Metric dimension of fullerene graphs

A resolving set W is a set of vertices of a graph G(V,E) such that for every pair of distinct vertices u, v ∈ V (G), there exists a vertex w ∈ W satisfying d(u,w) 6= d(v, w). A resolving set with minimum number of vertices is called metric basis of G. The metric dimension of G, denoted by dim(G), is the minimum cardinality of a resolving set of G. In this paper, we consider (3, 6)fullerene and (4, 6)-fullerene graphs and compute the metric dimension for these fullerene graphs. We also give conjecture on the metric dimension of (3, 6)-fullerene and (4, 6)-fullerene graphs.


Introduction
The metric dimension was initially studied by Slater [16] and Harary and Melter [7].They characterized the metric dimension of trees.Metric dimension has several applications in robot navigation [9], chemistry [3], sonar [16] and combinatorical optimization [13].Let G be a molecular graph, that is, a representation of the structural formula of a chemical compound in terms of graph theory.The vertices and edges of G correspond to atoms and chemical bonds, respectively.For u, v ∈ V (G), the length of a shortest path from u to v is called the distance between u and v and is denoted by d(u, v).A graph G is said to be k-connected if there does not exist a set of less than k vertices whose removal disconnects the graph G.A planar graph G is a graph that can be drawn in such a way that no two edges cross each other.A cubic graph G is a graph in which all vertices have degree 3.
If every pair of distinct vertices of G have a distinct metric representation then the ordered set W is called a resolving set of G.A resolving set of minimum cardinality is called the metric basis for G and this cardinality is the metric dimension of G, denoted by dim(G).If dim(G) = k, then G is said to be k-dimensional.Several variations of metric dimension have been discussed in the literature, like resolving dominating sets [2], independent resolving sets [4], local metric sets [11], resolving partitions [5] and strong metric generators [13].
In 1985, Kroto et al. [8] discovered fullerene molecule and since then, scientists took a great interest in the fullerene graphs.A (k, 6)−fullerene graph is a 3-connected cubic plane graph whose faces have sizes k and 6.The only values of k for which a (k, 6)−fullerene graph exists are 3, 4 and 5.A (5, 6)−fullerene is an ordinary fullerene and constructed from pentagons and hexagons.Fowler et al. [6] discussed the mathematical properties of (5, 6)−fullerene.
A (3, 6)-fullerene graph have attained attention due to the similarity of its structure with ordinary fullerenes.The Eulers formula implies that a (3, 6)-fullerene graph has exactly four faces of size 3 and (n/2) − 2 hexagons.If the triangles in (3, 6)-fullerene have no common edge then it is called isolated triangular rules (ITR).
A (4, 6)-fullerene graph is a mathematical model of a boron-nitrogen fullerene.The Eulers formula implies that a (4, 6)-fullerene graph has exactly six square faces and (n/2) − 4 hexagons.If the six quadrangles in (4, 6)-fullerene don't have common edge, then it is called isolated square rules (ISR).

Metric dimension of (3,6)-fullerene graphs
are the graphs of (3, 6)-fullerene depicted in Figures ??-4 with order 8n + 4, 12n + 4, 16n − 32 and 24n, respectively.In this section, we find the metric dimension of Proof.Let {z 1 , z 2 , z 3 } and {z 4 , z 5 , z 6 } be the vertex sets of outer triangles of ).We show that W is a resolving set of F 1 [n].For this we give the representation of vertices in V (F 1 [n]) \ W with respect to W .The representation of vertices z 1 , z 4 and z 6 is given by: The representation of vertices of upper half of the fullerene graph F 1 [n] is given by: The representation of vertices of lower half of the fullerene graph F 1 [n] is given by: All the vertices of F 1 [n] have different representation with respect to W , this implies that W is a resolving set of F 1 [n].Thus the metric dimension of dim(F 1 [n]) ≤ 3.
On the other hand, we show that dim(F 1 [n]) ≥ 3 by proving that there is no resolving set W such that |W | = 2. Let A = {z 1 , z 2 , z 3 , z 4 , z 5 , z 6 } be the set of vertices of outer triangles of F 1 [n].Suppose on contrary that dim(F 1 [n]) = 2 and W is a resolving set with |W | = 2, then there are following cases: Case 1.If both vertices of W are in upper half of the fullerene graph F 1 [n], then the representations of pair of vertices z 4 , z 6 and z 1 , z 3 are the same.Thus W is not a resolving set of F 1 [n].Case 2. If both vertices of W are in lower half of the fullerene graph F 1 [n], then the representations of pair of vertices z 4 , z 5 and z 1 , z 2 are the same.Thus W is not a resolving set of then the pair of vertices z 1 , z 5 or z 2 , z 6 have the same representation with respect to W . Thus W is not a resolving set of F 1 [n].Case 4. If one vertex from upper half of fullerene graph F 1 [n] and one from the set of vertices A in W , then the representation of some vertices in {x 1 , x 2 , • • • , x 4n−1 } and {y 1 , y 2 , • • • , y 4n−1 } is the same.Therefore W is not a resolving set in this case.Case 5.If one vertex from lower half of fullerene graph F 1 [n] and one from the set of vertices A in W , then the representation of some vertices in {x 1 , x 2 , • • • , x 4n−1 } and {y 1 , y 2 , • • • , y 4n−1 } is the same.Therefore W is not a resolving set in this case.Case 6.If both vertices of W belongs to the set of vertices A, then we have the following subcases: • If W = {z 2 , z 5 } or W = {z 2 , z 6 }, then the representation of pair of vertices x 1 , z 1 or x 1 , z 3 are the same.Similarly if W = {z 3 , z 5 } or W = {z 3 , z 6 }, then the representation of pair of vertices y 1 , z 1 or y 1 , z 2 are the same.
• All other possible subsets of A, then the representation of remanning pair of vertices of set A is the same.Thus, in every subcase we get a contradiction.
Proof.Let {a 1 , a 2 , a 12 } and {a 6 , a 7 , a 11 } be the vertex sets of outer triangles of For this purpose, we give the representation of vertices in V (F 2 [n]) \ W with respect to W .The representation of outer vertices of the fullerene graph F 2 [n] is given below:

