EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 2, 2023, 1068-1083
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Forcing 2-Metric Dimension in the Join and Corona of
Graphs

Dennis B. Managbanag1,∗, Helen M. Rara2

1 Department of Mathematics and Statistics, College of Science and Mathematics,
Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines
2 Department of Mathematics and Statistics, College of Science and Mathematics,
Center of Graph Theory, Algebra, and Analysis-Premier Research Institute of Science and
Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City,
Philippines

Abstract. This study deals with the forcing subsets of 2-metric basis in graphs. Some main
results generated in this study include the characterization of a 2-metric basis in graphs and the
characterization of the forcing subsets of these 2-metric bases. These characterizations are used to
determine values for the forcing 2-metric dimension of graphs resulting from some binary operations
such as the join and corona of graphs.

2020 Mathematics Subject Classifications: 05C69
Key Words and Phrases: 2-resolving set, 2-metric basis, 2-metric dimension, forcing subsets,
forcing number, join, corona

1. Introduction

Metric dimension and resolving sets, concepts initially drafted for the metric spaces
introduced by Blumenthal [3] in 1953. Since then, the notion of metric dimension has
been broadened to encompass both metric and geometric spaces [2, 7]. Nearly 20 years
after in 1976, Harary, Melter [5] and Slater [11, 12] each separatedly discovered the idea
of resolving set.

In 2019, Bailey and Yero [1] demonstrated the construction of error-correcting codes
out of graphs using k-resolving sets and provided a decoding algorithm that used covering
designs. A study on the idea of the k-resolving set, also known as “On 2-resolving sets in
the join and corona of graphs” was published by J. Cabaro and H. Rara [4] in 2021.

The concept of forcing numbers, which was established in 1987 as a result of Klein and
Randic’s introduction of the study of molecular resonance structure, is another intrigu-
ing topic that has drawn the interest of several researchers [8]. Consequently, in 1991,

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4750

Email addresses: dennis.managbanag@g.msuiit.edu.ph (D. Managbanag),
helen.rara@g.msuiit.edu.ph (H. Rara)

https://www.ejpam.com 1068 © 2023 EJPAM All rights reserved.



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1069

Harary et. al [6] coined the term “forcing number” and presented the idea of forcing as
a perfect match. In 1999, Chartrand et. al [13] initiated the investigation on the relation
between forcing and dimension of a graph. The notions of a 2-resolving set and the forcing
dimension of a graph serve as the inspiration for this work. We believe that this study
will be tremendously beneficial to someone who is familiar with the theory of the metric
dimension. The findings of this work amplified previously-revealed notions to obtain new
applications in graph-to-code theory, much like the idea of a 2-resolving set, by developing
a novel method for producing error-correcting codes out of graphs.

2. Terminology and Notation

In this study, we only consider graphs that are finite, simple, undirected and connected.
Readers are referred to [1, 4, 9, 10] for elementary Graph Theoretic concepts.

Let G be a connected graph of order n. For an ordered set of vertices
W = {w1, w2, ..., wk} ⊆ V (G) and a vertex v in G, we refer to the k-vector (ordered
k-tuple) rG(v/W ) = (dG(v, w1), dG(v, w2), ..., dG(v, wk)) as the (metric) representation of
v with respect to W.

An ordered set of vertices W = {w1, w2, ..., wk} is a k-resolving set for G if, for any
distinct vertices u, v ∈ V (G), the (metric) representations rG(u/W ) and rG(v/W ) of u
and v, respectively differ in at least k positions. If k = 1, then the k-resolving set is called
a resolving set for G. If k = 2, then the k-resolving set is called a 2-resolving set for G.

The least size of a 2-resolving set is called a 2-metric dimension of G and we denote
it by dim2(G). A resolving set of size dim2(G) is called a 2-metric basis for G.

Let G be any nontrivial connected graph and S ⊆ V (G). A set S ⊆ V (G) is
2-locating set of G if it satisfies the following conditions: (i) |[(NG(x) \ NG(y)) ∩ S]
∪ [(NG(y) \NG(x))∩ S]| ≥ 2, for all x, y ∈ V (G) \ S with x ̸= y and (ii) (NG(v) \NG(w))
∩S ̸= ∅ or (NG(w)\NG[v])∩S ̸= ∅ for all v ∈ S and for all w ∈ V (G)\S. The 2-locating
number of G, denoted by ln2(G), is the smallest cardinality of a 2-locating set of G. A
2-locating set of G of cardinality ln2(G) is referred to as an ln2-set of G.

