Generalized power domination in claw-free regular graphs

TL;DR

This paper provides counterexamples to a conjecture on k-power domination in regular graphs and establishes tight bounds for k-power domination in connected claw-free regular graphs.

Contribution

It disproves a previous conjecture and derives new tight bounds for k-power domination in claw-free regular graphs.

Findings

Counterexamples negate the conjecture on k-power domination.

Bound of n/(k+l+2) for k-power domination in certain claw-free regular graphs.

The bounds are proven to be tight.

Abstract

In this paper, we give a series of couterexamples to negate a conjecture and hence answer an open question on the $k$-power domination of regular graphs (see [P. Dorbec et al., SIAM J. Discrete Math., 27 (2013), pp. 1559-1574]). Furthermore, we focus on the study of $k$-power domination of claw-free graphs. We show that for $l\in\{2,3\}$ and $k\ge l$, the $k$-power domination number of a connected claw-free $(k+l+1)$-regular graph on $n$ vertices is at most $\frac{n}{k+l+2}$, and this bound is tight.

Equations

$$|N(u^{\prime })\setminus P^j(S_{t+1})|\le k+l+1-|L_u\setminus u^{\prime }|-|\{u\}|\le l\le k.$$

$$|P^{\infty }(S_{t+1})|\ge |P^{\infty }(S_t)|+|L_u|+|F_u\cap \overline{M}|\ge |P^{\infty }(S_t)|+k+l+2.$$

$$|P^{\infty }(S_{t+1})|\ge |P^{\infty }(S_t)|+|L_x|+|L_u|\ge |P^{\infty }(S_t)|+2(k+1)\ge |P^{\infty }(S_t)|+k+l+2.$$

$$n=|P^{\infty }(S_q)|\ge |P^0(S_0)|+q(k+l+2)=(|S_0|+q)(k+l+2)=|S_q|(k+l+2).$$

Full text

Key words. power domination, electrical systems monitoring, domination, regular graphs, claw-free graphs

AMS subject classification. 05C69

1 Introduction

In this paper, we only consider simple graphs. Let $G=(V(G),E(G))$ (abbreviated as $G=(V,E)$) be a graph. The open neighborhood $N_{G}(v)$ of a vertex $v$ consists of the vertices adjacent to $v$ and its closed neighborhood is $N_{G}[v]=N_{G}(v)\cup\{v\}$. The open neighborhood of a subset $S\subseteq V$ is the set $N_{G}(S)=\bigcup_{v\in S}N_{G}(v)$ and its closed neighborhood is $N_{G}[S]=N_{G}(S)\cup S$. The degree of a vertex $v$, denoted $d_{G}(v)$, is the size of its open neighborhood $|N_{G}(v)|$. Let $v$ be a vertex of $G$ and $F$ be a subset of $V$. We denote $N_{F}(v)=N_{G}(v)\cap F$, $N_{F}[v]=N_{G}[v]\cap F$ and $d_{F}(v)=|N_{G}(v)\cap F|$. A graph $G$ is $k$-regular if $d_{G}(v)=k$ for every vertex $v\in V$. If the graph $G$ is clear from the context, we will omit the subscripts $G$ for convenience. The complete bipartite graph with partite sets of cardinality $i$ and $j$ is denoted by $K_{i,j}$. A claw-free graph is a graph that does not contain a claw, i.e. $K_{1,3}$, as an induced subgraph. For a set $S\subseteq V$, we let $G[S]$ denote the subgraph induced by $S$. We say a subset $S\subseteq V$ is a packing if the vertices in $S$ are pairwise at distance at least three apart in $G$.

Electric power systems must be monitored continually. One way of monitoring these systems is to place phase measurement units (PMUs) at selected locations. Since the cost of a PMU is very high, it is desirable to minimize the number of PMUs. The authors of [3, 18] introduced power domination to model the problem of monitoring electrical systems. Then, the problem was formulated as a graph theoretical problem by Haynes et al. in [14]. Some additional propagation in power domination is using the Kirschoff’s laws in electrical systems. The definition of power domination was simplified to the following definition independently in [9, 10, 13, 16], which originally asked the systems to monitor both edges and vertices.

Definition 1.1**.**

(Power Dominating Set). Let $G=(V,E)$ be a graph. A subset $S$ of $V$ is a power dominating set (abbreviated as PDS) of $G$ if and only if all vertices of $V$ are observed either by Observation Rule 1 (abbreviated as OR 1) initially or by Observation Rule 2 (abbreviated as OR 2) recursively.

OR 1.* all vertices in $N_{G}[S]$ are observed initially.*

OR 2.* If an observed vertex $v$ has all neighbors observed except one neighbor $u$, then $u$ is observed (by $v$).*

The power domination number $\gamma_{p}(G)$ is the minimum cardinality of a PDS of $G$. The power domination problem is known to be NP-complete (see [1, 2, 13, 14]). Linear-time algorithms for this problem were presented for trees, interval graphs and block graphs (see [14, 16, 22]). The Nordhaus-Gaddum problems for power domination were investigated in [4] and parameterized results were given in [15]. The exact values of the power domination numbers of some special graphs were studied in [9, 10]. The upper bounds for the power domination numbers of regular graphs were investigated (see, for example, [19, 21]).

Chang et al. [6] generalized power domination to $k$-power domination. In here, we use a definition of monitored set to define $k$-power dominating set.

Definition 1.2**.**

(Monitored Set). Let $G=(V,E)$ be a graph, let $S\subseteq V$, and let $k\geq 0$ be an integer. We define the sets $(P^{i}_{G}(S))_{i\geq 0}$ of vertices monitored by $S$ at step $i$ by the following rules:

(1) $P_{G}^{0}(S)=N_{G}[S]$;

(2) $P_{G}^{i+1}(S)=\cup\{N_{G}[v]:v\in P_{G}^{i}(S)$ such that $|N_{G}[v]\setminus P_{G}^{i}(S)|\leq k\}$.

It is clear that $P_{G}^{i}(S)\subseteq P_{G}^{i+1}(S)\subseteq V$ for any $i$. If $P_{G}^{i_{0}}(S)=P_{G}^{i_{0}+1}(S)$ for some $i_{0}$, then $P_{G}^{j}(S)=P_{G}^{i_{0}}(S)$ for every $j\geq i_{0}$ and we accordingly define $P^{\infty}_{G}(S)=P_{G}^{i_{0}}(S)$.

Definition 1.3**.**

($k$-Power Dominating Set). Let $G=(V,E)$ be a graph, let $S\subseteq V$, and let $k\geq 0$ be an integer. If $P^{\infty}_{G}(S)=V$, then $S$ is called a $k$-power dominating set of $G$, abbreviated $k$-PDS. The $k$-power domination number of $G$, denoted by $\gamma_{p,k}(G)$, is the minimum cardinality of a $k$-PDS in $G$.

The $k$-power domination problem is known to be NP-complete for chordal graphs and bipartite graphs [6]. Linear-time algorithms for this problem were presented for trees [6] and block graphs [20]. The bounds for the $k$-power domination numbers in regular graphs were obtained in [6, 7]. The relationship between the $k$-forcing and the $k$-power domination numbers of a graph was given in [12]. The authors of [8] studied the exact values for the $k$-power domination numbers in Sierpi$\acute{n}$ski graphs.

