Open Journal of Discrete Mathematics
Vol.09 No.03(2019), Article ID:93904,14 pages
10.4236/ojdm.2019.93008
Domination and Eternal Domination of Jahangir Graph
Ramy Shaheen, Mohammad Assaad, Ali Kassem
Department of Mathematics, Faculty of Science, Tishreen University, Lattakia, Syria
Copyright © 2019 by author(s) and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
Received: April 1, 2019; Accepted: July 23, 2019; Published: July 26, 2019
ABSTRACT
In the eternal dominating set problem, guards form a dominating set on a graph and at each step, a vertex is attacked. 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 position after each attack
Keywords:
Jahangir Graph, Graph Protection, Domination Number, Eternal Domination
1. Introduction
In graph protection, mobile agents or guards are placed on vertices in order to defend against a sequence of attacks on a network. See [1] [2] [3] [4] [5] for more background of the graph protection problem. The first idea for eternal domination was introduced by Burger et al. in 2004 [1]. The “all guards move model” or “multiple guards move version” of eternal domination was introduced by Goddard et al. [2]. General bounds of were determined in [2], where denotes the domination number of G and denotes independence number of G. The eternal domination number for cycles
and paths was found by Goddard et al. [2] as follows: and . Jahangir Graph for is a graph on vertices,
i.e. a graph consisting of a cycle with one additional vertex which is adjacent to m vertices of at distance s from each other on see [6] for more information on 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 R. So, . By definition, for , Jahangir Graph is the wheel graph and it was mentioned in [6] that for . The k-dominating graph was defined by Goldwasser et al. [7] as follows: Let G be a graph with a dominating set of cardinality k. The vertex set of the k-dominating graph , denoted , is the set of all subsets of of 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, if and only if has an induced subgraph such that for each vertex x of , the union of the vertices in the closed neighborhood of x in is equal to .
Proposition 1.1 [6] : for .
2. Main Results
Eternal Domination Number of
In this section, we give the exact eternal domination number of .
Lemma 2.1: Let us have a graph . For when m is even and
when m is odd, then a set of cardinality can’t dominate if .
Proof: Since that means all the vertices of S are vertices from the
outer cycle . We know that . So let’s find out the values of m for which: . This arbitrator is true for m is even with and for m is odd with . ■
Theorem 2.1.
Proof: We know from the definition of eternal domination that
.
Therefore from proposition 1.1, we have for . This
means we only need to prove that for . In order to do that we form the k-dominating graph on graph with and . We consider the following cases:
Case 1. : It was found in [3] that . Therefore . However, it is obvious that two vertices can dominate if and only if both vertices belong to the outer cycle . Therefore if the central vertex is attacked, then one of the two guards that are located on the two dominating vertices would have to move to making it impossible for the new distribution of guards to dominate the entire graph because doesn’t belong to any of the 2-dominating sets of . Therefore . We form , the 3-dominating graph on . Let be the induced subgraph of on vertices: ,,. Since are all adjacent and , therefore we have , which means , therefore .
Case 2. : We have . Let’s form to be the induced subgraph of on vertices ,,. Since are adjacent and therefore we have
which means .
Case 3. : We have . Let’s form to be the
induced subgraph of on the following vertices: ,,,. Since are all adjacent and , therefore
which means .
Case 4. : We have . Let’s form to be the
induced subgraph of on the following vertices: ,,,. Since are all adjacent and therefore
which means .
Case 5. : We have . Let’s form to be the
induced subgraph of on these vertices: ,,,. Since are adjacent and , therefore
which means .
Case 6. : We have . Let’s form to be the induced subgraph of on the following vertices: ,,,. Since are adjacent and , therefore
which means .
Case 7. : We have . Let’s form to be the
induced subgraph of on the vertices: ,,,. Since are all adjacent and , therefore
which means .
(See Figure 1 for ). ■
Lemma 2.2: for and .
Proof: From the definition of eternal domination, we already know that
. By proposition 1.1, for . We just need to prove that for and . We consider both cases:
Case 1: m is even for .
