# Unique identification and domination of edges in a graph: The vertex-edge
dominant edge metric dimension

Hafiz Muhammad Ikhlaq
Department of Mathematics,
COMSATS University Islamabad (CUI), Lahore Campus,
Lahore, 54000, Pakistan

Sakander Hayat
School of Mathematics and Information Sciences,
Guangzhou University, Guangzhou,
510006, Guangdong, P.R. China
e-mail<EMAIL_ADDRESS>

Hafiz Muhammad Afzal Siddiqui
Department of Mathematics,
COMSATS University Islamabad (CUI), Lahore Campus,
Lahore, 54000, Pakistan
e-mail<EMAIL_ADDRESS>

###### Abstract

Dominating sets and resolving sets have important applications in control
theory and computer science. In this paper, we introduce an edge-analog of the
classical dominant metric dimension of graphs. By combining the concepts of a
vertex-edge dominating set and an edge resolving set, we introduce the notion
of a vertex-edge dominant edge resolving set of a graph. We call the minimum
cardinality of such a set in a graph $\Gamma$, the vertex-edge dominant edge
metric dimension $\gamma_{emd}(\Gamma)$ of $\Gamma$. The new parameter
$\gamma_{emd}$ is calculated for some common families such as paths, cycles,
complete bipartite graphs, wheel and fan graphs. We also calculate
$\gamma_{emd}$ for some Cartesian products of path with path and path with
cycle. Importantly, some general results and bounds are presented for this new
parameter. We also conduct a comparative analysis of $\gamma_{emd}$ with the
dominant metric dimension of graphs. Comparison shows that these two
parameters are not comparable, in general. Upon considering the class of
bipartite graphs, we show that $\gamma_{emd}(T_{n})$ of a tree $T_{n}$ is
always less than or equal to its dominant metric dimension. However, we show
that for non-tree bipartite graphs, the parameter is not comparable just like
general graphs. Based on the results in this paper, we propose some open
problems at the end.

2020 Mathematics Subject Classification: 05C12, 05C69

> Keywords: Graph; Resolving set; Dominating set; Edge resolving set; Vertex-
> edge dominating set; Vertex-edge dominant edge resolving set

##  1 Introduction

Distance based graph-theoretic parameters have a long history. In particular,
every connected graph $\Gamma$ is a metric space i.e. $(V(\Gamma),d)$, where
the distance function $d:=d(x,y)$ is the length of a shortest path between $x$
and $y$. See more about this direction in the book by Deza & Laurent [10].
From application perspective, Harold Wiener [35] in 1947 defined path number
to be the sum of distances over all the unordered pairs of vertices and showed
that it correlates well with the boiling points of alkanes. See a recent
survey by Goddard & Oellermann [11] on distance in graphs.

Independently, Slater [32] in 1975 and Harary & Melter [14] in 1976 introduced
a distance related parameter known as the metric dimension of graphs. The
metric dimension is the cardinality of a minimum resolving set in a graph.
Resolving sets and metric dimension have numerous applications in various
scientific disciplines such as network discovery & verification [1],
mastermind game [6], robot navigation [20], coin weighing problems [31],
combinatorial search and optimization [31], pattern recognition & image
processing [24], among others. The success of the metric dimension motivated
researcher to propose other resolvability related parameters such as
independent resolving set and its minimality by Chartrand et al. [8], strong
resolving set in graph and digraphs and its minimality by Oellermann & Peters-
Fransen [25], local resolving set and metric dimension by Okamoto et al. [26],
among others. For more mathematical results regarding resolvability, we refer
the reader to a survey by Chartrand & Zhang [7].

Dominating sets and the domination number of graphs is another important
research direction of graph theory which has a rich history. Berge [2] in 1958
gave the idea of dominating sets and their applications. By Berge [2], the
domination number was referred as the “coefficient of external stability” in
network analysis. The name dominating set was assigned by Ore [27] in 1962. In
1977, Cockayne & Hedetniemi [9] made an interesting and extensive survey of
the results known at that time about dominating sets in graphs. Rao &
Sreenivansan [30] studied the split domination number of arithmetic graphs.
For the detailed mathematical treatment, see the book on domination theory of
graph by Haynes et al. [15].

Brigham et al. [3] combined the concepts of a resolving and a dominating set
to propose resolving dominating set of a graph. Cardinality of a minimum such
set is known as the dominant metric dimension $\gamma_{md}(\Gamma)$. Brigham
et al. [3] calculated the dominant metric dimension of some classical families
such as paths, cycles, complete bipartite etc. Moreover, they proved some
general results such as

$\max\\{\beta(\Gamma),\gamma(\Gamma)\\}\leq\gamma_{md}(\Gamma)\leq\beta(\Gamma)+\gamma(\Gamma).$

Henning & Oellermann [16], later on, combined the concepts of a locating
dominating set and a resolving set. The cardinality of such a minimum set in
called the metric locating-dominating number $\gamma_{mld}(\Gamma)$ of a graph
$\Gamma$. Henning & Oellermann [16] showed $\gamma_{mld}(\Gamma)$ satisfy
$\gamma(\Gamma)\leq\gamma_{mld}(\Gamma)\leq n-1$ for every $n$-vertex graph
$\Gamma$. Gonzalez et al. [12] further studied $\gamma_{mld}(\Gamma)$ and
showed that it satisfies

$\max\\{\beta(\Gamma),\gamma(\Gamma)\\}\leq\gamma_{mld}(\Gamma)\leq\beta(\Gamma)+\gamma(\Gamma).$

Susilowati et al. [33] showed that the lower bound on $\gamma_{mld}(\Gamma)$
by Henning & Oellermann is not sharp, whereas, the upper bound on
$\gamma_{md}(\Gamma)$ by Brigham et al. [3] and the upper bound on
$\gamma_{mld}(\Gamma)$ by Gonzalez et al. [12] are not sharp.

Kelenc et al. [19] proposed the idea of a edge resolving set i.e.
resolvability of edges in a graph by its vertices. The corresponding parameter
is known as the edge metric dimension $\beta_{e}(\Gamma)$ of a graph $\Gamma$.
They calculated $\beta_{e}(\Gamma)$ for classical families and provided
comparison between the metric dimension and edge metric dimension of graphs
and showed that they are not comparable, in general. Such variations has
attracted attention of several researches, see for instance, [18, 21, 36, 37].

Boutrig et al. [4] introduced an edge-analog of the dominating set i.e.
vertices which dominate edges of a graph. They called such a set a vertex-edge
dominating set and the cardinality of such a minimum set is called the vertex-
edge domination number. Lewis [22] in his PhD thesis studied the vertex-edge
domination in detail. Lewis et al. [23] provided comparison of domination
number and the vertex-edge domination number of graphs. Peters [28] in his PhD
thesis provided a detailed study on the vertex-domination in graphs and
provided several upper and lower bounds on the vertex-edge domination number.
Thakkar & Jamvecha [34] further studied the vertex-edge domination in graphs.

In this paper, we propose to study vertex-edge dominant edge resolving sets in
graphs. The minimum cardinality of such a set in a graph $\Gamma$ is called
the vertex-edge dominant edge metric dimension of $\Gamma$. In other words, we
introduce the edge version of the dominant metric dimension of graphs. This
paper is organized as follows: In Chapter 2, we introduce all the necessary
definitions and preliminary results which are used in subsequent sections. In
Chapter 3, we introduce the concept of a vertex-edge dominant edge resolving
set and the vertex-edge dominant edge metric dimension $\gamma_{emd}(\Gamma)$
of a graph $\Gamma$. We calculate $\gamma_{emd}$ for classical families such
as paths, cycles, complete graphs, complete bipartite graphs, wheel and fan
graphs. Chapter 4 computes $\gamma_{emd}$ for some Cartesian products such as
path vs. path and path vs. cycle. Chapter 5 presents some general results and
bounds on $\gamma_{emd}$. Chapter 6 draws a comparison between the dominant
metric dimension and its edge version introduced in this paper. Although,
these two parameters are not comparable in general, for trees specifically, we
show that $\gamma_{emd}(T_{n})$ of a tree $T_{n}$ is always less than or equal
to its dominant metric dimension. Chapter 7 concludes the findings and propose
some open problems which naturally arise from this study.

##  2 Preliminaries

All graphs in this paper are finite, simple, undirected and connected. A
simple graph $\Gamma$ is a pair $\Gamma=(V,E)$ consists of a vertex set $V$
and an edge set $E\subseteq{V\choose 2}$. The cardinality $\mid V\mid$ (resp.
$\mid E\mid$) of $V$ (resp. $E$) is called the order (resp. size) of $\Gamma$.
For a vertex $x\in V$, the open (resp. closed) neighborhood of $x$ is defined
as $N_{\Gamma}(x)=\\{y\in V:xy\in E\\}$ (resp.
$N_{\Gamma}[x]=N_{\Gamma}(x)\cup\\{x\\}$). For $x,y\in V$, let
$\ell\big{(}p(x,y)\big{)}$ be the length of a path $p(x,y)$ between $x$ & $y$.
The distance between $x,y\in V$ is defined to be the length of a shortest path
between $x$ & $y$ i.e. $d_{\Gamma}(x,y)=\min\\{\ell\big{(}p(x,y)\big{)}:x,y\in
V\\}$. For a vertex $w\in V$ and an edge $e=xy$, the distance between $w$ and
$e$ is defined to be
$d_{\Gamma}(w,e)=\min\\{d_{\Gamma}(w,x),d_{\Gamma}(w,y)\\}$. we usually omit
$\Gamma$ from $N_{\Gamma}(x)$, $N_{\Gamma}[x]$, and $d_{\Gamma}(x,y)$ and
write $N(x)$, $N[x]$, and $d(x,y)$ instead. The path (resp. cycle) graph on
$n$ vertices is usually denoted by $P_{n}$ (resp. $C_{n}$). Similarly, the
complete graph of order $n$ is denoted by $K_{n}$. The star graph of order
$n+1$ is denoted by $S_{1,n}$. Moreover, the wheel and fan graph on $n+1$
vertices are denoted by $W_{1,n}$ and $F_{1,n}$, respectively. Similarly, the
complete bipartite graph is denoted by $K_{n,m}$.

For a graph $\Gamma$, a subset $S\subseteq V$ is said to be a dominating set
if for every $x\in V\setminus S$, we have $N[x]\cap S\neq\emptyset.$ Minimum
cardinality of such a set in $\Gamma$ is knows as the dominating number
usually denoted by $\gamma(\Gamma)$. A vertex $w\in V$ is said to resolve a
pair of vertices $x,y\in V$, if $d(w,x)\neq d(w,y)$. Let
$B=\\{x_{1},x_{2},\ldots,x_{p}\\}\subset V$ be an ordered subset. The
representation/code $r(y|B)$ of a vertex $y$ with respect to $B$ is the
distance vector
$r(y|B)=\big{(}d(y,x_{1}),d(y,x_{2}),\ldots,d(y,x_{p})\big{)}$. Such a subset
$B$ is said to be a resolving set of $\Gamma$ if every $y\in V\setminus B$ has
a distinct code corresponding to $B$. Minimal cardinality of such a resolving
set in $\Gamma$ is called the metric dimension of $\Gamma$ usually denoted by
$\beta(\Gamma)$.

Combining the concepts of a dominating and a resolving set in a graph yields a
dominant resolving set. Formally, an ordered subset $D\subset V$ in $\Gamma$
is said to be a dominant resolving set, if $D$ is simultaneously a dominating
set and a resolving set. Minimum cardinality of such as set is called the
dominant metric dimension of $\Gamma$ and is denoted by $\gamma_{md}(\Gamma)$.
A dominant resolving set of cardinality $\gamma_{md}(\Gamma)$ in $\Gamma$ is
called a dominant basis of $\Gamma$. For classical families of graphs,
Susilowati et al. [33] showed the following result.

###### Theorem 1

_[33]_ The dominant metric dimension $\gamma_{md}$ of some classical families
are calculated as follows:

  * _(i)_

For $n>4$, we have $\gamma_{md}(P_{n})=\gamma(P_{n})=\lceil\frac{n}{3}\rceil$,

  * _(ii)_

For $n\geq 7$, we have
$\gamma_{md}(C_{n})=\gamma(C_{n})=\lceil\frac{n}{3}\rceil$,

  * _(iii)_

For $n\geq 2$, we have $\gamma_{md}(S_{1,n})=n$,

  * _(iv)_

For $n\geq 2$, we have $\gamma_{md}(K_{n})=\beta(K_{n})=n-1$,

  * _(v)_

For $n,m\geq 2$, we have $\gamma_{md}(K_{n,m})=\beta(K_{n,m})=n+m-2.$

Kelenc et al. [19] introduced an edge version of the metric dimension of
graphs. The vertex $x$ is said to resolves the edges $e,f\in E$, if
$d(e,x)\neq d(f,y)$. For an ordered subset
$A=\\{x_{1},x_{2},\ldots,x_{q}\\}\subset V$, the representation/code $r(e|A)$
of an edge $e$ with respect to $A$ is the edge distance vector
$r(e|A)=\big{(}d(e,x_{1}),d(e,x_{2}),\ldots,d(e,x_{q})\big{)}$. For such an
$A$, if every edge of $\Gamma$ has a different code, then $A$ is said to be an
edge-resolving set of $\Gamma$. Denoted by $\beta_{e}(\Gamma)$, the minimum
cardinality of an edge resolving set in $\Gamma$ is called the edge metric
dimension of $\Gamma$. An edge resolving set of cardinality
$\beta_{e}(\Gamma)$ in $\Gamma$ is known as the edge metric basis of $\Gamma$.
For classical families of graphs, Kelenc et al. [19] showed the following
results.

###### Theorem 2

_[19]_ The edge metric dimension of certain families are computed as follows:

  * _(i)_

For a graph $\Gamma$, $\beta_{e}(\Gamma)=1$ if and only if $\Gamma=P_{n}$.

  * _(ii)_

For $n\geq 3$, we have $\beta_{e}(C_{n})=2$.

  * _(iii)_

For $n\geq 2$, we have $\beta_{e}(K_{n})=n-1$.

  * _(iv)_

For $m,n\geq 2$, we have $\beta_{e}(K_{n,m})=n+m-2$.

  * _(v)_

For $r\geq t\geq 2$, we have $\beta_{e}(\Gamma=P_{r}\Box P_{t})=2$.

  * _(vi)_

For $n\geq 3$, we have

$\beta_{e}(W_{1,n})=\left\\{\begin{array}[]{ll}n,&\hbox{$n=3,4$;}\\\
n-1,&\hbox{$n\geq 5$.}\end{array}\right.$

  * _(vii)_

For $n\geq 1$, we have

$\beta_{e}(F_{1,n})=\left\\{\begin{array}[]{ll}n,&\hbox{$n=1,2,3$;}\\\
n-1,&\hbox{$n\geq 4$.}\end{array}\right.$

###### Lemma 1

_[19]_ Let $\Gamma$ be a graph with $\Delta(\Gamma)=n-1$. If there exist at
least two vertices with degree $n-1$, then $\beta_{e}(\Gamma)=n-1$.

###### Proposition 1

_[19]_ Let $\Gamma$ be a graph of order $n$. If there is a vertex $v\in
V(\Gamma)$ of degree $n-1$, then, either $\beta_{e}(\Gamma)=n-1$ or
$\beta_{e}(\Gamma)=n-2$.

Filipovic et al. [13] showed the following result.

###### Corollary 1

_[13]_ Let $\Gamma$ be a connected $r$-regular graph. Then we have
$\beta_{e}(\Gamma)\geq 1+\lceil\log_{2}r\rceil.$

Peterin and Yero, [29] computed the edge metric dimension of the Corona
product and Join of two graphs.