If $G$ is a connected $(k+1)$-regular graph, then $\gamma_{p,k}(G)=1$. Some scholars began to study the $k$-power domination number of $(k+2)$-regular graphs. Zhao et al. [19] showed that if $G$ is a $3$-regular claw-free graph on $n$ vertices, then $\gamma_{p,1}(G)\leq\frac{n}{4}$. Chang et al. [6] generalized this result to $(k+2)$-regular claw-free graphs. Dorbec et al. [7] removed the claw-free condition and show that $\gamma_{p,k}(G)\leq\frac{n}{k+3}$ if $G$ is a $(k+2)$-regular graph on $n$ vertices. Moreover, they presented the following conjecture and question.

Conjecture 1.4**.**

([7]) For $k\geq 1$ and $r\geq 3$, if $G\not\cong K_{r,r}$ is a connected $r$-regular graph of order $n$, then $\gamma_{p,k}(G)\leq\frac{n}{r+1}$.

Question 1.5**.**

([7]) For $r\geq 3$, let $G\neq K_{r,r}$ is a connected $r$-regular graph of order $n$. Determine the smallest positive value, $k_{\min}(r)$, of $k$ such that $\gamma_{p,k}(G)\leq\frac{n}{r+1}$.

The result of Dorbec et al. in [7] implies that Conjecture 1.4 holds for $k=1$ and $r=3$ and $k_{min}(r)\leq r-2$. Recently, Lu et al. [17] showed that Conjecture 1.4 does not always hold for each even $r\geq 4$ and $k=1$. In this paper, we show that $k_{\min}(r)=r-2$ for $r\geq 3$ and negate Conjecture 1.4 for each $r\geq 4$ and $1\leq k\leq r-3$. We also show that there exists a series of claw-free $r$-regular graphs $G$ of order $n$ such that $\gamma_{p,k}(G)>\frac{n}{r}$ if $k<\lfloor\frac{r}{2}\rfloor$. But Conjecture 1.4 may hold for claw-free $r$-regular graphs if $k\geq\lfloor\frac{r}{2}\rfloor$. The following theorem is the main result in this paper.

Theorem 1.6**.**

For $l\in\{2,3\}$ and $k\geq l$, if $G$ is a connected claw-free $(k+l+1)$-regular graph of order $n$, then $\gamma_{p,k}(G)\leq\frac{n}{k+l+2}$ and the bound is tight.

2 Counterexamples

Motivated by the concept of a fort proposed in [5], we define the concept of a $k$-fort, which is a natural generalization of a fort.

Definition 2.1**.**

($k$-fort). For an integer $k\geq 1$, a $k$-fort of a graph $G$ is a nonempty set $F\subseteq V$ such that each vertex of $N_{G}(F)\backslash F$ is adjacent to at least $k+1$ vertices in $F$.

If $F$ is a $k$-fort of $G$, then $|F|\geq k+1$. We immediately obtain the following proposition.

Proposition 2.2**.**

Let $G=(V,E)$ be a graph and $F$ be a $k$-fort of $G$. If $S$ is a $k$-PDS of $G$, then $S\cap N_{G}[F]\neq\emptyset$.

Observation 2.3**.**

For each $r\geq 4$ and $q\geq 2$, there exists a connected $r$-regular graph $D_{r,q}\neq K_{r,r}$ of order $n=2qr$ such that $\gamma_{p,r-3}(D_{r,q})=2q=\frac{n}{r}>\frac{n}{r+1}$.

Proof.

We define the graph $D_{r,q}$ as follows: Take $q$ disjoint copies $D_{i}\cong K_{r,r}-x_{i}y_{i}$, where $x_{i},y_{i}\in V(K_{r,r})$ and $i\in\{1,2,\cdots q\}$. Then add edges $y_{i}x_{i+1}$ for each $i\in\{1,2,\cdots,q\}$, where $x_{q+1}=x_{1}$ (see Figure 1). Suppose that $T=\bigcup_{i=1}^{q}\{x_{i},y_{i}\}$ and $k=r-3$. It is clear that $T$ is a $k$-PDS of $D_{r,q}$. Then, we have $\gamma_{p,k}(D_{r,q})\leq|T|\leq 2q$. Now, we show $\gamma_{p,k}(D_{r,q})\geq 2q$. Let $S$ be a $k$-PDS of $D_{r,q}$. Assume that $(X_{i},Y_{i})$ is the bipartition of $D_{i}$, where $x_{i}\in X_{i}$, $y_{i}\in Y_{i}$ and $i\in\{1,2,\cdots,q\}$. We claim that $|S\cap V(D_{i})|\geq 2$ for each $i\in\{1,2,\cdots,q\}$. Otherwise, without loss of generality, suppose that $|S\cap V(D_{1})|\leq 1$ and $S\cap Y_{1}=\emptyset$. Then $F=X_{1}\backslash(S\cup\{x_{1}\})$ is a $k$-fort and $N_{D_{r,q}}[F]\cap S=\emptyset$, contradicting Proposition 2.2. ∎

By Observation 2.3, we know that Conjecture 1.4 does not always hold for each $r\geq 4$ and $1\leq k\leq r-3$, and hence $k_{\min}(r)=r-2$ for $r\geq 3$. A natural problem is whether $\frac{n}{r}$ is always the upper bound of $\gamma_{p,k}(G)$ in Conjecture 1.4. We will discuss this problem using the relation between $k$-power domination and total domination in regular graphs.

A set $S$ of vertices in a graph $G$ is called a total domination set (abbreviated as TDS) of $G$ if every vertex of $G$ is adjacent to some vertex in $S$. The minimum cardinality of a TDS of $G$ is the total domination number of $G$, denoted by $\gamma_{t}(G)$. Now we present the following observation.

Observation 2.4**.**

For each $k\geq 1$ and $r\geq 1$, if $G$ is a connected $r$-regular graph of order $n$, then there exists a connected $r^{\prime}$-regular graph $G^{\prime}$ of order $n^{\prime}=(k+2)n$ such that $r^{\prime}=(k+2)r$ and $\gamma_{p,k}(G^{\prime})=\gamma_{t}(G)$.

Proof.

Let $V(G)=\{v_{1},v_{2},\cdots,v_{n}\}$. Let $G^{\prime}$ be the graph constructed from $G$ as follows. Take $n$ disjoint independent sets $V_{i}=\{v_{i}^{1},v_{i}^{2},\cdots,v_{i}^{k+2}\}$ corresponding to $v_{i}$, where $i\in\{1,2,\cdots,n\}$. For each edge $v_{i}v_{j}\in E(G)$, add the edges $v_{i}^{s}v_{j}^{q}$ for each $s,q\in\{1,2,\cdots,k+2\}$ (see Figure 2).