Let G be any nontrivial connected graph and S ⊆ V (G). S is a
(2,2)-locating (respectively (2,1)-locating) set in G if S is a 2-locating and |NG(y) ∩ S|
≤ |S| − 2 (|NG(y) ∩ S| ≤ |S| − 1, respectively), for all y ∈ V (G). The (2,2)-locating
(respectively (2,1)-locating) number of G, denoted by ln(2,2)(G) (respectively ln(2,1)(G)),
is the smallest cardinality of a (2,2)-locating (respectively (2,1)-locating) set in G. A
(2,2)-locating (respectively (2,1)-locating) set in G of cardinality ln(2,2)(G) (respectively
ln(2,1)(G)) is referred to as an ln(2,2)-set (respectively ln(2,1)-set) in G.

Let W be a 2-metric basis of a graph G. A subset S of W is said to be a forcing subset
for W if W is the unique 2-metric basis containing S. The forcing 2-metric dimension of
W is given by fdim2(W ) = min{|S| : S is a forcing subset for W}. The forcing 2-metric
dimension of G is given by

fdim2(G) = min{fdim2(W ) : W is a 2-metric basis for G}.

Let W be an ln2-set of a graph G. A subset S of W is said to be a forcing subset for



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1070

W if W is the unique ln2-set containing S. The forcing 2-locating number of W is given
by fln2(W ) = min{|S| : S is a forcing subset for W}. The forcing 2-locating number of G
is given by

fln2(G) = min{fln2(W ) : W is a ln2-set of G}.

Let W be an ln(2,2)-set of a graph G. A subset S of W is said to be a forcing subset for
W if W is the unique ln(2,2)-set containing S. The forcing (2, 2)-locating number of W is
given by fln(2,2)(W ) = min{|S| : S is a forcing subset for W}. The forcing (2, 2)-locating
number of G is given by

fln(2,2)(G) = min{fln(2,2)(W ) : W is a ln(2,2)-set of G}.

3. Known Results

The following known results are taken from [4].

Remark 1. For any connected nontrivial graph G of order n ≥ 2, 2 ≤ ln2(G) ≤ n.
Moreover, ln2(Kn) = n, for n ≥ 2.

Theorem 1. Let G be a connected nontrivial graph. Then ln2(G) = 2 if and only if
G ∼= P2 or G ∼= P3.

Proposition 1. dim2(G) = 2 if and only if G ∼= Pn, n ≥ 2.

Example 1. Let n be a positive integer. Then Pn and Cn

ln2(Pn) =


n

2
+ 1, if n ≥ 2 and n is even,

n+ 1

2
, if n ≥ 3 and n is odd, and

ln2(Cn) =


n

2
, if n ≥ 6 and n is even,

n+ 1

2
, if n ≥ 5 and n is odd.

Example 2. The formulas below give the (2,2)-locating number of the path Pn and cycle
Cn.

ln(2,2)(Pn) =



4, if n = 5,

n

2
+ 1, if n ≥ 6 and n is even,

n+ 1

2
, if n ≥ 7 and n is odd, and

ln(2,2)(Cn) =


n

2
, if n ≥ 8 and n is even,

n+ 1

2
, if n ≥ 7 and n is odd.



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1071

Theorem 2. Let G be a connected graph of order greater than 3 and let K1 = ⟨v⟩. Then
S ⊆ V (K1 + G) is a 2-resolving set of K1 + G if and only if either v /∈ S and S is a
(2,2)-locating set in G or S = {v} ∪ T is (2,1)-locating set in G.

Theorem 3. Let G and H be nontrivial connected graphs. A proper subset S of V (G+H)
is a 2-resolving set in G + H if and only if SG = V (G) ∩ S and SH = V (H) ∩ S are
2-locating sets in G and H, respectively, where SG or SH is (2,2)-locating set or SG and
SH are (2,1)-locating sets.

Theorem 4. Let G and H be nontrivial connected graphs. A set
S ⊆ V (G ◦H) is a 2-resolving set of G ◦H if and only if S = A ∪ B, where A ⊆ V (G)
and B =

⋃
{Sv : Sv is a 2-resolving set of Hv, for all v ∈ V (G)}.

Remark 2. Let G and H be non-trivial connected graphs, C ⊆ V (G ◦ H) and
Sv = V (Hv) ∩ C where v ∈ V (G). For each x ∈ V (Hv) \ Sv and z ∈ Sv,

dG◦H(x, z) =

®
1, if z ∈ NHv(x),

2, otherwise.

