Computer networks are graphs with vertices representing hosts, servers, or hubs, and edges as connecting mediums between them. Vertex is actually a possible location to find faults or some damaged devices in a computer network. This idea somehow created an urge in Slater and independently in Harary and Melter (1976) to uniquely recognize each vertex of a graph in a network so that fault could be controlled efficiently. Thus, the basis for the notion of locating sets and locating the number of graphs came into existence. Since then, the resolving sets have been extensively investigated (Buczkowski et al., 2003; Caceres et al., 2005; Javaid et al., 2008). The resolving set contributes to various areas such as network discovery (Khuller et al., 1994, 1996), connected joins in graphs, strategies for the mastermind games (Chvatal, 1983), applications of pattern recognition, combinatorial optimization, image processing (Melter and Tomescu, 1984), pharmaceutical chemistry, and game theory.

A moving point in a graph may be located by finding the distance from the point to the collection of sonar stations that have been properly positioned in the graph. Thus finding a minimal but sufficiently large set of labeled vertices such that a robot can find its position, is a well-established problem known as robot navigation. This sufficiently large set of labeled vertices is a resolving set of the graph space and the corresponding cardinality is the metric dimension. Similarly, on another node, a real-world problem is the study of networks whose structure has not been imposed by a central authority but has arisen from local and distributed processes. It is very difficult and expensive to obtain a map of all nodes and the links between them. A commonly used technique is to obtain a local view of the network from various locations and combine them to obtain a good approximation for the real network. Metric dimension also has some applications in this aspect as well.

Consider a simple, connected graph G, and metric d_G : V(G) × V(G) → ℕ ∪ {0}, where ℕ is the set of positive integers and d_G(x, y) is the minimum number of edges in any path between x and y. Let W = {w₁, w₂, ..., w_k} be an ordered set of vertices of G and let v be a vertex of G. The representation r(v|W) of v with respect to W is the k−tuple (d_G(v, w₁), d_G(v, w₂), ..., d_G(v, w_k)). If distinct vertices of G have a distinct representation with respect to W, then W is called a resolving set of G (see Khuller et al., 1994, 1996; Buczkowski et al., 2003; Javaid et al., 2008). Such a resolving set with minimum cardinality is a basis of G and the metric dimension of G, denoted by β(G) is its cardinality.

Buczkowski et al. (2003) established the metric dimension of wheel W_n to be ⌊2n+25⌋ for n ≥ 7. Caceres et al. (2005) found that the metric dimension of fan to be ⌊2n+25⌋ for n ≥ 7. Tomescu and Javaid (2007) determined the dimension of Jahangir graphs J_2n to be ⌊2n3⌋ for all n ≥ 4. A particular metric-feature of the family of graphs is independent of metric dimension on the particular element of the family. A connected graph has a constant metric dimension if β(G) = k, where k ∈ ℕ is fixed. This feature has been presented in Imran et al. (2012) study. Imran et al. (2013) computed metric dimension of flower graph and some families of convex polytopes. Chartrand et al. (2000) proved that a graph has a constant metric dimension of 1 if it is a path. Ali et al. (2012) computed partial results of the metric dimension of the Mobius ladder, whereas Munir et al. (2017) computed exact and complete results for the metric dimension of the Mobius Ladders. Hussain et al. (2010) computed upper bounds for the metric dimension and partition dimension of generalized Mobius ladders. Poisson and Zhang (2002) computed the metric dimension of uni-cyclic graphs. For detailed review of some variants of dimensions, please see Tomescu and Imran (2009), Manuel (2010), Afzal and Imran (2014), Raza et al. (2018).

Recent development in this context has paved way for a new related concept known as fault-tolerance in the metric dimension. Suppose that, in a network, n processing units are interlinked, and of these units, forming a chain of maximal length are used to solve some task. To have a fault-tolerant self-stable system, it is necessary that in the case of failure of any single unit, another chain of units not containing the faulty unit can replace the originally used chain. Thus, a fault-tolerant design enables a system to continue its intended operation, possibly at a reduced level, rather than failing completely. The units and the links are represented by graphs refer to Slater (1975, 1998, 2002) and Hayes (1985). A resolving set Ś is considered fault-tolerant if Ś\{v} is also a resolving set, for each v ∈ Ś, and the fault-tolerant metric dimension, β′(G), is the minimum cardinality of such Ś. A family G of connected graphs is said to have a constant fault-tolerant metric dimension if it is independent of any choice of a member of that family. Fault-tolerant designs are widely used in engineering and computer sciences (Hayes, 1985). Slater (2002) introduced the study of fault-tolerant locating-dominating sets. Hernando et al. (2008) introduced the idea of a single fault-tolerant metric dimension. The authors discussed the single fault-tolerant metric dimension of trees. They also proved that fault-tolerant metric dimension is bounded by a function of the metric dimension irrespective of the choice of the graph given be β(G)́≤β(G)(1+2.5β(G)-1). Javed et al. discussed fault tolerance in resolvability (Chaudhry et al., 2010) and computed fault tolerant metric dimension of some graphs (Javaid et al., 2009). It is easy to gather that β(G)́≥β(G)+1 (Javaid et al., 2009). Shabbir and Zumfererscu (2016) discussed fault tolerance in triangular lattice networks.

