Abstract
Let be a simple connected graph, be a vertex, and be an edge. The distance between the vertex and edge e is given by , A vertex distinguishes two edges , if . A set S is said to be resolving if every pair of edges of G is distinguished by some vertices of S. A resolving set with minimum cardinality is the basis for G, and this cardinality is the edge metric dimension of G, denoted by . It has already been proved that the edge metric dimension is an NP-hard problem. The main objective of this article is to study the edge metric dimension of some families of wheel-related graphs and prove that these families have unbounded edge metric dimension. Moreover, the results are compared with the metric dimension of these graphs.
1. Introduction and Preliminary
The molecular structure of a chemical compound is known as molecular graph (or chemical graph). There are different structures of chemical compounds, and every structure correlated with many chemical properties, and these properties can be calculated through some specific mathematical formulas. Graph theory helps us to study and analyze these dense structures in detail. In graph theory, we consider every atom as a vertex and covalent bound between atoms as an edge. A graph G consists of a set of objects called vertices, set of connections between vertices called edges , and graph G is usually denoted as . The number of vertices in G denoted by is often called the order of G, while the number of edges is known as size of G and denoted by . The concept of metric dimension was introduced by Slater in 1975 [1], and this concept independently was elaborated for graphs by Harary and Melter in 1976 [2]. After these two papers, lot of research papers have been published related to its theoretical properties and applications. The theoretical studies on metric dimension are highly contributed by several authors like, for instance, independent resolving sets [3], resolving dominating sets [4], strong resolving sets [5], local resolving sets [6], k-metric generators [7, 8], simultaneous metric generators [9], resolving partitions [10], strong resolving partitions [11], and k-antiresolving sets [12]. The metric dimension has many applications in the digital world and in mathematical chemistry such as network verification [13], robot navigation, image processing and pattern recognition [14], mastermind game [15], chemistry [16, 17, 18], and pharmaceutical chemistry [18].
If G is a simple connected graph and u and are any two vertices of G, then the distance is the length of a shortest path between u and . Let be an ordered set of vertices of G and let be a vertex of G. The representation of with respect to S is the ordered k-tuple . If , then we shall say that vertex distinguishes two vertices x and y. If distinct vertices of G have distinct representations with respect to S, then S is called a resolving set for G. A resolving set of minimum cardinality is called a basis for G and the cardinality is known as metric dimension of G, and it is denoted as .
Let S be a set of vertices of a simple connected graph G. Then, S is called an edge metric generator for G if every two edges of G are distinguished by some vertices of S. The minimum cardinality of S is called the edge metric dimension, and it is denoted by . For an ordered set of vertices of a graph G and any edge e in G, we refer to the k-vector (ordered k-tuple) as the edge metric representation of e with respect to S. The set S is an edge metric generator for G if and only if for every pair of different edges , of G, .
Let be a family of connected graphs of order for which . If there exists a constant such that for every , then we shall say that has a bounded edge metric dimension, otherwise has unbounded edge metric dimension. Moreover, if , , then has a constant edge metric dimension.
There is a controversy in the definition of edge resolving set and edge metric dimension. In the literature, another version of edge metric dimension exists, which is based on the distance between edges, denoted as . This version of edge metric dimension follows the following steps. Step 1. Convert a graph G into a line graph . Step 2. Find metric dimension, of line graph .
According to the above steps, , for reference see [19]. But in our case, the distance is based on vertex to edge, , as defined by Kelenc et al. [20].
In 1979, Garey and Johnson [21] pointed out that the finding of the metric dimension of a graph is an NP-hard problem. In 1994, Khuller et al. [22] also showed by another construction that the metric dimension of a graph is an NP-hard problem.
Recently, Kelenc et al. [20] have introduced the notion of edge metric dimension of graph, denoted by , and they presented some results relating the notions of metric dimension and edge metric dimension for some families of graph such as for path : where , for cycle : , for complete graph : , where , for complete bipartite graph : , where , for tree which is not a path: , for grid graph (where ): , for wheel graph (): , and for the grid graph : . They also proved that the edge metric dimension is an NP-hard problem. Moreover, they have raised many open problems related to edge metric dimension. Furthermore, in 2019, Zubrilina has classified some graphs on the same topic which has the in her paper [23]. Recently, in 2019, Rafiullah et al. studied the edge metric dimensions of wheel-related convex polytopes in [24] and characterized these graphs.
