On (2-d)-Kernels in Two Generalizations of the Petersen Graph

Abstract

A subset J is a $(2$-$d)$-kernel of a graph if J is independent and 2-dominating simultaneously. In this paper, we consider two different generalizations of the Petersen graph and we give complete characterizations of these graphs which have $(2$-$d)$-kernel. Moreover, we determine the number of $(2$-$d)$-kernels of these graphs as well as their lower and upper kernel number. The property that each of the considered generalizations of the Petersen graph has a symmetric structure is useful in finding $(2$-$d)$-kernels in these graphs.

1. Introduction

In general, we use the standard terminology and notation of graph theory (see [1]). Let G be an undirected, connected, and simple graph with the vertex set $V\left(G\right)$ and the edge set $E\left(G\right)$. The order of the graph G is the number of vertices in G. The size of the graph G is its number of edges. By $P_n$, $n\ge 1$ and $C_n$, $n\ge 3$, we mean a path and a cycle of order n, respectively.

Let $G=(V,E)$ and $G^{\prime }=(V^{\prime },E^{\prime })$ be two graphs. If $V^{\prime }\subseteq V$ and $E^{\prime }\subseteq E$, then $G^{\prime }$ is a subgraph of G, written as $G^{\prime }\subseteq G$. If $G^{\prime }\subseteq G$ and $G^{\prime }$ contain all the edges $xy\in E$ with $x,y\in V^{\prime }$, then $G^{\prime }$ is an induced subgraph of G and we write $G^{\prime }:={⟨V^{\prime }⟩}_G$. Graphs G and $G^{\prime }$ are called isomorphic, and denoted by $G\cong G^{\prime }$, if there exists a bijection $\varphi :V\to V^{\prime }$ with $xy\in E$⇔$\varphi \left(x\right)\varphi \left(y\right)\in E^{\prime }$ for all $x,y\in V$. The complement of the graph G is a graph $\overline{G}$ such that $V\left(G\right)=V\left(\overline{G}\right)$ and two distinct vertices of $\overline{G}$ are adjacent if and only if they are not adjacent in G. A graph G is called bipartite if $V\left(G\right)$ admits a partition into two classes such that every edge has its ends in different classes.

A subset $D\subseteq V\left(G\right)$ is a dominating set of G if each vertex of G not belonging to D is adjacent to at least one vertex of D. A subset $S\subseteq V\left(G\right)$ is called an independent set of G if no two vertices of S are adjacent in G. A subset J being independent and dominating is a kernel of G.

The concept of kernels was initiated in 1953 by von Neumann and Morgenstern in digraphs with regard to game theory (see [2]). One of the pioneers studying the kernels in digraphs was C. Berge (see [3,4,5]). In literature, we can find many types and generalizations of kernels in digraphs (for results and applications, see, for example, [6,7,8,9,10,11]). The problem of the existence of kernels in undirected graphs is trivial because every maximal independent set is a kernel. Currently, distinct kind of kernels in undirected graphs are being studied quite intensively and many papers are available. For results and application, see, for example, [12,13,14,15,16,17,18]. Among many types of kernels in undirected graphs, there are kernels related to multiple domination, introduced by Fink and Jacobson in [19]. Let $p\ge 1$ be an integer. A subset S is said to be p-dominating if every vertex outside S has at least p neighbors in S. If $p=1$, then we obtain a dominating set in the classical sense. If $p=2$, we get a 2-dominating set. A set which is 2-dominating and independent is named a 2-dominating kernel ($(2$-$d)$-kernel in short). The concept of $(2$-$d)$-kernels was introduced by A. Włoch in [20]. Some properties of $(2$-$d)$-kernels were studied in [21,22,23,24]. In particular, in [23], it was proved that the problem of the existence of $(2$-$d)$-kernels is $\mathcal{NP}$-complete for general graphs. In [25], Nagy extended the concept of $(2$-$d)$-kernels to k-dominating kernels. He considered a k-dominating set instead of the 2-dominating set, which he called k-dominating independent sets. Some properties of these sets were studied in [26,27].

The number of $(2$-$d)$-kernels in the graph G is denoted by $\sigma \left(G\right)$. Let G be a graph with the $(2$-$d)$-kernel. The minimum cardinality of the $(2$-$d)$-kernel of G is called a lower $(2$-$d)$-kernel number and denoted by ${\gamma }_{(2-d)}\left(G\right)$. The maximum cardinality of the $(2$-$d)$-kernel of G is called an upper $(2$-$d)$-kernel number and is denoted by ${\mathsf{\Gamma }}_{(2-d)}\left(G\right)$.