The subject matter of this article is fault tolerance in metric dimension of lattices of two types of boron nanotubes, alpha boron α_m×n and triangular boron nanotubes T_m×n. These lattices are formed by cutting both tubes vertically. Kwun et al. (2018) computed M-polynomials and related topological indices of these tubes and drew some nice comparative remarks about these tubes. Recently, Hussain et al. (2018) computed the metric dimension of α_m×n. Figure 1 presents a triangular boron tube lattice.

Figure 2 presents alpha boron tube lattice.

This section discusses main results. At first, we compute the fault-tolerant metric dimension of triangular boron tubes and then we focus on alpha boron tubes.

Theorem 2.1. Let T_k,l denote the graph of k × l, 2D-lattice of triangular boron nano tubes. Then β′(Tk,l)≤4.

Proof. The vertex set of T_k,l is {v_1,1, v_1,2, v_1,3, ...., v_1,l, v_2,1, v_2,2, v_2,3, ...., v_2,l, v_3,1, v_3,2, v_3,3, ......, v_3,l, ......, v_k,1, v_k,2, v_k,3, ...., v_k,l}. Let F = {v_1,1, v_1,l, v_k,1, andv_k,l}. We prove that F is an FTRS for T_k,l. In general, the tube is “cut” vertically, and the upper left triangle is pointing up as shown in Figure 3. This figure presents labeling of vertices that becomes optimal for the basis of fault tolerance.

Case I: Let k ≤ 2l, k is odd.

We give the distance vectors of all vertices v_a,b of T_k,l relative to F.

For a = 1

For a = 2

In general, if k is odd, x = k − a, a is odd, and 3≤a≤⌈k2⌉ and then

If k is odd, a is odd, and ⌈k2⌉+1≤a≤k then

If k is odd, a is even, and 3<a≤⌈k2⌉ then

If k is odd, a is even, and ⌈k2⌉+1≤a≤k then

For a = 2l − 1, r(V_a,b|S) = (2l − 2, 2l − 2, q − 1, l − q), if 1 ≤ q ≤ l

If k is even, a is odd, and 3≤a≤⌈k2⌉ then

If a is odd, k is even, and ⌈k2⌉+1≤a≤k then

If k is even, a is even, and 3<a≤⌈k2⌉ then

If k is even, a is even, and ⌈k2⌉+1≤a≤k then

For a = 2l, r(V_a,b|S) = (2l − 1, 2l − 1, b − 1, l − b), if 1 ≤ b ≤ l

For a = 2l − 1, r(V_a,b|S) = (2l − 2, 2l − 2, q − 1, l − q), if 1 ≤ q ≤ l

If k is even, a is odd, and 3≤a≤⌈k2⌉ then

If k is even, a is odd, and ⌈k2⌉+1≤a≤k then

If k is even, a is even, and 3<a≤⌈k2⌉ then

If k is even, a is even, and ⌈k2⌉+1≤a≤k then

For a = 2l, r(V_a,b|S) = (2l − 1, 2l − 1, b − 1, l − b), if 1 ≤ b ≤ l

Case II: Let k > 2l.

Let y = rl + p − 1. If r ≥ 2, rl < k ≤ (r + 1)l, and k is odd.

For a = rl + p, where 1 ≤ p ≤ l and a is odd

If a is even

If k is even, r ≥ 2, and rl < k ≤ (r + 1)l then

Let y = rl + p − 1. For a = rl + p where 1 ≤ p ≤ l and a is odd

If a is even

These vectors are distinct in at least two coordinates. So F is an FTRS for T_k,l. Therefore, β′(Tk,l)≤4.

Theorem 2.2. Let T_k,l denote the graph of k × l, 2D-lattice of triangular boron nano tubes. Then β′(Tk,l)>3.

Proof. For the lower bound on β′(Tk,l), we discuss the following cases:

Case I: k ≤ l.

