Abstract
In this paper, we establish bounds on the domination number and the metric dimension of the co-normal product graph of two simple graphs G and H in terms of parameters associated with G and H. We also give conditions on the graphs G and H for which the domination number of is 1, 2, and the domination number of G. Moreover, we give formulas for the metric dimension of the co-normal product of some families of graphs G and H as a function of associated parameters of G and H.
Keywords: Dominating set, Resolving set, Adjacency resolving set, Co-normal product of graphs
Introduction
The domination number is a parameter that has appeared in numerous location problems [19] and in the analysis of social network problems [4]. The adjacency and non-adjacency relation between two vertices u, v in a graph G is denoted by and , respectively. A set , is a dominating set [22] of G if for every , we have or for some . The minimum cardinality of a dominating set in a graph G is called the domination number of G, denoted by . The problem of finding a minimum size dominating set of a graph is in general NP-hard [13].
The metric dimension is a parameter that has appeared in robot navigation problems [20], strategies for the mastermind game [8], drug discovery problems [7, 17, 18], coin weighing problems [26], network discovery and verification problems [3]. The notation or simply denotes the distance between two vertices , which is the length of a shortest path between them. For an ordered set and a vertex , the k-vector , is called the metric representation of v with respect to W, denoted by . A set is a resolving (locating) set [14, 27] of G if for any two distinct vertices , , which means that there exists at least one vertex for which . A minimum resolving set of G is called a metric basis of G and its cardinality is called the metric dimension of G, denoted by . Gary and Johnson [13] noted that the problem of finding the metric dimension of a graph is NP-hard; however, its explicit construction is given by Khuller et al. [20]. The problem of finding the metric dimension of a graph is formulated as an integer programming problem by Chartrand et al. [7]. Relations between the domination number and the metric dimension of a graph are given in [1].
It is found in [2] that there are 256 possible products of any two graphs using the adjacency and the non-adjacency relations of these graphs. Several interesting types of graph products have been studied extensively in the literature. For instance, Caceres et al. [6], Yero et al. [29], Rodriguez-Velazquez et al. [24], Saputroa et al. [25], and Jannesari and Omoomi [16] investigated the metric dimension of the cartesian product, the corona product, the strong product, and the lexicographic product of graphs, respectively.
Out of product graphs, there is another well-known product graph introduced by Ore in 1962 [22], with the name cartesian sum of graphs. It was named co-normal product of graphs in [12]. Different properties and results regarding coloring and the chromatic number of the co-normal product of graphs are discussed in [5, 9, 11, 12, 23, 28]. In [21], Kuziak et al. studied the strong metric dimension of the co-normal product of graphs using the strong metric dimension of its components. In this paper, we have studied the domination number and the metric dimension of the co-normal product of graphs.
All considered graphs in this paper are non-trivial, simple and finite. In the next section, we describe some structural properties of the co-normal product of graphs. In Sect. 3, we study the domination number of the co-normal product of graphs and describe conditions on the graphs G and H so that the domination number of is 1, 2, and . We also give bounds on the domination number of the co-normal product of graphs. In Sect. 4, we describe some properties of resolving sets in the co-normal product of graphs and give bounds on the metric dimension of the co-normal product of graphs. Moreover, we establish formulas for the metric dimension of some families of graphs.
Methods
We use the combinatorial computing, combinatorial inequalities and graph theoretic analytic methods to prove the main results. The aim of this research is to provide bounds on the domination number and the metric dimension of the co-normal product of graphs and to give exact formulas for the metric dimension of some families of graphs.
Co-normal product of graphs
The co-normal product (the terminology we have adopted) of a graph G of order m with the vertex set and a graph H of order n with the vertex set , is the graph with the vertex set and the adjacency relation defined as if in G or in H. All results given in this paper for also hold for due to the commutativity of this product. Figure 1 shows the co-normal product graph of two path graphs.
Figure 1.
The co-normal product graph of and
A graph having n vertices in which each vertex is adjacent to all other vertices is called a complete graph, denoted by . In [12], Frelih and Miklavic discussed the connectivity of and proved the following theorem.
Theorem 1
(Frelih and Miklavic)
is connected if and only if one of the following holds:
for some and G is connected.
for some and H is connected.
G and H are not null graphs and at least one of G or H is without isolated vertices.
The diameter of a graph G, denoted by , is the maximum distance between any two vertices of G. If G is a disconnected graph then . A graph having n vertices and no edges is called a null graph, denoted by . In [21], Kuziak et al. discussed the diameter of and proved the following theorem.
Theorem 2
(Kuziak, Yero, Rodriguez-Velazquez)
Let G and H be two non-trivial graphs such that at least one of them is non-complete and let be an integer. Then the following assertions hold:
.
