Optimal Multi-Level Fault-Tolerant Resolving Sets of Circulant Graph C(n : 1, 2)

Abstract

Let $G=\left(V\right(G),E(G\left)\right)$ be a simple connected unweighted graph. A set $R\subset V\left(G\right)$ is called a fault-tolerant resolving set with the tolerance level k if the cardinality of the set $S_{x,y}=\{w\in R:d(w,x)\ne d(w,y)\}$ is at least k for every pair of distinct vertices $x,y$ of G. A k-level metric dimension refers to the minimum size of a fault-tolerant resolving set with the tolerance level k. In this article, we calculate and determine the k-level metric dimension for the circulant graph $C(n:1,2)$ for all possible values of k and $n.$ The optimal fault-tolerant resolving sets with k tolerance are also delineated.

1. Introduction

Every network can be represented by a graph. The identification of every unique vertex (node) is of great importance so as to maintain the security of the network. Now, the question that needs to be posed is “What should be an identifying method for a given graph?” In a connected graph, distances play a vital role in the identification of unique vertices. Let $G=\left(V\right(G),E(G\left)\right)$ be a simple connected unweighted and undirected graph with the vertex set $V\left(G\right)$ and edge set $E\left(G\right)$. Two vertices u and v are called adjacent if there is an edge between u and v. An edge e is called adjacent to a vertex v if e has v as one end. The degree of a vertex $v\in V\left(G\right)$ is the number of edges adjacent to v. The distance between two vertices u and v, denoted by $d_G(u,v)$ (or simply $d(u,v)$), is the length of the shortest path between them. Let $S=\{s_1,s_2,\dots ,s_m\}$ be a subset (ordered) of the vertices of G. The code of a vertex v with respect to S, denoted by ${\mathrm{code}}_S\left(v\right)$, is the m-tuple

$$cod{e}_S\left(v\right)=\left(d(s_1,v),d(s_2,v),\dots ,d(s_m,v)\right).$$

The set $S\subset V\left(G\right)$ is called a resolving set if $cod{e}_S\left(v\right)\ne cod{e}_S\left(w\right)$ for every pair of distinct vertices v and w. A resolving set S is said to be a fault-tolerant resolving set with the tolerance level k if $S\setminus X$ is also a resolving set of G, where X is any subset of S consisting of k elements. The minimum size of a fault-tolerant resolving set with the tolerance level k is called a k-level metric dimension. We denote the k-level metric dimension for a graph G by ${\beta }_k\left(G\right).$ A fault-tolerant resolving set with tolerance k consisting of ${\beta }_k\left(G\right)$ elements is called the k-level metric basis. In the literature, the k-level metric dimension is known as the metric dimension and fault-tolerant metric dimension for $k=0$ and $k=1$, respectively. The concept of metric dimensions was first put forward by Slater [1] and Harary and Melter [2]. The problem of finding the metric dimension for a general graph is NP-hard. Khuller et al. [3] presented a construction that proves that the metric dimension of a graph is NP-hard. For a general k, the k-level metric dimension problem was introduced by Estrada-Moreno et al. [4], and they determined the exact value of the k-metric dimension for paths and cycle graphs. Yero et al. [5] showed that the problem of computing the k-level metric dimension of graphs is NP-hard. However, they solved the problem in linear time for the particular case of trees. Recently, Hakanen et al. [6] obtained some results on this topic.

Although the applications of metric bases arise in many various platforms, such as Robot Navigation [3], Network Optimization [7], and sensor networks [8], it has some limitations because, if some detectors (elements of the metric basis) are faulty, then it is not possible to identify the unique nodes. In order to improve the accuracy of the detection or the robustness of the system, Estrada et al. in [4] took the initiative to introduce a generalized metric basis, namely, the k-level metric basis. In the k-level metric basis, one can tolerate up to k detector faults.

Hernendo et al. [9] characterized all fault-tolerant resolving sets for any tree T. In the same article, they show the relation ${\beta }_1\left(G\right)\le {\beta }_0\left(G\right)\left(1+2·5^{{\beta }_0\left(G\right)-1}\right)$ for any graph G. For a positive integer n and any list $\mathcal{L}$ chosen from $\left\{1,2,\dots ,⌊\frac{n}{2}⌋\right\}$, a circulant graph, denoted by $C(n:\mathcal{L})$, is a graph on the vertices $v_0,v_1,\dots ,v_{n-1}$ such that each $v_i$, $0\le i\le n-1$, is adjacent to $v_{i+j}$ and $v_{i-j}$ (subscripts are taken modulo n) for every j in the list $\mathcal{L}$. The circulant graph $C\left(n:1,2,\dots ,⌊\frac{n}{2}⌋\right)$ is the complete graph $K_n$, and the graph $C(n:1)$ is the cycle $C_n$. For the cycle $C_n$, the fault-tolerant metric dimension was determined by Javaid et al. in [10] as ${\beta }_1\left(C_n\right)=3$. Basak at el. [11] determined the fault-tolerant metric dimension ($k=1)$ of a special type of circulant graph $C(n:\mathcal{L})$ when $\mathcal{L}=\{1,2,3\}$, and Saha et al. [12] determined the same for when $\mathcal{L}=\{1,2,3,4\}$. The exact value of ${\beta }_0\left(C(n:1,2)\right)$ (i.e., when $\mathcal{L}=\{1,2\}$) was determined by Javaid et al. [10]. In [13], the authors determined the metric dimension of the power of finite paths. The increased connectivity of circulant networks makes them ideal for parallel computing and signal processing [14]. The applications of circulant graphs in graph theory have appeared in coding theory, VLSI design, Ramsey theory, and many other areas [15]. Despite the widespread application of circulant networks, there are no results on fault-tolerant metrics with tolerance k. In this article, we determine the k-level metric dimension for the circulant graph $C(n:1,2)$ for all possible values of k and $n.$ We also delineate the optimal fault-tolerant resolving sets with the tolerance level k. In

Table 1. Existing results and our results on ${\beta }_k\left(C(n:\mathcal{L})\right)$ for different values of k and $\mathcal{L}$.

Author(s)	Values of k	The list $\mathcal{L}$	${\mathit{\beta }}_{\mathit{k}}\left(\mathit{C}(\mathit{n}:\mathcal{L})\right)$
Harary et al. in [2]	0	$\left\{1\right\}$	2
Javaid et al. in [10]	1	$\left\{1\right\}$	3
Borchert et al. in [16]	0	$\{1,2\}$	4 if $n\equiv 1(mod4)$, and 3 otherwise.
Borchert et al. in [16]	0	$\{1,2,3\}$	5 if $n\equiv 1(mod6)$, and 4 otherwise.
Basak et al. in [11]	1	$\{1,2,3\}$	7 if $n\equiv 1(mod6)$, and 6 otherwise.
Grigorious et al. in [17]	0	$\{1,2,3,4\}$	6 if $n\equiv 1,0(mod8)$, 5 if $n\equiv 2,3(mod8)$, and 4 otherwise.
Saha et al. in [12]	1	$\{1,2,3,4\}$	9 if $n\equiv 1(mod8)$, and 8 otherwise.
Current article	All possible values of k	$\{1,2\}$	$2k+3$ if $n\equiv 1(mod4)$, and $2(k+1)$ otherwise.

, we summarize the existing results and our results on ${\beta }_k\left(C(n:\mathcal{L})\right)$ for different values of k and $\mathcal{L}$.

2. Preliminaries

Henceforth, we denote the vertex set $V\left(C_n\right)$ by $\{v_0,v_1,\dots ,v_{n-1}\}$. As we are aware, $C(n:1,2)$ is a graph with the vertex set $V\left(C_n\right)$, and two vertices $v_i,v_j$ are adjacent in $C(n:1,2)$ if and only if $d_{C_n}(v_i,v_j)\le 2$, where $d_{C_n}(v_i,v_j)$ denotes the distance between $v_i$ and $v_j$ in the cycle graph $C_n$. Thus, the size of the maximum clique in $C(n:1,2)$ is 3. Additionally, the diameter of $C(n:1,2)$ is $⌈\frac{n-1}{4}⌉$. More explicitly, the diameter of $C(n:1,2)$ is $s+1$ if n is of the form $4s+r$, where $r\in \{2,3,4,5\}$. Every vertex $v_i$ has exactly $r-1$ antipodal vertices (the vertices that are at diametral distances). In particular, each vertex in $C(4s+2:1,2)$ has a unique antipodal vertex. We let $S_{v_i,v_j}\left(G\right)$ be the set of all vertices of G that resolve $v_i$ and $v_j$. Clearly, both $v_i,v_j$ are in $S_{v_i,v_j}$. Throughout the article, we use $S_{i,j}$ instead of $S_{v_i,v_j}$. We let $\mathcal{G}\left(k\right)$ be the set of all graphs in which there exists at least one fault-tolerant resolving set with the tolerance level k. Note that $\mathcal{G}(k+1)\subset \mathcal{G}\left(k\right)$. We let $k^{*}\left(G\right)$ (or simply $k^{*}$) be the maximum value of k such that G has a fault-tolerant resolving set with the tolerance level k. We call the vertex $v_i$ of $C(n:1,2)$ an odd vertex if the index i is an odd number; otherwise, we call it an even vertex. Two vertices $v_i$ and $v_j$ are said to have the same parity if $i+j$ is even, i.e., if both i and j are either even or odd. Throughout the article, the indices of the vertices of $C(n:1,2)$ are taken to be modulo n.

The workflow of the rest of the paper is presented below. To find the exact value of ${\beta }_k\left(C(n:1,2)\right)$, we first present a lower bound for this in Section 3. To find a lower bound for ${\beta }_k\left(C(n:1,2)\right)$, first, we identify the vertices that resolve the two consecutive pairs $\left(v_{\ell },v_{\ell +1}\right)$ and $\left(v_{\ell +1},v_{\ell +2}\right)$ (Lemma 5). Using this lemma, we show that any fault-tolerant resolving set with the tolerance level k must contain at least $2(k+1)$ or $2k+3$ elements accordingly as $n\not\equiv 1(mod4)$ or $n\equiv 1(mod4)$. In Section 4, we explore an optimal fault-tolerant resolving set with the tolerance level k. To construct an optimal fault-tolerant resolving set with the tolerance level k, we first find the number of vertices of a set S (of consecutive vertices) that may resolve the set $U=\{v_a,v_{a+1}\}$ in Lemmas 8 and 9. We prove the results (Lemma 8 and Lemma 9) by considering two complementary cases such that S contains at most one element of U or U is a proper subset of S. Then, we construct a set F that is the union of two sets $F_1$ and $F_2$ of consecutive vertices. The set $F_1$ is chosen from the first $k+1$ consecutive vertices, whereas the set $F_2$ is chosen from $k+1$ consecutive vertices starting from the middle vertex or from a vertex adjacent to the middle vertex/vertices of $C(n:1,2)$.