The representation of vertices of upper half of the fullerene graph
The representation of middle vertices of the fullerene graph F 2 [n] for n = 1 is given below: The representation of middle vertices of the fullerene graph F 2 [n] for n ≥ 2 is given below: The representation of vertices of lower half of the fullerene graph F 2 [n] is given below: However all pair of vertices can easily be resolved by the set W . Thus the set W is a resolving set of Thus in every case we get a contradiction.Thus we conclude that there is no resolving set W containing two vertices of F 2 [n].Thus the metric dimension of F 2 [n] is 3.This completes the proof.Proof.Let {z 1 , z 2 , z 3 } and {z 4 , z 5 , z 6 } be the vertex sets of outer triangles and {a 1 , a 2 , a 3 , a 4 , a 5 , a 6 } be the vertices of outer hexagonal of ).We need to show that W is a resolving set of F 3 [n].For this purpose, first we give the representation of vertices in The representation of vertices of outer triangles in F 3 [n] is given by: The representation of vertices of upper half of the fullerene graph F 3 [n] is given by: The representation of middle vertices of the fullerene graph The representation of middle vertices of the fullerene graph F 3 [n] for n ≥ 2 is given by: The representation of middle vertices of the fullerene graph F 3 [n] is given by: The representation of vertices of lower half of the fullerene graph F 3 [n] is given by: Therefore the set W resolves all the vertices in For this we show that there does not exists any resolving set W with two vertices.Let A = {a 1 , a 2 , a 3 , a 4 , a 5 , a 6 }, Thus, in every case we get a contradiction.From above cases, we conclude that there is no resolving set W with exactly two vertices of F 3 [n].Thus the metric dimension of F 3 [n] is 3.This completes the proof.The representation of vertices of upper half of the fullerene graph F 4 [n] is given below: The representation of middle vertices of the fullerene graph F 4 [n] is given below:

The representation of vertices of lower half of the fullerene graph
All vertices of V (F 4 [n]) \ W can be resolved with respect to W . Thus W is a resolving set of F 4 [n] and dim(F 4 [n]) ≤ 3.
On the other hand, we show that dim(F 4 [n]) ≥ 3 by proving that there is no resolving set W with cardinality 2. Suppose on contrary that dim(F 4 [n]) = 2 and W is a resolving set of The pairs of vertices a 1 , a 3 and a 5 , a 7 have the same distance with the vertices of set B. The pairs of vertices a 2 , a 4 and a 6 , a 8 have the same distance with the vertices of set C. The pairs of vertices a 2 , a 3 and a 6 , a 7 have the same distance with the vertices of set D. Similarly, The pairs of vertices a 1 , a 4 and a 5 , a 8 have the same distance with the vertices of set C. Therefore, there is no a resolving set W with cardinality This completes the proof.5-7 with order 8n, 8n + 4 and 12n + 12 respectively.In this subsection we find the metric dimension of

