Abstract
A set of a graph is called a co-independent liar’s dominating set of if (i) for all , , (ii) for every pair of distinct vertices, , and (iii) the induced subgraph of on has no edge. The minimum cardinality of vertices in such a set is called the co-independent liar’s domination number of , and it is denoted by . In this paper, we introduce the concept of co-independent liar’s domination number of the middle graph of some standard graphs such as path and cycle graphs, and we propose some bounds on this new parameter.
1. Introduction
For notations and nomenclature, we refer [1]. Specifically, let be a graph with vertex set of order and edge set of size . The diameter of is the greatest distance between any two vertices of . The middle graph is the derived graph obtained from by inserting a new vertex into every edge of and then joining these new vertices by edges which lie on the adjacent edges of [2]. Haynes et al. introduced the concept of domination in graphs [3].
A topological index is a real number related to a graph, which must be a structural invariant. The topological indices are a vital tool for quantitative structure activity relationship and quantitative structure property relationship. For more work on topological indices of a graph, refer recent papers [4, 5].
The concept of liar’s domination was introduced by Panda and Paul in [6]. A graph admits a liar’s dominating set if each of its connected components has at least three vertices. Several different domination parameters were studied in [7–12]. For references on liar’s domination, see, for instance, [2, 13]. A subset of a graph is called a co-independent liar’s dominating set of if (i) for all , , (ii) for every pair of distinct vertices, , and (iii) the induced subgraph of on has no edge. The minimum cardinality of vertices in such a set is called the co-independent liar’s domination number of , and it is denoted by . In this paper, we initiate the study of co-independent liar’s domination in graphs.
2. Co-Independent Liar’s Domination in Graphs
In this section, we first strengthen the co-independent liar’s domination number of the middle graphs of some standard graphs. Eventually, some bounds will be obtained.
Theorem 1. Let be the middle graph of a path graph of order . Then,
Proof. Let be the vertices of and also the vertices in be . Let . We prove that all the vertices of get a co-independent liars dominating set arising in four cases:(1)Case (i): let . Recall that . So, in and . Therefore, , for all , and every set consisting of a vertex of should be a component of and . So, for . Therefore, . Hence, , and it is independent.(2)Case (ii): let be an element of . Then, and for . Let for and . Then, Therefore, , and also, we get for all . Next, we demonstrate that for every pair of distinct vertices. Note that , , and . So, we have for . Therefore, , and no two elements are adjacent in .(3)Case (iii): let . Then, , and we can write . We have , and therefore, . Let . Then, and . We get and similarly .(4)Case (iv): let be an element of and . Then,and we obtain for . Therefore, , and for all , . Hence, is a co-independent liar’s dominating set of and .
Theorem 2. Let be the middle graph of a cycle graph. Then,
Proof. Let the vertices of be and be the vertices of . We investigate all vertices of to get a co-independent liar’s dominating set in two cases:(1)Case (i): let where , and let be an element of . Then, contains at least two vertices of and it double dominates for . Hence, If , then we consider . Let , where . So, we have Therefore, double dominates for . Clearly, we have as and for the vertices and , where for . Likewise, we have as and for the vertices and , where for . Therefore, triple dominates for . Hence, , and no two elements in can form an edge.(2)Case (ii): let , where , and let be an element of .We obtain for . Therefore, triple dominates for . So, . For every , . Hence, is a co-independent liar’s dominating set of and .
Note 1. Let be the middle graph of a wheel graph, then a co-independent liar’s dominating set does not exist.
We have the following lower bound on co-independent liar’s domination number in terms of the diameter of a graph.
Theorem 3. Let be a graph of order . Then,
Proof. Let be a set. We employ an induction on the number of components of to indicate that , and hence the result will follow: suppose that has precisely one component, that is, is connected and . We show that the distance between any pair of vertices in is atmost . Let and be two distinct vertices of . If , then . Next, assume that and . It may be verified that . Since , there are at least two vertices and in . Similarly, it may be concluded that there are two vertices and in . If , then since . Suppose that . Assume, without loss of generality, that . It is easy to verify that . Now,Next, we take and . As before, there are two vertices and in . If , then as . Let . Without loss of generality, we might assume that . Then, . Suppose that the result is true for the number of components of which are less than . Let , where is the component of for . Let be the set of all vertices of with at least two neighbors in and . If for every , then . In order to maximize the diameter, without loss of generality, we may assume that, for , and for every , . Let for . Let be two distinct vertices of with . Then,By induction, we find that . Hence,Hence, . Therefore, .
