Metric Dimensions of Bicyclic Graphs

Abstract

The distance d(va,vb) between two vertices of a simple connected graph G is the length of the shortest path between va and vb. Vertices va,vb of G are considered to be resolved by a vertex v if d(va,v)≠d(vb,v). An ordered set W={v1,v2,v3,…,vs}⊆V(G) is said to be a resolving set for G, if for any va,vb∈V(G),∃vi∈W∋d(va,vi)≠d(vb,vi). The representation of vertex v with respect to W is denoted by r(v|W) and is an s-vector(s-tuple) (d(v,v1),d(v,v2),d(v,v3),…,d(v,vs)). Using representation r(v|W), we can say that W is a resolving set if, for any two vertices va,vb∈V(G), we have r(va|W)≠r(vb|W). A minimal resolving set is termed a metric basis for G. The cardinality of the metric basis set is called the metric dimension of G, represented by dim(G). In this article, we study the metric dimension of two types of bicyclic graphs. The obtained results prove that they have constant metric dimension.

Full text

Citation: Khan, A.; Haidar, G.; Abbas,
N.; Khan, M.U.I.; Niazi, A.U.K.; Khan,
A.U.I. Metric Dimensions of Bicyclic
Graphs. Mathematics 2023,11, 869.
https://doi.org/10.3390/math11040869
Academic Editor: Rasul Kochkarov
Received: 24 November 2022
Revised: 2 February 2023
Accepted: 6 February 2023
Published: 8 February 2023
Copyright: © 2023 by the authors.
Licensee MDPI, Basel, Switzerland.
This article is an open access article
distributed under the terms and
conditions of the Creative Commons
Attribution (CC BY) license (https://
creativecommons.org/licenses/by/
4.0/).
mathematics
Article
Metric Dimensions of Bicyclic Graphs
Asad Khan 1,†, Ghulam Haidar 2,†, Naeem Abbas 2,† , Murad Ul Islam Khan 2, *,† , Azmat Ullah Khan Niazi 3
and Asad Ul Islam Khan 4
1School of Computer Science and Cyber Engineering, Guangzhou University, Guangzhou 510006, China
2Department of Mathematics and Statistics, The University of Haripur, Haripur 22620, Pakistan
3Department of Mathematics and Statistics, The University of Lahore, Sargodha 40100, Pakistan
4Economics Department, Ibn Haldun University, Istanbul 34480, Turkey
*Correspondence: muradulislam@uoh.edu.pk
† These authors contributed equally to this work.
Abstract:
The distance
d(va
,
vb)
between two vertices of a simple connected graph
G
is the length of
the shortest path between
va
and
vb
. Vertices
va
,
vb
of
G
are considered to be resolved by a vertex
v
if
d(va
,
v)6=d(vb
,
v)
. An ordered set
W={v1
,
v2
,
v3
,
. . .
,
vs} ⊆ V(G)
is said to be a resolving set for
G
,
if for any
va
,
vb∈V(G)
,
∃vi∈W3d(va
,
vi)6=d(vb
,
vi)
. The representation of vertex
v
with respect
to
W
is denoted by
r(v|W)
and is an
s
-vector(
s
-tuple)
(d(v
,
v1)
,
d(v
,
v2)
,
d(v
,
v3)
,
. . .
,
d(v
,
vs))
. Using
representation
r(v|W)
, we can say that
W
is a resolving set if, for any two vertices
va
,
vb∈V(G)
, we
have
r(va|W)6=r(vb|W)
. A minimal resolving set is termed a metric basis for
G
. The cardinality of
the metric basis set is called the metric dimension of
G
, represented by
dim(G)
. In this article, we
study the metric dimension of two types of bicyclic graphs. The obtained results prove that they have
constant metric dimension.
Keywords: graph theory; bicyclic graph; metric basis; resolving set; metric dimensions
MSC: 05C12; 05C90
1. Introduction
While studying the problem of finding out the location of an intruder in a network,
Slater in [
1
], and later in [
2
], introduced the term “locating set”. He termed the minimum
resolving set a “reference set” and referred to the cardinality of a minimum resolving set
(reference set) as the “location number”. Harary and Melter [
3
] also independently discussed
these concepts. They used the nomenclature “metric dimension” instead of location number.
In this article, we use the terminology developed by Harary and Melter. Hence,
the metric dimension,
dim(G)
, is the cardinality of the minimum resolving set. Fol-
lowing the convention in [
4
], we call the minimum resolving set the basis for
G
. Let
W={w1
,
w2
,
w3
,
· · ·
,
ws} ⊂ V(G)
be a basis for a simple graph
G
, then the
s
-tuple
(d(u
,
w1)
,
d(u
,
w2)
,
d(u
,
w3)
,
. . .
,
d(u
,
ws))
is termed a distance vector of
u
corresponding
to/with respect to
W
and is denoted by
r(u|W)
. It is worthwhile to mention, distinct
vertices have a distinct representation with respect to the basis vertices W.
This graph invariant has garnered a lot of attention from researchers. Chartrand
et al. [
4
] characterized graphs with metric dimensions 1,
n−
1, and
n−
2. Klein and Yi [
5
]
compared the metric dimensions of a graph and its line graph. Shao et al. [
6
] calculated the
metric dimensions of generalized Peterson graphs of type
(
2
k
,
k)
and
(
3
k
,
k)
and showed
that they have constant metric dimensions. Applications of metric dimensions to vari-
ous fields, e.g., navigation of robots [
7
], chemistry [
8
,
9
], coin-weighing, and mastermind
game [
10
] have been presented in the literature. Further studies on metric dimension and
metric basis were conducted in [11–20].
Many other variants of metric dimension have been defined to further study the
structure of a graph. Okamoto et al. [
21
] defined local metric dimensions and characterized
Mathematics 2023,11, 869. https://doi.org/10.3390/math11040869 https://www.mdpi.com/journal/mathematics