We claim that any set of cardinality 3 is not an FTRS for T_k,l. Let F = {v_i,j, v_a,b, v_r,s} where i < a < r and j < b < s be an FTRS for T_k,l. Then F₁ = F\{v} is a R.S. for each v ∈ F. We discuss the following possibilities:

Possibility 1: When all vertices in F lie on the same row.

(i) If all vertices in F lie in the first row, then i = a = r = 1. Let F₁ = {v_1,j, v_1,b} then r(v_1,b+1|F₁) = r(v_2,b+1|F₁), a contradiction.

(ii) If all vertices in F lie in p − th row and 1 < p. Let F₁ = {v_p,j, v_p,b} then either r(v_k,b+1|F₁) = r(v_k−1,b+1|F₁) or r(v_p,b+1|F₁) = r(v_p+1,b+1|F₁), a contradiction.

Possibility 2: When two vertices in F lie on the same row.

WLOG we may suppose that i = a = l and r ≠ l. Let F₁ = {v_l,j, v_l,q} then either r(v_l,q+1|F₁) = r(v_l−1,q+1|F₁) or r(v_l,q+1|F₁) = r(v_l+1,q+1|F₁), a contradiction.

Possibility 3: When the vertices in F lie on three different rows.

(i) Two vertices in F lie on same column, let j = b. If j = b = 1 and F₁ = {v_1,1, v_a,1} then r(v_2,1|F₁) = r(v_2,2|F₁), a contradiction.

(ii) If F₁ = {v_i,1, v_j,1}, 1 ≤ i < j < k then either r(v_j,3|F₁) = r(v_j+1,3|F₁) or r(v_j,3|F₁) = r(v_j+1,2|F₁), a contradiction.

(iii) If F₁ = {v_i,b, v_j,b}, 1 ≤ i < j < k, and 1 < q < l then either r(v_j+1,b|F₁) = r(v_j+1,b+1|F₁) or r(v_j+1,b|F₁) = r(v_j+1,b−1|F₁), a contradiction.

(iv) If all vertices in S lie on different columns. Let S₁ = {V_i,j, V_a,b}, i < a then

(a) If j < b then either r(v_a,b+1|F₁) = r(v_{a + 1, b+1}|F₁) or r(v_a,b+1|F₁) = r(v_{a + 1, b}|F₁), a contradiction.

(b) If j > b then either r(v_a−1,b|F₁) = r(v_a,b+1|F₁) or r(v_a−1,b+1|F₁) = r(v_a,b+1|F₁), a contradiction.

From the above discussion, we conclude that F is not an FTRS for T_m,n. So β′(Tm,n)>3 in this case. Hence β′(Tk,l)=4.

Case II: k > l

In order to prove that there is no FTRS for T_k,l having cardinality 3, it is enough to prove that any set of vertices having two elements is not an R.S. for T_k,l. Let F₁ = {v_i,j, v_a,b} be an R.S. for T_k,l. There are three possibilities

Possibility 1: If v_i,j, v_a,b are taken from same row, i.e., i = a then

(i) If F₁ = {v_1,1, v_1,l} then r(vl+1,l2|F1)=r(vl+1,l+22|F1) if l is even and r(vl+1,l+12|F1)=r(vl+1,l+32|F1) if l is odd, a contradiction.

(ii) If F₁ = {v_i,j, v_i,b}, 1 ≤ i < k, and 1 ≤ j < b < l then r(v_i,b+1|F₁) = r(v_{i + 1, b+1}|F₁), a contradiction.

(iii) If F₁ = {v_i,j, v_i,b}, 1 ≤ i < k, and 1 < j < b = l then r(v_i,j−1|F₁) = r(v_{i + 1, j}|F₁), a contradiction.

(iv) If F₁ = {v_k,j, v_k,b} and 1 ≤ j < b < l then r(v_k,b+1|F₁) = r(v_k−1,b+1|F₁) or r(v_k,b+1|F₁) = r(v_k−1,b|F₁), a contradiction.

(v) If F₁ = {v_k,j, v_k,b} and 1 < j < b = l then r(v_k,j−1|F₁) = r(v_k−1,j|F₁), a contradiction.

(vi) If F₁ = {v_i,1, v_i,l} and 1 < i < k then r(v_i−1,1|F₁) = r(v_i+1,1|F₁), a contradiction.

(vii) If F₁ = {v_k,1, v_k,l} then r(v_{k − l, 3}|F₁) = r(v_{k − l, 4}|F₁), a contradiction.

