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Topological Indices, and Applications of Graph Theory

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Volume 2021 |Article ID 6689816 | https://doi.org/10.1155/2021/6689816

G. Muhiuddin, N. Sridharan, D. Al-Kadi, S. Amutha, M. E. Elnair, "Reinforcement Number of a Graph with respect to Half-Domination", Journal of Mathematics, vol. 2021, Article ID 6689816, 7 pages, 2021. https://doi.org/10.1155/2021/6689816

Reinforcement Number of a Graph with respect to Half-Domination

Academic Editor: Ismail Naci Cangul
Received17 Dec 2020
Revised01 Feb 2021
Accepted30 Mar 2021
Published14 Apr 2021

Abstract

In this paper, we introduce the concept of reinforcement number with respect to half-domination and initiate a study on this parameter. Furthermore, we obtain various upper bounds for this parameter. AMS subject classification: 05C38, 05C40, 05C69.

1. Introduction

Throughout our discussion, we consider only simple finite graphs. For graph theoretic terminologies, we refer the readers to [1]. If is a graph to each vertex of , denotes the set of all vertices of which are adjacent to . For any subset , . For a vertex of , and . A vertex is said to dominate itself and its adjacent vertices. In other words, a vertex dominates a vertex if . A subset of is said to be a dominating set of if . The minimum cardinality of a dominating set of is denoted by and is called the domination number of . The total domination number of a graph G, denoted by , is the minimum cardinality of a total dominating set of G and, for their properties, we refer the reader to [26]. In [7], Kulli introduced the concept of the cobondage number of a graph, which is the minimum number of edges to be added to reduce the domination number. The same concept has been independently introduced and studied earlier by others under the name “reinforcement number” (refer to chapter 17 of Haynes et al.’s work [8]). Total reinforcement number of a graph has been studied in [912]. Recently, Muhiuddin et al. have studied various related concepts on graphs (see, e.g., [1317]).

In [18, 19], a new domination parameter , where , was introduced and a study on had been initiated. In this paper, we introduce the concept of reinforcement number with respect to half-domination and initiate a study on this parameter.

2. Definition and Examples

If is a graph for every vertix of , we define a map as follows:for all . The map is called the half-domination factor of .

A subset of is said to be a -dominating set for if, for each , . The minimum cardinality of a -dominating set is called the -domination number of and is denoted by . A -dominating set of with the minimum cardinality is said to be a -set of . The parameter was introduced in [18] and bounds for are obtained in [18].

Theorem 1 (see [18]). If is a graph on vertices and , then

Theorem 2 (see [18]). If is a connected graph with , then . Moreover, trees with , and are characterized in [18].

Definition 1 (reinforcement number with respect to half-domination). Let be a graph with . The reinforcement number with respect to half-domination is denoted by and is given by . For graphs , with , is not defined.

Remark 1. If , then (let be a vertex of with deg. If , then is a -dominating set for ).
for some standard graphs: here, we discuss path on vertices. As , if , it follows that , , , and .

Theorem 3. If , then .

Proof. for all . Let be . If or , then is a -set of . If , then and is a half-dominating set for . Thus, if or , .
If or , then is a -set of . Let and . Then, is a half-dominating set for , where . Thus, . We claim that . If , then there is a subset of and an edge such that and is a half-dominating set for . With respect the path , either or (otherwise, cannot be a half-dominating set of ). In (by ignoring the contribution by the edge ), . So, . But the contribution by the edge will be at most 2. Thus, cannot be a half-dominating set for , which is a contradiction. Thus, , if or .
One can easily prove the following theorem.

Theorem 4. If is a cycle on -vertices , then

Remark 2. We compare , the reinforcement number of with respect to -domination, and , the reinforcement number with respect to the total-domination for some graphs, to show that there is no correlation between and .(a)If , the path on eight vertices, and , while and (b)If , the path on nine vertices, and , while and (c)If , the cycle on eight vertices, and , while and

Theorem 5. If is a tree with , then .

Proof. As , diam. Let be a diagonal path in (so ). Let be a -set of . We can assume that . If , let for some . Then is a half-dominating set for . Assume that . If , select . Let . Then is a half-dominating set of . If , find , such that . Let , , and . Then is a half-dominating set for , where . Thus, , for all trees with .

Remark 3. An example for a tree , with , is given in Figure 1.

3. Bounds for

In this section, we obtain certain bounds for . If , then . First, we have the following theorem.

Theorem 6. If , then .

Proof. Let be a graph with . Let be a vertex of degree . As , contains at least vertices. Select a subset of such that and . Let and . Then is a dominating set of . Thus, and , as .

Corollary 1. If , if .