Mathematics 2023,11, 869 2 of 17
all nontrivial connected graphs of order
n
having local metric dimension 1,
n−
2, or
n−
1.
Kelenc et al. [
22
] defined mixed metric dimensions and showed that a graph is a path graph
if and only if its mixed metric dimension is 2. They also characterized complete graphs in
terms of mixed metric dimensions. Sedlar and Skrekovski [
23
] determined that for every
Theta graph
G
, the mixed metric dimension equals 3 or 4, with 4 being attained if and only
if
G
is a balanced Theta graph. Moreno et al. [
24
] defined
k
-metric dimensions and proved
that a graph
G
is
n
-metric dimensional if and only if
G'K2
. They also characterized
(n−1)-metric dimensional graphs.
Khuller et al. [
7
] showed that a graph
G
with metric dimension 2 can not have
K5
or
K3,3
as a subgraph. They also showed that there exist non-planar graphs with metric
dimension 2. In light of this information, characterizing all graphs with metric dimension 2
is a daunting task. On the other hand, if we only consider the problem of characterizing
planar graphs of metric dimension 2, the problem becomes more manageable.
It is a well known result that
dim(Cn) =
2. Further studies on metric dimension of
unicyclic graphs were conducted in [
25
,
26
]. Armed with the knowledge from these articles,
we can easily determine unicyclic graphs
G
, for which,
dim(G) =
2. We can also easily
deduce that, if a planar graph
G
contains a cycle as a subgraph, then
dim(G)≥
2. This raises
a question about metric dimensions of graphs having two or more cycles. In this article,
we will discuss the metric dimension of bicyclic graphs. Using these bicyclic graphs as
our building blocks, we can then move on to tricyclic and
n
-cyclic graphs and consider the
same problem in that context. The ultimate goal of this line of questioning is to determine
all planar graphs with metric dimensions 2.
2. Preliminaries
The order of a graph
G
is defined to be the cardinality of its vertex set. In what follows,
we will use the terms
Pn
for a path of order
n
,
Cn
for cycle,
Kn
for complete graph, and
¯
G
for the complement of G. Other notations will be defined when they are needed.
Definition 1.
A simple connected graph
G
with
|V(G)|=n
is said to be bicyclic, if
|E(G)|=
n+1.
It is well known that
|E(G)|=n−
1 when
G'Tn
. A bicyclic graph can be obtained
from this Tnby adding any two new edges.
Let
G
be a bicyclic graph, then the base bicyclic graph of
G
, denoted as
˜
G
is the unique
minimal bicyclic subgraph of
G
. It is easily concluded that
˜
G
is unique and contains no
vertices of degree 1 (a pendant vertex).
There are three types of bicyclic graphs containing no pendant vertices. These are
given in the following.
I. Cn,m
obtained from two disjoint cycles
Cn
and
Cm
, where
Cn
and
Cm
share a single
vertex. Let us label the vertices as given in Figure 1.
vnvn+m
CnCm
v1
v2
v3
vn−1
vn−2
vn+1
vn+2
vn+m−2
vn+m−1
Figure 1. Bicyclic graph of type-I.
The vertices
vn
of
Cn
and
vn+m
of
Cm
are identified together as the common vertex in
this labeling. Note that the vertices of
Cn
are labeled anti-clockwise, while vertices of
Cmare labeled clockwise.

Mathematics 2023,11, 869 3 of 17
II. Cn,r,m
obtained from two disjoint cycles
Cn
and
Cm
, by adding a path
Pr(r≥
1
)
,
from any vertex of
Cn
to any vertex of
Cm
. Let us consider the labeling given in
Figure 2.
vnvn+r
CnCm
v1
v2
v3
vn−1
vn−2
vn+1
vn+r+2
vn+r+m−2
vn+r+m−1
vn+r−1
vn+r+1
Pr
Figure 2. Bicyclic graph of type-II.
In this labeling, the vertex
vn
of
Cn
is attached to the vertex
vn+r
of
Cm
, by a path of
length r.
III. Ck,l,m
obtained from three pairwise internal disjoint paths
Pk
,
Pl
, and
Pm
, by joining
starting vertices of
Pk
and
Pm
to the starting vertex of
Pl
, and ending vertices of
Pk
and
Pm
, to the ending vertex of
Pl
. Let us denote the vertices of this graph as
v1,v2,· · · ,vk+l+m, then this type of bicyclic graph is given in Figure 3.
v1v2v3vk−1vk
vk+2vk+3vk+4vk+l−2vk+l−1
vK+l+1
vk+l+2
vk+l+3vk+l+m−1
vk+l+m
vk+1vk+l
Figure 3. Bicyclic graph of type-III.
Note that the starting vertices of paths, i.e.,
v1
of
Pk
,
vk+1
of
Pl
, and
vk+l+1
of
Pm
, are
joined together. The same is applied to the ending vertices of paths.
Let
Bn
be the class of all bicyclic graphs of order
n
. Using the three types of bases
given above, bicyclic graphs were divided into three classes, in [27], as follows.
•B1(n) = G∈ Bn|˜
G=C(k,l)for some k,l≥3
•B2(n) = G∈ Bn|˜
G=C(k,r,l)for some k,l≥3 and r≥1
•B3(n) = G∈ Bn|˜
G=C(Pk,Pl,Pm)for some 1 ≤m≤min{k,l}
It is obvious that
Bn=B1(n)∪ B2(n)∪ B3(n)
. Henceforth, we will use the term base
bicyclic graph to refer to the unique minimal bicyclic graph having no pendant vertices.
Let us use the notation
G∪H
to denote the disjoint union of graphs
G
and
H
, and
G+H
to denote the graph, obtained from
G∪H
, by joining every vertex of
G
with every
vertex of
H
. We also use
Kr,s
to denote a complete bipartite graph with partitions of order
r
and
s
. Using these notations, we state the following theorem, which gives the dimensions
of some well known graphs.
Theorem 1 ([4,7]).Given a connected simple graph G of order n ≥2, we have:
(a) dim(G) = 1if and only if G 'Pn.
(b) dim(G) = n−1if and only if G 'Kn.
(c) For n ≥3, dim(Cn) = 2.
(d)
For
n≥
4,
dim(G) = n−
2if and only if
G'Kr,s(r
,
s≥
1,
r+s=n)
,
G'Kr+
¯
Ks(r≥1, s≥2, r+s=n), or G 'Kr+ (K1∪Ks)(r,s≥1, r+s=n−1).
3. Results on Bicyclic Graphs of Type I
In what follows, let
Cn,m
be a base bicyclic graph of type I, also known as “
∞
-graph” [
28
].
The vertices are labeled as in Figure 1.