In this paper, we consider two different generalizations of the Petersen graph. Various types of domination in the class of generalized Petersen graphs have been extensively studied in the literature (see [28,29,30,31,32]). Referring to this research, we will consider $(2$-$d)$-kernels for two different generalizations of the Petersen graph. We solve the problem of the existence of $(2$-$d)$-kernels, their number, and their cardinality in these graphs. Moreover, we determine a lower and an upper kernel number in these graphs. It is worth noting that each of presented generalizations of the Petersen graph has a symmetric structure. This property is useful in finding $(2$-$d)$-kernels in these graphs.

2. Main Results

In this section, we consider the problem of the existence of $(2$-$d)$-kernels in two different generalizations of the Petersen graph. In particular, we give complete characterizations of these generalizations, which have the $(2$-$d)$-kernel. We determine the number of $(2$-$d)$-kernels in these graphs as well as the lower and the upper $(2$-$d)$-kernel number.

In the further part of the paper, we will use green color to mark vertices belonging to the $(2$-$d)$-kernel, and red color to indicate vertices that cannot belong to it.

2.1. Generalized Petersen Graph

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. The graph $P(n,k)$ is called the generalized Petersen graph, if $V\left(P(n,k)\right)={\displaystyle \bigcup _{i=0}^{n-1}}\{u_i,v_i\}$ and $E\left(P(n,k)\right)={\displaystyle \bigcup _{i=0}^{n-1}}\{u_i{u}_{i+1},u_i{v}_i,v_i{v}_{i+k}\}$, where subscripts are reduced modulo n. These graphs were first defined by Watkins in [33]. Figure 1 shows generalized Petersen graphs $P(10,3)$, $P(5,2)$ and examples of $(2$-$d)$-kernels in these graphs.

Figure 1. Examples of $(2$-$d)$-kernels in $P(10,3)$ and $P(5,2)$.

We start with the problem of existence of $(2$-$d)$-kernels. At the beginning, we give a sufficient condition, emerging from the property of bipartite graphs. We have the following complete characterization of bipartite generalized Petersen graphs.

Proposition 1

([34]). Let $n\ge 3$, $k<\frac{n}{2}$ be integers. The graph $P(n,k)$ is bipartite if and only if n is even and k is odd.

From this characterization we directly obtain the sufficient condition for the existence of $(2$-$d)$-kernels.

Proposition 2.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. If n is even and k is odd, then the graph $P(n,k)$ has at least two $(2$-$d)$-kernels which are a partition of the vertex set.

Proof. 

Let n, k be as in the statement of the proposition. From Proposition 1, it follows that the graph $P(n,k)$ is a bipartite graph. Thus, there exist two independent sets of vertices $V_1$, $V_2$ that are a partition of the set $V\left(P\right(n,k\left)\right)$. Moreover, the graph $P(n,k)$ is a 3-regular graph. Therefore, sets $V_1$, $V_2$ are $(2$-$d)$-kernels of the graph $P(n,k)$. □

Now, we improve the above proposition to obtain the complete characterization of the generalized Petersen graph having $(2$-$d)$-kernel.

Theorem 1.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. The graph $P(n,k)$ has a $(2$-$d)$-kernel if and only if

(i) 

n is even and k is odd or

(ii) 

$n\equiv 0\left(\mathrm{mod}5\right)$ and $k\equiv 2\left(\mathrm{mod}5\right)$ or

(iii) 

$n\equiv 0\left(\mathrm{mod}5\right)$ and $k\equiv 3\left(\mathrm{mod}5\right)$.

Proof. 

If $n=3,4$, then the result is obvious. Let $n\ge 5$, $k<\frac{n}{2}$ be integers. If n is even and k is odd, then by Proposition 2, (i) follows. Let $n\equiv 0\left(\mathrm{mod}5\right)$, $k\equiv j\left(\mathrm{mod}5\right)$, $j=2,3$. We will show that the set $J=\{u_i;i\in \{0,5,\dots ,n\}\}\cup \{u_{i+2};i\in \{0,5,\dots ,n\}\}\cup \{v_{i+3};i\in \{0,5,\dots ,n\}\}\cup \{v_{i+4};i\in \{0,5,\dots ,n\}\}$ is a $(2$-$d)$-kernel of $P(n,k)$. The independence of J follows from the definition of $P(n,k)$. Let us assume that $x\in V\left(P\right(n,k\left)\right)\backslash J$. Then, either $x=u_s$, $s\in \{0,1,\dots ,n-1\}$, $s\equiv a\left(\mathrm{mod}5\right)$, $a=1,3,4$ or $x=v_t$, $t\in \{0,1,\dots ,n-1\}$, $t\equiv b\left(\mathrm{mod}5\right)$, $b=0,1,2$. We consider two cases.

1. $x=u_s$.