G and H have isolated vertices, then .
If neither G nor H has isolated vertices, then .
If , then .
If , H has no isolated vertices and G is not a null graph having at least one isolated vertex, then .
The set of all vertices adjacent with a vertex , is called the open neighborhood of v in G, denoted by or simply . The cardinality of is called the degree of v in G, denoted by or simply . In the next two observations, we give formulas for the degree and the neighborhood of a vertex in using the structure of the co-normal product of graphs.
Observation 1
For any vertex ,
Observation 2
For any vertex ,
Two vertices having the same neighbors are called false twins. In the next theorem, we describe conditions for any two distinct vertices in to be false twins.
Theorem 3
For any two distinct vertices and in , if and only if in G and in H.
Proof
Let in , then, by Observation 2, we have , which shows that in G and in H. The converse follows from the definition of the co-normal product of graphs. □
Let , the set , is an equivalence class of false twins in . Using Observation 2, we have the following straightforward lemma.
Lemma 1
For any vertex , we have , where , are equivalence classes of false twins in G and H, respectively.
Domination in co-normal product of graphs
A vertex of a graph G is a dominating vertex if its degree is . Throughout this section and the next section, the graphs G, H and are as described in Sect. 2. We define vertex sets, and for and . In Fig. 1, we represent such classes. In the next two results, we give conditions on G and H for which have domination numbers 1 or 2.
Lemma 2
A vertex is a dominating vertex in if and only if and are dominating vertices in G and H, respectively.
Proof
Let be a dominating vertex in . To show that , are dominating in G and H, respectively, assume contrary that is not dominating in G so there exists such that , then a contradiction.
Now suppose that and are dominating vertices in G and H, respectively, then, by Observation 1, we have . □
Lemma 3
If G has a dominating vertex and H has no dominating vertex, then .
Proof
Suppose is a dominating vertex of G, so using the definition of co-normal product, for all with . Also, H has no dominating vertex so there must be a vertex such that , which shows that . Now for any vertex , the set , is a dominating set for for any chosen vertex . Hence, . □
A set is a total dominating set [10] of G, if every vertex is adjacent to an element of D. The total domination number, denoted by , is the cardinality of a minimum total dominating set for G. In the next theorem, we give conditions on G and H so that , by using the total domination number of G.
Theorem 4
For any two connected graphs G and H with , if and only if .
Proof
Let and be a minimum total dominating set of G. Consider the set where , and . To prove that , we only need to prove that D is a minimum dominating set for . First, we show that D is a dominating for . Clearly, . Now for if , then and if with , then there exists such that because is a total dominating set of G so . Suppose , then there exists such that so . Hence, D is dominating set for .
Now to prove that D is a minimum dominating set, assume contrarily that be a minimum dominating set such that . Consider the sets and then and are not dominating sets for G and H, respectively, which shows that there exists and such that , and , a contradiction. Hence, D is a minimum dominating set for .
Conversely, suppose and D be a minimum dominating set for . Let and . Since , we have also by given condition. For , there exist and such that , and and for with is not a dominating set for G a similar argument shows that D is not a dominating set for . If is a minimum dominating set for G, we are to prove that . Assume to the contrary that , then there exist such that and such that , which shows that , a contradiction to the assumption that . Hence, . □
Lemma 2, shows that if and only if . In the next theorem, we give general bounds on the domination number of .
Theorem 5
For any two connected graphs G and H, .
Proof
Let , be dominating sets for G, H, respectively and . To show that D is a dominating set for , consider a vertex , we have following cases:
Case 1: If and , then .
Case 2: If and , then there exists such that . As so .
Case 3: If and , then there exists such that . As so .
Case 4: Let and , then there exist and such that and so for . Hence, D is a dominating set for and .
Now for lower bound, consider . Suppose that and , then, by Lemma 2, . Also for and , Lemma 3, shows that . Suppose and be any set such that . To prove lower bound, we need to prove that D is not a dominating set for . Let and . Since , and are not dominating sets of G and H, respectively, which shows that there exist vertices and such that and . Using the definition of the co-normal product of graphs and . Hence, D is not a dominating set for . □
Note that the lower bound given in Theorem 5, is attainable when .
Metric dimension in co-normal product of graphs
In this section, we study the properties of resolving sets in and establish formulas for the co-normal product of some families of graphs. In Theorem 10, we give bounds on the metric dimension of the co-normal product of a connected graph G and a graph H (not necessarily connected). In the rest of this paper, we assume G and H such that is connected. Moreover, has diameter at most two unless otherwise stated. In the next lemma, we will prove that, for every , and an ordered set , the distance of to the vertices of is equal if and in G.