Mathematics 2023,11, 869 4 of 17
Theorem 2. Let Cn,mbe a base bicyclic graph of type 1, n,m≥3. Then,
dim(Cn,m)) = 




2when n, m are odd
2when n is even, m is odd
3when n, m are even
Proof. We will prove all three cases as different parts.
Part 1. Let Cn,mbe a bicyclic graph of type I where n,mare odd. Let us consider the set
W=nv1,vn+bm
2co.
Let Vkbe as given in the following:
Vk=

























nv1,v2, . . . , vbn
2co, for k=1
nvbn
2c+1o, for k=2
nvbn
2c+1, . . . , vno, for k=3
nvn+1, . . . , vn+bm
2co, for k=4
nvn+bm
2c+1, . . . , vn+m−1o, for k=5.
(1)
Then these Vkform a partition for V(Cn,m).
We observe that
r(va|W) =















a−1, m
2+ava∈V1
a−1, n+m
2−ava∈V2
n+1−a,n+m
2−ava∈V3
a+1−n,n+m
2−ava∈V4
n+m+1−a,a−n−m
2 va∈V5.
(2)
We will show that any two distinct vertices of
Cn,m
have distinct representation with
respect to the set
W
. Let
va
,
vb
be two distinct vertices of
Cn,m
. It is straightforward to prove
that when both va,vbare in the same partition, then
r(va|W)6=r(vb|W).
For all other cases, we proceed as follows.
Case 1. When va∈V1and vb∈V2.
We claim that
r(va|W)6=r(vb|W)
. If this is not the case, then
r(va|W) = r(vb|W)
gives
us, a−1=b−1 and m
2+a=n+m
2−b.
⇒a=band a+b=n.
Solving the above equations for bgives, b=n
2, which is a contradiction.
Case 2. When va∈V1and vb∈V3.
If we take
r(va|W) = r(vb|W)
, we get,=
a−
1
=n+
1
−b
and
m
2+a=n+m
2−b
.
Solving for
a+b
, we get,
a+b=n+
2 and
a+b=n
, which is a contradiction. Hence,
va,vbhave distinct representations.
Case 3. When va∈V1and vb∈V4.
If we consider
r(va|W) = r(vb|W)
and use the representation given in Equation sets
(2), we get a+b=n. This is a contradiction, since b∈n+1, · · · ,n+bm
2c.

Mathematics 2023,11, 869 5 of 17
Case 4. When va∈V1and vb∈V5.
We again claim that
r(va|W)6=r(vb|W)
. If this is not the case, then
r(va|W) = r(vb|W)
⇒(a−1, bm
2c+a) = (n+m+1−b,b−n−m
2)
⇒a−1=n+m+1−band bm
2c+a=b−n− b m
2c
⇒a+b=n+m+2 and a−b=−(m−1)−n, since 2bm
2c=m−1.
Solving the above equations for agives a=3
2. Again, this is a contradiction.
Case 5. When va∈V2and vb∈V3.
Assuming r(va|W) = r(vb|W)gives us
(a−1, n+m
2−a)=(n+1−b,n+m
2−b)
⇒a−1=n+1−band n+bm
2c − a=n+bm
2c − b
⇒a+b=n+2 and a=b
⇒b=n
2+1,
but b∈bn
2c+2, · · · ,n. Hence, r(va|W)6=r(vb|W).
Case 6. When va∈V2and vb∈V4.
For contradiction, let r(va|W) = r(vb|W). This gives,
(a−1, n+m
2−a)=(b+1−n,n+m
2−b)
⇒a−1=b+1−nand n+bm
2c − a=n+bm
2c − b
⇒b−a=n−2 and a=b
⇒n−2=0.
This is the desired contradiction.
Case 7. When va∈V2and vb∈V5.
If we take r(va|W) = r(vb|W), we obtain
(a−1, n+m
2−a)=(n+m+1−b,b−n−m
2)
⇒a−1=n+m+1−band n+bm
2c − a=b−n− b m
2c
⇒a+b=n+m+2 and a+b=2n+2bm
2c, which is a contradiction.
Case 8. When va∈V3and vb∈V4.
For contradiction, let us suppose that r(va|W) = r(vb|W), then
(n+1−a,n+bm
2c − a) = (b+1−n,n+m
2−b)
⇒n+1−a=b+1−nand n+bm
2c − a=n+bm
2c − b
⇒a+b=2nand a−b=0
⇒b=n
but b∈n+1, · · · ,n+bm
2c, hence r(va|W)6=r(vb|W).
Case 9. When va∈V3and vb∈V5.
We claim that r(va|W)6=r(vb|W). If this is not true, then r(va|W) = r(vb|W)
⇒(n+1−a,n+bm
2c − a) = (n+m+1−b,b−n− b m
2c)
⇒n+1−a=n+m+1−band n+bm
2c − a=b−n− b m
2c
⇒b−a=mand a+b=2n+m−1.
Solving these, we get b=n+m−1
2, which is a contradiction.
Case 10. When va∈V4and vb∈V5.
To obtain a contradiction, let r(va|W) = r(vb|W)
⇒(a+1−n,n+bm
2c − a) = (n+m+1−b,b−n− b m
2c)
⇒a+1−n=n+m+1−band n+bm
2c − a=b−n− b m
2c
⇒a+b=2n+mand a+b=2n+m−1, a contradiction.
From the above discussion, we get that
dim(Cn,m)≤
2. By Theorem 1(c), we have
dim(Cn,m)≥2 and hence, dim(Cn,m) = 2.
Part 2. Let Cn,mbe a bicyclic graph of type I, where nis even and mis odd. Let
W=nv1,vn+bm
2co.