3. Lower Bound for ${\mathit{\beta }}_{\mathit{k}}\left(\mathit{C}(\mathit{n}:\mathit{1},\mathit{2})\right)$

Let $S_{i,j}^c$ be the complement of the set $S_{i,j}$ for two non-consecutive vertices $v_i$ and $v_j$ in $V\left(G\right)$. Then, $S_{i,j}^c=V\left(G\right)\setminus S_{i,j}$, that is,

$$S_{i,j}^c=\left\{v_k\in V\left(G\right):d_G(v_k,v_i)=d_G(v_k,v_j)\right\}.$$

We obtain the formula for $S_{i,j}^c$ in the following lemma.

Lemma 1.

Let $v_i,v_j$ ($i<j$) be two vertices such that $d_{C_n}(v_i,v_j)\ge 2$. Then, the set $S_{i,j}^c$ is the following:

$\left(a\right)$

When $n\equiv 0(mod4)$, $S_{i,j}^c=\left\{\begin{array}{cc}\left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathit{if}j-i\equiv 1(mod4),\hfill \\ \left\{v_{\frac{i+j}{2}},v_{\frac{i+j+n}{2}}\right\}\hfill & \mathit{if}j-i\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j-1}{2}},v_{\frac{i+j+1}{2}}\right\}\hfill & \mathit{if}j-i\equiv 3(mod4).\hfill \end{array}\right.$

$\left(b\right)$

When $n\equiv 1(mod4)$, $S_{i,j}^c=\left\{\begin{array}{cc}\left\{v_{\frac{i+j}{2}}\right\}\hfill & \mathit{if}j-i\equiv 0(mod4),\hfill \\ \left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathit{if}j-i\equiv 1(mod4),\hfill \\ \left\{v_{\frac{i+j}{2}},v_{\frac{i+j+n-1}{2}}{v}_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathit{if}j-i\equiv 2(mod4),\hfill \\ \left\{v_{\frac{i+j-1}{2}},v_{\frac{i+j+1}{2}},v_{\frac{i+j+n}{2}}\right\}\hfill & \mathit{if}j-i\equiv 3(mod4).\hfill \end{array}\right.$

$\left(c\right)$

When $n\equiv 2(mod4)$, $S_{i,j}^c=\left\{\begin{array}{cc}\varnothing \hfill & \mathit{if}j-i\equiv 1(mod4),\hfill \\ \left\{v_{\frac{i+j}{2}},v_{\frac{i+j+n}{2}}\right\}\hfill & \mathit{if}j-i\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j-1}{2}},v_{\frac{i+j+1}{2}},v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathit{if}j-i\equiv 3(mod4).\hfill \end{array}\right.$

$\left(d\right)$

When $n\equiv 3(mod4)$, $S_{i,j}^c=\left\{\begin{array}{cc}\left\{v_{\frac{i+j}{2}},v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathit{if}j-i\equiv 0(mod4),\hfill \\ \left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathit{if}j-i\equiv 1(mod4),\hfill \\ \left\{v_{\frac{i+j}{2}}\right\}\hfill & \mathit{if}j-i\equiv 2(mod4),\hfill \\ \left\{v_{\frac{i+j-1}{2}},v_{\frac{i+j+1}{2}},v_{\frac{i+j+n}{2}}\right\},\hfill & \mathit{if}j-i\equiv 3(mod4).\hfill \end{array}\right.$

Proof. 

Let $v_{\ell }\in S_{i,j}^c$ be an internal vertex between $v_i$ and $v_j$ $(i\le \ell \le j)$ in G. We consider the following three cases:

$\mathbf{Case}\mathbf{1}.$$\ell <⌊\frac{\mathbf{i}+\mathbf{j}}{\mathbf{2}}⌋$. Then, $\ell -i<j-\ell$, which implies that $v_j$ cannot lie in the shortest path of $v_i$ and $v_{\ell }$; i.e., we have $d_{C_n}(v_{\ell },v_i)=\ell -i$. Note that $d_{C_n}(v_{\ell },v_j)=min\{j-\ell ,n-(j-\ell )\}$. Now, if the shortest path between $v_{\ell }$ and $v_j$ is via $v_i$, that is, $d_{C_n}(v_{\ell },v_j)=n-(j-\ell )$, then $d_{C_n}(v_{\ell },v_j)\ge d_{C_n}(v_{\ell },v_i)+2$, and hence, $d_G(v_{\ell },v_j)>d_G(v_{\ell },v_i)$. Again, if $d_{C_n}(v_{\ell },v_j)=j-\ell$, then $\ell -i<j-\ell$ gives $d_{C_n}(v_{\ell },v_j)<d_{C_n}(v_{\ell },v_i)$. Thus, in any case, $d_G(v_{\ell },v_i)<d_G(v_{\ell },v_j)$, so $v_{\ell }$ resolves $v_i$ and $v_j$ when $\ell <⌊\frac{i+j}{2}⌋$. Therefore, $v_{\ell }\in S_{i,j}$, a contradiction to $v_{\ell }\in S_{i,j}^c$.

$\mathbf{Case}\mathbf{2}.$$\ell >⌈\frac{\mathbf{i}+\mathbf{j}}{\mathbf{2}}⌉$. Then, $\ell -i>j-\ell$. Since $d_{C_n}(v_i,v_j)\ge 2$, the shortest path is not via $v_i$, and hence, we have $d_{C_n}(v_{\ell },v_j)=j-\ell$. Additionally, $d_{C_n}(v_{\ell },v_i)=min\{\ell -i,n-(\ell -i)\}$. By the aforementioned argument, $d_{C_n}(v_i,v_{\ell })\ge d_{C_n}(v_{\ell },v_j)+2$, and hence, $d_G(v_i,v_{\ell })>d_G(v_{\ell },v_j)$. So, $v_{\ell }$ resolves $v_i$ and $v_j$. Again, $v_{\ell }\in S_{i,j}$, a contradiction to $v_{\ell }\in S_{i,j}^c$.

$\mathbf{Case}\mathbf{3}.$$⌊\frac{\mathbf{i}+\mathbf{j}}{\mathbf{2}}⌋\le \ell \le ⌈\frac{\mathbf{i}+\mathbf{j}}{\mathbf{2}}⌉$. Let $W=\left\{⌊\frac{i+j}{2}⌋,⌈\frac{i+j}{2}⌉\right\}$. Then, $W=\left\{\frac{i+j}{2}\right\}$ or $\left\{\frac{i+j-1}{2},\frac{i+j+1}{2}\right\}$ according to whether $i+j$ is even or odd. Now, we check whether or not the vertex $v_{\ell }\in W$ resolves $v_i$ and $v_j$. For an even integer $i+j$, $v_{\ell }\in W$, $d_G(v_{\ell },v_i)=d_G(v_j,v_{\ell })$. For an odd integer $i+j$, we calculate the distances of $v_i$ and $v_j$ from each element of W in

Table 2. Distances of $v_i$ and $v_j$ from each element of W.

Values of ℓ	${\mathit{d}}_{{\mathit{C}}_{\mathit{n}}}({\mathit{v}}_{\mathit{i}},{\mathit{v}}_{\mathit{\ell }})$	${\mathit{d}}_{{\mathit{C}}_{\mathit{n}}}({\mathit{v}}_{\mathit{j}},{\mathit{v}}_{\mathit{\ell }})$	${\mathit{d}}_{\mathit{G}}({\mathit{v}}_{\mathit{i}},{\mathit{v}}_{\mathit{\ell }})$	${\mathit{d}}_{\mathit{G}}({\mathit{v}}_{\mathit{j}},{\mathit{v}}_{\mathit{\ell }})$
$\frac{i+j-1}{2}$	$\frac{j-i-1}{2}$	$\frac{j-i+1}{2}$	$⌈\frac{j-i-1}{4}⌉=m$ or $m+1$ accordingly $j-i=4m+1$ or $j-i=4m+3$	$⌈\frac{j-i+1}{4}⌉=m+1$ if $j-i=4m+r$, $r\in \{1,3\}$
$\frac{i+j+1}{2}$	$\frac{j-i+1}{2}$	$\frac{j-i-1}{2}$	$⌈\frac{j-i+1}{4}⌉=m+1$ if $j-i=4m+r$, $r\in \{1,3\}$	$⌈\frac{j-i-1}{4}⌉=m$ or $m+1$ accordingly $j-i=4m+1$ or $j-i=4m+3$.

.