Lemma 4
Let has diameter 2 and be an ordered subset of for some . If for some , then, for every in G, .
Proof
To show that , we will show that for each . Let , for some . Since has diameter 2, we have . First suppose that , then in H and in . Hence, . Now suppose that , which shows that in H and in . Hence, . □
For and ; , clearly and gives a partition of W. For any vertex , the code of with respect to W can be represented as:
In the next lemma, we give conditions on an ordered set to be a resolving set for .
Lemma 5
A set is a resolving set for if and only if for any two distinct vertices there exists at least one vertex such that , where .
Proof
Suppose W is a resolving set for and there exist two distinct vertices , in such that, for every in G, we have . Then for every because has diameter two and because , a contradiction.
Conversely, suppose for any two distinct vertices , there exists at least one vertex such that . Since has diameter at most 2, we have and hence showing that W is a resolving set for . □
In [15], the authors proved the following corollary, which gives the relation between resolving sets and false twins of a graph.
Corollary 1
(Hernando, Mora, Pelaya, Seara, Wood)
Suppose u, v are twins in a connected graph G and W resolves G. Then u or v is in W. Moreover, if and , then also resolves G.
Using Corollary 1, and Lemma 1, if H has false twins then, for every resolving set W of , for each . In the next theorem, we give conditions on G and H for which there exists a resolving set W of such that for some .
Theorem 6
Let G be a connected graph and H be an arbitrary graph such that . There exists a resolving set W for such that for some if and only if H has no false twins.
Proof
Let W be a resolving set of such that , for some . Assume contrary that for two distinct vertices , then, by Lemma 3, in for each so in . As so by Corollary 1, W is not a resolving set for , a contradiction.
Conversely, consider a set such that , for some , where is not a dominating vertex in G. To prove the converse, we only need to prove that W is a resolving set for . Let be two distinct vertices for some . Since H have no false twins and diameter at least 2, there exists at least one vertex, say , such that or . Now for every in G, we have or , which shows that . Hence, W is a resolving set for . □
The following corollary directly follows from Theorem 6, which gives the relation between dominating sets and resolving sets of , when both G, H are connected.
Corollary 2
For any two connected graphs G and H if at least one of G, H has false twins, then every resolving set of is a dominating set of .
In the next theorem, we give conditions on G and H for which the metric dimension of is the order of G times the metric dimension of H.
Theorem 7
Let be the distinct equivalence classes of false twins in a connected graph H with the property that for each and G be a connected graph having no false twins, then .
Proof
Since , for any two distinct vertices , G has distinct equivalence classes of false twins. Lemma 1, shows that the co-normal product has equivalence classes of false twins such that no class has cardinality 1, so . Also for each and , which shows that . Hence, . □
Let ; be a path graph and ; for each i, be a complete multipartite graph have k distinct equivalence classes of false twins. Since have no false twins, by Theorem 7, we have the following corollary.
Corollary 3
If ; and , then .
In [16], Jannesari and Omoomi introduced the concept of the adjacency metric dimension of a graph and used it to find the metric dimension of lexicographic product of graphs. A function defined as:
for , is called the adjacency function of G. The k-vector for a vertex , is called the adjacency metric representation of v with respect to W, denoted by . A set W is an adjacency resolving set for G if for any two distinct vertices , or . A minimum adjacency resolving set of G is called an adjacency basis of G and its cardinality is called the adjacency metric dimension of G, denoted by . They also gave that if G is a connected graph with diameter 2, then but the converse is not true because , while . Our next lemma directly follows from the definition of adjacency basis and the fact that the induced subgraph of is isomorphic to H, for each .
Lemma 6
If has diameter at most 3 and is an adjacency basis for H, then, for any , the vertices of are resolved by its subset .
Now consider a path graph having the vertex set such that ; and a star graph having the vertex set such that ; . The co-normal product graph of and is shown in Fig. 2. Note that, for every adjacency basis of , and is not a resolving set for . Let 1 represents a vector whose each entry is 1 and 2 represents a vector whose each entry is 2, i.e. and . In the next theorem, we provide conditions under which is a resolving set for , where is an adjacency basis of H and .
Figure 2.
The co-normal product graph of and
Theorem 8
Let G be a connected graph having no false twins and H be a graph such that has diameter at most three. If there exists an adjacency basis of H such that for all , then .
Proof
Let and or . By Lemma 6, resolves all the vertices of . To show that W is a resolving set for , consider two distinct vertices such that . Since G has no false twins, we have for all and for , W resolves , . Now for , we have also , which shows that W resolves , . Hence, by Lemma 5, is a resolving set for . □
Corollary 4
Let G be a complete graph and be an arbitrary graph. If H has an adjacency basis such that for all , then .