Mathematics 2023,11, 869 6 of 17
Let Vkbe as given in the following:
Vk=

















nv1,v2, . . . , vbn
2co, for k=1
nvbn
2c+1, . . . , vno, for k=2
nvn+1, . . . , vn+bm
2co, for k=3
nvn+bm
2c+1, . . . , vn+m−1o, for k=4.
(3)
Then, these Vkform a partition for V(Cn,m). We see that
r(va|W) = 










a−1, m
2+ava∈V1
n+1−a,n+m
2−ava∈V2
a+1−n,n+m
2−ava∈V3
n+m+1−a,a−n−m
2 va∈V4,
(4)
for all
va∈V(Cn,m)
. We will show that
va
and
vb
have distinct representations for all
va6=vb∈Cn,m. It is obvious that when va,vbare both in the same partition, then
r(va|W)6=r(vb|W)
When va,vbare in different partitions, the following cases arise.
Case 1. When va∈V1and vb∈V2.
We claim that r(va|W)6=r(vb|W). If this is not the case, then r(va|W) = r(vb|W)
⇒(a−1, bm
2c+a) = (n+1−b,n+bm
2c − b).
This give us a+b=n+2 and a+b=n, which is a contradiction.
Case 2. When va∈V1and vb∈V3.
If we assume that r(va|W) = r(vb|W), we get,
(a−1, bm
2c+a) = (b+1−n,n+bm
2c − b)
⇒a−1=b+1−nand bm
2c+a=n+bm
2c − b
⇒a−b=2−nand a+b=n,
but b∈V3=⇒a+b6=n, and hence r(va|W)6=r(vb|W).
Case 3. When va∈V1and vb∈V4.
Assuming r(va|W) = r(vb|W), we obtain
(a−1, bm
2c+a) = (n+m+1−b,b−n− b m
2c)
⇒a−1=n+m+1−band bm
2c+a=b−n− b m
2c
⇒a+b=n+m+2 and a−b=−(m−1)−n
⇒a+b=n+m+2 and a−b=−m+1−n.
Solving the above for agives a=3
2, a contradiction. Hence, r(va|W)6=r(vb|W).
Case 4. When va∈V2and vb∈V3.
We claim that r(va|W)6=r(vb|W). If this is not true, then r(va|W) = r(vb|W)
⇒(n+1−a,n+bm
2c − a) = (b+1−n,n+bm
2c − b)
⇒n+1−a=b+1−nand n+bm
2c − a=n+bm
2c − b
⇒a+b=2nand a−b=0⇒b=n,
but b∈V3=⇒b6=n, a contradiction again. Hence, r(va|W)6=r(vb|W).
Case 5. When va∈V2and vb∈V4.
Proceeding in the same way as before and considering r(va|W) = r(vb|W), we get
(n+1−a,n+bm
2c − a) = (n+m+1−b,b−n− b m
2c)
⇒n+1−a=n+m+1−band n+bm
2c − a=b−n− b m
2c
⇒b−a=mand a+b=2n+m−1.

Mathematics 2023,11, 869 7 of 17
Solving these, we get
b=n+m−1
2
, which is contradiction here, hence,
r(va|W)6=
r(vb|W).
Case 6. When va∈V3and vb∈V4.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a+1−n,n+bm
2c − a) = (n+m+1−b,b−n− b m
2c)
⇒a+1−n=n+m+1−band n+bm
2c − a=b−n− b m
2c
⇒a+b=2n+mand a+b=2n+m−1, a contradiction.
From the above cases, we see that
W=nv1,vn+bm
2co
is indeed a resolving set of
Cn,m
.
Hence,
dim(Cn,m)≤
2. Together with Theorem 1(c), this gives us,
dim(Cn,m) =
2 for
n
even
and modd.
Part 3.
Let
n
and
m
be even. Consider the set
W=nv1,vbn
2c,vn+1o
and consider the parti-
tions of
V(Cn,m)
as given in Equation set (3). Noting the representations of all
va∈V(Cn,m)
from the vertices of W, and observing the fact that n
2=n
2, we see that
r(va|W) = 










a−1, n
2−a,a+1va∈V1
n+1−a,a−n
2,n+1−ava∈V2
a+1−n,a−n
2,a−n−1va∈V3
n+m+1−a,3n
2+m−a,n+m+1−ava∈V4.
(5)
To prove that
W
is a resolving set for
Cn,m
, we show that no two distinct vertices of
Cn,m
have same representations with respect to
W
. It is obvious that when
va
,
vb
are either
both in V1or V2, then r(va|W)6=r(vbW), since
d(va,v1)6=d(vb,v1)or d(va,vn
2)6=d(vb,vn
2).
Similarly, when va,vb∈V3or V4, it can be easily observed that
r(va|W)6=r(vb|W).
When va,vbare from different partitions, the following cases arise.
Case 1. When va∈V1and vb∈V2.
We claim that
r(va|W)6=r(vb|W)
. If this is not the case, then
r(va|W) = r(vb|W)
⇒(a−1, n
2−a,a+1) = (n+1−b,b−n
2,n+1−b)
⇒a−1=n+1−b,n
2−a=b−n
2and a+1=n+1−b
⇒a+b=n+2 and a+b=n.
This is a contradiction. Hence, r(va|W)6=r(vb|W).
Case 2. When va∈V1and vb∈V3.
If we assume that r(va|W) = r(vb|W), we get
(a−1, n
2−a,a+1) = (b+1−n,b−n
2,b−n−1)
⇒a−1=b+1−n,n
2−a=b−n
2and a+1=b−n−1
⇒a−b=2−nand a−b=−n−2.
This is a contradiction. Hence, r(va|W)6=r(vb|W).
Case 3. When va∈V1and vb∈V4.
We claim that r(va|W)6=r(vb|W), otherwise r(va|W) = r(vb|W)
⇒(a−1, n
2−a,a+1) = (n+m+1−b,3n
2+m−b,n+m+1−b)
⇒a−1=n+m+1−b,n
2−a=3n
2+m−band a+1=n+m+1−b
⇒a+b=n+m+2 and a+b=n+m, which is a contradiction.
Case 4. When va∈V2and vb∈V3.
Considering r(va|W) = r(vb|W), we obtain