If $s\equiv 1\left(\mathrm{mod}5\right)$, then $\{u_{s-1},u_{s+1}\}\subseteq N\left(u_s\right)$ and $u_{s-1},u_{s+1}\in J$. If $s\equiv 3\left(\mathrm{mod}5\right)$, then $\{u_{s-1},v_s\}\subseteq N\left(u_s\right)$ and $u_{s-1},v_s\in J$. If $s\equiv 4\left(\mathrm{mod}5\right)$, then $\{u_{s+1},v_s\}\subseteq N\left(u_s\right)$ and $u_{s+1},v_s\in J$.

2. $x=v_t$.

Let $t\equiv 0\left(\mathrm{mod}5\right)$. If $k\equiv 2\left(\mathrm{mod}5\right)$, then $\{u_t,v_{t-2}\}\subseteq N\left(v_t\right)$ and $u_t,v_{t-2}\in J$. If $k\equiv 3\left(\mathrm{mod}5\right)$, then $\{u_t,v_{t+3}\}\subseteq N\left(v_t\right)$ and $u_t,v_{t+3}\in J$. If $t\equiv 1\left(\mathrm{mod}5\right)$, then $\{v_{t-k},v_{t+k}\}\subseteq N\left(v_t\right)$ and $v_{t-k},v_{t+k}\in J$, $k\equiv j\left(\mathrm{mod}5\right)$, $j=2,3$. Let $t\equiv 2\left(\mathrm{mod}5\right)$. If $k\equiv 2\left(\mathrm{mod}5\right)$, then $\{u_t,v_{t+2}\}\subseteq N\left(v_t\right)$ and $u_t,v_{t+2}\in J$. If $k\equiv 3\left(\mathrm{mod}5\right)$, then $\{u_t,v_{t-3}\}\subseteq N\left(v_t\right)$ and $u_t,v_{t-3}\in J$.

Summing up all the above cases we obtain that every vertex $x\in V\left(P\right(n,k\left)\right)\backslash J$ is 2-dominated by J. Hence, J is a $(2$-$d)$-kernel of $P(n,k)$.

Conversely, let $n\ge 5$, $k<\frac{n}{2}$, $i\in \{0,1,\dots ,n-1\}$ be integers and let J be a $(2$-$d)$-kernel of $P(n,k)$. If $u_i,u_{i+1},u_{i+2}\notin J$, then the vertex $u_{i+1}$ is not 2-dominated by J. Thus, each connected component of the graph ${⟨{\displaystyle \bigcup _{i=0}^{n-1}}{u}_i⟩}_{P(n,k)}\backslash J$ is isomorphic to either $P_1$ or $P_2$. We will show that in the graph $P(n,k)$ having a $(2$-$d)$-kernel, the configurations of these paths $P_1$, $P_2$ on the outer cycle, which are shown in the Figure 2 are forbidden.

Figure 2. Forbidden configurations of the paths $P_1$, $P_2$ for the graph $P(n,k)$ with the $(2$-$d)$-kernel.

Let us consider the following cases.

1. First, we will prove that the configuration of the paths $P_1$, $P_2$ shown on the left side of the Figure 2 is forbidden. Suppose that $u_i,u_{i+3},u_{i+6}\in J$ for some i, as in Figure 3. Then, $v_{i+1},v_{i+2},v_{i+4},v_{i+5}\in J$; otherwise, vertices $u_{i+1},u_{i+2},u_{i+4},u_{i+5}$ are not 2-dominated by J. Therefore, for every k vertices $v_{i+1+k},v_{i+2+k},v_{i+4+k},v_{i+5+k}\notin J$.

Figure 3. The case when $u_i,u_{i+3},u_{i+6}\in J$.

We have the next two possibilities.

1.1. $v_{i+3+k}\notin J$ for some i (see Figure 4).

Figure 4. The case when $u_i,u_{i+3},u_{i+6}\in J$ (the first subcase).

Since $v_{i+3+k}\notin J$, the vertex $u_{i+3+k}\in J$ and $u_{i+2+k},u_{i+4+k}\notin J$. Then $v_{i+2+2k},v_{i+3+2k},v_{i+4+2k}\in J$. This means that $u_{i+2+2k},u_{i+3+2k},u_{i+4+2k}\notin J$. Hence, the vertex $u_{i+3+2k}$ is not 2-dominated by J, a contradiction.

1.2. $v_{i+3+k}\in J$ for some i (see Figure 5).

Figure 5. The case when $u_i,u_{i+3},u_{i+6}\in J$ (the second subcase).