Thus, the set S of all intermediate vertices between $v_j$ and $v_j$ with $i<j$ that do not resolve $v_i$ and $v_j$ is given by

$$\begin{array}{c}\hfill S=\left\{\begin{array}{cc}\left\{v_{\frac{i+j}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j-1}{2}},v_{\frac{i+j+1}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 3(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv 1(mod4).\hfill \end{array}\right.\end{array}$$

(1)

Again, if $v_{\ell }$ is a vertex such that $\ell \notin \{i,i+1,\dots ,j\}$, then $v_{\ell }$ will be an intermediate vertex between $v_{i^{\prime }}$ and $v_{j^{\prime }}$, where $i^{\prime }=j$ and $j^{\prime }=i+n$. Then, applying (1) to $v_{i^{\prime }}$ and $v_{j^{\prime }}$, we obtain the set $S^{\prime }$ of all vertices that lie in $\{v_j,v_{j+1},\dots ,v_{i+n}\}$ and do not resolve the vertices $v_i$ and $v_j$, where $S^{\prime }$ is given by

$$\begin{array}{ccc}\hfill S^{\prime }& =& \left\{\begin{array}{cc}\left\{v_{\frac{i^{\prime }+j^{\prime }}{2}}\right\}\hfill & \mathrm{if}{j}^{\prime }-i^{\prime }\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i^{\prime }+j^{\prime }-1}{2}},v_{\frac{i^{\prime }+j^{\prime }+1}{2}}\right\}\hfill & \mathrm{if}{j}^{\prime }-i^{\prime }\equiv 3(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}{j}^{\prime }-i^{\prime }\equiv 1(mod4).\hfill \end{array}\right.\hfill \end{array}$$

(2)

Putting $i^{\prime }=j$ and $j^{\prime }=i+n$ in (2), we obtain

$$\begin{array}{ccc}\hfill S^{\prime }& =& \left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}i+n-j\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathrm{if}i+n-j\equiv 3(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}i+n-j\equiv 1(mod4).\hfill \end{array}\right.\hfill \\ & =& \left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}j-i\equiv r,r-2(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathrm{if}j-i\equiv r-3(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv r-1(mod4),\hfill \end{array}\right.\hfill \end{array}$$

(3)

where $n\equiv r(mod4)$. We denote $S^{\prime }$ by $S_r$ for $n\equiv r(mod4)$, $r\in \{0,1,2,3\}$. Then, from Equation (3), we have the following:

$$S_0=\left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 1(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv 3(mod4),\hfill \end{array}\right.$$

$$S_1=\left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 1,3(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}}{v}_{\frac{i+j+n+1}{2}}\right\},\hfill & \mathrm{if}j-i\equiv 2(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv 0(mod4),\hfill \end{array}\right.$$

$$S_2=\left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 0,2(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 3(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv 1(mod4),\hfill \end{array}\right.$$

$$S_3=\left\{\begin{array}{cc}\left\{v_{\frac{i+j+n}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 1,3(mod4),\hfill \\ \left\{v_{\frac{i+j+n-1}{2}},v_{\frac{i+j+n+1}{2}}\right\}\hfill & \mathrm{if}j-i\equiv 0(mod4),\hfill \\ \varnothing \hfill & \mathrm{if}j-i\equiv 2(mod4).\hfill \end{array}\right.$$

We obtain the result by taking $S\cup S_r$ for $r\in \{0,1,2,3\}$. □

Corollary 1.

For any two vertices $v_i$ and $v_j$ of $C(n:1,2)$, the following hold:

(a)

$|S_{i,j}|\ge n-4$, provided that $v_i$ and $v_j$ are non-consecutive vertices.

(b)

If $v_i$ and $v_j$ are two consecutive vertices, then the minimum value of $|S_{i,j}|$ is $⌊\frac{n}{2}⌋$.

Lemma 2.

Let $H\in \mathcal{G}\left(k\right)$ be a simple connected graph. A set $F\subset V\left(H\right)$ is a fault-tolerant resolving set with the tolerance level k if and only if $|F\cap S_{i,j}|\ge k+1$ for every pair of vertices $v_i$ and $v_j$ of H.

Proof. 

Let $v_i$ and $v_j$ be two arbitrary vertices of H. First, we assume that $F=\{w_1,w_2,\dots ,w_m\}$ (ordered set) is a fault-tolerant resolving set with the tolerance level k. Then, ${\mathrm{code}}_F\left(v_i\right)$ and ${\mathrm{code}}_F\left(v_j\right)$ differ by at least $k+1$ positions. Since ${\mathrm{code}}_F\left(v_i\right)=\left(d_H(v_i,w_{\ell }):1\le \ell \le m\right)$ and ${\mathrm{code}}_F\left(v_j\right)=\left(d_H(v_j,w_{\ell }):1\le \ell \le m\right)$, the set $W=\{w_{\ell }\in F:d_H(w_{\ell },v_i)\ne d_H(w_{\ell },v_j)\}$ consists of at least $k+1$ elements, which implies that $|F\cap S_{i,j}|\ge k+1$ for every pair of vertices $v_i$ and $v_j$ of H.

Conversely, we assume that $|F\cap S_{i,j}|\ge k+1$ for every pair of vertices $v_i$ and $v_j$ of H. Then, $F\cap S_{i,j}=\{w\in F:d_H(w,v_i)\ne d_H(w,v_j)\}$ consists of at least $k+1$ elements, and hence, $\left(d_H(v_i,w):w\in F\right)$ and $\left(d_H(v_j,w):w\in F\right)$ differ by at least $k+1$ positions. Since $v_i$ and $v_j$ are two arbitrary vertices of H, $F\subset V\left(H\right)$ is a fault-tolerant resolving set with the tolerance level k. □

Lemma 3.

For the circulant graph $C(n:1,2)$, $k^{*}\le ⌊\frac{n}{2}⌋-1$.

Proof. 

Let F be a fault-tolerant resolving set with the tolerance level $k^{*}$. Then, by Lemma 2, every pair of vertices must be resolved by at least $k^{*}+1$ elements of F. Thus, from Corollary 1, $k^{*}+1\le ⌊\frac{n}{2}⌋$. □

Lemma 4.

For a positive integer n, exactly one of the following is true:

(a)

$⌈\frac{n}{2}⌉=⌈\frac{n+1}{2}⌉\ne ⌈\frac{n+2}{2}⌉$

(b)

$⌈\frac{n}{2}⌉\ne ⌈\frac{n+1}{2}⌉=⌈\frac{n+2}{2}⌉$.

Proof. 

First, we assume that n is odd. Then, $n=2m+1$ (say). Thus, $⌈\frac{n}{2}⌉=m+1=⌈\frac{n+1}{2}⌉$ and $⌈\frac{n+2}{2}⌉=m+2$. Thus, only $\left(a\right)$ is true when n is an odd integer. Next, we assume that n is even. Then, $n=2m$ (say). Thus, $⌈\frac{n}{2}⌉=m$ and $⌈\frac{n+1}{2}⌉=m+1=⌈\frac{n+2}{2}⌉$. Thus, only $\left(b\right)$ is true when n is an even integer. □

Lemma 5.

For a circulant graph $C(n:1,2)$,

$$S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}=\left\{\begin{array}{cc}\left\{v_{\ell +1}\right\}\hfill & \mathit{if}n\not\equiv 2(mod4),\hfill \\ \{v_{\ell +1},v_{\frac{n}{2}+\ell +1}\}\hfill & \mathit{if}n\equiv 2(mod4).\hfill \end{array}\right.$$

Proof. 

Clearly, $v_{\ell +1}\in S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}$ and $v_{\ell },v_{\ell +2}\notin S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}$. We may write $n=4s+r$ for some positive integer s and $r\in \{2,3,4,5\}$. Let $v_j$ be a vertex in $C(n:1,2)$. For $j\in \left\{\ell +3,\dots ,2s+\ell +⌊\frac{r}{2}⌋\right\}$, we have $d_{C_n}(v_j,v_{\ell +1})=d_{C_n}(v_j,v_{\ell +2})+1$ and $d_{C_n}(v_j,v_{\ell })=d_{C_n}(v_j,v_{\ell +2})+2$. For $j\in \{0,\dots ,\ell -1\}\cup \left\{2s+\ell +⌊\frac{r}{2}⌋+3,\dots ,n-1\right\}$, we obtain $d_{C_n}(v_j,v_{\ell +1})=d_{C_n}(v_j,v_{\ell })+1$ and $d_{C_n}(v_j,v_{\ell +2})=d_{C_n}(v_j,v_{\ell })+2$. Therefore, for $j\in \left\{\ell +3,\dots ,2s+\ell +⌊\frac{r}{2}⌋\right\}$ or $j\in \{0,\dots ,\ell -1\}\cup \left\{2s+\ell +⌊\frac{r}{2}⌋+3,\dots ,n-1\right\}$, the distances of $v_{\ell }$, $v_{\ell +1}$, and $v_{\ell +2}$ from $v_j$ are of the form $⌈\frac{a}{2}⌉$, $⌈\frac{a+1}{2}⌉$, and $⌈\frac{a+2}{2}⌉$ or in reverse order, respectively, where a is a positive integer. Applying Lemma 4, it is clear that $v_j\notin S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}$, provided that $j\in \{0,\dots ,\ell -1\}\cup \left\{\ell +3,\dots ,2s+\ell +⌊\frac{r}{2}⌋\right\}\cup \left\{2s+\ell +⌊\frac{r}{2}⌋+3,\dots ,n-1\right\}$. Now, in Table 1, we calculate the distances of $v_{\ell }$, $v_{\ell +1}$, and $v_{\ell +2}$ from $v_j$, where $j\in \left\{2s+\ell +⌊\frac{r}{2}⌋+1,2s+\ell +⌊\frac{r}{2}⌋+2\right\}$ (

Table 3. The distances of $v_{\ell }$, $v_{\ell +1}$, and $v_{\ell +2}$ from $v_j$, where $j\in \left\{2s+\ell +⌊\frac{r}{2}⌋+1,2s+\ell +⌊\frac{r}{2}⌋+2\right\}$.

Values of r	Values of n	Values of j	${\mathit{d}}_{\mathit{G}}({\mathit{v}}_{\mathit{j}},{\mathit{v}}_{\mathit{\ell }})$	${\mathit{d}}_{\mathit{G}}({\mathit{v}}_{\mathit{j}},{\mathit{v}}_{\mathit{\ell }+1})$	${\mathit{d}}_{\mathit{G}}({\mathit{v}}_{\mathit{j}},{\mathit{v}}_{\mathit{\ell }+2})$
$r=2$	$n=4s+2$	$2s+\ell +2$	s	$s+1$	s
		$2s+\ell +3$	s	s	$s+1$
$r=3$	$n=4s+3$	$2s+\ell +2$	$s+1$	$s+1$	s
		$2s+\ell +3$	s	$s+1$	$s+1$
$r=4$	$n=4s+4$	$2s+\ell +3$	$s+1$	$s+1$	$s+1$
		$2s+\ell +4$	s	$s+1$	$s+1$
$r=5$	$n=4s+5$	$2s+\ell +3$	$s+1$	$s+1$	$s+1$
		$2s+\ell +4$	$s+1$	$s+1$	$s+1$

).

From Table 2, it is clear that if $j\in \left\{2s+\ell +⌊\frac{r}{2}⌋+1,2s+\ell +⌊\frac{r}{2}⌋+2\right\}$, then $v_j\in S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}$ only when $n=4s+2$ and $j=2s+\ell +⌊\frac{r}{2}⌋+1$. This completes the proof of the result. □

Example 1.