Proof. For a set
There are following cases: Case 1.If both vertices are in the upper half of G 1 [n] and the resolving set is W = {x s , x t }, 1 ≤ s ≤ t ≤ 4n.Then the representations of x i and y i+1 , 2n + 1 ≤ i ≤ 4n − 1 are the same.Similarly the representations of x i+1 and y i , 2 ≤ i ≤ 2n − 1 are the same.Therefore the resolving set of If both vertices are in the lower half of G 1 [n] and the resolving set is W = {y s , y t }, 1 ≤ s ≤ t ≤ 4n.Then the representations of x i+1 and y i , 2n + 1 ≤ i ≤ 4n − 1 are the same.Similarly the representations of x i and y i+1 , 2 ≤ i ≤ 2n − 1 are the same.Therefore the resolving set is not a subset of {y 1 , y 2 , • • • , y 4n }.Case 3. If one vertex belongs to the set of vertices {x 1 , x 2 , • • • , x 4n } and other is in the set of vertices {y 1 , y 2 , • • • , y 4n }.Without loss of generality, we can suppose that the resolving set is W = {x s , y t }, 1 ≤ s ≤ 4n and 1 ≤ t ≤ 4n.If s = t, then the representation of pairs of vertices x s+1 , x s−1 and y t−1 , y t+1 are the same.If s < t, then the representation of x i , i > s and y j , j < 4n are the same.If s > t, then the representation of x i , i < 4n and y j , j > t are the same.
Thus, in every subcase we get a contradiction.From above cases, we conclude that there is no resolving set W with |W | = 2. Thus dim(G 1 [n]) ≥ 3. Now we show that dim(G 1 [n] ≤ 3.For this purpose we give the representation of the vertices in V (G 1 [n]) \ W with respect to W .The representation of vertices of upper half of the fullerene graph G 1 [n] is given below: The representation of vertices of lower half of the fullerene graph G 1 [n] is given below: This implies that all vertices of V (G 1 [n]) \ W can be resolved with respect to W . Thus W is a resolving set of G 1 [n].Therefore dim(G 1 [n]) = 3 for n ≥ 2. This completes the proof.
First we give the representation of vertices in V (G 2 [n]) \ W with respect to W .The representation of vertices of upper half of the fullerene graph G 2 [n] is given by: The representation of vertices of lower half of the fullerene graph G 2 [n] is given by: If s = t, then the representation of pair of vertices x s+1 , x s−1 and y t−1 , y t+1 is the same.If s < t, then the representation of x i , i > s and y j , j < 4n is the same.If s > t, then the representation of x i , i < 4n and y j , j > t is the same.
From above cases, we conclude that there is no resolving set W for G 2 [n] with |W | = 2. Thus dim(G 2 [n]) = 3.This completes the proof.) \ W with respect to W for n ≥ 2 is given by: r(x i |W ) = (i − 1, i + 1, i + 1), if 2 ≤ i ≤ 2n + 2, r(y i |W ) = (i, i + 2, i), The representation of vertices of V (G 3 [n]) \ W with respect to W for n = 1 is given by: Hence there are no two vertices having the same representations.Thus W = {x 1 , c 1 , a 1 } is a resolving set of G 3 [n].Therefore dim(G 3 [n]) = 3.This completes the proof.
and W is a resolving set with |W |.Then there are following possibilities: Case 1.The pair of vertices a 1 , a 2 and a 7 , a 8 have the same distance from the vertices of set B, C and D. Then any subset of B, C and D is not a resolving set of F 2 [n].Case 2. If both vertices of W are from set of vertices A, then some vertices of A have same representation.Therefore W is not a resolving set of F 2 [n].
be the sets of vertices of F 3 [n].Then there are following cases: Case 1.If both vertices of W are in the set of vertices C, then the vertices of A and D have the same representation.Therefore W is not a resolving set of F 3 [n].Case 2. If both vertices of W are in the set of vertices D, then the vertices of B and C have the same representation.Therefore W is not a resolving set of F 3 [n].Case 3. The vertices of set A have same distance from the vertices of B. Therefore the resolving set is not the subset of A or B. Case 4. If both vertices of W are in the set of vertices F , then the vertices of A and E have the same representation.Therefore W is not a resolving set of F 3 [n].Case 5.If both vertices of W are in the set of vertices E, then the vertices of B and F have the same representation.Therefore W is not a resolving set of F 3 [n].
This implies that W is a resolving set of G 2 [n] and dim(G 2 [n]) ≤ 3. Now we show that dim(G 2 [n]) ≥ 3 by proving that W is a resolving set of G 2 [n] with |W | = 2.There are following possibilities: Case 1.If both vertices are in the upper half of G 2 [n] and the resolving set is W = {x s , x t }, 1 ≤ s ≤ t ≤ 4n.Then the representation of x i and y i+1 , 2n + 2 ≤ i ≤ 4n − 1 is the same.Similarly the representation of x i+1 and y i , 2 ≤ i ≤ 2n is the same.Therefore the resolving set ofG 2 [n] is not a subset of {x 1 , x 2 , • • • , x 4n }.Case 2. If both vertices are in the lower half of G 1 [n] and the resolving set is W = {y s , y t }, 1 ≤ s ≤ t ≤ 4n.Then the representation of x i+1 and y i , 2n + 2 ≤ i ≤ 4n − 1 is the same.Similarly the representation of x i and y i+1 , 2 ≤ i ≤ 2n is the same.Therefore the resolving set of G 2 [n] is not a subset of {y 1 , y 2 , • • • , y 4n }.Case 3. If one vertex belongs to the set of vertices {x 1 , x 2 , • • • , x 4n } and other is in the set of vertices {y 1 , y 2 , • • • , y 4n }.Without loss of generality, we can suppose that the resolving set of