# On Metric Dimension of Some Rotationally Symmetric Graphs

Authors: M. Ali; M. T. Rahim; G. Ali; U. Ali
Journal: ISRN Combinatorics
 (2013)

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## Abstract

A family𝒢 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 An∗ and the graph Anp obtained from the antiprism graph have constant metric dimension.


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## Body

## 1. Notation and Preliminary Results

For a connected graphG, 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,…,wk} 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,wk)). 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 verticesW={w1,w2,…,wk} of a graph G, the ith component of r(v∣W) is 0 if and only if v=wi. 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 fn, and Javaid et al. [3] found the metric dimension of Jahangir graph J2n.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
(1)dim(pm×Cn)={2,ifn,isodd;3,otherwise.
Prisms Dn are the trivalent plane graphs obtained by the cartesian product of the path P2 with a cycle Cn. In [4], Javaid et al. proved that this family has constant metric dimension. In [4], Javaid et al. also proved that the antiprism graph An, constitutes a family of regular graphs with constant metric dimension as dim(An)=3, for every n≥5.In this paper, we extend this study to antiprism-related graphs (see Figure1). The graph An* is defined as follows: for each vertex bi, i=1,2,…,n of the outer cycle of the antiprism graph, we introduce a new vertex ai, i=1,2,…,n and join ai to bi and bi-1, i=1,2,…,n, with n+i taken modulo n. Thus, V(An*)=⋃i=1n‍{ai,bi,ci}. Here {ci}  i=1,2,…,n are the inner cycle vertices, {bi}  i=1,2,…,n are the outer cycle vertices, and {ai}  i=1,2,…,n are adjacent vertices to the outer cycle. We define the graph Anp as follows: for each vertex bi, i=1,2,…,n of the outer cycle of the antiprism graph, we introduce a new vertex ai and join ai to bi, i=1,2,…,n. Thus, V(Anp)=⋃i=1n‍{ai,bi,ci}. Here, {ci} are the inner cycle vertices, {bi} are the outer cycle vertices,  {ai}i=1,2,…,n are the pendant vertices adjacent to the outer cycle vertices.GraphsA8* and A8p.
(a)
(b)


