On Metric Dimension of Some Rotationally Symmetric Graphs

Abstract

A family 𝒢 of connected graphs is a family with constant metric dimension if dim(𝐺) is finite and does not depend upon the choice of 𝐺 in 𝒢. In this paper, we show that the graph 𝐴^∗_𝑛 and the graph 𝐴^𝑝_𝑛 obtained from the antiprism graph have constant metric dimension.

1. Notation and Preliminary Results

For a connected graph 𝐺, the distance 𝑑(𝑢,𝑣) between two vertices 𝑢,𝑣∈𝑉(𝐺) is the length of a shortest path between them. A vertex 𝑤 of a graph 𝐺 is said to resolve two vertices, 𝑢 and 𝑣, of 𝐺 if 𝑑(𝑤,𝑢)≠𝑑(𝑤,𝑣). Let 𝑊={𝑤₁,𝑤₂,…,𝑤_𝑘} be an ordered set of vertices of 𝐺 and let 𝑣 be a vertex of 𝐺. The representation of the vertex 𝑣 with respect to 𝑊 denoted by 𝑟(𝑣∣𝑊) is the 𝑘-tuple (𝑑(𝑣,𝑤₁),𝑑(𝑣,𝑤₂),…,𝑑(𝑣,𝑤_𝑘)). If distinct vertices of 𝐺 have distinct representations with respect to 𝑊, then 𝑊 is called a resolving set, for 𝑉(𝐺) [1]. A resolving set of minimum cardinality is called a metric basis for 𝐺, and the cardinality of this set is the metric dimension of 𝐺, denoted by dim(𝐺).

For a given ordered set of vertices 𝑊={𝑤₁,𝑤₂,…,𝑤_𝑘} of a graph 𝐺, the 𝑖th component of 𝑟(𝑣∣𝑊) is 0 if and only if 𝑣=𝑤_𝑖. Thus, to show that 𝑊 is a resolving set it suffices to verify that 𝑟(𝑥∣𝑊)≠𝑟(𝑦∣𝑊) for each pair of distinct vertices 𝑥,𝑦∈𝑉(𝐺)⧵𝑊.

Caceres et al. [2] found the metric dimension of fan 𝑓_𝑛, and Javaid et al. [3] found the metric dimension of Jahangir graph 𝐽_2𝑛.

In [1], Chartrand et al. proved that a graph has metric dimension 1 if and only if it is a path; hence paths on 𝑛 vertices constitute a family of graphs with constant metric dimension. They also showed that cycles with 𝑛≥3 vertices also constitute such a family of graphs as their metric dimension is 2. In [2], Caceres et al. proved that dim𝑝_𝑚×𝐶_𝑛=2,if𝑛,isodd;3,otherwise. (1) Prisms 𝐷_𝑛 are the trivalent plane graphs obtained by the cartesian product of the path 𝑃₂ with a cycle 𝐶_𝑛. In [4], Javaid et al. proved that this family has constant metric dimension. In [4], Javaid et al. also proved that the antiprism graph 𝐴_𝑛, constitutes a family of regular graphs with constant metric dimension as dim(𝐴_𝑛)=3, for every 𝑛≥5.

In this paper, we extend this study to antiprism-related graphs (see Figure 1). The graph 𝐴^∗_𝑛 is defined as follows: for each vertex 𝑏_𝑖, 𝑖=1,2,…,𝑛 of the outer cycle of the antiprism graph, we introduce a new vertex 𝑎_𝑖, 𝑖=1,2,…,𝑛 and join 𝑎_𝑖 to 𝑏_𝑖 and 𝑏_𝑖−1, 𝑖=1,2,…,𝑛, with 𝑛+𝑖 taken modulo 𝑛. Thus, 𝑉(𝐴^∗_𝑛)=⋃^𝑛_𝑖=1{𝑎_𝑖,𝑏_𝑖,𝑐_𝑖}. Here {𝑐_𝑖}  𝑖=1,2,…,𝑛 are the inner cycle vertices, {𝑏_𝑖}  𝑖=1,2,…,𝑛 are the outer cycle vertices, and {𝑎_𝑖}  𝑖=1,2,…,𝑛 are adjacent vertices to the outer cycle. We define the graph 𝐴^𝑝_𝑛 as follows: for each vertex 𝑏_𝑖, 𝑖=1,2,…,𝑛 of the outer cycle of the antiprism graph, we introduce a new vertex 𝑎_𝑖 and join 𝑎_𝑖 to 𝑏_𝑖, 𝑖=1,2,…,𝑛. Thus, 𝑉(𝐴^𝑝_𝑛)=⋃^𝑛_𝑖=1{𝑎_𝑖,𝑏_𝑖,𝑐_𝑖}. Here, {𝑐_𝑖} are the inner cycle vertices, {𝑏_𝑖} are the outer cycle vertices, {𝑎_𝑖}𝑖=1,2,…,𝑛 are the pendant vertices adjacent to the outer cycle vertices.