As a corollary, we get a better upper bound for which is given by inequality (1).

Corollary 2. For all graphs with , .

Proof. Let be a connected graph with . Let be a -set of . Let , , and . Note that if , then for some , . Then, is a partition of . As is a -set of , to each , is not a half-dominating set of and hence the set , called private neighbors of , consisting of all which are not half-dominated by the set , is not empty. Note that , where and and . As , either or . We write , , and for , , and , respectively. Now, we observe the following.

Observation 1. .

Proof. As is a -set of , for each , either or , so for each , . Hence, is a dominating set of . Thus, .

Observation 2. If and for at least one -set of , then .

Proof. If , then is also a -set of and the maximum degree of the induced subgraph (if, for some , deg in , then is also a half-dominating set of , which is a contradiction as ). If is an edge in , select a vertex (note that ). Then is a half-dominating set for , where . Thus, in this case, . If is an independent set in , select . Let . Then, is a half-dominating set for . Thus, .

Observation 3. If for some -set of , then .

Proof. If , and hence is a half-dominating set of .

Observation 4. If is a connected graph with , then .

Proof. If for any , (with respect to some -set of ), then, by our Observation 1, and so (as is connected, ). If , then also and hence the result is true. So assume that and for all , . If , for some , then for exactly one . Thus, each is counted twice in . As , . So, and hence we can find one such that . To each vertex , select a vertex such that is an edge in , and select a vertex such that (since , there is a vertex such that ), and let . Let , where and . Then, and hence .

Observation 5. If , for some in a -set , then .

Observation 6. For a connected graph , with radius of and , (radius of to each vertex , there is a vertex such that ).

Proof. Let . Let be a vertex of with deg. Let be a vertex such that . Then, . Let . Then, contains at least vertices. Select vertices in . Let . Note that . Now, is a half-dominating set for . So , where .

Remark 4. The upper bound given in Observation 6 is attained by the graph in Figure 2. For this graph, , , , , and .

Remark 5. For a connected graph with radius and , Observation 6 yields a better bound for than the bound given by inequality (1). . In the following theorem, we obtain an upper bound for in terms of .

Theorem 7. For any graph , with , .

Proof. Let be a -set of . Select three vertices in . Let . Then, and is a half-dominating set of ; thus, .

Remark 6. There are graphs , for which and .

Example 1. (Example for a graph with and ). Consider the graph given in Figure 3. First, we note that for this graph , , , and . ( attains its lower bound given in (1). The set is the unique -set of (for any vertex , and if and for all . If is any three-element subset of , other than , then ). , where , , and . Assume that is a two-element subset of and such that is a half-domination set of . In order to show that , it is enough to prove that . Claim: .
In our further discussion, we consider the following notations.
If , we write , where , and the number . We say that the edge contributes the value to a vertex of if . Similarly, we say that contributes the value to if . If , we define and . We note that for all and . We can also assume that every edge in has one end in . Now, we establish our claim by considering various possibilities for (for each possibility, we consider one subcase; other subcases can be proved similarly):Case (i): . Assume that . Then if . If, for an edge , for one of the elements in , then for and is one of the ends of . Similarly, if for one , then for and is one of the ends of . Thus, .Case (ii): . Assume that . Then if and if . Let be such that ; for ; and for every , where . Then ; only if . If , contains at least two edges and so that and . So, in this case, . If , contains a set of at least three edges such that , where . Thus, in this case, also .Case (iii): . Assume that . Then if and if . If such that for , then ( contains at least two edges with one end at and at least two edges with one end at ). contains an edge with one end at and and contains another edge with one end at and . Thus, .Case (iv): and . Assume that . Then, if and if . The set contains a subset , with , for , and each edge in has one end at . Note that , either or , and either or . So, there is a set contained in with such that for . Thus, .Case (v): and . Assume that . Then, if and if . There are edges and in with one end at and . Note that and . There are edges and in , with one end at and ; . Now either or , and either or . As and for and for , we have .Case (vi): and . Let . Then for and if . The set should contain at least three edges to attribute positive values to ; it should contain at least two edges with one end at to contribute positive values to and and one edge with one end at to contribute a positive value to . Thus, . The upper bound for can be reduced further if .

Theorem 8. If and , then

Proof. Let . Let be a -set of . Let . Let . If , select one such that . If , then either or . If , then, for each vertex , . So, we can select one such that (this is possible since ). Thus, to each , we have selected one . If , we can select such that . Let . If , is a half-dominating set of . If , then is a half-dominating set of , where for some (as is not a half-dominating set of , ). Thus, if and if . In the following theorem, we obtain a lower bound for of some classes of graphs.

Theorem 9. If is a connected graph with and the order