|}{4}=\frac{n}{4}$.

Suppose that $T^{\prime}$ is a triangle. Let $W=V(T)\cup V(T^{\prime})$. We have $x\notin W$, $y\in N(x)\cap V(T)$, and $v\in N(x)\cap V(T^{\prime})$. Also, the components of $G-\{x\}$ are the components of $G[W]$ and the members of $\mathcal{H}_{x}^{*}$. By the induction hypothesis, each member $H$ of $\mathcal{H}_{x}^{*}$ has a cycle isolating set $D_{H}$ with $|D_{H}|\leq\frac{|V(H)|}{4}$.

Suppose that $G[W]-(N(x)\cap W)$ contains no cycle. Then, $\{x\}\cup\bigcup_{H\in\mathcal{H}_{x}^{*}}D_{H}$ is a cycle isolating set of $G$. Thus, $\iota_{\rm c}(G)\leq 1+\sum_{H\in\mathcal{H}_{x}^{*}}\frac{|V(H)|}{4}=1+\frac{n-|\{x\}\cup W|}{4}=1+\frac{n-7}{4}<\frac{n}{4}$.

Now suppose that $G[W]-(N(x)\cap W)$ contains a cycle $A$. Since $v,y\in N(x)\cap W$ and $|W|=6$, either $A\simeq C_{3}$ or $A\simeq C_{4}$.

Suppose $A\simeq C_{4}$. Then, $N(x)\cap W=\{v,y\}$, $V(A)=(V(T)\backslash\{y\})\cup(V(T^{\prime})\backslash\{v\})$, and hence $uw\in E(A)\subseteq E(G)$ for some $u\in V(T)\backslash\{y\}$ and some $w\in V(T^{\prime})\backslash\{v\}$. Let $Z=\{w\}\cup V(T)$ and let $x^{\prime}$ be the member of $V(T^{\prime})\backslash\{v,w\}$. We have $V(G-Z)=\{v,x,x^{\prime}\}\cup\bigcup_{H\in\mathcal{H}_{x}^{*}}V(H)$ and $x,x^{\prime}\in N(v)$. Thus, since the members of $\mathcal{H}_{x}^{*}$ are linked to $x$, $G-Z$ is connected. Since $N(x)\cap W=\{v,y\}$, we have $xx^{\prime}\notin E(G-Z)$, so $G-Z$ is not a triangle. By the induction hypothesis, $\iota_{\rm c}(G-Z)\leq\frac{n-|Z|}{4}=\frac{n-4}{4}$. Since $Z\subseteq N[u]$, Lemma 2.1 (with $X=\{u\}$) gives us $\iota_{\rm c}(G)\leq 1+\iota_{\rm c}(G-Z)\leq\frac{n}{4}$.

Now suppose $A\simeq C_{3}$. Since $V(A)\subseteq W\backslash N(x)\subseteq(V(T)\cup V(T^{\prime}))\backslash\{v,y\}$, $V(A)$ contains either the two vertices in $V(T)\backslash\{y\}$ and one of the two vertices in $V(T^{\prime})\backslash\{v\}$ or the two vertices in $V(T^{\prime})\backslash\{v\}$ and one of the two vertices in $V(T)\backslash\{y\}$. Suppose $|V(A)\cap(V(T^{\prime})\backslash\{v\})|=1$. Then, $V(T)\backslash\{y\}\subseteq V(A)$. Let $x^{\prime}$ be the member of $V(A)\cap(V(T^{\prime})\backslash\{v\})$. Let $Z=(V(T)\backslash\{y\})\cup V(T^{\prime})$. Since $V(A)\cup V(T^{\prime})\subseteq N[x^{\prime}]$, $Z\subseteq N[x^{\prime}]$. We have $V(G-Z)=\{x,y\}\cup\bigcup_{H\in\mathcal{H}_{x}^{*}}V(H)$. Since $xy\in E(G)$ and the members of $\mathcal{H}_{x}^{*}$ are linked to $x$, $G-Z$ is connected. Since $N(y)\cap\left(\bigcup_{H\in\mathcal{H}_{x}^{*}}V(H)\right)=\emptyset$, $G-Z$ is not a triangle. By the induction hypothesis, $\iota_{\rm c}(G-Z)\leq\frac{n-|Z|}{4}=\frac{n-5}{4}$. Since $Z\subseteq N[x^{\prime}]$, Lemma 2.1 (with $X=\{x^{\prime}\}$) gives us $\iota_{\rm c}(G)\leq 1+\iota_{\rm c}(G-Z)<\frac{n}{4}$. Similarly, $\iota_{\rm c}(G)<\frac{n}{4}$ if $|V(A)\cap(V(T)\backslash\{y\})|=1$.