(a)

(b)

Figure 1 

Graphs 𝐴∗8 and 𝐴𝑝8.

2. Antiprism-Related Graphs with Constant Metric Dimension

In this section, we show that 𝐴^∗_𝑛, 𝐴^𝑝_𝑛 have constant metric dimension.

Theorem 1. Let 𝑛≥6 be an integer, then dim(𝐴^∗_𝑛)=3.

Proof   
Case  1. Let 𝑛=2𝑘, 𝑘≥3, 𝑘∈ℕ. Consider the set 𝑊={𝑐₁,𝑐₂,𝑐_𝑘+1}⊂𝑉(𝐴^∗_𝑛). We show that 𝑊 is a resolving set for 𝑉(𝐴^∗_𝑛). We find the representations of the vertices of 𝑉(𝐴^∗_𝑛)⧵𝑊 with respect to 𝑊: 𝑟𝑐_𝑖=∣𝑊𝑖−1,𝑖−2,𝑛+22−𝑖,for3≤𝑖≤𝑘,𝑛−𝑖+1,𝑛+2−𝑖,𝑖−𝑛+22𝑟𝑏,for𝑘+2≤𝑖≤𝑛,_𝑖=𝑛∣𝑊1,1,2,for𝑖=1,𝑖,𝑖−1,𝑛+22𝑛−𝑖,for2≤𝑖≤𝑘,2,𝑛2𝑛,1,for𝑖=𝑘+1,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−2𝑟𝑎,for𝑘+2≤𝑖≤𝑛,_𝑖=∣𝑊2,2,𝑛+22𝑛,for𝑖=1,2,2,2,for𝑖=2,𝑖,𝑖−1,𝑛+42−𝑖,for3≤𝑖≤𝑘,𝑛+12,𝑛2𝑛,2,for𝑖=𝑘+1,𝑛+2−𝑖,𝑛+3−𝑖,𝑖−2,for𝑘+2≤𝑖≤𝑛. (2) We note that there are no two vertices having the same representation, which implies that dim(𝐴^∗_𝑛)≤3.
To prove the theorem, it is sufficient to show that dim(𝐴^∗_𝑛)≥3. Contrarily, assume that there exists a resolving set 𝑊^′ with |𝑊^′|=2. We have the following possibilities.(1)Both vertices of 𝑊^′ belong to {𝑐_𝑖}⊂𝑉(𝐴^∗_𝑛). Without loss of generality, we suppose that one resolving vertex is 𝑐₁ and the other is 𝑐_𝑡(2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑐_𝑛∣𝑊^′)=𝑟(𝑏_𝑛∣𝑊^′)=(1,𝑡). For 𝑡=𝑘+1, 𝑟(𝑏₁∣𝑊^′)=𝑟(𝑏_𝑛∣𝑊^′)=(1,𝑘), a contradiction.(2)Both vertices of 𝑊^′ belong to {𝑏_𝑖}⊂𝑉(𝐴^∗_𝑛). Without loss of generality, we suppose that one resolving vertex is 𝑏₁ and the other is 𝑏_𝑡(2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑏_𝑛∣𝑊^′)=𝑟(𝑎₁∣𝑊^′)=(1,𝑡). For 𝑡=𝑘+1, 𝑟(𝑎₁∣𝑊^′)=𝑟(𝑎₂∣𝑊^′)=(1,𝑘), a contradiction.(3)Both vertices of 𝑊^′ belong to {𝑎_𝑖}⊂𝑉(𝐴^∗_𝑛). We suppose that one resolving vertex is 𝑎₁ and the other is 𝑎_𝑡(2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑐_𝑛∣𝑊^′)=𝑟(𝑎_𝑛∣𝑊^′)=(2,𝑡+1). For 𝑡=𝑘+1, 𝑟(𝑏₁∣𝑊^′)=𝑟(𝑏_𝑛∣𝑊^′)=(1,𝑘), a contradiction.