Mathematics 2023,11, 869 8 of 17
⇒(n+1−a,a−n
2,n+1−a) = (b+1−n,b−n
2,b−n−1)
⇒n+1−a=b+1−n,a−n
2=b−n
2and n+1−a=b−n−1
⇒a+b=2nand a+b=2n+2, which is again a contradiction.
Hence, r(va|W)6=r(vb|W).
Case 5. When va∈V2and vb∈V4.
We claim that r(va|W)6=r(vb|W), for if not, then r(va|W) = r(vb|W)
⇒(n+1−a,a−n
2,n+1−a) = (n+m+1−b,3n
2+m−b,n+m+1−b)
⇒n+1−a=n+m+1−b,a−n
2=3n
2+m−band n+1−a=n+m+1−b
⇒b−a=mand b+a=2n+m.
Solving for b, we get b=n+m.
This is a contradiction, since Cn,mcontains vertices only up to n+m−1.
Case 6. When va∈V3and vb∈V4.
We claim that r(va|W)6=r(vb|W). If this is not the case, then r(va|W) = r(vb|W)
⇒(a+1−n,a−n
2,a−n−1) = (n+m+1−b,3n
2+m−b,n+m+1−b)
⇒a+1−n=n+m+1−band a−n
2=3n
2+m−band a−n−1=n+m+1−b
⇒a+b=2n+mand a+b=2n+m+2, again a contradiction.
The above discussion ensures that
W
is a resolving set for
Cn,m
, for even
n
and
m
. We
now prove that
W
is indeed a minimal resolving set. For this, consider the set
W1=W−
v1=nvbn
2c,vn+1o
and consider the vertices,
v1
,
vn−1/∈W1
. It is easily observable that
W1
does not resolve
v1
and
vn−1
. Similarly, considering the set
W2=W−vbn
2c={v1,vn+1}
and taking the vertices
vn−1
,
vn+m−1
, we see that
W2
does
not resolve these two. Lastly, considering
W3=nv1,vbn
2co
, we see that it does not resolve
vn+1and vn+m−1. This concludes our result for this part.
4. Results on Bicyclic Graphs of Type II
In this section, we will work with the metric dimensions of base bicyclic graph of type
II. Ahmad et al. used the term “Kayak Paddles graph”
KP(l
,
m
,
n)
to represent these graphs
and calculated their metric dimension [
29
]. They showed that whenever
G=KP(l
,
m
,
n)
,
then
dim(G) =
2. For completeness, we also provide a proof for these graphs. The com-
binatorial approach used herein, differs from their proof, and serves as a verification for
their result.
Let Cn,r,mbe a base bicyclic graph of type II. Let the vertices be labeled as in Figure 2.
Theorem 3. Let Cn,r,mbe a base bicyclic graph of type II, n,m≥3and r ≥1. Then,
dim(Cn,r,m) = 2.
Proof.
We will discuss the proof for different cases of
n
,
m
, namely, when both are odd/even
or one is odd and the other is even.
Part 1. Let n,mboth be even in Cn,r,m. Let us consider the set W={v1,vn+r+1}.
Let Vkbe as given in the following:
Vk=















nv1,v2, . . . , vbn
2cofor k=1
nvbn
2c+1, . . . , vnofor k=2
nvn+1, . . . , vn+r+bm
2cofor k=3
nvn+r+bm
2c+1, . . . , vn+m−1ofor k=4
(6)
Then these Vkform a partition for the vertices of the given graph.

Mathematics 2023,11, 869 9 of 17
We observe that
r(va|W) = 










(a−1, r+a+1)va∈V1
(n+1−a,n+r+1−a)va∈V2
(a+1−n,|a−n−r−1|)va∈V3
(n+m+2r+1−a,n+m+r+1−a)va∈V4.
(7)
Proceeding as before and assuming that
va
,
vb
are either both in
V1
or
V2
, we can easily
see that
r(va|W)6=r(vb|W)
. Similarly, when
va
,
vb
are both from
V3
or
V4
, we again see
that r(va|W)6=r(vb|W).
When va,vbare in different partitions Vk, we proceed as follows.
Case 1. When va∈V1and vb∈V2.
Assuming r(va|W) = r(vb|W), we get
(a−1, r+a+1) = (n+1−b,n+r+1−b).
This produces a contradiction, whereas, we get a+b=n+2 and a+b=n.
Hence, r(va|W)6=r(vb|W).
Case 2. When va∈V1and vb∈V3.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a−1, r+a+1) = (b+1−n,|b−n−r−1|)
If b>n+r+1, we get,
a−1=b−n+1 and r+a+1=b−n−r−1
⇒a−b=2−nand a−b=−n−2r−2, a contradiction.
If
b<n+r+
1, we get,
a−b=
2
−n
and
a+b=n
. Solving these for
b
gives the
contradiction b=n
2−1.
Case 3. When va∈V1and vb∈V4.
We claim that
va
and
vb
have distinct representations with respect to
W
. If this is not
the case, then
r(va|W) = r(vb|W)
gives us the contradiction
a+b=n+m+2r+1
and
a+b=n+m.
Case 4. When va∈V2and vb∈V3.
If we consider r(va|W) = r(vb|W), we get,
(n+1−a,n+r+1−a) = (b+1−n,|b−n−r−1|).
If b>n+r+1, we get the contradiction a+b=2nand a+b=2n+2r+2.
If b<n+r+1, we get a+b=2nand a=b, but a6=b.
Hence, r(va|W)6=r(vb|W).
Case 5. When va∈V2and vb∈V4.
Proceeding as before and considering r(va|W) = r(vb|W), we get
(n+
1
−a
,
n+r+
1
−a) = (n+m+
2
r+
1
−b
,
n+m+r+
1
−b)
. This again produces
a contradiction. Hence, r(vi|W)6=r(vj|W).
Case 6. When va∈V3and vb∈V4.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a+1−n,|a−n−r−1|) = (n+m+2r+1−b,n+m+r+1−b).
If a>n+r+1, we obtain,
a+1−n=n+m+2r+1−band a−n−r−1=n+m+r+1−b
⇒a+b=m+2n+2rand a+b=m+2n+2r+2, which is a contradiction.
If a>n+r+1, we obtain the contradiction b=n+m+r.
The above discussion concludes that
dim(G)≤
2. Together with Theorem 1(c), we
obtain dim(G) = 2.