For the circulant graph $C(34:1,2)$, $n=34\equiv 2(mod4)$. Let us find $S_{\ell ,\ell +1}\cap S_{\ell +1,\ell +2}$ for $\ell =0$; i.e., we show that $S_{0,1}\cap S_{1,2}=\{v_1,v_{18}\}$. Then, each vertex $u\in \{v_1,v_{18}\}$ resolves both pairs $(v_0,v_1)$ and $(v_1,v_2)$, as the distances are given by

$$\begin{array}{cc}& \begin{array}{ccc}{v}_0& v_1& v_2\end{array}\\ \begin{array}{l}{v}_1\\ v_{18}\end{array}& \left(\begin{array}{ccc}1& 0& 1\\ 8& 9& 8\end{array}\right).\end{array}$$

Thus, $\{v_1,v_{18}\}\subseteq S_{0,1}\cap S_{1,2}$. One can easily show that any vertex $v\notin \{v_1,v_{18}\}$ cannot resolve both pairs $(v_0,v_1)$ and $(v_1,v_2)$.

Remark 1.

It is noted that, when $n\equiv 2(mod4)$, each vertex of $C(n:1,2)$ has a unique antipodal vertex. Indeed, $v_{\frac{n}{2}+\ell +1}$ is the antipodal vertex of $v_{\ell }$. Hence, $v_{\frac{n}{2}+\ell +1}$ resolves both pairs $(v_{\ell },v_{\ell +1})$ and $(v_{\ell +1},v_{\ell +2})$.

Below, we give a lower bound for the k-level fault-tolerant metric dimension for the circulant graph $C(n:1,2)$.

Theorem 1.

Every fault-tolerant resolving set with tolerance k of $C(n:1,2)$ consists of at least $2(k+1)$ elements. Moreover, if $n\equiv 1(mod4)$, then every fault-tolerant resolving set with tolerance k contains at least $2k+3$ elements.

Proof. 

Let F be an arbitrary fault-tolerant resolving set with the tolerance level k. First, we assume $n\not\equiv 2(mod4)$. If F contains every vertex of $C(n:1,2)$, then the theorem holds trivially. So, we assume that there exists a vertex $u\notin F$. Then, $u=v_i$ for some i, where $1\le i\le n-1$. Then, using Lemma 5 with $\ell =i-1$, we obtain $\left(F\cap S_{i-1,i}\right)\cap \left(F\cap S_{i,i+1}\right)=\varnothing$. Again, Lemma 2 states that each of $|F\cap S_{i-1,i}|$ and $|F\cap S_{i,i+1}|$ must be at least $k+1$, so $\left|F\right|\ge |F\cap S_{i-1,i}|+|F\cap S_{i,i+1}|\ge 2(k+1)$ (for a better understanding, refer to Example 3). Now, we assume $n\equiv 2(mod4)$. If F contains neither $v_i$ nor $v_{\frac{n}{2}+i}$ for some i, then, using a similar argument to that above and Lemma 5, one can easily prove that $\left(F\cap S_{i-1,i}\right)\cap \left(F\cap S_{i,i+1}\right)=\varnothing$ and, consequently, $\left|F\right|\ge |F\cap S_{i-1,i}|+|F\cap S_{i,i+1}|\ge 2(k+1)$, as each of the sets $F\cap S_{i-1,i}$ and $F\cap S_{i,i+1}$ consists of at least $k+1$ elements due to Lemma 2 (for a better understanding, refer to Example 2). Again, if F contains at least one element from $\left\{v_i,v_{\frac{n}{2}+i}\right\}$, then $\left|F\right|\ge ⌊\frac{n}{2}⌋$. Thus, in any case, $\left|F\right|\ge 2(k+1)$, provided that $2(k+1)\le ⌊\frac{n}{2}⌋$.

We now prove that if $n=4s+5$, then every resolving set with tolerance k contains at least $2k+3$ elements. Assume to the contrary that there is a k-tolerance resolving set F that contains exactly $2(k+1)$ elements. Our claim is:

$$\begin{array}{c}\hfill \mathrm{If}{v}_i\in F,\mathrm{then}{v}_{2s+3+i}\in F.\end{array}$$

(4)

Let $v_i\in F$. Note that $v_{2s+2+i},v_{2s+3+i}$, and $v_{2s+4+i}$ are antipodal vertices of $v_i$. So, $v_i$ cannot resolve the two pairs $(v_{2s+2+i},v_{2s+3+i})$ and $(v_{2s+3+i},v_{2s+4+i})$. If $v_{2s+3+i}\notin F$, then, by a similar argument to that above, we need at least $2(k+1)$ elements, except $v_i$, to resolve these pairs. Thus, we must have $v_{2s+3+i}\in F$ whenever $v_i\in F$. Without loss of generality, we assume that $v_0\in F$. Then, from (4), we have $v_{2s+3}\in F$. So, once $v_{2s+3}\in F$, then, again, (4) gives $v_1\in F$. Applying (4) repeatedly, we obtain $\left|F\right|>2(k+1)$, which is a contradiction. □

Example 2.

For the circulant graph $C(22:1,2)$, $n=22\equiv 2(mod4)$. Let us take $k=2$ and $i=4$. Then, $S_{i-1,i}$ and $S_{i,i+1}$, i.e., $S_{3,4}$ and $S_{4,5}$, are given by

$$\begin{array}{c}\hfill S_{3,4}=\{v_1,v_3,v_4,v_6,v_8,v_{10},v_{12},v_{14},v_{15},v_{17},v_{19},v_{21}\}\\ \hfill S_{4,5}=\{v_0,v_2,v_4,v_5,v_7,v_9,v_{11},v_{13},v_{15},v_{16},v_{18},v_{20}\}\end{array}$$

Note that $S_{3,4}\cap S_{4,5}=\{v_4,v_{15}\}=\{v_i,v_{\frac{n}{2}+i}\}$. If F contains neither $v_4$ nor $v_{15}$, then $\left(F\cap S_{3,4}\right)\cap \left(F\cap S_{4,5}\right)=\varnothing$ and, consequently, $\left|F\right|\ge |F\cap S_{3,4}|+|F\cap S_{4,5}|\ge 6$ because $\left|F\cap S_{3,4}\right|\ge 3$ and $\left|F\cap S_{4,5}\right|\ge 3$ due to Lemma 2.

Example 3.

For the circulant graph $C(23:1,2)$, $n=23\equiv 3(mod4)$. Let us take $k=3$ and $i=6$. Then, $S_{i-1,i}$ and $S_{i,i+1}$, i.e., $S_{5,6}$ and $S_{6,7}$, are given by

$$\begin{array}{c}\hfill S_{5,6}=\{v_1,v_3,v_5,v_6,v_8,v_{10},v_{12},v_{14},v_{16},v_{18},v_{20},v_{22}\}\\ \hfill S_{6,7}=\{v_0,v_2,v_4,v_6,v_7,v_9,v_{11},v_{13},v_{15},v_{17},v_{19},v_{21}\}.\end{array}$$

Note that $S_{5,6}\cap S_{6,7}=\left\{v_6\right\}$. So, $\left(F\cap S_{5,6}\right)\cap \left(F\cap S_{6,7}\right)=\varnothing$, provided that $v_6\notin F$. Therefore, if $v_6\notin F$, then $\left|F\right|\ge |F\cap S_{5,6}|+|F\cap S_{6,7}|\ge 8$ because $\left|F\cap S_{5,6}\right|\ge 4$ and $\left|F\cap S_{6,7}\right|\ge 4$ due to Lemma 2.

4. Optimal Fault-Tolerant Resolving Set with Tolerance $\mathit{k}$

For two subsets $U_1$ and $U_2$ of vertices of G, let us define $d_G(U_1,U_2)=min\left\{d_G(u_1,u_2):u_1\in U_1,u_2\in U_2\right\}$. For a set U, we denote the set of all common antipodal vertices of the elements in U by $A\left(U\right)$. A vertex x resolves the vertices u and v if $d_G(x,u)\ne d_G(x,v)$. From here onward, by $U=\{v_a$, $v_{a+1}\}$, we mean that U is a set of any two consecutive vertices of $C(n:1,2)$, so a is a non-negative integer and less than n.

Lemma 6.

Let S be a set of consecutive vertices of $C(n:1,2)$, $n\not\equiv 2(mod4)$ such that $A\left(U\right)\cap S=\varnothing$, where $U=\{v_a$, $v_{a+1}\}$. If S contains at most one vertex of U, then for each $u\in U$, $d_{C_n}(u,v)\ne d_{C_n}(u,w)$ for all distinct $v,w\in S$.

Proof. 

For symmetries of $C(n:1,2)$, without loss of generality, we can assume that $a=0$, i.e., $U=\{v_0,v_1\}$. Assume to the contrary that there are $v_p,v_q\in S$ $(p<q)$ such that $d_{C_n}(v_{\ell },v_p)=d_{C_n}(v_{\ell },v_q)$ for some $v_{\ell }\in U$. Then, $p-\ell =n-(q-\ell )$, i.e., $q=n+2\ell -p$ with $p\le ⌊\frac{n}{2}⌋$. Thus, $\{v_p,v_{p+1},\dots ,v_{n+2\ell -p}\}\subset S$. Since $n\not\equiv 2(mod4)$, we may write $n=4s+r$ for some positive integer s and $r\in \{3,4,5\}$. Then, $A\left(U\right)=A\left(v_0\right)\cap A\left(v_1\right)=\{v_{2s+2},\dots ,v_{2s+r-1}\}$ and $\{v_p,v_{p+1},\dots ,v_{4s+r+\ell -p-1}\}\subset S$, where $\ell \in \{0,1\}$. Since $p\le 2s+⌊\frac{r}{2}⌋$, $A\left(U\right)\cap S\ne \varnothing$, which is a contradiction. This contradiction leads to the result. □

The following lemma can be proved using a similar argument to that in the proof of Lemma 6.

Lemma 7.

Let S be a set of consecutive vertices of $C(4s+2:1,2)$ such that $\left|S\right|\le 2s$. If S contains at most one vertex of $U=\{v_a$, $v_{a+1}\}$, then for each $u\in U$, $d_{C_n}(u,y)\ne d_{C_n}(u,z)$ for all distinct $y,z\in S$.

Lemma 8.