Let $S=\{v_{i_{1}},v_{i_{2}},\cdots,v_{i_{h}}\}$ be a TDS of $G$ with $h=\gamma_{t}(G)$. It is easy to check that $\{v_{i_{1}}^{1},v_{i_{2}}^{1},\cdots,v_{i_{h}}^{1}\}$ is a $k$-PDS of $G^{\prime}$. Hence, $\gamma_{p,k}(G^{\prime})\leq\gamma_{t}(G)$. On the other hand, let $S^{\prime}$ be a $k$-PDS of $G^{\prime}$ with $|S^{\prime}|=\gamma_{p,k}(G^{\prime})$. We can change some vertices of $S^{\prime}$ such that $|S^{\prime}\cap V_{i}|\leq 1$ for each $i\in\{1,2,\cdots,n\}$. Otherwise, without loss of generality, assume that $|S^{\prime}\cap V_{1}|\geq 2$. If there exists $j\in\{2,3,\cdots,n\}$ such that $S^{\prime}\cap V_{j}\neq\emptyset$ and $V_{j}\subseteq N_{G}(v_{1}^{1})$, then $S^{\prime\prime}=(S^{\prime}\backslash V_{1})\cup\{v_{1}^{1}\}$ is also a $k$-PDS of $G^{\prime}$ and $|S^{\prime\prime}|<|S^{\prime}|=\gamma_{p,k}(G^{\prime})$, a contradiction. Now we assume $S^{\prime}\cap V_{j}=\emptyset$ for each $V_{j}\subseteq N_{G}(v_{1}^{1})$, where $j\in\{2,3,\cdots,n\}$. Let $S^{\prime\prime}=(S^{\prime}\backslash V_{1})\cup\{v_{1}^{1},v_{j}^{1}\}$. Thus, $S^{\prime\prime}$ is also a $k$-PDS of $G^{\prime}$ such that $|S^{\prime\prime}\cap V_{1}|=1$. Let $S^{\prime}=S^{\prime\prime}$. Hence, we find a $k$-PDS $S^{\prime}$ of $G^{\prime}$ such that $|S^{\prime}\cap V_{i}|\leq 1$ for each $i\in\{1,2,\cdots,n\}$. Let $S=\emptyset$. For each $i\in\{1,2,\cdots,n\}$, if $|S^{\prime}\cap V_{i}|=1$, we add $v_{i}$ to $S$. Then $S$ is a TDS of $G$ with $|S|=\gamma_{p,k}(G^{\prime})$, implying that $\gamma_{p,k}(G^{\prime})\geq\gamma_{t}(G)$. ∎

The authors of [11] constructed 3-regular graphs $F_{0,q}$ of order $4q$ such that $\gamma_{t}(F_{0,q})=2q$ (see Figures 4-4). By Observation 2.4, we can construct $F_{k,q}$ $(=G^{\prime})$ from $F_{0,q}$ $(=G)$, and so $\gamma_{p,k}(F_{k,q})=\gamma_{t}(F_{0,q})=2q=\frac{3}{2}\frac{n^{\prime}}{3k+6}=\frac{3n^{\prime}}{2r^{\prime}}$. Hence, $\frac{n^{\prime}}{r^{\prime}}$ is not the upper bound of $\gamma_{p,k}(G^{\prime})$ in Conjecture 1.4.

Now, an interesting problem is whether $\frac{n}{r}$ is always the upper bound of $\gamma_{p,k}(G)$ when $G$ is claw-free. We will discuss this problem in next section.

3 Claw-free regular graphs

First, we establish the relation between $k$-power domination and domination by presenting Observation 3.1. Then, we use Observation 3.1 to construct a series of regular claw-free graphs satisfying that $\gamma_{p,k}(G)=\frac{4n}{3(r+1)}>\frac{n}{r}$, where $r>3$.

A set $S$ of vertices in a graph $G$ is called a domination set (abbreviated as DS) of $G$ if every vertex of $V\setminus S$ is adjacent to some vertex of $G$. The minimum cardinality of a DS of $G$ is the domination number of $G$, denoted by $\gamma(G)$.

Observation 3.1**.**

For each $k\geq 1$ and $r\geq 1$, if $G$ is a connected $r$-regular claw-free graph of order $n$, then there exists a connected $r^{\prime}$-regular claw-free graph $G^{\prime}$ of order $n^{\prime}=(k+1)n$ such that $r^{\prime}=kr+r+k$ and $\gamma_{p,k}(G^{\prime})=\gamma(G)$.

Proof.

Let $V(G)=\{v_{1},v_{2},\cdots,v_{n}\}$. Let $G^{\prime}$ be the graph constructed from $G$ as follows. Take $n$ disjoint cliques $V_{i}=\{v_{i}^{1},v_{i}^{2},\cdots,v_{i}^{k+1}\}$ corresponding to $v_{i}$. For each edge $v_{i}v_{j}\in E(G)$, add the edges $v_{i}^{s}v_{j}^{q}$ for each $s,q\in\{1,2,\cdots,k+1\}$ (see Figure 5). It is easy to check that $G^{\prime}$ is a claw-free graph.

Let $S=\{v_{i_{1}},v_{i_{2}},\cdots,v_{i_{t}}\}$ be a DS of $G$ with $t=\gamma(G)$. Then $\{v_{i_{1}}^{1},v_{i_{2}}^{1},\cdots,v_{i_{t}}^{1}\}$ is a $k$-PDS of $G^{\prime}$, implying that $\gamma_{p,k}(G^{\prime})\leq\gamma(G)$. On the other hand, let $S^{\prime}=\{v_{i_{1}}^{j_{1}},v_{i_{2}}^{j_{2}},\cdots,v_{i_{t}}^{j_{t}}\}$ be a $k$-PDS of $G^{\prime}$ with $t=\gamma_{p,k}(G^{\prime})$. If there exists $i\in\{1,2,\cdots,n\}$ such that $|S^{\prime}\cap V_{i}|\geq 2$, then $S^{\prime\prime}=(S^{\prime}\backslash V_{i})\cup\{v_{i}^{1}\}$ is also a $k$-PDS of $G^{\prime}$ with $|S^{\prime\prime}|<|S^{\prime}|$, a contradiction. Hence, $|S^{\prime}\cap V_{i}|\leq 1$ for each $i\in\{1,2,\cdots,n\}$. Thus, $\{v_{i_{1}},v_{i_{2}},\cdots,v_{i_{t}}\}$ is a DS of $G^{\prime}$, implying that $\gamma_{p,k}(G^{\prime})\geq\gamma(G)$. ∎

Let $H$ be the graph of order 6 as drawn in Figure 7. We define the graph $H_{0,q}$ as follows. Take $q$ disjoint copies $H_{i}\cong H$, where $i=1,2,\cdots,q$. For each $i\in\{1,2,\cdots,q\}$, let $x_{i},y_{i}\in V(H_{i})$ such that $d_{H_{i}}(x_{i})=d_{H_{i}}(y_{i})=2$. Add the edges $y_{i}x_{i+1}$, where $i=1,2,\cdots,q$ and $x_{q+1}=x_{1}$ (see Figure 7). It is clear that $H_{0,q}$ is a connected $3$-regular claw-free graph of order $6q$. By Observation 3.1, we can construct $H_{k,q}$ $(=G^{\prime})$ from $H_{0,q}$ $(=G)$.

Let $S=\bigcup_{i=1}^{q}\{x_{i},y_{i}\}$. Then $S$ is a DS of $H_{0,q}$, implying that $\gamma(H_{0,q})\leq 2q$. Since $\gamma(C_{4})=2$, we get $\gamma(H_{0,q})\geq 2q$. So $\gamma(H_{0,q})=2q$. By Observation 3.1, $\gamma_{p,k}(H_{k,q})=\gamma(H_{0,q})=2q=\frac{4}{3}\frac{n^{\prime}}{4k+4}=\frac{4n^{\prime}}{3(r^{\prime}+1)}>\frac{n^{\prime}}{r^{\prime}}$. Hence, $\frac{n^{\prime}}{r^{\prime}}$ is not always the upper bound of $\gamma_{p,k}(G^{\prime})$ when $G^{\prime}$ is claw-free.

