On Eternal Domination of Generalized Js , m

An eternal dominating set of a graph G 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 m-eternal domination model. The size of the smallestm-eternal dominating set is called the m-eternal domination number and is denoted by γm ðGÞ. In this paper, we find the domination number of Jahangir graph Js,m for s ≡ 1, 2 ðmod 3Þ, and the m-eternal domination numbers of Js,m for s,m are arbitraries.


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][2][3][4][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 γðGÞ ≤ γ ∞ m ðGÞ ≤ αðGÞ, where αðGÞ denotes the independence number of G and γðGÞ denotes the domination number of G: The m-eternal domination numbers for cycles C n and paths P n were found by Goddard et al. [3] as follows: γ ∞ m ðC n Þ = dn/3e and γ ∞ m ðP n Þ = dn/2e. For further information on eternal domination, see survey [1]. The k-dominating graph HðG, kÞ was defined by Goldwasser, Klostermeyer, and Mynhardt in [6] as follows: let G be a graph with a dominating set of cardinality k. The vertex set of the k-dominating graph HðG, kÞ, denoted VðHÞ, is the set of all subsets of VðGÞ of the size k which are dominating sets, and two vertices of H are adjacent if and only if the k guards occupying the vertices of G of one can move (at most distance one each) to the vertices of the other, γ ∞ m ðGÞ ≤ k if and only if HðG, kÞ has an induced subgraph SðG, kÞ such that for each vertex x of SðG, kÞ, the union of the vertices in the closed neighborhood of x in SðG, kÞ is equal to VðGÞ.
A generalized Jahangir graph J s,m for m ≥ 2 is a graph on sm + 1 vertices, i.e., a graph consisting of a cycle C sm with one additional vertex which is adjacent to m vertices of C sm at the distance s from each other on C sm , see [7] for more information on the Jahangir graph. Let v sm+1 be the label of the central vertex and v 1 , v 2 , ⋯, v sm be the labels of the vertices that incident clockwise on cycle C sm so that deg ðv 1 Þ = 3. We will use this labeling for the rest of the paper. The vertices that are adjacent to v sm+1 have the labels v 1 , v 1+s , v 1+2s , :: v 1+ðm−1Þs : We denote the set fv 1 , v 1+s , ::v 1+ðm−1Þs g by R: So, R = fv 1+is : i = 0, 1, ::, m − 1g: Mtarneh, Hasni, Akhbari, and Movahedi claim to have found the domination number of J s,m in [8] as follows: Theorem 1 [8]. For s, m ≥ 2,γðJ s,m Þ = dsm/3e: However, in this paper, we will prove that this theorem does not hold when s ≡ 1, 2ðmod 3Þ: In a previous paper, we found that Theorem 3 [9].

Main Results
In this section, we prove that Theorem 1 does not hold if s ≡ 1, 2ðmod 3Þ, and we find γðJ s,m Þ in these cases; then, we find γ ∞ m ðJ s,m Þ for s ≡ 0, 1ðmod 3Þ and arbitrary m. We also give an upper and a lower bound for the m-eternal domination number of J s,m for s ≡ 2ðmod 3Þ and arbitrary m: Proof. Let D 0 be a dominating set of J s,m with cardinality sm/3. Since s ≡ 0 ðmod 3Þ, then sm ≡ 0 ðmod 3Þ; therefore, dsm/3e = sm/3. We imply that the vertices of R that are dominated by v sm+1 separate the cycle C sm into m paths each of which has s − 1 vertices and needs ds − 1/3e vertices to dominate its vertices. If v sm+1 ∈ D 0 , then we will need at least mds − 1/3e = sm/3 vertices to dominate the m non-dominated paths of C sm which means we will be needing ðsm/3Þ + 1 vertices, and that is a contradiction. This means D 0 ⊂ VðC sm Þ. We have three sets of cardinality sm/3 that can dominate the cycle C sm when sm ≡ 0 ðmod 3Þ. They are However, D 0 must contain at least one vertex adjacent to vertex v sm+1 to dominate it. Since s ≡ 0 ðmod 3Þ, then s = 3k : k = 1, 2, ⋯, ðs/3Þ . Since s = 3k, then we can obviously see that R = fv 1+3ik : i = 0, 1, ⋯, m − 1 and k = 1, 2, ⋯, m − 1g. However, considering ik > i for k > 1, then it is obvious that the set R ⊂ S 1 , therefore S 1 ∩ R = R, S 2 ∩ R = ∅, and S 3 ∩ R = ∅: Which means only S 1 dominates both the cycle C sm and the vertex v sm+1 at the same time. Therefore, D 0 = S 1 , and this set is unique. Proof. Let S be a set of cardinality mðs − 1/3Þ + 1. Let v sm+1 ∈ S. We imply that the vertices of R that are dominated by v sm+1 separate the cycle C sm into m paths each of which has s − 1 vertices and needs ds − 1/3e vertices to dominate its vertices. We denote these paths by T i : i = 1, ::, m: Since s ≡ 1 ðmod 3Þ, then s − 1 ≡ 0 ðmod 3Þ; therefore, mds − 1/3e = mðs − 1/3Þ and jT i j ≡ 0 ðmod 3Þ. Let S i : i = 1, ::, m be the family of dominating sets of paths T i : i = 1, ::, m, respectively. We know that for an arbitrary path T k , the dominating set is defined as S k = fv sðk−1Þ+3 , v sðk−1Þ+6 , ::, v ks−1 g, which means S = S m i=1 S i ∪ fv sm+1 g is a dominating set of J s,m with cardinality m ðs − 1/3Þ + 1: It is known that the k/3 − dominating set of a path P k : k ≡ 0 ðmod 3Þ is unique; therefore, S is unique because each of S i : i = 1, ::, m is unique. Let us prove that whatever set B which with cardinality mðs − 1/3Þ is not enough to dominate J s,m . We consider the following cases, as illustrated in Figure 1 for J 7,4 .