(4)One vertex belong to {𝑐_𝑖}⊂𝑉(𝐴^∗_𝑛) and the other belong to {𝑏_𝑖}⊂𝑉(𝐴^∗_𝑛). Consider one resolving vertex is 𝑐₁ and the other is 𝑏_𝑡(1≤𝑡≤𝑘+1). For 1≤𝑡≤𝑘, we have 𝑟(𝑏_𝑘+1∣𝑊^′)=𝑟(𝑐_𝑘+1∣𝑊^′)=(𝑘,𝑘+1−𝑡). For 𝑡=𝑘+1,  𝑟(𝑐_𝑘+1∣𝑊^′)=𝑟(𝑎_𝑘+2∣𝑊^′)=(𝑘,1), a contradiction.(5)One vertex belong to {𝑐_𝑖}⊂𝑉(𝐴^∗_𝑛) and the other belong to {𝑎_𝑖}⊂𝑉(𝐴^∗_𝑛). Consider one resolving vertex is 𝑐₁ and the other is 𝑎_𝑡(1≤𝑡≤𝑘+1). For 1≤𝑡≤𝑘, we have 𝑟(𝑎_𝑘+1∣𝑊^′)=𝑟(𝑐_𝑘+1∣𝑊^′)=(𝑘,𝑘−𝑡+2). For 𝑡=𝑘+1,𝑟(𝑏_𝑘∣𝑊^′)=𝑟(𝑏_𝑘+1∣𝑊^′)=(𝑘,1), a contradiction.(6)One vertex belong to {𝑏_𝑖}⊂𝑉(𝐴^∗_𝑛) and the other belong to {𝑎_𝑖}⊂𝑉(𝐴^∗_𝑛). Consider one resolving vertex is 𝑏₁ and the other is 𝑎_𝑡(1≤𝑡≤𝑘+1). For this we have 𝑟(𝑎_𝑛∣𝑊^′)=𝑟(𝑐_𝑛∣𝑊^′)=(2,𝑡+1), a contradiction.
Hence, from above it follows that there is no resolving set with two vertices for 𝑉(𝐴^∗_𝑛). Therefore, dim(𝐴^∗_𝑛)≥3, which implies that dim(𝐴^∗_𝑛)=3.
Case  2. Let 𝑛=2𝑘+1, 𝑘≥3𝑘∈ℕ. Consider the set 𝑊={𝑐₁,𝑐₂,𝑐_𝑘+1}⊂𝑉(𝐴^∗_𝑛). We show that 𝑊 is a resolving set for 𝑉(𝐴^∗_𝑛). We compute the representations of the vertices of 𝑉(𝐴^∗_𝑛)⧵𝑊 with respect to 𝑊: 𝑟𝑐_𝑖=∣𝑊𝑖−1,𝑖−2,𝑛+12−𝑖,for3≤𝑖≤𝑘,𝑛−12,𝑛−12,1,for𝑖=𝑘+2,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−𝑛+12𝑟𝑏,for𝑘+3≤𝑖≤𝑛,_𝑖=∣𝑊1,1,𝑛−12,for𝑖=1,𝑖,𝑖−1,𝑛+12−𝑖,for2≤𝑖≤𝑘,𝑛+12,𝑛−12,1,for𝑖=𝑘+1,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−𝑛−12𝑟𝑎,for𝑘+2≤𝑖≤𝑛,_𝑖=∣𝑊2,2,𝑛+12,for𝑖=1,2,2,𝑛−12,for𝑖=2,𝑖,𝑖−1,𝑛+32−𝑖,for3≤𝑖≤𝑘,𝑛+22,𝑛−12,2,for𝑖=𝑘+1,𝑛+12,𝑛+12,2,for𝑖=𝑘+2,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−𝑛−12,for𝑘+3≤𝑖≤𝑛. (3) We observe that there are no two vertices having the same representation, which implies that dim(𝐴^∗_𝑛)≤3.
We now show that dim(𝐴^∗_𝑛)≥3. Suppose contrarily 𝑊^′ with |𝑊^′|=2. Proceeding on the same way as in Case  1, it can be shown that no such 𝑊^′ can be a resolving set. Finally, from Cases  1 and 2, we get dim(𝐴^∗_𝑛)=3, which completes the proof.