Now we know that in Conjecture 1.4, if $r-k$ is sufficiently large, then $\frac{n}{r}$ is not always the upper bound of $\gamma_{p,k}(G)$. For each $r\geq 4$ and $k=\lfloor\frac{r}{2}\rfloor-1$, we will show that Conjecture 1.4 does not always hold for claw-free $r$-regular graphs by presenting Observations 3.2 and 3.3. It means that $k_{min}(r)\geq\lfloor\frac{r}{2}\rfloor$ even restricted to claw-free regular graphs in the Question 1.5.

Observation 3.2**.**

For each odd $r\geq 5$ and $q\geq 1$, there exists a connected claw-free $r$-regular graph $G_{r,q}$ of order $n=|V(G_{r,q})|$ such that $\gamma_{p,\frac{r-3}{2}}(G_{r,q})=\frac{n+2}{r+1}>\frac{n}{r+1}$.

Proof.

We define $A_{i}=\{a_{i}^{1},\cdots,a_{i}^{(r-1)/2}\}$, $B_{i}=\{b_{i}^{1},\cdots,b_{i}^{(r-1)/2}\}$ and $U_{i}=\{u_{i}^{1},u_{i}^{2}\}$ for each $i\in\{0,1,\cdots,q\}$. Then, we construct $G_{r,q}$ by the following steps. Firstly, let $V(G_{r,q})=(A_{0}\cup B_{0})\cup\left(\bigcup_{i=1}^{q}{(U_{i}\cup A_{i}\cup B_{i})}\right)$. Secondly, add the edges such that $A_{q}\cup B_{q}$, $A_{i}\cup B_{i}$, $B_{i}\cup U_{i+1}$ and $U_{i+1}\cup A_{i+1}$ are cliques for each $i\in\{0,1,\cdots,q-1\}$. Finally, add the edges $a_{0}^{j}b_{q}^{j}$ and $a_{0}^{j}b_{q}^{j+1}$ for each $j\in\{1,\cdots,\frac{r-1}{2}\}$, where $b_{q}^{\frac{r+1}{2}}=b_{q}^{1}$ (see Figures 10-10).

It is easy to check that $G_{r,q}$ is a connected $r$-regular claw-free graph of order $n=(q+1)(r+1)-2$. Let $k=\frac{r-3}{2}$. Since $\{a_{0}^{1},\cdots,a_{q}^{1}\}$ is a $k$-PDS of $G_{r,q}$, we have $\gamma_{p,k}(G_{r,q})\leq q+1$. On the other hand, let $S$ be a $k$-PDS of $G_{r,q}$. It is clear that $A_{q}$ is a $k$-fort and $B_{i}$ is also a $k$-fort for each $i\in\{0,\cdots,q-1\}$. By Propostion 2.2, $|S\cap(A_{q}\cup B_{q}\cup U_{q})|\geq 1$ and $|S\cap(A_{i}\cup B_{i}\cup U_{i+1})|\geq 1$ for each $i\in\{0,\cdots,q-1\}$. It leads to $|S|\geq q$. Moreover, if $|S|=q$, then $|S\cap U_{i}|=1$ for each $i\in\{1,\cdots,q\}$. In this case, $P_{G_{r,q}}^{\infty}(S)=V\backslash(A_{0}\cup B_{q})$, contradicting that $S$ is a $k$-PDS of $G_{r,q}$. Hence, $\gamma_{p,k}(G_{r,q})=q+1=\frac{n+2}{r+1}>\frac{n}{r+1}$.

∎

Observation 3.3**.**

For each even $r\geq 4$ and $q\geq 1$, there exists a connected claw-free $r$-regular graph $G_{r,q}$ of order $n=|V(G_{r,q})|$ such that $\gamma_{p,\frac{r-2}{2}}(G_{r,q})=\frac{n+1}{r+1}>\frac{n}{r+1}$.

Proof.

We consider a graph $G_{r,q}$ which was presented by Lu et al. in [17] and was noted by $Q_{r,k}$ in their paper. Let $A_{i}=\{a_{i}^{1},\cdots,a_{i}^{r/2}\}$, $B_{i}=\{b_{i}^{1},\cdots,b_{i}^{r/2}\}$ and $U_{i}=\{u_{i}\}$ for each $i\in\{0,1,\cdots,q\}$. Now we redefine $G_{r,q}$ by the following steps. Firstly, let $V(G_{r,q})=(A_{0}\cup B_{0})\cup\left(\bigcup_{i=1}^{q}{(U_{i}\cup A_{i}\cup B_{i})}\right)$. Secondly, add the edges such that $A_{q}\cup B_{q}$, $A_{i}\cup B_{i}$, $B_{i}\cup U_{i+1}$ and $U_{i+1}\cup A_{i+1}$ are cliques for each $i\in\{0,\cdots,q-1\}$. Finally, add the edges $a_{0}^{j}b_{q}^{j}$ for each $j\in\{1,\cdots,\frac{r}{2}\}$ (see Figures 13-13).

It is easy to check that $G_{r,q}$ is a connected claw-free $r$-regular graph. Similar to the proof of Observation 3.2, we have $\gamma_{p,\frac{r-2}{2}}(G_{r,q})=q+1=\frac{n+1}{r+1}>\frac{n}{r+1}$. ∎

Hence, we will consider Conjecture 1.4 when $G$ is a connected claw-free $r$-regular graph and $k\geq\lfloor\frac{r}{2}\rfloor$. It means that $k\geq\frac{r-1}{2}$. If we let $r=k+l+1$, we have $k\geq\frac{k+l}{2}$, implying that $k\geq l$. Chang et al. [6] studied the case that $l=1$. We further studied the cases $l=2$ and $l=3$ by proving Theorem 1.6.

If the statement of Theorem 1.6 fails, then we suppose that $G$ is a counterexample with minimal $|V(G)|$, i.e, $G$ is a connected claw-free $(k+l+1)$-regular graph of minimal order $n$ and $\gamma_{p,k}(G)>\frac{n}{k+l+2}$ for $l\in\{2,3\}$ and $k\geq l$.

Before giving the proof of Theorem 1.6, we define an important structure, which is an $L$-configuration in $G$.

Definition 3.4**.**

($L$-configuration). The subgraph $H\cong G[N[L]]$ is an $L$-configuration if $L$ is both a clique and a $k$-fort of $G$.

Let $j\leq k$ be a positive integer and $A_{j}$ be the graph obtained from $K_{k+j+2}$ by removing $j$ edges which share a common vertex in $K_{k+j+2}$ (see Figures 15-15). Remark that $A_{j}$ is an $L$-configuration in $G$.

Then, we present three useful lemmas.

Lemma 3.5**.**

Let $H$ be an $L$-configuration of $G$. If $S\subseteq L$ and $|S|\geq|L|-k$, then $N[S]=V(H)$.

Proof.

Suppose that $S\subseteq L$ and $|S|\geq|L|-k$. It is clear that $L\subseteq N[S]\subseteq V(H)$. For each $v\in V(H)\setminus L$, we have $|N_{L}(v)\cap S|\geq 1$ since $L$ is a $k$-fort of $G$ and $|L|-|S|\leq k$. Hence, $v\in N[S]$, implying that $V(H)\subseteq N[S]$. ∎

Lemma 3.6**.**

Let $H$ be an $L$-configuration of $G$ and $H^{\prime}$ be an $L^{\prime}$-configuration of $G$. If $V(H)\cap V(H^{\prime})\neq\emptyset$, then $V(H)=V(H^{\prime})$.