In this case the sets
and
are the only two minimum dominating sets (γ-dominating set) of where both and are similar by symmetry. We study an arbitrary attack on a vertex from a graph protected by and we prove that fails to eternally protect . Let the attacked vertex have an odd (index) label, . The only guard protecting in this case is the guard occupying the central vertex (which is adjacent to all the odd vertices of ). This means the guard on has to move to to defend the attack. However, that would leave the vertices: unprotected. To try to avoid that we have two strategies:
Strategy 1: We move another guard (occupying an odd vertex : ) to to keep protected. However, that would leave at least one of the two vertices unprotected and this strategy fails, see Figure 2.
Strategy 2: We don’t move any other guard to which would leave
guards on the vertices of cycle to protect . By Lemma 2.1
Figure 1. .
Figure 2. Illustrating strategy 1 when m = 8.
these guards can’t protect if therefore this strategy fails as well, see Figure 3.
Since both of these strategies fail then for &
. Without loss of generality, the same argument can be followed to
prove that guards can’t eternally protect in case the minimum dominating set is .
Case 2: m is odd for .
In this case the minimum dominating sets (γ-dominating sets) of are:
Figure 3. Illustrating strategy 2 when m = 8.
We study an arbitrary attack on a vertex from three cases of protected by of respectively. We prove that fail to eternally protect these graphs. Let the attacked vertex have an odd (index) label, . The only guard protecting in this case is the guard occupying the central vertex (which is adjacent to all the odd vertices of ). This means the guard on has to move to to defend the attack. However, that would leave the vertices: unprotected. To try to avoid that we have two strategies:
Strategy 1: We move another guard (occupying an odd vertex ) to vertex to keep protected. However, that would leave at least one of the two neighboring vertices to unprotected therefore this strategy fails, see Figure 4.
Strategy 2: We don’t move any other guard to which would leave guards on the vertices of cycle to protect . By Lemma 2.1 these guards can’t protect if , therefore this strategy fails as well, see Figure 5.
Since both strategies fail then for m is odd and .
Without loss of generality, the same argument can be followed to prove that
guards cannot eternally protect in case the minimum dominating set is or . From cases 1 and 2 we conclude that:
for and .
However, we know for , therefore:
for and . ■
Figure 4. For Strategy 1 on .
Figure 5. For Strategy 2 on .
Theorem 2.2: for and .
Proof: From Lemma 2.2 It is enough to prove the existence of one eternal
dominating family of the vertices of with cardinality in order to prove that . We consider both cases:
Case a. m is even:
We start by forming the k-dominating graph denoted on with
. is a dominating set of . We form
Y the family of dominating sets as follows : . Hence the cardinality of is . Therefore each set of the family Y is a vertex of . It is obvious that the union of these vertices is . We now need to prove that these vertices are all adjacent in . There are two types of sets depending on the label of the vertex :
Type 1:
.
Type 2:
.
When an arbitrary unoccupied vertex is attacked we consider the following cases:
Case a.1. : we consider the following cases:
Case a.1.1. is an unoccupied odd vertex : In this case the guard on the central vertex moves to to defend the attack and the guard on moves to to protect the remaining odd vertices without disturbing the protection of the even vertices (which are protected by the
guards on the vertices of ) (see Figure 6), which means guards are
enough to protect in this case. After defending the attack and since the resulting dominating set . We will now use the same argument to prove the results for all cases of when m is even, taking into consideration that the same path can be followed to prove all the possible cases of when m is odd.
Case a.1.2. is an even vertex : In this case the neighboring odd vertex has the only available guard to defend . So the guard on (or ) moves to to defend the attack leaving (or ) respectively unprotected, so the guard on moves to (or ) respectively, and the guard on moves to to protect the remaining vertices of M. While the guards on the vertices of the set keep protecting those vertices and the even vertices of leaving protected, see Figure 7.
After defending the attack and since the resulting dominating set .
Figure 6. .
Figure 7. .
Case a.2: , we consider the following cases:
Case a.2.1: is an unoccupied odd vertex . In this case the guard on moves to , either or . So the guard on (or ) moves to and the guard on moves to (or ) respectively, keeping the entire graph protected. After defending the attack and since the resulting dominating set . Figure 8, illustrates the
process in which guards can successfully defend when the
attacked vertex has an odd index label ( ) while the additional guard besides has an even index label ( ).
Case a.2.2: is an even index vertex : In this case , so the guard on (or ) moves to . The guard on moves to (or ) respectively. The guard on (or ) moves to and the guard on moves to (or ) respectively leaving the graph fully protected, see Figure 9.