Let S be a set of consecutive vertices of $C(n:1,2)$, $n\not\equiv 2(mod4)$, such that $A\left(U\right)\cap S=\varnothing$, where $U=\{v_a$, $v_{a+1}\}$. Then, the elements of U are resolved by at least $⌈\frac{\left|S\right|}{2}⌉$ or at least $⌊\frac{\left|S\right|}{2}⌋$ elements of S according to whether $d_{C_n}(U,S)$ is even or odd.

Proof. 

Let $d_{C_n}(U,S)=d_{C_n}(x,y)$, where $x\in U$ and $y\in S$. Then, $d_{C_n}(U\setminus \left\{x\right\},s)=d_{C_n}(x,s)+1$. Since S is a set of consecutive vertices and $A\left(U\right)\cap S=\varnothing$, applying Lemma 7, we obtain $d_{C_n}(x,v)\ne d_{C_n}(x,w)$ for all $v,w\in S$.

$\mathbf{Case}\mathbf{1}.$$\mathit{S}$ contains at most one element of $\mathit{U}$. Since $d_G(u,w)=⌈\frac{d_{C_n}(u,w)}{2}⌉$ for all $u,w\in V\left(C\right(n:1,2\left)\right)$, $v_a$ and $v_{a+1}$, i.e., x and $U\setminus \left\{x\right\}$, are resolved by s in $C(n:1,2)$ if and only if $⌈\frac{d_{C_n}(x,s)}{2}⌉\ne ⌈\frac{d_{C_n}(U\setminus \left\{x\right\},s)}{2}⌉$. Again, since $d_{C_n}(U\setminus \left\{x\right\},s)=d_{C_n}(x,s)+1$ for all $s\in S$, the last inequality holds only when $d_{C_n}(x,s)$ is even. Since for distinct v and w of S, $d_{C_n}(x,v)\ne d_{C_n}(x,w)$, there are $⌈\frac{\left|S\right|}{2}⌉$ or $⌊\frac{\left|S\right|}{2}⌋$ elements in S when $d_{C_n}(x,s)$ is even, according to whether $d_{C_n}(x,y)$ is even or odd.

$\mathbf{Case}\mathbf{2}.$$\mathit{U}$ is a proper subset of $\mathit{S}$. If $v_a$ and $v_{a+1}$ are the first two elements or last two elements of S, then $S^{\prime }=S\setminus \{v_a,v_{a+1}\}$ (say). Therefore, $S^{\prime }$ contains no elements of U and, by a similar argument to that in $\mathbf{Case}\mathbf{1}$, the elements of U are resolved by $⌊\frac{|S^{\prime }|}{2}⌋$ elements of $S^{\prime }$ because $d_{C_n}(U,S^{\prime })=1$. Since both $v_a$ and $v_{a+1}$ resolve the elements of U, there are at least $⌊\frac{|S^{\prime }|}{2}⌋+2=⌊\frac{\left|S\right|-2}{2}⌋+2$ elements of S that resolve U. Now, we assume that both $v_a$ and $v_{a+1}$ are internal vertices of S. We divide S into two disjoint segments $S_1$ and $S_2$ that contain $v_a$ and $v_{a+1}$, respectively. Then, applying the same arguments as those in $\mathbf{Case}\mathbf{1}$ to both $S_1$ and $S_2$, we see that $v_a$ and $v_{a+1}$ are resolved by $⌈\frac{|S_1|}{2}⌉$ elements of $S_1$ and $⌈\frac{|S_2|}{2}⌉$ elements of $S_2$. Thus, $v_a$ and $v_{a+1}$ are resolved by $⌈\frac{|S_1|}{2}⌉+⌈\frac{|S_2|}{2}⌉$ elements of $S=S_1\cup S_2$. □

Lemma 9.

Let S be a set of consecutive vertices of $C_{4s+2}$ such that $\left|S\right|\le 2s$. Then, the elements of $U=\{v_a$, $v_{a+1}\}$ are resolved by at least $⌈\frac{\left|S\right|}{2}⌉$ or at least $⌊\frac{\left|S\right|}{2}⌋$ elements of S according to whether $d_{C_n}(U,S)$ is even or odd.

Proof. 

By a similar argument to that in the proof of Lemma 8, we can prove the result. □

Lemma 10.

Suppose that $F_1=\{v_0,v_1,\cdots ,v_k\}$ and $F_2=\{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+2}\}$ are two subsets of $V\left(C\right(n:1,2\left)\right)$, where $n=4s+2$ or $n=4s+4$ and $k\le \frac{n}{2}-2$. Then, any two consecutive vertices in $F_1\cup F_2$ are resolved by at least $k+1$ elements of $F_1\cup F_2$.

Proof. 

We consider the following two cases:

$\mathbf{Case}\mathbf{1}.$$\mathbf{n}=\mathbf{4}\mathbf{s}+\mathbf{4}$. For symmetries of $C(4s+4:1,2)$ and the properties of the sets $F_1$ and $F_2$, it is sufficient to prove the result for two consecutive vertices $v_i$ and $v_{i+1}$ of $F_2$. Then, $i=2s+t$ for some t with $2\le t\le k+1.$ Let $U=\{v_{2s+t},v_{2s+t+1}\}$. Then, $A\left(U\right)=A\left(v_{2s+t}\right)\cap A\left(v_{2s+t+1}\right)=\{v_{t-2},v_{t-1}\}$. We now consider the following cases.

$\mathbf{Case}\mathbf{1.1}.$$\mathbf{t}\in \{\mathbf{2},\mathbf{k}+\mathbf{1}\}$. First, we assume that $t=2$. Then, applying Lemma 8 to both $F_2\setminus \left\{v_{2s+2}\right\}$ and $F_1\setminus \{v_0,v_1\}$ and combining them, the vertices $v_i$ and $v_{i+1}$ are resolved by at least $⌈\frac{k}{2}⌉+⌊\frac{k}{2}⌋=k$ elements. Again, $v_i$ and $v_{i+1}$ are also resolved by $v_{2s+2}$. Thus, there are at least $k+1$ elements of $F_1\cup F_2$ that resolve $v_i$ and $v_{i+1}$ when $t=2$. Next, we assume that $t=k+1$. Then, applying Lemma 8 to the sets $F_2\setminus \left\{v_{2s+k+2}\right\}$ and $F_1\setminus \{v_{k-1},v_k\}$ and a applying similar argument, we obtain the result.

$\mathbf{Case}\mathbf{1.2}.$$\mathbf{3}\le \mathbf{t}\le \mathbf{k}$. In this case, $A\left(U\right)\cap S_1=\varnothing$ and $A\left(U\right)\cap S_2=\varnothing$, where $S_1=\{v_0,\dots ,v_{t-3}\}$ and $S_2=\{v_t,\dots ,v_k$}. Note that $d_{C_n}(U,S_1)=d_{C_n}(v_{2s+t+1},v_0)=2s-t+3$ and $d_{C_n}(U,S_1)=d_{C_n}(v_{2s+t},v_k)=2s+t-k$. If t is even, then applying Lemma 8 to both $S_1$ and $S_2$, we have the following:

(a)

The elements of U are resolved by at least $\frac{t-2}{2}$ elements of $S_1$;

(b)

The elements of U are resolved by at least $\frac{k+2-t}{2}$ or at least $\frac{k+1-t}{2}$ elements of $S_1$ according to whether k is even or odd.

Thus, when t is even, then the elements of U are resolved by at least $\frac{k}{2}$ elements or at least $\frac{k-1}{2}$ elements of $F_1$ according to whether k is even or odd. Again, applying the same argument as is used in the proof in $\mathbf{Case}\mathbf{2}$ of Lemma 8, the elements of U are resolved by at least $⌈\frac{t-1}{2}⌉+⌈\frac{k+2-t}{2}⌉$ elements of $F_2$. If t is even, then the value of $⌈\frac{t-1}{2}⌉+⌈\frac{k+2-t}{2}⌉$ is $\frac{k+2}{2}$ or $\frac{k+3}{2}$ according to whether k is even or odd. Thus, combining all of the above facts, there are at least $k+1$ elements in $F_1\cup F_2$ that resolve the elements of U when t is even. Using a similar argument, we can prove the result for an odd integer t.

$\mathbf{Case}\mathbf{2}.$$\mathbf{n}=\mathbf{4}\mathbf{s}+\mathbf{2}$. Let $v_i,v_{i+1}\subset F_1\cup F_2$ and $U=\{v_i,v_{i+1}\}$. Since $k\le 2s-1$, $d_{C_n}(F_1,F_2)\ge 2$, $U\subset F_1\cup F_2$ gives either $U\subset F_1$ or $U\subset F_2$. We prove the result for $U\subset F_1$. The proof for $U\subset F_2$ will be similar. Since $U\subset F_1$, $0\le i\le k-1$. For $i\ne 0$, we divide $F_1$ into two parts, $S_1=\{v_0,\dots ,v_i\}$ and $S_2=\{v_{i+1},\dots ,v_k\}$. Then, applying Lemma 9 twice for $S=S_1$ and for $S=S_2$, we see that $v_i$ and $v_{i+1}$ are resolved by at least $⌈\frac{i+1}{2}⌉+⌈\frac{k-i}{2}⌉$ elements of $F_2$. For $i\notin \{0,1\}$, we partition $F_2$ as $F_2=W_1\cup \{v_{2s+1+i},v_{2s+2+i}\}\cup W_2$, where $W_1=\{v_{2s+2},\dots ,v_{2s+i}\}$ and $W_2=\{v_{2s+3+i},\dots ,v_{2s+k+2}\}$. Since every vertex in $C(4s+2:1,2)$ has a unique antipodal vertex, and $v_{2s+i+1}$ and $v_{2s+i+2}$ are antipodal vertices of $v_i$ and $v_{i+1}$, respectively, each vertex in $\{v_{2s+1+i},v_{2s+2+i}\}$ resolves $v_i$ and $v_{i+1}$. Again, applying Lemma 9 to $S=W_1,W_2$ and combining them, we see that $v_i$ and $v_{i+1}$ are resolved by at least $⌊\frac{i-1}{2}⌋+⌊\frac{k-i}{2}⌋$ elements from $W_1\cup W_2$. Thus, $v_i$ and $v_{i+1}$ are resolved by at least $⌈\frac{i+1}{2}⌉+⌈\frac{k-i}{2}⌉+⌊\frac{i-1}{2}⌋+⌊\frac{k-i}{2}⌋+2$ elements of $F_1\cup F_2$ when $i\notin \{0,1\}$. Now, $⌈\frac{i+1}{2}⌉+⌈\frac{k-i}{2}⌉+⌊\frac{i-1}{2}⌋+⌊\frac{k-i}{2}⌋+2=k-i+i+2=k+2$, which shows that the result is true when $i\notin \{0,1\}$. For the remaining values of i, i.e., $i\in \{0,1\}$, we can prove the result by a similar argument. □

Example 4.