Proof
Since G is complete, G has no false twins. Also, satisfies the condition of Theorem 8, so . Now for some , consider and note that, for any , will not resolves the vertices of , because is an adjacency basis of H, so there exists such that in H, which shows that for all because G is complete. Hence, . □
In the next theorem, we give a formula for the metric dimension of when G is complete and H is a graph for which each adjacency basis has one vertex such that .
Theorem 9
Let G be a complete graph and be an arbitrary graph. If for each adjacency basis of H, there exists a vertex such that , then .
Proof
By using Lemma 6, will resolve the vertices of . Since G is complete, for all . Also for each . Hence, is not a resolving set for . Also the induced subgraph of the vertex set is isomorphic to G and G is complete. Hence, . □
Since is complete if and only if G and H are complete, . Also, if G is trivial and if H is trivial. Note that if and only if one of G or H is trivial. In the next theorem, we give bounds for the metric dimension of when G and H are non-trivial and at least one is not a complete graph.
Theorem 10
Let G be a connected graph and be an arbitrary graph, then
Proof
Let , where , are adjacency basis of G and H respectively. Let for and for . For any vertex , the metric representation is of the form or . For any two distinct vertices , we have and . To prove that W is a resolving set for , we discuss the following cases:
Case 1: Let and . Lemma 6, shows that resolves the vertices of also shows that . Hence, .
Case 2: Let and . resolves the vertices of , which shows that . Hence, .
Case 3: Let and . Since and are adjacency bases for G and H, respectively, we have and . Also, shows that , which implies W is a resolving set for .
For the lower bound, let and be adjacency basis for G and H, respectively and . We consider the following cases:
Case 1: Suppose G or H has false twins. Since, for every , we can have , by Theorem 6, W is not a resolving set for if H has false twins. A similar argument holds if G has false twins.
Case 2: Suppose neither G nor H have false twins. As is an adjacency basis for H so there exists at least one vertex such that for some . Also, for and the definition of the co-normal product graph gives . Hence, W is not a resolving set for . □
For a complete graph G and a null graph H, Theorem 2(1) shows that and the metric dimension of is given in the next theorem.
Theorem 11
If G is a complete graph and H is a null graph, then .
Proof
Let and . It is clear from the definition of co-normal product that, for each , for all . So any resolving set must contain at least vertices from each , which shows that . Since H is a null graph, we have for each i and for each , which shows that any subset of containing vertices from each will be a resolving set for . Hence, . □
In the next theorem, we give formula for the metric dimension of when G is a path graph and H is a star graph.
Theorem 12
For any two integers , if G is a path graph and H is a star graph having order m and respectively, then .
Proof
Let and , where in H. Also, for all , by using Lemma 3, we have for each i. So, any resolving set W for G must contain at least vertices from each . Since , by the definition of a co-normal product for all and , which means that the vertices of are not resolved by any of , , . Also, in and induced subgraph of is isomorphic to G so we must choose vertices from , which shows that . □
Conclusions
To study the product graphs with respect to graph theoretic parameters is always an important problem. In this paper, we have studied two parameters, the domination number and the metric dimension of the co-normal product of two graphs G and H. These two parameters have a lot of applications in networks and facility location problems. We have given conditions on G and H under which the graph has the domination number 1, 2 and . We also proved that, for any two connected graphs G and H, . We described some properties of resolving sets of and gave conditions on G and H such that . We have also given conditions on G and H under which . For a complete graph G and a non-complete graph H, we have given conditions on H under which and . For a connected graph G and a non-complete graph H, we proved that . We have also given explicit formulas for the metric dimension of the co-normal product of a path graph and a complete multipartite graph, a complete graph and a null graph, a path graph and a star graph for the first time. Our derived inequality relations can be very helpful in the characterizations of graphs with given metric dimension or given domination number.
Acknowledgements
The authors are thankful to the anonymous referees for their valuable suggestions, which led to the improvement of the paper.
Authors’ contributions
All authors contributed equally and significantly in conducting this research work and writing this paper. All authors read and approved the final manuscript.
Funding
This research is supported by the Start-up Research Grant 2016 of United Arab Emirates University (UAEU), Al Ain, United Arab Emirates via Grant No. G00002233 and UPAR Grant of UAEU via Grant No. G00002590.
Competing interests
The authors declare that they have no competing interests.
Footnotes
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Contributor Information
Imran Javaid, Email: imran.javaid@bzu.edu.pk.
Shahid ur Rehman, Email: shahidurrehman1982@gmail.com.
Muhammad Imran, Email: imrandhab@gmail.com.
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