This paper is based on the question raised in [20] to characterize the families of graphs which observe one of the relations , , or .
We consider Jahangir graph , helm graph , sunflower graph , and friendship graph for edge metric dimension (NP-hard problem). Our aim is to characterize these families with respect to the nature of edge metric dimension. Moreover, we present closed formulas for edge metric dimension of these graphs. For further reading on metric dimension and edge metric dimension, we refer [19, 24, 25, 26, 27].
2. Main Results
2.1. Jahangir Graph ()
The Jahangir graph obtained from wheel graph by alternately deleting n spokes. The order of is , and the size is . The Jahangir graph is also referred to as gear graph [28]. The metric dimension of Jahangir graph is given in [29]. Now, we present some observations for in the form of lemmas which help us to compute its edge metric dimension.
Lemma 1. The central vertex does not belong to any edge metric basis S of , for .
Proof. Let be a gear graph of order . The vertices of are labeled as , and edges of are labeled as and for all and as shown in Figure 1. The vertices and are named as major and minor vertices, respectively. Since and , does not belong to any basis of .
Let be a basis of and number of vertices on for . We say that the pairs of vertices for and are pair of neighboring vertices. We define the gap for as the set of vertices and . The number of gaps is r and some of which may be empty. If or , then the gaps and are known as the neighboring gaps. Let and or and with . Then, the gap between u and is denoted by . The number of vertices between u and is known as the size of gap. There are three types of gaps observed on of , which are , , and . We represent the major vertices by in the following proofs.
(a)
(b)
Lemma 2. If S is a basis of , then every , , and gap of S contains at most 5, 4, and 3 vertices, respectively.
Proof. Let be edges on where and be spokes edges of where . Suppose that there is a gap of S containing seven consecutive vertices of , determined by and with . Then, we have , a contradiction. The existence of a gap containing six consecutive vertices of : , determined by and with and , it follows that and .
If there is a gap of S containing five consecutive vertices of and determined by and with would imply and , a contradiction.
Hence, it is proved that every , , and gap of S contains at most 5, 4, and 3 vertices, respectively.
Note that these , , and gaps are the known major gaps.
Lemma 3. If S is a basis of (), then it contains at most one major gap.
Proof. Suppose that S contains two distinct major gaps and .(i) and : and , it follows and (ii) and : and ; in this case (iii) and : and ; we have and (iv) and : and ; in this case, , , and (v) and : and ; in this case, , , and (vi) and : and ; in this case, , , and , which contradicts the hypothesis.
Lemma 4. If and S be a basis of . Then, any two neighboring gaps, one of which being a major gap, contain together at most six vertices.
Proof. If we take a gap which contains three vertices, then its neighboring gap should contain three vertices that is only possible when we take as a neighboring gap, which is not possible because Lemma 2 says that the basis only contains one major gap.
If we take major gap with five vertices, then the neighboring cannot be a minor gap with three vertices or minor gap with two vertices, we show them below case wise: Case 1. The vertices of gaps and are and , respectively, and determined by , , and , then we have and , a contradiction. Case 2. The vertices of gaps and are and , respectively, determined by , , and , then we have and .If the gap is gap with four vertices, then Lemma 2 is sufficient to show the neighboring gap cannot be minor gap with three vertices. In this case, the gaps and vertices are and , respectively, so and .
Lemma 5. If S is a basis of (), then any two minor neighboring gaps contain together at most four vertices.
Proof. Since by Lemma 2 any minor , , and gap contains 3, 2, and 1 vertexes, respectively, it is sufficient to consider the following cases: Case 1. A gap with 3 vertices has a neighboring gap with 3 vertices. Case 2. A gap with 3 vertices having a neighboring gap with 2 vertices cannot occur.If Case 1 holds, then and , and if Case 2 holds, then and , so there are contradictions.
Theorem 1. If , then .
