#### Abstract

Let be a finite, connected graph of order of, at least, 2 with vertex set and edge set . A set of vertices of the graph is a doubly resolving set for if every two distinct vertices of are doubly resolved by some two vertices of . The minimal doubly resolving set of vertices of graph is a doubly resolving set with minimum cardinality and is denoted by . In this paper, first, we construct a class of graphs of order , denoted by , and call these graphs as the layer Sun graphs with parameters , , and . Moreover, we compute minimal doubly resolving sets and the strong metric dimension of the layer Sun graph and the line graph of the layer Sun graph .

#### 1. Introduction

In this paper, suppose is a finite, simple connected graph of order of, at least, 2, with vertex set and edge set . If and are vertices in the graph , then the distance from in is denoted by or simply , where is the length of the shortest path from to . The line graph of a graph is denoted by , with vertex set and where two edges of are adjacent in if and only if they are incident in , see . Vertices of the graph are said to doubly resolve vertices of if . A set of vertices of the graph is a doubly resolving set of if every two distinct vertices of are doubly resolved by some two vertices of . The minimal doubly resolving set of vertices of graph is a doubly resolving set with minimum cardinality and is denoted by . The notion of a doubly resolving set of vertices of the graph was introduced by Cáceres et al. . A vertex strongly resolves two vertices and if belongs to a shortest path or belongs to a shortest path. A vertex set of the graph is a strong resolving set of if every two distinct vertices of are strongly resolved by some vertex of . A strong metric basis of is denoted by defined as the minimum cardinality of a strong resolving set of . The notion of a strong metric dimension problem set of vertices of the graph was introduced by A. Sebö and E. Tannier  and further investigated by O. R. Oellermann and Peters-Fransen . The minimal doubly resolving sets for jellyfish and cocktail party graphs have been obtained in . For more results related to these concepts, see . In this paper, first, we construct a class of graphs of order , denoted by , and call these graphs as the layer Sun graphs with parameters , , and , which is defined as follows:

Let be integers such that , and be a graph with vertex set , where are called the layers of such that , , and for , we have , and let , such that every is a vertex in the layer , and in the layer , , , , where is the complement of the complete graph on vertices. Now, suppose that every vertex in the cycle is adjacent to exactly one vertex in the layer , say , and every vertex in the layer is adjacent to exactly vertices , in particular for , every vertex is adjacent to exactly vertices , and then, the resulting graph is called the layer Sun graph with parameters , , and . Also, for , we recall that as the components of the layer , , . In particular, we say that two components , , are fundamental if and . It is natural to consider its vertex set of the layer Sun graph as partitioned into layers. The layers and consist of the vertices and , respectively. In particular, each layer consists of the vertices. Note that, for each vertex in the layer and every vertex , , , we have . In this paper, we consider the problem of determining the cardinality of minimal doubly resolving sets of the layer Sun graph . First, we find the metric dimension of the layer Sun graph ; in fact, we prove that if and , then the metric dimension of the layer Sun graph is . Moreover, we consider the problem of determining the cardinality of minimal doubly resolving sets of and the strong metric dimension for the layer Sun graph and the line graph of the layer Sun graph . The graph is shown in Figure 1.

#### 2. Definitions and Preliminaries

Definition 1. Let be a graph. A vertex is said to resolve a pair if . For an ordered subset of vertices in the graph and a vertex of , the metric representation of with respect to is the -vector . If every pair of distinct vertices of has different metric representations, then the ordered set is called a resolving set of . If the set is as small as possible, then it is called a metric basis of the graph . We recall that the metric dimension of , denoted by , is defined as the minimum cardinality of a resolving set for .

Proposition 1. Let be a graph. It is well known that a doubly resolving set is also a resolving set and . In particular, every strong resolving set is a resolving set and .

#### 3. Main Results

##### 3.1. Minimal Doubly Resolving Sets and the Strong Metric Dimension for the Layer Sun Graph

Theorem 1. Let be the layer Sun graph which is defined already. Suppose that are integers such that and . Then, the metric dimension of is .

Proof. Let , where are called the layers of vertices in the layer Sun graph , which is defined already. It is clear that if is an ordered subset of the layers , then is not a resolving set in . We may assume that the layer is equal towhere , , . In the following cases, it can be shown that the metric dimension of the layer Sun graph is .Case 1: let be an ordered subset of the layer in the layer Sun graph such thatHence,We know that the cardinality of is because , . Therefore, the metric representation of all the vertices in the component is the same as -vector with respect to . Thus, is not a resolving set in Case 2: let be an ordered subset of the layer in the layer Sun graph such thatHence,We know that . Therefore, the metric representation of two vertices in the component is the same as -vector with respect to . Thus, is not a resolving set in .Case 3: let be an ordered subset of the layer in the layer Sun graph such thatHence,We know that . We can show that all the vertices in have different representations with respect to . Let be the vertex of the layer . We can assume without loss of generality that , . Hence, , where , , ; otherwise, if , then . Now, let . We can assume without loss of generality that , . Hence, , where , , ; otherwise, if , then . In a similar way, we can show that all the vertices in the layers have different representations with respect to . In particular, for every vertex , we have , , if ; otherwise, if and , then . Therefore, all the vertices in have different representations with respect to . This implies that is a resolving set in . From the abovementioned cases, we can be concluded that the minimum possible cardinality of a resolving set in is .

Theorem 2. Let be the layer Sun graph which is defined already. Suppose that are integers such that and . Then, the cardinality of minimum doubly resolving set of the is .

