Research Article | Open Access

Ramy Shaheen, Mohammad Assaad, Ali Kassem, "On Eternal Domination of Generalized ", *Journal of Applied Mathematics*, vol. 2021, Article ID 8882598, 7 pages, 2021. https://doi.org/10.1155/2021/8882598

# On Eternal Domination of Generalized

**Academic Editor:**Frank Werner

#### Abstract

An eternal dominating set of a graph is a set of guards distributed on the vertices of a dominating set so that each vertex can be occupied by one guard only. These guards can defend any infinite series of attacks, an attack is defended by moving one guard along an edge from its position to the attacked vertex. We consider the “all guards move” of the eternal dominating set problem, in which one guard has to move to the attacked vertex, and all the remaining guards are allowed to move to an adjacent vertex or stay in their current positions after each attack in order to form a dominating set on the graph and at each step can be moved after each attack. The “all guards move model” is called the -eternal domination model. The size of the smallest -eternal dominating set is called the -eternal domination number and is denoted by . In this paper, we find the domination number of Jahangir graph for , and the -eternal domination numbers of for are arbitraries.

#### 1. Introduction

The term graph protection refers to the process of placing guards or mobile agents in order to defend against a sequence of attacks on a network. Go to [1–5] for more background of the graph protection problem. In 2004, Burger et al. [2] introduced the concept of eternal domination. Goddard et al. [3] introduced the “all guards move model” and determined general bounds of where denotes the independence number of and denotes the domination number of The -eternal domination numbers for cycles and paths were found by Goddard et al. [3] as follows: and . For further information on eternal domination, see survey [1]. The -dominating graph was defined by Goldwasser, Klostermeyer, and Mynhardt in [6] as follows: let be a graph with a dominating set of cardinality . The vertex set of the -dominating graph , denoted , is the set of all subsets of of the size which are dominating sets, and two vertices of are adjacent if and only if the guards occupying the vertices of of one can move (at most distance one each) to the vertices of the other, if and only if has an induced subgraph such that for each vertex of , the union of the vertices in the closed neighborhood of in is equal to .

A generalized Jahangir graph is a graph on vertices, i.e., a graph consisting of a cycle with one additional vertex which is adjacent to vertices of at the distance from each other on , see [7] for more information on the Jahangir graph. Let be the label of the central vertex and be the labels of the vertices that incident clockwise on cycle so that . We will use this labeling for the rest of the paper. The vertices that are adjacent to have the labels We denote the set by So,

Mtarneh, Hasni, Akhbari, and Movahedi claim to have found the domination number of in [8] as follows:

Theorem 1 [8]. * *

However, in this paper, we will prove that this theorem does not hold when

Theorem 2 [3].

In a previous paper, we found that

Theorem 3 [9].

Theorem 4 [9].

#### 2. Main Results

In this section, we prove that Theorem 1 does not hold if , and we find in these cases; then, we find for and arbitrary . We also give an upper and a lower bound for the -eternal domination number of for and arbitrary

Proposition 5. *The *

*Proof. *Let be a dominating set of with cardinality . Since , then ; therefore, . We imply that the vertices of that are dominated by separate the cycle into paths each of which has vertices and needs vertices to dominate its vertices. If , then we will need at least vertices to dominate the non-dominated paths of which means we will be needing vertices, and that is a contradiction. This means . We have three sets of cardinality that can dominate the cycle when . They are
However, must contain at least one vertex adjacent to vertex to dominate it. Since , then . Since , then we can obviously see that . However, considering for , then it is obvious that the set and Which means only dominates both the cycle and the vertex at the same time. Therefore, , and this set is unique.

Theorem 6.

*Proof. *Let be a set of cardinality . Let . We imply that the vertices of that are dominated by separate the cycle into paths each of which has vertices and needs vertices to dominate its vertices. We denote these paths by Since , then ; therefore, and . Let be the family of dominating sets of paths , respectively. We know that for an arbitrary path , the dominating set is defined as , which means is a dominating set of with cardinality It is known that the dominating set of a path is unique; therefore, is unique because each of is unique. Let us prove that whatever set which with cardinality is not enough to dominate . We consider the following cases, as illustrated in Figure 1 for .

*Case 1. *.

In this case, we only have vertices to dominate which is impossible because it leaves one path with dominating vertices, which means three vertices of will not be dominated, and that is a contradiction.

*Case 2. *.

In this case, . We know that . However, is one of the following:

It is obvious that , which means cannot dominate all the vertices of . Therefore, cannot dominate entirely. From Case 1 and Case 2, we imply that Hence, which means . However, ; therefore, .

Theorem 7.

*Proof. *We know that a set of vertices can dominate . Let us identify as Since , therefore dominates which means it dominates entirely. However, let us consider another dominating set of cardinality , where =. We consider the two following cases:

*Case 1. * is even.

Let us prove that we can derive a dominating set of cardinality from . Since , it is obvious that dominates all the vertices of ; therefore, dominates . Let us consider the first two paths identified in Theorem 6. We notice that , while and We need to dominate and so we include it in our new set . However, since are dominated by , respectively, then we do not need to add to ; therefore, we can start with and recreate the same distribution of dominating sets we used on the unit {}. This distribution spares one vertex () from the need of being protected by a dominating vertex from Without loss of generality, we can use this distribution for every two consecutive paths of , since we have paths, and is even, and this distribution spares vertices from the need of being dominated by a dominating vertex from ; therefore,

*Case 2. * is odd.

We use the same discussion in Case 1 on paths (), but since we have one additional path (), we will have one less vertex to spare from the need of being dominated by a dominating vertex from ; therefore, in this case.