Proof. We can easily see that , in this case, the possible resolving sets are , , and . Similarly, we can also see that . First, we show that by constructing a resolving set in with vertices. We consider three cases according to the residue classes of modulo 3, where n belongs to the following cases. Case 1. . Let , where k is even, , and . In this case, . Case 2. . Let , where k is even, , and . We define . Case 3. . Let , where k is odd, , and and .The set S contains only one major gap, and remaining gaps are minor gaps which are , , and gaps containing 3, 2, and 1 number of vertices, respectively. The total number of vertices in any two neighboring gaps, one of which is major, are at most 6 (Lemma 4). Any two neighboring gaps contain at most 4 vertices (Lemma 5). Now, we show that . Let S be a basis of and . Then, S induces r gaps on , denoted as such that and for every and also and are neighboring gaps. By Lemma 3, at most one of them, say is a major gap. By Lemma 4, we can write , , and by Lemma 5, the relation for the remaining neighboring gaps is for every .
So, we haveSince , for each , we obtain .
2.2. Helm Graph ()
The helm graph is obtained from wheel graph by adding a single pendant edge at each vertex of cycle . The number of vertices in the helm graph are , and the number of edges are . We labeled the central vertex, cycle vertices, and pendant vertices as , , and , respectively, where . There are three types of edges, and we labeled them in such a way , , and , see Figure 2.
(a)
(b)
In [30], it was proved that the metric dimension of the helm graph is unbounded, i.e., .
Theorem 2. Let be a helm graph. Then,
Proof. If or 4, then the proof is simple. Let and , where , vertices are on and they have degree 4, i.e., , and the are the pendant vertices having . To prove that S is a minimum edge resolving set, we consider the following three cases:(1)If , then has distance 0 to and distance 1 to , where and , distance 1 to and distance 2 to , where and .(2)If , then has distance 0 to and , distance 1 to , distance 2 to all other vertices on , distance 1 to , distance 2 to , and distance 3 to all other pendant vertices.(3)If , then has distance 0 to and , distance 1 to , distance 2 to , and distance 3 to all other pendant vertices.The above cases show that the edge metric representation with any vertices or less number of vertices of is same for some edges. So, S is the edge metric generator with vertices, and therefore,We can also show that the by considering S as a basis of with . For this we have the following cases:(i)If we take , then .(ii)If we suppose , then .(iii)If we consider and , then and , respectively.So, set S is not the basis of . This implies thatHence, from inequalities (3) and (4), we have .
Note. The total number of edge resolving sets for the helm graph () is .
2.3. Sunflower Graph ()
The sunflower graph obtained from wheel graph () having a cycle () and by adding n new vertices with new edges () and () where are shown in Figure 3. The number of vertices in the sunflower graph are , and the number of edges are . We labeled the central vertex, cycle vertices, and the newly added vertices by , , and , respectively, where . We denote the cycle by . There are three types of edges and we label them in the following manner, , , and or , consider Figure 3 for reference. The metric dimension of the sunflower graph for ,
Now, we present some observations for to prove its edge metric dimension.(i)There are two edge disjoint cycles: and .(ii)Every gap of S contains one vertex or three vertices of and , respectively.(iii)There are three types of gaps, namely, , , and with the maximum gap of sizes 3, 2, and 0, respectively.(iv)The resolving set S contains two gaps of sizes 2 and 0, when .(v)The resolving set S mostly contains gaps.(vi)Any two neighboring gaps contain together at most 4 number of vertices.(vii)The neighboring and gaps contain together at most 3 number of vertices.(viii)Only one vertex of degree 5 from belongs to edge resolving set S of , when and all remaining vertices belong to . If we consider any two consecutive vertices of the degree 5 in basis set S such as and then and So, the gap is not possible in the sunflower graph for edge metric dimension.(ix)Any two neighboring gaps contain together at most 2 vertices. If there are 4 vertices in neighboring gaps for where , then , , and .(x)The central vertex does not belong to any edge metric basis S of because , , and where and .
We can also define the selection of basis S of by the following Claims.
Claim 1. The resolving set S mostly contains gaps having one vertex.
If a gap of the size 1 with vertex on and , then contains two edges and on . If thenwhere the normal part of the code and overlined part of the code represent the distance of z from and types of vertices, respectively.
By observation and Claim 1, there is no other edge in having these types of codes.
Claim 2. At most one gap of S has at most 3 number of vertices.
If a gap of the size 3 with vertices , on and , , , , , then the contains 5 number of edges, one of them on and 4 of them , , , on . If , thenwhere the normal part of the code and overlined part of the code represent the distance of z from and types of vertices, respectively.