Proof. In the following cases, it can be shown that the cardinality of minimum doubly resolving set of the layer Sun graph is .Case 1: we know that the ordered subset W of vertices in equation (8) in the layer of is a resolving set for of cardinality . We show that this subset is not a doubly resolving set for . Because if the vertex , , is adjacent to a vertex , then for every , we have .Case 2: now, let the subset of vertices in beIn a similar fashion as in Case 3 of Theorem 1, we can show that all the vertices in the layers of and the vertex in the layer of have different representations with respect to . So, is a resolving set in of cardinality . Note that, in this case, by a similar way as in Case 1, we can show that this subset is not a doubly resolving set for .Case 3: finally, let the subset of vertices in beIn a similar fashion as in Theorem 1, we can show that all the vertices in the layers of have different representations with respect to . So, this subset is also a resolving set in of cardinality . We show that this subset is a doubly resolving set for . It is sufficient to prove that for two vertices and in , there are vertices such that . Consider two vertices and in . Then, we have the following.Case 3.1: suppose that both vertices and lie in the layer . Hence, there are such that and . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is some vertex such as in the component , in the layer , at distance from ; in fact, . In the same way, there is some vertex such as in the component , in the layer , at distance from . In particular, it is easy to prove that because . Also, because .Case 3.2: now, suppose that both vertices and lie in the layer . In a similar way as in Case 3.1, we can show that there are vertices such that .Case 3.3: suppose that both vertices and lie in the layer , such that these vertices lie in the one component of the layer , say , , . In this case, . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is a component of the layer , say , , such that for any vertex , we have . In the same way, for the vertex in the layer , there is a component of the layer , say , , , such that, for any vertex , we have . Thus, because and .Case 3.4: suppose that both vertices and lie in the layer , such that these vertices lie in the two distinct components of the layer . We can assume without loss of generality that and , , and . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is a component of the layer , say , such that for any vertex , we have . In the same way, for the vertex in the layer , there is a component of the layer , say , , such that for any vertex , we have . In the following, let two components and be fundamental; indeed, . Hence, , . Thus, . Now, let two components and not be fundamental; indeed, . Hence, , . Thus, .Case 3.5: suppose that vertices and lie in distinct layers , respectively. Note that if and , and , or and , there is nothing to do. Now, let . Hence, there is a component of the layer , say , , such that . Also, there is a component of the layer , say , , such that . In particular, there is a component of the layer , say , , such that for any vertex , we have . Now, let ; if we consider , , , and , then we have . Because , and . Note that if , then there is a component of the layer , say , , such that for any vertex , we have , and then, we have because and . Thus, from the abovementioned cases, we can conclude that the cardinality of the minimum doubly resolving set of the layer Sun graph is .

Theorem 3. Let be the layer Sun graph which is defined already. Suppose that are integers such that and . Then, the strong metric dimension of is .

Proof. In the following cases, it can be seen that the cardinality of the minimum strong resolving set of the layer Sun graph is .Case 1: we know that the ordered subset W of vertices in equation (11) in the layer of the layer Sun graph is a resolving set for of cardinality . Now, let By considering distinct vertices , we can show that there is not a vertex such that belongs to a shortest path or belongs to a shortest path because the valency of every vertex in the layer is one. So, this subset is not a strong resolving set for . Thus, we can be conclude that if is a strong resolving set for graph , then because must be less than .Case 2: on the other hand, we can show that the subset W of vertices in equation (12) in the graph is a resolving set for graph . We show that this subset is a strong resolving set in graph . It is sufficient to prove that every two distinct vertices are strongly resolved by a vertex . Then, we have the following:Case 2.1: suppose that both vertices and lie in the layer . Hence, there are such that and . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is a component , in the layer such that, for every vertex such as , we have and , and hence, belongs to a shortest path.Case 2.2: now suppose that both vertices and lie in the layer . In a similar way as in Case 2.1, we can show that the vertices and are strongly resolved by a vertex .Case 2.3: suppose that both vertices and lie in the layer , such that these vertices lie in the one component of the layer , say , , . In this case, . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is a component of the layer , say , , such that for any vertex , we have , and hence, belongs to a shortest path.Case 2.4: suppose that both vertices and lie in the layer , such that these vertices lie in the two distinct components of the layer . We can assume without loss of generality that and , , and . Moreover, we know that the layer Sun graph has the property that, for each vertex in the layer , there is a component of the layer , say , such that for any vertex , we have , and hence, belongs to a shortest path.Case 2.5: suppose that vertices and lie in distinct layers , respectively. Note that if and , and , or and , there is nothing to do. Now, let . Hence, there is a component of the layer , say , , such that . Also, there is a component of the layer , say , , such that . In particular, there is a component of the layer , say , , such that, for any vertex , we have , and hence, belongs to a shortest path.Case 2.6: let be two distinct vertices in such that and , . So, there is a component in the layer , and such that . Thus, there is some vertex in the component , say such that , , and belongs to a shortest path.Case 2.7: let be two distinct vertices in such that and . So, there is some such that . If , indeed , and then, there is a component in the layer , and such that, for every vertex such as in the component , we have , , and belongs to a shortest path. Now, let ; indeed, , and hence, there is a component in the layer , such that, for every vertices such as in the components , we have , , and belongs to a shortest path.Thus, from the abovementioned cases, we can be concluded that the cardinality of minimum strong resolving set of the layer Sun graph is .

##### 3.2. Minimal Doubly Resolving Sets and the Strong Metric Dimension for the Line Graph of Layer Sun Graph

Let be the layer Sun graph which is defined already. Now, let be a graph with vertex set , where are called the layers of which is defined as follows:

Let and , and for , we haveand let such that every is a vertex in the layer and in the layer , , , , where is the complete graph on vertices. Now, suppose that every vertex in the cycle or the layer is adjacent to exactly two vertices in the layer say