Mathematics 2023,11, 869 10 of 17
Part 2. Let G=Cn,r,mbe a bicyclic graph of type II with n,modd. Considering the set
W={v1,vn+r+1}.
Considering Vkas given in the following,
Vk=



























nv1,v2, . . . , vbn
2cofor k=1
nvbn
2c+1ofor k=2
nvbn
2c+2, . . . , vnofor k=3
nvn+1, . . . , vn+r+bm
2cofor k=4
nvn+r+bm
2c+1ofor k=5
nvn+r+bm
2c+2, . . . , vn+r+m−1ofor k=6
(8)
we see that these Vkform a partition for V(Cn,r,m).
r(va|W) =





















(a−1, r+a+1)va∈V1
(a−1, n+r+1−a)va∈V2
(n+1−a,n+r+1−a)va∈V3
(a+1−n,|a−n−r−1|)va∈V4
(n+m+2r+1−a,a−n−r−1)va∈V5
(n+m+2r+1−a,n+m+r+1−a)va∈V6.
(9)
Case 1. When va∈V1and vb∈V2.
We claim that
r(va|W)6=r(vb|W)
. If this is not the case then,
r(va|W) = r(vb|W)
gives
us, a−1=b−1 and r+a+1=n+r+1−b⇒a=band a+b=n.
Solving the above equations for bgives b=n
2, which is a contradiction.
Case 2. When va∈V1and vb∈V3.
Assuming r(va|W) = r(vb|W), we get (a−1, r+a+1) = (n+1−b,n+r+1−b).
This produces a contradiction, whereas, we get a+b=n+2 and a+b=n.
Hence, r(va|W)6=r(vb|W).
Case 3. When va∈V1and vb∈V4.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a−1, r+a+1) = (b+1−n,|b−n−r−1|).
If b>n+r+1, we get a−1=b−n+1 and r+a+1=b−n−r−1
⇒a−b=2−nand a−b=−n−2r−2, a contradiction.
If b<n+r+1, we obtain a−b=2−nand a+b=n.
This produces the contradiction b=n−1.
Case 4. When va∈V1and vb∈V5.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a−1, r+a+1) = (n+m+2r+1−b,b−n−r−1)
⇒a−1=n+m+2r+1−band r+a+1=b−n−r−1
⇒a+b=n+m+2r+2 and b−a=n+2r+2.
Solving the above equations for agives b=m
2, which is a contradiction.
Case 5. When va∈V1and vb∈V6.
We claim that
va
and
vb
have distinct representations with respect to
W
. If this is not
the case, then
r(va|W) = r(vb|W)
gives us the contradiction
a+b=n+m+2r+1
and
a+b=n+m.

Mathematics 2023,11, 869 11 of 17
Case 6. When va∈V2and vb∈V3.
We claim that r(va|W)6=r(vb|W), otherwise r(va|W) = r(vb|W)gives us,
a−1=n+1−band n+r+1−a=n+r+1−b.
⇒a+b=n+2 and a=b.
Solving the above equations for bgives b=n
2+1, an obvious contradiction.
Case 7. When va∈V2and vb∈V4.
If we consider r(va|W) = r(vb|W), we get,
(a−1, n+r+1−a) = (b+1−n,|b−n−r−1|).
Two possibilities arise for b. If b>n+r+1, we get,
b−a=n−2 and a+b=2n+2r+2.
Solving the above equations for agives a=n
2+r+2, which is a contradiction.
On the other hand, if
b<n+r+
1, we obtain
b−a=n−
2 and
a=b
. Solving these,
we get, n=2, which is again a contradiction.
Hence, r(va|W)6=r(vb|W).
Case 8. When va∈V2and vb∈V5.
Considering r(va|W) = r(vb|W), we get,
(a−1, n+r+1−a) = (n+m+2r+1−b,b−n−r−1).
This gives us the contradiction a+b=n+m+2r+2 and a+b=2n+2r+2.
Case 9. When va∈V2and vb∈V6.
If we consider r(va|W) = r(vb|W), we get,
(a−1, n+r+1−a) = (n+m+2r+1−b,m+n+r+1−b)
=⇒a+b=n+m+2r+2 and b−a=m.
Solving the above equations for agives, a=n
2+r+1, an obvious contradiction.
Case 10. When va∈V3and vb∈V4.
Assuming
r(va|W) = r(vb|W)
, gives us the contradiction
a+b=
2
n
and
a+b=
2
n+
2
r+
2, when
b>n+r+
1. On the other hand, if
b<n+r+
1, we obtain
a+b=
2
n
and a=b. This is the desired contradiction since va,vbare distinct vertices.
Case 11. When va∈V3and vb∈V5.
Proceeding as before and considering r(va|W) = r(vb|W), we get,
(n+1−a,n+r+1−a) = (n+m+2r+1−b,b−n−r−1)
=⇒b−a=m+2rand a+b=2n+2r+2.
Solving the above equations for
b
gives
b=n+
2
r+
1
+m
2
, which is a contradiction.
Case 12. When va∈V3and vb∈V6.
If we assume r(va|W) = r(vb|W), we get,
(n+
1
−a
,
n+r+
1
−a) = (n+m+
2
r+
1
−b
,
n+m+r+
1
−b)
. This produces the
contradiction b−a=m+2rand b−a=m.
Case 13. When va∈V4and vb∈V5.
We claim that r(va|W)6=r(vb|W). If not, then r(va|W) = r(vb|W)
⇒(a+1−n,|a−n−r−1|) = (n+m+2r+1−b,b−n−r−1).
If a>n+r+1, we get,
a+1−n=n+m+2r+1−band a−n−r−1=b−n−r−1
⇒a+b=m+2n+2rand a=b.
Solving the above equations for bgives, b=n+r+m
2, which is a contradiction.
If
a<n+r+
1, we get,
a+b=m+
2
n+
2
r
and
a+b=
2
n+
2
r+
2. This produces
the contradiction m=2. Hence, r(va|W)6=r(vb|W).
Case 14. When va∈V4and vb∈V6.
If we assume that r(va|W) = r(vb|W), we obtain,
(a+1−n,|a−n−r−1|) = (n+m+2r+1−b,n+m+r+1−b).