Then, $u_{i+3+k}\notin J$ and $u_{i+2+k},u_{i+4+k}\in J$; otherwise, they are not 2-dominated by J. Because J is an independent set, $u_{i+1+k},u_{i+5+k}\notin J$. Moreover, $u_{i+k},u_{i+6+k}\in J$ to 2-dominate $u_{i+1+k},u_{i+5+k}$. Hence, $v_{i+k},v_{i+6+k}\notin J$. To 2-dominate $v_{i+1+k},v_{i+5+k}$, we must have $v_{i+1+2k},v_{i+5+2k}\in J$. Moreover, $u_{i+1+2k},u_{i+5+2k},v_{i+3+2k}\notin J$. Since $v_{i+k},v_{i+6+k}$ have exactly one neighbour in J, vertices $v_{i+2k},v_{i+6+2k}\in J$ and $u_{i+2k},u_{i+6+2k}\notin J$. Next, $u_{i+2+2k},u_{i+4+2k}\in J$ to 2-dominate $u_{i+1+2k},u_{i+5+2k}$ and $v_{i+2+2k},v_{i+4+2k},u_{i+3+2k}\notin J$. Thus, $v_{i+2+3k},v_{i+3+3k},v_{i+4+3k}\in J$ to 2-dominate $v_{i+2+2k},v_{i+3+2k},v_{i+4+2k}$. Therefore, $u_{i+2+3k},u_{i+3+3k},u_{i+4+3k}\notin J$. This means that $u_{i+3+3k}$ is not 2-dominated, a contradiction.

Hence, for each n and k, it is not possible that the vertices $u_i,u_{i+3},u_{i+6}$ belong to a $(2$-$d)$-kernel of $P(n,k)$.

2. Now, we will prove that the configuration of the paths $P_1$, $P_2$ shown on the right side of the Figure 2 is forbidden. Suppose that $u_i,u_{i+2},u_{i+4},u_{i+7}\in J$ for some i, as in Figure 6. Then, $v_{i+5},v_{i+6}$, which causes $v_i,v_{i+1},v_{i+2},v_{i+3},v_{i+4},v_{i+5},v_{i+6},v_{i+7}\notin J$.

Figure 6. The case when $u_i,u_{i+2},u_{i+4},u_{i+7}\in J$.

We consider four subcases.

2.1. $v_{i+1},v_{i+3}\notin J$ for some i (see Figure 7).

Figure 7. The case when $u_i,u_{i+2},u_{i+4},u_{i+7}\in J$ (the first subcase).

Then, $v_{i+1-k},v_{i+3-k},v_{i+1+k},v_{i+3+k}\in J$. Since $v_{i+2}$ must be 2-dominated, so $v_{i+2-k}\in J$ or $v_{i+2+k}\in J$. Without loss of generality, assume that $v_{i+2+k}\in J$. Thus, $u_{i+1+k},u_{i+2+k},u_{i+3+k}\notin J$. Hence, the vertex $u_{i+2+k}$ is not 2-dominated, a contradiction.

2.2. $v_{i+1}\notin J$ and $v_{i+3}\in J$ for some i (see Figure 8).

Figure 8. The case when $u_i,u_{i+2},u_{i+4},u_{i+7}\in J$ (the second subcase).

Then, $v_{i+3+k},v_{i+5+k},v_{i+6+k}\notin J$. Since $v_{i+7}$ must be 2-dominated, we obtain that $v_{i+7-k}\in J$ or $v_{i+7+k}\in J$. Without loss of generality, assume that $v_{i+7+k}\in J$. Thus, $u_{i+7+k}\notin J$ and $u_{i+6+k}\in J$. Because J is an independent set and $u_{i+6+k}\in J$, $u_{i+5+k}\notin J$. Therefore, $u_{i+4+k}\in J$, which causes $u_{i+3+k},v_{i+4+k}\notin J$. Moreover, $v_{i+3+2k},v_{i+4+2k},v_{i+5+2k}\in J$, and finally $u_{i+3+2k},u_{i+4+2k},u_{i+5+2k}\notin J$. Hence, the vertex $u_{i+4+2k}$ is not 2-dominated, a contradiction.

2.3. $v_{i+1}\in J$ and $v_{i+3}\notin J$ for some i (see Figure 9).

Figure 9. The case when $u_i,u_{i+2},u_{i+4},u_{i+7}\in J$ (the third subcase).

Then, $v_{i+3+k}\in J$ and $v_{i+1+k},v_{i+5+k},v_{i+6+k},u_{i+3+k}\notin J$. Since $v_{i+4}$ must be 2-dominated, $v_{i+4-k}\in J$ or $v_{i+4+k}\in J$. Without loss of generality, assume that $v_{i+4+k}\in J$. Thus, $u_{i+4+k}\notin J$. Moreover, $u_{i+2+k},u_{i+5+k}\in J$, which causes $u_{i+1+k},u_{i+6+k},v_{i+2+k}\notin J$. To 2-dominate $u_{i+1+k}$, we must have $u_{i+k}\in J$. Then, $v_{i+k}\notin J$ and $v_{i+2k},v_{i+1+2k},v_{i+2+2k}\in J$. From the independence of the set J, we get that $u_{i+2k},u_{i+1+2k},u_{i+2+2k}\notin J$. Hence, the vertex $u_{i+1+2k}$ is not 2-dominated, a contradiction.