Possibility 2: If v_i,j, v_a,b are taken from same column, i.e., j = b then

(i) If F₁ = {v_1,1, v_k,1} then r(v_2,1|F₁) = r(v_2,2|F₁), a contradiction.

(ii) If F₁ = {v_i,1, v_j,1} where 1 ≤ i < j < m then r(v_j,3|F₁) = r(v_j+1,3|F₁) or r(v_j,3|F₁) = r(v_j+1,2|F₁), a contradiction.

(iii) If F₁ = {v_i,1, v_j,1} and 1 < i < j = k then r(v_i,3|F₁) = r(v_i−1,2|F₁) or r(v_i,3|F₁) = r(v_i−1,3|F₁), a contradiction.

(iv) If F₁ = {v_i,b, v_j,b}, 1 ≤ i < j < k, and 1 < b < l then r(v_j+1,b|F₁) = r(v_j+1,b+1|F₁) or r(V_j+1,b|S₁) = r(V_j+1,b−1|S₁), a contradiction.

(v) If F₁ = {v_i,b, v_j,b}, 1 < i < j ≤ k, and 1 < b < l then r(v_i−1,j−1|F₁) = r(v_i−1,j|F₁) or r(v_i−1,j|F₁) = r(v_i−1,j+1|F₁), a contradiction.

(vi) If F₁ = {v_i,b, v_j,b}, i = 1, j = k, and 1 < q < l then r(v_2,j|F₁) = r(v_2,j+1|F₁), a contradiction.

(vii) If F₁ = {v_i,l−1, v_j,l−1} and i < j < k then r(v_j,l−2|F₁) = r(v_{j+1,l − 2}|F₁) or r(v_j,l−3|F₁) = r(v_{j+1,l − 2}|F₁), a contradiction.

(viii) If F₁ = {v_1,l, v_k,l} then r(v_k−1,l−1|F₁) = r(v_k−1,l|F₁) or r(v_k−2,l−1|F₁) = r(v_k−2,l|F₁), a contradiction.

Possibility 3: If v_i,j, v_a,b are taken from different rows and different columns, i.e., i ≠ p,j ≠ q. Let F₁ = {v_i,j, v_a,b} and i < a

(i) If j < b then r(v_a,b+1|F₁) = r(v_{a + 1, b+1}|F₁) or r(v_a,b+1|F₁) = r(v_{a + 1, b}|F₁), a contradiction.

(ii) If j > b then r(v_a−1,b|F₁) = r(v_a,b+1|F₁) or r(v_a−1,b+1|F₁) = r(v_a,b+1|F₁), a contradiction.

(iii) If a = k and i = 1 then r(v_2,j|F₁) = r(v_2,j+1|F₁) or r(v_1,j+1|F₁) = r(v_2,j+1|F₁) or r(v_k−1,b+1|F₁) = r(v_k,b−1|F₁), a contradiction.

Thus no set having two vertices is an R.S. for T_k,l. So β′(Tk,l)>3 in this case. Hence β′(Tk,l)=4.

The next theorem gives β′(αk,l).

Theorem 2.3. Let α_k,l denote the graph of k × l 2D-lattice of α boron nano tubes. Then β′(αk,l)=4

Proof. Consider k × l 2D-lattice of α-boron Nanotubes. We denote this graph by α_k,l. In general, the tube is “cut” vertically and the upper left triangle is pointing up as shown in Figure 4. This figure presents labeling of vertices that becomes optimal for the basis of fault tolerance. The vertex set of α_k,l is partitioned as {v_1,1, v_1,2, v_1,3, ...., v_1,l, v_2,1, v_2,2, ...., v_2,l, v_4,1, v_4,2, ......, v_4,l, v_5,1, v_5,2, ......, v_5,l, v_7,1, v_7,2, ......, v_7,l, ......, v_k,1, v_k,2, v_k,3, ...., v_k,l} ∪ {v_3,1, v_3,2, v_3,4, v_3,5, v_3,7, v_3,8, ...., v_3,l, v_6,1, v_6,3, v_6,4, v_6,6, v_6,7, v_6,9, ...., v_6,l, v_9,1, v_9,2, v₉₄,, v_9,5, v_9,7, v_9,8, ....,_9,l, ......}.

Case I: When k < l.

Let F = {v_1,1, v_1,l, v_k,1, v_k,l}. We show that F is an FTRS for α_k,l. We give the distance vectors of all vertices v_a,b of α_k,l relative to F.

Let k be odd and k ≠ 6q + 1, q ∈ Z⁺.

For a = 1

For a = 2

In general, if a is odd, a ≠ 3p, p ∈ Z⁺, and 3≤a≤⌈k2⌉ then