Mathematics 2023,11, 869 12 of 17
If a>n+r+1, we get,
a+1−n=n+m+2r+1−band a−n−r−1=n+m+r+1−b
⇒a+b=m+2n+2rand a+b=m+2n+2r+2, an obvious contradiction.
On the other hand, if
a<n+r+
1, we obtain
a+b=m+
2
n+
2
r
and
b−a=m
.
Solving for bgives, b=n+m+r, a contradiction.
Case 15. When va∈V5and vb∈V6.
If we take r(va|W) = r(vb|W), we get,
(n+m+2r+1−a,a−n−r−1) = (n+m+2r+1−b,n+m+r+1−b)
⇒n+m+2r+1−a=n+m+2r+1−band a−n−r−1=n+m+r+1−b
⇒a=band a+b=m+2n+2r+2.
Solving the above equations for bgives b=n+r+1+m
2, a contradiction.
All the above cases ensure that
dim(Cn,r,m)≤
2. This, together with Theorem 1(c),
gives dim(Cn,r,m) = 2 for n,modd.
Part 3.
Let us take
Cn,r,m
,
n
odd, and
m
even. Let us consider the set
W={v1
,
vn+r+1}
.
If we take Vkas given in the following,
Vk=





















nv1,v2, . . . , vbn
2cofor k=1
nvbn
2c+1ofor k=2
nvbn
2c+2, . . . , vnofor k=3
nvn+1, . . . , vn+r+bm
2cofor k=4
nvn+r+bm
2c+2, . . . , vn+r+m−1ofor k=5,
(10)
we see that these
Vk
form a partition of
V(Cn,r,m)
. Noting the distances of these
Vk
from
W
,
we get,
r(va|W) =















(a−1, r+a+1)va∈V1
(a−1, n+r+1−a)va∈V2
(n+1−a,n+r+1−a)va∈V3
(a+1−n,a−n−r−1)va∈V4
(n+m+2r+1−a,n+m+r+1−a)va∈V5.
(11)
The rest of the proof pattern is again similar to Part 1 and/or Part 2.
5. Perturbation in Metric Dimension of Bicyclic Graphs After Edge/Vertex Deletion
In this section, we consider the change in the metric dimension of bicyclic graphs type
I and II, when an edge is removed. We only consider the cases when such a graph is still
connected. Before proceeding further, we introduce some notations and definitions used in
this section. The degree of a vertex
v
of a simple connected graph
G
, denoted by
d(v)
, is
the number of edges incident at
v
. By
G−e
, we mean that the edge
e
has been removed
from the graph
G
. By a “pendant path”, we mean a path
x1x2· · · xk
such that every
xi
is of
degree 2 except
xk
, which is of degree 1, and the vertex
x1
is attached to a vertex
v
of
G
,
where d(v)≥3. These ideas are evident in Figure 4.

Mathematics 2023,11, 869 13 of 17
v1v2
v3
x1
G
x2
x3
v1v2
v3
x1
G − {v2v3}
x2
x3
Figure 4. Graph Gwith a pendant path x1x2x3and Graph G−ewhere e={v2v3}.
Let
G
be a bicyclic graph of type I. Let us now consider the graph
G−e
. If
e
is an
edge incident with a vertex of degree 4,
G−e
is a unicyclic graph with one pendant path.
Meanwhile, when
e
is any other edge,
G−e
is a unicyclic graph with two pendant paths,
we name these configurations Aand B, respectively; these are shown in Figure 5.
vn
vn+1vn+m−2vn+m−1
AB
v1
v2
vn−2
vn−1
vn
v1
v2
vn−2
vn−1
vn+1
vn+r
vn+m−1
vn+r+1
Figure 5. Configurations Aand Bof G−ewhen G=Cn,m.
Note that the connected graph
G−v(v∈V(G))
gives us the same configurations. We
will only discuss the case of
G−e
here, keeping in mind that the same results are applicable
to G−v.
We denote the cycle in
G−e
by
Cn
, even interchanging
n
and
m
if necessary. We now
introduce the following lemma.
Lemma 1. For a unicyclic graph G with a pendant path of length m −1, dim(G) = 2.
Proof.
Let us denote the cycle in
G
by
Cn
. Let the vertices be labeled as in Figure 5,
Configuration A. Let W=nv1,vbn
2co.
When nis even, we have,
r(vi|W) = 






i+1−n,n
2+i−n∀vi∈ {vn+1,vn+2,· · · ,vn+m−1}
i−1, n
2−i∀vi∈nv1,v2,· · · ,vbn
2co
n+1−i,i−n
2 ∀vi∈nvbn
2c+1,· · · ,vno,
and when nis odd, we obtain,
r(vi|W) = 












i+1−n,n
2+i−n∀vi∈ {vn+1,vn+2,· · · ,vn+m−1}
i−1, n
2−i∀vi∈nv1,v2,· · · ,vbn
2co
i−1, i−n
2 vi=vbn
2c+1
n+1−i,i−n
2 ∀vi∈nvbn
2c+2,· · · ,vno.