We consider the circulant graph $C(28:1,2)$ (i.e., $n=4s+4$) and $k=3$. For this graph, $F_1=\{v_0,v_1,v_2,v_3\}$ and $F_2=\{v_{14},v_{15},v_{16},v_{17}\}$. The codes of the vertices in $F_1\cup F_2$ are given in the matrix below. In this matrix, the columns corresponding to the vertex $v_i$ represent the code of $v_i$ with respect to $F_1\cup F_2$.

$$\begin{array}{cc}& \begin{array}{cccccccc}{v}_0& v_1& v_2& v_3& v_{14}& v_{15}& v_{16}& v_{17}\end{array}\\ \begin{array}{l}{v}_0\\ v_1\\ v_2\\ v_3\\ v_{14}\\ v_{15}\\ v_{16}\\ v_{17}\end{array}& \left(\begin{array}{cccccccc}0& 1& 1& 2& 7& 7& 6& 6\\ 1& 0& 1& 1& 7& 7& 7& 6\\ 1& 1& 0& 1& 6& 7& 7& 7\\ 2& 1& 1& 0& 6& 6& 7& 7\\ 7& 7& 6& 6& 0& 1& 1& 2\\ 7& 7& 7& 6& 1& 0& 1& 1\\ 6& 7& 7& 7& 1& 1& 0& 1\\ 6& 6& 7& 7& 2& 1& 1& 0\end{array}\right).\end{array}$$

It is easy to observe that any two consecutive columns differ by at least four places.

Lemma 11.

For $n=4s+3$ and an odd integer $k\le 2s-1$, let $F_1=\{v_0,v_1,\cdots ,v_k\}$ and $F_2=\left\{v_{2s+1},v_{2s+2},\dots ,v_{2s+k+1}\right\}$ be two sets of vertices of $C(4s+3,1,2)$. Then, any two consecutive vertices of $C(4s+3:1,2)$ are resolved by at least $k+1$ elements of $F_1\cup F_2$.

Proof. 

Let $v_i$ and $v_{i+1}$ be two consecutive vertices of $C(4s+3:1,2)$ and $U=\{v_i,v_{i+1}\}$. Recall that $A\left(U\right)=A\left(v_i\right)\cap A\left(v_{i+1}\right)=\left\{v_{2s+2+i}\right\}$, where the indices are taken to be modulo $4s+3$.

$\mathbf{Case}\mathbf{1}.$$\mathbf{U}\not\subset {\mathbf{F}}_{\mathbf{1}}\cup {\mathbf{F}}_{\mathbf{2}}$. In this case, $k\le i\le 2s$ or $2s+k+1\le i\le n-1$. If $i\ne 2s+k+1$, then $A\left(U\right)\cap F_i=\varnothing$ for $i=1,2$. Since k is odd and both $F_1$ and $F_2$ contain $k+1$ elements, applying Lemma 8 twice for $S=F_1,F_2$ and combining them, we see that the vertices of U are resolved by at least $k+1$. Otherwise, $i=2s+k+1$. Then, $A\left(U\right)=A\left(v_{2s+k+1}\right)\cap A\left(v_{2s+k+2}\right)=\left\{v_k\right\}$. Let $F_1^{\prime }=F_1\setminus \left\{v_k\right\}$. Then, $F_1^{\prime }\cap A\left(U\right)=\varnothing$, and also $d_{C_n}(U,F_1^{\prime })=d_{C_n}(v_{2s+k+2},v_0)=n-2s-k-2=2s+1-k$ and $d_{C_n}(U,F_2)=0$. Thus, both $d_{C_n}(U,F_1^{\prime })$ and $d_{C_n}(U,F_2)$ are even, so applying Lemma 8, we obtain the result when $i=2s+k+1$.

$\mathbf{Case}\mathbf{2}.$$\mathbf{U}\subset {\mathbf{F}}_{\mathbf{1}}\cup {\mathbf{F}}_{\mathbf{2}}$. As $k\le 2s-1$ and $d_{C_n}(F_1,F_2)\ge 2$, $U\subset F_1\cup F_2$ implies either $U\subset F_1$ or $U\subset F_2$. First, we assume that $U\subset F_1$. Then, $0\le i\le k-1$. Then, by a similar argument to that in the proof in $\mathbf{Case}\mathbf{2}$ of Lemma 8, we see that the vertices $v_i$ and $v_{i+1}$ are resolved by at least $⌈\frac{i+1}{2}⌉+⌈\frac{k-i}{2}⌉$ elements of $F_1$. Now, for $i\ne k-1$, we divide $F_2$ into two segment $S_1=\{v_{2s+1},\dots ,v_{2s+1+i}\}$ and $S_2=\{v_{2s+3+i},\dots ,v_{2s+k+1}\}$. Then, $A\left(U\right)\cap S_{\ell }=\varnothing$ for $\ell =1,2$, and $d_{C_n}(U,S_1)=d_{C_n}(v_{i+1},v_{2s+1})=2s+1-i$ and $d_{C_n}(U,S_2)=d_{C_n}(v_i,v_{2s+k+1})=n+i-2s-k-1=2s+2+i-k$. Applying Lemma 8 to $S_1$ and $S_2$, the vertices are resolved by $\frac{k+1}{2}$ or $\frac{k-1}{2}$ elements of $S_1\cup S_2\subset F_2$ according to whether i is odd or even. Again, $v_i$ and $v_{i+1}$ are resolved by at least $⌈\frac{i+1}{2}⌉+⌈\frac{k-i}{2}⌉$ elements of $F_1$, i.e., by $\frac{k+1}{2}$ or $\frac{k+3}{2}$ elements of $F_1$ according to whether i is odd or even. Hence, we obtain the result when $i\ne k-1$. For $i=k-1$, it is easy to observe that the elements of U are resolved by at least $k+1$ elements of $F_1\cup F_2$. Next, we assume that $U\subset F_1$. Similarly, we prove the result. □

Example 5.

In this example, we consider the circulant graph $C(23:1,2)$ and $k=5$. For this graph, $F_1=\{v_0,v_1,\cdots ,v_5\}$ and $F_2=\{v_{11},v_{12},\cdots ,v_{16}\}$. The codes of the vertices in $F_1\cup F_2$ are given in the matrix below. In this matrix, the columns corresponding to the vertex $v_i$ represent the code of $v_i$ with respect to $F_1\cup F_2$.

$$\begin{array}{cc}& \begin{array}{cccccccccccc}{v}_0& v_1& v_2& v_3& v_4& v_5& v_{11}& v_{12}& v_{13}& v_{14}& v_{15}& v_{16}\end{array}\\ \begin{array}{l}{v}_0\\ v_1\\ v_2\\ v_3\\ v_4\\ v_5\\ v_{11}\\ v_{12}\\ v_{13}\\ v_{14}\\ v_{15}\\ v_{16}\end{array}& \left(\begin{array}{cccccccccccc}0& 1& 1& 2& 2& 3& 6& 6& 5& 5& 4& 4\\ 1& 0& 1& 1& 2& 2& 5& 6& 6& 5& 5& 4\\ 1& 1& 0& 1& 1& 2& 5& 5& 6& 6& 5& 5\\ 2& 1& 1& 0& 1& 1& 4& 5& 5& 6& 6& 5\\ 2& 2& 1& 1& 0& 1& 4& 4& 5& 5& 6& 6\\ 3& 2& 2& 1& 1& 0& 3& 4& 4& 5& 5& 6\\ 6& 5& 5& 4& 4& 3& 0& 1& 1& 2& 2& 3\\ 6& 6& 5& 5& 4& 4& 1& 0& 1& 1& 2& 2\\ 5& 6& 6& 5& 5& 4& 1& 1& 0& 1& 1& 2\\ 5& 5& 6& 6& 5& 5& 2& 1& 1& 0& 1& 1\\ 4& 5& 5& 6& 6& 5& 2& 2& 1& 1& 0& 1\\ 4& 4& 5& 5& 6& 6& 3& 2& 2& 1& 1& 0\end{array}\right).\end{array}$$

It is easy to observe that any two consecutive columns differ by at least six places.

Remark 2.

If we drop the condition on k, i.e., if we consider k to be an even integer, then the result of Lemma 11 may not hold. For example, we take the circulant graph $C(23:1,2)$ and $k=4$. Then, $F_1=\{v_0,v_1,v_2,v_3,v_4\}$ and $F_2=\{v_{11},v_{12},v_{13},v_{14},v_{15}\}$. The codes of the vertices $v_9$ and $v_{10}$ (two consecutive vertices outside of $F_1\cup F_2$) with respect to $F_1\cup F_2$ are given by

$$\begin{array}{ccc}\hfill \mathrm{code}\left(v_9|F_1\cup F_2\right)& =& \left(5,4,4,3,3,1,2,2,3,3\right)\hfill \\ \hfill \mathrm{code}\left(v_{10}|F_1\cup F_2\right)& =& \left(5,5,4,4,3,1,1,2,2,3\right).\hfill \end{array}$$

Note that $\mathrm{code}\left(v_9|F_1\cup F_2\right)$ and $\mathrm{code}\left(v_{10}|F_1\cup F_2\right)$ differ by at exactly four places not in $k+1$ places, where $k=4$.

By a similar argument to that described in Lemma 11, we can prove the following result.

Lemma 12.

For an even integer k, let $F_1=\{v_0,v_1,\cdots ,v_{k+1}\}$ and $F_2=\left\{v_{2s+1},v_{2s+2},\dots ,v_{2s+k}\right\}$ be two sets of vertices of $C(4s+3:1,2)$. Then, any two consecutive vertices of $C(4s+3:1,2)$ are resolved by at least $k+1$ elements of $F_1\cup F_2$.

Lemma 13.