In this case, we only have mðs − 1/3Þ − 1 vertices to dominate T i : i = 1, ::, m which is impossible because it leaves one path T k with ðs − 1/3Þ − 1 dominating vertices, which means three vertices of T k will not be dominated, and that is a contradiction.
However, let us consider another dominating set D 1 of cardinality dsm/3e + 1, where D 1 =D 0 ∪ fv sm+1 g. We consider the two following cases: Case 1. m is even.
Let us prove that we can derive a dominating set D 2 of cardinality dsm − m/2/3e + 1 from D 1 . Since v sm+1 ∈ D 1 , it is obvious that v sm+1 dominates all the vertices of R; therefore, v sm+1 dominates v 1 , v 1+s , v 1+2s . Let us consider the first two paths T 1 , T 2 identified in Theorem 6. We notice that D 1 ∩ 2 Journal of Applied Mathematics , v s+7 , ::, v 2s−1 g and v 1+s ∈ D 1 : We need v 1+s to dominate v s and v 2+s so we include it in our new set D 2 . However, since v 2s , v 1+2s are dominated by v 2s−1 , v sm+1 , respectively, then we do not need to add v 2+2s to D 2 ; therefore, we can start D 2 ∩ VðT 3 Þ with v 3+2s and recreate the same distribution of dominating sets we used on the unit {T 1 , v 1+s , T 2 }. This distribution spares one vertex (v 1+2s ) from the need of being protected by a dominating vertex from C sm : Without loss of generality, we can use this distribution for every two consecutive paths of C sm , since we have m paths, and m is even, and this distribution spares m/2 vertices from the need of being dominated by a dominating vertex from C sm ; therefore, jD 2 j = dsm − m/ 2/3e + 1: We use the same discussion in Case 1 on paths (T 1 , ::, T m−1 ), but since we have one additional path (T m ), we will have one less vertex to spare from the need of being dominated by a dominating vertex from C sm ; therefore, jD 2 j = dsm − m − 1/2/3e + 1 in this case.
From Cases 1 and 2, we imply that dsm − bm/2c/3e + 1 vertex can dominate J s,m if s ≡ 2 ðmod 3Þ; therefore, Let X be a set of cardinality dsm − bm/2c/3e, and we consider the following cases: In this case, there will be at least two nondominated vertices of C sm even after applying the strategy we showed earlier.
Proof. From Theorem 2, we have γ ∞ m ðC sm Þ = dsm/3e. With the right distribution of the required dsm/3e guards on the vertices of the outer cycle C sm and placing one additional   Journal of Applied Mathematics guard on the central vertex v sm+1 , this additional guard would not have to move to defend any attack on the vertices of C sm since dsm/3e guards are enough to eternally dominate the vertices of C sm . The vertex v sm+1 cannot be attacked because it is occupied which means γ ∞ m ðJ s,m Þ ≤ dsm/3e + 1.
Proof. From Theorem 1, we know that γðJ s,m Þ = ðsm/3Þfor s ≡ 0 ðmod 3Þ. We imply that Theorem 7 that S 0 = fv 1 , v 4 , ⋯, v sm−2 g is the unique sm/3 − dominating set of J s,m : We consider an attack on an arbitrary vertex v 2 ∈ VðC sm Þ − S 0 . The only guard that can move to v 2 to defend the attack is the guard located on v 1 . We can only consider the two following strategies: Strategy 1: we move the guard from v 1 to v 2 without moving any other guard and that would leave v sm unprotected.