Proof.

For each $u\in V(H)\cap V(H^{\prime})$, we define $S_{u}=N[u]\cap(L\cap L^{\prime})$. Then $|S_{u}|=|N[u]\cap L|+|N[u]\cap L^{\prime}|-|N[u]\cap(L\cup L^{\prime})|$ according to the inclusion and exclusion principle.

It is clear that $|L|-|N[u]\cap(L\cup L^{\prime})|\geq(k+1)-(k+l+2)\geq-k-1$. We claim that the equation can’t hold. Otherwise, suppose the equation holds. Then, we have $|L|=k+1$ and $N[u]\subseteq L\cup L^{\prime}$. Without loss of generality, assume $u\in L$, and so $N[u]\backslash L\subseteq N[L]\setminus L$. Since $L$ is a $k$-fort, $N(v)\cap L=L$ for each $v\in N[u]\backslash L$. Since $L^{\prime}$ is a clique and $N[u]\setminus L\subseteq L^{\prime}$, we have $N[u]\setminus L$ is a clique. It means that $N[u]$ is a clique, and so $G\cong K_{k+l+2}$, contradicting that $G$ is a counterexample. So, $|L|-|N[u]\cap(L\cup L^{\prime})|\geq-k$.

We claim that $L\cap L^{\prime}\neq\emptyset$. Otherwise, suppose that $L\cap L^{\prime}=\emptyset$. If $u\notin L\cup L^{\prime}$ for each $u\in V(H)\cap V(H^{\prime})$, then $d_{G}(u)\geq|L|+|L^{\prime}|\geq 2(k+1)>k+l+1$, a contradiction. Without loss of generality, we assume $u\in L$. Then $|S_{u}|=|N[u]\cap L^{\prime}|+|L|-|N[u]\cap(L\cup L^{\prime})|\geq|N[u]\cap L^{\prime}|-k\geq 1$. It means that $|L\cap L^{\prime}|\geq 1$, a contradiction. Hence, $L\cap L^{\prime}\neq\emptyset$.

Let $v\in L\cap L^{\prime}$. Then $|S_{v}|=|L|+|L^{\prime}|-|N[v]\cap(L\cup L^{\prime})|$. It means that $|S_{v}|\geq|L|-k$ and $|S_{v}|\geq|L^{\prime}|-k$. By Lemma 3.5, $V(H)=N[S_{v}]=V(H^{\prime})$. ∎

Lemma 3.7**.**

Let $H$ be an $L$-configuration of $G$. Then, we have $V(H)\subseteq P^{\infty}(u)$ for each $u\in L$.

Proof.

Let $u\in L$. If $|L|=k+1$, then $N[u]=V(H)$ by Lemma 3.5, implying that $V(H)\subseteq P^{\infty}(u)$. Now suppose that $|L|\geq k+2$. Since $G$ is a $(k+l+1)$-regular graph and $l\leq k$, $V(H)\subseteq P^{\infty}(u)$. ∎

We give the following method to choose a vertex subset $\mathcal{P}_{0}$ for $G$. First, let $\mathcal{P}_{0}=\emptyset$. Then, we process the following step. If $G$ contains an $L$-configuration and none vertex of $L$ is contained in $P^{\infty}(\mathcal{P}_{0})$, then we add one vertex of $L$ to $\mathcal{P}_{0}$. Process the step till $G$ contains no such an $L$-configuration.

By Lemmas 3.6 and 3.7, it is clear that $\mathcal{P}_{0}$ is a packing of $G$. We extend the packing $\mathcal{P}_{0}$ of $G$ to a maximal packing and denote the resulting packing by $S_{0}$.

Lemma 3.8**.**

For $l\in\{2,3\}$ and $k\geq l$, $G$ has a sequence $S_{0},S_{1},\cdots,S_{q}$ such that the following holds:

(a) For all $t\geq 0$, $|S_{t+1}|=|S_{t}|+1$ and $|P^{\infty}(S_{t+1})|\geq|P^{\infty}(S_{t})|+k+l+2$.

(b) $P^{\infty}(S_{q})=V(G)$.

Proof.

We prove part (a) and part (b) by induction on $t$. If $P^{\infty}(S_{0})=V(G)$, then there is nothing to prove. Hence, we may assume that $P^{\infty}(S_{0})\neq V(G)$. Let $t\geq 0$ and suppose that $S_{t}$ exists and $P^{\infty}(S_{t})\neq V(G)$. Denote $M=P^{\infty}(S_{t})$ and $\overline{M}=V(G)\setminus M$. Let $\mathcal{U}=\{u~{}|~{}u\in M$ and $N_{G}(u)\cap\overline{M}\neq\emptyset\}$. For each vertex $u\in\mathcal{U}$, since $N_{G}[u]\not\subseteq M$, we note that $d_{M}(u)\geq 1$ and $k+1\leq d_{\overline{M}}(u)\leq k+l$. Moreover, for each $u\in\mathcal{U}$, we define $L_{u}=N_{G}(u)\cap\overline{M}=\{u_{1},u_{2},\dots,u_{d_{\overline{M}}(u)}\}$, $F_{u}=N_{G}(L_{u})\backslash L_{u}$ and $F^{\prime}_{u}=F_{u}\setminus\{u\}$. Hence, $k+1\leq|L_{u}|\leq k+l$.

We claim that for each vertex $x\in\overline{M}$, $N_{G}(x)\cap\mathcal{U}\neq\emptyset$. Otherwise, suppose to the contrary that there exists $y\in\overline{M}$ such that $N_{G}(y)\cap\mathcal{U}=\emptyset$. Then $S_{0}\cup\{y\}$ is also a packing, contradicting that $S_{0}$ is a maximal packing. Now we present seven useful claims.

Claim 1**.**

*If $H$ is an $L$-configuration of $G$, then $V(H)\subseteq M$. *

Proof.

By the choose of $S_{0}$ and Lemma 3.7, we immediately obtain the Claim 1. ∎

Claim 2**.**

For each $u\in\mathcal{U}$, $L_{u}$ induces a clique in $G$.

Proof.

Suppose $x_{1}$ and $x_{2}$ are two neighbors of $u$ in $L_{u}$ and $u$ is observed by $v$ in $M$. Then $x_{1}v,x_{2}v\notin E(G)$. If $x_{1}x_{2}\notin E(G)$, then $\{u,x_{1},x_{2},v\}$ induces a claw, a contradiction. Therefore, $L_{u}$ induces a clique in $G$. ∎

Claim 3**.**

Let $u\in\mathcal{U}$. If $|L_{u}|+|F_{u}\cap\overline{M}|\geq k+l+2$, then for $S_{t+1}=S_{t}\cup\{u_{1}\}$, we have $|P^{\infty}(S_{t+1})|\geq|P^{\infty}(S_{t})|+k+l+2$.

Proof.