###### Theorem 3

_[29]_ Let $\Gamma$ and $\Omega$ be two graphs where $\Gamma$ is connected and
$|V(\Omega)|\geq 2$. Then we have

$\beta_{e}(\Gamma\odot\Omega)=|V(\Gamma)|\cdot(|V(\Omega)|-1).$

###### Theorem 4

_[29]_ Let $\Gamma$ and $\Omega$ be two connected graphs. Then we have
$\beta_{e}(\Gamma\vee\Omega)=|V(\Gamma)|+|V(\Omega)|-1$.

Zubrilina [38] showed the following results regarding the edge metric
dimension of graphs.

###### Theorem 5

_[38]_ Let $\Gamma$ be a graph with order $n$. Then $\beta_{e}(\Gamma)=n-1$ if
and only if for any distinct $v_{1},v_{2}\in V(\Gamma)$, there exists $u\in
V(\Gamma)$ such that $v_{1}u,v_{2}u\in E(\Gamma)$ and $u$ is adjacent to all
non-mutual neighbors of $v_{1}$ and $v_{2}$.

The following corollary is followed from Theorem 5.

###### Corollary 2

_[38]_ Let $\Gamma$ be a graph of order $n$. Suppose $\beta_{e}(\Gamma)=n-1$.
Then, $\mathrm{diam}(\Gamma)=2$ and every edge lies on a triangle.

Let $T_{n}=(V,E)$ be a tree and let $v\in V$. Define the equivalence relation
$R_{v}$ in the following way: for every two edges $e,~{}f$ we let $eR_{v}f$ if
and only if there is a path in $T_{n}$ including $e$ and $f$ that does not
have $v$ as an internal vertex. The subgraphs induced by the edges of the
equivalence classes of $E$ are called the bridges of $T_{n}$ relative to $v$.
Furthermore, for each vertex $v\in V$, the legs at $v$ are the bridges which
are paths. We denote by $l_{v}$ the number of legs at $v$.

For a tree $T_{n}$, Kelenc et al. [19] showed the following.

###### Remark 1

_[19]_ Let $T_{n}=(V,E)$ be a tree. If $T_{n}$ is not a path, then
$\beta(T_{n})=\beta_{e}(T_{n})=\sum_{v\in V,l_{v}>1}(l_{v}-1).$

Boutrig et al. [4] introduced the concept of vertex-edge domination in graphs.
A vertex $x\in V$ in $\Gamma$ is said to $ve-$dominate an edge if it dominates
the edges incident to it as well as the edges adjacent to these incident
edges. In other words, a vertex $w\in V$ in a graph $ve-$dominates an edge
$xy\in E$ in $\Gamma$ if

  * (i)

$w=x$ or $w=y$ ($w$ is incident to $xy$), or

  * (ii)

$wx$ or $wy$ is an edge in $\Gamma$ ($w$ is incident to an edge adjacent to
$xy$).

A set $T\subseteq V(\Gamma)$ is a $ve-$dominating set if for all edges $e\in
E(\Gamma)$, there exists a vertex $x\in T$ such that $x$ dominates $e$. The
minimum cardinality of a $ve-$dominating set in $\Gamma$ is called the
$ve-$domination number and is denoted by $\gamma_{ve}(\Gamma)$. A
$ve-$dominating set $T$ of cardinality $\gamma_{ve}(\Gamma)$ is known as a
$\gamma_{ve}$-set. Peters [28] in his PhD thesis showed the following results.

###### Proposition 2

_[28]_ The vertex-edge domination number of some graph families is computed as
follows:

  * _(i)_

For $m,n\geq 2$, we have $\gamma_{ve}(Kn)=\gamma_{ve}(K_{m,n})=1$.

  * _(ii)_

For $n\geq 2$, we have $\gamma_{ve}(P_{n})=\lfloor\frac{n+2}{4}\rfloor$.

  * _(iii)_

For $n\geq 3$, we have $\gamma_{ve}(C_{n})=\lfloor\frac{n+3}{4}\rfloor$.

###### Proposition 3

_[28]_ Let $\Gamma$ be a graph of order $n$. Then, $\gamma_{ve}(\Gamma)=1$ if
and only if there exists a vertex $x\in V(\Gamma)$ such that every vertex of
$\Gamma$ is within distance two of $x$ and if $Y=\\{y\in
V(\Gamma):d(x,y)=2\\}$ then $Y$ is an independent set of vertices.

###### Proposition 4

_[28]_ For any graph $\Gamma$ of order $n$, size $m$ and maximum degree
$\Delta(\Gamma)$,

$\Big{\lceil}\frac{m}{(\Delta(\Gamma))^{2}}\Big{\rceil}\leq\gamma_{ve}(\Gamma).$

###### Proposition 5

_[28]_ For any graph $\Gamma$ of size $m$, maximum degree $\Delta(\Gamma)$,
and minimum degree $\delta(\Gamma)$,

$\gamma_{ve}(\Gamma)\leq
m-\Delta(\Gamma)-\frac{\Delta(\Gamma)(\delta(\Gamma)-1)}{2}+1.$

Lewis [22] in his PhD thesis showed the following relation.

###### Proposition 6

_[22]_ For any graph $\Gamma$, we have
$\gamma_{ve}(\Gamma)\leq\gamma(\Gamma)$.

##  3 The vertex-edge dominant edge metric dimension of graphs

By combining the ideas of a vertex-edge dominating set and an edge resolving
set, we introduce a vertex-edge dominant edge resolving set. A subset
$W\subset V(\Gamma)$ is said to be a $ve$-dominant edge resolving set, if $W$
is a $ve-$dominating set and an edge resolving set simultaneously. Minimal
cardinality of such a set in $\Gamma$ is called the vertex-edge dominant edge
metric dimension of $\Gamma$. We denote this parameter by
$\gamma_{emd}(\Gamma)$.

In order to explain the idea, we present an example. We consider $\Gamma$ as
the graph with vertex set $V(\Gamma)=\\{v_{1},v_{2},\dots,v_{8}\\}$ and edge
set $E(\Gamma)=\\{a,b,c,d,e,f,g,h\\}$ as shown in Figure 1. Note that the set
$W=\\{v_{1},v_{4},v_{5}\\}$ is a vertex-edge dominant edge resolving set of
$\Gamma$.

Figure 1: A graph with the $\gamma_{emd}(G)=3$.

The vertex $v_{1}$ will dominate the edges $a,b,g,f$ and $e$ and the vertex
$v_{4}$ will dominate the edges $b,c,d,e$ and $h$. However, the vertices
$v_{1}$ and $v_{4}$ do not resolve all the edges $\Gamma$ since, for example,
$r(a|\\{v_{1},v_{4}\\})=r(f|\\{v_{1},v_{4}\\})$,
$r(b|\\{v_{1},v_{4}\\})=r(e|\\{v_{1},v_{4}\\})$ and
$r(c|\\{v_{1},v_{4}\\})=r(d|\\{v_{1},v_{4}\\})$. Table 1 exhibits the
representations of all the edges of $\Gamma$ corresponding to $W$. Routine
calculations show that no set of cardinality smaller than three is a vertex
edge dominant edge resolving set of $\Gamma$. This leads us to the fact that
$\gamma_{emd}(G)=3$.

Edges | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ | $g$ | $h$
---|---|---|---|---|---|---|---|---
${\small\gamma_{emd}r(..|T)}$ | $\\{0,2,2\\}$ | $\\{1,1,2\\}$ | $\\{2,0,1\\}$ | $\\{2,0,0\\}$ | $\\{1,1,0\\}$ | $\\{0,2,1\\}$ | $\\{0,3,1\\}$ | $\\{3,0,1\\}$
Table 1: The representations of the edges of $\Gamma$ with respect to the set
$W=\\{v_{1},v_{4},v_{5}\\}$.

Next, we compute the vertex-edge dominant edge metric dimension of classical
families of graphs.

###  3.1 Path and cycle graphs

The following theorems compute the vertex-edge dominant edge metric dimension
of path graph $P_{n}$ and cycle graph $C_{n}$.

###### Theorem 6

Let $P_{n}$ be a path graph, where $n\geq 2$. Then

$\gamma_{emd}(P_{n})=\begin{cases}\beta_{e}(P_{n})=1,&n=2,3;\\\
\beta_{e}(P_{n})+1=2,&n=4,5;\\\
\gamma_{ve}(P_{n})=\lfloor\frac{n+2}{4}\rfloor,&n\geq 6.\end{cases}$

Proof: Let $V(P_{n})=\\{v_{1},v_{2},\ldots,v_{n}\\}$ with $n-1$ edges and
$E(P_{n}=\\{e_{i}=v_{i}v_{i+1};1\leq i\leq n-1\\})$. Recall that, by Theorem
_(i)_(i) (resp. Proposition 2) we have $\beta_{e}(P_{n})=1$ (resp.
$\gamma_{ve}(P_{n})=\lfloor\frac{n+2}{4}\rfloor$). We divide the proof into a
number of cases:

Case 1: If $n=2$, then $W=\\{v_{1}\\}$ will dominate and gave a unique code to
the edge $e_{1}$.

Case 2: If $n=3$, then $W=\\{v_{1}\\}$ will dominate and gave a unique code to
the edges $e_{1}$ and $e_{2}$.

Case 3: If $n=4$, then there are four vertices and three edges. Without loss
of generality, we assume that $v_{1},v_{2},v_{3},v_{4}$ be the four vertices
and $e_{1},e_{2},e_{3}$ be the three edges. In this case, the vertex $v_{2}$
will dominate all the edges, however, does not give a unique code to the all
edges. As, $r(e_{1}|v_{2})=r(e_{2}|v_{2})$, the edge $e_{1}$ and the edge
$e_{2}$ have same code or representation. Similarly, the vertex $v_{3}$ will
dominate all the edges, however, does not give a unique code to the all edges.
As, $r(e_{2}|v_{3})=r(e_{3}|v_{3})$, the edge $e_{2}$ and the edge $e_{3}$
have same code or representation. So, the set $W=\\{v_{1},v_{2}\\}$ will
dominate and provide a unique code to all the edges of $P_{4}$ and we obtain
that $\gamma_{emd}(P_{4})=\beta_{e}(P_{4})+1=2$.

Case 4: For $n=5$, there are five vertices, say,
$v_{1},v_{2},v_{3},v_{4},v_{5}$ and four edge, say, $e_{1},e_{2},e_{3},e_{4}$.
In this case, the vertex $v_{3}$ will dominate all edges, however, does not
yield a unique code to the all edges. The edge $e_{2}$ and the edge $e_{3}$
would have the same representation. So, the set $W=\\{v_{1},v_{3}\\}$ will
dominate and gives a unique code to the all edges of $P_{5}$ and we obtain
that $\gamma_{emd}(P_{5})=\beta_{e}(P_{5})+1=2$.

Case 5: Assume $n\geq 6$. Let $W\subset V(P_{n})$, such that

$W=\begin{cases}\\{v_{2+4i}:0\leq i<\lfloor\frac{n+2}{4}\rfloor\\},&n\equiv
0(\mod 4);\\\ \\{v_{3+4i}:0\leq i<\lfloor\frac{n+2}{4}\rfloor\\},&n\equiv
1(\mod 4);\\\ \\{v_{1+4i}:0\leq i<\lfloor\frac{n+2}{4}\rfloor\\},&n\equiv
2,3(\mod 4).\end{cases}$

Since $|W|=\lfloor\frac{n+2}{4}\rfloor$, by Proposition 2, the set $W$
dominates all the edges of $P_{n}$ and no two edges have the same
representation with respect to $W$. We obtain the desired result.

###### Theorem 7

Let $C_{n}$ ba a cycle graph with $n\geq 3$. Then

$\gamma_{emd}(C_{n})=\begin{cases}\beta_{e}(C_{n})=2,&3\leq n\leq 7;\\\
\beta_{e}(C_{n})+1=3,&n=8;\\\
\gamma_{ve}(C_{n})=\lfloor\frac{n+3}{4}\rfloor,&n\geq 9.\end{cases}$

Proof: Let $V(C_{n})=\\{u_{1},u_{2},\ldots,u_{n}\\}$ and
$E(C_{n})=\\{e_{i}=u_{i}u_{i+1};1\leq i\leq n,u_{n+1}=u_{1}\\}$. Recall that,
by Theorem _(i)_(ii) (resp. Proposition 2) we have $\beta_{e}(C_{n})=2$ (resp.
$\gamma_{ve}(C_{n})=\lfloor\frac{n+3}{4}\rfloor$). We divide the proof into a
number of cases:

Case 1: If $n=3$, then the vertex $u_{1}$, will dominate all edges of $C_{3}$,
however, the edges $e_{1}$ and $e_{2}$ would have the same representation as
$\beta_{e}(C_{3})=2$. The set $W=\\{u_{1},u_{2}\\}$ is a vertex-edge dominant
edge metric basis. So, we get $\gamma_{emd}(C_{3})=2$.

Case 2: If $n=4$, then the vertex $u_{1}$ will dominate all edges of $C_{4}$
but the edges $e_{1}$ (resp. $e_{2}$) would have the same representation with
$e_{4}$ (resp. $e_{3}$) because of the fact that $\beta_{e}(C_{4})=2$. The set
$W=\\{u_{1},u_{2}\\}$ is a vertex-edge dominant edge metric basis. Therefore,
we obtain $\gamma_{emd}(C_{4})=2$.

Case 3: If $n=5$, then any vertex $u\in V(C_{5})$ will not dominate all edges
of $C_{5}$ as $\gamma_{ve}(C_{5})=2$ by Proposition 2. However, the set
$W=\\{u_{1},u_{4}\\}$ is a vertex-edge dominant edge metric basis as it would
generate distinct codes for all the edges of $C_{5}$. This implies that
$\gamma_{emd}(C_{5})=2$.

Case 4: If $n=6$ or $n=7$, then similar to Case 3, a single vertex $u\in
V(C_{6})$ or $u\in V(C_{7})$ can never dominate all edges of $C_{6}$ or
$C_{7}$ as $\gamma_{ve}(C_{6})=\gamma_{ve}(C_{7})=2$ by Proposition 2.
However, the set $W=\\{u_{1},u_{5}\\}$ is a vertex-edge dominant edge metric
basis as it would generate distinct codes for all the edges of $C_{6}$ and
$C_{7}$. This implies that $\gamma_{emd}(C_{6})=\gamma_{emd}(C_{7})=2$.

Case 5: If $n=8$, then the vertices $u_{1}$ and $u_{5}$ dominate all edges of
$C_{8}$. However, they do not give different codes of representation to all
the edges as $r(e_{j}|\\{u_{1},u_{5}\\})=r(e_{n+j-1}|\\{u_{1},u_{5}\\})$. Note
that $W=\\{u_{1},u_{3},u_{5}\\}$ is a vertex-edge dominant edge metric basis
as it would generate distinct codes for all the edges of $C_{8}$. This implies
that $\gamma_{emd}(C_{8})=3$.

Case 6: Assume $n\geq 9$. Let $W\subset V(C_{n})$, such that

$W=\begin{cases}\\{u_{1+5i}:0\leq i<\lfloor\frac{n+3}{4}\rfloor\\},&n\equiv
0,2,3(\mod 4);\\\ \\{u_{n-1},u_{1+5i}:0\leq
i<\lfloor\frac{n+3}{4}\rfloor-1\\},&n\equiv 1(\mod 4).\end{cases}$

Since, $|W|=\lfloor\frac{n+3}{4}\rfloor$, by Proposition 2, the set $W$
dominates all the edges of $C_{n}$ and no two edges have the same
representation with respect to $W$. We obtain the desired result.

###  3.2 Complete graphs and complete bipartite graphs

The following theorem computes the vertex-edge dominant edge metric dimension
of the complete graph $K_{n}$ and the complete bipartite graph $K_{n,m}$.

###### Theorem 8

Let $K_{n}$ be the complete graph with order $n\geq 2$. Then
$\gamma_{emd}(K_{n})=\beta_{e}(K_{n})=n-1$.