Mathematics 2023,11, 869 14 of 17
In both cases, it can easily be concluded that no two vertices of
G
have the same
representation with respect to
W
. This together with Theorem 1(c) gives us,
dim(G) =
2.
For configuration
B
, we see that there are two pendant paths attached at
vn
. One can
see that the vertices on both paths equidistant to
vn
, e.g., vertices
vn+1
and
vn+m−1
can not
be distinguished from any vertex of the cycle
Cn
. Similarly, vertices of
Cn
at equal distance
from
vn
, e.g.,
v1
and
vn−1
, can not be distinguished from the vertices of attached paths.
This brings us to the following lemma.
Lemma 2.
Let
G
be a unicyclic graph with two pendant paths attached to a vertex of the cycle in
G, then dim(G) = 3.
Proof.
Let us denote the vertices of the cycle
Cn
by
v1
,
v2
,
· · · vn
. Let the two pendant paths
attached at
vn
be
vn+1vn+2· · · vn+r
and
vn+m−1vn+m−2· · · vn+r+1
. This representation of
vertices is given in Figure 5, Configuration B.
Since the cycle
Cn
in
G
has dimension 2, and no vertices of
Cn
distinguish vertices of
both the pendant paths,
dim(G)>
2
=⇒dim(G)≥
3. Let
W={vi
,
vj} ⊂ V(Cn)
be the
basis for
Cn
and let
W0=W∪ {vn+r}
. Since
W
resolves
Cn
, and
vn+r
resolves both pendant
paths; W0resolves all vertices of G. Hence, dim(G)≤3 and we obtain our result.
Combining the results of Lemmas 1and 2, we obtain the following.
Theorem 4. Let G −e be a graph obtained by deleting an edge from G =Cn,m. Then,
dim(G−e) = (2if e is incident with a vertex of degree 4
3otherwise.
Let us now consider the bicyclic graph of type II, i.e.,
G=Cn,r,m
.
G−e
again gives us
two configurations depending on whether
e
is incident with a vertex of degree 2 or 3. One
of the configurations we obtain is similar to
A
, while we name the other one as
C
, given in
Figure 6.
vn
vn+1vn+r
vn+r+s+1
vn+r+m−1
C
vn+r+1
vn+r+s
Figure 6.
Configuration
C
when an edge incident at vertices of degree 2 is removed from
G=Cn,r,m
.
We again mention here that the connected graph
G−v
gives us the same configura-
tions, and the results for G−ecan easily be applied to G−v.
We now present the following lemma for configuration C.
Lemma 3.
Let
G
be a unicyclic graph with two pendant paths
Pk
and
Pl
, of lengths
k
and
l
,
respectively, attached to a vertex of the cycle in G. Let |V(Pk)∩V(Pl)|>1, then dim(G) = 3.
Proof.
Let us label the vertices of the cycle
Cn
in
G
by
v1
,
v2
,
· · ·
,
vn
. Let
|V(Pk)∩V(Pl)|=r
and let these vertices be labeled as
vn+1
,
vn+2
,
· · ·
,
vn+r
. For an easier proof, we relabel the
remaining vertices from Figure 6as
u1
,
u2
,
· · ·
,
uk−r
and
w1
,
w2
,
· · ·
,
wl−r
. This representa-
tion is given in Figure 7.

Mathematics 2023,11, 869 15 of 17
vn
vn+1vn+r
wl−r
w1
C
u1
uk−r
Figure 7. Configuration Cof Cn,r,m−erelabeled.
Let
G1
be the induced subgraph of
G
on the vertices
v1
,
v2
,
· · ·
,
vn+r
, and let
G2
be the
induced subgraph on the vertices
V(G)−V(G1)∪ {vn+r}
. Since
G1
is a unicyclic graph
with an attached pendant path, by Lemma 1,
dim(G1) =
2. Following a similar argument to
Lemma 2, we conclude that
dim(G)≥
3. Let
W={vi
,
vj} ⊂ V(G1)
be the basis for
G1
and
let
W0=W∪ {uk−r}
. Since
W
resolves
G1
, and
uk−r
resolves
G2
,
W0
resolves all vertices of
G. Hence, dim(G)≤3 and we obtain our result.
Lemmas 1and 3give us the following theorem.
Theorem 5.
Let
G−e
be a connected graph, obtained by deleting an edge from
G=Cn,r,m
. Then,
dim(G−e) = (2if e is incident with a vertex of degree 3
3otherwise.
If we continue to remove edges from
G−e
to obtain a connected graph, equivalently,
if we remove more than one edge from
G
in such a way that the final graph is connected
(
G
is bicyclic graph of type I or type II), we either arrive at one of the configurations
A
,
B
,
or
C
, or we obtain a tree. In both cases, their metric dimension can easily be concluded by
already established results of Lemmas 1–3, or by ([4] Theorem 5).
6. Summary
We studied the resolving set and metric dimension of base bicyclic graphs and showed
that they are constant for type I and II base bicyclic graphs. Particularly,
dim(Cn,m) = 




2 when n,mare odd
2 when nis even and mis odd
3 when n,mare even
where n,m≥3 and
dim(Cn,r,m) = 2 for all n≥3, m≥3, r≥1.
We also considered the problem of removing an edge/vertex from these graphs and
obtained results for dim(G−e).
7. Conclusions
Unicyclic graphs have been studied extensively under varying graph invariants related
to metric dimensions. Bicyclic graphs have not enjoyed the same level of interest from
researchers till now. In this article, we showed that the base bicyclic graphs of type I and II
have constant metric dimensions. This opens up new avenues for researchers to discuss
other graph invariants for these types of graphs.
In the future, the following problems are an effective way of extending this research.

Mathematics 2023,11, 869 16 of 17
Problem 1: Studying the bicyclic graph of type III and providing a generalized proof
that they have constant metric dimension.
Problem 2: Studying various other graph invariants, e.g., local metric dimensions,
mixed metric dimensions, and k-metric dimensions for bicyclic graphs.
Problem 3: Studying the characterization of a bigger class of graphs with metric
dimension 2.
It was observed in [
30
] that the metric dimension problem is NP-complete for pla-
nar graphs. Diaz et al. also proposed an algorithm to calculate the metric dimension of
outerplanar graphs in polynomial time. Since bicyclic graphs of type I and type II are
also outerplanar graphs, their metric dimension can also be calculated in polynomial time.
Following the results from this article, there is no need to apply a generalized algorithm to
calculate the metric dimension of bicyclic graphs of type I and II. This effectively reduces
the computational time for anyone who wants to use the metric dimension of these graphs
in their applications/research.
Author Contributions:
A.K., G.H., N.A. and M.U.I.K. contributed equally to this work. Conceptual-
ization, methodology, software, validation, formal analysis, investigation, A.K., N.A. and M.U.I.K.;
funding acquisition, A.K.; data curation, visualization, writing—original draft preparation, writing—
review and editing, G.H., A.U.K.N. and A.U.I.K. All authors have read and agreed to the published
version of the manuscript.
Funding:
This research was sponsored by the Guangzhou Government Project under Grant No.
62216235, and the National Natural Science Foundation of China (Grant No. 12250410247).
Acknowledgments:
The authors acknowledge the huge contribution provided by the reviewer’s
suggestions in improving the quality of this article. Their valuable input not only helped in expanding
the results, but also helped in formatting the article for an overall better presentation.
Conflicts of Interest: The authors declare no conflict of interest.
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