Suppose $|L_{u}|+|F_{u}\cap\overline{M}|\geq k+l+2$. By Claim 2, $L_{u}$ induces a clique in $G$. We define $S_{t+1}=S_{t}\cup\{u_{1}\}$ and we let $j$ be the minimum integer such that $P^{j}(S_{t})=P^{\infty}(S_{t})$. Then, $N[u_{1}]\subseteq P^{0}(S_{t+1})\subseteq P^{j}(S_{t+1})$, and so $L_{u}\cup\{u\}\subseteq P^{j}(S_{t+1})$. For each $u^{\prime}\in L_{u}\setminus\{u_{1}\}$, we have

[TABLE]

It means that $N[u^{\prime}]\subseteq P^{j+1}(S_{t+1})$. Therefore,

[TABLE]

∎

Claim 4**.**

Let $u\in\mathcal{U}$. If there exists a vertex $w\in F_{u}\cap\overline{M}$ such that $|L_{u}|-d_{L_{u}}(w)\leq k$ and $vw\notin E$ for each $v\in M\cap F_{u}$, then for $S_{t+1}=S_{t}\cup\{w\}$, we have $|P^{\infty}(S_{t+1})|\geq|P^{\infty}(S_{t})|+k+l+2$.

Proof.

Suppose there exists a vertex $w\in F_{u}\cap\overline{M}$ such that $|L_{u}|-d_{L_{u}}(w)\leq k$ and $vw\notin E$ for each $v\in M\cap F_{u}$. By Claim 2, $L_{u}$ induces a clique in $G$. Since $N_{G}(w)\cap\mathcal{U}\neq\emptyset$, there exists $x\in\mathcal{U}$ such that $w\in L_{x}$. We claim that $L_{x}\cap L_{u}=\emptyset$. Otherwise, without loss of generality, assume $u_{1}\in L_{x}\cap L_{u}$. Then, $u_{1}x\in E$, and so $x\in F_{u}\cap M$. It leads to $xw\notin E$, a contradiction. Hence, $L_{x}\cap L_{u}=\emptyset$. We define $S_{t+1}=S_{t}\cup\{w\}$ and we let $j$ be the minimum integer such that $P^{j}(S_{t})=P^{\infty}(S_{t})$. Then, $N[w]\subseteq P^{0}(S_{t+1})\subseteq P^{j}(S_{t+1})$. By Claim 2, $L_{x}\subseteq P^{j}(S_{t+1})\setminus P^{j}(S_{t})$. Since $|L_{u}|-d_{L_{u}}(w)\leq k$, we have $L_{u}\subseteq P^{j+1}(S_{t+1})$. Therefore, we obtain

[TABLE]

∎

Claim 5**.**

If there is a vertex $u\in\mathcal{U}$ such that $|L_{u}|=k+l$, part (a) follows as desired.

Proof.

Suppose there is a vertex $u\in\mathcal{U}$ such that $|L_{u}|=k+l$. By Claim 2, $L_{u}$ induces a clique in $G$. If there is a vertex $w\in F^{\prime}_{u}$ such that $d_{L_{u}}(w)\geq k+1$, then $G[\{u,w\}\cup L_{u}]$ is an $L$-configuration where $L=N_{G}(w)\cap L_{u}$, contradicting Claim 1.

Now we assume that $d_{L_{u}}(w)\leq k$ for each $w\in F^{\prime}_{u}$. Then, $|F^{\prime}_{u}|\geq 2$. If there is a vertex $w\in F^{\prime}_{u}$ such that $w\in M$, without loss of generality, suppose $u_{1}\in L_{w}$. Since $|L_{w}|\geq k+1$ and $d_{L_{u}}(w)\leq k$, there is a vertex $w^{\prime}\in L_{w}\setminus L_{u}$. By Claim 2, $u_{1}w^{\prime}\in E$. It leads to $d(u_{1})\geq|L_{u}\setminus\{u_{1}\}|+|\{u,w,w^{\prime}\}|\geq k+l+2$, a contradiction. Now suppose $F^{\prime}_{u}\subseteq\overline{M}$. Then, $|L_{u}|+|F_{u}\cap\overline{M}|=|L_{u}|+|F^{\prime}_{u}|\geq k+l+2$. By Claim 3, part (a) follows as desired. ∎

Claim 6**.**

When $l=3$, if $|L_{u}|=k+2$ for each $u\in\mathcal{U}$, part (a) follows as desired.

Proof.

When $l=3$, suppose $|L_{u}|=k+2$ for each $u\in\mathcal{U}$. By Claim 2, $L_{u}$ induces a clique in $G$. Since $G$ is a connected claw-free $(k+l+1)$-regular graph, $|N(u_{1})\setminus(L_{u}\cup\{u\})|=k+l+1-(k+2)=2$, implying that $|F^{\prime}_{u}|\geq 2$. We claim that $|F^{\prime}_{u}|\geq 3$. Otherwise, we suppose $F^{\prime}_{u}=\{w_{1},w_{2}\}$, implying that $d_{L_{u}}(w_{1})=d_{L_{u}}(w_{2})=k+2$. Then, $G[L_{u}\cup F_{u}]$ is an $L$-configuration where $L=L_{u}$, contradicting Claim 1. Hence, $|F^{\prime}_{u}|\geq 3$. If $F^{\prime}_{u}\cap M=\emptyset$, then $|L_{u}|+|F_{u}\cap\overline{M}|=|L_{u}|+|F^{\prime}_{u}|\geq k+l+2$. By Claim 3, part (a) follows as desired.

Now suppose that $F^{\prime}_{u}\cap M\neq\emptyset$. If there is a vertex $w\in F^{\prime}_{u}\cap M$ such that $d_{L_{u}}(w)\leq k$, without loss of generality, suppose that $u_{1}\in L_{w}$. Since $|L_{w}|=k+2$, there are two vertices $w^{\prime},w^{\prime\prime}\in L_{w}\setminus L_{u}$. By Claim 2, $u_{1}w^{\prime},u_{1}w^{\prime\prime}\in E$. It leads to $d(u_{1})\geq|L_{u}\backslash\{u_{1}\}|+|\{u,w,w^{\prime},w^{\prime\prime}\}|=k+5$, a contradiction.

If there is a vertex $w\in F^{\prime}_{u}\cap M$ such that $d_{L_{u}}(w)=k+1$ , without loss of generality, suppose $N_{L_{u}}(w)=\{u_{1},u_{2},\cdots,u_{k+1}\}$. Since $|L_{w}|=k+2$, there is a vertex $w^{\prime}\in L_{w}\setminus L_{u}$. By Claim 2, $\{u_{1},u_{2},\cdots,u_{k+1},w^{\prime}\}$ induces a clique in $G$. Then, $G[L_{u}\cup\{u,w,w^{\prime}\}]$ is an $L$-configuration where $L=N_{G}(w)\cap L_{u}$, contradicting Claim 1.

Finally, we consider the case that there is a vertex $w\in F^{\prime}_{u}\cap M$ such that $d_{L_{u}}(w)=k+2$. Let $F^{\prime\prime}_{u}=F^{\prime}_{u}\setminus\{w\}$. If $F^{\prime\prime}_{u}\cap M\neq\emptyset$, let $w^{\prime}\in F^{\prime\prime}_{u}\cap M$. By the above argument, we deduce that $d_{L_{u}}(w^{\prime})=k+2$. Hence, $G[L_{u}\cup\{u,w,w^{\prime}\}]$ is an $L$-configuration where $L=L_{u}$, contradicting Claim 1. Now suppose $F^{\prime\prime}_{u}\subseteq\overline{M}$. If $|F^{\prime\prime}_{u}|=1$, let $F^{\prime\prime}_{u}=\{w^{\prime\prime}\}$ and we have $d_{L_{u}}(w^{\prime\prime})=k+2$. Similar to the above proof, we obtain a contradiction. If $|F^{\prime\prime}_{u}|=2$, let $F^{\prime\prime}_{u}=\{w_{1},w_{2}\}$ and $w_{1},w_{2}\in\overline{M}$. Since $d_{L_{u}}(w_{1})+d_{L_{u}}(w_{2})=k+2$, without loss of generality, we assume that $d_{L_{u}}(w_{1})\geq 2$. Since $|L_{w}|=|L_{u}|=k+2$, we obtain $|L_{u}|-d_{L_{u}}(w_{1})\leq k$, $uw_{1}\notin E$ and $ww_{1}\notin E$. By Claim 4, we have proved part (a). If $|F^{\prime\prime}_{u}|\geq 3$, then $|L_{u}|+|F_{u}\cap\overline{M}|=|L_{u}|+|F^{\prime\prime}_{u}|\geq k+5$. By Claim 3, part (a) follows as desired. ∎

Claim 7**.**

If there is a vertex $u\in\mathcal{U}$ such that $|L_{u}|=k+1$, part (a) follows as desired.

