On the Characterization of a Minimal Resolving Set for Power of Paths

Abstract

For a simple connected graph $G=(V,E)$, an ordered set $W\subseteq V$, is called a resolving set of G if for every pair of two distinct vertices u and v, there is an element w in W such that $d(u,w)\ne d(v,w)$. A metric basis of G is a resolving set of G with minimum cardinality. The metric dimension of G is the cardinality of a metric basis and it is denoted by $\beta \left(G\right)$. In this article, we determine the metric dimension of power of finite paths and characterize all metric bases for the same.

1. Introduction

The study of the metric dimension or the resolving set of a simple connected graph using the distance between vertices is very popular among research scholars as it is applicable to many areas such as networks, robotic navigation and drug design. The concept of the metric dimension of a graph was first introduced by Slater [1]. The introduction of this invariant was motivated by its application to the placement of a minimum number of sonar/loran detecting devices in a network so that the position of every vertex in the network can be uniquely described in terms of its distances to the devices in the set. They used a location set in place of a resolving set. Harary and Melter [2] in 1976 also introduced the same concept as a metric dimension rather than a location number.

Throughout this article, $G=(V,E)$ denotes a simple connected graph with vertex set V and edge set E. The distance between two vertices u and v in G, denoted by $d(u,v)$, is the length of the shortest $u-v$ path. For an ordered subset $W=\{w_1,w_2,\dots ,w_k\}\subset$V and a vertex v of G, the distance code of v with respect to W is a k-vector given by

$${\displaystyle cod{e}_W\left(v\right)=(d(v,w_1),d(v,w_2),\dots ,d(v,w_k)).}$$

If $cod{e}_W\left(u\right)\ne cod{e}_W\left(v\right)$ for all distinct vertices u and v, then W is called a resolving set for the graph G. Every simple connected graph G has a resolving set, as the vertex set V forms a resolving set. The metric dimension of graph G is the minimum cardinality of a resolving set for G and it is denoted by $\beta \left(G\right)$. A resolving set with cardinality $\beta \left(G\right)$ is called a metric basis and elements of it are called basis elements. Several papers studied resolving sets and metric dimension in the literature [3,4,5,6,7,8,9,10,11,12]. Throughout this article, the distance code $cod{e}_W\left(v\right)$ is simply denoted by $code\left(v\right)$.

For a path $P_n$ with vertex set $V\left(P_n\right)=\{v_0,v_1,\dots ,v_{n-1}\}$, the $r$-th power of $P_n$, denoted by $P_n^r$, is a simple graph with the same vertex set $V\left(P_n\right)$, and the two vertices $v_i$ and $v_j$ are adjacent if $d_{P_n}(v_i,v_j)\le r$. It is clear to observe that $d_{P_n^r}(v_i,v_j)=⌈\frac{d_{P_n}(v_i,v_j)}{r}⌉=⌈\frac{|i-j|}{r}⌉$. In this paper, we determine the metric dimension of power of finite paths and characterize all metric bases for the same.

2. Preliminaries

In this section, we give some basic definitions, lemmas and a proposition that will be used in the sequel. Note that the vertex set $V\left(P_n^r\right)=\{v_0,v_1,\dots ,v_{n-1}\}$. By the division algorithm, we can write $n=rq+s$ for some positive integer q and $s\in \{0,1,\dots ,r-1\}$. Keeping in mind this form of n, we partitioned $V\left(P_n^r\right)$ into the sets $B_0,B_1,\dots ,B_{\lceil \frac{n}{r}\rceil -1}$, where $B_i$’s are to be defined as:

(a)

$B_i=\left\{v_{ir+j}\in V\left(P_n^r\right):0\le j\le r-1\right\}$ for $0\le i\le \lceil \frac{n}{r}\rceil -2$.

(b)

$B_{\lceil \frac{n}{r}\rceil -1}=\left\{v_{r\left(\lceil \frac{n}{r}\rceil -1\right)},\dots ,v_{n-1}\right\}$ or ∅ according to $n\not\equiv 0(modr)$ or $n\equiv 0(modr)$.

We call the above $B_i$’s a block. For example, the vertex set of $P_{17}^3$ can be divided into six blocks, $B_0$, $B_1,$ $B_2$, $B_3$, $B_4$ and $B_5$, where $B_i=\{v_{3i},v_{3i+1},v_{3i+2}\}$ for $0\le i\le 4$ and $B_5=\{v_{15},v_{16}\}$. Again, the vertex set of $P_{15}^5$ has three blocks, $B_0=\{v_0,v_1,v_2,v_3,v_4\}$, $B_1=\{v_5,v_6,v_7,v_8,v_9\}$ and $B_2=\{v_{10},v_{11},v_{12},v_{13},v_{14}\}$.

Proposition 1.

For$P_n^r$, the blocks$B_i$’s have the following properties:

(a) 

For each $i\in \{0,1,\dots ,\lceil \frac{n}{r}\rceil -2\}$, the block $B_i$ consists r elements, namely, $v_{ir},v_{ir+1},v_{ir+2},\dots ,v_{ir+(r-1)}$. Moreover, the block $B_{\lceil \frac{n}{r}\rceil -1}$ consists r when $n\equiv 0(modr)$ and consists of ℓ elements when $n\equiv \ell (modr)$ with $\ell \ne 0$.

(b) 

For each $i\in \left\{0,1,\dots ,\lceil \frac{n}{r}\rceil -1\right\}$, the induced subgraph of $B_i$ forms a clique.

Definition 1.

For an integer i satisfying $0\le i\le r-1$, by a class $\left[i\right]$ we mean the set $\left[i\right]=\{j:j\equiv i(modr),0\le j\le n-1\}.$ For $0\le t\le r-1$, a vertex $v_k$ is called a $t\text{-}class$ element if $k\in \left[t\right]$ and $v_k$ is called the largest $t\text{-}class$ element if k is the largest element in $\left[t\right]$. For example, $v_2$, $v_5$, $v_8$ and $v_{11}$ are the 2-class elements in $P_{13}^3$, whereas $v_{11}$ is the largest 2-class element. From here onward, we denote the set of all $t\text{-}class$ elements by $S_{\left[t\right]}$. It is noted that ${\displaystyle V\left(P_n^r\right)=\bigcup _{t=0}^{r-1}{S}_{\left[t\right]}}$ and $S_{\left[x\right]}\ne S_{\left[y\right]}$ for $x\ne y$.

Example 1.

For $r=4$ and $n=23$, the classes $\left[0\right]$, $\left[1\right]$, $\left[2\right]$ and $\left[3\right]$ are given by $\left[0\right]=\{0,4,\dots ,20\}$, $\left[1\right]=\{1,5,\dots ,21\}$, $\left[2\right]=\{2,6,\dots ,22\}$ and $\left[3\right]=\{3,7,\dots ,19\}$. From the definition, the largest $\left[0\right]$-class, $\left[1\right]$-class and $\left[3\right]$-class elements are $v_{20}$, $v_{21}$ and $v_{19}$, respectively. For $P_{23}^4$, $S_{\left[0\right]}=\{v_0,v_4,\dots ,v_{20}\}$, $S_{\left[1\right]}=\{v_1,v_5,\dots ,v_{21}\}$, $S_{\left[2\right]}=\{v_2,v_6,\dots ,v_{22}\}$ and $S_{\left[3\right]}=\{v_3,v_7,\dots ,v_{19}\}$. Note that $S_{\left[x\right]}\ne S_{\left[y\right]}$ for distinct $x,y\in \{0,1,2,3\}$ and also $S_{\left[0\right]}\cup S_{\left[1\right]}\cup S_{\left[2\right]}\cup S_{\left[3\right]}=\{v_0,v_1,\dots ,v_{22}\}=V\left(P_{23}^4\right)$.

Lemma 1.

For any two vertices $u=v_{ir+r_1}$ and w = $v_{jr+r_2}$ of $P_n^r$,

$$d_{P_n^r}(u,w)=\left\{\begin{array}{cc}|i-j|\hfill & if{r}_2\le r_1;\hfill \\ |i-j|+1\hfill & otherwise.\hfill \end{array}\right.$$

Proof. 

Without loss of generality, we assume that u is left of w, that is, $ir+r_1<jr+r_2$. If $r_2\le r_1$, then

$$\begin{array}{ccc}\hfill d_{P_n^r}(u,v)& =& ⌈\frac{jr+r_2-ir-r_1}{r}⌉\hfill \\ & =& ⌈\frac{(j-i)r+r_2-r_1}{r}⌉\hfill \\ & =& ⌈\frac{(j-i-1)r+\left(r-(r_1-r_2)\right)}{r}⌉=j-i.\hfill \end{array}$$

Otherwise, $r_2>r_1$. Then, ${\displaystyle d_{P_n^r}(u,v)=⌈\frac{(j-i)r+(r_2-r_1)}{r}⌉=j-i+1}$. □

Remark 1.

For $0\le s\le r-1$, $d_{P_n^r}(v_s,u)=d_{P_n^r}(v_s,w)$ implies $d_{P_n^r}(v_{ir+s},u)=d_{P_n^r}(v_{ir+s},w).$

3. Metric Dimension of ${\mathit{P}}_{\mathit{n}}^{\mathit{r}}$

In this section, we present a lower bound for the metric dimension of $P_n^r$ and then we build up a resolving set with cardinality that is the same as that of the lower bound. It is noted that $\{v_i,v_{i+1},\dots ,v_{i+r}\}$ forms a clique in $P_n^r$ for every $i\in \{0,1,\dots ,n-r-1\}$. The following lemma represents an effective result to determine a lower bound for the metric dimension of $P_n^r$.

Lemma 2.

Let $A\subset \{v_i,v_{i+1},\dots ,v_{i+r}\}$ with $\left|A\right|=\ell$ for $2\le \ell \le r+1$. If X resolves A then $\left|X\right|\ge \ell -1$.

Proof. 

We prove this lemma by induction on ℓ. The result obviously is true for $\ell =2$. Now we show that it is true for $\ell =k$ with an assumption that it holds for $\ell =k-1$. Let the elements of A in order be $v_{i+a_1},v_{i+a_2},\dots ,v_{i+a_k}$ where $0\le a_1<a_2<\cdots <a_k\le r$. As X resolves A, there is an element $x\in X$ such that $d(x,v_{i+a_1})\ne d(x,v_{i+a_2})$. If $x=v_{i+a_2}$, then $d(x,v_{i+a_s})=1$ for all $s\in \{1,3,\dots ,k\}$ and hence $X\backslash \left\{x\right\}$ resolves $A\backslash \left\{v_{i+a_2}\right\}$. Otherwise, $x\ne v_{i+a_2}$. Then, $d(x,v_{i+a_2})=d(x,v_{i+a_s})$ for all $s\in \{2,3,\dots ,k\}$ and $X\backslash \left\{x\right\}$ resolves $A\backslash \left\{v_{i+a_1}\right\}$. Thus, in both cases $X\backslash \left\{x\right\}$ resolves a subset of $\{v_i,v_{i+1},\dots ,v_{i+r}\}$ with cardinality $k-1$. Therefore, by assumption, $|X\backslash \{x\left\}\right|\ge k-2$ and hence $\left|X\right|\ge k-1$. □

With the help of Lemma 2, the following result represents a lower bound of $\beta \left(P_r^n\right)$ for all values of n and $r<\frac{n}{2}$.

Lemma 3.

For two integers n and r with $r<\frac{n}{2}$, $\beta \left(P_n^r\right)\ge r.$

Proof. 

Let B be a resolving set of $P_n^r$ and $v_i\in B$. Then, $A=\{v_{i+1},v_{i+2},\dots ,v_{i+r}\}$ is not resolved by $v_i$ as $d(v_i,v_{i+s})=1$ for all $s\in \{1,2,\dots ,r\}$. By Lemma 2, we obtain $|B\backslash v_i|\ge r-1$ and hence $\left|B\right|\ge r$. □

Now our aim is to construct a resolving set for $P_n^r$ with cardinality r.

Lemma 4.

Let i and ℓ be two integers such that $0\le i<i+\ell \le r-1$. Then, $A=\{v_i,v_{i+1},\dots ,v_{i+\ell }\}\subset V\left(P_n^r\right)$ resolves the set ${\cup }_{j=1}^{\lceil \frac{n}{r}\rceil -1}\{v_{jr+i},\dots ,v_{jr+i+\ell +1}\}\subseteq V\left(P_n^r\right)$, that is, A resolves the set ${\cup }_{j=i}^{i+\ell +1}{S}_{\left[j\right]}$.

Proof. 

To prove this lemma, it is sufficient to show that for any pair of vertices $u,w\in {\cup }_{j=1}^{\lceil \frac{n}{r}\rceil -1}\{v_{jr+i},\dots ,v_{jr+i+\ell +1}\}$ there exists at least one vertex $x\in A$ such that $d(u,x)\ne d(w,x)$. Let $u,w\in {\cup }_{j=1}^{\lceil \frac{n}{r}\rceil -1}\{v_{jr+i},\dots ,v_{jr+i+\ell +1}\}$ be two distinct vertices. Then, we may write $u=v_{ar+r_1}$ and $w=v_{br+r_2}$ for some $0\le a,b\le \lceil \frac{n}{r}\rceil -1$ and $i\le r_1,r_2\le i+\ell +1$. Without loss of generality, we can assume that u is in the left side of w, that is, $ar+r_1<br+r_2$. Now we take $x=v_{r_1}\in A$. Then, $d(v_{r_1},u)=a$ and $d(v_{r_1},w)=b$ or $b+1$ according to $r_2\le r_1$ or $r_2>r_1$. If $r_2>r_1$, then $d(u,v_{r_1})\ne d(w,v_{r_1})$. Otherwise, $r_2\le r_1$. Then, $d(u,v_{r_1})\ne d(w,v_{r_1})$, as $ar+r_1<br+r_2$. Therefore, $v_{r_1}\in A$ resolves $u,w$ and hence A resolves ${\cup }_{j=1}^{\lceil \frac{n}{r}\rceil -1}\{v_{jr+i},\dots ,v_{jr+i+\ell +1}\}$. □

By a similar argument as in Lemma 4, we have the following result.

Lemma 5.

Let i and ℓ be two integers such that $0\le i<i+\ell \le r-1$. Then $B=\{v_{n-r+i},v_{n-r+i+1},\dots ,v_{n-r+i+\ell }\}\subset V\left(P_n^r\right)$ resolves the set ${\cup }_{j=1}^{\lceil \frac{n}{r}\rceil -1}\{v_{n-jr+i-1},\dots ,v_{n-jr+i+\ell }\}\subseteq V\left(P_n^r\right)$, that is, B resolves the set ${\cup }_{j=i-1}^{i+\ell }{S}_{\left[j\right]}$.

The following theorem gives the exact value of the metric dimension of $P_n^r$ for all values of n and r.

Theorem 1.

For two integers n and r with $r<\frac{n}{2}$, $\beta \left(P_n^r\right)=r$.

Proof. 

From Lemma 3, we have $\beta \left(P_n^r\right)\ge r$. Thus, to prove the theorem it is sufficient to construct a resolving set A with cardinality r. Consider $A=\{v_0,v_1,\dots ,v_{r-1}\}$ to be the set of the first r consecutive vertices of $P_n^r$. Now we apply Lemma 4 to A, and then A resolves ${\cup }_{t=0}^{r-1}{S}_{\left[t\right]}=V\left(P_n^r\right)$. Therefore, A is a resolving set for $P_n^r$ with cardinality r and consequently, $\beta \left(P_n^r\right)\le r$. This completes the proof of the theorem. □

From the above theorem, it is clear that any metric basis $\mathcal{B}$ consists of r elements. A metric basis $\mathcal{B}$ may have a non-empty intersection with all $S_{\left[i\right]}$ or may have an empty intersection with some $S_{\left[i\right]}$, where $0\le i\le r-1$. In the next two sections we determine all metric bases. First, we characterize all metric bases $\mathcal{B}$ with $\mathcal{B}\cap S_{\left[i\right]}\ne \varphi$ for all $i\in \{0,\dots ,r-1\}$.

4. Metric Bases Consisting of Elements from All Classes

Definition 2.

For two distinct integers $a,b$ with $0\le a,b\le r-1$, by an intermediate class between $\left[a\right]$ and $\left[b\right]$, we mean any one of the classes $[a+1],[a+2],\dots ,[b-1]$ if $a<b$ and for $a>b$, we mean it by any class of $[a+1],[a+2],\dots ,[r-1],\left[0\right],\left[1\right],\dots ,[b-1]$. We denote the set of all intermediate classes between $\left[a\right]$ and $\left[b\right]$ by $I_{(a,b)}$. We also define $I_{[a,b]}$ = $I_{(a,b)}\cup \{\left[a\right],\left[b\right]\}$, $I_{[a,b)}$ = $I_{(a,b)}\cup \left\{\left[a\right]\right\}$ and $I_{(a,b]}$ = $I_{(a,b)}\cup \left\{\left[b\right]\right\}$. For better understanding, see Figure 1.

Definition 3.

For two vertices $v_a$ and $v_{a+\ell }$ in $P_n^r$ with $1\le \ell \le r-1$, we define $W_{a,a+\ell }^L$ to be the set of all left-side vertices of $v_a$ that resolve the vertices $v_a$ and $v_{a+\ell }$. Similarly, $W_{a,a+\ell }^R$ is the set of all right-side vertices of $v_{a+\ell }$ that resolve the vertices $v_a$ and $v_{a+\ell }$.

Example 2.

We consider $P_{20}^5$ ($r=5$, $n=20$). Then it is clear that $v_2$ resolves the pair $(v_{12},v_{15})$ as $d(v_2,v_{12})=2$ and $d(v_2,v_{15})=3$. Therefore, $v_2\in W_{12,15}^L$. Again, $v_{18}\in W_{12,15}^R$ as $d(v_{18},v_{12})=2$ and $d(v_{18},v_{15})=1$.

The lemma below gives explicit forms for both $W_{a,a+\ell }^L$ and $W_{a,a+\ell }^R$ for any two vertices $v_a$ and $v_{a+\ell }$ of $P_n^r$.

Lemma 6.

For any two vertices $v_a$ and $v_{a+\ell }$ with $1\le \ell \le r-1$, the following hold:

(a) 

$W_{a,a+\ell }^L=\left\{v_j:j\in \{\left[a\right],[a+1],\dots ,[a+\ell -1]\right\}$,

(b) 

$W_{a,a+\ell }^R=\left\{v_j:j\in \{[a+1],[a+2],\dots ,[a+\ell ]\right\}$.

Proof. 

(a)

Let $v_j\in V\left(P_n^r\right)$ with $j\le a$, that is, $v_j$ is a left-side vertex of $v_a$. Then, the distances of the two vertices $v_a$ and $v_{a+\ell }$ from $v_j$ are $\lceil \frac{a-j}{r}\rceil$ and $\lceil \frac{a-j+\ell }{r}\rceil$, respectively. Let $a-j=mr+r^{\prime }$, where $0\le r^{\prime }\le r-1$. Then, $\lceil \frac{a-j}{r}\rceil \ne \lceil \frac{a-j+\ell }{r}\rceil$ implies that $\lceil \frac{mr+r^{\prime }}{r}\rceil \ne \lceil \frac{mr+r^{\prime }+\ell }{r}\rceil$ and this is true only when $r^{\prime }=0$ or $r+1\le r^{\prime }+\ell$, that is, $r^{\prime }\in \{0,r+1-\ell ,r+2-\ell ,\dots ,r-1\}=S$ (say). Therefore,

$$\begin{array}{ccc}\hfill W_{a,a+\ell }^L& =& \{v_j:j\le a\mathrm{and}a-j\equiv r^{\prime }(modr),r^{\prime }\in S\}\hfill \\ & =& \left\{v_j:j\le a\mathrm{and}j\in \{\left[a\right],[a+1],\dots ,[a+\ell -1]\}\right\}.\hfill \end{array}$$

(b)

Let $v_j\in V\left(P_n^r\right)$ with $j\ge a+\ell$, that is, $v_j$ is a right-side vertex of $v_{a+\ell }$. Then, $d(v_j,v_a)=\lceil \frac{j-a}{r}\rceil$ and $d(v_j,v_{a+\ell })=\lceil \frac{j-a-\ell }{r}\rceil$. Let $j-a-\ell =mr+r^{\prime }$, where $0\le r^{\prime }\le r-1$. Then $\lceil \frac{a-j}{r}\rceil \ne \lceil \frac{a-j+\ell }{r}\rceil$ implies $\lceil \frac{mr+r^{\prime }}{r}\rceil \ne \lceil \frac{mr+r^{\prime }+\ell }{r}\rceil$ and this is true only when $r^{\prime }\in \{0,r+1-\ell ,r+2-\ell ,\dots ,r-1\}=S$ (say). Therefore,

$$\begin{array}{ccc}\hfill W_{a,a+\ell }^R& =& \{v_j:j\ge a+\ell \mathrm{and}j-a-l\equiv r^{\prime }(modr),r^{\prime }\in S\}\hfill \\ & =& \left\{v_j:j\ge a+\ell \mathrm{and}j\in \{[a+1],[a+2],\dots ,[a+\ell ]\}\right\}.\hfill \end{array}$$

 □

Example 3.

We consider $P_{29}^5$. We now calculate $W_{12,15}^L$ and $W_{12,15}^R$. Since $\left[12\right]=\left[2\right]$ and $\left[15\right]=\left[0\right]$, $W_{12,15}^L$ and $W_{12,15}^R$ are given by:

$$\begin{array}{ccc}\hfill W_{12,15}^L& =& \{v_j:j\le 12\mathit{and}j\in \{\left[12\right],\left[13\right],\left[14\right]\}\}\hfill \\ & =& \{v_j:j\le 12\mathit{and}j\in \{\left[2\right],\left[3\right],\left[4\right]\}\}\hfill \\ & =& \{v_2,v_7,v_{12},v_3,v_8,v_4,v_9\},\hfill \\ \hfill W_{12,15}^R& =& \{v_j:j\ge 15\mathit{and}j\in \{\left[13\right],\left[14\right],\left[15\right]\}\}\hfill \\ & =& \{v_j:j\ge 15\mathit{and}j\in \{\left[3\right],\left[4\right],\left[0\right]\}\}\hfill \\ & =& \{v_{15},v_{20},v_{25},v_{18},v_{23},v_{28},v_{19},v_{24}\}.\hfill \end{array}$$

Corollary 1.

For $\ell =1$, Lemma 6 implies that $W_{a,a+1}^L=\{v_j:j\le a\mathit{and}j\in \left[a\right]\}$ and $W_{a,a+1}^R=\{v_j:j\ge a+1\mathit{and}j\in [a+1]\}$. For example, in $P_{29}^5$, $W_{12,13}^L=\{v_j:j\le 12\mathit{and}j\in \left[12\right]\}=\{v_j:j\le 12\mathit{and}j\in \left[2\right]\}=\{v_2,v_7,v_{12}\}$.

Remark 2.

If $\mathcal{B}$ is a metric basis such that $\mathcal{B}\cap S_{\left[a\right]}=\varphi$ and $\mathcal{B}$ does not contains any $(a+1)$-class element in the right side of $v_{a+1}$, then $v_{a+1}\in \mathcal{B}$. Similarly, if a metric basis $\mathcal{B}$ with $\mathcal{B}\cap S_{[a+1]}=\varphi$ and $\mathcal{B}$ does not contain any a-class element in the left side of $v_a$, then $v_a\in \mathcal{B}$.

Example 4.

If $\mathcal{B}$ be a matric basis of $P_{25}^5$ such that $\mathcal{B}\cap S_{\left[0\right]}=\varphi$, then $v_{24}\in \mathcal{B}$, as there is no 24-class element, that is, a 4-class element in the right side of $v_{24}$.

Using Lemma 6, we have the following corollaries.

Corollary 2.

For $a\equiv \alpha (modr)$, we have $\left(a\right)W_{a,a+\ell }^L=I_{[\alpha ,\ell )}$ and $\left(b\right)W_{a,a+\ell }^R=I_{(\alpha ,\ell ]}$.

Corollary 3.

For $a+\ell \equiv 0(modr)$, we have $\left(a\right)W_{a,a+\ell }^L=I_{[r-\ell ,r)}$ and $\left(b\right)W_{a,a+\ell }^R=I_{(r-\ell ,r]}$.

Lemma 7.

Let a and m be two positive integers such that $0<a<a+m\le r-1$. Additionally, let $\mathcal{B}$ be a metric basis for $P_n^r$ with $|\mathcal{B}\cap S_{\left[t\right]}|\le 1$ for all $t\in \{a,\dots ,a+m\}$. Then, the following hold:

(a) 

If $\mathcal{B}\cap S_{[a+m]}=\varphi$, then $\{v_a,\dots ,v_{a+m-1}\}\subset \mathcal{B}$. Moreover, if $a\ge 2$ then $v_{a-1}\in \mathcal{B}$.

(b) 

If $\mathcal{B}\cap S_{\left[a\right]}=\varphi$, then $\mathcal{B}$ must contain the largest $s\text{-}$class elements for all $s\in \{a+1,\dots ,a+m\}$. Moreover, if $\mathcal{B}\cap S_{[a+m+1]}\ne \varphi$, then $\mathcal{B}$ must contain the largest $(a+m+1)\text{-}class$ element.

(c) 

Suppose that $|\mathcal{B}\cap S_{\left[s\right]}|=1$ for $s\in \{r-1,r-2,\dots ,t\}$. If $\mathcal{B}\cap S_{\left[0\right]}=\varphi$, then $\{v_t,v_{t+1},\dots ,v_{r-1}\}\subset \mathcal{B}$.

(d) 

Suppose that $|\mathcal{B}\cap S_{\left[s\right]}|=1$ for $s\in \{0,r-1,r-2,\dots ,t\}$. If $\mathcal{B}\cap S_{\left[0\right]}=\left\{v_0\right\}$, then $\{v_t,v_{t+1},\dots ,v_{r-1}\}\subset \mathcal{B}$.

Proof. 

(a)

Since $\mathcal{B}\cap S_{[a+m]}=\varphi$ and there are no $(a+m-1)\text{-}class$ elements of $V\left(P_n^r\right)$ in the left side of $v_{a+m-1}$, applying Remark 2, we have $v_{a+m-1}\in \mathcal{B}$. Now we have to prove that $\{v_a,\dots ,v_{a+m-1}\}\subset \mathcal{B}$. Assume to the contrary that there are at least one of $v_a,\dots ,v_{a+m-2}$ not in $\mathcal{B}$. Let s be the largest positive integer such that $v_{a+s}\notin \mathcal{B}$, where $0\le s\le m-2$. Then, all $v_{a+j}\in \mathcal{B}$ for all $j\in \{s+1,\dots ,m-1\}$. Applying Lemma 6 to $v_{a+s}$ and $v_{a+m}$, we have:

$$\begin{array}{cc}\hfill & W_{a+s,a+m}^L=\left\{v_j:\left[j\right]\in \{[a+s],\dots ,[a+m-1]\}\right\}\hfill \\ \hfill \mathrm{and}& W_{a+s,a+m}^R=\left\{v_j:\left[j\right]\in \{[a+s+1],\dots ,[a+m]\}\right\}.\hfill \end{array}$$

However, $S=\{v_{a+j}:s+1\le j\le a+m-1\}\subset \mathcal{B}$ cannot resolve $v_{a+s}$ and $v_{a+m}$ as all the vertices of $S\cup \{v_{a+s},v_{a+m}\}$ are in a same clique and $|\mathcal{B}\cap S_{\left[j\right]}|=1$ for all $j\in \{a+s+1,\dots ,a+m-1\}$. Therefore, $v_{a+s}$ and $v_{a+m}$ will be resolved by $\mathcal{B}$ only if $v_{a+s}\in \mathcal{B}$ or $v_{a+m}\in \mathcal{B}$, which is a contradiction. Hence, $\{v_a,\dots ,v_{a+m-1}\}\subset \mathcal{B}$. Since there is no $(a-1)\text{-}class$ element to the left of $v_{a-1}$ and $\mathcal{B}\cap S_{\left[a\right]}=\left\{v_a\right\}$, applying Corollary 1 to $v_a$ and $v_{a+r}$ and by a similar argument as above, one can easily see that $v_{a-1}\in \mathcal{B}$.

(b)

Let $n\equiv \ell (modr)$. Here the last block contains elements from each class of $\left[0\right],\dots ,[\ell -1]$. Now we consider three cases according to $\ell >a+m$, $a<\ell <a+m$ and $\ell <a$.

Case I.$\ell >a+m$. In this case the last block $B_{\lceil \frac{n}{r}\rceil -1}$ contains the largest $s\text{-}$class element for each $s\in \{a,\dots ,a+m\}$. Since $B\cap S_{\left[a\right]}=\varphi$ and there is no $(a+1)\text{-}class$ element of $V\left(P_n^r\right)$ in the right side of $v_{n-\ell +a+1}$, applying Remark 2, we have $v_{n-\ell +a+1}\subset \mathcal{B}$. Now we have to prove that $\{v_{n-\ell +a+1},\dots ,v_{n-\ell +a+m}\}\subset \mathcal{B}$. Assume to the contrary that there is at least one of $v_{n-\ell +a+1},\dots ,v_{n-\ell +a+m}$ not in $\mathcal{B}$. Let s be the smallest positive integer such that $v_{n-\ell +a+s}\notin \mathcal{B}$, where $2\le s\le m$. Then, all $v_{n-\ell +a+j}\in \mathcal{B}$ for all $j\in \{1,\dots ,s-1\}$. Applying Lemma 6 to $v_{n-\ell +a}$ and $v_{n-\ell +a+s}$, we have $W_{n-\ell +a,n-\ell +a+s}^L=\left\{v_j:\left[j\right]\in \{\left[a\right],\dots ,[a+s-1]\}\right\}$ and $W_{n-\ell +a,n-\ell +a+s}^R=\left\{v_j:\left[j\right]\in \{[a+1],\dots ,[a+s]\}\right\}$. However, $S=\{v_{n-\ell +a+j}:1\le j\le s-1\}\subset \mathcal{B}$ cannot resolve $v_{n-\ell +a}$ and $v_{n-\ell +a+s}$, as all of the vertices of $S\cup \{v_{n-\ell +a},v_{n-\ell +a+s}\}$ are in a same clique and $|\mathcal{B}\cap S_{\left[j\right]}|=1$ for all $j\in \{a+1,\dots ,a+s-1\}$. Therefore, $v_{n-\ell +a}$ and $v_{n-\ell +a+s}$ will be resolved by $\mathcal{B}$ only if $v_{n-\ell +a}\in \mathcal{B}$ or $v_{n-\ell +a+s}\in \mathcal{B}$, which is a contradiction. Hence, $\{v_{n-\ell +a+1},\dots ,v_{n-\ell +a+m}\}\subset \mathcal{B}$.

Case II.$a<\ell <a+m$. Here the last block $B_{\lceil \frac{n}{r}\rceil -1}$ contains the largest $s\text{-}$class element for all $s\in \{a,\dots ,\ell -1\}$ and the second-last block $B_{\lceil \frac{n}{r}\rceil -2}$ contains the largest $s$-class element for all $s\in \{\ell ,\dots ,a+m\}$. By a similar argument as in Case I, one can easily prove that $\{v_{n-\ell +a+1},\dots ,v_{n-1}\}\subset \mathcal{B}$. Now we have to prove that $\{v_{n-r},\dots ,v_{n-\ell -r+a+m}\}\subset \mathcal{B}$. The two vertices $v_{n-r-1}$ and $v_{n-r}$ cannot be resolved by $\left\{v_{n-1}\right\}$, as these three vertices are on the same clique. Again, since $\mathcal{B}\cap S_{[\ell -1]}=\left\{v_{n-1}\right\}$ and $|\mathcal{B}\cap S_{[\ell -1]}|=1$, applying Corollary 1 to $v_{n-r-1},v_{n-r}$ we have only $v_{n-r}$ or a $\ell \text{-}class$ element to right the of $v_{n-r}$ that can resolve $v_{n-r-1}$ and $v_{n-r}$. However, there is no $\ell \text{-}class$ element to the right of $v_{n-r}$. Thus, $v_{n-r}\in \mathcal{B}$. Assume to the contrary that at least one of $v_{n-r+1},\dots ,v_{n-\ell -r+a+m}$ is not in $\mathcal{B}$. Let s be the smallest positive integer such that $v_{n-\ell -r+s}\notin \mathcal{B}$, where $\ell +1\le s\le a+m$. Then, all $v_{n-\ell -r+j}\in \mathcal{B}$ for all $j\in \{\ell +1,\dots ,s-1\}$. Applying Lemma 6 to $v_{n-r-1}$ and $v_{n-\ell -r+s}$, we have $W_{n-r-1,n-\ell -r+s}^L=\left\{v_j:\left[j\right]\in \{[\ell -1],\dots ,[s-1]\}\right\}$ and $W_{n-r-1,n-\ell -r+s}^R=\left\{v_j:\left[j\right]\in \{[\ell ],\dots ,\left[s\right]\}\right\}$. However, $S=\{v_{n-\ell -r+j}:\ell \le j\le s-1\}\cup \left\{v_{n-1}\right\}\subset \mathcal{B}$ cannot resolve $v_{n-r-1}$ and $v_{n-\ell -r+s}$, as all of these vertices are in the same clique and $|\mathcal{B}\cap S_{\left[j\right]}|=1$ for all $j\in \{\ell -1,\dots ,s\}$. Therefore, $\mathcal{B}$ cannot resolve $v_{n-r-1}$ and $v_{n-\ell -r+s}$, which is a contradiction. Hence, $\{v_{n-r},\dots ,v_{n-\ell -r+a+m}\}\subset \mathcal{B}$.

Case III.$\ell <a$. For all $s\in \{a,\dots ,a+m\}$, the largest s-class element is in block $B_{\lceil \frac{n}{r}\rceil -1}$ and by a similar argument as in Case I, one can easily prove that $\{v_{n-\ell -r+a},v_{n-\ell -r+a+1},\dots ,v_{n-\ell -r+a+j}\}\subset \mathcal{B}$.

Now we have to prove that if $\mathcal{B}\cap S_{[a+m+1]}\ne \varphi$, and then $\mathcal{B}$ must contain the largest $(a+m+1)$-class element. Since $|\mathcal{B}\cap S_{[a+m]}|\le 1$, the largest $(a+m+1)$-class element and the largest $(a+m)$-class element are resolved only by an $(a+m+1)$-class element, which is not to the left of the largest $(a+m)$-class element. Again, there is no $(a+m+1)$-class element to the right of the largest $(a+m)$-class element. Therefore, the largest $(a+m+1)$-class element must be in $\mathcal{B}$.

(c)

As $\mathcal{B}\cap S_{\left[0\right]}=\varphi$, applying Corollary 1 to vertices $v_{r-1}$ and $v_r$, we have only $v_{r-1}$ or an $(r-1)$-class element to the left of $v_{r-1}$ that can resolve $v_{r-1}$ and $v_r$. However, there is no $(r-1)$-class elements to the left of $v_{r-1}$. Thus, $v_{r-1}\in \mathcal{B}$. Now we have to prove that $\{v_t,v_{t+1},\dots ,v_{r-1}\}\subset \mathcal{B}$. Assume to the contrary that there is at least one of $v_t,\dots ,v_{r-1}$ not in $\mathcal{B}$. Let s be the smallest positive integer such that $v_{r-1-s}\notin \mathcal{B}$. Then, all $v_{a+j}\in \mathcal{B}$ for all $j\in \{s+1,\dots ,m-1\}$. Applying Lemma 6 to the vertices $v_{r-1-s}$ and $v_r$, we have $W_{r-1-s,r}^L=\left\{v_j:\left[j\right]\in \{[r-1-s],\dots ,[r-1]\}\right\}$ and $W_{r-1-s,r}^R=\left\{v_j:\left[j\right]\in \{[r-s],\dots ,[r-1],\left[0\right]\}\right\}$. However, $S=\{v_j:r-s\le j\le r-1\}$ cannot resolve $v_{r-1-s}$ and $v_r$ as all of the vertices of $S\cup \{v_{r-1-s},v_r\}$ are in a same clique and $|\mathcal{B}\cap S_{\left[j\right]}|=1$ for all $j\in \{r-s,\dots ,r-1\}$. Therefore, $v_{r-1-s}$ and $v_r$ will be resolved by $\mathcal{B}$ only if $v_{r-1-s}\in \mathcal{B}$ or $v_r\in \mathcal{B}$, which is a contradiction. Hence, $\{v_t,v_{t+1},\dots ,v_{r-1}\}\subset \mathcal{B}$.

(d)

Similar proof as of (c).

 □

Lemma 8.

Let $\mathcal{B}$ be a basis of $P_n^r$ such that $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{0,1,\dots ,r-1\}$. In addition, let $\mathcal{B}\cap S_{\left[0\right]}=\left\{v_{jr}\right\}$ and ℓ be a positive integer such that $1\le \ell \le r-1$.

(a) 

For $s=1,2$, $S_{[r-s]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$.

(b) 

If $S_{[r-s]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$ for all $s\in \{1,\dots ,\ell \}$, then $S_{[r-\ell -1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$.

Proof. 

(a)

First we have to prove that $S_{[r-1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$. By Corollary 1, for the two vertices $v_{jr+r-1}$ and $v_{jr+r}$, we have:

$$W_{jr+r-1,jr+r}^L=\left\{v_j:\left[j\right]\in \left\{[r-1]\right\}\right\}\mathrm{and}{W}_{jr+r-1,jr+r}^R=\left\{v_j:\left[j\right]\in \left\{\left[0\right]\right\}\right\}.$$

However, $\mathcal{B}$ is already taken in the 0-class element $v_{jr}$ and $|\mathcal{B}\cap S_{\left[0\right]}|=1$, so to resolve $v_{jr+r-1},v_{jr+r}$, we have to take an element from $W_{jr+r-1,jr+r}^L$, and hence $S_{[r-1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$.

We now prove that $S_{[r-2]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$. Since $S_{[r-1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$, we have either $S_{[r-1]}\cap \mathcal{B}\in {\cup }_{t=0}^{j-1}{B}_t$ or $S_{[r-1]}\cap \mathcal{B}=\left\{v_{jr+r-1}\right\}$. If $S_{[r-1]}\cap \mathcal{B}=\{v_{jr+r-1}$}, then the 0-class element $v_{jr}$ and the $(r-1)$-class element $v_{jr+r-1}$ cannot resolve $v_{jr+r-2},v_{jr+r}$ due to the fact that these four vertices form a clique. By Corollary 3 to the vertices $u=v_{jr+r-2}$ and $w=v_{jr+r}$, the set of all vertices that can resolve u and w are $W_{jr+r-1,jr+r}^L\cup W_{jr+r-1,jr+r}^R$. Thus, $S_{[r-2]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$. Otherwise, $S_{[r-1]}\cap \mathcal{B}\in {\cup }_{t=0}^{j-1}{B}_t$. Then, by a similar argument to the one above, the two vertices $u=v_{jr+r-2}$ and $w=v_{jr+r-1}$ can only be resolved by the $(r-2)$-class element u and $u\in R_{jr+r-2,jr+r-1}^L$. Therefore, $S_{[r-2]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$.

(b)

We have $S_{[r-j]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$ for all $j\in \{1,\dots ,\ell \}$. Here we consider two cases.

Case I.$S_{[r-i]}\cap \mathcal{B}=\left\{v_{jr+r-t}\right\}$for all$i\in \{1,\dots ,\ell \}$. Then, applying Corollary 3 to the vertices $u=v_{jr+r}$ and $w=v_{jr+r-\ell -1}$, we have $W_{jr+r-\ell -1,jr+r}^L=\left\{v_j:\left[j\right]\in \{[r-\ell -1],\dots ,[r-1]\}\right\}$ and $W_{jr+r-\ell -1,jr+r-t}^R=\left\{v_j:\left[j\right]\in \{[r-\ell ],\dots ,[r-1],\left[0\right]\}\right\}$. Since $|\mathcal{B}\cap S_{\left[i\right]}|=1$ for all $i\in \{0,1,\dots ,r-1\}$, and the two vertices $u,w$ are not resolved by $\{v_{jr+r-t}:1\le t\le \ell \}$ as $\{u,v\}\cup \{v_{jr+r-t}:1\le t\le \ell \}$ forms a clique. Therefore, either $v_{jr+r-\ell -1}\in \mathcal{B}$, or an $(r-\ell -1)$-class element that is to the left of $v_{jr+r-\ell -1}$ must be in $\mathcal{B}$. Thus, $S_{[r-\ell -1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$.

Case II.$S_{[r-i]}\cap \mathcal{B}\ne \left\{v_{jr+r-i}\right\}$for at least one$i\in \{1,\dots ,\ell \}$. Let m be the smallest positive integer such that $S_{[r-m]}\cap \mathcal{B}\in {\cup }_{t=0}^{j-1}{B}_t$. Then, for each $s\in \{t+1,\dots ,\ell \}$, $S_{[r-s]}\cap \mathcal{B}\subset B_j$. Now we consider the two vertices $u=v_{jr+r-t}$ and $w=v_{jr+r-\ell -1}$. For each $s\in \{m+1,\dots ,\ell \}$, these two vertices are resolved by the $(r-s)$-class elements of $\mathcal{B}$. By Corollary 3 to the vertices u and w, we have $W_{jr+r-\ell -1,jr+r-t}^L=\left\{v_j:\left[j\right]\in \{[r-\ell -1],\dots ,[r-t-1]\}\right\}$ and $W_{jr+r-\ell -1,jr+r-t}^R=\left\{v_j:\left[j\right]\in \{[r-\ell ],\dots ,[r-t]\}\right\}$. Since the $(r-m)$-class elements of $\mathcal{B}$ are to the left of w and for each $s\in \{m+1,\dots ,\ell \}$, $[r-s]$ cannot resolve u and w, we have from $W_{jr+r-\ell -1,jr+r-t}^L$ and $W_{jr+r-\ell -1,jr+r-t}^R$ that the $(r-\ell -1)$-class element is needed to resolve u and w and this class element must be to the left of w. Therefore, $S_{[r-\ell -1]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$ whenever $S_{[r-j]}\cap \mathcal{B}\in {\cup }_{t=0}^j{B}_t$ for all $j\in \{1,2,\dots ,\ell \}$.

 □

Lemma 9.

Let $\mathcal{B}$ be a basis of $P_n^r$ such that $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{0,1,\dots ,r-1\}$ and $v_{jr}\in \mathcal{B}$.

(a) 

For $s=1,2$, $S_{\left[s\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$.

(b) 

If $S_{\left[s\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$ for all $s\in \{1,\dots ,\ell -1\}$, then $S_{[\ell ]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$.

Proof. 

(a)

First we prove that $S_{\left[1\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$. Applying Corollary 1 to the two vertices $v_{(j-1)r}$ and $v_{(j-1)r+1}$, we have $W_{(j-1)r,(j-1)r+1}^L=\left\{v_j:\left[j\right]\in \left\{\left[0\right]\right\}\right\}$ and $W_{(j-1)r,(j-1)r+1}^R$ = $\left\{v_j:\left[j\right]\in \left\{\left[1\right]\right\}\right\}$. However, $\mathcal{B}$ has already taken the 0-class element $v_{jr}$ and $|\mathcal{B}\cap S_{\left[0\right]}|=1$ so to resolve $v_{(j-1)r}$ and $v_{(j-1)r+1}$, we have to take an element from $W_{(j-1)r,(j-1)r+1}^R$. Thus, $S_{\left[1\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$. We now prove that $S_{\left[2\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$. Since $S_{\left[1\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$, we have either $S_{\left[1\right]}\cap \mathcal{B}=\left\{v_{(j-1)r+1}\right\}$ or $S_{\left[1\right]}\cap \mathcal{B}\in {\cup }_{t=j}^{\lceil \frac{n}{r}\rceil }{B}_t$. If $S_{\left[1\right]}\cap \mathcal{B}=\left\{v_{(j-1)r+1}\right\}$, then the $0\text{-}class$ element $v_{jr}$ and $1\text{-}class$ element $v_{(j-1)r+1}$ cannot resolve the two vertices $u=v_{(j-1)r}$ and $w=v_{(j-1)r+2}$ due to the fact that all four are in the same clique. Again, by Corollary 2, the two vertices $u=v_{(j-1)r}$ and $w=v_{(j-1)r+2}$ can only be resolved by $\left\{v_{(j-1)r+2}\right\}$ or a 2-class element that is on the right of $\left\{v_{(j-1)r+2}\right\}$. Therefore, $S_{\left[2\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$. Otherwise, $S_{\left[1\right]}\cap \mathcal{B}\in {\cup }_{t=j}^{\lceil \frac{n}{r}\rceil }{B}_t$. Then, by a similar argument as above, the two vertices $u=v_{(j-1)r+1}$ and $w=v_{(j-1)r+2}$ can be resolved by either $\left\{v_{(j-1)r+2}\right\}$ or a 2-class element that is on the right of $\left\{v_{(j-1)r+2}\right\}$. Therefore, $S_{\left[2\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$.

(b)

We have $S_{\left[j\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$ for all $j\in \{1,\dots ,\ell -1\}$. Here we consider the two following cases.

Case I.$S_{\left[i\right]}\cap \mathcal{B}=\left\{v_{(j-1)r+i}\right\}$for all$i\in \{1,\dots ,\ell -1\}$. Then, applying Corollary 2 to the vertices $u=v_{(j-1)r}$ and $w=v_{(j-1)r+\ell }$, we have $W_{(j-1)r,(j-1)r+\ell }^L=\left\{v_j:\left[j\right]\in \{\left[0\right],\dots ,[\ell -1]\}\right\}$ and $W_{(j-1)r,(j-1)r+\ell }^R=\left\{v_j:\left[j\right]\in \{\left[1\right],\dots ,[\ell ]\}\right\}$. Since the two vertices $u,w$ are not resolved by $\{v_{(j-1)r+t}:1\le \ell -1\}$, therefore either $\left\{v_{(j-1)r+\ell }\right\}$ or an ℓ-class element to the right of $\left\{v_{(j-1)r+\ell }\right\}$ can resolve $u,w$ and $S_{[\ell ]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$.

Case II.$S_{\left[i\right]}\cap \mathcal{B}\ne \left\{v_{(j-1)r+i}\right\}$for at least one$i\in \{1,\dots ,\ell -1\}$. Let m be the largest positive integer such that $S_{\left[m\right]}\cap \mathcal{B}\in {\cup }_{t=j}^{\lceil \frac{n}{r}\rceil }{B}_t$. Then, $S_{\left[s\right]}\cap \mathcal{B}=\left\{v_{(j-1)r+s}\right\}$ for all $s\in \{m+1,\dots ,\ell -1\}$. Now we consider the two vertices $u=v_{(j-1)r}$ and $w=v_{(j-1)r+\ell }$. Then, for each $s\in \{m+1,\dots ,\ell -1\}$, the two vertices $u,w$ are not resolved by an s-class element of $\mathcal{B}$. Applying Corollary 2 to u and w, we have $W_{(j-1)r+m,(j-1)r+\ell }^L=\left\{v_j:\left[j\right]\in \{\left[m\right],\dots ,[\ell -1]\}\right\}$ and $W_{(j-1)r+m,(j-1)r+\ell }^R=\left\{v_j:\left[j\right]\in \{[m+\ell ],\dots ,\left[1\right]\}\right\}$. Therefore, $u,w$ are resolved by either $\left\{v_{(j-1)r+\ell }\right\}$ or an ℓ-class element to the right of $\left\{v_{(j-1)r+\ell }\right\}$ and $S_{[\ell ]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$ whenever $S_{\left[j\right]}\cap \mathcal{B}\in {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$ for all $j\in \{1,\dots ,\ell -1\}$.

 □

Theorem 2.

Let $\mathcal{B}$ be a metric basis of $P_n^r$ with $|\mathcal{B}\cap S_{\left[i\right]}|=1$ for all $i\in \{0,\dots ,r-1\}$. Then, $\mathcal{B}=\{v_0,v_1,\dots ,v_{r-1}\}$ or $\{v_{n-r},v_{n-r+1},\dots ,v_{n-1}\}$.

Proof. 

Since $|\mathcal{B}\cap S_{\left[i\right]}|=1$ for all $i\in \{0,\dots ,r-1\}$, we can assume that $v_{jr}\in \mathcal{B}$ for some j, $0\le j\le \lceil \frac{n}{r}\rceil -1$. Then, applying Lemmas 8 and 9, we have $\mathcal{B}\subset {\cup }_{t=0}^j{B}_t$ and $\mathcal{B}\subset {\cup }_{t=j-1}^{\lceil \frac{n}{r}\rceil }{B}_t$. Therefore, $\mathcal{B}\in B_{j-1}\cup B_j=\{v_{(j-1)r},\dots ,v_{jr+r-1}\}$. Now our aim is to prove that the elements of $\mathcal{B}$ are consecutive vertices of $P_n^r$. If $\mathcal{B}=B_{j-1}$ or $\mathcal{B}=B_j$ then we have nothing to prove. Again, since $\left|\mathcal{B}\right|=r$, $\mathcal{B}$ does not contain all of the vertices in $B_{j-1}\cup B_j$. Let $u=v_{(j-1)r+\alpha }$ be the last vertex in $B_{j-1}$ and $w=v_{jr+\beta }$ be the first vertex in $B_j$ such that both vertices are not in $\mathcal{B}$. Then, $\{v_{(j-1)r+\alpha +1},\dots ,v_{(j-1)r+r-1},v_{jr},\dots ,v_{jr+\beta -1}\}\subset \mathcal{B}$ and in this case the condition $\beta =\alpha +1$ is sufficient to show that basis elements are consecutive. Since $v_{(j-1)r+\alpha +1}\in \mathcal{B}$ and $\mathcal{B}$ contains each class element exactly once, $v_{jr+\alpha +1}\notin \mathcal{B}$ and hence $\beta \le \alpha +1$. Assume to the contrary that $\beta \le \alpha$. Then, applying Corollary 3 to vertices u and w, we have $W_{(j-1)r+\alpha ,jr+\beta }^L=\{v_j:\left[j\right]\in I_{[\alpha ,\beta )}\}$ and $W_{(j-1)r+\alpha ,jr+\beta }^R=\{v_j:\left[j\right]\in I_{(\alpha ,\beta ]}\}$. However, $\left({\cup }_{\left[t\right]\in I_{(\alpha ,\beta )}}{S}_{\left[t\right]}\right)\cap \mathcal{B}=\{v_{(j-1)r+\alpha +1},\dots ,v_{(j-1)r+r-1},v_{jr},\dots ,v_{jr+\beta -1}\}$ cannot resolve u and w. Therefore, there is an element $x\in S_{\left[\alpha \right]}$ to the left of u or an element $x\in S_{\left[\beta \right]}$ that is to the right of w, and only they can resolve u and w. However, there is no such element in $\mathcal{B}$, as $\mathcal{B}\subset B_{j-1}\cup B_j$ and $u,w\notin \mathcal{B}$, which contradicts the fact that $\mathcal{B}$ is a metric basis. Therefore, $\beta =\alpha +1$, and hence basis elements are consecutive, that is, $\mathcal{B}=\{v_a,v_{a+1},\dots ,v_{a+r-1}\}$ for some a, $0\le a\le n-r$. If $a>0$ or $a+r<n$, then $d(u,v_{a-1})=d(u,v_{a+r})=1$ for all $u\in \mathcal{B}$. Therefore, the codes of $v_{a-1}$ and $v_{a+r}$ are the same. Thus, either $a=0$ or $a+r=n$, but both cannot occur simultaneously (except for $n=r$ and in this case we obtain a complete graph) as basis elements that are consecutive. Therefore, $\mathcal{B}=\{v_0,v_1,\dots ,v_{r-1}\}$ or $\mathcal{B}=\{v_{n-r},\dots ,v_{n-1}\}$ according to $a=0$ or $a=n-r$, respectively. By Lemmas 4 and 5, we conclude that $\mathcal{B}$ forms a resolving set of $P_n^r$. □

5. Metric Bases with Missing Classes

In this section, we determine all metric bases of $P_n^r$ that miss one or more classes. Throughout this section we assume that a metric basis $\mathcal{B}$ misses the classes $\left[{\alpha }_1\right],\left[{\alpha }_2\right],\dots ,\left[{\alpha }_k\right]$. Additionally, we denote $\mathcal{B}({\alpha }_1,\dots ,{\alpha }_k)$ as a metric basis of $P_n^r$ with missing classes $\left[{\alpha }_1\right],\left[{\alpha }_2\right],\dots ,\left[{\alpha }_k\right]$, where ${\alpha }_1<\cdots <{\alpha }_k$. A natural question arises: can any metric basis miss two consecutive classes? In the following lemma, we give an answer to this question and also determine the classes for which a metric basis consists of at least two elements from these classes.

Lemma 10.

Let $\mathcal{B}({\alpha }_1,\dots ,{\alpha }_k)$ be a metric basis for $P_n^r$ with missing classes ${\alpha }_1,\dots ,{\alpha }_k$, and let $S_{\left[{\beta }_i\right]}$ denote the set of all ${\beta }_i\text{-}class$ elements of $P_n^r.$ Then:

(a) 

${\alpha }_{\ell +1}-{\alpha }_{\ell }\ge 2$ for $\ell \in \{1,\dots ,k-1\}$, provided that $k\ge 2$;

(b) 

There exists only one class $\left[{\beta }_t\right]\in I_{({\alpha }_t,{\alpha }_{t+1})}$ such that $|\mathcal{B}\cap S_{\left[{\beta }_t\right]}|=2$;

(c) 

There exists only one class $\left[{\beta }_k\right]\in I_{({\alpha }_k,{\alpha }_1)}$ such that $|\mathcal{B}\cap S_{\left[{\beta }_k\right]}|=2$.

Proof. 

Since $\mathcal{B}({\alpha }_1,\dots ,{\alpha }_k)$ is a metric basis for $P_n^r$ with missing classes ${\alpha }_1,\dots ,{\alpha }_k$, $\mathcal{B}\cap S_{\left[{\alpha }_{\ell }\right]}=\varphi$ for all $\ell \in \{1,\dots ,k\}$, where $S_{\left[{\alpha }_{\ell }\right]}$ denotes the set of all ${\alpha }_{\ell }\text{-}class$ elements of $P_n^r.$

(a)

Assume to the contrary that there exists a positive integer t such that $\mathcal{B}\cap S_{\left[{\alpha }_t\right]}=\varphi$ and $\mathcal{B}\cap S_{[{\alpha }_t+1]}=\varphi$. Then, the codes of $v_{{\alpha }_t}$ and $v_{{\alpha }_{t+1}}$ are the same due to Corollary 1, which contradicts the fact that $\mathcal{B}$ is a metric basis. Again, assume to the contrary that $\mathcal{B}\cap S_{[r-1]}=\varphi$ and $\mathcal{B}\cap S_{\left[0\right]}=\varphi$. Again, by Corollary 1, codes of $v_{r-1}$ and $v_r$ are the same and we arrive at a contradiction. Hence, $\mathcal{B}$ cannot miss two consecutive classes, that is, ${\alpha }_{\ell +1}-{\alpha }_{\ell }\ge 2$.

(b)

Assume to the contrary that there exists no $\left[{\beta }_i\right]\in I_{({\alpha }_i,{\alpha }_{i+1})}$ such that $|\mathcal{B}\cap S_{\left[{\beta }_i\right]}|=2$. Then, $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{{\alpha }_i+1,\dots ,{\alpha }_{i+1}-1\}$ and $|\mathcal{B}\cap S_{\left[{\alpha }_{i+1}\right]}|=0$. Applying Lemma 7 (a), we have $v_{{\alpha }_i},v_{{\alpha }_i+1}\dots ,v_{{\alpha }_{i+1}-1}\in \mathcal{B}$, which contradicts the fact that $\mathcal{B}\cap S_{\left[{\alpha }_i\right]}=\varphi$, and hence there is a ${\beta }_i$ such that $|\mathcal{B}\cap [{\beta }_i\left]\right|=2$.

(c)

Assume to the contrary that there exists no $\left[{\beta }_k\right]\in I_{({\alpha }_k,{\alpha }_1)}$ such that $|\mathcal{B}\cap S_{\left[{\beta }_k\right]}|=2$. Then, $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{{\alpha }_k+1,\dots ,r-1,0,\dots ,{\alpha }_1-1\}$ with $|\mathcal{B}\cap S_{\left[{\alpha }_k\right]}|=0$ and $|\mathcal{B}\cap S_{\left[{\alpha }_1\right]}|=0$. From $\left(a\right)$, $\left[{\alpha }_1\right]$ and $\left[{\alpha }_k\right]$ are not consecutive, that is, both cannot be in $\left\{\right[0],[r-1\left]\right\}$. Thus, we consider the following three cases:

Case I.${\alpha }_1=0$. Then, ${\alpha }_k\ne r-1$. Since $|\mathcal{B}\cap S_{\left[t\right]}|\le 1$ for all $t\in \{{\alpha }_k,\dots ,r-1\}$ and $|\mathcal{B}\cap S_{\left[0\right]}|=0$, applying Lemma 7 (a), we have $\{v_{{\alpha }_k},v_{{\alpha }_k+1},\dots ,v_{r-1}\}\subset \mathcal{B}$, which contradicts the fact that $\mathcal{B}\cap S_{\left[{\alpha }_k\right]}=\varphi$.

Case II.${\alpha }_k=r-1$. Then, ${\alpha }_1\ne 0$. Here $|\mathcal{B}\cap S_{\left[t\right]}|\le 1$ for all $t\in \{0,1,\dots ,{\alpha }_1-1\}$. From Lemma 7 $\left(a\right)$, we have $\{v_0,v_1\dots ,v_{{\alpha }_1-1}\}\subset \mathcal{B}$. Then, $v_{r-1}$ and $v_r$ can only be resolved by $0\text{-}class$ elements that are to the right of $v_r$, so $|\mathcal{B}\cap S_{\left[0\right]}|=2$, which contradicts the fact that $|\mathcal{B}\cap S_{\left[0\right]}|=1$.

Case III.${\alpha }_k\ne r-1$and${\alpha }_1\ne 0$. We have $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{0,1,\dots ,{\alpha }_1-1\}\cup \{{\alpha }_k+1,\dots ,r-1\}$. Since $\mathcal{B}\cap S_{\left[{\alpha }_1\right]}=\varphi$ and $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{0,1,\dots ,{\alpha }_1-1\}$, applying Lemma 7 $\left(a\right)$, we have $\{v_0,v_1\dots ,v_{{\alpha }_1-1}\}\subset \mathcal{B}$. Therefore, $\mathcal{B}\cap S_{\left[0\right]}=\left\{v_0\right\}$. Now applying Lemma 7 $\left(d\right)$, we have $\{v_{{\alpha }_k},v_{{\alpha }_k+1}\dots ,v_{r-1}\}\subset \mathcal{B}$, which contradicts the fact that $v_{{\alpha }_k}\notin \mathcal{B}$. Therefore, there exists a class $\left[{\beta }_k\right]$ with ${\beta }_k\in \{0,1,\dots ,{\alpha }_1-1\}\cup \{{\alpha }_k,\dots ,r-1\}$ such that $|\mathcal{B}\cap S_{\left[{\beta }_k\right]}|=2$.

Now we prove the uniqueness of ${\beta }_i$ where $i\in \{1,\dots ,k\}$. Since $\left|\mathcal{B}\right|=r$ and $\mathcal{B}$ miss exactly k classes and between every two missing classes $\left[{\alpha }_i\right]$ and $\left[{\alpha }_{i+1}\right]$ there exists a class $\left[{\beta }_i\right]$ with $|\mathcal{B}\cap S_{\left[{\beta }_i\right]}|=2$ and two missing classes $\left[{\alpha }_k\right]$ and $\left[{\alpha }_1\right]$, there exists a class $\left[{\beta }_k\right]$ such that $|\mathcal{B}\cap S_{\left[{\beta }_k\right]}|=2$. Hence, we conclude that each ${\beta }_i$ is unique (otherwise, $\left|\mathcal{B}\right|\ge r+1$).

 □

Corollary 4.

Any metric basis cannot miss consecutive classes.

Theorem 3.

Let $0\le {\alpha }_1<{\alpha }_2<\cdots <{\alpha }_k\le r-1$ and $\left[{\beta }_i\right]$ be an intermediate class between $\left[{\alpha }_i\right]$ and $\left[{\alpha }_{i+1}\right]$ if there be any. Let $X=\left\{v_t:0\le t\le r-1,\left[t\right]\in {\cup }_{i=1}^{k-1}{I}_{[{\beta }_i,{\alpha }_{i+1})}\cup I_{[{\beta }_k,{\alpha }_1)}\right\}$ and Y be the set of all largest s-class elements, where $\left[s\right]\in {\cup }_{i=1}^{k-1}{I}_{({\alpha }_i,{\beta }_i]}\cup I_{({\alpha }_k,{\beta }_k]}$. Then, $X\cup Y$ forms a resolving for $P_n^r$.

Proof. 

Here, $X\subset \{v_0,\dots ,v_{r-1}\}$ and $Y\subset \{v_{n-r},\dots ,v_{n-1}\}$, as Y contains all largest s-class elements. Let u and w be two distinct vertices from $V\left(P_n^r\right)\backslash (X\cup Y)$. Then, we can write $u=v_{ar+r_1}$ and $w=v_{br+r_2}$ for some $0\le a,b\le \lceil \frac{n}{r}\rceil -1$, $0\le r_1,r_2\le r-1$. Without loss of generality, we can assume that u is in the left side of w, that is, $ar+r_1<br+r_2$. If $d(u,w)\ge 2$, then $b-a\ge 1$ and equality occurs only when $r_1<r_2$. The distances of u and w from $v_{{\beta }_1}$ are given by $d(u,v_{{\beta }_1})=a$ or $a+1$ and $d(w,v_{{\beta }_1})=b$ or $b+1$ according to $r_i\le {\beta }_1$ or $r_i>{\beta }_1$, where $i=1,2$. Therefore, the equality, $d(u,v_{{\beta }_1})=d(w,v_{{\beta }_1})$ gives $b=a+1$ and $r_2\le {\beta }_1<r_1$. This contradicts the fact that $r_1<r_2$ and hence $v_{{\beta }_1}$ resolves u and w when $d(u,w)\ge 2$. It remains to be shown that u and w are resolved by a vertex $x\in X\cup Y$ when $d(u,w)=1$. Now we consider the following three cases:

• Case I.$|I_{(r_1,r_2)}|\ge 2$. Since $X\cup Y$ does not miss two consecutive classes, there always exists at least $\left[r^{\prime }\right]\in I_{(r_1,r_2)}$ such that $S_{\left[r^{\prime }\right]}\cap (X\cup Y)\ne \varphi$. Let $z\in S_{\left[r^{\prime }\right]}\cap (X\cup Y)$. If $z\in X$, then $d(z,w)=d(z,u)+1=a+1$, that is, z resolves u and w. Similarly, if $z\in Y$, then we can show that u and w are resolved by z. Therefore, u and w are resolved by $z\in S_{\left[r^{\prime }\right]}\cap (X\cup Y)$.
• Case II.$|I_{(r_1,r_2)}|=1$. Let $\left[r^{\prime }\right]\in I_{(r_1,r_2)}$. If $S_{\left[r^{\prime }\right]}\cap (X\cup Y)\ne \varphi$, then with a similar argument as in Case I, u and w are resolved by an element $z\in S_{\left[r^{\prime }\right]}\cap (X\cup Y)$. Otherwise, $S_{\left[r^{\prime }\right]}\cap (X\cup Y)=\varphi$. Then, $S_{\left[r_1\right]}\cap X=\left\{v_{r_1}\right\}$ and $S_{\left[r_2\right]}\cap Y\ne \varphi$. We have $d(v_{r_1},w)=d(v_{r_1},u)+1=a+1$. Therefore, u and w are resolved by $v_{r_1}\in X$.
• Case III.$|I_{(r_1,r_2)}|=0$. In this case u and w are consecutive vertices. From the construction of X and Y, we have $S_{\left[r_1\right]}\cap X\ne \varphi$ or $S_{\left[r_2\right]}\cap Y\ne \varphi$. First we assume that $S_{\left[r_2\right]}\cap Y\ne \varphi$. Let $z\in S_{\left[r_2\right]}\cap Y$. Then, $d_{P_n}(z,u)=d_{P_n}(z,w)+1$. Since both $z,w\in S_{\left[r_2\right]}$, $d_{P_n}(z,w)\equiv 0(modr)$ and hence $d(z,u)=d(z,w)+1$ in $P_n^r$. Thus, z resolves u and w. Next we assume that $S_{\left[r_1\right]}\cap X\ne \varphi$. Then, $S_{\left[r_1\right]}\cap X=\left\{v_{r_1}\right\}$. By a similar argument as above it is clear that $v_{r_1}$ resolves u and w.

Thus, from the above, it follows that $X\cup Y$ forms a resolving set for $P_n^r$. □

Corollary 5.

Let X be the set of all smallest s-class elements for all $s\in \{0,1,\dots ,{\alpha }_1-1\}\cup \{{\beta }_1,\dots ,r-1\}$ or $s\in \{{\beta }_1,\dots ,{\alpha }_1-1\}$ according to ${\alpha }_1<{\beta }_1\le r-1$ or $0\le {\beta }_1<{\alpha }_1$. Additionally, let Y be the set of all largest t-class elements for all $t\in \{{\alpha }_1+1,\dots ,{\beta }_1\}$ or $t\in \{{\alpha }_1+1,\dots ,r-1\}\cup \{0,\dots ,{\beta }_1\}$ according to ${\alpha }_1<{\beta }_1\le r-1$ or $0\le {\beta }_1<{\alpha }_1$, where $0\le {\alpha }_1,{\beta }_1\le r-1$. Then, $X\cup Y$ forms a resolving set for $P_n^r$.

Example 5.

For $P_{35}^8$, we consider $k=2$, ${\alpha }_1=2$, ${\alpha }_2=5,{\beta }_1=4$ and ${\beta }_2=7$. Then, applying Theorem 3, $\mathcal{B}=\{v_0,v_1,v_4,v_7,v_{27},v_{28},v_{30},v_{31}\}$ forms a resolving set. Here $X=\{v_0,v_1,v_4,v_7\}$ and $Y=\{v_{27},v_{28},v_{30},v_{31}\}$.

Example 6.

For $P_{35}^8$ with ${\alpha }_1=2,{\alpha }_2=5,{\beta }_1=3,$ and ${\beta }_2=1$, $\mathcal{B}=\{v_1,v_3,v_4,v_5,v_{27},v_{31},v_{32},v_{33}\}$ forms a resolving set.

From Theorem 3, we see that for $P_n^r$, there is a resolving set that does not take an element from some $S_{\left[{\alpha }_i\right]}$ for $i=1,2,\dots ,k$. The following theorem gives all metric bases $\mathcal{B}$ with $\mathcal{B}\cap S_{\left[{\alpha }_i\right]}=\varphi$.

Theorem 4.

Any metric basis $\mathcal{B}({\alpha }_1,\dots ,{\alpha }_k)$ is of the form $X\cup Y$, where $X=\left\{v_t:0\le t\le r-1,\left[t\right]\in \left({\cup }_{i=1}^{k-1}{I}_{[{\beta }_i,{\alpha }_{i+1})}\right)\cup I_{[{\beta }_k,{\alpha }_1)}\right\}$ and Y is the set of all largest s-class elements, where $\left[s\right]\in {\cup }_{i=1}^k{I}_{({\alpha }_i,{\beta }_i]}$ for some $\left[{\beta }_i\right]\in I_{({\alpha }_i,{\alpha }_{i+1})}$ with $i=1,\dots ,k-1$ and $\left[{\beta }_k\right]\in I_{({\alpha }_k,{\alpha }_1)}$.

Proof. 

Here we consider two cases according to the position of ${\beta }_k$.

Case I.${\alpha }_k<{\beta }_k\le (r-1)$. Here $\mathcal{B}\cap S_{\left[{\alpha }_1\right]}=\varphi$ and $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{0,1,\dots ,{\alpha }_1-1\}$. Hence, applying Lemma 7, we have $\{v_0,v_1,\dots ,v_{{\alpha }_1-1}\}\subset \mathcal{B}$ if ${\alpha }_1\ne 0$. Again, for every $i\in \{1,2,\dots ,k-1\}$, $\mathcal{B}\cap S_{\left[{\alpha }_{i+1}\right]}=\varphi$ and $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{{\beta }_i+1,\dots ,{\alpha }_{i+1}-1\}$, so applying Lemma 7, we have $\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\subset \mathcal{B}$ and hence ${\cup }_{i=1}^{k-1}\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\subset \mathcal{B}$. Again, $\mathcal{B}\cap S_{\left[0\right]}=\left\{v_0\right\}$ and $|\mathcal{B}\cap S_{\left[t\right]}|=1$ for all $t\in \{{\beta }_k+1,\dots ,r-1\}$, so applying Lemma 7, we have $\{v_{{\beta }_k},\dots ,v_{r-1}\}\subset \mathcal{B}$. Therefore, $X\subset \mathcal{B}$, where $X=\{v_0,\dots ,v_{{\alpha }_1-1}\}\cup \left({\cup }_{i=1}^{k-1}\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\right)\cup \{v_{{\beta }_k},\dots ,v_{r-1}\}$, that is,

$$X=\left\{v_t:0\le t\le r-1,\left[t\right]\in \left({\cup }_{i=1}^{k-1}{I}_{[{\beta }_i,{\alpha }_{i+1})}\right)\cup I_{[{\beta }_k,{\alpha }_1)}\right\}.$$

For every $i\in \{1,\dots ,k\}$, $S_{\left[{\alpha }_i\right]}\cap \mathcal{B}=\varphi$ and $|S_{\left[j\right]}\cap \mathcal{B}|=1$, $j={\alpha }_i+1,\dots ,{\beta }_i-1$, so applying Lemma 7, we have that $\mathcal{B}$ contains all largest s-class elements for all $s\in \{{\alpha }_i+1,\dots ,{\beta }_i\}$, where $i=1,\dots ,k$. Therefore, $Y\subset \mathcal{B}$, where Y is the set of all largest s-class elements for all $s\in \{{\alpha }_i+1,\dots ,{\beta }_i\}$, where $i=1,\dots ,k$.

Since $X\subset \{v_0,\dots ,v_{r-1}\}$, $Y\subset \{v_{n-r},\dots ,v_{n-1}\}$, and thus $X\cap Y=\varphi$ and $|X\cup Y|=\left|X\right|+\left|Y\right|={\alpha }_1+{\sum }_{i=1}^{k-1}({\alpha }_{i_1}-{\beta }_i)+(r-{\beta }_k)+{\sum }_{i=1}^k({\beta }_i-{\alpha }_i)=r$. Again, from Theorem 3, $X\cup Y$ forms a resolving set. Therefore, we write $\mathcal{B}=X\cup Y$.

Case II.$0\le {\beta }_k<{\alpha }_1$. Here $\mathcal{B}\cap \left[{\alpha }_1\right]=\varphi$, $|\mathcal{B}\cap [t\left]\right|=1$ for all $t\in \{{\beta }_k+1,\dots ,{\alpha }_1-1\}$, so applying Lemma 7, we have $\{v_{{\beta }_k},\dots ,v_{{\alpha }_1-1}\}\subset \mathcal{B}$. Again, for every $i\in \{1,2,\dots ,k-1\}$, $\mathcal{B}\cap \left[{\alpha }_{i+1}\right]=\varphi$ and $|\mathcal{B}\cap [t\left]\right|=1$ for all $t\in \{{\beta }_i+1,\dots ,{\alpha }_{i+1}-1\}$, so applying Lemma 7, we have $\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\subset \mathcal{B}$ for all $i=\{1,\dots ,k-1\}$. Thus, ${\cup }_{i=1}^{k-1}\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\subset \mathcal{B}$. Therefore, $X\subset \mathcal{B}$, where $X=\{v_{{\beta }_k},\dots ,v_{{\alpha }_1-1}\}\cup \left({\cup }_{i=1}^{k-1}\{v_{{\beta }_i},\dots ,v_{{\alpha }_{i+1}-1}\}\right)$, that is,

$$X=\left\{v_t:0\le t\le r-1,\left[t\right]\in \left({\cup }_{i=1}^{k-1}{I}_{[{\beta }_i,{\alpha }_{i+1})}\right)\cup I_{[{\beta }_k,{\alpha }_1)}\right\}.$$

For every $i\in \{1,\dots ,k-1\}$, $S_{\left[{\alpha }_i\right]}\cap \mathcal{B}=\varphi$ and $|S_{\left[j\right]}\cap \mathcal{B}|=1$ for all $j={\alpha }_i+1,\dots ,{\beta }_i-1$, so applying Lemma 7, we have that $\mathcal{B}$ contains all largest s-class elements for all $s\in \{{\alpha }_i+1,\dots ,{\beta }_i\}$, where $i=1,\dots ,k-1$. Additionally, $\mathcal{B}\cap \left[{\alpha }_k\right]=\varphi$, $|\mathcal{B}\cap [t\left]\right|=1$ for all $t\in \{{\alpha }_k+1,\dots ,r-1\}\cup \{0,\dots ,{\beta }_k\}$, so applying Lemma 7, we have that $\mathcal{B}$ contains all largest s-class elements for all $s\in \{{\alpha }_k+1,\dots ,r-1\}\cup \{0,\dots ,{\beta }_k\}$. Therefore, $Y\subset \mathcal{B}$, where Y is the set of all largest s-class elements for all $s\in \{0,\dots ,{\beta }_k\}\cup \{{\alpha }_k+1,\dots ,r-1\}\cup \left({\cup }_{i=1}^{k-1}\{{\alpha }_i+1,\dots ,{\beta }_i\}\right)$, that is, Y is the set of all largest s-class elements, where $\left[s\right]\in {\cup }_{i=1}^{k-1}{I}_{({\alpha }_i,{\beta }_i]}\cup I_{({\alpha }_k,{\beta }_k]}$.

Since $X\subset \{v_0,\dots ,v_{r-1}\}$, $Y\subset \{v_{n-r},\dots ,v_{n-1}\}$, so $X\cap Y=\varphi$ and $|X\cup Y|=\left|X\right|+\left|Y\right|=({\alpha }_1-{\beta }_k)+{\sum }_{i=1}^{k-1}({\alpha }_{i_1}-{\beta }_i)+(r-{\alpha }_k+{\beta }_k)+{\sum }_{i=1}^{k-1}({\beta }_i-{\alpha }_i)=r$. Again, from Theorem 3, $X\cup Y$ forms a resolving set. Therefore, we write $\mathcal{B}=X\cup Y$. □

6. Conclusions

In this paper we have determined all metric bases for the power of any n-vertex path $P_n$ in the two main Theorems 2 and 4. Theorem 2 determines all metric bases that contain all class elements, whereas Theorem 4 gives the forms of all metric bases that miss one or more classes. Here we pose two related problems.

Problem 1.

A characterization of all metric bases for the power of any n-vertex cycle.

Problem 2.

A characterization of all metric bases for the power of any trees.