By observation iii and Claim 2, there is no other edge in with these types of codes.
Claim 3. At most one gap of S has 2 vertices.
If a gap of the size 2 with vertices ,, and , then the contains 4 number of edges, one of them on and 3 of them , , on . If z , thenwhere the normal part of the code and overlined part of the code represent the distance of z from and types of vertices, respectively.
By observations iii and iv and Claim 3, there is no other edge in with these type of codes.
Claim 4. Any two, neighboring gaps contain together at most 4 number of vertices.
Let two neighboring gaps and have 1 and 3 number of vertices, respectively. The gaps and are determined by , , and . The gap has one vertex and two edges , on . The has 3 vertices , , and and 5 number of edges, one of them on and 4 of them , , , on . If and z , thenwhere the normal part of the code and overlined part of the code represent the distance of z from and types of vertices, respectively.
By observations iii, v, and vi and Claim 4, there is no other edge in having these types of codes.
Claim 5. The neighboring and gaps contain together at most 3 number of vertices.
Let and be and neighboring gaps having 1 and 2 vertices, respectively. These gaps can be determined by , , and . The gap has one vertex on or two edges , on . The has 2 vertices , and 4 number of edges, one of them on and 3 of them , , and on . If and , thenwhere the normal part of the code and overlined part of the code represent distance of z from and types of vertices, respectively.
By observations iii, v, and vii and Claim 5, there is no other edge in with these types of codes.
Theorem 3. If , then .
Proof. The resolving set for is .
First, we show that for by constructing a resolving set for with vertices. We consider three cases according to the residue class modulo 3 to which n belongs. Case 1. . Let , where k is even, , and . In this case, . Case 2. . Let , where k is even, , and . We define . Case 3. . Let , where k is odd, , and and .To show that , we consider the aforementioned observations.
Let S be a basis of and . Then, S induces r gaps which are denoted as such that and for every and also and are neighboring gaps. We can write , and the relation for the remaining neighboring gaps is for every .
So, we haveSince , for each , we obtain Hence, the theorem is proved.
2.4. Friendship Graph ()
The friendship graph can be obtained from the wheel graph by removing the alternative edges of the cycle of . The number of vertices in the friendship graph are , and the number of edges are . We label the central vertex and remaining vertices as and , respectively, where . There are two types of edges, and we label them in such a way and where , for reference, see Figure 4
The metric dimension of friendship graph for , computed by Mulyono and Wulandari in [31].
Theorem 4. Let be a friendship graph, then for .
Proof. Let and , where and , where . To prove that S is a minimum edge resolving set, we consider the following three cases: Case 1. If , then has distance 0 to and distance 1 to all other vertices, where . Case 2. If , then has distance 0 to and distance 1 to , distance 2 to all other vertices.The above cases show that the edge metric representation of edges of with respect to a set with vertices is not distinct. So S can be edge metric generator if we take number of vertices. Therefore, the edge metric dimension of the friendship graph is .
We also show that the by considering the resolving set, S of the size . Let , then or . So set S is not the basis of . Hence, it is proved that .
Note. The total number of edge resolving sets for a fan graph () is .
3. Conclusion
In this article, the edge metric dimension for certain families of wheel-related graphs, like Jahangir graph , helm graph , sunflower graph , and friendship graph , have been studied and proved that all these families have unbounded edge metric dimensions. We also compared the results with their metric dimensions and concluded that if , , and if , .
4. Open Problems
It is natural to ask about the characterization of the graph families on the bases of the nature of their edge metric dimension. To answer the aforementioned question, the readers are invited to study the following problems.(1)Characterize the nanostructures, nanocones, nanotubes, nanotorus, and chemical trees on the basis of edge metric dimension.(2)Is it the case that the other families of graphs generated by wheel also have unbounded edge metric dimension?
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare no conflicts of interest.
Authors’ Contributions
All the authors contributed equally to preparing this article.
Acknowledgments
The authors are grateful to Prof. Jia-Bao Liu for his useful suggestions and help to improve and complete this manuscript. This study was supported by Major University Science Research Project of Anhui Province (KJ2016A605) and Major Nature Science Project of Hefei University Research and Development Foundation (16ZR13ZDA).