Proof.

Suppose there is a vertex $u\in\mathcal{U}$ such that $|L_{u}|=k+1$. By Claim 2, $L_{u}$ induces a clique in $G$. If $M\cap F^{\prime}_{u}=\emptyset$, then $F^{\prime}_{u}\subseteq\overline{M}$. Since $G$ is a connected claw-free $(k+l+1)$-regular graph, $|N(u_{1})\setminus(L_{u}\cup\{u\})|=k+l+1-|L_{u}|=l$, implying that $|F^{\prime}_{u}|\geq l$. We claim that $|F^{\prime}_{u}|\geq l+1$. Otherwise, suppose $F^{\prime}_{u}=\{v_{1},v_{2},\cdots,v_{l}\}$, implying that $L_{u}\subseteq N_{G}[v_{i}]$ for each $i\in\{1,2,\cdots,l\}$. Then, $G[L_{u}\cup F_{u}]$ is an $L$-configuration where $L=L_{u}$, contradicting Claim 1. So, $|F^{\prime}_{u}|\geq l+1$ and $|L_{u}|+|F_{u}\cap\overline{M}|=|L_{u}|+|F^{\prime}_{u}|\geq k+l+2$. By Claim 3, part (a) follows as desired.

Now assume that $M\cap F^{\prime}_{u}\neq\emptyset$. If there is a vertex $w\in M\cap F^{\prime}_{u}$ such that $d_{L_{u}}(w)\leq k-l+1$, without loss of generality, suppose that $u_{1}\in N_{G}(w)\cap L_{u}$. Since $|L_{w}|\geq k+1$, we have $|L_{w}\setminus L_{u}|\geq l$. Assume that $\{x_{1},x_{2},\cdots,x_{l}\}\subseteq(L_{w}\setminus L_{u})$. By Claim 2, $u_{1}x_{i}\in E$ for each $i\in\{1,2,\cdots,l\}$. It leads to $d(u_{1})\geq|L_{u}\setminus\{u_{1}\}|+|\{u,w,x_{1},x_{2},\cdots,x_{l}\}|\geq k+l+2$, a contradiction.

Then, we suppose $d_{L_{u}}(w)\geq k-l+2$ for each $w\in M\cap F^{\prime}_{u}$. If there exists a vertex $w_{1}\in F_{u}\cap\overline{M}$ such that $vw_{1}\notin E$ for each $v\in M\cap F_{u}$, by Claim 4, part (a) follows as desired. Otherwise, we can assume that for each $w_{1}\in F_{u}\cap\overline{M}$, there is a vertex $v\in M\cap F_{u}$ such that $vw_{1}\in E$. By Claim 2, $N_{G}(v)\cap L_{u}\subseteq N_{G}(w_{1})\cap L_{u}$, and so $d_{L_{u}}(w_{1})\geq d_{L_{u}}(v)\geq k-l+2$. Hence, $d_{L_{u}}(w_{1})\geq k-l+2$ for each $w_{1}\in F_{u}$. If $d_{L_{u}}(w)=k+1$ for each $w\in M\cap F^{\prime}_{u}$, then for each $w^{\prime}\in F^{\prime}_{u}\cap\overline{M}$, there is a vertex $w^{\prime\prime}\in M\cap F_{u}$ such that $w^{\prime\prime}w^{\prime}\in E$ and $d_{L_{u}}(w^{\prime\prime})=k+1$. By the above argument, we deduce that $d_{L_{u}}(w^{\prime})\geq d_{L_{u}}(w^{\prime\prime})=k+1$ and $|F^{\prime}_{u}|=l$. Then, $G[L_{u}\cup F_{u}]$ is an $L$-configuration where $L=L_{u}$, contradicting Claim 1.

If there is a vertex $w\in M\cap F^{\prime}_{u}$ such that $d_{L_{u}}(w)=k$, without loss of generality, suppose that $N_{G}(w)\cap L_{u}=\{u_{1},u_{2},\cdots,u_{k}\}$. Since $|L_{w}|\geq k+1$, there is a vertex $w_{1}\in L_{w}\setminus L_{u}$. By Claim 2, $u_{i}w_{1}\in E$ for each $i\in\{1,2,\cdots,k\}$. Let $F^{\prime\prime}_{u}=F^{\prime}_{u}\setminus\{w,w_{1}\}$. It is clear that $F^{\prime\prime}_{u}\neq\emptyset$. For $l=2$, let $w_{2}\in F^{\prime\prime}_{u}$. Then $d_{L_{u}}(w_{2})=1<k=k-l+2$, contradicting that $d_{L_{u}}(x)\geq k-l+2$ for each $x\in F_{u}$. For $l=3$, if there is a vertex $w_{2}\in F^{\prime\prime}_{u}$ such that $\{u_{1},u_{2},\cdots,u_{k}\}\subseteq N_{G}(w_{2})\cap L_{u}$, we can similarly get a contradiction. Now we assume that for each vertex $v^{\prime}\in F^{\prime\prime}_{u}$, $\{u_{1},u_{2},\cdots,u_{k}\}\not\subseteq N_{G}(v^{\prime})\cap L_{u}$. If $F^{\prime\prime}_{u}\cap M\neq\emptyset$, suppose $w_{2}\in F^{\prime\prime}_{u}\cap M$. Since $d_{L_{u}}(w_{2})\geq k-l+2\geq k-1\geq l-1\geq 2$, we have $N_{L_{u}}(w)\cap N_{G}(w_{2})\neq\emptyset$. Let $x\in N_{L_{u}}(w)\cap N_{G}(w_{2})$. Since $d(x)=k+4$ and Claim 2, $\{u_{1},u_{2},\cdots,u_{k}\}\subseteq N_{G}(w_{2})\cap L_{u}$, a contradiction. So, $F^{\prime\prime}_{u}\subseteq\overline{M}$. Let $y\in F^{\prime\prime}_{u}$. It is clear that $uy\notin E$. We claim that $wy\notin E$. Otherwise, suppose $wy\in E$. By Claim 2, $\{u_{1},u_{2},\cdots,u_{k}\}\subseteq N_{G}(y)\cap L_{u}$, a contradiction. Hence, $|L_{u}|-d_{L_{u}}(y)\leq k$ and $vy\notin E$ for each $v\in M\cap F_{u}$. By Claim 4, part (a) follows as desired.

