Abstract
For any given graph , we say is a resolving set or resolves the graph if every vertex of is uniquely determined by its vector of distances to the vertices in . The metric dimension of is the minimum cardinality of all the resolving sets. The study of metric dimension of chemical structures is increasing in recent times and it has application about the topology of such structures. The carbon atoms can bond together in various ways, called allotropes of carbon, one of which is crystal cubic carbon structure . The aim of this article is to find the metric dimension of .
1. Introduction
Let be a simple connected graph and let be an ordered subset of the set of vertices of . The distance of two vertices of is the length of shortest path between and . The representation of a vertex of with respect to is the -vector and it is denoted as . The set is called the resolving set or to resolve if the representation of distinct vertices is distinct. That is, if and are two distinct vertices, then . The metric dimension of a graph is the cardinality of the minimal resolving set and it is denoted as . As there may be many different resolving subsets in of different sizes, the study of the minimal one is important and it has been studied over the years. Some authors also use the term basis for which is a resolving set with minimum cardinal number (see [1]). This work is about a study of resolving sets in chemical structural graphs.
The metric dimension of a general metric space was introduced in 1953 in [2], but at that time, it attracted little attention. Then, about twenty years later, it was applied to the distances between vertices of a graph [3–5]. Since then, it has been frequently used in graph theory, chemistry, biology, robotics, and many other disciplines. For some literature studies, see [6–9].
From many parameters for the study of graphs, the metric dimension is one of those that has many applications, and these applications are diverse like in pharmaceutical chemistry [10, 11], robot navigation [12], and combinatorial optimization [13]. A chemical compound or material can be represented by many graph structures, but only one of them may express its topological properties. The chemists require mathematical forms for a set of chemical compounds to give distinct representations to distinct compound structures. The structure of chemical compounds or materials can be represented by a labeled graph whose vertex and edge labels specify the atom and bond types, respectively. Thus, a graph theoretic interpretation of this problem is to provide representations for the vertices of a graph in such a way that distinct vertices have distinct representations.
At very high pressures of above 1000 GPa (gigapascal), one of the forms of carbon, namely, diamond, is predicted to transform into the so-called structure, a body-centered cubic structure with 8 atoms in the unit cell. This cubic carbon phase might have importance in astrophysics. Its structure is known in one of the metastable phases of silicon and is similar to cubane. The structure of this phase was proposed in 2012 as carbon sodalite [14]. In 2017, Baig et al. [15] modified and extended this structure and named it crystal cubic carbon . We are taking all the notations as they were in [15]. The structure of crystal cubic carbon consists of cubes.
The molecular graph of crystal cubic carbon for the second level is depicted in Figure 1. Its structure starts from one unit cube and then by attaching cubes at each vertex of the unit cube by an edge. For the third level, the is constructed by attaching cube to each vertex of cubes of having degree 3 or you can say by attaching cubes by an edge to all the white vertices of . So, at each level, a new set of cubes is attached by edges to the white vertices of cubes of the preceding level. The third level of is displayed in Figure 2 which is constructed and presented in a most suitable manner to explain the structure of .
All the new attached cubes, at each level, will be called the outermost layer of cubes or outermost level of cubes, or you can say at each level, the cubes with white vertices will be called the outermost layer. As in , the outermost layer of cubes consists of 8 cubes. Because there are vertices of degree 3, so in , the outermost layer of cubes will consist of cubes. Similarly, this procedure is repeated to get the next level. The cardinality of vertices and edges in is given below, respectively.
There are some articles that describe the different topological properties of structure, the famous of those topological indices are Randic, ABC, and Zagreb indices and other degree-based indices of which are computed in [15–18]. In the articles [19, 20], theauthors calculated eccentricity and Szeged-type topological indices of . The aim of this article is to compute the metric dimension of . Note that if is the ordered set of vertices of a graph , then component of is . Thus, in order to show that is a resolving set, it suffices to verify that for each pair of distinct vertices .
2. Main Result
In this section, we will present the main result about the . But before going further, let us discuss the very simple case of which is just a cube. We claim that indeed is true, let us see how.
Assume that , and because of symmetry, we can take any vertex of cube to be the resolving set as in Figure 3(a), say , then , which is a contradiction. So, . Assume that . Then, there are two possibilities for the elements of the resolving set of because of its symmetric shape. The possible cases are as follows:(I)The two elements of are the vertices on the main diagonal of .(II)The two elements of are on the same face of the cube. In this case, the both elements are either on the main diagonal of a face or on the same edge of a face.
(a)
(b)
(c)
(d)
(e)
Without loss of generality, we can assume that for case (I). For case (II) without loss of generality, we can assume and , respectively. Then, Figures 3(b)–3(d) show that ; the ordered pairs in these Figures denote the representations of the vertices. Thus, from Figure 3(e), it is proved that .
Now, we will prove the main result of this article.
Theorem 1. The metric dimension of crystal cubic carbon structure is , for all , that is, .