After defending the attack and since the resulting dominating set .
After discussing all possible cases we find that for any : are
adjacent in because the guards occupying can move to occupy in one move and vice versa. Therefore we form on
Figure 8. .
Figure 9. .
the vertices , therefore these vertices are adjacent in the induced subgraph . It is obvious that , therefore
for m is even and . However, from Lemma 2.2 . Therefore for m is even and .
Case b. m is odd and :
We begin by forming the k-dominating graph on with
. is a dominating set of . We
form Y the family of dominating sets : .
Hence the cardinality of is . Therefore each set of the family Y is a vertex of . It is obvious that the union of these vertices is . We now need to prove that these vertices are all adjacent in .
There are two types of dependeng on the label of the vertex :
Type 1: O = { where and is an unoccupied odd vertex of }.
Type 2: Q = { where and is an even vertex of }.
By following the same argument that we followed in case a, we conclude that:
for m is odd and .
From case a and case b we conclude that:
for and . ■
3. Domination and Eternal Domination Numbers of
In this section we consider the graph . So, we found the exact domination and eternal domination numbers of .
Theorem 3.1: for and the γ-dominating set is unique.
Proof: For , let . Since it is easy to verify that the set of vertices is a dominating set of cardinality m for . Therefore for . Let such that and is a dominating set of . We consider the following cases:
Case a: Let then the vertex clearly dominates m vertices of the cycle , which leaves the remaining guards in the set to dominate the remaining 2m vertices of cycle .
are m subsets of cardinality 2, each consists of two non-dominated vertices of . In order to dominate each of these subsets we need a vertex , which means we need at least m vertices to dominate these remaining vertices. Therefore since , there are at least four vertices of that no vertices of can dominate which is a contradiction.
Case b: . In this case is a dominating set of vertices that
dominates a cycle which creates a contradiction since .
Therefore, . Finally, by case a and case b we conclude that is the unique dominating set of cardinality m for . ■
Lemma 3.2: for .
Proof: In Theorem 3.1, we found that . Since , we conclude that . Let be the m-dominating set of and let’s assume that each vertex of is occupied by a guard. When an unoccupied vertex is attacked we consider the following cases:
Case a. The attacked vertex : In this case the only guard that can move to to defend the attack is or because .
Case a.1: If the guard is situated on then it moves to to defend the attack. Therefore all the guards of the exterior cycle should move one edge (counter clockwise) to keep the cycle protected making the new dominating set . However, according to Theorem 3.1 the vertex won’t be protected anymore, see Figure 10.
Case a.2: If the guard is situated on then it moves to and all the guards on the vertices of should move one edge clockwise to keep the cycle protected making the new dominating set . However, according to Theorem 3.1 the vertex won’t be protected anymore, see Figure 11.
Case b: The attacked vertex is . In this case, there are m guards on the vertices of each one qualifies to move to . Let have the guard that moves to . This leaves the two vertices unprotected and there are no available guards on the cycle to protect them without leaving gaps of unprotected vertices. From cases a and b we conclude that . Hence . ■
Theorem 3.3: for .
Proof: We form the (k-dominating Graph) on with . Let the dominating sets be defined as follows:
Each of is a dominating set of the cardinality for therefore they are vertices of and they are all adjacent in because they are reachable from each other in one step only. With the guard on the central vertex staying in place, the dominating sets can result from each other as follows:
We form the induced subgraph from on the previous vertices . Since then .
Now, by last results together with Lemma 3.2 that . Therefore . See Figure 12. ■
Figure 10. .
Figure 11. .
Figure 12. .
4. Conclusion
In this paper, we studied the eternal domination number of Jahangir graph for =2, 3 and arbitrary m. We also find the domination number for . By using the same approach, we will work to find the eternal domination number of Jahangir graph for arbitraries s and m.
Conflicts of Interest
The authors declare no conflicts of interest regarding the publication of this paper.
Cite this paper
Shaheen, R., Assaad, M. and Kassem, A. (2019) Domination and Eternal Domination of Jahangir Graph. Open Journal of Discrete Mathematics, 9, 68-81. https://doi.org/10.4236/ojdm.2019.93008
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