Proof: By Proposition 2, we have $\gamma_{ve}(K_{n})=1$. Any vertex $u\in
V(K_{n})$ will dominate all edges of the graph $K_{n}$ but, by Theorem
_(iii)_(iii), the edge dimension of $K_{n}$ is $n-1$ i.e.
$\beta_{e}(K_{n})=n-1$ . Thus, any vertex from an edge metric basis for the
graph $K_{n}$ will dominate all edges of the complete graph $K_{n}$.
Therefore, we obtain $\gamma_{emd}(K_{n})=\beta_{e}(K_{n})=n-1$.

###### Theorem 9

Let $K_{n,m}$ be the complete bipartite graph, where $n,m\geq 2$. Then
$\gamma_{emd}(K_{n,m})=\beta_{e}(K_{n,m})=n+m-2$.

Proof: By Proposition 2, we have $\gamma_{ve}(K_{n,m})=1$. Any vertex $u\in
V(K_{n,m})$ will dominate all edges of the graph $K_{n,m}$ but, by Theorem
_(iv)_(iv), the edge dimension of complete bipartite graph $K_{n,m}$ is
$n+m-2$ i.e. $\beta_{e}(K_{n,m})=n+m-2$. Thus, any vertex from an edge metric
basis for the graph $K_{n,m}$ will dominate all edges of the complete
bipartite graph $K_{n,m}$. Therefore, we obtain
$\gamma_{emd}(K_{n,m})=\beta_{e}(K_{n,m})=n+m-2$.

As a special case to Theorem 9, we obtain the following.

###### Corollary 3

Let $S_{1,n}$ be the star graph, where $n\geq 3$. Then
$\gamma_{emd}(S_{1,n})=\beta_{e}(S_{1,n})=n-1$.

###  3.3 Wheel and fan graphs

For the ($n+1$)-dimensional wheel and fan graphs, first we compute its
dominant metric dimension and the vertex-edge domination number.

###### Proposition 7

Let $\Gamma$ be a wheel graph $W_{1,n}$ or fan graph $F_{1,n}$. Then,

$\gamma_{md}(\Gamma)=\left\\{\begin{array}[]{ll}\beta(\Gamma),&\hbox{$n\equiv
0,2,4(\mod 5)$;}\\\ \beta(\Gamma)+1,&\hbox{$n\equiv 1,3(\mod
5)$.}\end{array}\right.$

Proof: Let $V(\Gamma)=\\{x,y_{1},y_{2},\ldots,y_{n}\\}$, where the vertex $x$
has degree $n$ and the vertices $y_{1},y_{2},\ldots,y_{n}$ induce a cycle
$C_{n}$ in wheel graph $\Gamma=W_{1,n}$ ( or induce a path $P_{n}$ in fan
graph $\Gamma=F_{1,n}$). Let $S=\\{x\\}$. Then, every vertex of $\Gamma$ is
dominated by $S$. This implies that $\gamma(\Gamma)=1$. Buczkowski et al. [5]
showed that $\beta(\Gamma=W_{1,n})=\lfloor\frac{2n+2}{5}\rfloor$, where $n\geq
7$ and Hernando et al. [17] showed that
$\beta(\Gamma=F_{1,n})=\lfloor\frac{2n+2}{5}\rfloor$, where $n\geq 7$. Based
on resolvability of vertices in $\Gamma$, we further divide the proof into
five cases:

Case 1: Assume that $n\equiv 0(\mod 5)$.

Let $n=5k$ where $k\geq 2$ and $\lfloor\frac{2n+2}{5}\rfloor=2k$. Note that
for $D=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}$, $V(\Gamma)\backslash D=N(D)$.
Since $|D|=\lfloor\frac{2n+2}{5}\rfloor$. This implies that $D$ is a dominant
resolving set for $\Gamma$ where $n\equiv 0(\mod 5)$. Thus,
$\gamma_{md}(\Gamma)=\beta(\Gamma)$ for $n\equiv 0(\mod 5)$.

Case 2: Assume $n\equiv 1(\mod 5)$.

Let $n=5k+1$ where $k\geq 2$ and $\lfloor\frac{2n+2}{5}\rfloor=2k$. Note that
for $D_{1}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}$, we have
$V(\Gamma)\backslash D_{1}\neq N(D_{1})$, because the vertex $y_{5k+1}$ is not
dominated by any vertex of the set $D_{1}$. The set $D_{1}$ is resolving set
but is not dominating set for $\Gamma$. This implies that $D_{1}$ is not a
dominant resolving set for $\Gamma$, where $n\equiv 1(\mod 5)$. Note that for
$D_{2}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-2\\}\cup\\{y_{5k-3},y_{5k+1}\\}$, we
have $V(\Gamma)\backslash D_{2}\neq N(D_{2})$, because the vertex $y_{5k-1}$
is not dominated by any vertex of the set $D_{2}$. The set $D_{2}$ is
resolving set but is not dominating set for $\Gamma$. This implies that
$D_{2}$ is not a dominant resolving set for $\Gamma$, where $n\equiv 1(\mod
5)$. Note that for $D_{3}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq
k-2\\}\cup\\{y_{5k-3},y_{5k}\\}$, we have $V(\Gamma)\backslash
D_{3}=N(D_{3})$. The set $D_{3}$ is dominating set but is not resolving set
for $\Gamma$ as $r(y_{5k-1}|D_{3})=r(y_{5k+1}|D_{3})$. This implies that
$D_{3}$ is not a dominant resolving set of $\Gamma$, where $n\equiv 1(\mod
5)$. Note that for $D=S\cup D_{1}=\\{x\\}\cup\\{y_{5i+2},y_{5i+4}:0\leq i\leq
k-1\\}$, we have $V(\Gamma)\backslash D=N(D)$, the set $D$ is dominating and
resolving set for $\Gamma$. This implies that $D$ is a dominant resolving set
for $\Gamma$, where $n\equiv 1(\mod 5)$. Thus, we obtain
$\gamma_{md}(\Gamma)=\beta(\Gamma)+1$, where $n\equiv 1(\mod 5)$.

Case 3: Assume $n\equiv 2(\mod 5)$.

Let $n=5k+2$ where $k\geq 1$ and $\lfloor\frac{2n+2}{5}\rfloor=2k+1$. For
$D=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}\cup\\{y_{5k+2}\\}$ we have
$V(\Gamma)\backslash D=N(D)$. This implies that $D$ is a dominant resolving
set of $\Gamma$ where $n\equiv 2(\mod 5)$. Therefore, we obtain
$\gamma_{md}(\Gamma)=\beta(\Gamma)$, where $n\equiv 2(\mod 5)$.

Case 4: Assume $n\equiv 3(\mod 5)$.

Let $n=5k+3$ where $k\geq 1$ and $\lfloor\frac{2n+2}{5}\rfloor=2k+1$. For
$D_{1}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}\cup\\{y_{5k+3}\\}$ we have
$V(\Gamma)\backslash D_{1}\neq N(D_{1})$, because the vertex $y_{5k+1}$ is not
dominated by any vertex of the set $D_{1}$. The set $D_{1}$ is resolving set
but is not dominating set for $\Gamma$. This implies that the set $D_{1}$ is
not a dominant resolving set for $\Gamma$ where $n\equiv 3(\mod 5)$. For
$D_{2}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}\cup\\{y_{5k+1}\\}$ we have
$V(\Gamma)\backslash D_{2}\neq N(D_{2})$, because the vertex $y_{5k+3}$ is not
dominated by any vertex of the set $D_{2}$. The set $D_{2}$ is resolving set
but is not dominating set for $\Gamma$. This implies that the set $D_{2}$ is
not a dominant resolving set for $\Gamma$ where $n\equiv 3(\mod 5)$. For
$D_{3}=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}\cup\\{y_{5k+2}\\}$ we have
$V(\Gamma)\backslash D_{3}=N(D_{3})$. The set $D_{3}$ is dominating set but is
not resolving set for $\Gamma$ as $r(y_{5k+1}|D_{3})=r(y_{5k+3}|D_{3})$. This
implies that $D_{3}$ is not a dominant resolving set for $\Gamma$, where
$n\equiv 1(\mod 5)$. Note that for $D=S\cup
D_{1}=\\{x\\}\cup\\{y_{5i+2},y_{5i+4}:0\leq i\leq k-1\\}\cup\\{y_{5k+3}\\}$,
we have $V(\Gamma)\backslash D=N(D)$. This implies that the set $D$ is a
dominant resolving set for $\Gamma$ where $n\equiv 3(\mod 5)$. Thus, we obtain
$\gamma_{md}(\Gamma)=\beta(\Gamma)+1$, where $n\equiv 3(\mod 5)$.

Case 5: Assume $n\equiv 4(\mod 5)$.

Let $n=5+4k$ where $k\geq 1$ and $\lfloor\frac{2n+2}{5}\rfloor=2k+2$. Note
that if $D=\\{y_{5i+2},y_{5i+4}:0\leq i\leq k\\}$, then we have
$V(\Gamma)\backslash D=N(D)$. This implies that $D$ is a dominant resolving
set for $\Gamma$ where $n\equiv 4(\mod 5)$. Therefore,
$\gamma_{md}(\Gamma)=\beta(\Gamma)$, where $n\equiv 4(\mod 5)$. This completed
the proof.

The following proposition computes the vertex-edge domination number of wheel
graphs.

###### Proposition 8

For the wheel graph $W_{1,n}$, fan graph $F_{1,n}$ and star graph $S_{1,n}$,
we have

$\gamma_{ve}(W_{1,n})=\gamma_{ve}(F_{1,n})=\gamma_{ve}(S_{1,n})=1.$

Proof: Assume that
$V(W_{1,n})=V(F_{1,n})=V(S_{1,n})=\\{x,y_{1},y_{2},\ldots,y_{n}\\}$, where the
vertex $x$ has degree $n$ and the vertices $y_{1},y_{2},\ldots,y_{n}$ induce a
cycle $C_{n}$ in wheel graph and induce a path $P_{n}$ in the fan graph. In
$S_{1,n}$, all the edges are incident to the vertex $x$ so, the set
$T=\\{x\\}$ dominate all edges of star graph $S_{1,n}$. In the wheel and the
fan graph, all edges are incident to $N(x)=\\{y_{1},y_{2},\ldots,y_{n}\\}$ so,
the set $T=\\{x\\}$ dominate all edges of $W_{1,n}$ and $F_{1,n}$. Therefore,
the set $T=\\{x\\}$ dominate all edges of the wheel graph $W_{1,n}$, the fan
graph $F_{1,n}$ and the star graph $S_{1,n}$. Thus, we obtain
$\gamma_{ve}(W_{1,n})=\gamma_{ve}(F_{1,n})=\gamma_{ve}(S_{1,n})=1$.

The next theorem calculates the vertex-edge dominant edge metric dimension of
wheel graphs.

###### Theorem 10

Let $W_{1,n}$ be a wheel graph $W_{1,n}$ and $F_{1,n}$ be a fan graph with
$n\geq 5$. Then

$\gamma_{emd}(W_{1,n})=\gamma_{emd}(F_{1,n})=n-1.$

Proof: For $n\geq 5$, the edge metric dimension of wheel graph is
$\beta_{e}(W_{1,n})=n-1$ by Theorem _(vi)_(vi) and the edge metric dimension
of fan graph is $\beta_{e}(F_{1,n})=n-1$ by Theorem _(vii)_(vii). The set
$W=\\{y_{1},y_{2},\ldots,y_{n-1}\\}$ is an edge metric basis/generator. The
set $W$ dominates all the edges of wheel graph $W_{1,n}$ and fan graph
$F_{1,n}$. The set $W$ is also the vertex-edge dominant set. Therefore, we
obtain that
$\gamma_{emd}(W_{1,n})=\gamma_{emd}(F_{1,n})=\beta_{e}(W_{1,n})=\beta_{e}(W_{1,n})=n-1$,
if $n\geq 5$.

##  4 The vertex-edge dominant edge metric dimension of Cartesian products

In this section, we study the vertex-edge dominant edge metric dimension in
some standard Cartesian products such as $P_{n}\Box P_{r}$ and $C_{n}\Box
P_{r}$.

The graph $\Gamma=P_{n}\Box P_{2}$ is the Cartesian product of the graphs
$P_{n}$ and $P_{2}$. The vertex set and edge set of $\Gamma$ are $V(P_{n}\Box
P_{2})=\\{u_{i},vj;1\leq i,j\leq n\\}$ and $E(P_{n}\Box
P_{2})=\\{a_{i}=u_{i}u_{i+1},b_{j}=v_{j}v_{j+1};1\leq i,j\leq
n-1\\}\cup\\{c_{i}=u_{i}v_{j};i=j,1\leq i,j\leq n\\}$, respectively.

###### Proposition 9

Let $\Gamma=P_{n}\Box P_{2}$ be the Cartesian product of $P_{n}$ and $P_{2}$.
Then, we have

$\gamma_{ve}(\Gamma)=\Big{\lceil}\frac{n}{3}\Big{\rceil}.$

Proof: Note that the order (resp. size) of $\Gamma$ is $2n$ (resp. $3n-2$).
Furthermore, degrees of all vertices of $\Gamma$ are 3 except the vertices
$u_{1},u_{n},v_{1},v_{n}$ which are of degree 2. We divide the proof into a
number of cases:

Case 1: Assume that $n=1,2,3$.

For $n=1,2$, the singleton set $T=\\{u_{1}\\}$ is a minimum vertex-edge
dominant set, whereas for $n=3$, the singleton set $T=\\{u_{2}\\}$ is a
minimum vertex-edge dominant set.

Case 2: Assume that $n=4,5,6$.

The set $T=\\{u_{1},v_{4}\\}$ (resp. $T=\\{u_{2},v_{5}\\}$) is a minimum
vertex-edge dominant set if $n=4,5$ (resp. $n=6$).

Case 3: Assume that $n\geq 7$ and $n\equiv 0(\mod 6)$. The minimum vertex-edge
dominant set in this case is

$T=\\{u_{2+6i};0\leq
i<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}\cup\\{v_{5+6j};0\leq
j<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}.$

Case 4: Assume that $n\equiv 1,2(\mod 6)$ and $n\geq 7$. The minimum vertex-
edge dominant set in this case is

$T=\\{u_{1+6i};0\leq
i<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}\cup\\{v_{4+6j};0\leq
j<\Big{\lceil}\frac{n}{6}\Big{\rceil}-1\\}.$

Case 5: Assume that $n\equiv 3(\mod 6)$ and $n\geq 7$. The minimum vertex-edge
dominant set in this case is

$T=\\{u_{2+6i};0\leq
i<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}\cup\\{v_{5+6j};0\leq
j<\Big{\lceil}\frac{n}{6}\Big{\rceil}-1\\}.$

Case 6: Assume that $n\equiv 4,5(\mod 6)$ and $n\geq 7$. The set

$T=\\{u_{1+6i};0\leq
i<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}\cup\\{v_{4+6j};0\leq
j<\Big{\lceil}\frac{n}{6}\Big{\rceil}\\}.$

provides the minimum vertex-edge dominant set of $\Gamma$. From Case 3, Case
4, Case 5 and Case 6 it is clear that the
$\gamma_{ve}(\Gamma)\leq\lceil\frac{n}{3}\rceil$. Since, the maximum degree of
$\Gamma$ is $3$ i.e. $\Delta(\Gamma)=3$, by Proposition 4, we obtained that
$\gamma_{ve}(\Gamma)\geq\lceil\frac{n}{3}\rceil$. This show that
$\gamma_{ve}(\Gamma)=\lceil\frac{n}{3}\rceil$.