2.4. $v_{i+1},v_{i+3}\in J$ for some i.

Proving analogously as in subcase 2.3., we obtain a contradiction with the assumption that J is a $(2$-$d)$-kernel.

Therefore, for each n and k, it is not possible that the vertices $u_i,u_{i+2},u_{i+4},u_{i+7}$ belong to a $(2$-$d)$-kernel of $P(n,k)$.

Hence, for the graph with the $(2$-$d)$-kernel, the configurations of $P_1$, $P_2$ shown in the Figure 10 are the only ones that may be possible. Now, we will show that they are indeed possible.

Figure 10. Possible configurations of the paths $P_1$, $P_2$ for the graph $P(n,k)$ with the $(2$-$d)$-kernel.

3. Suppose that $u_i,u_{i+2},u_{i+4}\in J$ for some i, as in Figure 11. Then, $u_{i+1},u_{i+3},v_i,v_{i+2},v_{i+4}\notin J$.

Figure 11. The case when $u_i,u_{i+2},u_{i+4}\in J$.

We consider four subcases.

3.1. $v_{i+1},v_{i+3}\notin J$ for some i (see Figure 12).

Figure 12. The case when $u_i,u_{i+2},u_{i+4}\in J$ (the first subcase).

Since $v_{i+2}$ must be 2-dominated, we obtain that $v_{i+2+k}\in J$ or $v_{i+2-k}\in J$. Without loss of generality, assume that $v_{i+2+k}\in J$. Moreover, $v_{i+1+k},v_{i+3+k}\in J$ and $u_{i+1+k},u_{i+2+k},u_{i+3+k}\notin J$. Hence, the vertex $u_{i+2+k}$ is not 2-dominated, a contradiction.

3.2. $v_{i+1}\notin J$ and $v_{i+3}\in J$ for some i (see Figure 13).

Figure 13. The case when $u_i,u_{i+2},u_{i+4}\in J$ (the second subcase).

Then, $v_{i+1+k}\in J$ and $v_{i+3+k}\notin J$. Since $v_{i+2}$ must be 2-dominated, $v_{i+2+k}\in J$ or $v_{i+2-k}\in J$. Without loss of generality, assume that $v_{i+2+k}\in J$. Thus, $u_{i+1+k},u_{i+2+k}\notin J$ and $u_{i+k},u_{i+3+k}\in J$, which causes $v_{i+k},u_{i+4+k}\notin J$ and $v_{i+4+k}\in J$. Moreover, $v_{i+1+2k},v_{i+2+2k},v_{i+4+2k}\notin J$, $v_{i+2k}\in J$, $u_{i+2k}\notin J$, $u_{i+1+2k}\in J$, $u_{i+2+2k}\notin J$, $u_{i+3+2k}\in J$ and $u_{i+4+2k},v_{i+3+2k}\notin J$. Finally, $v_{i+2+3k},v_{i+3+3k},v_{i+4+3k}\in J$ and $u_{i+2+3k},u_{i+3+3k},u_{i+4+3k}\notin J$. Hence, the vertex $u_{i+3+3k}$ is not 2-dominated, a contradiction.

3.3. $v_{i+1}\in J$ and $v_{i+3}\notin J$ for some i.

Proving analogously as in subcase 3.2., we obtain a contradiction with the assumption that J is a $(2$-$d)$-kernel.

3.4. $v_{i+1},v_{i+3}\in J$ for some i (see Figure 14).

Figure 14. The case when $u_i,u_{i+2},u_{i+4}\in J$ (the fourth subcase).

Then, $v_{i+1+k},v_{i+3+k}\notin J$. First, we will show that $v_{i+k}$ and $v_{i-k}$ must belong to a $(2$-$d)$-kernel J. Suppose on contrary that $v_{i+k}\notin J$. Since $v_{i+k}$ must be 2-dominated, $u_{i+k}\in J$. Thus, $u_{i+1+k}\notin J$ and $u_{i+2+k}\in J$. Moreover, $v_{i+2k},v_{i+1+2k},v_{i+2+2k}\in J$ and $u_{i+2k},u_{i+1+2k},u_{i+2+2k}\in J$. Hence, the vertex $u_{i+1+2k}$ is not 2-dominated, a contradiction.

This means that $v_{i+k},v_{i-k}\in J$ and also $v_{i+2+k},v_{i+4+k},u_{i+1+k},u_{i+3+k}$ belong to a $(2$-$d)$-kernel (see Figure 15).