Theorem 2. Let 𝑛≥6 be an integer, then dim(𝐴^𝑝_𝑛)=3.

Proof   
Case  1. Let 𝑛=2𝑘,  𝑘≥3, 𝑘∈ℕ. Consider the set 𝑊={𝑏₁,𝑏₂,𝑏_𝑘+1}⊂𝑉(𝐴^𝑝_𝑛). We show that 𝑊 is a resolving set for 𝑉(𝐴^𝑝_𝑛). We compute the representations of the vertices of 𝑉(𝐴^𝑝_𝑛)⧵𝑊 with respect to 𝑊: 𝑟𝑏_𝑖=∣𝑊𝑖−1,𝑖−2,𝑛+22−𝑖,for3≤𝑖≤𝑘,𝑛−𝑖+1,𝑛+2−𝑖,𝑖−𝑛+12𝑟𝑎,for𝑘+2≤𝑖≤𝑛,_𝑖=∣𝑊1,2,𝑛+22,for𝑖=1,𝑖,𝑖−1,𝑛+42𝑛−𝑖,for2≤𝑖≤𝑘+1,𝑛+2−𝑖,𝑛+3−𝑖,𝑖−2𝑟𝑐,for𝑘+2≤𝑖≤𝑛,_𝑖=𝑛∣𝑊1,1,2,for𝑖=1,𝑖,𝑖−1,𝑛+22𝑛−𝑖,for2≤𝑖≤𝑘,2,𝑛2𝑛,1,for𝑖=𝑘+1,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−2,for𝑘+2≤𝑖≤𝑛. (4)
We observe that there are no two vertices having the same representations implying that dim(𝐴^𝑝_𝑛)≤3.
To prove the theorem, it is sufficient to show that dim(𝐴^𝑝_𝑛)≥3. Contrarily, assume that there exists a resolving set 𝑊^′ with |𝑊^′|=2. We have the following possibilities.(1)Both vertices of 𝑊^′ belong to {𝑐_𝑖}⊂𝑉(𝐴^𝑝_𝑛). Without loss of generality, we suppose that one resolving vertex is 𝑐₁ and the other is 𝑐_𝑡(2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑐_𝑛∣𝑊^′)=𝑟(𝑏₁∣𝑊^′)=(1,𝑡). For 𝑡=𝑘+1  𝑟(𝑐₂∣𝑊^′)=𝑟(𝑐_𝑛∣𝑊^′)=(1,𝑘−1), a contradiction.(2)Both vertices of 𝑊^′ belong to {𝑏_𝑖}⊂𝑉(𝐴^𝑝_𝑛). Without loss of generality, we suppose that one resolving vertex is 𝑏₁ and the other is 𝑏_𝑡  (2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑐_𝑛∣𝑊^′)=𝑟(𝑎₁∣𝑊^′)=(1,𝑡). For 𝑡=𝑘+1  𝑟(𝑐_𝑘∣𝑊^′)=𝑟(𝑐_𝑘+1∣𝑊^′)=(𝑘,1), a contradiction.(3)Both vertices of 𝑊^′ belong to {𝑎_𝑖}⊂𝑉(𝐴^𝑝_𝑛). We suppose that one resolving vertex is 𝑎₁ and the other is 𝑎_𝑡(2≤𝑡≤𝑘+1). For 2≤𝑡≤𝑘, we have 𝑟(𝑐_𝑛∣𝑊^′)=𝑟(𝑏_𝑛∣𝑊^′)=(2,𝑡+1). For 𝑡=𝑘+1𝑟(𝑐_𝑘∣𝑊^′)=𝑟(𝑐_𝑘+1∣𝑊^′)=(𝑘+1,2), a contradiction.(4)One vertex belong to {𝑐_𝑖}⊂𝑉(𝐴^𝑝_𝑛) and the other vertex belong to {𝑏_𝑖}⊂𝑉(𝐴^𝑝_𝑛). Consider one resolving vertex is 𝑐₁ and the other is 𝑏_𝑡(1≤𝑡≤𝑘+1). For 1≤𝑡≤𝑘, we have 𝑟(𝑎₁∣𝑊^′)=𝑟(𝑏_𝑛∣𝑊^′)=(2,𝑡). For 𝑡=𝑘+1  𝑟(𝑐_𝑘+1∣𝑊^′)=𝑟(𝑏_𝑘+2∣𝑊^′)=(𝑘,1), a contradiction.