###### Theorem 11

Let $\Gamma=P_{n}\Box P_{2}$ be the Cartesian product of $P_{n}$ and $P_{2}$.
Then, we have

$\gamma_{emd}(\Gamma)=\begin{cases}\beta_{e}(\Gamma)=1&n=1;\\\
\beta_{e}(\Gamma)=2&n=2;\\\ \beta_{e}(\Gamma)+1=3&n=3,4,5,6;\\\
\gamma_{ve}(\Gamma)=\lceil\frac{n}{3}\rceil&n\geq 7.\end{cases}$

Proof: By Theorem _(v)_ (resp. Proposition 9), we have that
$\beta_{e}(\Gamma=P_{n}\Box P_{2})=2$ (resp.
$\gamma_{ve}(\Gamma)=\lceil\frac{n}{3}\rceil$). Next, we divide the proof into
a number of cases:

Case 1: Assume that $n=1,2,3$.

Note that we have $\beta_{e}(\Gamma)=1$ (resp. $\beta_{e}(\Gamma)=2$) for
$n=1$ (resp. $n=2,3$). Moreover, $\gamma_{ve}(\Gamma)=1$ for $n=1,2,3$. Then,
$W=\\{u_{1}\\}$ provide unique representations to every edge of $\Gamma$ if
$n=1$. Similarly, $T=\\{u_{1},u_{2}\\}$ (resp. $T=\\{u_{1},u_{2},u_{3}\\}$))
provide distinct codes to the edges of $\Gamma$ if $n=2$ (resp. $n=3$). Thus,
we obtain

$\gamma_{emd}(\Gamma)=\begin{cases}1,&n=1;\\\ 2,&n=2;\\\ 3,&n=3.\end{cases}$

Case 2: Assume that $n=4,5,6$.

Note that, for $n=4$, the set $A=\\{u_{1},u_{4}\\}$ is a minimum edge
resolving set of $\Gamma$. In a similar manner, the set $A=\\{u_{1},u_{5}\\}$
(resp. $A=\\{u_{1},u_{6}\\}$) provides a minimum edge resolving set for $n=5$
(resp. $n=6$). Therefore, we obtain that $\beta_{e}(\Gamma)=2$, if $n=4,5,6$.
Moreover, the set $T=\\{u_{1},v_{4}\\}$ is a minimum vertex-edge dominating
set for $P_{4}\Box P_{2}$. And, the set $T=\\{u_{1},v_{5}\\}$ (resp.
$T=\\{u_{2},v_{5}\\}$) is a minimum vertex-edge dominating set for $P_{5}\Box
P_{2}$ (resp. $P_{6}\Box P_{2}$). This implies that $\gamma_{ve}(\Gamma)=2$ if
$n=4,5,6$. However, the set $A$ can not dominate the edge $b_{2}=v_{2}v_{3}$
if $n=4$, the edges $b_{2}=v_{2}v_{3},b_{3}=v_{3}v_{4}$ if $n=5$ and the edges
$a_{3}=u_{3}u_{4},b_{2}=v_{2}v_{3},b_{3}=v_{3}v_{4},b_{4}=v_{4}v_{5},c_{3}=u_{3}v_{3},c_{4}=u_{4}v_{4}$
if $n=6$. Consequently, the set $T$ does not resolve all the edges of $\Gamma$
for $n=4,5,6$. Additionally, the set

$W=\left\\{\begin{array}[]{ll}\\{u_{1},u_{4},v_{4}\\},&\hbox{$n=4$;}\\\
\\{u_{1},u_{5},v_{5}\\},&\hbox{$n=5$;}\\\
\\{u_{2},u_{6},v_{5}\\},&\hbox{$n=6$.}\end{array}\right.$

dominates and delivers unique codes to all the edges of $P_{n}\Box P_{2}$, if
$n=4,5,6$. This implies that $\gamma_{emd}(\Gamma)=\beta_{e}(\Gamma)+1=3$, for
$n=4,5,6$.

Case 3: Finally, we assume that $n\geq 7$.

The set $A=\\{u_{1},u_{n}\\}$ is a minimum edge resolving set. Thus, we have
$\beta_{e}(\Gamma)=2$. By Proposition 9, the vertex-edge domination number of
$\Gamma$ is $\gamma_{ve}(\Gamma)=\lceil\frac{n}{3}\rceil$. Note that the
minimum vertex-edge dominating set is also a minimum edge resolving set for
$P_{n}\Box P_{2}$. Thus, we obtain that
$\gamma_{emd}(\Gamma)=\gamma_{ve}(\Gamma)=\lceil\frac{n}{3}\rceil$, if $n\geq
7$.

###### Proposition 10

Let $\Gamma=C_{n}\Box P_{2}$ be the Cartesian product of $C_{n}$ and $P_{2}$,
with $n\geq 3$. Then, $\beta_{e}(\Gamma)=3$.

Proof: Let $V(\Gamma)=\\{x_{i},y_{i};1\leq i\leq n\\}$ and
$E(\Gamma)=\\{e_{i}=x_{i}x_{i+1},f_{i}=y_{i}y_{i+1},g_{i}=x_{i}y_{i};1\leq
i\leq n\\}$. Clearly the order (resp. size) of $\Gamma$ is $2n$ (resp. $3n$).
Let $A=\\{x_{1},x_{2},x_{l+1}\\}$ where $l=\lceil\frac{n}{2}\rceil$ be an edge
metric basis. We divide the proof further into some cases:

Case 1: Assume that $n\equiv 0(\mod 2)$. In this case, the representations of
edges of $\Gamma$ is as follow:

Subcase 1.1: Assume $e_{i}=x_{i}x_{i+1}$ for $i=1,2,3,\ldots,n$. Then,

$r(e_{i}|A)=\begin{cases}(0,0,l-i)&i=1;\\\ (i-1,i-2,l-i)&2\leq i\leq l;\\\
(l-1,l-1,i-l-1)&i=l+1;\\\ (n-i,n-i+1,i-l-1)&l+2\leq i\leq n.\end{cases}$

Subcase 1.2: Assume $f_{i}=y_{i}y_{i+1}$ for $i=1,2,3,\ldots,n$. Then,

$r(f_{i}|A)=\begin{cases}(1,1,l)&i=1;\\\ (i,i-1,l-i+1)&2\leq i\leq l;\\\
(l,l,i-l)&i=l+1;\\\ (n-i+1,n-i+2,i-l)&l+2\leq i\leq n.\end{cases}$

Subcase 1.3: Assume $g_{i}=x_{i}y_{i}$ for $i=1,2,3,\ldots,n$. Then,

$r(g_{i}|A)=\begin{cases}(0,1,l)&i=1;\\\ (i-1,i-2,l-i+1)&2\leq i\leq l+1;\\\
(n-i+1,n-i+2,i-l-1)&l+2\leq i\leq n.\end{cases}$

For the case where $n$ is even, it is easily observed that no two edges have
the same representation. Therefore, $\beta_{e}(\Gamma)\leq 3$, if $n\equiv
0(\mod 2)$.

Case 2: Assume that $n\equiv 1(\mod 2)$. The representations for the edges of
$\Gamma$ in this cases are as follow:

Subcase 2.1: Assume $e_{i}=x_{i}x_{i+1}$ for $i=1,2,3,\ldots,n$. Then,

$r(e_{i}|A)=\begin{cases}(0,0,l-i)&i=1;\\\ (i-1,i-2,l-i)&2\leq i\leq l;\\\
(n-i,n-i+1,i-l-1)&l+1\leq i\leq n.\end{cases}$

Subcase 2.2: Assume $f_{i}=y_{i}y_{i+1}$ for $i=1,2,3,\ldots,n$. Then,

$r(f_{i}|A)=\begin{cases}(1,1,l)&i=1;\\\ (i,i-1,l-i+1)&2\leq i\leq l;\\\
(n-i+1,n-i+2,i-l)&l+1\leq i\leq n.\end{cases}$

Subcase 2.3: Assume $g_{i}=x_{i}y_{i}$ for $i=1,2,3,\ldots,n$. Then,

$r(g_{i}|A)=\begin{cases}(0,1,l-1)&i=1;\\\ (i-1,i-2,l-i+1)&2\leq i\leq l;\\\
(l-1,l-1,i-l-1)&i=l+1;\\\ (n-i+1,n-i+2,i-l-1)&l+2\leq i\leq n.\end{cases}$

Thus, for the case where $n$ is odd, it can be seen that no two edges have the
same representation. Therefore, $\beta_{e}(\Gamma)\leq 3$, if $n\equiv
0,1(\mod 2)$. Since $\Gamma$ is 3-regular graph i.e. $r=3$, by Corollary 1, we
obtain that $\beta_{e}(\Gamma)\geq 3$. This shows that $\beta_{e}(\Gamma)=3$
for any $n$. This completes the proof.

###### Proposition 11

Let $\Gamma=C_{n}\Box P_{2}$ be the Cartesian product of $C_{n}$, $n\geq 4$
and $P_{2}$. Then,
$\gamma_{ve}(\Gamma)=\Big{\lceil}\frac{|E(\Gamma)|}{9}\Big{\rceil}$.

Proof: Note that $\Gamma$ is cubic graph, i.e. every vertex has exactly three
edges incident with it. Therefore, by definition of vertex-edge domination,
each vertex of $\Gamma$ will dominate nine of its edges. For instance, the
vertex $x_{i}$ (resp. $y_{i}$) will dominate the 9 edges
$e_{i-2},e_{i-1},e_{i},e_{i+1},f_{i-1},f_{i},g_{i-1},g_{i},g_{i+1}$ (resp.
$f_{i-2},f_{i-1},f_{i},f_{i+1},e_{i-1},e_{i},g_{i-1},g_{i},g_{i+1}$).
Moreover, all these edges are incident to the vertices of $N(x_{i})$ and
$N(y_{i})$, respectively. This implies that the vertex-edge domination number
of $\Gamma$ is at least $\lceil\frac{|E(\Gamma)|}{9}\rceil$. We divide the
vertex-edge domination set of $\Gamma$ into a number of cases:

Case 1: Let $4\leq n\leq 9$.

If $n=4,5,6$ (resp. $n=7,8,9$), then $T=\\{x_{1},y_{4}\\}$ (resp.
$T=\\{x_{1},y_{4},x_{7}\\}$) is a minimum vertex-edge domination set of
$\Gamma$.

Case 2: Let $n\geq 10$. We divide the case further into two subcases:

Subcase 2.1: Assume that $n\equiv 0,4,5(\mod 6)$. The minimum vertex-edge
dominant set in this case is

$T=\\{x_{1+6i},y_{4+6i};0\leq i\leq\lceil\frac{n}{6}\rceil-1\\}.$

Subcase 2.2: Assume that $n\equiv 1,2,3(\mod 6)$. The set

$T=\\{x_{1+6i};0\leq i\leq\lceil\frac{n}{6}\rceil-1\\}\cup\\{y_{4+6i};0\leq
i\leq\lceil\frac{n}{6}\rceil-2\\}.$

is a minimum vertex-edge domination set of $\Gamma$. Since $\mid
T\mid=\lceil\frac{|E(\Gamma)|}{9}\rceil$, we obtain that the vertex-edge
domination set is at most $\lceil\frac{|E(\Gamma)|}{9}\rceil$. This completes
the proof.

###### Theorem 12

Let $\Gamma=C_{n}\Box P_{2}$ be the Cartesian product of $C_{n}$, $n\geq 4$
and $P_{2}$. Then, we have

$\gamma_{emd}(\Gamma)=\begin{cases}\big{\lceil}\frac{|E(\Gamma)|}{9}\big{\rceil}+1,&4\leq
n\leq 9;\\\ \big{\lceil}\frac{|E(\Gamma)|}{9}\big{\rceil},&n\geq
10.\end{cases}$

Proof: By Proposition 10, we obtain that $\beta_{e}(\Gamma)=3$. Similarly, by
Proposition 11, the vertex-edge dominant edge metric dimension
$\gamma_{ve}(\Gamma)$ of $\Gamma$ is
$\Big{\lceil}\frac{|E(\Gamma)|}{9}\Big{\rceil}.$ We divide the proof into a
number of cases:

Case 1: Let $n=4,5$.

Since $\beta_{e}(\Gamma)=3$ and $\gamma_{ve}(\Gamma)=1$. The vertex $x_{1}$
will dominate all the edges but would not resolve all edges as $\Gamma$ is not
a path graph. However, the set $W=\\{x_{1},x_{2},x_{3}\\}$ will dominate and
assign unique codes simultaneously to all the edges $\Gamma$. Thus, we obtain
$\gamma_{emd}(\Gamma)=3$.

Case 2: Let $n=6$.

Note that, in this case, $\beta_{e}(\Gamma)=3$ and the
$\gamma_{ve}(\Gamma)=2$. The set $W=\\{x_{1},x_{3},x_{5}\\}$ will dominate and
assign unique codes simultaneously to all the edges $\Gamma$. Thus, we obtain
$\gamma_{emd}(\Gamma)=3$.

Case 3: Let $n=7,8,9$.

Note that, in this case, $\beta_{e}(\Gamma)=3$ and the
$\gamma_{ve}(\Gamma)=3$. However, neither the edge resolving set by
Proposition 10 nor the vertex-edge dominating set by Proposition 11 will
dominate all edges $\Gamma$, as some of the edges would have same codes. The
set $T=\\{x_{i},x_{i+6},y_{i+3}\\}$ where $1\leq i\leq n$ and
$x_{n+i}=x_{i},y_{n+i}=y_{i}$ is vertex-edge dominating set and the set
$A=\\{x_{i},x_{i+1},x_{i+l}\\}$ where $l=\lceil\frac{n}{2}\rceil$ is an edge
resolving set for $\Gamma$. For $n=7$, the set $T$ is vertex-edge dominating
set but is not edge resolving set as
$r(y_{i}y_{i+1}|T)=r(y_{i+1}x_{i+1}|T)=r(x_{i+1}x_{i+2}|T)=(1,2,2)$ and
$r(y_{i+1}y_{i+2}|T)=r(y_{i+2}x_{i+2}|T)=r(x_{i+2}x_{i+3}|T)=(2,3,1)$. The set
$A$ is an edge resolving set but is not vertex-edge dominant set as the edges
$y_{i+2}y_{i+3}$ and $y_{i+l+1}y_{i+l+2}$ are not dominated by any vertex of
$A$. For $n=8$, the set $T$ is vertex-edge dominating set but is not edge
resolving set as $r(x_{i+2}y_{i+2}|T)=r(x_{i+2}x_{i+3}|T)=(2,4,1)$,
$r(x_{i+3}y_{i+3}|T)=r(y_{i+2}y_{i+3}|T)=(3,3,0)$ and
$r(x_{i+6}y_{i+6}|T)=r(x_{i+7}y_{i+6}|T)=(2,0,3)$. The set $A$ is an edge
resolving set but is not vertex-edge dominant set as the edges
$y_{i+2}y_{i+3}$, $y_{i+l+1}y_{i+l+2}$, $y_{i+l+2}y_{i+l+3}$ and
$y_{i+l+2}x_{i+l+2}$ are not dominated by any vertex of $A$. For $n=9$, the
set $T$ is vertex-edge dominating set but is not edge resolving set as
$r(y_{i}y_{i+1}|T)=r(y_{i+1}x_{i+1}|T)=r(x_{i+1}x_{i+2}|T)=(1,4,2)$ and
$r(x_{l+i-1}x_{l+i}|T)=r(x_{l+i}y_{l+i}|T)=r(y_{l+i}y_{l+i+1}|T)=(4,1,2)$. The
set $A$ is an edge resolving set but is not vertex-edge dominant set as the
edges $y_{i+2}y_{i+3}$ and $y_{i+l+1}y_{i+l+2}$. The set $A$ is an edge
resolving set but is not vertex-edge dominant set as the edges
$y_{i+2}y_{i+3}$, $y_{i+3}y_{i+4}$, $x_{i+3}y_{i+3}$, $y_{i+l+1}y_{i+l+2}$ and
$y_{i+l+2}y_{i+l+3}$ are not dominated by any vertex of $A$.