Proof. Let be the crystal cubic carbon structure and . To show that the firstly, we will show that . Let be a cube on the outermost layer of , as depicted in Figure 4 (note that there are no cubes attached to the vertices ). In other words, all these vertices are of degree 3 and they belong to only one cube which is . Observe that the red vertex of cube is attached with red edge to a cube of the preceding level at its blue vertex. Also, note that and and .
Let be a resolving set of . We claim that at least two vertices of belong to . Suppose on contrary that no vertex of belongs to and let be a representation of vertex . Note that all the shortest paths from any vertex of to any vertex of contain the vertex of . So, we can say that all such paths pass through vertex (path may end at it). Then,this is a contradiction. Now, assume that exactly one vertex from the set belongs to . Without loss of generality, we can assume that this common vertex is . Case 1. If , then Case 2. If , then Similar contradictions appear for and , let us look at it. Case 3. If , then Case 4. If , then Case 5. If , then Case 6. If , then Case 7. If , then Case 8. If , then The contradiction in all the cases proved our claim. So, at least two vertices from the vertex set of are in the resolving set of . Since was taken arbitrary, so contains at least two vertices from each of the cube in the outermost layer of cubes of . By the construction of , we can see that at each step or at each level, the cubes in are increased by a number equal to 7 multiplied by the number of cubes in the outermost layer of the previous level. For example, in , we have 8 cubes in the outer layer, and in , we have cubes in the outermost layer. Thus, there are exactly cubes in the outermost layer of . Since from each such cube there are at least two vertices in , so .
2.1. Second Part of Proof
In this part, we will show that . Let be the collection of all the vertices of type and just like we have discussed in part one of the proof and depicted in Figure 4. Then, . We claim that is a resolving set of . The representations of the two arbitrary vertices of can be compared in five different cases and they are discussed as follows:(1)The two arbitrary selected vertices are on the same cube in the outermost level of (see Figure 4).(2)The two arbitrary selected vertices are on the same cube, but this cube is not the outer most cube and neither the central cube (i.e., ), as depicted in Figure 5.(3)The two arbitrary selected vertices are on the central cube, as displayed in Figure 6.(4)The two arbitrary selected vertices are on a chain of cubes with one end being the cube of the outermost level (see Figure 7).(5)The two arbitrary selected vertices are on distinct chains of cubes and those chains are connecting at a cube which we can call a branching cube. As explained in Figure 8, in which B cube is the branching cube, -cube and -cube are on different chains each containing one of the selected vertices. Case (1). This can be proved by a direct computation for the representation of all the vertices in this cube (Figure 4). Without loss of generality, we can assume that , then and We can see from the above that these representations are all distinct in this case. Case (2). Let the two arbitrary selected vertices be on the same cube and this cube is not on the outermost cube and neither is it the central cube. A visualization of such cube is given in Figure 5. We can label the vertices of this cube , as shown in Figure 5. Without loss of generality, we can assume that are on the cube in the outermost layer of cubes and that cube is connected to cube at vertex by a chain of cubes. Similarly, we can assume that are on the cube in the outermost layer of cubes and those cubes are connected to cube at vertices , , by a chain of cubes, respectively. Also, and . All these computations show that for and for . This completes the proof in this case. Case (3). Assume that the two arbitrary selected vertices are on the central cube, as displayed in Figure 6, where just like in the previous case (2), we have labeled all 8 vertices with . Again, without loss of generality, we assume that , are on the cube in the outermost layer of cubes and those outermost cubes containing are connected to the central cube at vertices , , by a chain of cubes, respectively. These assumptions imply that So, we get the conclusion that, in this case, again for and , . Case (4). Now, we are going to discuss case (4). Assume that the two arbitrary selected vertices are on two distinct cubes and those cubes are on a chain of cubes, see Figure 7. Assume that one end of this chain is the outermost cube containing two arbitrary resolving elements, say (without loss of generality, we can assume that those vertices are ), and the other end is the central cube. As depicted in Figure 7, let be a vertex of cube and be a vertex of cube , then , and therefore, . This completes the proof in this case. Case (5). Finally, suppose that the two arbitrary selected vertices are on distinct chains of cubes and those chains are connecting at a cube which we can call a branching cube; this branching cube can also be the central cube. As explained in Figure 8, in which cube is the branching cube, cube and cube are on different chains each containing one of the selected vertices, that is, and . Both of the two cubes and or any one of these cube can also be the cubes in the outermost level of cubes.
Note: in the idea of case (4), we can say that someone can select two vertices on different cubes such that there is chain of cube connecting them and both ends of this chain are the cubes on the outermost level of cubes. But then, there must be a cube (which we call branching cube) in this chain that connects to the central cube by the chain of cubes.) Without loss of generality, we can assume that and . We can see that the length of the shortest path from vertex to vertex of cube is greater than the length of the shortest path from vertex to vertex of cube . Thus, , so this implies that .
All these five cases prove that is a resolving set. Since there are number of elements in , therefore the proof of theorem concludes.
3. Conclusion
In this article, we have studied the metric dimension of the crystal cubic carbon structure and we gave a formula for its metric dimension. We have found that the metric dimension of is not constant and find its closed form.
Data Availability
All the proofs and exemplary data of this study are included in the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the National Key Research and Development Program under Grant 2018YFB0904205.