We now study some bounds on co-independent liar’s domination number in terms of the components of a cut vertex deleted graph.
Theorem 4. Let be a cut vertex of a graph and be the components of . If and for , then
Proof. First, we show that . Let be a set and for . If , we have and if , then . Clearly, for every vertex, , , , and . Moreover, for any pair , , , and also the set is independent. We have to consider following two cases: Case (i): let . For any , there is at least one vertex , so . Since each vertex should be double dominated by , there is at least a vertex , . Therefore, for each vertex , Hence, let and . So, for every vertex , and for every , . Hence, is a co-independent liar’s dominating set for each , and therefore, For each vertex , . Repeating these, each vertex should be double dominated by . In such a component, there is at least one vertex . Let for each vertex , . Hence, for every vertex , Also, and no two vertices can be adjacent in . There is at least one vertex . As before, for , we have Let . For each vertex , we have Thus, for each pair of vertices in for , is a co-independent liar’s dominating set and Thus, in any case, as , is a co-independent liar’s dominating set for . Thus, So, . Case (ii): let . Because , there is at least one component. Without loss of generality, we might assume that . Let , where is an arbitrary vertex in . As for every vertex , , there are at least two vertices . So, for each , we haveand , and for every , . Therefore,Similarly, the second inequality can be proven.
We now characterize the graphs according to the sensitivity of a co-independent liar’s dominating set versus a cut edge.
Theorem 5. Let be a cut edge (bridge) in a graph and and be the components of . If and , then
Proof. Let be a co-independent liar’s dominating set and , . When , deleting the edge does not change the sizes of co-independent liar’s dominating sets of and . Assume that and . Then, clearly, for every component of having a minimum of four vertices, we can assume that there are two vertices and and has no edge. Similarly, there are two vertices , , and also, the subgraph induced by is independent. Let and . Then, and form co-independent liar’s dominating sets for and andNext, suppose that both and . Then, there is a vertex or . Without loss of generality, we might assume that it is . Let , where and , where and . Therefore, form co-independent liar’s dominating sets of and andThe right side inequality follows from the fact that the union of the set and the set forms a co-independent liar’s dominating set for .
Theorem 6. Let and be connected and , , then .
Proof. The upper bound is obvious. We established the lower bound. Let be a connected graph of order such that is connected. Clearly, and . The result is obvious if min. Hence, let min. Without loss of generality, assume that . We show that . Let be a set and be a set. We partition the set into two sets and . Note that . Since is connected and none of the vertices in are adjacent to any vertex in in , we can deduce that . Hence, . Let , where is not adjacent in for . Since is connected, we may think that . Now, the only vertex adjacent to in is , so . In addition, according to the co-independent liar’s domination which has been discussed, . Therefore, .
Theorem 7. Let be the middle graph of a tree with . Then, .
Proof. Let the vertices of be and be the vertices of . We prove that all vertices of arise in four cases to get a co-independent liars dominating set:(i)Case (i): if , that is, is a pendant vertex in , then in and . Therefore, , , and all the vertices of should be components of , . So, for . Therefore, .(ii)Case (ii): if in , then for . Let , where and . Then, . Therefore, and . Next, we demonstrate that for every pair of distinct vertices. We see that , , and . So, we have , where . Therefore, , and no two elements are adjacent in .(iii)Case (iii): if in , then . We have and . Let be . Then, and . We get and .(iv)Case (iv): if in , thenWe obtain , where . Therefore, , and for all , . Hence, is the co-independent liar’s dominating set of and .
Remark 1. (i) A tree is a co-independent liar’s dominating set. Indeed in Figure 1(a), let . If we take , then which is a co-independent liar’s dominating set. (ii) Co-independent liar’s dominating set need not exist for all trees. For example, in Figure 1(b), let . Suppose we take the co-independent liar’s dominating set . But which satisfies conditions (i) and (ii) but not (iii).(i)For all , (ii)For every pair of distinct vertices, (iii)The induced subgraph of on has no edge
(a)
(b)
3. Conclusion
In this paper, the co-independent liar’s domination number of the middle graphs of some graph classes such as path and cycle graphs is calculated. Also, some general results and bounds on the co-independent liar’s domination number of graphs are obtained. It has been shown that no general result can be obtained for trees, unicyclic, bicyclic, and tricyclic graphs in terms of co-independent liar’s domination number.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.