Hence, the set $W=\\{x_{1},x_{3},x_{5},x_{7}\\}$ will dominate and assign
unique codes simultaneously to all the edges $\Gamma$. Thus, we obtain
$\gamma_{emd}(\Gamma)=4$.

Case 4: Let $n\geq 10$.

By Proposition 10, the set $A=\\{x_{1},x_{2},x_{l+1}\\}$ is an edge resolving
set. Moreover, by Proposition 11, we have
$\gamma_{ve}(\Gamma)=\Big{\lceil}\frac{|E(\Gamma)|}{9}\Big{\rceil}$. The
corresponding vertex-edge resolving set is also an edge resolving set of
$\Gamma$. Therefore, we obtain
$\gamma_{emd}(\Gamma)=\gamma_{ve}(\Gamma)=\Big{\lceil}\frac{|E(C_{n}\Box
P_{2})|}{9}\Big{\rceil}$.

##  5 Some general results and bounds

In this section, we provide some results for general graphs. First, we define
some necessary terminologies. For a graph $\Gamma$, two vertices $x,y\in
V(\Gamma)$ are said to be true twins (resp. false twins) if $N[x]=N[y]$ (resp.
$N(x)=N(y)$). In a similar fashion, a vertex $x\in V(\Gamma)$ is called true
twin (resp. false twin) in $\Gamma$, if there exists a vertex $y\neq x$ such
that $x,~{}y$ are true twins (resp. false twins).

###### Lemma 2

Let $\Gamma$ be a connected graph. Then there exists a minimum vertex-edge
dominating set for $\Gamma$ which does not have any pair of false twin
vertices.

Proof: Let $T$ be a minimum vertex-edge dominating set for $\Gamma$ with
minimum number of false twin pairs of vertices and $u,v$ be any arbitrary
false twin pair in $T$. Since $u$ and $v$ dominate the same edges in $\Gamma$,
by definition of vertex-edge domination, $T\setminus\\{u\\}$ and
$T\setminus\\{v\\}$ are vertex-edge dominating sets in $\Gamma$. This implies
that there exists a minimum vertex-edge dominating set for $\Gamma$ which does
not have any pair of false twin vertices. This shows the lemma.

###### Theorem 13

For every connected graph $\Gamma$ of order $n$, we have
$\beta_{e}(\Gamma)\leq n-\gamma_{ve}(\Gamma)$. In particular, if $T$ is a
minimum vertex-edge dominating set for $\Gamma$ with no false twin pair of
vertices, then $V(\Gamma)\setminus T$ is an edge resolving set for $\Gamma$.

Proof: By Lemma 1, $\Gamma$ has a minimum vertex-edge dominating set $S$ with
no pair of false twin vertices. Suppose, on the contrary, that
$V(\Gamma)\setminus T$ is not an edge resolving set for $\Gamma$. By
definition of edge metric dimension, there exist edges $e$ and $f$ such that
$r(e|V(\Gamma)\setminus T)=r(f|V(\Gamma)\setminus T)$ if edge $e$ is incident
to vertex $u$ and edge $f$ is incident to vertex $v$, then
$r(u|V(\Gamma)\setminus T)=r(v|V(\Gamma)\setminus T)$. This implies that all
neighboring vertices of $u$ and $v$ in $V(\Gamma)\setminus T$ are the same.
So, vertices $u$ and $v$ have no neighbor in the set $T$. On the other hand,
we can remove one of the vertices $u$ and $v$ from the set $T$ and get a
dominating set with order of $\mid T\mid-1$. Hence, $u$ and $v$ are false twin
vertices, which is a contradiction. Thus, $V(G)\setminus T$ is a resolving set
for $\Gamma$. Accordingly, $\beta_{e}(\Gamma)\leq n-\gamma_{ve}(\Gamma)$.

###### Lemma 3

Let $\Gamma$ be a connected graph. If $\Gamma$ has no vertex-edge dominant
edge resolving set with cardinality $l$, then $K\subseteq V(\Gamma)$ with
$|K|<l$ is not a vertex-edge dominant edge resolving set.

Proof:  Let $\Gamma$ be a connected graph. Assume that $\Gamma$ has no vertex-
edge dominant edge resolving set with cardinality $l$ and suppose that there
exist a vertex-edge dominant edge resolving set $X\subseteq V(\Gamma)$ with
$|X|<l$. So, that for every $e_{i},e_{j}\in E(\Gamma)$, we have
$r(e_{i}|X)\neq r(e_{j}|X)$ and every edge $e\in E(\Gamma)$, is dominated by
any vertex of set $X$. Moreover, there exist a set $Y$ and $Y\subseteq
V(\Gamma)\setminus X$ such that $|X\cup Y|=l$. The set $X\cup Y$ is an edge
resolving set for $\Gamma$ as $X$ is an edge resolving set for $\Gamma$ and
$X\cup Y$ is vertex-edge dominant set for $\Gamma$ as $X$ is vertex-edge
dominant set for $\Gamma$. So, that the set $X\cup Y$ is vertex-edge dominant
edge resolving set for $\Gamma$, which is contradiction and the proof is
complete.

The following proposition bounds the vertex-edge domination number of a
connected graph $\Gamma$.

###### Proposition 12

Let $\Gamma$ be a connected graph of order $n$ and size $m$. Then

$\max\\{\gamma_{ve}(\Gamma),\beta_{e}(\Gamma)\\}\leq\gamma_{emd}(\Gamma)\leq\gamma_{ve}(\Gamma)+\beta_{e}(\Gamma).$

Proof:  Assume that the set $T\subseteq V(\Gamma)$ is vertex-edge dominating
set and the set $A\subseteq V(\Gamma)$ is an edge resolving set of a graph
$\Gamma$. If $T\cup A=T$ or $T\cup A=A$ then
$\gamma_{emd}(\Gamma)=\max\\{\gamma_{ve}(\Gamma),\beta_{e}(\Gamma)\\}$. If
$T\cap A\neq\phi$ and $T\subset T\cup A$ or $A\subset T\cup A$, then
$\gamma_{emd}(\Gamma)>\max\\{\gamma_{ve}(\Gamma),\beta_{e}(\Gamma)\\}$. Since,
the vertex-edge dominant edge metric dimension of $\Gamma$ is greater than its
vertex-edge domination number and its edge metric dimension, we obtain that
$\gamma_{emd}(\Gamma)\geq\max\\{\gamma_{ve}(\Gamma),\beta_{e}(\Gamma)\\}$. If
the vertex-edge dominating set and the edge resolving set of $\Gamma$ do not
intersect, i.e. $T\cap A=\phi$, then
$\gamma_{emd}(\Gamma)\leq\gamma_{ve}(\Gamma)+\beta_{e}(\Gamma)$. This
completes the proof.

Next theorem presents sharp upper and lower bounds on the
$\gamma_{emd}(\Gamma)$ of $\Gamma$ in terms of its order.

###### Theorem 14

Let $\Gamma$ be a graph of order $n$. Then, the vertex-edge dominant edge
metric dimension of $\Gamma$ satisfies

$\Big{\lfloor}\frac{n+2}{4}\Big{\rfloor}\leq\gamma_{emd}(\Gamma)\leq n-1.$

Proof:  Let $\Gamma$ be a graph of order $n$ and size $m$. Assume
$\Delta(\Gamma)$ is the maximum degree of $\Gamma$ and $\Gamma$ satisfies
$2\leq\Delta(\Gamma)\leq n-1$. Since $\Delta(\Gamma)=2$ implies that $\Gamma$
is either path $P_{n}$ or cycle $C_{n}$. We know that $\beta_{e}(P_{n})=1$ and
$\gamma_{ve}(P_{n})=\big{\lfloor}\frac{n+2}{4}\big{\rfloor}$. This implies
that, we obtain $\gamma_{emd}(P_{n})=\big{\lfloor}\frac{n+2}{4}\big{\rfloor}$.
Moreover, we know that $\beta_{e}(C_{n})=2$ and
$\gamma_{ve}(C_{n})=\big{\lfloor}\frac{n+3}{4}\big{\rfloor}$. We obtain
$\gamma_{emd}(C_{n})=\big{\lfloor}\frac{n+3}{4}\big{\rfloor}$. If maximum
degree of $\Gamma$ is $\Delta(\Gamma)>2$, then
$\gamma_{emd}(\Gamma)>\lfloor\frac{n+2}{4}\rfloor$. Moreover, if the maximum
degree of $\Gamma$ is $n-1$ i.e. $\Delta(\Gamma)=n-1$ then, either
$\beta_{e}(\Gamma)=n-1$ or $\beta_{e}(\Gamma)=n-2$ by Proposition 1 but
$\gamma_{ve}(\Gamma)=1$ because $\mathrm{diam}(\Gamma)\leq 2$. In this case,
the edge resolving set will dominate all edges and we obtained that
$\gamma_{emd}(\Gamma)\leq n-1$. Proved that

$\Big{\lfloor}\frac{n+2}{4}\Big{\rfloor}\leq\gamma_{emd}(\Gamma)\leq n-1.$

A vertex $v\in V(\Gamma)$ of an $n$-vertex graph $\Gamma$ is said to be
universal, if $d_{v}=n-1$. The following theorem computes the vertex-edge
dominant edge metric dimension of $\Gamma$, if it has at least one universal
vertex.

###### Theorem 15

Let $\Gamma$ be a graph of order $n$ and $\Delta(\Gamma)=n-1$. Then, the
vertex-edge dominant edge metric dimension $\gamma_{emd}(\Gamma)$ of $\Gamma$
is $\gamma_{emd}(\Gamma)=n-1~{}\mathrm{or}~{}n-2.$

Proof:  Let $\Gamma$ be a graph of order $n$ and size $m$. Suppose that,
$\Delta(\Gamma)$ is maximum degree of $\Gamma$ and $\Delta(\Gamma)=n-1$. By
Proposition 1, if there exists a vertex $v\in V(\Gamma)$ of degree $n-1$,
then, either $\beta_{e}(\Gamma)=n-1$ or $\beta_{e}(\Gamma)=n-2$. By Corollary
2, if $\beta_{e}(\Gamma)=n-1$, then $\mathrm{diam}(\Gamma)=2$ and every edge
belongs to a triangle. Assume that the $\beta_{e}(\Gamma)=n-2$, then for any
$v_{i}\neq v_{j}$ with $1\leq i<j\leq n$, there is a vertex $u\in V(\Gamma)$
such that $v_{i}u\in E(\Gamma)$ and $v_{j}u\in E(\Gamma)$, so
$\mathrm{diam}(\Gamma)=2$. If $\mathrm{diam}(\Gamma)\leq 2$, then Proposition
3 implies that the vertex-edge domination number is $1$ i.e.
$\gamma_{ve}(\Gamma)=1$. As the edge metric dimension of graph $\Gamma$ is
either $\beta_{e}(\Gamma)=n-1$ or $\beta_{e}(\Gamma)=n-2$ the vertex-edge
domination is $1$ i.e. $\gamma_{ve}(\Gamma)=1$. This implies that the edge
resolving set of $\Gamma$ will dominate and resolve all the edges of graph
$\Gamma$ and we obtain that the vertex-edge dominant edge dimension of
$\Gamma$ is $\gamma_{emd}(\Gamma)=n-1~{}\mathrm{or}~{}n-2.$ This shows the
result.

The next theorem refines Theorem 15.

###### Theorem 16

Let $\Gamma$ be an $n$-vertex graph such that $\Delta(\Gamma)=n-1$. Assume
that there exist at least two vertices with degree $n-1$. Then,
$\gamma_{emd}(\Gamma)=n-1$.

Proof:  Let $\Gamma$ be a graph of order $n$, size $m$, maximum degree
$\Delta(\Gamma)=n-1$. If there exist at least two vertices with degree $n-1$,
then, Lemma 1 implies that the $\beta_{e}(\Gamma)=n-1$. Moreover, by Theorem
5, we have $\beta_{e}(\Gamma)=n-1$ if and only if for any distinct
$v_{1},v_{2}\in V(\Gamma)$, there exists $u\in V(\Gamma)$ such that $v_{1}u\in
E(\Gamma),v_{2}u\in E(\Gamma)$ and $u$ is adjacent to all non-mutual neighbors
of $v_{1},v_{2}$. This property restricts $\Gamma$ to have
$\mathrm{diam}(\Gamma)\leq 2$. By Corollary 2 if $\beta_{e}(\Gamma)=n-1$, then
$\mathrm{diam}(\Gamma)=2$ and every edge lies on a cycle of length 3. If
$\mathrm{diam}(\Gamma)\leq 2$, then $\gamma_{ve}(\Gamma)=1$ by Proposition 3.
The edge metric dimension of graph $\Gamma$ is $\beta_{e}(\Gamma)=n-1$ and the
vertex-edge domination is $1$ i.e. $\gamma_{ve}(\Gamma)=1$. This implies that
the edge resolving set of cardinality $n-1$ will dominate and resolve all the
edges of $\Gamma$ and we obtain that the vertex-edge dominant edge dimension
of $\Gamma$ is $\gamma_{emd}(\Gamma)=n-1$. This shows the result.

Next, we show some result on the vertex-edge dominant edge metric dimension of
trees. The following proposition shows upper and lower bounds on the vertex-
edge domination number of trees.

###### Proposition 13

Let $T_{n}$ be a tree of order $n$, size $m$, maximum degree $\Delta(T_{n})$,
and minimum degree $\delta(T_{n})$. Then,

$\Big{\lceil}\frac{m}{4}\Big{\rceil}\leq\gamma_{ve}(T_{n})\leq
n-\Delta(T_{n}).$

Proof:  Note that $2\leq\Delta(T_{n})\leq n-1$ and $\delta(T_{n})=1$. As we
know that,
$\big{\lceil}\frac{m}{(\Delta(T_{n}))^{2}}\big{\rceil}\leq\gamma_{ve}(T_{n})$
by Proposition 4. It can easily be verified that
$\lceil\frac{m}{(\Delta(T_{n}))^{2}}\rceil\leq\gamma_{ve}(T_{n})$ holds for
every tree $T_{n}$. By Proposition 5, we know that

$\gamma_{ve}(T_{n})\leq
m-\Delta(T_{n})-\frac{\Delta(T_{n})(\delta(T_{n})-1)}{2}+1.$

Thus, we obtain $\gamma_{ve}(T_{n})\leq n-\Delta(T_{n})$ satisfied by $T_{n}$.
We obtain that

$\Big{\lceil}\frac{m}{4}\Big{\rceil}\leq\gamma_{ve}(T_{n})\leq
n-\Delta(T_{n}).$

This shows the proposition.

Based on Proposition 13, we present upper and lower bounds on the vertex-edge
dominant edge metric dimension of trees.

###### Theorem 17

Let $T_{n}$ be a tree with order $n\geq 2$ and size $m$. Then, the vertex-edge
dominant edge metric dimension of $T_{n}$ satisfies

$\Big{\lceil}\frac{m}{4}\Big{\rceil}\leq\gamma_{emd}(T_{n})\leq n-1.$

Proof:  Note that $T_{n}$ satisfies $2\leq\Delta(T_{n})\leq n-1$. If maximum
degree of tree $T_{n}$ is $\Delta(T_{n})=2$, then Proposition 4 implies that
$\big{\lceil}\frac{m}{4}\big{\rceil}\leq\gamma_{emd}(T_{n})$. If maximum
degree of $T_{n}$ is $\Delta(T_{n})=n-1$, then Theorem 15 implies that the
vertex-edge dominant edge metric dimension $\gamma_{emd}(T_{n})$ of $T_{n}$ is
$\gamma_{emd}(T_{n})=n-1$ or $\gamma_{emd}(T_{n})=n-2$. Thus, if the maximum
degree of tree $T_{n}$ satisfies $2<\Delta(T_{n})<n-1$, then, this implies
that the vertex-edge dominant edge metric dimension of $T_{n}$ satisfies
$\Big{\lceil}\frac{m}{4}\Big{\rceil}\leq\gamma_{emd}(T_{n})\leq n-1.$ This
shows the theorem.