4. Forcing 2-Metric Dimension of Some Special Graphs

Remark 3. Let G be a nontrivial connected graph. Then

0 ≤ fdim2(G) ≤ dim2(G).

Remark 4. Let G be a connected graph. Then
(i) fdim2(G) = 0 if and only if G has a unique 2-metric basis, and
(ii) fdim2(G) = 1 if and only if G has at least two 2-metric bases, one of which, say

B, that contains an element not in any 2-metric basis for G.

Theorem 5. Let G be a connected graph. Then fdim2(G) = dim2(G) if and only if
for all 2-metric basis D for G and for each u ∈ D, there exists vu ∈ V (G) \D such that[
D \ {u}

]
∪ {vu} is a 2-metric basis for G.

Proof: Suppose that fdim2(G) = dim2(G). Let D be a 2-metric basis for G such
that fdim2(G) = |D| = dim2(G), that is, D is the only forcing subset for itself. Let
u ∈ D. Since D \ {u} is not a forcing subset for D, there exists a vu ∈ V (G) \D such that[
D \ {u}

]
∪ {vu} is a 2-metric basis for G.

Conversely, suppose that every 2-metric basis for G satisfies the given condition. Let
D be a 2-metric basis for G such that fdim2(G) = fdim2(D). Suppose further that D
has a forcing subset J with |J | < |D|, that is, D = J ∪K where K = {w ∈ B : w /∈ J}.
Pick w ∈ K. By assumption, there exists vw ∈ V (G) \D such that

[
D \ {w}

]
∪ {vw} = T

is a 2-metric basis for G. Hence, T = J ∪M, where M =
[
K \ {w}

]
∪ {vw}, is a 2-metric

basis containing J, a contradiction. Hence, D is the only forcing subset for D. Therefore,
fdim2(G) = dim2(G).



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1072

Proposition 2. For any complete graph Kn with n ≥ 1 vertices,

fdim2(Kn) = 0.

Proof: By definition of Kn, V (Kn) is the only minimum 2-resolving set for Kn. By
Remark 4 (i), fdim2(Kn) = 0.

Proposition 3. For any path Pn with n ≥ 2 vertices,

fdim2(Pn) = 0.

Proof: Suppose that Pn = [v1, v2, . . . , vn]. By Proposition 1, dim2(Pn) = 2 for all n ≥ 2.
We claim that S = {v1, vn} is the unique 2-metric basis of Pn. Suppose S

′ = {vi, vj} where
1 ≤ i < j ≤ n and S′ ̸= S. Consider the following cases:
Case 1. Suppose i = 1.

If j = 2, then rPn(v1/S
′) = (0, 1) and rPn(v3/S

′) = (2, 1). If 2 < j < n, then
rPn(vj−1/S

′) = (j − 2, 1) and rPn(vj+1/S
′) = (j, 1).

Case 2. Suppose 1 < i < j < n.
If j = i + 1, then rPn(vi/S

′) = (0, 1), rPn(vj+1/S
′) = (2, 1), rPn(vi−1/S

′) = (1, 2),
and rPn(vj/S

′) = (1, 0). If j > i + 1, then rPn(vi−1/S
′) = (1, j − i + 1)

and rPn(vi+1/S
′) = (1, j − i− 1).

By Cases 1 and 2, S′ is not a 2-resolving set for Pn. Thus, S is unique. Hence, by
Remark 4 (i), fdim2(Pn) = 0 for all n ≥ 4. Therefore, fdim2(Pn) = 0 for all n ≥ 2
vertices.

Proposition 4. For any cycle Cn with n ≥ 3 vertices,

fdim2(Cn) =

®
0, if n = 3, 4

3, if n ≥ 5.

Proof: Suppose that Cn = [u1, u2, . . . , un]. If n = 3 or 4, then V (Cn) is the only
2-resolving set for Cn. By Remark 4 (i), fdim2(Cn) = 0. Let n ≥ 5. By Proposition
1, dim2(Cn) > 2. Since {u1, u2, u3} is a 2-resolving set for Cn, dim2(Cn) = 3. Let
S = {ua, ub, uc} be a 2-metric basis for Cn. Hence, for all 2-metric basis S for Cn and for
each uk ∈ S, there exists ul ∈ V (Cn) \S such that (S \ {uk})∪{ul} is a 2-metric basis for
Cn. By Theorem 5, fdim2(Cn) = dim2(Cn) = 3.