For an odd integer k, let $F_1=\{v_0,v_1,\cdots ,v_k\}$ and $F_2=\left\{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+3}\right\}$ be two sets of vertices $C(4s+5:1,2)$. Then, any two consecutive vertices of $C(4s+5:1,2)$ are resolved by at least $k+1$ elements of $F_1\cup F_2$.

Proof. 

Let $v_i$ and $v_{i+1}$ be two consecutive vertices of $C(4s+5:1,2)$ and $U=\{v_i,v_{i+1}\}$. Recall that $A\left(U\right)=A\left(v_i\right)\cap A\left(v_{i+1}\right)=\{v_{2s+2+i},v_{2s+3+i},v_{2s+4+i}\}$, where the indices are taken to be modulo $4s+5$. Thus,

$$A\left(U\right)\subset \left\{\begin{array}{cc}{F}_1\hfill & \mathrm{if}U\subset F_2,\hfill \\ F_2\hfill & \mathrm{if}U\subset F_1.\hfill \end{array}\right.$$

We consider the following two cases depending on the positions of $v_i$ and $v_{i+1}$.

$\mathbf{Case}\mathbf{1}.$Neither $v_i$ nor $v_{i+1}$ is in $F_1\cup F_2$. From the above, it is clear that $A\left(U\right)\cap F_i=\varnothing$ for $i=0,1$. Thus, applying Lemma 8 to both $F_1$ and $F_2$, we see that the elements of U are resolved by at least $\frac{k+1}{2}$ elements from each of $F_1$ and $F_2$.

$\mathbf{Case}\mathbf{2}.$Exactly one of $v_i$ and $v_{i+1}$ is in $F_1\cup F_2$. Without loss of generality, we can assume that either $v_i$ or $v_{i+1}$ is in $F_1$. Since $F_1$ is a set of consecutive vertices and exactly one of $v_i$ and $v_{i+1}$ is in $F_1$, either $U\cap F_1=\left\{v_0\right\}$ or $U\cap F_1=\left\{v_k\right\}$. For the first one, $i=4s+4$, and for the second, $i=k$.

$\mathbf{Case}\mathbf{2.1}.$$\mathbf{U}\cap {\mathbf{F}}_{\mathbf{1}}=\left\{{\mathbf{v}}_{\mathbf{0}}\right\}$. In this case, $U=\{v_{4s+4},v_0\}$ and $A\left(U\right)=A\left(v_{4s+4}\right)\cap A\left(v_0\right)=\{v_{2s+1},v_{2s+2},v_{2s+3}\}$. Let $S=F_2\setminus \{v_{2s+2},v_{2s+3}\}$. Then, $A\left(U\right)\cap S=\varnothing$ and $d_{C_n}(U,S)=d_{C_n}(v_{4s+4},v_{2s+k+3})=2s-k+1$, an even integer. Thus, applying Lemma 8 to both $F_1$ and S, we see that the vertices of U are resolved by at least $\frac{k+1}{2}$ elements from each of $F_1$ and S.

$\mathbf{Case}\mathbf{2.2}.$$\mathbf{U}\cap {\mathbf{F}}_{\mathbf{1}}=\left\{{\mathbf{v}}_{\mathbf{k}}\right\}$. In this case, $U=\{v_k,v_{k+1}\}$ and $A\left(U\right)=${$v_{2s+k+2}$,$v_{2s+k+3}$,$v_{2s+k+4}$}. By applying the same argument as is used in $\mathbf{Case}\mathbf{2.1}$ and applying Lemma 8 to both $F_1$ and $F_2\setminus \{v_{2s+k+2},v_{2s+k+3}\}$, we obtain the result.

$\mathbf{Case}\mathbf{3}.$Both $v_i$ and $v_{i+1}$ are in $F_1\cup F_2$. Since $U=\{v_i,v_{i+1}\}$ and $d_{C_n}(F_1,F_2)\ge 2$, in this case, either $U\subset F_1$ or $U\subset F_2$. We assume that $U\subset F_1$. The proof will be similar to when $U\subset F_2$. Here, $A\left(U\right)=A\left(v_i\right)\cap A\left(v_{i+1}\right)=\{v_{2s+2+i},v_{2s+3+i},v_{2s+4+i}\}$. Divide the set $F_2$ into two segments, $S_1=\{v_{2s+2},\dots ,v_{2s+1+i}\}$ and $S_2=\{v_{2s+4+i},\dots ,v_{2s+k+3}\}$. Note that $d_{C_n}(U,S_1)=d_{C_n}(v_{i+1},v_{2s+2})=2s+1-i$ and $d_{C_n}(U,S_2)=d_{C_n}(v_i,v_{2s+k+3})=2s+2+i-k$. Then, using Lemma 8 and a similar argument to that used in $\mathbf{Case}\mathbf{3}$ of Lemma 10, we obtain the result. □

Example 6.

For the circulant graph $C(21:1,2)$ and $k=5$, $F_1=\{v_0,v_1,v_2,v_3,v_4,v_5\}$ and $F_2=\{v_{10},v_{11},\dots ,v_{16}\}$. The codes of the vertices in $F_1\cup F_2$ are given in the matrix below. In this matrix, the columns corresponding to the vertex $v_i$ represent the code of $v_i$ with respect to $F_1\cup F_2$. It is easy to observe that any two consecutive columns differ by at least six places.

$$\begin{array}{cc}& \begin{array}{ccccccccccccc}{v}_0& v_1& v_2& v_3& v_4& v_5& v_{10}& v_{11}& v_{12}& v_{13}& v_{14}& v_{15}& v_{16}\end{array}\\ \begin{array}{l}{v}_0\\ v_1\\ v_2\\ v_3\\ v_4\\ v_5\\ v_{10}\\ v_{11}\\ v_{12}\\ v_{13}\\ v_{14}\\ v_{15}\\ v_{16}\end{array}& \left(\begin{array}{ccccccccccccc}0& 1& 1& 2& 2& 3& 5& 5& 5& 4& 4& 3& 3\\ 1& 0& 1& 1& 2& 2& 5& 5& 5& 5& 4& 4& 3\\ 1& 1& 0& 1& 1& 2& 4& 5& 5& 5& 5& 4& 4\\ 2& 1& 1& 0& 1& 1& 4& 4& 5& 5& 5& 5& 4\\ 2& 2& 1& 1& 0& 1& 3& 4& 4& 5& 5& 5& 5\\ 3& 2& 2& 1& 1& 0& 3& 3& 4& 4& 5& 5& 5\\ 5& 5& 4& 4& 3& 3& 0& 1& 1& 2& 2& 3& 3\\ 5& 5& 5& 4& 4& 3& 1& 0& 1& 1& 2& 2& 3\\ 5& 5& 5& 5& 4& 4& 1& 1& 0& 1& 1& 2& 2\\ 4& 5& 5& 5& 5& 4& 2& 1& 1& 0& 1& 1& 2\\ 4& 4& 5& 5& 5& 5& 2& 2& 1& 1& 0& 1& 1\\ 3& 4& 4& 5& 5& 5& 3& 2& 2& 1& 1& 0& 1\\ 3& 3& 4& 4& 5& 5& 3& 3& 2& 2& 1& 1& 0\end{array}\right).\end{array}$$

Remark 3.

If we drop the condition on k, i.e., if we consider k to be an even integer, then the result of Lemma 13 may not hold. For example, we consider the circulant graph $C(21:1,2)$ and $k=4$. Then, $F_1=\{v_0,v_1,v_2,v_3,v_4,v_5\}$ and $F_2=\{v_{10},v_{11},\dots ,v_{15}\}$. The codes of the vertices $v_{11}$ and $v_{12}$ (two consecutive vertices in $F_2$) with respect to $F_1\cup F_2$ are given by

$$\begin{array}{ccc}\hfill \mathrm{code}\left(v_{11}|F_1\cup F_2\right)& =& \left(5,5,5,4,4,1,0,1,1,2,2\right)\hfill \\ \hfill \mathrm{code}\left(v_{12}|F_1\cup F_2\right)& =& \left(5,5,5,5,4,1,1,0,1,1,2\right).\hfill \end{array}$$

Note that $\mathrm{code}\left(v_{11}|F_1\cup F_2\right)$ and $\mathrm{code}\left(v_{12}|F_1\cup F_2\right)$ differ by exactly four places not in $k+1$ places, where $k=4$.

Lemma 14.

Let $k\le 2s-2$ be an even integer and $F_1=\{v_0,v_1,\cdots ,v_{k-1}\}\cup \{v_{k+1},v_{k+2}\}$ and $F_2=\left\{v_{2s+2},v_{2s+3},\cdots ,v_{2s+k+1}\right\}\cup \left\{v_{2s+k+4}\right\}$ be two subsets of $V\left(C\right(4s+5:1,2\left)\right)$. Then, any two consecutive vertices in $C(4s+5:1,2)$ are resolved by at least $k+1$ elements of $F_1\cup F_2$.

Proof. 

By a similar argument to that stated in the proof of Lemma 13, we can prove the result. □

Example 7.

For the circulant graph $C(21:1,2)$ and $k=4$, $F_1=\{v_0,v_1,v_2,v_3\}\cup \{v_5,v_6\}$ and $F_2=\{v_{10},v_{11},v_{12},v_{13}\}\cup \left\{v_{16}\right\}$. The codes of the vertices in $F_1\cup F_2$ are given in the matrix below. In this matrix, the columns corresponding to the vertex $v_i$ represent the code of $v_i$ with respect to $F_1\cup F_2$. It is easy to observe that any two consecutive columns differ by at least five places.

$$\begin{array}{cc}& \begin{array}{ccccccccccc}{v}_0& v_1& v_2& v_3& v_5& v_6& v_{10}& v_{11}& v_{12}& v_{13}& v_{16}\end{array}\\ \begin{array}{l}{v}_0\\ v_1\\ v_2\\ v_3\\ v_5\\ v_{10}\\ v_{11}\\ v_{12}\\ v_{13}\\ v_{16}\end{array}& \left(\begin{array}{ccccccccccc}0& 1& 1& 2& 3& 3& 5& 5& 5& 4& 3\\ 1& 0& 1& 1& 2& 3& 5& 5& 5& 5& 3\\ 1& 1& 0& 1& 2& 2& 4& 5& 5& 5& 4\\ 2& 1& 1& 0& 1& 2& 4& 4& 5& 5& 4\\ 3& 2& 2& 1& 0& 1& 3& 3& 4& 4& 5\\ 3& 3& 2& 2& 1& 0& 2& 3& 3& 4& 5\\ 5& 5& 4& 4& 3& 2& 0& 1& 1& 2& 3\\ 5& 5& 5& 4& 3& 3& 1& 0& 1& 1& 3\\ 5& 5& 5& 5& 4& 3& 1& 1& 0& 1& 2\\ 4& 5& 5& 5& 4& 4& 2& 1& 1& 0& 2\\ 3& 3& 4& 4& 5& 5& 3& 3& 2& 2& 0\end{array}\right).\end{array}$$

Remark 4.