Case 2: $|\mathcal{H}_{x}^{\prime}|=2$. Let $T_{1}$ and $T_{2}$ be the two members of $\mathcal{H}_{x}^{\prime}$.

Suppose that $T_{2}$ is linked to a member of $L\backslash\{x\}$. Since $T_{1}$ is linked to $x$, $xy\in E(G)$ for some $y\in V(T_{1})$. Let $Y=\{x\}\cup V(T_{1})$. Then, no component of $G-Y$ is a triangle ($G-Y$ has a component $A$ with $(N[v]\backslash\{x\})\cup V(T_{2})\cup\bigcup_{H\in\mathcal{H}^{\prime}\backslash\{T_{1},T_{2}\}}V(H)\subseteq V(A)$, and the other components of $G-Y$ are the members of $\mathcal{H}_{x}^{*}$). Also, $Y\subseteq N[y]$. By Lemma 2.1 (with $X=\{y\}$), Lemma 2.2, and the induction hypothesis, we have

[TABLE]

Similarly, $\iota_{\rm c}(G)\leq\frac{n}{4}$ if $T_{1}$ is linked to a member of $L\backslash\{x\}$.

Now suppose that, for each $i\in\{1,2\}$ and each $x^{\prime}\in L\backslash\{x\}$, $T_{i}$ is not linked to $x^{\prime}$. Then, $T_{1}$ and $T_{2}$ are components of $G-\{x\}$. Let $Y=\{x\}\cup V(T_{1})\cup V(T_{2})$.

Suppose that no component of $G-\{x\}$ other than $T_{1}$ and $T_{2}$ is a triangle. Then, $\iota_{\rm c}(G-Y)\leq\frac{n-|Y|}{4}$ by Lemma 2.2 and the induction hypothesis. Let $D_{G-Y}$ be a cycle isolating set of $G-Y$ of size $\iota_{\rm c}(G-Y)$. Since $N(x)\cap V(T_{1})\neq\emptyset\neq N(x)\cap V(T_{2})$, $\{x\}\cup D_{G-Y}$ is a cycle isolating set of $G$, so $\iota_{\rm c}(G)\leq 1+\frac{n-|Y|}{4}=1+\frac{n-7}{4}<\frac{n}{4}$.

Now suppose that $G-\{x\}$ has a component $T_{3}$ such that $T_{3}\notin\{T_{1},T_{2}\}$ and $T_{3}$ is a triangle. Since $T_{1}$ and $T_{2}$ are the only members of $\mathcal{H}^{\prime}$ that are linked to $x$, it follows that $V(T_{3})=N[v]\backslash\{x\}$. Let $Y^{\prime}=Y\cup V(T_{3})$. Since $G-Y^{\prime}$ contains no triangle, $\iota_{\rm c}(G-Y^{\prime})\leq\frac{n-|Y^{\prime}|}{4}$ by Lemma 2.2 and the induction hypothesis. Let $D_{G-Y^{\prime}}$ be a cycle isolating set of $G-Y^{\prime}$ of size $\iota_{\rm c}(G-Y^{\prime})$. Since $v\in N(x)\cap V(T_{3})$ and $N(x)\cap V(T_{1})\neq\emptyset\neq N(x)\cap V(T_{2})$, $\{x\}\cup D_{G-Y^{\prime}}$ is a cycle isolating set of $G$. Thus, $\iota_{\rm c}(G)\leq 1+\frac{n-|Y^{\prime}|}{4}=1+\frac{n-10}{4}<\frac{n}{4}$.  $\Box$