Strategy 2: we make a full rotation (clockwise) from the vertices of set S 0 to the vertices of the only set that contains v 2 and can dominate C sm as well. This required set is the set S 2 = fv 2 , v 5 , ⋯, v sm−1 g = fv 2+3i : i = 0, 1 ⋯ , sm/3 − 1g: But from Theorem 7, we imply that S 2 ∩ R = ∅; therefore, the central vertex v sm+1 will not be protected anymore. Without loss of generality, both strategies fail for any vertex v i ∈ VðC sm Þ − S 0 , and we conclude that Proof. It is obvious that any path P k : k ≡ 0ðmod 3Þ can be divided into k/3 subsets each of which contains three vertices. To dominate each subset, we need only one guard located on the middle vertex, so each vertex v i ∈ S dominates itself, v i−1 , v i+1 : If v i−1 is attacked, then the guard on v i moves to v i−1 which would leave v i+1 unprotected. However, the guard on v i+3 moves to v i+2 , and so all the guards need to move one step to the left which means the guard on v k−1 moves to v k−2 leaving v k unprotected.
If v i+1 is attacked, then the guard on v i moves to v i+1 which would leave v i−1 unprotected. However, the guard on v i−3 moves to v i−2 , and so all the guards need to move one step to the right which means the guard on v 2 moves to v 3 leaving v 1 unprotected.
Theorem 11. γ ∞ m ðJ s,m Þ = mbs/3c + 2 for s ≡ 1ðmod 3Þ: Proof. Let J s,m be a Jahangir graph with s ≡ 1ðmod 3Þ, and let S be the unique dominating set of cardinality mbs/3c + 1 for J s,m that was identified in Theorem 9. We position a guard on each vertex of S: Since S is unique, then it will fail at defending J s,m against the first attack. We study a series of arbitrary attacks on J s,m denoted by (A 1 , A 2 , ⋯, A i , ⋯), and we assume that A 1 occurs on v l ∈ R; then, the only guard capable of defending the attack is located on v sm+1 but moving this guard to v l would leave the remaining vertices of R unprotected. And since we cannot bring any other guard to v sm+1 to protect them which means γ ∞ m ðJ s,m Þ ≠ mbs/3c + 1, but from the definition of Jahangir graph and also from Theorem 8, we can easily find that γ ∞ m ðJ s,m Þ ≥ mbs/3c + 1 which means We add a guard on an arbitrary vertex y ∈ R so the new set of guards consisting of mbs/3c + 2 guards is S ∪ fyg. We will denote this distribution of guards by M 0 : We notice that the vertices of R partition J s,m are into m subgraph of s + 2 vertices so that v sm+1 belongs to every subgraph. We denote these subgraphs by SJ i : i = 1, ::, m. We have We assume A 1 occurs on a vertex v i ∈ SJ 1 , and that y ∉ SJ 1 ; then, the guard set on SJ 1 is fv 3 , v 6 , ⋯, v s−1 g, and we consider the following cases: In this case, the only guard capable of moving to v i is located on v i−1 which leaves v i−2 unprotected and in order to protect it, the guard on v i−4 moves to v i−3 , as in Lemma 10 which all the guards located on vertices of T 1 move one step towards the higher index which leaves the end vertex with the lowest index v 2 unprotected, and the guard on v sm+1 is the only one capable of moving directly to v 1 to protect v 2 . This is possible because of v 1 ∈ R, and this movement will make R − fv 1 , yg unprotected. To avoid that, we move the guard on y to v sm+1 to protect all the vertices of R. So, the new set of guards located on SJ 1 is We assume the attacks A 2 , A 3 , ⋯ occur on SJ 1 as well. We notice that the previous distribution W 1 with the two following distributions: Form an eternal dominating family on SJ 1 for the following reasons: It is obvious that W 3 is a special case of M 0 , because y ∈ fv 1 , v s+1 g in W 3 , while y ∈ R in M 0 .
After a series of attacks on SJ 1 , we assume an attack A l occurs on v k ∈ SJ k : k ≠ 1, because each one of W 1 , W 2 , W 3 contains v sm+1 and a vertex of R (v 1 or v s+1 ); then, it is possible to repeat the same previous strategy to defend the attack A 1 and apply it to SJ k : When trying to defend A l , the guard of R that moves to v 1+sðk−1Þ to protect v 2+sðk−1Þ is the vertex y ∈ fv 1 , v s+1 g taking into consideration that the distribu-tions W 1 − y, W 2 − y can move back to M 0 in one step while W 3 is already a special case of M 0 : Without loss of generality, this strategy applies to every subgraph SJ k , and any series of attacks taking into consideration that the proof does not change if y ∈ SJ 1 when A 1 occurs (in this case, we start at W 3 instead of M 0 ).
By following the same process in Case 1 with one difference which is moving the guards of T 1 towards the lower index when defending A 1 , v s becomes unprotected, as in Case 1, and the guard on v sm+1 moves to v s+1 to protect v s , and y moves to v sm+1 forming W 2 , in a similar way to Case 1. W 1 ,