5. Forcing 2-Locating and (2,2)-Locating Numbers of Some Special
Graphs

Remark 5. Let G be a connected graph. Then
(i) fln2(G) = 0 if and only if G has a unique ln2-set, and
(ii) fln2(G) = 1 if and only if G has at least two ln2-sets, one of which, say B, that

contains an element not in any ln2-sets of G.



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1073

Theorem 6. Let G be a connected graph. Then fln2(G) = ln2(G) if and only if for all
ln2-set S of G and for each u ∈ S, there exists vu ∈ V (G) \ S such that

[
S \ {u}

]
∪ {vu}

is a 2-locating set of G.

Proof: Suppose that fln2(G) = ln2(G). Let S be an ln2-set of G such that
fln2(G) = |S| = ln2(G), that is, S is the only forcing subset for itself. Let u ∈ S. Since
S \{u} is not a forcing subset for S, there exists a vu ∈ V (G)\S such that

[
S \{u}

]
∪{vu}

is an ln2-set of G.
Conversely, suppose that every ln2-set of G satisfies the given condition. Let S be a

2-locating set of G such that fln2(G) = fln2(S). Suppose further that S has a forcing
subset H with |H| < |S|, that is, S = H ∪ I where I = {w ∈ S : w /∈ H}. Pick w ∈ I. By
assumption, there exists vw ∈ V (G) \S such that

[
S \ {w}

]
∪{vw} = J is an ln2-set of G.

Hence, J = H∪T , where T =
[
I \{w}

]
∪{vw}, is an ln2-set containing H, a contradiction.

Hence, S is the only forcing subset for S. Therefore, fln2(G) = ln2(G).

Proposition 5. For any complete graph Kn with n > 1 vertices,

fln2(Kn) = 0.

Proof: By Remark 1, V (Kn) is the only ln2-set of Kn. By Remark 5 (i),
fln2(Kn) = 0.

Remark 6. Let S be a 2-locating set of Pn = [u1, u2, . . . , un−1, un] where n ≥ 2. Then
(i) {u1, u2} ∩ S ̸= ∅.
(ii) {un−1, un} ∩ S ̸= ∅.
(iii) {ui, ui+1, ui+2} ∩ S ̸= ∅ where 1 ≤ i < n.

Proposition 6. For any path Pn with n ≥ 2 vertices,

fln2(Pn) =


0, if n = 2, 3 and n ≥ 7 is odd,

1, if n = 5,

2, if n ≥ 6 is even,

3, if n = 4.

Proof: Suppose that Pn = [u1, u2, . . . , un]. By Theorem 1, if n = 2, then V (P2) is the
only ln2-set of P2 and if n = 3, then M = {u1, u3} is the only ln2-set of P3. Thus, by
Remark 5 (i), fln2(P2) = fln2(P3) = 0. Suppose that n = 5. By Example 1, ln2(P5) = 3.
Then by Remark 6, the ln2-sets of P5 are N1 = {u1, u3, u5} and N2 = {u2, u3, u4} with
u1 ∈ N1 and u1 /∈ N2. Thus, by Remark 5 (ii), fln2(N1) = 1 = fln2(P5). If n = 4, then
by Remark 6, R1 = {u1, u2, u4}, R2 = {u1, u3, u4}, R3 = {u1, u2, u3} and R4 = {u2, u3, u4}
are the ln2-sets of P4. Clearly, none of the singletons and doubletons is a forcing subset
for an ln2-set. Thus, fln2(P4) = 3. Suppose that n = 6. By Remark 6, the ln2-sets of P6

are

S1 = {u1, u2, u4, u6},



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1074

S2 = {u1, u3, u4, u5},
S3 = {u1, u3, u4, u6},
S4 = {u1, u3, u5, u6},
S5 = {u2, u3, u4, u5} and

S6 = {u2, u3, u4, u6}.

It can be verified that {u1, u2} is the forcing subset of S1 and the minimum forcing
subset of P6. Thus, fln2(P6) = 2. Next, suppose that n ≥ 7 and n is odd. By Remark
6, T = {u1, u3, u5, . . . , un−2, un} is the only ln2-set of Pn. Thus, fln2(T ) = fln2(Pn) = 0,
by Remark 5 (i).