From Cases 1 and 2, we imply that vertex can dominate ; therefore,

Let be a set of cardinality , and we consider the following cases:

*Case 1. *

In this case, there will be at least two nondominated vertices of even after applying the strategy we showed earlier.

*Case 2. *

In this case, and since and , we have a contradiction.

We conclude from both Case 1 and Case 2 that

From (5) and (6), we conclude that .

Theorem 8.

*Proof. *From Theorem 2, we have . With the right distribution of the required guards on the vertices of the outer cycle and placing one additional guard on the central vertex this additional guard would not have to move to defend any attack on the vertices of since guards are enough to eternally dominate the vertices of . The vertex cannot be attacked because it is occupied which means .

Theorem 9. *.*

*Proof. *From Theorem 1, we know that . We imply that Theorem 7 that is the unique set of We consider an attack on an arbitrary vertex . The only guard that can move to to defend the attack is the guard located on . We can only consider the two following strategies:

Strategy 1: we move the guard from to without moving any other guard and that would leave unprotected.

Strategy 2: we make a full rotation (clockwise) from the vertices of set to the vertices of the only set that contains and can dominate as well. This required set is the set But from Theorem 7, we imply that ; therefore, the central vertex will not be protected anymore. Without loss of generality, both strategies fail for any vertex , and we conclude that From Theorem 8, we know the result that Then, see Figures 2 and 3.

Lemma 10. *For any path, with the dominating set and for any vertex if is attacked, then there is a way to move the guards to keep all vertices of protected with the exception of , respectively.*

*Proof. *It is obvious that any path can be divided into subsets each of which contains three vertices. To dominate each subset, we need only one guard located on the middle vertex, so each vertex dominates itself,

If is attacked, then the guard on moves to which would leave unprotected. However, the guard on moves to , and so all the guards need to move one step to the left which means the guard on moves to leaving unprotected.

If is attacked, then the guard on moves to which would leave unprotected. However, the guard on moves to , and so all the guards need to move one step to the right which means the guard on moves to leaving unprotected.

Theorem 11. * for *

*Proof. *Let be a Jahangir graph with , and let be the unique dominating set of cardinality for that was identified in Theorem 9. We position a guard on each vertex of Since is unique, then it will fail at defending against the first attack. We study a series of arbitrary attacks on denoted by (), and we assume that occurs on ; then, the only guard capable of defending the attack is located on but moving this guard to would leave the remaining vertices of unprotected. And since we cannot bring any other guard to to protect them which means , but from the definition of Jahangir graph and also from Theorem 8, we can easily find that which means

We add a guard on an arbitrary vertex so the new set of guards consisting of guards is . We will denote this distribution of guards by

We notice that the vertices of partition are into subgraph of vertices so that belongs to every subgraph. We denote these subgraphs by . We have

We assume occurs on a vertex , and that ; then, the guard set on is , and we consider the following cases:

*Case 1. * is occupied.

In this case, the only guard capable of moving to is located on which leaves unprotected and in order to protect it, the guard on moves to , as in Lemma 10 which all the guards located on vertices of move one step towards the higher index which leaves the end vertex with the lowest index unprotected, and the guard on is the only one capable of moving directly to to protect . This is possible because of and this movement will make unprotected. To avoid that, we move the guard on to to protect all the vertices of . So, the new set of guards located on is

We assume the attacks occur on as well. We notice that the previous distribution with the two following distributions: Form an eternal dominating family on for the following reasons: (i)Each two sets of are reachable from each other in one step, see Figure 4

It is obvious that is a special case of , because in , while in .

After a series of attacks on , we assume an attack occurs on because each one of contains and a vertex of (); then, it is possible to repeat the same previous strategy to defend the attack and apply it to When trying to defend , the guard of that moves to to protect is the vertex taking into consideration that the distributions , can move back to in one step while is already a special case of Without loss of generality, this strategy applies to every subgraph , and any series of attacks taking into consideration that the proof does not change if when occurs (in this case, we start at instead of ).

*Case 2. * is occupied.

By following the same process in Case 1 with one difference which is moving the guards of towards the lower index when defending , becomes unprotected, as in Case 1, and the guard on moves to to protect , and moves to forming , in a similar way to Case 1. form an eternal dominating family on , and the same argument about can be followed here.

*Case 3. *.

When occurs, we consider the following subcases:*Case 3(a)*. The current distribution is , and then the guard set is The guard on moves to , the guard on moves to , and the new set of guards is ; therefore, we remain in .*Case 3(b)*. The current distribution is in the previously attacked subgraph, and let us assume this subgraph is while the remaining subgraphs fall under ; then, the guard on moves to , the guard on moves to to defend the attack, and the guards move back to so the new set of guards is therefore, we remain in . Without loss of generality, this strategy applies no matter which subgraph was attacked in the previous attack.*Case 3(c)*. The current distribution is in the previously attacked subgraph, and we use the same argument in Case 3(b).

From all the previous cases, we can see that guards can eternally dominate Therefore,

From (8) and (13), we conclude that for . Figure 4 also demonstrates how 10 guards can move directly from any guard distribution of to the distribution of , and this direct transportation can occur from every guard distribution of any subgraph to all other distributions of every subgraph .

Corollary 12.

*Proof. *We conclude 1 by combining Theorems 9 and 11.

We know that ; therefore, from Theorems 7 and 8, we conclude 2.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

This work was supported by the Faculty of Science, Tishreen University, Syria.

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#### Copyright

Copyright © 2021 Ramy Shaheen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.