## 2. Antiprism-Related Graphs with Constant Metric Dimension

In this section, we show thatAn*, Anp have constant metric dimension.Theorem 1.
Letn≥6 be an integer, then dim(An*)=3.Proof
Case  1.Let n=2k, k≥3, k∈ℕ. Consider the set W={c1,c2,ck+1}⊂V(An*). We show that W is a resolving set for V(An*). We find the representations of the vertices of V(An*)∖W with respect to W:
(2)r(ci∣W)={(i-1,i-2,n+22-i),for3≤i≤k,(n-i+1,n+2-i,i-n+22),fork+2≤i≤n,r(bi∣W)={(1,1,n2),fori=1,(i,i-1,n+22-i),for2≤i≤k,(n2,n2,1),fori=k+1,(n+1-i,n+2-i,i-n2),fork+2≤i≤n,r(ai∣W)={(2,2,n+22),fori=1,(2,2,n2),fori=2,(i,i-1,n+42-i),for3≤i≤k,(n+12,n2,2),fori=k+1,(n+2-i,n+3-i,i-n2),fork+2≤i≤n.
We note that there are no two vertices having the same representation, which implies that dim(An*)≤3.
To prove the theorem, it is sufficient to show thatdim(An*)≥3. Contrarily, assume that there exists a resolving set W′ with |W′|=2. We have the following possibilities.(1)
Both vertices ofW′ belong to {ci}⊂V(An*). Without loss of generality, we suppose that one resolving vertex is c1 and the other is ct(2≤t≤k+1). For 2≤t≤k, we have r(cn∣W′)=r(bn∣W′)=(1,t). For t=k+1,   r(b1∣W′)=r(bn∣W′)=(1,k), a contradiction.(2)
Both vertices ofW′ belong to {bi}⊂V(An*). Without loss of generality, we suppose that one resolving vertex is b1 and the other is bt(2≤t≤k+1). For 2≤t≤k, we have r(bn∣W′)=r(a1∣W′)=(1,t). For t=k+1, r(a1∣W′)=r(a2∣W′)=(1,k), a contradiction.(3)
Both vertices ofW′ belong to {ai}⊂V(An*). We suppose that one resolving vertex is a1 and the other is at(2≤t≤k+1). For 2≤t≤k, we have r(cn∣W′)=r(an∣W′)=(2,t+1). For t=k+1, r(b1∣W′)=r(bn∣W′)=(1,k), a contradiction.(4)
One vertex belong to{ci}⊂V(An*) and the other belong to {bi}⊂V(An*). Consider one resolving vertex is c1 and the other is bt(1≤t≤k+1). For 1≤t≤k, we have r(bk+1∣W′)=r(ck+1∣W′)=(k,k+1-t). For t=k+1,  r(ck+1∣W′)=r(ak+2∣W′)=(k,1), a contradiction.(5)
One vertex belong to{ci}⊂V(An*) and the other belong to {ai}⊂V(An*). Consider one resolving vertex is c1 and the other is at(1≤t≤k+1). For 1≤t≤k, we have r(ak+1∣W′)=r(ck+1∣W′)=(k,k-t+2). For t=k+1,r(bk∣W′)=r(bk+1∣W′)=(k,1), a contradiction.(6)
One vertex belong to{bi}⊂V(An*) and the other belong to {ai}⊂V(An*). Consider one resolving vertex is b1 and the other is at(1≤t≤k+1). For this we have r(an∣W′)=r(cn∣W′)=(2,t+1), a contradiction.
Hence, from above it follows that there is no resolving set with two vertices forV(An*). Therefore, dim(An*)≥3, which implies that dim(An*)=3.
Case  2. Let n=2k+1, k≥3k∈ℕ. Consider the set W={c1,c2,ck+1}⊂V(An*). We show that W is a resolving set for V(An*). We compute the representations of the vertices of V(An*)∖W with respect to W:
(3)r(ci∣W)={(i-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+12),fork+3≤i≤n,r(bi∣W)={(1,1,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-12),fork+2≤i≤n,r(ai∣W)={(2,2,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.
We observe that there are no two vertices having the same representation, which implies that dim(An*)≤3.
We now show thatdim(An*)≥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(An*)=3, which completes the proof.Theorem 2.
Letn≥6 be an integer, then dim(Anp)=3.Proof
Case  1. Let n=2k,  k≥3, k∈ℕ. Consider the set W={b1,b2,bk+1}⊂V(Anp). We show that W is a resolving set for V(Anp). We compute the representations of the vertices of V(Anp)∖W with respect to W:
(4)r(bi∣W)={(i-1,i-2,n+22-i),for3≤i≤k,(n-i+1,n+2-i,i-n+12),fork+2≤i≤n,r(ai∣W)={(1,2,n+22),fori=1,(i,i-1,n+42-i),for2≤i≤k+1,(n+2-i,n+3-i,i-n2),fork+2≤i≤n,r(ci∣W)={(1,1,n2),fori=1,(i,i-1,n+22-i),for2≤i≤k,(n2,n2,1),fori=k+1,(n+1-i,n+2-i,i-n2),fork+2≤i≤n.
We observe that there are no two vertices having the same representations implying thatdim(Anp)≤3.
To prove the theorem, it is sufficient to show thatdim(Anp)≥3. Contrarily, assume that there exists a resolving set W′ with |W′|=2. We have the following possibilities.(1)
Both vertices ofW′ belong to {ci}⊂V(Anp). Without loss of generality, we suppose that one resolving vertex is c1 and the other is ct(2≤t≤k+1). For 2≤t≤k, we have r(cn∣W′)=r(b1∣W′)=(1,t). For t=k+1  r(c2∣W′)=r(cn∣W′)=(1,k-1), a contradiction.(2)
Both vertices ofW′ belong to {bi}⊂V(Anp). Without loss of generality, we suppose that one resolving vertex is b1 and the other is bt  (2≤t≤k+1). For 2≤t≤k, we have r(cn∣W′)=r(a1∣W′)=(1,t). For t=k+1  r(ck∣W′)=r(ck+1∣W′)=(k,1), a contradiction.(3)
Both vertices ofW′ belong to {ai}⊂V(Anp). We suppose that one resolving vertex is a1 and the other is at(2≤t≤k+1). For 2≤t≤k, we have r(cn∣W′)=r(bn∣W′)=(2,t+1). For t=k+1r(ck∣W′)=r(ck+1∣W′)=(k+1,2), a contradiction.(4)
One vertex belong to{ci}⊂V(Anp) and the other vertex belong to {bi}⊂V(Anp). Consider one resolving vertex is c1 and the other is bt(1≤t≤k+1). For 1≤t≤k, we have r(a1∣W′)=r(bn∣W′)=(2,t). For t=k+1  r(ck+1∣W′)=r(bk+2∣W′)=(k,1), a contradiction.(5)
One vertex belong to{ci}⊂V(Anp) and the other vertex belong to {ai}⊂V(Anp). Consider one resolving vertex is c1 and the other is at(1≤t≤k+1). For 1≤t≤k-1, we have r(an∣W′)=r(bn-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{bi}⊂V(Anp) and the other vertex belong to {ai}⊂V(Anp). Consider one resolving vertex is b1 and the other is at(1≤t≤k+1). For 1≤t≤k, we have r(bn∣W′)=r(cn∣W′)=(1,t+1). For t=k+1r(ck∣W′)=r(ck+1∣W′)=(k,2), a contradiction.
Hence, from above it follows that there is no resolving set with two vertices forV(Anp) implying that dim(Anp)=3.
Case  2. For n=2k+1, k≥3, k∈ℕ. Consider the set W={c1,c2,ck+1}⊂V(Anp). We show that W is a resolving set for V(Anp). We compute the representations of the vertices of V(Anp)∖W with respect to W:
(5)r(ci∣W)={(i-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+12),fork+3≤i≤n,r(bi∣W)={(1,2,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+12),fork+3≤i≤n,r(ai∣W)={(2,3,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.
We observe that there are no two vertices having the same representation. Thus,dim(Anp)≤3.
We now show thatdim(Anp)≥3. Assume that dim(Anp)=2. Proceeding on the same lines as in Case  1, we can show that dim(Anp)≥3. Finally, from Cases  1 and 2, we get dim(Anp)=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.



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*Source: 724049-2013-02-24.xml*