Now, suppose that n ≥ 8 and n is even. Then the ln2-sets of Pn are

M1 = {u1, u2, u4, . . . , un−2, un},
M2 = {u1, u3, u4, . . . , un−2, un},
M3 = {u1, u3, u5, . . . , un−2, un−1},
M4 = {u1, u3, u5, . . . , un−2, un},
M5 = {u1, u3, u5, . . . , un−1, un} and

M6 = {u2, u3, u4, . . . , un−2, un}

by Remark 6. Since {u2, u3} ⊆ M6 and not contained in any other ln2-sets of Pn, {u2, u3} is
the forcing subset of M6 and the minimum forcing subset of Pn. Hence, fln2(Pn) = 2.

Remark 7. Let W be a 2-locating set of Cn = [v1, v2, . . . , vn, v1] where n ≥ 3. Then
(i) S ∩W ̸= ∅ for all S ⊆ V (Cn) with ⟨S⟩ = P3.
(ii) If vj , vj+1 ∈ W , then vj+3, vj+5, . . . , vj−2 ∈ W where 1 ≤ j ≤ n and n + k ≡ k

(mod n) for any positive integer k.

Proposition 7. For any cycle Cn with n ≥ 3 vertices,

fln2(Cn) =


0, if n = 3, 4,

1, if n ≥ 6 is even,

2, if n ≥ 7 is odd,

3, if n = 5.

Proof: Suppose that Cn = [u1, u2, . . . , un, u1]. Note that C3 = K3 and by Proposition
5, fln2(C3) = 0. If n = 4, then V (C4) is the only 2-locating set of C4. Thus, by
Remark 5 (i), fln2(C4) = 0. Suppose that n = 5. By Example 1, ln2(C5) = 3. Then
Bi,j = V (C5) \ {vi, vj} for all i, j ∈ {1, 2, . . . , 5} and i ̸= j are the ln2-sets of C5. Thus, for
every vk ∈ Bi,j where k ̸= i, j there exists vi ∈ Bj,k such that [Bi,j \ {vk}] ∪ {vi} = Bj,k is
an ln2-set of C5. Hence, by Theorem 6, fln2(C5) = 3.

Next, suppose n ≥ 6 and n is even. Then by Example 1, ln2(Cn) =
n

2
. Thus, Cn has

ln2-sets D1 = {u1, u3, u5, . . . , un−1} and D2 = {u2, u4, u6, . . . , un}. It can be seen that D1



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1075

is the only ln2-set containing the vertex u1. Thus, by Remark 5 (ii), fln2(Cn) = 1. Now,

suppose that n ≥ 7 and n is odd. By Example 1, ln2(Cn) =
n+ 1

2
. Hence, by Remark 7,

the ln2-sets of Cn is of the form

Si = {ui, ui+1, ui+3, ui+5, . . . , ui−4, ui−2}

where 1 ≤ i ≤ n and n+k ≡ k (mod n) for any positive integer k. Observe that no single
vertex is contained in a unique ln2-set of Cn. Thus, fln2(Cn) > 1. It can be verified that
{ui, ui+1} is uniquely contained in Si. Hence, fln2(Si) = 2 = fln2(Cn).

Remark 8. Let G be a connected graph. Then
(i) fln(2,2)(G) = 0 if and only if G has a unique ln(2,2)-set, and
(ii) fln(2,2)(G) = 1 if and only if G has at least two ln(2,2)-sets, one of which, say B,

that contains an element not in any ln(2,2)-sets of G.

Theorem 7. Let G be a connected graph. Then fln(2,2)(G) = ln(2,2)(G) if and only if
for all (2, 2)-locating set S of G and for each u ∈ S, there exists vu ∈ V (G) \ S such that[
S \ {u}

]
∪ {vu} is an ln(2,2)-set of G.

Proof: Suppose that fln(2,2)(G) = ln(2,2)(G). Let S be an ln(2,2)-set of G such that
fln(2,2)(G) = |S| = ln(2,2)(G) that is, S is the only forcing subset for itself. Let u ∈ S. Since
S \{u} is not a forcing subset for S, there exists a vu ∈ V (G)\S such that

[
S \{u}

]
∪{vu}

is an ln(2,2)-set of G.
Conversely, suppose that every ln(2,2)-set of G satisfies the given condition. Let S

be an ln(2,2)-set of G such that fln(2,2)(G) = fln(2,2)(S). Suppose further that S has
a forcing subset D with |D| < |S|, that is, S = D ∪ I where I = {w ∈ S : w /∈
D}. