The fractional $k$-metric dimension of graphs

TL;DR

This paper introduces the fractional $k$-metric dimension of graphs, exploring its properties, bounds, and specific cases, and establishing foundational results and open problems in this new area of graph theory.

Contribution

It initiates the study of fractional $k$-metric dimension, providing bounds, characterizations, and exact values for certain graph classes, and proposing open problems for future research.

Findings

Bounds for fractional $k$-metric dimension established

Characterizations for graphs with extremal fractional $k$-metric dimensions

Examples showing the gap between $ ext{dim}_f^k(G)$ and $k ext{dim}_f(G)$

Abstract

Let $G$ be a graph with vertex set $V(G)$. For any two distinct vertices $x$ and $y$ of $G$, let $R\{x, y\}$ denote the set of vertices $z$ such that the distance from $x$ to $z$ is not equal to the distance from $y$ to $z$ in $G$. For a function $g$ defined on $V(G)$ and for $U \subseteq V(G)$, let $g(U)=\sum_{s \in U}g(s)$. Let $\kappa(G)=\min\{|R\{x,y\}|: x\neq y \mbox{ and } x,y \in V(G)\}$. For any real number $k \in [1, \kappa(G)]$, a real-valued function $g: V(G) \rightarrow [0,1]$ is a \emph{$k$-resolving function} of $G$ if $g(R\{x,y\}) \ge k$ for any two distinct vertices $x,y \in V(G)$. The \emph{fractional $k$-metric dimension}, $\dim^k_f(G)$, of $G$ is $\min\{g(V(G)): g \mbox{ is a $k$-resolving function of } G\}$. In this paper, we initiate the study of the fractional $k$-metric dimension of graphs. For a connected graph $G$ and $k \in [1, \kappa(G)]$, it's easy to see that $k \le \dim_f^k(G) \le \frac{k|V(G)|}{\kappa(G)}$; we characterize graphs $G$ satisfying $\dim_f^k(G)=k$ and $\dim_f^k(G)=|V(G)|$, respectively. We show that $\dim_f^k(G) \ge k \dim_f(G)$ for any $k \in [1, \kappa(G)]$, and we give an example showing that $\dim_f^k(G)-k\dim_f(G)$ can be arbitrarily large for some $k \in (1, \kappa(G)]$; we also describe a condition for which $\dim_f^k(G)=k\dim_f(G)$ holds. We determine the fractional $k$-metric dimension for some classes of graphs, and conclude with two open problems, including whether $\phi(k)=\dim_f^k(G)$ is a continuous function of $k$ on every connected graph $G$.

Equations

$$\left\{\begin{array}[]{c}R\{a_{i,s},b_{i,s}\}=V(P^{i,a})\cup V(P^{i,b}),\\ R\{a_{i,s},c_{i,s}\}=V(P^{i,a})\cup V(P^{i,c}),\\ R\{b_{i,s},c_{i,s}\}=V(P^{i,b})\cup V(P^{i,c}).\end{array}\right.$$

$$\left\{\begin{array}[]{c}|S\cap R\{a_{i,s},b_{i,s}\}|=|S\cap V(P^{i,a})|+|S\cap V(P^{i,b})|\geq k,\\ |S\cap R\{a_{i,s},c_{i,s}\}|=|S\cap V(P^{i,a})|+|S\cap V(P^{i,c})|\geq k,\\ |S\cap R\{b_{i,s},c_{i,s}\}|=|S\cap V(P^{i,b})|+|S\cap V(P^{i,c})|\geq k.\end{array}\right.$$

$$h(v)=\left\{\begin{array}[]{ll}0&\mbox{ if }\deg(v)\geq 3,\\ \frac{k}{2s}&\mbox{ if }\deg(v)\leq 2.\end{array}\right.$$

$$\dim_{f}(G)=\left\{\begin{array}[]{ll}\frac{n-1}{2}&\mbox{if }s=1,\\ \frac{n}{2}&\mbox{otherwise}.\end{array}\right.$$

$$\dim_{f}^{k}(P_{n})=\left\{\begin{array}[]{ll}k&\mbox{ if }k\in[1,2],\\ 2+(k-2)\frac{n-2}{n-3}&\mbox{ if }k\in(2,n-1].\end{array}\right.$$

$$g(u_{i})=\left\{\begin{array}[]{ll}1&\mbox{ if }i\in\{1,n\},\\ \frac{k-2}{n-3}&\mbox{ otherwise}.\end{array}\right.$$

$$\kappa (T)=\min \bigcup _{w\in M^{\ast }(T)}\{d({\ell }_i,{\ell }_j):{\ell }_i\text{mbox}and{\ell }_j\text{mbox}aretwodistinctterminalverticesofw\}.$$

$$\dim_{f}^{k}(T)=\left\{\begin{array}[]{ll}\frac{ka}{2}&\mbox{ for }k\in[1,2d(v,\ell_{1})],\\ (a-1)k-(a-2)d(v,\ell_{1})&\mbox{ for }k\in(2d(v,\ell_{1}),\kappa(T)].\end{array}\right.$$

$$g(u)=\left\{\begin{array}[]{ll}0&\mbox{ if }u=v,\\ \frac{k}{2d(v,\ell_{i})}&\mbox{ for each vertex }u\in V(P^{i}),\mbox{ where }i\in\{1,2,\ldots,a\}.\end{array}\right.$$

$$g(u)=\left\{\begin{array}[]{ll}0&\mbox{ if }u=v,\\ 1&\mbox{ for each vertex }u\in V(P^{1}),\\ \frac{k-d(v,\ell_{1})}{d(v,\ell_{j})}&\mbox{ for each vertex }u\in V(P^{j}),\mbox{ where }j\in\{2,\ldots,a\}.\end{array}\right.$$

$$d(x,u)=d(y,u)\text{mbox}andd(x,v)=d(y,v).$$

$$d(x,w^{\prime })=d(y,w^{\prime })\text{mbox}andd(x,t)=d(y,t),\text{mbox}wheret\in V(T_w).$$

$${\dim }_f^k(T)=k|M_2(T)|+\sum _{v\in M_3(T)}{\dim }_f^k(T_v).$$

$$g(V(T))\ge \sum _{v\in M_2(T)}g(V(T_v))+\sum _{w\in M_3(T)}g(V(T_w))\ge k|M_2(T)|+\sum _{w\in M_3(T)}{\dim }_f^k(T_w).$$

$$g(u)=\left\{\begin{array}[]{ll}g_{w}(u)&\mbox{if }u\in V(T_{w})\mbox{ for }w\in M_{3}(T),\\ h_{w}(u)&\mbox{if }u\in V(T_{w})\mbox{ for }w\in M_{2}(T),\\ 0&\mbox{otherwise}.\end{array}\right.$$

$$\dim_{f}^{k}(C_{n})=k\dim_{f}(C_{n})=\left\{\begin{array}[]{ll}\frac{kn}{n-2}&\mbox{ if $n$ is even and }k\in[1,n-2],\\[3.0pt] \frac{kn}{n-1}&\mbox{ if $n$ is odd and }k\in[1,n-1].\end{array}\right.$$

$$\dim_{f}^{k}(W_{n})=k\dim_{f}(W_{n})=\left\{\begin{array}[]{ll}2k&\mbox{ if $n=5$ and }k\in[1,2],\\ \frac{3k}{2}&\mbox{ if $n=6$ and }k\in[1,4],\\[3.0pt] \frac{k(n-1)}{4}&\mbox{ if $n\geq 7$ and }k\in[1,4].\end{array}\right.$$

$$h(u)=\left\{\begin{array}[]{ll}\frac{1}{r_{i}}&\mbox{ for each }u\in V(P^{i})\mbox{ if $C^{i}$ is an odd cycle},\\ \frac{1}{r_{j}-1}&\mbox{ for each }u\in V(P^{j})-\{u_{j,\lceil\frac{r_{j}}{2}\rceil}\}\mbox{ if $C^{j}$ is an even cycle},\\ 0&\mbox{ otherwise}.\end{array}\right.$$

$$\dim^{k}_{f}(G)=k\dim_{f}(G)=\left\{\begin{array}[]{ll}\frac{k(n-1)}{2}&\mbox{if $s=1$},\\ \frac{kn}{2}&\mbox{otherwise}.\end{array}\right.$$

$$g(v)=\left\{\begin{array}[]{ll}\frac{k}{s+t-2}&\mbox{ if }v\in L,\\ 0&\mbox{ otherwise. }\end{array}\right.$$

Full text

**Keywords: fractional metric dimension, fractional $k$-metric dimension, $k$-metric dimension, trees, cycles, wheel graphs, the Petersen graph, a bouquet of cycles, complete multi-partite graphs, grid graphs

2010 Mathematics Subject Classification: 05C12, 05C38, 05C05

**

1 Introduction

Let $G$ be a finite, simple, undirected, and connected graph with vertex set $V(G)$ and edge set $E(G)$. For $v\in V(G)$, the open neighborhood of $v$ is $N(v)=\{u\in V(G):uv\in E(G)\}$, and the closed neighborhood of $v$ is $N[v]=N(v)\cup\{v\}$. The degree of a vertex $v\in V(G)$, denoted by $\deg(v)$, is $|N(v)|$; a leaf is a vertex of degree one, and a major vertex is a vertex of degree at least three. The distance between two vertices $x,y\in V(G)$, denoted by $d(x,y)$, is the length of a shortest path between $x$ and $y$ in $G$. The diameter, $\textnormal{diam}(G)$, of a graph $G$ is $\max\{d(x,y):x,y\in V(G)\}$. The complement of $G$, denoted by $\overline{G}$, is the graph whose vertex set is $V(G)$ and $xy\in E(\overline{G})$ if and only if $xy\not\in E(G)$ for $x,y\in V(G)$. We denote by $K_{n}$ and $P_{n}$ the complete graph and the path on $n$ vertices, respectively.

For two distinct vertices $x,y\in V(G)$, let $R\{x,y\}=\{z\in V(G):d(x,z)\neq d(y,z)\}$. A subset $S\subseteq V(G)$ is called a resolving set of $G$ if $|S\cap R\{x,y\}|\geq 1$ for any two distinct vertices $x$ and $y$ in $G$. The metric dimension, $\dim(G)$, of $G$ is the minimum cardinality of $S$ over all resolving sets of $G$. Since metric dimension is suggestive of the dimension of a vector space in linear algebra, sometimes a minimum resolving set of $G$ is called a basis of $G$. The concept of metric dimension was introduced independently by Slater [28], and by Harary and Melter [20]. Applications of metric dimension can be found in network discovery and verification [7], robot navigation [24], sonar [28], combinatorial optimization [27], chemistry [25], and strategies for the mastermind game [9]. It was noted in [19] that determining the metric dimension of a graph is an NP-hard problem. Metric dimension has been extensively studied. For a survey on metric dimension in graphs, see [4, 8]. The effect of the deletion of a vertex or of an edge on the metric dimension of a graph was raised as a fundamental question in graph theory in [8]; the question is essentially settled in [11].

If a minimum number of requisite robots are installed in a network to identify the exact location of an intruder in the network, one malfunctioning robot can lead to failure of detection. Thus, it is natural to build a certain level of redundancy into the detection system. As a generalization of metric dimension, $k$-metric dimension was introduced first by Estrada-Moreno et al. [12] and, independently, by Adar and Epstein [1] soon afterwards. Let $\kappa(G)=\min\{|R\{x,y\}|:x\neq y\mbox{ and }x,y\in V(G)\}$. For a positive integer $k\in\{1,2,\ldots,\kappa(G)\}$, a set $S\subseteq V(G)$ is called a $k$-resolving set of $G$ if $|S\cap R\{x,y\}|\geq k$ for any two distinct vertices $x$ and $y$ in $G$. The $k$-metric dimension, $\dim^{k}(G)$111In fact, the notation of this parameter has been $\dim_{k}(G)$ in previous works. However, we rather prefer to use $\dim^{k}(G)$ here, to facilitate the notation of $\dim_{f}^{k}(G)$., of $G$ is the minimum cardinality over all $k$-resolving sets of $G$. It was shown in [12] that $k$-metric dimension of a connected graph $G$ exists for every $k\in\{1,2,\ldots,\kappa(G)\}$, and $G$ is called $\kappa(G)$-metric dimensional. For an application of $k$-metric dimension to error-correcting codes, see [5]. For other articles on the $k$-metric dimension of graphs, see [6, 13, 14].

The fractionalization of various graph parameters has been extensively studied (see [26]). Currie and Oellermann [10] defined fractional metric dimension as the optimal solution to a linear programming problem, by relaxing a condition of the integer programming problem for metric dimension. A formulation of fractional metric dimension as a linear programming problem can be found in [15]. Arumugam and Mathew [2] officially studied the fractional metric dimension of graphs. For a function $g$ defined on $V(G)$ and for $U\subseteq V(G)$, let $g(U)=\sum_{s\in U}g(s)$. A real-valued function $g:V(G)\rightarrow[0,1]$ is a resolving function of $G$ if $g(R\{x,y\})\geq 1$ for any two distinct vertices $x,y\in V(G)$. The fractional metric dimension of $G$, denoted by $\dim_{f}(G)$, is $\min\{g(V(G)):g\mbox{ is a resolving function of }G\}$. Notice that $\dim_{f}(G)$ reduces to $\dim(G)$, if the codomain of resolving functions is restricted to $\{0,1\}$. For more articles on the fractional metric dimension, as well as the closely related fractional strong metric dimension, of graphs, see [3, 16, 17, 18, 21, 22, 23, 29].

Next, we introduce fractional $k$-metric dimension, which can be viewed as a generalization of $\dim_{f}(G)$ as well as a fractionalization of $\dim^{k}(G)$. For any real number $k\in[1,\kappa(G)]$, a real-valued function $h:V(G)\rightarrow[0,1]$ is a $k$-resolving function of $G$ if $h(R\{x,y\})\geq k$ for any two distinct vertices $x,y\in V(G)$. The fractional $k$-metric dimension of $G$, denoted by $\dim_{f}^{k}(G)$, is \min\{h(V(G)):h\mbox{ is a k-resolving function of }G\}; notice that $\dim^{1}_{f}(G)=\dim_{f}(G)$. Note that $\dim_{f}^{k}(G)$ reduces to $\dim^{k}(G)$ when the codomain of $k$-resolving functions is restricted to $\{0,1\}$ and $k\in[1,\kappa(G)]$ is restricted to positive integers.

In this paper, we initiate the study of the fractional $k$-metric dimension of graphs. For a connected graph $G$, let $\kappa(G)=\min\{|R\{x,y\}|:x\neq y\mbox{ and }x,y\in V(G)\}$. The paper is organized as follows. In section 2, we compare $\dim^{k}_{f}(G)$ with $\dim^{k}(G)$ for certain $k$, and we recall some results on the fractional metric dimension of graphs. In section 3, we prove that $\dim_{f}^{k}(G)\geq k\dim_{f}(G)$ for any $k\in[1,\kappa(G)]$; we describe a condition for which $\dim_{f}^{k}(G)=k\dim_{f}(G)$ holds for all $k\in[1,\kappa(G)]$. For $k\in[1,\kappa(G)]$, we show that $k\leq\dim_{f}^{k}(G)\leq\frac{k}{\kappa(G)}|V(G)|$, which implies $k\leq\dim_{f}^{k}(G)\leq|V(G)|$; we characterize graphs $G$ satisfying $\dim_{f}^{k}(G)=k$ and $\dim_{f}^{k}(G)=|V(G)|$, respectively. In section 4, for $k\in[1,\kappa(G)]$, we determine the fractional $k$-metric dimension of trees, cycles, wheel graphs, the Petersen graph, a bouquet of cycles (i.e., the vertex sum of cycles at one common vertex), complete multi-partite graphs, and grid graphs (i.e., the Cartesian product of two paths). Along the way, we give an example showing that $\dim_{f}^{k}(G)-k\dim_{f}(G)$ can be arbitrarily large for some $k\in(1,\kappa(G)]$. We conclude with some open problems.

2 Preliminaries

In this section, we make some observations involving $\dim_{f}^{k}(G)$ for $k\in[1,\kappa(G)]$, or $\dim^{k}(G)$ for $k\in\{1,2,\ldots,\kappa(G)\}$. We also recall some results on the fractional metric dimension of graphs. We begin with some observations. Two distinct vertices $x,y\in V(G)$ are called twin vertices if $N(x)-\{y\}=N(y)-\{x\}$.

Observation 2.1**.**

Let $G$ be a connected graph and let $k\in[1,\kappa(G)]$. If two distinct vertices $x$ and $y$ are twin vertices in $G$, then $R\{x,y\}=\{x,y\}$, and thus $\kappa(G)=2$ and $g(x)+g(y)\geq k$ for any $k$-resolving function $g$ of $G$.

Observation 2.2**.**

Let $G$ be a connected graph.

• (a)

For $k\in[1,\kappa(G)]$, $\dim_{f}(G)\leq\dim_{f}^{k}(G)$.

• (b)

For $k\in\{1,2,\ldots,\kappa(G)\}$, $\dim_{f}^{k}(G)\leq\dim^{k}(G)$.

• (c)

[2, 12]* For $k\in\{1,2,\ldots,\kappa(G)\}$, $\dim_{f}(G)\leq\dim(G)\leq\dim^{k}(G)$.*

Observation 2.2 provides inequalities between any two graph parameters among $\dim_{f}(G)$, $\dim(G)$, $\dim_{f}^{k}(G)$ for $k\in[1,\kappa(G)]$, and $\dim^{k}(G)$ for $k\in\{1,2,\ldots,\kappa(G)\}$, excluding the relation between $\dim(G)$ and $\dim_{f}^{k}(G)$. So, it is natural to compare $\dim(G)$ and $\dim_{f}^{k}(G)$ for $k\in[1,\kappa(G)]$.

Remark 2.3**.**

(a) The value of $\dim_{f}^{k}(G)-\dim(G)$ can be arbitrarily large, as $G$ varies, for some $k\in(1,\kappa(G)]$. For $n\geq 4$, note that $\dim(P_{n})=1\leq\dim_{f}^{k}(P_{n})$ for $k\in[1,n-1]$ and $\dim_{f}^{n-1}(P_{n})=n$ (see Proposition 4.1). Thus, $\dim_{f}^{n-1}(P_{n})-\dim(P_{n})=n-1$ can be arbitrarily large.

(b) The value of $\dim(G)-\dim_{f}^{k}(G)$ can be arbitrarily large, as $G$ varies, for some $k\in[1,\kappa(G)]$. For $n\geq 3$, note that $\dim(K_{n})=n-1$ and $\dim_{f}^{k}(K_{n})=\frac{kn}{2}$ for $k\in[1,2]$ (see Proposition 4.12). Now, let $k\in[1,2)$; then $\dim(K_{n})-\dim_{f}^{k}(K_{n})=\frac{(2-k)n}{2}-1$ becomes arbitrarily large as $n\rightarrow\infty$.

In light of Observation 2.2(b), we have the following

Theorem 2.4**.**

The value of $\dim^{k}(G)-\dim_{f}^{k}(G)$ can be arbitrarily large, as $G$ varies, for some $k\in\{1,2,\ldots,\kappa(G)\}$.

Proof.

Let $H$ be a connected graph with vertex set $V(H)=\{u_{1},u_{2},\ldots,u_{n}\}$, where $n\geq 2$. Let $G$ be the graph obtained from $H$ as follows:

• (i)

for each $i\in\{1,2,\ldots,n\}$, add three vertices $a_{i,1},b_{i,1},c_{i,1}$ and three edges $u_{i}a_{i,1},u_{i}b_{i,1},u_{i}c_{i,1}$;

• (ii)

for each $i\in\{1,2,\ldots,n\}$, subdivide the edge $u_{i}a_{i,1}$ ($u_{i}b_{i,1}$ and $u_{i}c_{i,1}$, respectively) exactly $s-1$ times so that the edge $u_{i}a_{i,1}$ ($u_{i}b_{i,1}$ and $u_{i}c_{i,1}$, respectively) in (i) becomes the $u_{i}-a_{i,1}$ path given by $u_{i},a_{i,s},a_{i,s-1},\ldots,a_{i,1}$ (the $u_{i}-b_{i,1}$ path given by $u_{i},b_{i,s},b_{i,s-1},\ldots,b_{i,1}$ and the $u_{i}-c_{i,1}$ path given by $u_{i},c_{i,s},c_{i,s-1},\ldots,c_{i,1}$, respectively).

For each $i\in\{1,2,\ldots,n\}$, let $T_{i}$ be the subtree of $G$ consisting of the $u_{i}-a_{i,1}$ path, the $u_{i}-b_{i,1}$ path, and the $u_{i}-c_{i,1}$ path; further, let $P^{i,a}$ be the $a_{i,s}-a_{i,1}$ path, $P^{i,b}$ the $b_{i,s}-b_{i,1}$ path, and $P^{i,c}$ the $c_{i,s}-c_{i,1}$ path. Then, for each $i\in\{1,2,\ldots,n\}$,

[TABLE]

We determine $\kappa(G)$, $\dim^{k}(G)$ for $k\!\in\!\{1,2,\ldots,\kappa(G)\}$, and $\dim_{f}^{k}(G)$ for $k\in[1,\kappa(G)]$.

Claim 1: $\kappa(G)=2s$.

Proof of Claim 1. Let $x$ and $y$ be two distinct vertices of $G$. First, let $x,y\in V(T_{i})$ for some $i\in\{1,2,\ldots,n\}$. If $d(u_{i},x)\neq d(u_{i},y)$, then $R\{x,y\}\supseteq V(T_{j})$ with $|R\{x,y\}|\geq|V(T_{j})|=3s+1$, where $j\in\{1,2,\ldots,n\}-\{i\}$. If $d(u_{i},x)=d(u_{i},y)$, say $x\in V(P^{i,a})$ and $y\in V(P^{i,b})$ without loss of generality, then $R\{x,y\}=V(P^{i,a})\cup V(P^{i,b})$ with $|R\{x,y\}|=|V(P^{i,a})|+|V(P^{i,b})|=2s$. Second, let $x\in V(T_{i})$ and $y\in V(T_{j})$ for distinct $i,j\in\{1,2,\ldots,n\}$, say $x$ lies on the $u_{i}-a_{i,1}$ path and $y$ lies on the $u_{j}-a_{j,1}$ path, without loss of generality. Then $|R\{x,y\}|\geq 2s$, since at most one vertex lying on a $a_{i,1}-a_{j,1}$ geodesic is at equal distance from both $x$ and $y$. So, $\kappa(G)=2s$. $\Box$

Claim 2: For $k\in\{1,2,\ldots,2s\}$, \dim^{k}(G)=\left\{\begin{array}[]{ll}\frac{3kn}{2}&\mbox{ if kis even},\\[3.0pt] \frac{(3k+1)n}{2}&\mbox{ ifk is odd}.\end{array}\right.

Proof of Claim 2. Let $k\in\{1,2,\ldots,2s\}$. First, we show that $\dim^{k}(G)\geq\frac{3kn}{2}$ for an even $k$, and $\dim^{k}(G)\geq\frac{(3k+1)n}{2}$ for an odd $k$. Let $S$ be a minimum $k$-resolving set of $G$. For each $i\in\{1,2,\ldots,n\}$, (1) implies

[TABLE]

Suppose $k$ is even. By summing over the three inequalities in (2), we obtain $|S\cap V(P^{i,a})|+|S\cap V(P^{i,b})|+|S\cap V(P^{i,c})|\geq\frac{3k}{2}$, and thus $|S\cap V(T_{i})|\geq\frac{3k}{2}$ for each $i\in\{1,2,\ldots,n\}$. So, $|S\cap V(G)|\geq\sum_{i=1}^{n}\frac{3k}{2}=\frac{3kn}{2}$, and hence $\dim^{k}(G)\geq\frac{3kn}{2}$. Now, suppose $k$ is odd. If $|S\cap V(P^{i,a})|\leq\lfloor\frac{k}{2}\rfloor$ and $|S\cap V(P^{i,b})|\leq\lfloor\frac{k}{2}\rfloor$ for some $i\in\{1,2,\ldots,n\}$, then $|S\cap R\{a_{i,s},b_{i,s}\}|=|S\cap V(P^{i,a})|+|S\cap V(P^{i,b})|\leq\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{2}\rfloor=k-1$, contradicting the assumption that $S$ is a $k$-resolving set of $G$; thus, $|S\cap V(P^{i,t})|\leq\lfloor\frac{k}{2}\rfloor$ for at most one $t\in\{a,b,c\}$ for each $i\in\{1,2,\ldots,n\}$. Let $|S\cap V(P^{i,c})|=\alpha\leq\min\{|S\cap V(P^{i,a})|,|S\cap V(P^{i,b})|\}$ for $i\in\{1,2,\ldots,n\}$. Then $|S\cap V(P^{i,a})|\geq k-\alpha$ and $|S\cap V(P^{i,b})|\geq k-\alpha$ from (2), and thus $|S\cap V(T_{i})|\geq 2k-\alpha\geq 2k-\lfloor\frac{k}{2}\rfloor=\frac{3k+1}{2}$. So, $|S\cap V(G)|\geq\sum_{i=1}^{n}\frac{3k+1}{2}=\frac{(3k+1)n}{2}$, and hence $\dim^{k}(G)\geq\frac{(3k+1)n}{2}$.

Second, we show that $\dim^{k}(G)\leq\frac{3kn}{2}$ for an even $k$, and $\dim^{k}(G)\leq\frac{(3k+1)n}{2}$ for an odd $k$. If $k$ is even, let $W_{0}=\cup_{i=1}^{n}(\{a_{i,1},a_{i,2},\ldots,a_{i,\frac{k}{2}}\}\cup\{b_{i,1},b_{i,2},\ldots,b_{i,\frac{k}{2}}\}\cup\{c_{i,1},c_{i,2},\ldots,c_{i,\frac{k}{2}}\})$. If $k$ is odd, let $W_{1}=\cup_{i=1}^{n}(\{a_{i,1},a_{i,2},\ldots,a_{i,\lceil\frac{k}{2}\rceil}\}\cup\{b_{i,1},b_{i,2},\ldots,b_{i,\lceil\frac{k}{2}\rceil}\}\cup\{c_{i,1},c_{i,2},\ldots,c_{i,\lfloor\frac{k}{2}\rfloor}\})$. Note that $|W_{0}|=\frac{3kn}{2}$, $|W_{1}|=\frac{(3k+1)n}{2}$, $|W_{0}\cap V(P^{i,a})|=|W_{0}\cap V(P^{i,b})|=|W_{0}\cap V(P^{i,c})|=\frac{k}{2}$ and $|W_{1}\cap V(P^{i,a})|=|W_{1}\cap V(P^{i,b})|=\lceil\frac{k}{2}\rceil=1+|W_{1}\cap V(P^{i,c})|$ for $i\in\{1,2,\ldots,n\}$. It suffices to show that $W_{0}$ ($W_{1}$, respectively) is a $k$-resolving set of $G$ when $k$ is even (odd, respectively). Let $x$ and $y$ be distinct vertices of $G$. Suppose $x,y\in V(T_{i})$ for some $i\in\{1,2,\ldots,n\}$. If $d(u_{i},x)\neq d(u_{i},y)$, then $|W_{\ell}\cap R\{x,y\}|\geq|V(T_{j})|\geq\frac{3k}{2}\geq k$ for $j\neq i$ and for $\ell\in\{0,1\}$. If $d(u_{i},x)=d(u_{i},y)$, say $x\in V(P^{i,a})$ and $y\in V(P^{i,c})$ (other cases can be handled similarly), then $|W_{\ell}\cap R\{x,y\}|=|W_{\ell}\cap(V(P^{i,a})\cup V(P^{i,c}))|\geq k$ for $\ell\in\{0,1\}$. Now, let $x\in V(T_{i})$ and $y\in V(T_{j})$ for distinct $i,j\in\{1,2\ldots,n\}$; suppose $d(u_{i},x)\leq d(u_{j},y)$, without loss of generality. Since $R\{x,y\}\supseteq V(T_{i})$, $|W_{\ell}\cap R\{x,y\}|\geq|W_{\ell}\cap V(T_{i})|\geq\frac{3k}{2}$ for $\ell\in\{0,1\}$. So, $W_{0}$ ($W_{1}$, respectively) is a $k$-resolving set of $G$ when $k$ is even (odd, respectively). $\Box$

Claim 3: For $k\in[1,2s]$, $\dim_{f}^{k}(G)=\frac{3kn}{2}$.

Proof of Claim 3. Let $k\in[1,2s]$. First, we show that $\dim_{f}^{k}(G)\geq\frac{3kn}{2}$. Let $g:V(G)\rightarrow[0,1]$ be any $k$-resolving function of $G$. From (1), for each $i\in\{1,2,\ldots,n\}$, we have $g(R\{a_{i,s},b_{i,s}\})\!=\!g(V(P^{i,a}))+g(V(P^{i,b}))\geq k$, $g(R\{a_{i,s},c_{i,s}\})\!=\!g(V(P^{i,a}))+g(V(P^{i,c}))\geq k$, and $g(R\{b_{i,s},c_{i,s}\})=g(V(P^{i,b}))+g(V(P^{i,c}))\geq k$. By summing over the three inequalities, we obtain $g(V(P^{i,a}))+g(V(P^{i,b}))+g(V(P^{i,c}))\geq\frac{3k}{2}$, and thus $g(V(T_{i}))\geq\frac{3k}{2}$ for each $i\in\{1,2,\ldots,n\}$. So, $g(V(G))\geq\sum_{i=1}^{n}\frac{3k}{2}=\frac{3kn}{2}$, and hence $\dim_{f}^{k}(G)\geq\frac{3kn}{2}$.

Second, we show that $\dim_{f}^{k}(G)\leq\frac{3kn}{2}$. Let $h:V(G)\rightarrow[0,1]$ be a function defined by

[TABLE]

Notice that $h(V(P^{i,a}))=h(V(P^{i,b}))=h(V(P^{i,c}))=\frac{k}{2}$ for each $i\in\{1,2,\ldots,n\}$, and $h(V(G))=\frac{3kn}{2}$. It suffices to show that $h$ is a $k$-resolving function of $G$. Let $x$ and $y$ be distinct vertices of $G$. Suppose $x,y\in V(T_{i})$ for some $i\in\{1,2,\ldots,n\}$. If $d(u_{i},x)\neq d(u_{i},y)$, then $h(R\{x,y\})\geq h(V(T_{j}))\geq\frac{3k}{2}$ for $j\neq i$. If $d(u_{i},x)=d(u_{i},y)$, say $x\!\in\!V(P^{i,a})$ and $y\!\in\!V(P^{i,b})$ without loss of generality, then $h(R\{x,y\})\!=\!h(V(P^{i,a}))+h(V(P^{i,b}))=k$. Now, let $x\in V(T_{i})$ and $y\in V(T_{j})$ for distinct $i,j\in\{1,2\ldots,n\}$; suppose $d(u_{i},x)\leq d(u_{j},y)$, without loss of generality. Then $h(R\{x,y\})\geq h(V(T_{i}))\geq\frac{3k}{2}$. So, $h$ is a $k$-resolving function of $G$. $\Box$

By Claims 2 and 3, we see that, for each odd $k\in\{1,2,\ldots,2s\}$, $\dim^{k}(G)-\dim_{f}^{k}(G)=\frac{(3k+1)n}{2}-\frac{3kn}{2}=\frac{n}{2}$, which can be arbitrarily large.  ∎

Next, we recall some results on the fractional metric dimension of graphs. One can easily see that, for any connected graph $G$ of order at least two, $1\leq\dim_{f}(G)\leq\frac{|V(G)|}{2}$ (see [2]). For the characterization of graphs $G$ achieving the lower bound, see Theorem 2.7(a). Regarding the characterization of graphs $G$ achieving the upper bound, the following result is stated in [2] and a correct proof is provided in [21].

Theorem 2.5**.**

[2, 21]* Let $G$ be a connected graph of order at least two. Then $\dim_{f}(G)=\frac{|V(G)|}{2}$ if and only if there exists a bijection $\alpha:V(G)\rightarrow V(G)$ such that $\alpha(v)\neq v$ and $|R\{v,\alpha(v)\}|=2$ for all $v\in V(G)$.*

An explicit characterization of graphs $G$ satisfying $\dim_{f}(G)=\frac{|V(G)|}{2}$ is given in [3]. We recall the following construction from [3]. Let $\mathcal{K}=\{K_{n}:n\geq 2\}$ and $\overline{\mathcal{K}}=\{\overline{K}_{n}:n\geq 2\}$. Let $H[\mathcal{K}\cup\overline{\mathcal{K}}]$ be the family of graphs obtained from a connected graph $H$ by (i) replacing each vertex $u_{i}\in V(H)$ by a graph $H_{i}\in\mathcal{K}\cup\overline{\mathcal{K}}$, and (ii) each vertex in $H_{i}$ is adjacent to each vertex in $H_{j}$ if and only if $u_{i}u_{j}\in E(H)$.

Theorem 2.6**.**

[3]* Let $G$ be a connected graph of order at least two. Then $\dim_{f}(G)=\frac{|V(G)|}{2}$ if and only if $G\in H[\mathcal{K}\cup\overline{\mathcal{K}}]$ for some connected graph $H$.*

Now, we recall the fractional metric dimension of some classes of graphs. We begin by recalling some terminologies. Fix a graph $G$. A leaf $u$ is called a terminal vertex of a major vertex $v$ if $d(u,v)<d(u,w)$ for every other major vertex $w$. The terminal degree, $ter_{G}(v)$, of a major vertex $v$ is the number of terminal vertices of $v$. A major vertex $v$ is an exterior major vertex if it has positive terminal degree. Let $ex(G)$ denote the number of exterior major vertices of $G$, $ex_{a}(G)$ the number of exterior major vertices $u$ with $ter_{G}(u)=a$, and $\sigma(G)$ the number of leaves of $G$.

Theorem 2.7**.**

[23]* For any graph $G$ of order $n\geq 2$, $\dim_{f}(G)=1$ if and only if $G\cong P_{n}$.*

[29]* For a tree $T$, $\dim_{f}(T)=\frac{1}{2}(\sigma(T)-ex_{1}(T))$.*

[2]* For the Petersen graph $\mathcal{P}$, $\dim_{f}(\mathcal{P})=\frac{5}{3}$.*

[2]* For an $n$-cycle $C_{n}$, \dim_{f}(C_{n})=\left\{\begin{array}[]{ll}\frac{n}{n-2}&\mbox{if nis even},\\ \frac{n}{n-1}&\mbox{ifn is odd}.\end{array}\right.*

[2]* For the wheel graph $W_{n}$ of order $n\geq 5$, \dim_{f}(W_{n})=\left\{\begin{array}[]{ll}2&\mbox{if }n=5,\\ \frac{3}{2}&\mbox{if }n=6,\\ \frac{n-1}{4}&\mbox{if }n\geq 7.\end{array}\right.*

[23]* If $B_{m}$ is a bouquet of $m$ cycles with a cut-vertex (i.e., the vertex sum of $m$ cycles at one common vertex), where $m\geq 2$, then $\dim_{f}(B_{m})=m$.*

[29]* For $m\geq 2$, let $G=K_{a_{1},a_{2},\ldots,a_{m}}$ be a complete $m$-partite graph of order $n=\sum_{i=1}^{m}a_{i}$, and let $s$ be the number of partite sets of $G$ consisting of exactly one element. Then*

[TABLE]

[2]* For the grid graph $G=P_{s}\square P_{t}$ ($s,t\geq 2$), $\dim_{f}(G)=2$.*

3 Some general results on fractional $k$-metric dimension

In this section, we show that $\dim_{f}^{k}(G)\geq k\dim_{f}(G)$ for any $k\in[1,\kappa(G)]$. We also describe a condition for which $\dim_{f}^{k}(G)=k\dim_{f}(G)$ holds for all $k\in[1,\kappa(G)]$. For all $k\in[1,\kappa(G)]$, we show that $k\leq\dim_{f}^{k}(G)\leq\frac{k}{\kappa(G)}|V(G)|$, which implies $k\leq\dim_{f}^{k}(G)\leq|V(G)|$; we characterize graphs $G$ satisfying $\dim_{f}^{k}(G)=k$ and $\dim_{f}^{k}(G)=|V(G)|$, respectively. We conclude with an example such that two non-isomorphic graphs $H_{1}$ and $H_{2}$ satisfy $\dim^{k}_{f}(H_{1})=\dim_{f}^{k}(H_{2})$ for all $k\in[1,\kappa]$, where $\kappa(H_{1})=\kappa(H_{2})=\kappa$.

We begin by comparing the fractional metric dimension and the fractional $k$-metric dimension of graphs.

Lemma 3.1**.**

For any connected graph $G$ and for any $k\in[1,\kappa(G)]$, $\dim_{f}^{k}(G)\geq k\dim_{f}(G)$.

Proof.

Let $g:V(G)\rightarrow[0,1]$ be a minimum $k$-resolving function of $G$. Then $g(R\{x,y\})\geq k$ for any two distinct vertices $x,y\in V(G)$. Now, let $h:V(G)\rightarrow[0,1]$ be a function defined by $h(u)=\frac{1}{k}g(u)$ for each $u\in V(G)$. Then $h(R\{x,y\})=\frac{1}{k}g(R\{x,y\})\geq 1$ for any distinct vertices $x,y\in V(G)$; thus, $h$ is a resolving function of $G$. So, $h(V(G))=\frac{1}{k}\dim^{k}_{f}(G)\geq\dim_{f}(G)$, i.e., $\dim^{k}_{f}(G)\geq k\dim_{f}(G)$.  ∎

Next, we examine the conditions for which $\dim_{f}^{k}(G)=k\dim_{f}(G)$ holds.

Lemma 3.2**.**

Let $G$ be a connected graph and let $k\in[1,\kappa(G)]$. If there exists a minimum resolving function $g:V(G)\rightarrow[0,1]$ such that $g(v)\leq\frac{1}{k}$ for each $v\in V(G)$, then $\dim_{f}^{k}(G)=k\dim_{f}(G)$ for any $k\in[1,\kappa(G)]$.

Proof.

Let $g:V(G)\rightarrow[0,1]$ be a minimum resolving function of $G$ satisfying $g(v)\leq\frac{1}{k}$ for each $v\in V(G)$. Let $h:V(G)\rightarrow[0,1]$ be a function defined by $h(v)=kg(v)$ for each $v\in V(G)$. Then $h$ is a $k$-resolving function of $G$: (i) for each $v\in V(G)$, $0\leq h(v)=kg(v)\leq 1$; (ii) for any two distinct $x,y\in V(G)$, $h(R\{x,y\})=kg(R\{x,y\})\geq k$, since $g(R\{x,y\})\geq 1$ by the assumption that $g$ is a resolving function of $G$. So, $\dim_{f}^{k}(G)\leq h(V(G))=kg(V(G))=k\dim_{f}(G)$. Since $\dim_{f}^{k}(G)\geq k\dim_{f}(G)$ by Lemma 3.1, $\dim_{f}^{k}(G)=k\dim_{f}(G)$.  ∎

Next, we obtain the lower and upper bounds of $\dim_{f}^{k}(G)$ in terms of $k$, $\kappa(G)$, and the order of $G$.

Proposition 3.3**.**

Let $G$ be a connected graph of order $n$. For any $k\in[1,\kappa(G)]$, $\displaystyle k\leq\dim_{f}^{k}(G)\leq\frac{kn}{\kappa(G)},$ where both bounds are sharp.

Proof.

The lower bound is trivial. For the upper bound, let $g:V(G)\rightarrow[0,1]$ be a function such that $g(v)=\frac{k}{\kappa(G)}$ for each $v\in V(G)$. Since $\kappa(G)=\min\{|R\{x,y\}|:x\neq y\mbox{ and }x,y\in V(G)\}$, $g(R\{u,v\})\geq k$ for any distinct vertices $u,v\in V(G)$. So, $g$ is a $k$-resolving function of $G$, and hence $\dim_{f}^{k}(G)\leq g(V(G))=\sum_{i=1}^{n}\frac{k}{\kappa(G)}=\frac{kn}{\kappa(G)}$.

For the sharpness of the lower bound, see Proposition 3.5(a); for the sharpness of the upper bound, we refer to Proposition 4.7.  ∎

As an immediate consequence of Proposition 3.3, we have the following.

Corollary 3.4**.**

For a connected graph $G$ of order $n$ and for $k\in[1,\kappa(G)]$, $\displaystyle k\leq\dim_{f}^{k}(G)\leq n$.

Next, we characterize graphs $G$ achieving the lower bound and the upper bound, respectively, of Corollary 3.4. Let $\mathcal{R}_{\kappa}(G)=\displaystyle\bigcup_{x,y\in V(G),x\neq y,|R\{x,y\}|=\kappa}R\{x,y\}$, where $\kappa=\kappa(G)$.

Proposition 3.5**.**

For a connected graph $G$ of order $n\geq 2$ and for $k\in[1,\kappa(G)]$,

• (a)

$\dim_{f}^{k}(G)=k$* if and only if $G\cong P_{n}$ and $k\in[1,2]$,*

• (b)

$\dim_{f}^{k}(G)=n$* if and only if $k=\kappa(G)=\kappa$ and $V(G)=\mathcal{R}_{\kappa}(G)$.*

Proof.

(a) ($\Leftarrow$) If $G\cong P_{n}$ and $k\in[1,2]$, then $\dim_{f}^{k}(G)=k$ by Proposition 4.1.

($\Rightarrow$) Suppose that $\dim_{f}^{k}(G)=k$. Since $\dim_{f}^{k}(G)\geq k\dim_{f}(G)\geq k$ by Lemma 3.1, $\dim_{f}^{k}(G)=k$ implies $\dim_{f}(G)=1$; thus $G\cong P_{n}$ by Theorem 2.7(a) and $k\in[1,2]$ from Proposition 4.1.

(b) ($\Leftarrow$) Let $k=\kappa(G)=\kappa$ and $V(G)=\mathcal{R}_{\kappa}(G)$. Then, for any vertex $v\in V(G)$, there exist two distinct vertices $x,y\in V(G)$ such that $v\in R\{x,y\}$ with $|R\{x,y\}|=\kappa$. Since any $\kappa$-resolving function $g$ of $G$ must satisfy $g(R\{x,y\})\geq\kappa$ and $g(v)\leq 1$ for each $v\in V(G)$, $g(u)=1$ for each $u\in R\{x,y\}$ with $|R\{x,y\}|=\kappa$. Since $V(G)=\mathcal{R}_{\kappa}(G)$, $g(v)=1$ for each $v\in V(G)$. Thus, $g(V(G))=n$ and hence $\dim_{f}^{k}(G)=n$.

($\Rightarrow$) Let $\dim_{f}^{k}(G)=n$. By Proposition 3.3, $k=\kappa$. Suppose that $\mathcal{R}_{\kappa}(G)\subsetneq V(G)$ and let $w\in V(G)-\mathcal{R}_{\kappa}(G)$. Then, for any vertex $w^{\prime}\in V(G)-\{w\}$, $|R\{w,w^{\prime}\}|\geq\kappa(G)+1$. If $h:V(G)\rightarrow[0,1]$ is a function defined by $h(w)=0$ and $h(v)=1$ for each $v\in V(G)-\{w\}$, then $h$ is a $\kappa$-resolving function of $G$ with $h(V(G))=n-1$, which contradicts the assumption that $\dim_{f}^{k}(G)=n$.  ∎

Remark 3.6**.**

Let $G\in H[\mathcal{K}\cup\overline{\mathcal{K}}]$ for some connected graph $H$, as described in Theorem 2.6. Then $\kappa(G)=2$ and $V(G)=\mathcal{R}_{\kappa}(G)$; thus $\dim_{f}^{2}(G)=|V(G)|$ by Proposition 3.5(b). More generally, $\dim_{f}^{k}(G)=k\dim_{f}(G)=\frac{k}{2}|V(G)|$ for $k\in[1,2]$ by Lemma 3.2, since $g:V(G)\rightarrow[0,1]$ defined by $g(u)=\frac{1}{2}$, for each $u\in V(G)$, forms a minimum resolving function of $G$.

We conclude this section with an example showing that two non-isomorphic graphs can have the same $\kappa$ and identical $k$-fractional metric dimension for all $k\in[1,\kappa]$.

Remark 3.7**.**

There exist non-isomorphic graphs $H_{1}$ and $H_{2}$ such that $\dim_{f}^{k}(H_{1})=\dim_{f}^{k}(H_{2})$ for all $k\in[1,\kappa]$, where $\kappa=\kappa(H_{1})=\kappa(H_{2})$. For example, let $H_{1}\cong K_{2,2,3}$ and $H_{2}\cong K_{3,4}$; then $\dim_{f}(H_{1})=\dim_{f}(H_{2})=\frac{7}{2}$ by Theorem 2.7(g). Since both $H_{1}$ and $H_{2}$ have twin vertices, $\kappa(H_{1})=\kappa(H_{2})=2$. Also note that a function $g_{i}:V(H_{i})\rightarrow[0,1]$ defined by $g_{i}(u)=\frac{1}{2}$ for each $u\in V(H_{i})$ forms a minimum resolving function for $H_{i}$, where $i\in\{1,2\}$. By Lemma 3.2, $\dim_{f}^{k}(H_{1})=\dim_{f}^{k}(H_{2})=\frac{7}{2}k$ for every $k\in[1,\kappa]$, whereas $H_{1}\not\cong H_{2}$.

4 The fractional $k$-metric dimension of some graphs

In this section, we determine $\dim_{f}^{k}(G)$ for $k\in[1,\kappa(G)]$ when $G$ is a tree, a cycle, a wheel graph, the Petersen graph, a bouquet of cycles, a complete multi-partite graph, or a grid graph (the Cartesian product of two paths). Along the way, we provide an example showing that $\dim_{f}^{k}(G)-k\dim_{f}(G)$ can be arbitrarily large for some $k\in(1,\kappa(G)]$. First, we determine $\dim_{f}^{k}(G)$ when $G$ is a path.

Proposition 4.1**.**

Let $P_{n}$ be an $n$-path, where $n\geq 2$. Then $\dim_{f}^{k}(P_{2})=k$ for $k\in[1,2]$ and, for $n\geq 3$,

[TABLE]

Proof.

Let $P_{n}$ be an $n$-path given by $u_{1},u_{2},\ldots,u_{n}$, where $n\geq 2$; then $\kappa(P_{2})=2$ and $\kappa(P_{n})=n-1$ for $n\geq 3$. Since a function $h$ defined on $V(P_{2})$ by $h(u_{1})=h(u_{2})=\frac{1}{2}$ is a minimum resolving function of $P_{2}$, $\dim_{f}^{k}(P_{2})=k\dim_{f}(P_{2})=k$ for $k\in[1,2]$ by Theorem 2.7(a) and Lemma 3.2. So, let $n\geq 3$ and we consider two cases.

Case 1: $k\in[1,2]$. If $g:V(P_{n})\rightarrow[0,1]$ is a function defined by $g(u_{1})=g(u_{n})=\frac{1}{2}$ and $g(u_{i})=0$ for each $i\in\{2,\ldots,n-1\}$, then $g$ is a minimum resolving function of $P_{n}$: (i) for any two distinct vertices $x,y\in V(P_{n})$, $R\{x,y\}\supseteq\{u_{1},u_{n}\}$, and hence $g(R\{x,y\})\geq g(u_{1})+g(u_{n})=\frac{1}{2}+\frac{1}{2}=1$; (ii) $g(V(P_{n}))=1=\dim_{f}(P_{n})$ by Theorem 2.7(a). Since $g(u_{i})\leq\frac{1}{2}\leq\frac{1}{k}$ for each $i\in\{1,2,\ldots,n\}$, we have $\dim_{f}^{k}(P_{n})=k\dim_{f}(P_{n})=k$ for any $k\in[1,2]$ by Lemma 3.2 and Theorem 2.7(a).

Case 2: $k\in(2,n-1]$. Note that $n\geq 4$ in this case since $\kappa(P_{3})=2$. Let $h:V(P_{n})\rightarrow[0,1]$ be a $k$-resolving function of $P_{n}$. Let $h(u_{1})+h(u_{n})=b$; then $0\leq b\leq 2$. Since $R\{u_{i},u_{i+2}\}=V(P_{n})-\{u_{i+1}\}$ for $i\in\{1,2,\ldots,n-2\}$, $h(R\{u_{i},u_{i+2}\})=h(V(P_{n}))-h(u_{i+1})\geq k$ for each $i\in\{1,2,\ldots,n-2\}$. By summing over the $(n-2)$ inequalities, we have $(n-2)h(V(P_{n}))-\sum_{j=2}^{n-1}h(u_{j})\geq(n-2)k$, i.e., $(n-3)h(V(P_{n}))+h(u_{1})+h(u_{n})\geq(n-2)k$. So, $h(V(P_{n}))\geq\frac{(n-2)k-b}{n-3}$ since $n>3$; note that the minimum of $h(V(P_{n}))$ is $\frac{(n-2)k-2}{n-3}$ when $b=h(u_{1})+h(u_{n})$ takes the maximum value 2. Thus, $\dim_{f}^{k}(P_{n})\geq\frac{(n-2)k-2}{n-3}=2+(k-2)\frac{n-2}{n-3}$.

Now, let $g:V(P_{n})\rightarrow[0,1]$ be a function defined by

[TABLE]

Then $g$ is a $k$-resolving function of $P_{n}$: (i) $0<\frac{k-2}{n-3}\leq 1$ for any $k\in(2,n-1]$; (ii) for any distinct $i,j\in\{1,2,\ldots,n\}$, $g(R\{u_{i},u_{j}\})\geq 2+(n-3)\frac{k-2}{n-3}=k$ since $R\{u_{i},u_{j}\}\supseteq\{u_{1},u_{n}\}$ and $|R\{u_{i},u_{j}\}|\geq n-1$. So, $\dim_{f}^{k}(P_{n})\leq g(V(P_{n}))=2+(n-2)\frac{k-2}{n-3}$.

Therefore, $\dim_{f}^{k}(P_{n})=2+(k-2)\frac{n-2}{n-3}$ for $k\in(2,n-1]$.  ∎

Second, we determine $\dim_{f}^{k}(T)$ when $T$ is a tree that is not a path. Let $M(T)$ be the set of exterior major vertices of a tree $T$. Let $M_{1}(T)=\{w\in M(T):ter_{T}(w)=1\}$, $M_{2}(T)=\{w\in M(T):ter_{T}(w)=2\}$, and $M_{3}(T)=\{w\in M(T):ter_{T}(w)\geq 3\}$; note that $M(T)=M_{1}(T)\cup M_{2}(T)\cup M_{3}(T)$. For any vertex $v\in M(T)$, let $T_{v}$ be the subtree of $T$ induced by $v$ and all vertices belonging to the paths joining $v$ with its terminal vertices. Let $M^{*}(T)=M_{2}(T)\cup M_{3}(T)$. We recall the following result on $\kappa(T)$.

Theorem 4.2**.**

[12]* For a tree $T$ that is not a path,*

[TABLE]

We begin by examining $\dim_{f}^{k}(T)$ when $T$ is a tree with exactly one exterior major vertex.

Proposition 4.3**.**

Let $T$ be a tree with $ex(T)=1$. Let $v$ be the exterior major vertex of $T$ and let $\ell_{1},\ell_{2},\ldots,\ell_{a}$ be the terminal vertices of $v$ in $T$ (note that $a\geq 3$). Suppose that $d(v,\ell_{1})\leq d(v,\ell_{2})\leq\ldots\leq d(v,\ell_{a})$. Then $\kappa(T)=d(\ell_{1},\ell_{2})$, and

• (a)

if $d(v,\ell_{1})=d(v,\ell_{2})$, then $\dim_{f}^{k}(T)=k\dim_{f}(T)=\frac{ka}{2}$ for $k\in[1,\kappa(T)]$;

• (b)

if $d(v,\ell_{1})<d(v,\ell_{2})$, then

[TABLE]

Proof.

For each $i\in\{1,2,\ldots,a\}$, let $s_{i}$ be the neighbor of $v$ lying on the $v-\ell_{i}$ path, and let $P^{i}$ denote the $s_{i}-\ell_{i}$ path in $T$. By Theorem 4.2, $\kappa(T)=d(\ell_{1},\ell_{2})$. Let $k\in[1,\kappa(T)]$.

(a) Let $d(v,\ell_{1})\!=\!d(v,\ell_{2})$. By Lemma 3.1, $\dim_{f}^{k}(T)\!\geq\!k\dim_{f}(T)$. We will show that $\dim_{f}^{k}(T)\!\leq\!k\dim_{f}(T)$. Let $g:V(T)\rightarrow[0,1]$ be a function defined by

[TABLE]

Note that, for each $i\in\{1,2,\ldots,a\}$, (i) $g(V(P^{i}))=\frac{k}{2}$; (ii) $0\leq\frac{k}{2d(v,\ell_{i})}\leq 1$ since $k\leq\kappa(T)=d(\ell_{1},\ell_{2})=2d(v,\ell_{1})\leq 2d(v,\ell_{i})$. If two distinct vertices $x$ and $y$ lie on the $v-\ell_{i}$ path for some $i\in\{1,2,\ldots,a\}$, then $g(R\{x,y\})\geq g(V(T)-V(P^{i}))\geq k$ since $a\geq 3$. For distinct $i,j\in\{1,2,\ldots,a\}$, if $x\in V(P^{i})$ and $y\in V(P^{j})$ with $d(v,x)\neq d(v,y)$, say $d(v,x)<d(v,y)$ without loss of generality, then $g(R\{x,y\})\geq g(V(T)-V(P^{j}))\geq k$; if $x\in V(P^{i})$ and $y\in V(P^{j})$ with $d(v,x)=d(v,y)$, then $g(R\{x,y\})=g(V(P^{i})\cup V(P^{j}))=k$. So, $g$ is a $k$-resolving function of $T$ with $g(V(T))=\frac{ka}{2}$. Thus $\dim^{k}_{f}(T)\leq\frac{ka}{2}=k\dim_{f}(T)$ by Theorem 2.7(b). Therefore, $\dim_{f}^{k}(T)=k\dim_{f}(T)=\frac{ka}{2}$ for $k\in[1,\kappa(T)]$.

(b) Let $d(v,\ell_{1})<d(v,\ell_{2})$, and we consider two cases.

Case 1: $k\in[1,2d(v,\ell_{1})]$. In this case, the function $g$ in (3) is a $k$-resolving function of $T$ as shown in the proof for (a); thus, $\dim_{f}^{k}(T)\leq g(V(T))=\frac{ka}{2}=k\dim_{f}(T)$. Since $\dim_{f}^{k}(T)\geq k\dim_{f}(T)$ by Lemma 3.1, $\dim_{f}^{k}(T)=k\dim_{f}(T)=\frac{ka}{2}$.

Case 2: $k\in(2d(v,\ell_{1}),\kappa(T)]$. Let $h:V(T)\rightarrow[0,1]$ be a minimum $k$-resolving function of $T$. Note that (i) $h(V(P^{1}))\leq d(v,\ell_{1})$ since $h(u)\leq 1$ for each $u\in V(T)$; (ii) for distinct $i,j\in\{1,2,\ldots,a\}$, $h(V(P^{i}))+h(V(P^{j}))\geq k$ since $R\{s_{i},s_{j}\}=V(P^{i})\cup V(P^{j})$. Let $h(V(P^{1}))=\beta$. From $h(V(P^{1}))+h(V(P^{j}))\geq k$ for each $j\in\{2,3,\ldots,a\}$, $h(V(P^{j}))\geq k-\beta$. So, $h(V(T))\geq h(V(P^{1}))+\sum_{i=2}^{a}h(V(P^{i}))\geq\beta+(a-1)(k-\beta)=(a-1)k-(a-2)\beta\geq(a-1)k-(a-2)d(v,\ell_{1})$ since $\beta\leq d(v,\ell_{1})$; thus $\dim_{f}^{k}(T)\geq(a-1)k-(a-2)d(v,\ell_{1})$.

Next, we show that $\dim_{f}^{k}(T)\leq(a-1)k-(a-2)d(v,\ell_{1})$. Let $g:V(T)\rightarrow[0,1]$ be a function defined by

[TABLE]

Note that (i) $g(V(P^{1}))=d(v,\ell_{1})$; (ii) for $j\in\{2,3,\ldots,a\}$, $g(V(P^{j}))=k-d(v,\ell_{1})>k-\frac{k}{2}=\frac{k}{2}$ since $k>2d(v,\ell_{1})$; (iii) $g(V(T))=d(v,\ell_{1})+(a-1)(k-d(v,\ell_{1}))=(a-1)k-(a-2)d(v,\ell_{1})$. Also note that $g$ is a $k$-resolving function of $T$: (i) $0\leq\frac{d(v,\ell_{1})}{d(v,\ell_{j})}\leq\frac{k-d(v,\ell_{1})}{d(v,\ell_{j})}\leq\frac{d(\ell_{1},\ell_{2})-d(v,\ell_{1})}{d(v,\ell_{j})}=\frac{d(v,\ell_{2})}{d(v,\ell_{j})}\leq 1$ for $j\in\{2,\ldots,a\}$; (ii) if two distinct vertices $x$ and $y$ lie on the $v-\ell_{i}$ path for some $i\in\{1,2,\ldots,a\}$, then $g(R\{x,y\})\geq g(V(T))-g(V(P^{i}))\geq\min\{k,2(k-d(v,\ell_{1}))\}=k$ since $a\geq 3$ and $k-d(v,\ell_{1})>\frac{k}{2}$; (iii) if $x\in V(P^{i})$ and $y\in V(P^{j})$ with $d(v,x)\neq d(v,y)$ for distinct $i,j\in\{1,2,\ldots,a\}$, then $g(R\{x,y\})\geq\min\{g(V(T)-V(P^{i})),g(V(T)-V(P^{j}))\}\geq k$, since at most one vertex in the $\ell_{i}-\ell_{j}$ path can be at equal distance from both $x$ and $y$ in $T$; (iv) if $x\in V(P^{i})$ and $y\in V(P^{j})$ with $d(v,x)=d(v,y)$ for distinct $i,j\in\{1,2,\ldots,a\}$, then $g(R\{x,y\})=g(V(P^{i}))+g(V(P^{j}))\geq\min\{k,2(k-d(v,\ell_{1}))\}=k$. Thus, $\dim_{f}^{k}(T)\leq g(V(T))=(a-1)k-(a-2)d(v,\ell_{1})$.

Therefore, $\dim_{f}^{k}(T)=(a-1)k-(a-2)d(v,\ell_{1})$.  ∎

Next, we determine $\dim_{f}^{k}(T)$ for a tree $T$ with $ex(T)\geq 1$. We begin with the following lemma, which, besides being useful for Theorem 4.5, bears independent interest.

Lemma 4.4**.**

Let $T$ be a tree with $ex(T)\geq 2$. For $w\in M^{*}(T)$, let $x\in V(T_{w})$ and $y\in V(T)-V(T_{w})$. Then either $R\{x,y\}\supseteq V(T_{w})$ or $R\{x,y\}\supseteq V(T_{w^{\prime}})$ for some $w^{\prime}\in M^{*}(T)-\{w\}$.

Proof.

Since $ex(T)\geq 2$, $|M^{*}(T)|\geq 2$. Let $x\in V(T_{w})$ for some $w\in M^{*}(T)$. First, suppose $y\in V(T_{w^{\prime}})$ for some $w^{\prime}\in M^{*}(T)-\{w\}$. Assume, for contradiction, that there exist $u\in V(T_{w})$ and $v\in V(T_{w^{\prime}})$ such that

[TABLE]

Put $a=d(y,v),b=d(v,w^{\prime}),c=d(w^{\prime},w),d=d(w,u),e=d(u,x)$. Let us call a path leading from $w$ to any of its leaves a “$w$-terminal path”. We may assume that $u$ and $x$ lie in the same $w$-terminal path, since $d(x,u)=d(y,u)$ implies $d(x,w)=d(y,w)$ if $u$ and $x$ lie in distinct $w$-terminal paths. Likewise, we assume $v$ and $y$ lie in the same $w^{\prime}$-terminal path. After writing the two equations (4) in terms of components $a,b,c,d,e$ and simplifying, we obtain $b+c+d=0$. This means that $b=c=d=0$, since all variables denote (nonnegative) distances. In particular, the distinctness of $w$ and $w^{\prime}$ is contradicted by $c=0$.

Now, suppose $y\notin V(T_{z})$ for any $z\in M^{*}(T)$; then either $y\in V(T_{y^{\prime}})$ for some $y^{\prime}\in M_{1}(T)$ or $y\not\in V(T_{z^{\prime}})$ for any $z^{\prime}\in M(T)$. Note that there exists a vertex $w^{\prime}\in M^{*}(T)-\{w\}$ such that either $y$ or $y^{\prime}$ lies in the $w-w^{\prime}$ path in $T$. Since $d(s,x)=d(s,y)$ for $s\in V(T_{w^{\prime}})$ is equivalent to $d(w^{\prime},x)=d(w^{\prime},y)$, we may assume, for contradiction, that

[TABLE]

Put $d(y,w^{\prime})=a+b$ and $d(w,w^{\prime})=c+b$, where $b\geq 0$ denotes the length of the path shared between the $w-w^{\prime}$ path and the $y-w^{\prime}$ path. Similarly, put $d(w,t)=d$ and $d(x,t)=e$. As before, we may assume that $x$ and $t$ lie in the same $w$-terminal path. After simplifying the two equations (5) in terms of $a,b,c,d,e$, we obtain $c=d=0$. This implies that either $y\in V(T_{w})$ (when $b>0$) or $w=w^{\prime}$ (when $b=0$); both possibilities contradict the present assumptions.  ∎

Theorem 4.5**.**

Let $T$ be a tree with $ex(T)\geq 1$. Then $\kappa(T)=\min\{\kappa(T_{v}):v\in M^{*}(T)\}$ and, for $k\in[1,\kappa(T)]$,

[TABLE]

Proof.

If $ex(T)=1$, then $|M_{2}(T)|=0$ and $|M_{3}(T)|=1$, and so (6) trivially holds; see Propostion 4.3 for explicit formulas for $\kappa(T)$ and $\dim_{f}^{k}(T)$. So, let $ex(T)\geq 2$; then $|M^{*}(T)|\geq 2$. By Theorem 4.2, $\kappa(T)=\displaystyle\min\{\kappa(T_{v}):v\in M^{*}(T)\}$. Let $k\in[1,\kappa(T)]$; notice that $\kappa(T)\leq\kappa(T_{z})$ for any $z\in M^{*}(T)$.

First, we show that $\dim_{f}^{k}(T)\geq k|M_{2}(T)|+\sum_{v\in M_{3}(T)}\dim_{f}^{k}(T_{v})$. For $v\in M_{2}(T)$, let $N(v)\cap V(T_{v})=\{r_{1},r_{2}\}$. For $w\in M_{3}(T)$, let $\ell_{1},\ell_{2},\ldots,\ell_{\sigma}$ be the terminal vertices of $w$ with $ter_{T}(w)=\sigma$, and let $s_{i}$ be the neighbor of $w$ lying on the $w-\ell_{i}$ path, where $i\in\{1,2,\ldots,\sigma\}$; further, let $P^{i}$ denote the $s_{i}-\ell_{i}$ path. Let $g:V(T)\rightarrow[0,1]$ be a minimum $k$-resolving function of $T$. If $M_{2}(T)\neq\emptyset$, then, for any $v\in M_{2}(T)$, $R\{r_{1},r_{2}\}=V(T_{v})-\{v\}$ and $g(R\{r_{1},r_{2}\})=g(V(T_{v})-\{v\})\geq k$; thus $\sum_{v\in M_{2}(T)}g(V(T_{v}))\geq\sum_{v\in M_{2}(T)}k=k|M_{2}(T)|$. If $M_{3}(T)\neq\emptyset$, then, for any $w\in M_{3}(T)$, notice $R\{s_{i},s_{j}\}=V(P^{i})\cup V(P^{j})$ for any distinct $i,j\in\{1,2,\ldots,\sigma\}$. This, together with the argument used in the proof of Proposition 4.3, we have $\sum_{w\in M_{3}(T)}g(V(T_{w}))\geq\sum_{w\in M_{3}(T)}\dim_{f}^{k}(T_{w})$. Thus, we have

[TABLE]

Next, we show that $\dim_{f}^{k}(T)\leq k|M_{2}(T)|+\sum_{v\in M_{3}(T)}\dim_{f}^{k}(T_{v})$. For each $w\in M_{3}(T)$, let $g_{w}$ be a minimum $k$-resolving function on $V(T_{w})$. For each $w\in M_{2}(T)$, define a function $h_{w}$ on $V(T_{w})$ such that $h_{w}(w)=0$ and $h_{w}(u)=\frac{k}{|V(T_{w})|-1}$ if $u\neq w$. For $k\in[1,\kappa(T)]$, let $g:V(T)\rightarrow[0,1]$ be the function defined by

[TABLE]

Note that (i) for each $w\in M_{2}(T)$, $g(V(T_{w}))=h_{w}(V(T_{w})-\{w\})=k$; (ii) for each $w\in M_{3}(T)$, $g(V(T_{w}))=g_{w}(V(T_{w}))=\dim_{f}^{k}(T_{w})\geq k$; (iii) $g(V(T))=k|M_{2}(T)|+\sum_{w\in M_{3}(T)}\dim_{f}^{k}(T)$. It suffices to show $g$ is a $k$-resolving function of $T$. Obviously, $0\leq g(u)\leq 1$ for each $u\in V(T)$. So, let $x$ and $y$ be distinct vertices of $T$; we will show that $g(R\{x,y\})\geq k$. Consider three cases: (1) there is a $w\in M^{*}(T)$ such that $\{x,y\}\subseteq V(T_{w})$; (2) there is a $w\in M^{*}(T)$ such that $x\in V(T_{w})$ and $y\notin V(T_{w})$; (3) $\{x,y\}\subseteq V(T)-\cup_{w\in M^{*}(T)}V(T_{w})$. In case (1), if $w\in M_{3}(T)$, then $g(R\{x,y\})\geq g_{w}(R\{x,y\}\cap V(T_{w}))\geq k$, since $g_{w}$ is a $k$-resolving function on $V(T_{w})$; if $w\in M_{2}(T)$ and $d(x,w)\neq d(y,w)$, then there is a $z\in M^{*}(T)-\{w\}$ such that $g(R\{x,y\})\geq g(V(T_{z}))\geq k$; if $w\in M_{2}(T)$ and $d(x,w)=d(y,w)$, then $g(R\{x,y\})=h_{w}(V(T_{w})-\{w\})=k$. In case (2), by Lemma 4.4, either $R\{x,y\}\supseteq V(T_{w})$ or $R\{x,y\}\supseteq V(T_{w^{\prime}})$ for some $w^{\prime}\in M^{*}(T)-\{w\}$; thus $g(R\{x,y\})\geq k$. So, we consider case (3). Note that $x\in V(T_{x^{\prime}})$ for some $x^{\prime}\in M_{1}(T)$ or $x\not\in V(T_{z})$ for any $z\in M(T)$; similarly, $y\in V(T_{y^{\prime}})$ for some $y^{\prime}\in M_{1}(T)$ or $y\not\in V(T_{z^{\prime}})$ for any $z^{\prime}\in M(T)$. If $\{x,y\}\subseteq V(T_{v})$ for some $v\in M_{1}(T)$, then $d(v,x)\neq d(v,y)$ and there exist distinct $v^{\prime},v^{\prime\prime}\in M^{*}(T)$ such that $v$ lies on the $v^{\prime}-v^{\prime\prime}$ path in $T$; thus $g(R\{x,y\})\geq g(V(T_{v^{\prime}}))+g(V(T_{v^{\prime\prime}}))\geq 2k$. If $\{x,y\}\not\subseteq V(T_{v})$ for any $v\in M(T)$, there exist distinct $w_{1},w_{2}\in M^{*}(T)$ such that both $x$ (or $x^{\prime}$) and $y$ (or $y^{\prime}$) lie on the $w_{1}-w_{2}$ path in $T$; then $d(w_{1},x)=d(w_{1},y)$ and $d(w_{2},x)=d(w_{2},y)$ imply either $x=y$ or $\{x,y\}\subseteq V(T_{s})$ for some $s\in M_{1}(T)$, where both possibilities contradict the present assumptions. Thus $R\{x,y\}\supseteq V(T_{w_{i}})$ for at least one $i\in\{1,2\}$, and $g(R\{x,y\})\geq g(V(T_{w_{i}}))\geq k$.  ∎

Next, we provide an example showing that $\dim_{f}^{k}(G)-k\dim_{f}(G)$ can be arbitrarily large for some $k\in(1,\kappa(G)]$.

Remark 4.6**.**

The value of $\dim_{f}^{k}(G)-k\dim_{f}(G)$ can arbitrarily large, as $G$ varies, for some $k\in(1,\kappa(G)]$. Let $T$ be a tree with $ex(T)=1$. Let $v$ be the exterior major vertex of $T$ and let $\ell_{1},\ell_{2},\ldots,\ell_{\alpha}$ be the terminal vertices of $T$ such that $d(v,\ell_{1})=1<\beta=d(v,\ell_{j})$ for each $j\in\{2,3,\ldots,\alpha\}$, where $\alpha\geq 3$. By Proposition 4.3, $\kappa(T)=\beta+1$ and $\dim_{f}^{\beta+1}(T)=(\alpha-1)(\beta+1)-(\alpha-2)=(\alpha-1)\beta+1$. Since $\dim_{f}(T)=\frac{\alpha}{2}$ by Theorem 2.7(b), $\dim_{f}^{\beta+1}(T)-(\beta+1)\dim_{f}(T)=(\alpha-1)\beta+1-(\beta+1)(\frac{\alpha}{2})=(\frac{\alpha}{2}-1)(\beta-1)$ can be arbitrarily large, as $\alpha$ or $\beta$ gets big enough.

Next, we determine the fractional $k$-metric dimension of cycles.

Proposition 4.7**.**

Let $C_{n}$ be an n-cycle, where $n\geq 3$. Then

[TABLE]

Proof.

Note that $\kappa(C_{n})=n-2$ for an even $n$, and $\kappa(C_{n})=n-1$ for an odd $n$. Let $k\in[1,\kappa(C_{n})]$. For an even $n\geq 4$, a function $g:V(C_{n})\rightarrow[0,1]$ defined by $g(u)=\frac{1}{n-2}$, for each $u\in V(C_{n})$, is a minimum resolving function of $C_{n}$: (i) $0<g(u)=\frac{1}{n-2}\leq\frac{1}{k}\leq 1$ since $n\geq 4$; (ii) for distinct $x,y\in V(C_{n})$, $|R\{x,y\}|\geq n-2$, and thus $g(R\{x,y\})\geq(n-2)(\frac{1}{n-2})=1$; (iii) $g(V(C_{n}))=\frac{n}{n-2}=\dim_{f}(C_{n})$ by Theorem 2.7(d). Similarly, for an odd $n\geq 3$, one can easily check that a function $h:V(C_{n})\rightarrow[0,1]$ defined by $h(u)=\frac{1}{n-1}$, for each $u\in V(C_{n})$, is a minimum resolving function of $C_{n}$ satisfying $h(u)\leq\frac{1}{k}$. By Lemma 3.2 and Theorem 2.7(d), (7) follows.  ∎

Remark 4.8**.**

Note that, for any fixed $k\in[1,\kappa(C_{n})]$, $\lim_{n\rightarrow\infty}\dim_{f}^{k}(C_{n})=k$ by Proposition 4.7 (c.f. Proposition 3.5(a)).

Next, we determine the fractional $k$-metric dimension of wheel graphs.

Proposition 4.9**.**

For the wheel graph $W_{n}$ of order $n\geq 5$,

[TABLE]

Proof.

For $n\geq 5$, the wheel graph $W_{n}=C_{n-1}+K_{1}$ is obtained from an $(n-1)$-cycle $C_{n-1}$ by joining an edge from each vertex of $C_{n-1}$ to a new vertex, say $v$; let the $C_{n-1}$ be given by $u_{1},u_{2},\ldots,u_{n-1},u_{1}$. Note that $\textnormal{diam}(W_{n})=2$ for $n\geq 5$.

Case 1: $n=5$. Note that $\kappa(W_{5})=2$ since $R\{u_{1},u_{3}\}=\{u_{1},u_{3}\}$. Let $k\in[1,2]$. Let $g:V(W_{5})\rightarrow[0,1]$ be a function defined by $g(v)=0$ and $g(u_{i})=\frac{1}{2}$ for each $i\in\{1,2,3,4\}$. Then $g$ is a minimum resolving function of $W_{5}$: (i) $0\leq g(x)\leq\frac{1}{k}\leq 1$ for each $x\in V(W_{5})$; (ii) for distinct $i,j\in\{1,2,3,4\}$, $g(R\{u_{i},u_{j}\})\geq g(u_{i})+g(u_{j})=1$; (iii) for $i\in\{1,2,3,4\}$, $g(R\{v,u_{i}\})\geq g(u_{i})+g(u_{\ell})=1$, where $u_{\ell}\in V(W_{5})-N[u_{i}]$; (iv) $g(V(W_{5}))=2=\dim_{f}(W_{5})$ by Theorem 2.7(e). By Lemma 3.2 and Theorem 2.7(e), $\dim_{f}^{k}(W_{5})=k\dim_{f}(W_{5})=2k$ for $k\in[1,2]$.

Case 2: $n\geq 6$. First, we show that $\kappa(W_{n})=4$ in this case. For each $i\in\{1,2,\ldots,n-1\}$, $R\{v,u_{i}\}=(V(W_{n})-N(u_{i}))\cup\{v\}$ with $|R\{v,u_{i}\}|=n-2\geq 4$. For distinct $i,j\in\{1,2,\ldots,n-1\}$, (i) if $u_{i}u_{j}\in E(W_{n})$, then $R\{u_{i},u_{j}\}=(N(u_{i})\cup N(u_{j}))-\{v\}$ with $|R\{u_{i},u_{j}\}|=4$; (ii) if $u_{i}u_{j}\not\in E(W_{n})$ and $|N(u_{i})\cap N(u_{j})|=1$, then $R\{u_{i},u_{j}\}=(N[u_{i}]\cup N[u_{j}])-\{v\}$ with $|R\{u_{i},u_{j}\}|=6$; (iii) if $u_{i}u_{j}\not\in E(W_{n})$ and $|N(u_{i})\cap N(u_{j})|=2$, then $R\{u_{i},u_{j}\}=(N[u_{i}]\cup N[u_{j}])-(N(u_{i})\cap N(u_{j}))$ with $|R\{u_{i},u_{j}\}|=4$. So, $\kappa(W_{n})=4$ for $n\geq 6$.

Second, we determine $\dim_{f}^{k}(W_{n})$ for $n\geq 6$. Let $k\in[1,4]$. For $n=6$, a function $g:V(W_{6})\rightarrow[0,1]$ defined by $g(x)=\frac{1}{4}$, for each $x\in V(W_{6})$, is a minimum resolving function of $W_{6}$: (i) $0<g(x)\leq\frac{1}{k}\leq 1$ for each $x\in V(W_{6})$; (ii) for any distinct $x,y\in V(W_{6})$, $g(R\{x,y\})\geq 4(\frac{1}{4})=1$; (iii) $g(V(W_{6}))=\frac{3}{2}=\dim_{f}(W_{6})$ by Theorem 2.7(e). For $n\geq 7$, let $h:V(W_{n})\rightarrow[0,1]$ be a function defined by $h(v)=0$ and $h(u_{i})=\frac{1}{4}$ for each $i\in\{1,2,\ldots,n-1\}$. Then $h$ is a minimum resolving function of $W_{n}$: (i) $0\leq h(x)\leq\frac{1}{k}\leq 1$ for each $x\in V(W_{n})$; (ii) for each $i\in\{1,2,\ldots,n-1\}$, $|R\{v,u_{i}\}|\geq n-2\geq 5$ since $n\geq 7$, and hence $h(R\{v,u_{i}\})\geq 4(\frac{1}{4})=1$; (iii) for distinct $i,j\in\{1,2,\ldots,n-1\}$, $|R\{u_{i},u_{j}\}|\geq 4$ and $v\not\in R\{u_{i},u_{j}\}$, and thus $h(R\{u_{i},u_{j}\})\geq 4(\frac{1}{4})=1$; (iv) $h(V(W_{n}))=\frac{n-1}{4}=\dim_{f}(W_{n})$ by Theorem 2.7(e). Therefore, by Lemma 3.2 and Theorem 2.7(e), (8) holds for $n\geq 6$ and $k\in[1,4]$.  ∎

Next, we determine the fractional $k$-metric dimension of the Petersen graph.

Proposition 4.10**.**

For the Petersen graph $\mathcal{P}$, $\dim_{f}^{k}(\mathcal{P})=k\dim_{f}(\mathcal{P})=\frac{5}{3}k$ for $k\in[1,6]$.

Proof.

Note that $\mathcal{P}$ is 3-regular and vertex-transitive. Since $\textnormal{diam}(\mathcal{P})=2$, any two distinct vertices in $\mathcal{P}$ are either adjacent or at distance two apart.

We first show that $\kappa(\mathcal{P})=6$. For any distinct $x,y\in V(\mathcal{P})$, $R\{x,y\}=N[x]\cup N[y]-(N(x)\cap N(y))$ and $|R\{x,y\}|=6$: (i) if $xy\in E(\mathcal{P})$, then $N(x)\cap N(y)=\emptyset$ and $x\in N[y]$ and $y\in N[x]$; (ii) if $xy\not\in E(\mathcal{P})$, then $|N(x)\cap N(y)|=1$. So, $\kappa(\mathcal{P})=6$.

Now, let $k\in[1,6]$. Since $\dim_{f}^{k}(\mathcal{P})\geq k\dim_{f}(\mathcal{P})$ by Lemma 3.1, it suffices to show that $\dim_{f}^{k}(\mathcal{P})\leq k\dim_{f}(\mathcal{P})$. Let $g:V(\mathcal{P})\rightarrow[0,1]$ be a function defined by $g(v)=\frac{k}{6}$ for each $v\in V(\mathcal{P})$. Since $0\leq g(v)\leq 1$ for each vertex $v\in V(\mathcal{P})$ and $g(R\{x,y\})=6(\frac{k}{6})\geq k$ for any two distinct $x,y\in V(\mathcal{P})$, $g$ is a $k$-resolving function of $\mathcal{P}$. So, $\dim_{f}^{k}(\mathcal{P})\leq|V(\mathcal{P})|(\frac{k}{6})=\frac{10k}{6}=k(\frac{5}{3})=k\dim_{f}(\mathcal{P})$ by Theorem 2.7(c).  ∎

Next, we determine the fractional $k$-metric dimension of a bouquet of cycles.

Proposition 4.11**.**

Let $B_{m}$ be a bouquet of $m$ cycles $C^{1},C^{2},\ldots,C^{m}$ with a cut-vertex (i.e., the vertex sum of m cycles at one common vertex), where $m\geq 2$; further, let $C^{1}$ be the cycle of the minimum length among the $m$ cycles of $B_{m}$. Then, for $k\in[1,\kappa(B_{m})]$, $\dim_{f}^{k}(B_{m})=k\dim_{f}(B_{m})=km$, where

\kappa(B_{m})=\left\{\begin{array}[]{ll}|V(C^{1})|-1&\mbox{ if C^{1}is an odd cycle},\\ |V(C^{1})|-2&\mbox{ ifC^{1} is an even cycle}.\end{array}\right.**

Proof.

Let $v$ be the cut-vertex of $B_{m}$. For each $i\in\{1,2,\ldots,m\}$, let $C^{i}$ be given by $v,u_{i,1},u_{i,2},\ldots,u_{i,r_{i}},v$ and let $P^{i}=C^{i}-v$; further, let $P^{i,1}$ be the $u_{i,1}-u_{i,\lceil\frac{r_{i}}{2}\rceil}$ geodesic. Without loss of generality, let $r_{1}\leq r_{2}\leq\ldots\leq r_{m}$.

Claim 1: If $C^{1}$ is an odd cycle, then $\kappa(B_{m})=r_{1}=|V(C^{1})|-1$; if $C^{1}$ is an even cycle, $\kappa(B_{m})=r_{1}-1=|V(C^{1})|-2$.

Proof of Claim 1. Let $x$ and $y$ be distinct vertices of $B_{m}$. First, let $x,y\in V(C^{i})$ for some $i\in\{1,2,\ldots,m\}$. If $d(v,x)\neq d(v,y)$, then $R\{x,y\}\supseteq V(C^{j})$ with $|R\{x,y\}|\geq|V(C^{j})|=r_{j}+1\geq r_{1}+1$ for $j\neq i$. If $d(v,x)=d(v,y)$ and $C^{i}$ is an odd cycle, then $R\{x,y\}=V(P^{i})$ with $|R\{x,y\}|=r_{i}\geq r_{1}$; notice, for an odd cycle $C^{1}$, $|R\{u_{1,1},u_{1,r_{1}}\}|=r_{1}$. If $d(v,x)=d(v,y)$ and $C^{i}$ is an even cycle, then $R\{x,y\}=V(P^{i})-\{u_{i,\lceil\frac{r_{i}}{2}\rceil}\}$ with $|R\{x,y\}|=r_{i}-1$, where $r_{i}-1\geq r_{1}$ if $C^{1}$ is an odd cycle, and $r_{i}-1\geq r_{1}-1$ if $C^{1}$ is an even cycle; notice, for an even cycle $C^{1}$, $|R\{u_{1,1},u_{1,r_{1}}\}|=r_{1}-1$.

Second, let $x\in V(P^{i})$ and $y\in V(P^{j})$ for distinct $i,j\in\{1,2,\ldots,m\}$; let $x\in V(P^{i,1})$ and $y\in V(P^{j,1})$, without loss of generality. If $d(v,x)=d(v,y)$, then $R\{x,y\}\supseteq V(P^{i,1})\cup V(P^{j,1})$ with $|R\{x,y\}|\geq\lceil\frac{r_{i}}{2}\rceil+\lceil\frac{r_{j}}{2}\rceil\geq r_{1}$. So, let $d(v,x)\neq d(v,y)$, say $d(v,x)<d(v,y)$ without loss of generality; then $d(u,x)\neq d(u,y)$ for each $u\in V(P^{i,1})$, and $r_{j}\geq 3$. If $r_{j}=3$, then $R\{x,y\}\supseteq V(P^{i,1})\cup V(P^{j})$. If $r_{j}\geq 4$, then at most two vertices in $C^{j}$ are at equal distance from $x$ and $y$; thus, $R\{x,y\}\supseteq V(P^{i,1})\cup(V(P^{j})-\{w_{1},w_{2}\})$ such that $w_{t}\in V(P^{j})$ with $d(w_{t},x)=d(w_{t},y)$, where $t\in\{1,2\}$. In each case, $|R\{x,y\}|\geq\lceil\frac{r_{i}}{2}\rceil+\lceil\frac{r_{j}}{2}\rceil\geq r_{1}$. $\Box$

Claim 2: For $k\in[1,\kappa(B_{m})]$, $\dim_{f}^{k}(B_{m})=k\dim_{f}(B_{m})=km$.

Proof of Claim 2. Let $k\in[1,\kappa(B_{m})]$, and let $h:V(B_{m})\rightarrow[0,1]$ be a function defined by

[TABLE]

Note that, for each $i\in\{1,2,\ldots,m\}$, $h(V(P^{i}))=1$ if $C^{i}$ is an odd cycle, and $h(V(P^{i}))=h(V(P^{i}))-h({u_{i,\lceil\frac{r_{i}}{2}\rceil}})=1$ if $C^{i}$ is an even cycle. We also note that $h$ is a minimum resolving function of $B_{m}$: (i) $0\leq h(u)\leq\frac{1}{k}\leq 1$ for each $u\in V(B_{m})$; (ii) if $x,y\in V(C^{i})$ with $d(v,x)\neq d(v,y)$, for some $i\in\{1,2,\ldots,m\}$, then $h(R\{x,y\})\geq h(V(C^{j}))\geq 1$ for $j\neq i$; (iii) if $x,y\in V(C^{i})$ with $d(v,x)=d(v,y)$ and $x\neq y$, for some $i\in\{1,2,\ldots,m\}$, then $h(R\{x,y\})=h(V(P^{i}))=1$ when $C^{i}$ is an odd cycle, and $h(R\{x,y\})=h(V(P^{i}))-h(u_{i,\lceil\frac{r_{i}}{2}\rceil})=1$ when $C^{i}$ is an even cycle; (iv) if $x\in V(P^{i})$ and $y\in V(P^{j})$ for distinct $i,j\in\{1,2,\ldots,m\}$, then $h(R\{x,y\})\geq\frac{1}{2}h(V(P^{i}))+\frac{1}{2}h(V(P^{j}))\geq 1$ using a similar argument used in the proof of Claim 1; (v) $h(V(B_{m}))\!=\!m\!=\!\dim_{f}(B_{m})$ by Theorem 2.7(f). So, by Lemma 3.2 and Theorem 2.7(f), $\dim_{f}^{k}(B_{m})=k\dim_{f}(B_{m})=km$ for $k\in[1,\kappa(B_{m})]$. $\Box$  ∎

Next, we determine the fractional $k$-metric dimension of complete multi-partite graphs.

Proposition 4.12**.**

For $m\geq 2$, let $G=K_{a_{1},a_{2},\ldots,a_{m}}$ be a complete $m$-partite graph of order $n=\sum_{i=1}^{m}a_{i}\geq 3$, and let $s$ be the number of partite sets of $G$ consisting of exactly one element. Then, for $k\in[1,2]$,

[TABLE]

Proof.

Let $V(G)$ be partitioned into $m$-partite sets $V_{1},V_{2},\ldots,V_{m}$ with $|V_{i}|=a_{i}$, where $i\in\{1,2,\ldots,m\}$. Without loss of generality, let $a_{1}\leq a_{2}\leq\ldots\leq a_{m}$. Note that $\kappa(G)=2$: (i) if $a_{m}\geq 2$, then, for two distinct $x,y\in V_{m}$, $R\{x,y\}=\{x,y\}$; (ii) if $a_{m}=1$, then, for $x\in V_{1}$ and $y\in V_{2}$, $R\{x,y\}=\{x,y\}$. Let $k\in[1,2]$.

First, let $s\neq 1$. A function $g:V(G)\rightarrow[0,1]$ defined by $g(v)=\frac{1}{2}$, for each $v\in V(G)$, is a minimum resolving function of $G$: (i) $0<g(v)\leq\frac{1}{k}\leq 1$ for each $v\in V(G)$; (ii) for any distinct vertices $x,y\in V(G)$, $g(R\{x,y\})\geq g(x)+g(y)=1$; (iii) $g(V(G))=\frac{n}{2}=\dim_{f}(G)$ by Theorem 2.7(g). So, $\dim_{f}^{k}(G)=k\dim_{f}(G)=\frac{kn}{2}$ by Lemma 3.2 and Theorem 2.7(g).

Second, let $s=1$. Let $h:V(G)\rightarrow[0,1]$ be a function defined by $h(u)=0$ for $u\in V_{1}$ and $h(v)=\frac{1}{2}$ for each $v\in V(G)-V_{1}$. Then $h$ is a minimum resolving function of $G$ : (i) $0\leq h(v)\leq\frac{1}{k}\leq 1$ for each $v\in V(G)$; (ii) for any two distinct vertices $x,y\in V(G)-V_{1}$, $h(R\{x,y\})\geq h(x)+h(y)=1$; (iii) for $x\in V_{1}$ and $y\in V_{i}\subseteq V(G)-V_{1}$, $h(R\{x,y\})\geq h(V_{i})\geq 1$, where $i\in\{2,\ldots,m\}$; (iv) $h(V(G))=\frac{n-1}{2}=\dim_{f}(G)$ by Theorem 2.7(g). So, $\dim_{f}^{k}(G)=k\dim_{f}(G)=\frac{k(n-1)}{2}$ for $k\in[1,2]$ by Lemma 3.2 and Theorem 2.7(g).  ∎

Now, we consider the fractional $k$-metric dimension of grid graphs (i.e., the Cartesian product of two paths). The Cartesian product of two graphs $G$ and $H$, denoted by $G\square H$, is the graph with the vertex set $V(G)\times V(H)$ such that $(u,w)$ is adjacent to $(u^{\prime},w^{\prime})$ if and only if either $u=u^{\prime}$ and $ww^{\prime}\in E(H)$, or $w=w^{\prime}$ and $uu^{\prime}\in E(G)$. See Figure 1 for the labeling of $P_{6}\square P_{4}$.

We recall the following result.

Theorem 4.13**.**

* [5]** For $s,t\geq 2$, $\kappa(P_{s}\square P_{t})=s+t-2$ and $\dim^{k}(P_{s}\square P_{t})=2k$, where $k\in\{1,2,\ldots,s+t-2\}$.*

Proposition 4.14**.**

For $k\in[1,s+t-2]$, $\dim_{f}^{k}(P_{s}\square P_{t})=k\dim_{f}(P_{s}\square P_{t})=2k$, where $s,t\geq 2$.

Proof.

Let $s\geq t\geq 2$, and let $G=P_{s}\square P_{t}$ and $L=\{v\in V(G):2\leq\deg(v)\leq 3\}$. By Theorem 4.13, $\kappa(G)=s+t-2$. Let $k\in[1,s+t-2]$. Since $\dim^{k}_{f}(G)\geq k\dim_{f}(G)=2k$ by Lemma 3.1 and Theorem 2.7(h), it suffices to show that $\dim_{f}^{k}(G)\leq 2k$. Let $g:V(G)\rightarrow[0,1]$ be a function defined by

[TABLE]

Note that $g(V(G))=2k$. We will show that $g$ is a $k$-resolving function of $G$. Clearly, $0\leq g(v)\leq 1$ for each $v\in V(G)$. Let $x=(a,b)$ and $y=(c,d)$ be two distinct vertices of $G$. We consider two cases.

Case 1: $a=c$ or $b=d$. If $a=c$, then $R\{x,y\}\cap L\supseteq\cup_{i=1}^{s}\{(i,1),(i,t)\}$, and thus $g(R\{x,y\})\geq(2s)(\frac{k}{s+t-2})\geq k$ since $s\geq t\geq 2$. So, let $b=d$; without loss of generality, let $a<c$. Let $z=(\alpha,\beta)\in L$. Note that (i) if $\alpha\leq a$, then $d(z,x)<d(z,y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{j=1}^{t}\{(1,j)\}\cup(\cup_{i=2}^{a}\{(i,1),(i,t)\})$; (ii) if $\alpha\geq c$, then $d(z,x)>d(z,y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{j=1}^{t}\{(s,j)\}\cup(\cup_{i=c}^{s-1}\{(i,1),(i,t)\})$; (iii) if $a<\alpha<c$ and $\beta=1$, then there exists at most one such $z\in L$ satisfying $d(z,x)=d(z,y)$, since $d(z,x)=d(z,y)$ implies $\alpha-a+b-1=c-\alpha+b-1$, i.e., $2\alpha=a+c$; similarly, if $a<\alpha<c$ and $\beta=t$, then there exists at most one such $z\in L$ satisfying $d(z,x)=d(z,y)$. Thus, $|R\{x,y\}\cap L|\geq 2s+2t-6$, and hence $g(R\{x,y\})\geq(2s+2t-6)(\frac{k}{s+t-2})=2k(\frac{s+t-3}{s+t-2})\geq k$, since $s+t\geq 4$.

Case 2: $a\neq c$ and $b\neq d$. Without loss of generality, let $a<c$; further, assume that $b<d$ (the case for $b>d$ can be handled similarly). Let $z^{\prime}=(\alpha^{\prime},\beta^{\prime})\in L$. Note that (i) if $\alpha^{\prime}\leq a$ and $\beta^{\prime}=1$, then $d(z^{\prime},x)<d(z^{\prime},y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{i=1}^{a}\{(i,1)\}$; (ii) if $\alpha^{\prime}\geq c$ and $\beta^{\prime}=t$, then $d(z^{\prime},x)>d(z^{\prime},y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{i=c}^{s}\{(i,t)\}$; (iii) if $a<\alpha^{\prime}<c$ (i.e., $c\neq a+1$) and $\beta^{\prime}=1$, then there exists at most one such $z^{\prime}\in L$ satisfying $d(z^{\prime},x)=d(z^{\prime},y)$, since $d(z^{\prime},x)=d(z^{\prime},y)$ implies $\alpha^{\prime}-a+b-1=c-\alpha^{\prime}+d-1$, i.e., $2\alpha^{\prime}=a-b+c+d$; similarly, if $a<\alpha^{\prime}<c$ and $\beta^{\prime}=t$, there exists at most one such $z^{\prime}\in L$ satisfying $d(z^{\prime},x)=d(z^{\prime},y)$. Likewise, we note that (i) if $\alpha^{\prime}=1$ and $\beta^{\prime}\leq b$, then $d(z^{\prime},x)<d(z^{\prime},y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{j=1}^{b}\{(1,j)\}$; (ii) if $\alpha^{\prime}=s$ and $\beta^{\prime}\geq d$, then $d(z^{\prime},x)>d(z^{\prime},y)$ and thus $R\{x,y\}\cap L\supseteq\cup_{j=d}^{t}\{(s,j)\}$; (iii) if $\alpha^{\prime}=1$ and $b<\beta^{\prime}<d$ (i.e., $d\neq b+1$), then there exists at most one such $z^{\prime}\in L$ satisfying $d(z^{\prime},x)=d(z^{\prime},y)$, since $d(z^{\prime},x)=d(z^{\prime},y)$ implies $a-1+\beta^{\prime}-b=c-1+d-\beta^{\prime}$, i.e., $2\beta^{\prime}=-a+b+c+d$; similarly, if $\alpha^{\prime}=s$ and $b<\beta^{\prime}<d$, then there exists at most one such $z^{\prime}\in L$ satisfying $d(z^{\prime},x)=d(z^{\prime},y)$. So, if $c=a+1$ or $d=b+1$, then $|R\{x,y\}\cap L|\geq s+t-2$; if $c\geq a+2$ and $d\geq b+2$, then $|R\{x,y\}\cap L|\geq a+(s-c+1)+2(c-a-2)+b+(t-d+1)+2(d-b-2)\geq s+t-2$. In each case, $g(R\{x,y\})\geq(s+t-2)(\frac{k}{s+t-2})=k$.

Thus, in each case, $g$ is a $k$-resolving function of $G$, and hence $\dim_{f}^{k}(G)\leq g(V(G))=2k$. Therefore, $\dim_{f}^{k}(G)=k\dim_{f}(G)=2k$ for $k\in[1,s+t-2]$ for $s\geq t\geq 2$.  ∎

5 Open Problems

We conclude this paper with two open problems.

Problem 1. Let $\phi(k)=\dim_{f}^{k}(G)$ be a function of $k$, for a fixed $G$, on domain $[1,\kappa(G)]$. Is $\phi$ a continuous function of $k$ on every connected graph $G$?

Problem 2. Suppose $\dim_{f}^{k}(G)$ is given by $\psi(k)$ for integral values of $k$. When and how can we interpolate $\psi$ and deduce $\dim_{f}^{k}(G)$ for any real number $k\in[1,\kappa(G)]$?

For example, let $G=P_{s}\square P_{t}$, where $s,t\geq 2$. Then $\dim_{f}^{k}(G)=2k$ for integers $k\in\{1,2,\ldots,\kappa(G)\}$ by Theorems 2.7(h), Lemma 3.1, Observation 2.2(b), and Theorem 4.13. In Proposition 4.14, we proved that $\dim_{f}^{k}(G)=2k$ for any real number $k\in[1,\kappa(G)]$, by using Lemma 3.1 and constructing a $k$-resolving function $g$ on $V(G)$ with $g(V(G))=2k$ for $k\in[1,\kappa(G)]$. The construction of $k$-resolving function for any real number $k\in[1,\kappa(G)]$ in determining $\dim_{f}^{k}(G)$ in Proposition 4.14 does not appear to carry to the construction of $k$-resolving set for any integral values $k\in\{1,2,\ldots,\kappa(G)\}$ in determining $\dim^{k}(G)$, and vice versa.

Acknowledgement. The authors greatly appreciate Dr. Douglas J. Klein for graciously hosting Dr. Ismael G. Yero during his visit to Texas A&M University at Galveston – this visit sparked an ongoing collaboration of which the present paper is a product. The authors also thank the anonymous referees for some helpful comments.