###  5.1 The Corona Product of two graphs

Let $\Gamma$ and $\Omega$ be two graphs of order $p$ and $q$, respectively.
The corona product graph $\Gamma\odot\Omega$ is defined as the graph obtained
from $\Gamma$ and $\Omega$, by taking one copy of $\Gamma$ and $p$ copies of
$\Omega$ and joining by an edge the $i^{th}$-vertex of $\Gamma$ with every
vertex from the $i^{th}$-copy of $\Omega$. Given a vertex $u\in V(\Gamma)$,
the copy of $\Omega$ whose vertices are adjacent to $u$ is denoted by
$\Omega_{u}$.

###### Proposition 14

Let $\Gamma$ and $\Omega$ be two connected graphs and $|V(\Omega)|\geq 2$.
Then, $\gamma_{ve}(\Gamma\odot\Omega)=|V(\Gamma)|$.

Proof:  Let the graph $\Gamma\odot\Omega$ is corona product of two graphs
$\Gamma$ and $\Omega$ and $|V(\Omega)|\geq 2$. If the vertex $x_{i}\in
V(\Gamma\odot\Omega)$, then all edges incident to $N(x_{i})$ are dominated by
the vertex $x_{i}$. The subset $V(\Omega_{x_{i}})\subset N(x_{i})$ is the
vertex set of $i^{th}$-copy of $\Omega$ and $U=\\{y\in V(\Gamma):x_{i}y\in
E(\Gamma)\\}\subset N(x_{i})$. The set $N(x_{i})=V(\Omega_{x_{i}})\cup U$
shows that the edges of $i^{th}$-copy of $\Omega$ are dominated by the vertex
$x_{i}$. So, that the set $T=V(\Gamma)$ will dominate all edges of
$\Gamma\odot\Omega$ and it is clear that
$\gamma_{ve}(\Gamma\odot\Omega)\leq|V(\Gamma)|$. The set
$T=V(\Gamma)=\\{x_{1},x_{2},\ldots,x_{p}\\}$ is the minimum vertex-edge
dominant set for $\Gamma\odot\Omega$. If the vertex $x_{i}\not\in T$, then all
edges of $i^{th}$-copy of $\Omega$ are not dominated by any vertex of $T$, it
is clear that $\gamma_{ve}(\Gamma\odot\Omega)\geq|V(\Gamma)|$. This implies
that $\gamma_{ve}(\Gamma\odot\Omega)=|V(\Gamma)|$.

###### Theorem 18

Let $\Gamma$ and $\Omega$ be two connected graphs and $|V(\Omega)|\geq 2$.
Then, $\gamma_{emd}(\Gamma\odot\Omega)=\beta_{e}(\Gamma\odot\Omega)$.

Proof:  Note that the graph $\Gamma\odot\Omega$ is corona product of two
graphs $\Gamma$ with order $p$ and $\Omega$ with order $q$, where
$|V(\Omega)|\geq 2$ and the vertex set of $\Gamma\odot\Omega$ is defined as
$V(\Gamma\odot\Omega)=\\{x_{i}|1\leq i\leq p\\}\cup\\{y_{i}^{j}|1\leq i\leq
p,1\leq j\leq q\\}$. By Theorem 3 (resp. Proposition 14), we have that
$\beta_{e}(\Gamma\odot\Omega)=|V(\Gamma)|\cdot(|V(\Omega)|-1)$ (resp.
$\gamma_{ve}(\Gamma\odot\Omega)=|V(\Gamma)|$). The set $W=\\{x_{i}|1\leq i\leq
p\\}\cup\\{y_{i}^{j}|1\leq i\leq p,1\leq j\leq q-1\\}$ is an edge resolving
set for $\Gamma\odot\Omega$ and $T=\\{x_{i}|1\leq i\leq p\\}\subset W$ is
vertex-edge dominant set for $\Gamma\odot\Omega$ by Proposition 14. Since,
$|W|=|V(\Gamma)|\cdot(|V(\Omega)|-1)$ and the set $W$ is vertex-edge dominant
edge resolving set for $\Gamma\odot\Omega$. We obtain the desired result.

###  5.2 The Join of two graphs

Noted that the $\Gamma$ and $\Omega$ are two graphs, the join graph
$\Gamma\vee\Omega$ is obtained from $\Gamma$ and $\Omega$ by adding an edge
between every vertex of $\Gamma$ and every vertex of $\Omega$. It is clearly
observed that the join graph $\Gamma\vee\Omega$ is always a connected
independently of the connectivity of the graphs $\Gamma$ and $\Omega$.

###### Proposition 15

Let $\Gamma$ and $\Omega$ be two connected graphs. Then,
$\gamma_{ve}(\Gamma\vee\Omega)=1~{}or~{}2$.

Proof:  Let the graph $\Gamma\vee\Omega$ is join of two connected graphs
$\Gamma$ and $\Omega$. The vertex set and edge set of $\Gamma\vee\Omega$ are
$V(\Gamma\vee\Omega)=V(\Gamma)\cup V(\Omega)$ and
$E(\Gamma\vee\Omega)=E(\Gamma)\cup E(\Omega)\cup\\{xu|x\in V(\Gamma),u\in
V(\Omega)\\}$, respectively. If the vertex $x\in V(\Gamma\vee\Omega)$, then
all edges incident to $N(x)$ are dominated by the vertex $x$.

Case 1: Assume That $\Gamma\cong K_{n}$ or $\Omega\cong K_{n}$.

For $\Gamma\vee\Omega$, the vertex $x\in V(\Gamma)$ will dominate all edges of
$\Gamma\vee\Omega$ as $N(x)=V(\Gamma)\cup V(\Omega)$. This implies that the
singleton set $T=\\{x:x\in V(\Gamma)\\}$ is vertex-edge dominating set for
$\Gamma\vee\Omega$ where $\Gamma\cong K_{n}$. Therefore, we obtain
$\gamma_{ve}(\Gamma\vee\Omega)=1$.
For $\Gamma\vee\Omega$, the vertex $u\in V(\Omega)$ will dominate all edges of
$\Gamma\vee\Omega$ as $N(u)=V(\Gamma)\cup V(\Omega)$. This implies that the
singleton set $T=\\{u:u\in V(\Omega)\\}$ is vertex-edge dominating set for
$\Gamma\vee\Omega$ where $\Gamma\cong K_{n}$. Therefore, we obtain
$\gamma_{ve}(\Gamma\vee\Omega)=1$.

Case 2: Assume That $\Gamma,\Omega\not\cong K_{n}$.

The set $T=\\{x,u:x\in V(\Gamma),u\in V(\Omega)\\}$ provides the minimum
vertex-edge dominant set of $\Gamma\vee\Omega$. It is clear that the
$\gamma_{ve}(\Gamma\vee\Omega)\leq 2$. The vertex $x\in V(\Gamma)$ will
dominate all edges incident to $N(x)=V(\Omega)\cup\\{y\in V(\Gamma):xy\in
E(\Gamma)\\}$. Similarly, the vertex $u\in V(\Omega)$ will dominate all edges
incident to $N(u)=V(\Gamma)\cup\\{v\in V(\Omega):uv\in E(\Gamma)\\}$. If set
$T_{1}=\\{x\\}$ (resp. $T_{2}=\\{u\\}$), then all edges of graph $\Gamma$
(resp. $\Omega$) are not dominated by $T_{1}$ (resp. $T_{2}$), it is clear
that $\gamma_{ve}(\Gamma\vee\Omega)\geq 2$. This implies that
$\gamma_{ve}(\Gamma\vee\Omega)=2$. This completes the proof.

###### Theorem 19

Let $\Gamma$ and $\Omega$ be two connected graphs. Then,
$\gamma_{emd}(\Gamma\vee\Omega)=\beta_{e}(\Gamma\vee\Omega)$.

Proof:  Note that the graph $\Gamma\vee\Omega$ is join of two graphs $\Gamma$
with order $p$ and $\Omega$ with order $q$ the vertex set of
$\Gamma\vee\Omega$ is defined as $V(\Gamma\vee\Omega)=\\{x_{i}|1\leq i\leq
p\\}\cup\\{u_{j}|1\leq j\leq q\\}$. By Theorem 4 (resp. Proposition 15), we
have that $\beta_{e}(\Gamma\vee\Omega)=|V(\Gamma)|+|V(\Omega)|-1$ (resp.
$\gamma_{ve}(\Gamma\vee\Omega)=1~{}or~{}2$). The set $W=\\{x_{i}|1\leq i\leq
p\\}\cup\\{y_{j}|1\leq j\leq q-1\\}$ is an edge resolving set for
$\Gamma\vee\Omega$ and $T=\\{x,u\\}\subset W$ is vertex-edge dominant set for
$\Gamma\vee\Omega$ by Proposition 15. Since, $|W|=|V(\Gamma)|+|V(\Omega)|-1$
and the set $W$ is vertex-edge dominant edge resolving set for
$\Gamma\vee\Omega$. We obtain the desired result.

##  6 Comparison between the dominant metric dimension and the vertex-edge
dominant edge metric dimension

Kelenc et al. [19] showed the metric dimension and the edge metric dimension
of a graph $\Gamma$ are not comparable, since there exist graphs satisfying
one of either $\beta(\Gamma)<\beta_{e}(\Gamma)$ or
$\beta(\Gamma)=\beta_{e}(\Gamma)$ or $\beta(\Gamma)>\beta_{e}(\Gamma)$. In
response, Knor et al. [21] presented further infinite families of graphs
satisfying $\beta(\Gamma)>\beta_{e}(\Gamma)$. Lewis [22] in his PhD thesis
showed that, for any graph $\Gamma$, the inequality
$\gamma_{ve}(\Gamma)\leq\gamma(\Gamma)$ holds, in general.

In view of the comparison by Kelenc et al. [19], it is natural to consider a
similar comparison between the dominant metric dimension and the vertex-edge
dominant edge metric dimension of graphs. This section presents families of
graph showing that these aforementioned parameters are also not comparable, in
general. Table 2 presents standard families of graphs with their dominant
metric dimension$\gamma_{md}(\Gamma)$ and the vertex-edge dominant edge metric
dimension $\gamma_{emd}(\Gamma)$. Note that some standard families such the
path graph $P_{n}$, the cycle graph $C_{n}$, the complete graph $K_{n}$ and
the complete bipartite graph $K_{n,m}$ showcase the behavior
$\gamma_{md}(\Gamma)\geq\gamma_{emd}(\Gamma)$. For instance, the star graph
$S_{1,n}$ exhibits $\gamma_{md}(S_{1,n})>\gamma_{emd}(S_{1,n})$, whereas, the
wheel and fan graphs satisfy $\gamma_{md}(\Gamma)<\gamma_{emd}(\Gamma)$.
Moreover, some path graphs satisfy the
$\gamma_{md}(\Gamma)=\gamma_{emd}(\Gamma)$ instead.

Families of graphs | $\gamma_{md}(\Gamma)$ | $\gamma_{emd}(\Gamma)$ | Remark
---|---|---|---
$P_{2}$ | $\beta(P_{2})=1$ | $\beta_{e}(P_{2})=1$ | $\gamma_{md}(P_{2})=\gamma_{emd}(P_{2})$
$P_{3}$ | $\beta(P_{3})+1=2$ | $\beta_{e}(P_{3})=1$ | $\gamma_{md}(P_{3})>\gamma_{emd}(P_{3})$
$P_{4}$ | $\beta(P_{4})+1=2$ | $\beta_{e}(P_{4})+1=2$ | $\gamma_{md}(P_{4})=\gamma_{emd}(P_{4})$
$P_{5}$ |  $\gamma(P_{5})=\lceil\frac{n}{3}\rceil$=2 | $\beta_{e}(P_{5})+1=2$ | $\gamma_{md}(P_{5})=\gamma_{emd}(P_{5})$
$P_{n}:n\geq 6$ | $\gamma(P_{n})=\lceil\frac{n}{3}\rceil$ | $\gamma_{ve}(P_{n})=\lfloor\frac{n+2}{4}\rfloor$ | $\gamma_{md}(P_{n})\geq\gamma_{emd}(P_{n})$
$C_{n}:n=3,4,5$ | $\beta(C_{n})=2$ | $\beta_{e}(C_{n})=2$ | $\gamma_{md}(C_{n})=\gamma_{emd}(C_{n})$
$C_{n}:n=6,7$ | $\beta(C_{n})+1=3$ | $\beta_{e}(C_{n})=2$ | $\gamma_{md}(C_{n})>\gamma_{emd}(C_{n})$
$C_{n}:n=8$ | $\beta(C_{8})+1=3$ | $\beta_{e}(C_{8})+1=3$ | $\gamma_{md}(C_{8})=\gamma_{emd}(C_{8})$
$C_{n}:n\geq 9$ | $\gamma(C_{n})=\lceil\frac{n}{3}\rceil$ | $\gamma_{ve}(C_{n})=\lfloor\frac{n+3}{4}\rfloor$ | $\gamma_{md}(C_{n})\geq\gamma_{emd}(C_{n})$
$S_{1,n}:n\geq 3$ | $\beta(S_{1,n})+1$ | $\beta_{e}(S_{1,n})=n-1$ | $\gamma_{md}(S_{1,n})>\gamma_{emd}(S_{1,n})$
$K_{n}:n\geq 2$ | $\beta(K_{n})$ | $\beta_{e}(K_{n})$ | $\gamma_{md}(K_{n})=\gamma_{emd}(K_{n})$
$K_{n,m}:n,m\geq 2$ | $\beta(K_{n,m})$ | $\beta_{e}(K_{n,m})$ | $\gamma_{md}(K_{n,m})=\gamma_{emd}(K_{n,m})$
$W_{1,n};n\geq 6$ | $\beta(W_{1,n})$ | $\beta_{e}(W_{1,n})$ | $\gamma_{md}(W_{1,n})<\gamma_{emd}(W_{1,n})$
$n\equiv 0,2,4(\mod 5)$ |  |  |
$W_{1,n};n\geq 6$ | $\beta(W_{1,n})+1$ | $\beta_{e}(W_{1,n})$ | $\gamma_{md}(W_{1,n})<\gamma_{emd}(W_{1,n})$
$n\equiv 1,3(\mod 5)$ |  |  |
$F_{1,n};n\geq 6$ | $\beta(F_{1,n})$ | $\beta_{e}(F_{1,n})$ | $\gamma_{md}(F_{1,n})<\gamma_{emd}(F_{1,n})$
$n\equiv 0,2,4(\mod 5)$ |  |  |
$F_{1,n};n\geq 6$ | $\beta(F_{1,n})+1$ | $\beta_{e}(F_{1,n})$ | $\gamma_{md}(F_{1,n})<\gamma_{emd}(F_{1,n})$
$n\equiv 1,3(\mod 5)$ |  |  |
Table 2: Comparison between the dominant metric dimension and the vertex-edge
dominant edge metric dimension of some families of graphs.

Based on the data in Table 2, there is no comparison between the dominant
metric dimension and the vertex-edge dominant edge metric dimension, in
general.

Next, we focus on bipartite graphs for comparison between the dominant metric
dimension and the vertex-edge dominant edge metric dimension.

###  6.1 Comparison between $\gamma_{md}$ and $\gamma_{emd}$ for bipartite
graphs

Although, general graphs are not comparable for $\gamma_{md}$ and
$\gamma_{emd}$, we show that trees are comparable.

###### Theorem 20

Let $T_{n}$ be a tree. Then, we have
$\gamma_{md}(T_{n})\geq\gamma_{emd}(T_{n})$.