Figure 15. The case when $u_i,u_{i+2},u_{i+4}\in J$ implies that $v_{i+k},v_{i+2+k},v_{i+4+k},u_{i+1+k},u_{i+3+k}\in J$.

Hence, n must be even, and from the definition of $P(n,k)$, we conclude that k must be odd, which proves (i).

4. Suppose that $u_i,u_{i+2},u_{i+5}\in J$ for some i. Then, $u_{i+1},u_{i+3},u_{i+4},v_i,v_{i+2},v_{i+5}\notin J$. Since $u_{i+3},u_{i+4}$ must be 2-dominated, $v_{i+3},v_{i+4}\in J$. First, we prove that $v_{i+1}\notin J$. Suppose on contrary that $v_{i+1}\in J$, as in Figure 16. Then, $v_{i+1+k},v_{i+3+k},v_{i+4+k}\notin J$. Since $v_i$ must be 2-dominated, $v_{i-k}\in J$ or $v_{i+k}\in J$. Without loss of generality, assume that $v_{i+k}\in J$. Thus, $u_{i+k}\notin J$, $u_{i+1+k}\in J$ and $u_{i+2+k}\notin J$. Moreover, $u_{i+3+k}\in J$, $u_{i+4+k}\notin J$, $u_{i+5+k}\in J$, $v_{i+5+k}\notin J$, and $v_{i+2+k}\in J$. Proving analogously as in subcase 3.3., we obtain a contradiction with the assumption that J is a $(2$-$d)$-kernel.

Figure 16. The case when $u_i,u_{i+2},u_{i+5},v_{i+1}\in J$.

Hence, $v_{i+1}\notin J$ (see Figure 17). Moreover, $v_{i+1+k}\in J$ and $u_{i+1+k},v_{i+3+k},v_{i+4+k}\notin J$. We consider two subcases.

Figure 17. The case when $u_i,u_{i+2},u_{i+5}\in J$, and $v_{i+1}\notin J$.

4.1. $v_{i+2+k}\notin J$ for some i (see Figure 18).

Figure 18. The case when $u_i,u_{i+2},u_{i+5}\in J$ (the first subcase).

Then, $u_{i+2+k}\in J$, $u_{i+3+k}\notin J$, $u_{i+4+k}\in J$, $u_{i+5+k}\notin J$, and $v_{5+i+k}\in J$. Moreover, $v_{i+k}\in J$ and $u_{i+k}\notin J$; otherwise, we obtain the same configuration as in subcase 3.3.

Hence, n must be divisible by 5, and from the definition of $P(n,k)$, we conclude that $k\equiv 2\left(\mathrm{mod}5\right)$, which proves (ii).

4.2. $v_{i+2+k}\in J$ for some i (see Figure 19).

Figure 19. The case when $u_i,u_{i+2},u_{i+5}\in J$ (the second subcase).

Then, $u_{i+2+k}\notin J$, $u_{i+k},u_{i+3+k}\in J$ and $v_{i+k},u_{i+4+k}\notin J$. Moreover, $u_{i+5+k}\in J$ and $v_{i+5+k}\notin J$

Hence, n must be divisible by 5, and from the definition of $P(n,k)$, we conclude that $k\equiv 3\left(\mathrm{mod}5\right)$, which proves (iii), which ends the proof. □

Basing on the proof of Theorem 1, the following corollaries are obtained. They concern the number of $(2$-$d)$-kernels in the generalized Petersen graph as well as the lower and upper $(2$-$d)$-kernel numbers. By a rotation of configurations shown on Figure 10, condition (i) of Theorem 1 gives two $(2$-$d)$-kernels in generalized Petersen graph and conditions (ii) and (iii) give five $(2$-$d)$-kernels. Therefore, if n and k satisfy more than one of these conditions, we obtain more $(2$-$d)$-kernels. Moreover, the proof of the Theorem 1 presents the constructions of the $(2$-$d)$-kernels in the generalized Petersen graph $P(n,k)$. Figure 20 shows the smallest and the largest $(2$-$d)$-kernel in the graph $P(20,7)$.

Figure 20. The largest (left side) and the smallest (right side) $(2$-$d)$-kernel in the graph $P(20,7)$.