If we drop the condition on k, i.e., if we consider k to be an odd integer, then the result of Lemma 14 may not hold. For example, we consider the circulant graph $C(21:1,2)$ and $k=5$. Then, $F_1=\{v_0,v_1,v_2,v_3,v_4\}\cup \{v_6,v_7\}$ and $F_2=\{v_{10},v_{11},v_{12},v_{13},v_{14}\}\cup \left\{v_{17}\right\}$. The codes of the vertices $v_{11}$ and $v_{12}$ (two consecutive vertices in $F_2$) with respect to $F_1\cup F_2$ are given by

$$\begin{array}{ccc}\hfill \mathrm{code}\left(v_{11}|F_1\cup F_2\right)& =& \left(5,5,5,4,4,3,2,1,0,1,1,2,3\right)\hfill \\ \hfill \mathrm{code}\left(v_{12}|F_1\cup F_2\right)& =& \left(5,5,5,5,4,3,3,1,1,0,1,1,3\right).\hfill \end{array}$$

Note that $\mathrm{code}\left(v_{11}|F_1\cup F_2\right)$ and $\mathrm{code}\left(v_{12}|F_1\cup F_2\right)$ differ by exactly five places not in $k+1$ places, where $k=5$.

Theorem 2.

Let n and k be integers such that $1\le k\le ⌊\frac{n}{2}⌋-3$. Then, the fault-tolerant metric dimension with the tolerance level k of the circulant graph $C(n:1,2)$ is $2(k+1)$ or $2k+3$ according to whether $n\not\equiv 1(mod4)$ or $n\equiv 1(mod4).$

Proof. 

In Theorem 1, it is shown that every fault-tolerant resolving set F with the tolerance level k of $C(n:1,2)$ must contain at least $2(k+1)$ or at least $2k+3$ elements according to whether $n\not\equiv 1(mod4)$ or $n\equiv 1(mod4)$. Thus, to prove the result, we need to construct a fault-tolerant resolving set F with the tolerance level k that contains $2(k+1)$ and $2k+3$ elements accordingly as $n\not\equiv 1(mod4)$ and $n\equiv 1(mod4)$. Our claim is that the set F forms a fault-tolerant resolving set with the tolerance level k, where F is defined in the following manner.

(a)

For $n=4s+4$ or $n=4s+2$, $F=\{v_0,v_1,\cdots ,v_k\}\cup \{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+2}\}$.

(b)

For $n=4s+3$ and odd integer k, $F=\{v_0,v_1,\dots ,v_k\}\cup \{v_{2s+1},v_{2s+2},\cdots ,v_{2s+k+1}\}$.

(c)

For $n=4s+3$ and even integer k, $F=\{v_0,v_1,\cdots ,v_{k+1}\}\cup \{v_{2s+1},v_{2s+2},\cdots ,v_{2s+k}\}$.

(d)

For $n=4s+5$ and odd integer k, $F=\{v_0,v_1,\cdots ,v_k\}\cup \left\{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+3}\right\}.$

(e)

For $n=4s+5$ and even integer k,

$F=\{v_0,v_1,\dots ,v_{k-1}\}\cup \{v_{k+1},v_{k+2}\}\cup \left\{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+1}\right\}\cup \left\{v_{2s+k+4}\right\}.$

Let $v_i$ and $v_j$ be any two vertices of $C(n:1,2)$. Since F contains at least $2(k+1)$ elements, applying Lemma 1, the vertices $v_i$ and $v_j$ are resolved by at least $k+1$ element of F, provided that $|i-j|\ge 2$. So, we assume that $|i-j|=1$, i.e., we take two consecutive vertices $v_i$ and $v_{i+1}$. Our claim is that $v_i$ and $v_{i+1}$ are resolved by at least $k+1$ elements of F. When $n=4s+3$, by Lemmas 11 and 12, our claim is true. Again, by Lemmas 13 and 14, our claim is also true when n is of the form $4s+5$. Thus, we consider the following cases.

$\mathbf{Case}\mathbf{1}.$$\mathbf{n}=\mathbf{4}\mathbf{s}+\mathbf{2}$. Here, $F=\{v_0,v_1,\cdots ,v_k\}\cup \{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+2}\}=F_1\cup F_2$, where $F_1=\{v_0,v_1,\cdots ,v_k\}$ and $F_2=\{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+2}\}$. Let $U=\{v_i,v_{i+1}\}$. If $U\subset F_1\cup F_2$, then applying Lemma 7, the vertices of U are resolved by at least $k+1$ elements of $F_1\cup F_2$. So, we now assume that $U\not\subset F_1\cup F_2$, i.e., either $k\le i\le 2s+1$ or $2s+k+2\le i\le n-1$.

$\mathbf{Case}\mathbf{1.1}.$$\mathbf{k}\le \mathbf{i}\le \mathbf{2}\mathbf{s}+\mathbf{1}$. Since $k\le 2s-1$, applying Lemma 8 twice for $S=F_1$ and $S=F_2$, the vertices $v_i$ and $v_{i+1}$ are resolved by at least $⌊\frac{|F_1|}{2}⌋+⌊\frac{|F_2|}{2}⌋$, i.e., at least $2⌊\frac{k+1}{2}⌋$ elements of F. Thus, the proof is complete when k is odd. For an even integer k, we calculate $d_{C_n}(U,F_1)=i-k$ and $d_{C_n}(U,F_2)=2s+1-i$. Then, $d_{C_n}(U,F_1)+d_{C_n}(U,F_2)=2s+1-k$, an odd integer when k is even. So, at least one of $d_{C_n}(U,F_1)$ and $d_{C_n}(U,F_1)$ is even, and hence, applying Lemma 7, we see that $v_i$ and $v_{i+1}$ are resolved by at least $⌊\frac{k+1}{2}⌋+⌈\frac{k+1}{2}⌉=k+1$ elements.

$\mathbf{Case}\mathbf{1.2}.$$\mathbf{2}\mathbf{s}+\mathbf{k}+\mathbf{2}\le \mathbf{i}\le \mathbf{n}-\mathbf{1}$. If k is odd, then by a similar argument to that in $\mathbf{Case}\mathbf{1.1}$, our claim is true. So, we assume that k is even integer. Then, $d_{C_n}(U,F_1)=d_{C_n}(v_{i+1},v_0)=n-i-1=4s+1-i$ and $d_{C_n}(U,F_2)=d_{C_n}(v_i,v_{2s+k+2})=i-2s-k-2$, so $d_{C_n}(U,F_1)+d_{C_n}(U,F_2)=2s-1-k$, an odd integer. Thus, exactly one of $d_{C_n}(U,F_1)$ and $d_{C_n}(U,F_2)$ is odd, so applying Lemma 7 to both $F_1$ and $F_2$, we see that our claim is true.

$\mathbf{Case}\mathbf{2}.$$\mathbf{n}=\mathbf{4}\mathbf{s}+\mathbf{4}$. In this case, $F=\{v_0,v_1,\dots ,v_k\}\cup \{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+2}\}$. Here, $A\left(U\right)=A\left(v_i\right)\cap A\left(v_{i+1}\right)=\{v_{2s+2+i},v_{2s+3+i}\}$. Now, $A\left(U\right)\cap F=\varnothing$, provided that $v_i\notin \{v_0,v_1,\dots ,v_{k-1}\}\cup \{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+1}\}$. Again, Lemma 10 states that our claim is true when $v_i\in \{v_0,v_1,\dots ,v_{k-1}\}\cup \{v_{2s+2},v_{2s+3},\dots ,v_{2s+k+1}\}$. This completes the proof of the theorem. □

Example 8.

In this example, we take $k\in \{2,3\}$ and the circulant graphs $C(n:1,2)$, where $n\in \{34,35,36,37\}$. Then, for $n\in \{34,36\}$, $F=\{v_0,v_1,v_2,v_{18},v_{19},v_{20}\}$ or $F=${$v_0$, $v_1$, $v_2$, $v_3$, $v_{18}$, $v_{19}$, $v_{20}$, $v_{21}$} according to whether $k=2$ or $k=3$. When $n=35$,

$$\begin{array}{c}\hfill F=\left\{\begin{array}{cc}\{v_0,v_1,v_2,v_3,v_{17},v_{18}\},\hfill & \mathit{if}k=2;\hfill \\ \{v_0,v_1,v_2,v_3,v_{17},v_{18},v_{19},v_{20}\},\hfill & \mathit{if}k=3.\hfill \end{array}\right.\end{array}$$

Finally, for $n=37$,

$$\begin{array}{c}\hfill F=\left\{\begin{array}{cc}\{v_0,v_1,v_3,v_4,v_{18},v_{19},v_{22}\},\hfill & \mathit{if}k=2;\hfill \\ \{v_0,v_1,v_2,v_3,v_{18},v_{19},v_{20},v_{21},v_{22}\},\hfill & \mathit{if}k=3.\hfill \end{array}\right.\end{array}$$

5. Conclusions

Circulant networks have a broad range of applications in parallel computing and signal processing because of their increased connectivity [14]. For all possible values of k, this article presents the k-level metric dimension and optimal fault-tolerant resolving set with tolerance level k for the circulant graphs $C(n:\mathcal{L})$ when $\mathcal{L}=\{1,2\}$ and n is any positive integer. Theorem 2 gives not only the fault-tolerant metric dimension of $C(n:1,2)$ with the tolerance level k but also an optimal fault-tolerant resolving set for the same. This fault-tolerant metric helps us to detect the optimal position of the sensors with the tolerance level k.

The same idea can be used to determine the following problem:

Problem: Determine the k-level metric dimension for the circulant graphs $C(n:\mathcal{L})$ when $\mathcal{L}=\{1,2,3\}$ or $\{1,2,3,4\}$ or any t-consecutive integers.