(5)One vertex belong to {𝑐_𝑖}⊂𝑉(𝐴^𝑝_𝑛) and the other vertex belong to {𝑎_𝑖}⊂𝑉(𝐴^𝑝_𝑛). Consider one resolving vertex is 𝑐₁ and the other is 𝑎_𝑡(1≤𝑡≤𝑘+1). For 1≤𝑡≤𝑘−1, we have 𝑟(𝑎_𝑛∣𝑊^′)=𝑟(𝑏_𝑛−1∣𝑊^′)=(3,𝑡+2). For 𝑘≤𝑡≤𝑘+1𝑟(𝑏₂∣𝑊^′)=𝑟(𝑐₂∣𝑊^′)=(1,𝑡−1), a contradiction.(6)One vertex belong to {𝑏_𝑖}⊂𝑉(𝐴^𝑝_𝑛) and the other vertex belong to {𝑎_𝑖}⊂𝑉(𝐴^𝑝_𝑛). Consider one resolving vertex is 𝑏₁ and the other is 𝑎_𝑡(1≤𝑡≤𝑘+1). For 1≤𝑡≤𝑘, we have 𝑟(𝑏_𝑛∣𝑊^′)=𝑟(𝑐_𝑛∣𝑊^′)=(1,𝑡+1). For 𝑡=𝑘+1𝑟(𝑐_𝑘∣𝑊^′)=𝑟(𝑐_𝑘+1∣𝑊^′)=(𝑘,2), a contradiction.
Hence, from above it follows that there is no resolving set with two vertices for 𝑉(𝐴^𝑝_𝑛) implying that dim(𝐴^𝑝_𝑛)=3.
Case  2. For 𝑛=2𝑘+1, 𝑘≥3, 𝑘∈ℕ. Consider the set 𝑊={𝑐₁,𝑐₂,𝑐_𝑘+1}⊂𝑉(𝐴^𝑝_𝑛). We show that 𝑊 is a resolving set for 𝑉(𝐴^𝑝_𝑛). We compute the representations of the vertices of 𝑉(𝐴^𝑝_𝑛)⧵𝑊 with respect to 𝑊: 𝑟𝑐_𝑖=∣𝑊𝑖−1,𝑖−2,𝑛+12−𝑖,for3≤𝑖≤𝑘,𝑛−12,𝑛−12,1,for𝑖=𝑘+2,𝑛+1−𝑖,𝑛+2−𝑖,𝑖−𝑛+12𝑟𝑏,for𝑘+3≤𝑖≤𝑛,_𝑖=∣𝑊1,2,𝑛+12,for𝑖=1,1,1,𝑛−12,for𝑖=2,𝑖−1,𝑖−2,𝑛+32−𝑖,for3≤𝑖≤𝑘+1,𝑛+12,𝑛−12,1,for𝑖=𝑘+2,𝑛+2−𝑖,𝑛+3−𝑖,𝑖−𝑛+12𝑟𝑎,for𝑘+3≤𝑖≤𝑛,_𝑖=∣𝑊2,3,𝑛+32,for𝑖=1,2,2,𝑛+12,for𝑖=2,𝑖,𝑖−1,𝑛+52−𝑖,for3≤𝑖≤𝑘+1,𝑛+32,𝑛+12,2,for𝑖=𝑘+2,𝑛+3−𝑖,𝑛+4−𝑖,𝑖−𝑛−12,for𝑘+2≤𝑖≤𝑛. (5)
We observe that there are no two vertices having the same representation. Thus, dim(𝐴^𝑝_𝑛)≤3.
We now show that dim(𝐴^𝑝_𝑛)≥3. Assume that dim(𝐴^𝑝_𝑛)=2. Proceeding on the same lines as in Case  1, we can show that dim(𝐴^𝑝_𝑛)≥3. Finally, from Cases  1 and 2, we get dim(𝐴^𝑝_𝑛)=3.

3. Conclusion

In this paper, we have studied the metric dimension of some families of graphs which are the extension of the antiprism graph. We have seen that the metric dimension of these graphs is finite and does not depend on the order of the graph and only three appropriately chosen vertices suffice to resolve all the vertices of these graphs.

Acknowledgments

The authors are indebted to the anonymous referees for their many valuable comments and suggestions on the earlier version of this paper. This research was partially supported by FAST-NU, Peshawar, and by the Higher Education Commission of Pakistan.