If there is a vertex $w\in M\cap F^{\prime}_{u}$ such that $d_{L_{u}}(w)=k-1$, then we obtain $l=3$ since $d_{L_{u}}(w)=k-1\geq k-l+2$. Without loss of generality, assume that $N_{G}(w)\cap L_{u}=\{u_{1},u_{2},\cdots,u_{k-1}\}$. Since $|L_{w}|\geq k+1$, there are two vertices $w_{1},w_{2}\in L_{w}\setminus L_{u}.$ By Claim 2, $u_{i}w_{1},u_{i}w_{2}\in E$ for each $i\in\{1,2,\cdots,k-1\}$. Let $F^{\prime\prime}_{u}=F^{\prime}_{u}\backslash\{w,w_{1},w_{2}\}$. It is clear that $F^{\prime\prime}_{u}\neq\emptyset$. Then, for each $w^{\prime}\in F^{\prime\prime}_{u}$, we have $d_{L_{u}}(w^{\prime})\leq 2$. Since $d_{L_{u}}(w^{\prime})\geq k-l+2$ and $k\geq l$, we obtain $k=3$ and $d_{L_{u}}(w^{\prime})=2$. If $F^{\prime\prime}_{u}\cap M=\emptyset$, then $F^{\prime\prime}_{u}\subseteq\overline{M}$. Let $z\in F^{\prime\prime}_{u}$. Then, $zu\notin E$. We claim that $zw\notin E$. Otherwise, suppose $zw\in E$. By Claim 2, $zu_{1}\in E$. It leads to $d(u_{1})\geq|L_{u}\setminus\{u_{1}\}|+|\{u,w,w_{1},w_{2},z\}|\geq k+5$, a contradiction. Since $|L_{u}|-d_{L_{u}}(z)\leq k$ and Claim 4, part (a) follows as desired. Then, we assume that $F^{\prime\prime}_{u}\cap M\neq\emptyset$ and $w_{3}\in F^{\prime\prime}_{u}\cap M$. If $w_{1}w_{3},w_{2}w_{3}\in E$, then $d_{L_{u}}(w_{1})=d_{L_{u}}(w_{2})=4$ by Claim 2. So, $G[L_{u}\cup F_{u}]$ is an $L$-configuration where $L=L_{u}\cup\{w_{1},w_{2}\}$, contradicting Claim 1. If $w_{1}w_{3},w_{2}w_{3}\notin E$, then there are two vertices $w_{4},w_{5}\in L_{w_{3}}\setminus L_{u}$. Since $w_{3}\in\mathcal{U}$ and Claim 2, we have $w_{4},w_{5}\in F_{u}$. Then, $|L_{u}|+|F_{u}\cap\overline{M}|\geq|L_{u}|+|\{w_{1},w_{2},w_{4},w_{5}\}|\geq k+l+2$. By Claim 3, part (a) follows as desired. Now we consider the last case. Without loss of generality, suppose $w_{1}w_{3}\in E$ and $w_{2}w_{3}\notin E$. Then, there is a vertex $w_{4}\in N(w_{3})\setminus(L_{u}\cup\{w_{1}\})$ such that $w_{4}\in\overline{M}$. By Claim 2, $\{u_{3},u_{4},w_{1},w_{4}\}$ induces a clique in $G$. So, $d(w_{1})\geq|L_{u}|+|\{w,w_{2},w_{3},w_{4}\}|=8>k+l+1=7$, a contradiction.

∎

Since $|L_{u}|\in\{k+1,k+2\}$ for $l=2$ and $|L_{u}|\in\{k+1,k+2,k+3\}$ for $l=3$, by Claims 5-7, part (a) follows as desired. Since $|V(G)|$ is finite, there exists an integer $q$ such that $P^{\infty}(S_{q})=V(G)$. Hence, we complete the proof. ∎

We are now in a position to prove our main result, namely, Theorem 1.6.

Proof.

Let $G$ be a counterexample such that $|V(G)|$ is minimal. Let $S_{0},S_{1},\cdots,S_{q}$ be a sequence satisfying properties (a)-(b) in the statement of Lemma 3.8 with $q$ as small as possible. By Lemma 3.8 (b), the set $S_{q}$ is a $k$-PDS in $G$, and so $\gamma_{p,k}(G)\leq|S_{q}|$. Since $S_{0}$ is a packing in $G$, we have that $|P^{0}(S_{0})|=|N[S_{0}]|=(k+l+2)|S_{0}|$. If $q=0$, then $(k+l+2)|S_{0}|\leq n$ and $\gamma_{p,k}(G)\leq|S_{0}|\leq\frac{n}{k+l+2}$, a contradiction. Now we suppose that $q\geq 1$. By Lemma 3.8 (a), $|S_{q}|=|S_{0}|+q$. By our choice of $q$, we decuce that $|P^{\infty}(S_{t+1})|\geq|P^{\infty}(S_{t})|+k+l+2$ for $0\leq t\leq q-1$. Thus,

[TABLE]

Hence, $\gamma_{p,k}(G)\leq|S_{q}|\leq\frac{n}{k+l+2}$, a contradiction. This proves the desired upper bound.

Next, we show this bound is tight. For positive integers $k\geq l$ and $t$, we define the graph $C_{k,t}$ as follows. Take $t$ disjoint copies $C_{i}\cong A_{l}$ and link any two copies $(C_{i},C_{i+1})$ with $l$ edges, where the subscripts are to be read as integers modulo $t$ and where $i=1,2,\cdots,t$. (see Figure 16). Then, $C_{k,t}$ is a connected claw-free $(k+l+1)$-regular graph of order $n=t(k+l+2)$. Suppose that $S$ is an arbitrary $k$-PDS in $C_{k,t}$. It is easy to check that $C_{i}$ contains a $k$-fort of $G$, where $i=1,2,\cdots,t$. By Proposition 2.2, $|S\cap V(C_{i})|\geq 1$ for each $i\in\{1,2,\cdots,t\}$. It means that $\gamma_{p,k}(C_{k,t})\geq t=\frac{n}{k+l+2}$. Since the above proof, we obtain $\gamma_{p,k}(C_{k,t})\leq\frac{n}{k+l+2}$. Hence, $\gamma_{p,k}(C_{k,t})=\frac{n}{k+l+2}$.∎

4 Conjecture and Question

We pose the following conjecture which is still open.

Conjecture 4.1**.**

For $l\geq 1$ and $k\geq l$, if $G$ is a connected claw-free $(k+l+1)$-regular graph of order $n$, then $\gamma_{p,k}(G)\leq\frac{n}{k+l+2}$ and the bound is tight.

Remark that if $l=1$, then the conjecture is true by the result of Chang et al. in [6]. If $l\in\{2,3\}$, the conjecture is true by our Theorem 1.6. When $l\geq 4$, the conjecture is still open. However, note that the bound of Conjecture 4.1 is tight since we can generalize the graph $C_{k,t}$ (defined in Section 3) to achieve this bound.

Now we pose the following question.

Question 4.2**.**

For $r\geq 3$, let $G$ be a connected claw-free $r$-regular graph of order $n$. Determine the smallest positive value, $k_{min}(r)$, of $k$ such that $\gamma_{p,k}(G)\leq\frac{n}{r+1}$.

By Observations 3.2 and 3.3, we deduce that $k_{min}(r)\geq\lfloor\frac{r}{2}\rfloor$. We remark that if Conjecture 4.1 is true, the answer of Question 4.2 is $k_{min}(r)=\lfloor\frac{r}{2}\rfloor$.