Corollary 1.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. Then,

$$\sigma \left(P(n,k)\right)=\left\{\begin{array}{cc}7& \mathit{for}n\equiv 0\left(\mathrm{mod}10\right)\mathit{and}k\equiv a\left(\mathrm{mod}10\right),a=3,7\hfill \\ 5& \mathit{for}n\equiv 5\left(\mathrm{mod}10\right)\mathit{and}k\equiv a\left(\mathrm{mod}5\right),a=2,3\mathit{or}\hfill \\ & \mathit{for}n\equiv 0\left(\mathrm{mod}10\right)\mathit{and}k\equiv a\left(\mathrm{mod}10\right),a=2,8\hfill \\ 2& \mathit{for}n\equiv 0\left(\mathrm{mod}10\right)\mathit{and}k\equiv a\left(\mathrm{mod}10\right),a=1,5,9\mathit{or}\hfill \\ & \mathit{for}\mathit{even}n,n\neg \equiv 0\left(\mathrm{mod}10\right)\mathit{and}\mathit{odd}k.\hfill \end{array}\right.$$

Corollary 2.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. If $n\equiv 0\left(\mathrm{mod}10\right)$ and $k\equiv a\left(\mathrm{mod}10\right)$, $a=3,7$, then

$${\gamma }_{(2-d)}\left(P(n,k)\right)=\frac{4}{5}nand{\mathsf{\Gamma }}_{(2-d)}\left(P(n,k)\right)=n.$$

Corollary 3.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. If $n\equiv 5\left(\mathrm{mod}10\right)$ and $k\equiv a\left(\mathrm{mod}5\right)$, $a=2,3$ or $n\equiv 0\left(\mathrm{mod}10\right)$ and $k\equiv a\left(\mathrm{mod}10\right)$, $a=2,8$, then

$${\gamma }_{(2-d)}\left(P(n,k)\right)={\mathsf{\Gamma }}_{(2-d)}\left(P(n,k)\right)=\frac{4}{5}n.$$

Corollary 4.

Let $n\ge 3$, $k<\frac{n}{2}$ be integers. If $n\equiv 0\left(\mathrm{mod}10\right)$ and $k\equiv a\left(\mathrm{mod}10\right)$, $a=1,5,9$ or n is even, $n\not\equiv 0\left(\mathrm{mod}10\right)$ and k is odd, then

$${\gamma }_{(2-d)}\left(P(n,k)\right)={\mathsf{\Gamma }}_{(2-d)}\left(P(n,k)\right)=n.$$

The above corollaries characterize all possible graphs $P(n,k)$, which have the $(2$-$d)$-kernel.

2.2. The Second Generalization of the Petersen Graph

Now, we consider another generalization of the Petersen graph. Let $n\ge 5$ be an integer. Let $C_n$ be a cycle and $\overline{C_n}$ its complement such that $V\left(C_n\right)=\{x_1,x_2,\dots ,x_n\}$, $V\left(\overline{C_n}\right)=\{x_1^c,x_2^c,\dots ,x_n^c\}$ with the numbering of vertices in the natural order. Let $G\left(n\right)$ be the graph such that $V\left(G\left(n\right)\right)=V\left(C_n\right)\cup V\left(\overline{C_n}\right)$ and $E\left(G\left(n\right)\right)=E\left(C_n\right)\cup E\left(\overline{C_n}\right)\cup \{x_i{x}_i^c;i\in \{1,2,\dots ,n\}\}$. Figure 21 shows an example of a $(2$-$d)$-kernel in $G\left(13\right)$. It is easy to check that if $n=5$, then $G\left(5\right)$ is isomorphic to the Petersen graph.

Figure 21. An example of a $(2$-$d)$-kernel in $G\left(13\right)$.

The next Theorem shows a complete characterization of graphs $G\left(n\right)$ with the $(2$-$d)$-kernel.

Theorem 2.

Let $n\ge 5$ be integer. The graph $G\left(n\right)$ has a $(2$-$d)$-kernel if and only if n is odd.

Proof. 

Let $n\ge 5$ be odd. We will show that $J=\{x_2^c,x_3^c,x_1,x_4,x_6,\dots ,x_{n-1}\}$ is the $(2$-$d)$-kernel of the graph $G\left(n\right)$. The independence of J is obvious. It is sufficient to show that J is a 2-dominating set. By the definition of the graph $G\left(n\right)$, we can assume that $x_{n+1}=x_1$. Suppose that $y\in V\left(G\right(n\left)\right)\backslash J$. Hence, $y\in V\left(C_n\right)$ or $y\in V\left(\overline{C_n}\right)$. Let $y\in V\left(C_n\right)$. Thus $y=x_k$, $k\in \{2,3,5,\dots ,n\}$. If $x_k^c\notin J$, then there exist vertices $x_{k-1},x_{k+1}\in J$ adjacent to $x_k$. If $x_k^c\in J$, then $k=2$ or $k=3$. For $k=2$, the vertex $x_2$ is adjacent to $x_1,x_2^c\in J$. Moreover, if $k=3$, then the vertex $x_3$ is adjacent to $x_4,x_3^c\in J$. Hence, every vertex from the set $V\left(C_n\right)$ is 2-dominated by the set J. Let now $y\in V\left(\overline{C_n}\right)$. Thus $y=x_k^c$, $k\in \{1,4,5,\dots ,n\}$. Then, the vertex $x_k^c$, $k\in \{5,6,\dots ,n\}$ is adjacent to $x_2^c,x_3^c\in J$. If $k=1$, then $x_1^c{x}_1,x_1^c{x}_3^c\in E\left(G\left(n\right)\right)$. Moreover, for $k=4$ the vertex $x_4^c$ is adjacent to $x_4,x_2^c$. Therefore, vertices from the set $V\left(\overline{C_n}\right)$ are 2-dominated by J and hence J is a $(2$-$d)$-kernel of $G\left(n\right)$.