Proof: By Proposition 6, we have $\gamma_{ve}(T_{n})\leq\gamma(T_{n})$ if
$T_{n}$ not a path. Moreover, by Remark 1, we have
$\beta(T_{n})=\beta_{e}(T_{n})$. Let $S\subseteq V(T_{n})$ be a dominating set
and $B\subseteq V(T_{n})$ be a resolving set of $T_{n}$. Moreover, let
$T\subseteq V(T_{n})$ be a vertex-edge dominating set and $A\subseteq
V(T_{n})$ be an edge resolving set of $T_{n}$. We distinguish the following
cases:

Case 1: If $S\subset B$ and $T\subset A$, then
$\gamma_{md}(T_{n})=\gamma_{emd}(T_{n})$.

Case 2: If $B\subset S$ and $A\subset T$, then
$\gamma_{md}(T_{n})\geq\gamma_{emd}(T_{n})$.

Case 3: If $B\nsubseteq S$ and $A\nsubseteq T$, then
$\gamma_{md}(T_{n})\geq\gamma_{emd}(T_{n})$.

It is clear that the dominant metric dimension of $T_{n}$ is always greater or
equal to its vertex-edge dominant edge metric dimension, i.e.
$\gamma_{md}(T_{n})\geq\gamma_{emd}(T_{n})$.

By Theorem 20, trees are comparable with respect to $\gamma_{md}$ and
$\gamma_{emd}$. However, for non-tree bipartite graphs, we show that there is
no comparison between $\gamma_{md}$ and $\gamma_{emd}$.

First, we construct a non-tree bipartite graph $\Gamma$ satisfying
$\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$. The graph $\Gamma$ in Figure 2
having $V(\Gamma)=\\{a,c,e,f,h\\}\cup\\{b,d,g,i,j\\}$ and
$E(\Gamma)=\\{e_{1},e_{2},\ldots,e_{14}\\}$. We show that
$\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$ for $\Gamma$ in Figure 2.

Figure 2: The non-tree bipartite graph $\Gamma$ satisfying
$\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$.

Next, we show that $\gamma_{md}(\Gamma)=5$ and $\gamma_{emd}(\Gamma)=3$, thus,
satisfying $\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$. Note that the set
$B=\\{a,c,e,f\\}$ is a resolving set of $\Gamma$. Since, $\Gamma$ is not a
path graph, we have $\beta(\Gamma)\geq 2$. The representations of all vertices
of $\Gamma$ with respect to the set $B=\\{a,c,e,f\\}$ are given in the Table
3.

Vertex | $a$ | $b$ | $c$ | $d$ | $e$
---|---|---|---|---|---
${\small r(.,.)}$ | $(0,2,2,2)$ | $(1,1,3,3)$ | $(2,0,2,2)$ | $(1,1,1,1)$ | $(2,2,0,2)$
Vertex | $f$ | $g$ | $h$ | $i$ | $j$
${\small r(.,.)}$ | $(2,2,2,0)$ | $(3,1,1,1)$ | $(2,2,2,2)$ | $(3,3,3,3)$ | $(3,3,3,1)$
Table 3: The representations of all vertices of $\Gamma$ with respect to the
set $B=\\{a,c,e,f\\}$.

Table 3 shows the representations of all the vertices of $\Gamma$ with respect
to $B=\\{a,c,e,f\\}$ are unique. Thus, we deduce that $\beta(\Gamma)\leq 4$.
Suppose to the contrary that $\beta(\Gamma)=3$, and no two vertices have the
same representation. Let $B$ be a resolving set in $\Gamma$ having cardinality
three. Assume $V(\Gamma)=X\cup Y$ and let $X=\\{a,c,e,f,h\\}$ and
$Y=\\{b,d,g,i,j\\}$ We divide this into the following cases:

Case 1: Assume $B\subset X$ is a resolving set and $|B|=3$. If $c,e\not\in B$,
then $r(c|X\setminus\\{c,e\\})=r(e|X\setminus\\{c,e\\})$. If $e,f\not\in B$,
then $r(e|X\setminus\\{e,f\\})=r(f|X\setminus\\{e,f\\})$. It is clear that
$c,e,f\in B$ but $B=\\{c,e,f\\}$ is not a resolving set for $\Gamma$ as
$r(a|B)=r(h|B)$, so, any pair of vertices $c,e$ and $f$ is compulsory for
resoling set. If $B_{1}=\\{a,c,f\\}$, then $r(e|B_{1})=r(h|B_{1})$, which is a
contradiction. If $B_{2}=\\{c,f,h\\}$, then $r(a|B_{2})=r(e|B_{2})$, which is
a contradiction. If $B_{3}=\\{a,e,f\\}$, then $r(c|B_{3})=r(h|B_{3}$, which is
a contradiction. If $B_{4}=\\{e,f,h\\}$, then $r(a|B_{4})=r(c|B_{4})$, which
is a contradiction. If $B_{5}=\\{a,c,e\\}$, then $r(h|B_{5})=r(f|B_{5})$,
which is a contradiction. If $B_{6}=\\{c,e,h\\}$, then
$r(a|B_{6})=r(f|B_{6})$, which is a contradiction. This shows that no
resolving set $B\subset X$, of cardinality three exists in $\Gamma$.

Case 2: Assume $B\subset Y$ is a resolving set and $|B|=3$. If $b\not\in B$,
then $r(c|Y\setminus\\{b\\})=r(e|Y\setminus\\{b\\})$. If $j\not\in B$, then
$r(e|Y\setminus\\{j\\})=r(f|Y\setminus\\{j\\})$. It is clear that vertex $b$
and vertex $j$ are compulsory for resoling set. If $B_{1}=\\{b,d,j\\}$, then
$r(a|B_{1})=r(c|B_{1})$, which is a contradiction. If $B_{2}=\\{b,g,j\\}$,
then $r(a|B_{2})=r(c|B_{2})$ and $r(d|B_{2})=r(i|B_{2})$, which is a
contradiction. If $B_{3}=\\{b,d,i\\}$, then $r(a|B_{3})=r(c|B_{3})$, which is
a contradiction. This shows that no resolving set $B\subset Y$, of cardinality
three exists in $\Gamma$.

Case 3: Assume $B\subset X\cup Y$ is a resolving set and $|B|=3$. From the
vertices $b,c$ and $e$, a vertex is compulsory for resolving set because
$r(c|V(\Gamma)\setminus\\{b,c,e\\})=r(e|V(\Gamma)\setminus\\{b,c,e\\})$.
Similarly, From the vertices $e,f$ and $j$, a vertex is compulsory for
resolving set because
$r(e|V(\Gamma)\setminus\\{e,f,j\\})=r(f|V(\Gamma)\setminus\\{e,f,j\\})$. If
$B_{1}=\\{b,f,x_{1}\\}$, and $x_{1}\in\\{d,e,g,h,i,j\\}$ then
$r(a|B_{1})=r(c|B_{1})$, which is a contradiction. If $B_{2}=\\{b,f,x_{2}\\}$,
and $x_{2}\in\\{a,c,h,i\\}$ then $r(g|B_{2})=r(j|B_{2})$, which is a
contradiction. If $B_{3}=\\{b,j,x_{3}\\}$, and $x_{3}\in\\{d,e,f,g,h,i\\}$
then $r(a|B_{3})=r(c|B_{3})$, which is a contradiction. If
$B_{4}=\\{b,f,x_{4}\\}$, and $x_{4}\in\\{a,c,h\\}$ then
$r(g|B_{4})=r(i|B_{4})$, which is a contradiction. If $B_{5}=\\{e,f,x_{5}\\}$,
and $x_{5}\in\\{b,d,h,i,j\\}$ then $r(a|B_{5})=r(c|B_{5})$, which is a
contradiction. If $B_{6}=\\{e,f,x_{6}\\}$, and $x_{6}\in\\{a,g\\}$ then
$r(c|B_{6})=r(h|B_{6})$, which is a contradiction. If $B_{7}=\\{c,e,f\\}$,
then $r(a|B_{7})=r(h|B_{7})$, which is a contradiction. If
$B_{8}=\\{c,f,x_{8}\\}$, and $x_{8}\in\\{d,g,h,i,j\\}$ then
$r(a|B_{8})=r(e|B_{8})$, which is a contradiction. If $B_{9}=\\{c,f,x_{9}\\}$,
and $x_{9}\in\\{b,e,j\\}$ then $r(d|B_{9})=r(g|B_{9})$, which is a
contradiction. This shows that no resolving set $B\subset X\cup Y$, of
cardinality three exists in $\Gamma$.

This shows that no resolving set of cardinality three exists in $\Gamma$.
Thus, we have $\beta(\Gamma)\geq 4$ and therefore, $\beta(\Gamma)=4$.
Moreover, note that the set $S=\\{h,d\\}$ is dominating set, as it dominate
all the vertices, however, it does not resolve all the vertices of $\Gamma$.
The set $A=\\{a,c,e,f\\}$ is a resolving set as it resolve all the vertices,
however, it does not dominate all the vertices. So, the set
$D=\\{a,c,e,f,h\\}$ is dominant resolving set and we have
$\gamma_{md}(\Gamma)=5$.

Next we show that $\gamma_{emd}(\Gamma)=3$, for $\Gamma$ given in Figure 2.
Note that the set $A=\\{b,g,j\\}$ is an edge resolving set of $\Gamma$. The
representations of all edges of $\Gamma$ with respect to the set $A$ are given
in the Table 4

Edge | $e_{1}$ | $e_{2}$ | $e_{3}$ | $e_{4}$ | $e_{5}$ | $e_{6}$ | $e_{7}$
---|---|---|---|---|---|---|---
${\small r(.,.)}$ | $(0,2,2)$ | $(1,2,2)$ | $(0,1,2)$ | $(1,1,2)$ | $(2,1,2)$ | $(2,0,2)$ | $(2,1,1)$
Edge | $e_{8}$ | $e_{9}$ | $e_{10}$ | $e_{11}$ | $e_{12}$ | $e_{13}$ | $e_{14}$
${\small r(.,.)}$ | $(1,0,2)$ | $(0,1,1)$ | $(2,0,1)$ | $(1,0,1)$ | $(2,1,0)$ | $(1,1,0)$ | $(1,1,1)$
Table 4: The representations of all edges of $\Gamma$ with respect to the set
$A$.

Since all the representations in Table 4 are unique, we deduce that
$\beta_{e}(\Gamma)\leq 3$. Suppose on the contrary that $\beta_{e}(\Gamma)=2$.
Let $A$ be an edge resolving set of cardinality two in $\Gamma$ such that no
two edges have the same representation. We divide our discussion into a number
of cases:

Case 1: Assume $A=\\{e,x\\}$ and $x\in\\{a,d,g,h,i\\}$. Then, we have
$r(e_{4}|A)=r(e_{7}|A)$ and $r(e_{8}|A)=r(e_{10}|A)$, thus, arising a
contradiction. Assume $A=\\{e,y\\}$ and $y\in\\{b,c,f,j\\}$. Then, we have
$r(e_{4}|A)=r(e_{8}|A)$, $r(e_{5}|A)=r(e_{6}|A)$ and $r(e_{7}|A)=r(e_{10}|A)$,
thus, arising a contradiction.

Case 2: Assume $A=\\{x,y\\}$ and $x,y\in\\{g,h,i\\}$ or $A=\\{a,d\\}$. Then,
we have $r(e_{4}|A)=r(e_{5}|A)=r(e_{7}|A)$ and
$(e_{6}|A)=r(e_{8}|A)=r(e_{10}|A)$, thus, arising a contradiction. Assume
$A=\\{c,b\\}$. Then, we have $r(e_{4}|A)=r(e_{8}|A)$ and
$(e_{5}|A)=r(e_{6}|A)=r(e_{7}|A)=r(e_{10}|A)$, thus, arising a contradiction.
Assume $A=\\{f,j\\}$. Then, we have $r(e_{7}|A)=r(e_{10}|A)$ and
$(e_{4}|A)=r(e_{5}|A)=r(e_{6}|A)=r(e_{8}|A)$, thus, arising a contradiction.

Case 3: Assume $A_{1}=\\{x,y_{1}\\}$ and $x\in\\{g,h,i\\}$,
$y_{1}\in\\{a,d\\}$. Then, we have
$r(e_{4}|A_{1})=r(e_{5}|A_{1})=r(e_{7}|A_{1})$ and
$(e_{6}|A_{1})=r(e_{8}|A_{1})=r(e_{10}|A_{1})$, thus, arising a contradiction.
Assume $A_{2}=\\{x,y_{2}\\}$ and $x\in\\{g,h,i\\}$, $y_{2}\in\\{b,c\\}$. Then,
we have $r(e_{5}|A_{2})=r(e_{7}|A_{2})$ and $(e_{6}|A_{2})=r(e_{10}|A_{2})$,
thus, arising a contradiction. Assume $A_{3}=\\{x,y_{3}\\}$ and
$x\in\\{g,h,i\\}$, $y_{3}\in\\{f,j\\}$. Then, we have
$r(e_{4}|A_{3})=r(e_{5}|A_{3})$ and $(e_{6}|A_{3})=r(e_{8}|A_{3})$, thus,
arising a contradiction.

Case 4: Assume $A_{1}=\\{x_{1},y_{1}\\}$ and $x_{1}\in\\{a,d\\}$,
$y_{1}\in\\{b,c\\}$. Then, we have $r(e_{5}|A_{1})=r(e_{7}|A_{1})$ and
$(e_{6}|A_{1})=r(e_{10}|A_{1})$, thus, arising a contradiction. Assume
$A_{2}=\\{x_{2},y_{2}\\}$ and $x_{2}\in\\{a,d\\}$, $y_{2}\in\\{f,j\\}$. Then,
we have $r(e_{4}|A_{2})=r(e_{5}|A_{2})$ and $(e_{6}|A_{2})=r(e_{8}|A_{2})$,
thus, arising a contradiction. Assume $A_{3}=\\{x_{3},y_{3}\\}$ and
$x_{3}\in\\{b,c\\}$, $y_{3}\in\\{f,j\\}$. Then, we have
$r(e_{4}|A_{3})=r(e_{8}|A_{3})$, $r(e_{5}|A_{3})=r(e_{6}|A_{3})$ and
$(e_{7}|A_{3})=r(e_{10}|A_{3})$, thus, arising a contradiction.

Thus, we have $\beta_{e}(\Gamma)\geq 3$, which altogether leads to
$\beta_{e}(\Gamma)=3$. The set $T=\\{h,d\\}$ is a vertex-edge dominating set,
it dominate all the edges but it does not resolve all the edges of $\Gamma$.
The set $A=\\{b,g,j\\}$ is an edge resolving set, as it resolves all the edges
as well as it dominate all the edges. So, the set $A=\\{b,g,j\\}$ is a vertex-
edge dominant edge resolving set and we obtain $\gamma_{emd}(\Gamma)=3$.

###### Remark 2

The non-tree bipartite graph $\Gamma$ in Figure 2 satisfies
$\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$. More precisely, we have
$\gamma_{md}(\Gamma)=5$ and $\gamma_{emd}(\Gamma)=3$.

Next, we construct a non-tree bipartite graph $\Omega$ satisfying
$\gamma_{md}(\Omega)<\gamma_{emd}(\Omega)$. Consider the graph $\Omega$ in
Figure 3 having
$V(\Omega)=\\{a_{1},a_{2},\ldots,a_{8}\\}\cup\\{b_{0},b_{1},\ldots,b_{8}\\}$
and $E(\Omega)=\\{a_{l}b_{0},a_{l}b_{l},a_{l+1}b_{l};1\leq l\leq 8\\}$.

Figure 3: The non-tree bipartite graph $\Omega$ satisfying
$\gamma_{md}(\Omega)<\gamma_{emd}(\Omega)$.

Next, we show that $\gamma_{md}(\Omega)=6$ and $\gamma_{emd}(\Omega)=7$, thus,
satisfying $\gamma_{md}(\Omega)<\gamma_{emd}(\Omega)$. Note that the set
$B=\\{a_{4},b_{1},b_{2},b_{5},b_{6}\\}$ is a resolving set of $\Omega$. Since,
$\Omega$ is not a path graph, we have $\beta(\Omega)\geq 2$. The
representations of all vertices of $\Omega$ with respect to the set
$B=\\{a_{4},b_{1},b_{2},b_{5},b_{6}\\}$ are given in the Table 5.

Vertices | $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$
---|---|---|---|---|---|---
${\small r(.,.)}$ | $(1,3,2,3,3)$ | $(1,1,2,3,3)$ | $(3,1,2,3,3)$ | $(3,3,0,3,3)$ | $(3,3,2,1,3)$ | $(3,3,2,1,1)$
Vertices | $a_{7}$ | $a_{8}$ | $b_{0}$ | $b_{1}$ | $b_{2}$ | $b_{3}$
${\small r(.,.)}$ | $(3,3,2,3,1)$ | $(3,3,2,3,3)$ | $(2,2,1,2,2)$ | $(0,2,3,4,4)$ | $(2,0,3,4,4)$ | $(4,2,1,4,4)$
Vertices | $b_{4}$ | $b_{5}$ | $b_{6}$ | $b_{7}$ | $b_{8}$ |
${\small r(.,.)}$ | $(4,4,1,2,4)$ | $(4,4,3,0,2)$ | $(4,4,3,2,0)$ | $(4,4,3,4,2)$ | $(2,4,3,4,4)$ |
Table 5: The representations of all vertices of $\Omega$ with respect to the
set $B$.

Table 5 shows the representations of all the vertices of $\Omega$ with respect
to $B=\\{a_{4},b_{1},b_{2},b_{5},b_{6}\\}$ are unique. Thus, we deduce that
$\beta(\Omega)\leq 5$. Moreover, the set
$B=\\{a_{4},b_{1},b_{2},b_{5},b_{6}\\}$ is not a dominating set of $\Omega$
because it does not dominate the vertices $a_{8},b_{7},b_{8}$, so,
$\gamma(\Omega)=6$. Furthermore, the set
$S=\\{a_{4},a_{8},b_{1},b_{2},b_{5},b_{6}\\}$ is a dominating set, as it
dominates all the vertices and it resolves all the vertices of $\Omega$. So,
the set $D=S=\\{a_{4},a_{8},b_{1},b_{2},b_{5},b_{6}\\}$ is a dominant
resolving set and, thus, we have $\gamma_{md}(\Omega)=6$.

Next, we consider the set $A=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}\\}$
which is an edge resolving set of $\Omega$. The representation of all the
edges of $\Omega$ with respect to the set
$A=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}\\}$ are given in the Table 6.

Edges | $a_{1}b_{0}$ | $a_{2}b_{0}$ | $a_{3}b_{0}$ | $a_{4}b_{0}$ | $a_{5}b_{0}$
---|---|---|---|---|---
${\small r(.,.)}$ | $(0,1,1,1,1,1,1)$ | $(1,0,1,1,1,1,1)$ | $(1,1,0,1,1,1,1)$ | $(1,1,1,0,1,1,1)$ | $(1,1,1,1,0,1,1)$
Edges | $a_{6}b_{0}$ | $a_{7}b_{0}$ | $a_{8}b_{0}$ | $a_{1}b_{1}$ | $a_{2}b_{2}$
${\small r(.,.)}$ | $(1,1,1,1,1,0,1)$ | $(1,1,1,1,1,1,0)$ | $(1,1,1,1,1,1,1)$ | $(0,1,2,2,2,2,2)$ | $(2,0,1,2,2,2,2)$
Edges | $a_{3}b_{3}$ | $a_{4}b_{4}$ | $a_{5}b_{5}$ | $a_{6}b_{6}$ | $a_{7}b_{7}$
${\small r(.,.)}$ | $(2,2,0,1,2,2,2)$ | $(2,2,2,0,1,2,2)$ | $(2,2,2,2,0,1,2)$ | $(2,2,2,2,2,0,1)$ | $(2,2,2,2,2,2,0)$
Edges | $a_{8}b_{8}$ | $b_{1}a_{2}$ | $b_{2}a_{3}$ | $b_{3}a_{4}$ | $b_{4}a_{5}$
${\small r(.,.)}$ | $(1,2,2,2,2,2,2)$ | $(1,0,2,2,2,2,2)$ | $(2,1,0,2,2,2,2)$ | $(2,2,1,0,2,2,2)$ | $(2,2,2,1,0,2,2)$
Edges | $b_{5}a_{6}$ | $b_{6}a_{7}$ | $b_{7}a_{8}$ | $b_{8}a_{1}$ |
${\small r(.,.)}$ | $(2,2,2,2,1,0,2)$ | $(2,2,2,2,2,1,0)$ | $(2,2,2,2,2,2,1)$ | $(0,2,2,2,2,2,2)$ |
Table 6: The representations of all edges of $\Omega$ with respect to the set
$A$.

Table 6 shows the representations of all the edges of $\Omega$ with respect to
$A$ are unique. We deduce that $\beta_{e}(\Omega)\leq 7$. Suppose that
$\beta_{e}(\Omega)=6$, such that no two edges have the same representation.
Let $A$ be an edge resolving set of cardinality 6 in $\Omega$. We discuss the
following case:

Case 1: Assume that
$A=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}\\}\setminus\\{a_{k}:k=1,2,\ldots,7\\}$.
Then, $r(a_{k}b_{0}|A)=r(a_{8}b_{0}|A)=(1,1,1,1,1,1)$ where $k=1,2,\ldots,7$,
which contradicts to our assumption.

This shows that $\beta_{e}(\Omega)\geq 7$, as $A$ can not be an edge resolving
set. This altogether leads to $\beta_{e}(\Omega)=7$.

Moreover, the set $T=\\{b_{0}\\}$ is a vertex-edge dominating set, as it
dominates all the edges but it does not resolve all the edges of $\Omega$. The
set $A=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}\\}$ is an edge resolving
set, it resolves all the edges and it dominates all the edges as well. So, the
set $A=\\{a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}\\}$ is a vertex-edge
dominant edge resolving set, and, we have $\gamma_{emd}(\Omega)=7$. Thus, we
have the following remark.

###### Remark 3

The non-tree bipartite graph $\Omega$ in Figure 3 satisfies
$\gamma_{md}(\Omega)<\gamma_{emd}(\Omega)$. More precisely, we have
$\gamma_{md}(\Omega)=6$ and $\gamma_{emd}(\Omega)=7$.

Finally, we construct a non-tree bipartite graph $\Pi$ satisfying
$\gamma_{md}(\Pi)=\gamma_{emd}(\Pi)$. Consider the graph $\Pi$ in Figure 4
having $V(\Pi)=\\{a_{1},a_{2},a_{3},a_{4}\\}\cup\\{b_{1},b_{2}\\}$ and
$E(\Pi)=\\{a_{i}b_{1},a_{i}b_{2}:1\leq i\leq 4\\}$.

Figure 4: The non-tree bipartite graph $\Pi$ satisfying
$\gamma_{md}(\Pi)=\gamma_{emd}(\Pi)$.

Next, we show that $\gamma_{md}(\Pi)=\gamma_{emd}(\Pi)=4$, thus, satisfying
$\gamma_{md}(\Pi)=\gamma_{emd}(\Pi)$. Note that the set
$B=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is a resolving set of $\Pi$. Since, $\Pi$ is
not a path graph, we have $\beta(\Pi)\geq 2$. The representations of all
vertices of $\Pi$ with respect to the set $B=\\{a_{1},a_{2},a_{3},b_{1}\\}$
are given in the Table 7.

Vertices | $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $b_{1}$ | $b_{2}$
---|---|---|---|---|---|---
${\small r(.,.)}$ | $(0,2,2,1)$ | $(2,0,2,1)$ | $(2,2,0,1)$ | $(2,2,2,1)$ | $(1,1,1,0)$ | $(1,1,1,2)$
Table 7: The representations of all vertices of $\Pi$ with respect to the set
$B=\\{a_{1},a_{2},a_{3},b_{1}\\}$.

Table 7 shows the representations of all the vertices of $\Pi$ with respect to
$B$ are unique. Thus, we deduce that $\beta(\Pi)\leq 4$. Suppose that
$\beta(\Pi)=3$, and let $B$ be a resolving set of cardinality 3. We divide our
discussion into a number of cases:

Case 1: Assume that $B=\\{a_{1},a_{2},a_{3}\\}$. Then,
$r(b_{1}|B)=r(b_{2}|B)=(1,1,1)$, which is contradicts to our assumption.

Case 2: Assume that $B=\\{a_{1},a_{2},a_{3},b_{1}\\}\setminus\\{a_{p}\\}$
where $p=1,2,3$. Then, we have $r(a_{p}|B)=r(a_{4}|B)=(2,2,1)$, which is
contradicts to our assumption.

Since $B$ could not be a resolving set, we obtain that $\beta(\Pi)\geq 4$,
which then leads to $\beta(\Pi)=4$. The set $S=\\{a_{1},b_{1}\\}$ is
dominating set, as it dominates all the vertices but it does not resolve all
the vertices of $\Pi$. The set $B=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is a
resolving set, as it resolve all the vertices as well as it dominates all the
vertices. So, the set $B=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is a dominant
resolving set and we obtain that $\gamma_{md}(\Pi)=4$.

Next, we show that $\gamma_{emd}(\Pi)=4$. Note that the set
$A=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is an edge resolving set of $\Pi$. The
representations of all edges of $\Pi$ with respect to the set
$A=\\{a_{1},a_{2},a_{3},b_{1}\\}$ are given in the Table 8.

Edges | $a_{1}b_{1}$ | $a_{2}b_{1}$ | $a_{3}b_{1}$ | $a_{4}b_{1}$ | $a_{1}b_{2}$ | $a_{2}b_{2}$ | $a_{3}b_{2}$ | $a_{4}b_{2}$
---|---|---|---|---|---|---|---|---
${\small r(.,.)}$ | $(0,1,1,0)$ | $(1,0,1,0)$ | $(1,1,0,0)$ | $(1,1,1,0)$ | $(0,1,1,1)$ | $(1,0,1,1)$ | $(1,1,0,1)$ | $(1,1,1,1)$
Table 8: The representations of all edges of $\Pi$ with respect to the set
$A=\\{a_{1},a_{2},a_{3},b_{1}\\}$.

Table 8 shows the representations of all the edges of $\Pi$ with respect to
$A$ are unique. Since no two edges have the same representation with respect
to $A$, we deduce that $\beta_{e}(\Pi)\leq 4$. Suppose that $\beta_{e}(\Pi)=3$
and let $A$ be the corresponding edge resolving set in $\Pi$. We divide our
discussion into the following two cases:

Case 1: Assume that $A=\\{a_{1},a_{2},a_{3}\\}$. Then we have
$r(a_{1}b_{1}|A)=r(a_{1}b_{2}|A)=(0,1,1)$,
$r(a_{2}b_{1}|A)=r(a_{2}b_{2}|A)=(1,0,1)$,
$r(a_{3}b_{1}|A)=r(a_{3}b_{2}|A)=(1,1,0)$ and
$r(a_{4}b_{1}|A)=r(a_{4}b_{2}|A)=(1,1,1)$, which clearly arise a
contradiction.

Case 2: Assume that $A=\\{a_{1},a_{2},a_{3},b_{1}\\}\setminus\\{a_{p}\\}$
where $p=1,2,3$. Then, $r(a_{p}b_{2}|A)=r(a_{4}b_{2}|A)=(1,1,1)$, which is a
contradiction to our assumption.

Since $A$ could not be an edge resolving set, we obtain that
$\beta_{e}(\Pi)\geq 4$, which then leads to $\beta_{e}(\Pi)=4$. The set
$T=\\{b_{1}\\}$ is a vertex-edge dominating set, as it dominates all the edges
but it does not resolve all the edges of $\Pi$. The set
$A=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is an edge resolving set, as it resolves all
the edges as well as it dominates all the edges. So, the set
$A=\\{a_{1},a_{2},a_{3},b_{1}\\}$ is vertex-edge dominant edge resolving set
and we obtain that $\gamma_{emd}(\Pi)=4$. Thus, we have the following remark:

###### Remark 4

The non-tree bipartite graph $\Pi$ in Figure 4 satisfies
$\gamma_{md}(\Pi)=\gamma_{emd}(\Pi)=4$.

Based on Remarks 2, 3 and 4, we conclude that non-tree bipartite graphs are
not comparable with respect to the dominant metric dimension and the vertex-
edge dominant edge metric dimension of graphs.

##  7 Concluding remarks and further open problems

This paper introduces a new parameter related to vertex-edge domination and
distances in a graph. The vertex-edge dominant edge metric dimension
$\gamma_{emd}(\Gamma)$, in fact, combines the theory of vertex-edge domination
and edge resolvability of graphs. The parameter $\gamma_{emd}(\Gamma)$ has
been calculated for classical families such as paths, cycles, complete graphs,
complete bipartite graphs, wheel graphs and fan graphs. Some general results
and bounds are provided. The most significant contribution is that the vertex-
edge dominant edge metric dimension $\gamma_{emd}(\Gamma)$ is not comparable,
in general, with its vertex-version counterpart i.e. the dominant metric
dimension $\gamma_{md}(\Gamma)$. In fact, for a graph $\Gamma$, all three
cases i.e. $\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$,
$\gamma_{md}(\Gamma)<\gamma_{emd}(\Gamma)$, or
$\gamma_{md}(\Gamma)=\gamma_{emd}(\Gamma)$ can occur. Upon considering the
class of bipartite graphs, we show that $\gamma_{emd}(T_{n})$ of a tree
$T_{n}$ is always less than or equal to its dominant metric dimension.
However, we show that for non-tree bipartite graphs, the parameter is not
comparable just like general graphs.

Based on the results in this paper, we would like to propose the following
open questions which arise naturally.

###### Problem 1

Characterize graphs $\Gamma$ satisfying one of the following three cases:

  * _(i)_

$\gamma_{md}(\Gamma)>\gamma_{emd}(\Gamma)$,

  * _(ii)_

$\gamma_{md}(\Gamma)=\gamma_{emd}(\Gamma)$,

  * _(iii)_

$\gamma_{md}(\Gamma)<\gamma_{emd}(\Gamma)$.

Theorem 20 shows that trees are comparable with respect to their vertex-edge
dominant edge metric dimension. However, it motivates the following problem.

###### Problem 2

Let $T_{n}$ be a tree on $n\geq 2$ vertices. Characterize all trees $T_{n}$
with $\gamma_{md}(T_{n})=\gamma_{emd}(T_{n})$.

Although, the dominant metric dimension and the vertex-edge dominant edge
metric dimension are not comparable, in general, there might exist a
relationship between these parameters for special classes of graph such trees,
unicyclic or bipartite graphs etc.

###### Problem 3

Is there any relationship between the dominant metric dimension
$\gamma_{md}(\Gamma)$ and $ve-$dominant edge metric dimension
$\gamma_{emd}(\Gamma)$ for special classes of graph, unicyclic or bipartite
graphs etc.?

###### Problem 4

Let $\Gamma$ be a graph with order $n$ and size $m$. Let $\delta(\Gamma)$
(resp. $\Delta(\Gamma)$) be the minimum degree (resp. maximum degree) of
$\Gamma$. Derive upper/lower bounds for $\gamma_{emd}(\Gamma)$ in terms of
$n,m,\delta(\Gamma)$ or $\Delta(\Gamma)$?

##  8 Declarations

###  8.1 Funding

Sakander Hayat is supported by the Higher Education Commission, Pakistan under
grant number 20-11682/NRPU/RGM/R&D/HEC/2020.

###  8.2 Competing Interests

There are no competing interests.

###  8.3 Author Contributions

All authors contributed equally to this paper.

###  8.4 Data Availability

There is no data associated with this paper.

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