Conversely, suppose that a graph $G\left(n\right)$ has a $(2$-$d)$-kernel J. We will show that n is odd. By the definition of the graph $G\left(n\right)$, we obtain that $J\cap V\left(\overline{C_n}\right)\ne \varnothing$. Otherwise, vertices from the set $V\left(\overline{C_n}\right)$ are not 2-dominated by the set J. Let $x_1^c\in J$. Then either $x_2^c\in J$ or $x_n^c\in J$. Otherwise, $x_2^c$ or $x_n^c$ is not 2-dominated. Hence, $|J\cap V(\overline{C_n}\left)\right|=2$. Without loss of generality assume that $x_1^c,x_2^c\in J$. This means that $x_i^c$, $i\in \{4,5,\dots ,n-1\}$ is 2-dominated by J and $x_3^c,x_n^c$ are dominated by J. Let $J^{*}=J\backslash \{x_1^c,x_2^c\}$. Then, $J^{*}\subset V\left(C\right)$. Since J is the $(2$-$d)$-kernel, $x_3,x_n\in J^{*}$; otherwise, $x_3^c,x_n^c$ are not 2-dominated by J. Therefore, the graph ${⟨\{x_3,x_4,\dots ,x_n\}⟩}_{G\left(n\right)}\cong P_{n-2}$ must have a $(2$-$d)$-kernel to 2-dominate vertices from $V\left(C_n\right)\backslash J^{*}$. This means that n must be odd. Thus, $J^{*}=\{x_3,x_5,\dots ,x_n\}$, which ends the proof. □

Finally, it turns out that if a graph $G\left(n\right)$ has $(2$-$d)$-kernel, then the number of $(2$-$d)$-kernels depends linearly on the number of vertices. Moreover, each $(2$-$d)$-kernel of $G\left(n\right)$ has the same cardinality.

Corollary 5.

If $n\ge 5$ is odd, then $\sigma \left(G\right(n\left)\right)=n$ and

$${\gamma }_{(2-d)}\left(G\left(n\right)\right)={\mathsf{\Gamma }}_{(2-d)}\left(G\left(n\right)\right)=⌊\frac{n}{2}⌋+2.$$

Proof. 

Let $n\ge 5$ be odd. From the construction of a $(2$-$d)$-kernel described in the proof of Theorem 2, we conclude that exactly two not adjacent vertices from the set $V\left(\overline{C_n}\right)\subset V\left(G\left(n\right)\right)$ belong to a $(2$-$d)$-kernel. The selection of these two vertices will determine the $(2$-$d)$-kernel in $G\left(n\right)$. Since two not adjacent vertices can be chosen on n ways, $\sigma \left(G\right(n\left)\right)=n$. Moreover, from the construction of $(2$-$d)$-kernels in $G\left(n\right)$, it follows that all of them have the same cardinality. Hence, ${\gamma }_{(2-d)}\left(G\left(n\right)\right)={\mathsf{\Gamma }}_{(2-d)}\left(G\left(n\right)\right)=⌊\frac{n}{2}⌋+2$, which ends the proof. □

3. Concluding Remarks

In this paper, we considered two different generalizations of the Petersen graph, and we discussed the problem of the existence of $(2$-$d)$-kernels in these graphs. In particular, we determined the number of $(2$-$d)$-kernels in these graphs and their lower and upper $(2$-$d)$-kernel number. The generalized Petersen graphs considered in this paper are special cases of I-graphs (see, for example, [35]). The I-graph $I(n,j,k)$ is a graph with a vertex set $V\left(I(n,j,k)\right)=\{u_1,u_2,\dots ,u_n,v_1,v_2,\dots ,v_n\}$ and an edge set $E\left(I(n,j,k)\right)=\{u_i{u}_{i+j},u_i{v}_i,v_i{v}_{i+k};i\in \{1,2,\dots ,n\}\}$, where subscripts are reduced modulo n. Because $P(n,k)=I(n,1,k)$, the results obtained could be a starting point to studying and counting $(2$-$d)$-kernels in I-graphs. It could also be interesting to investigate the number of $(2$-$d)$-kernels in other generalizations of generalized Petersen graphs. For more generalizations, see, for example, [36].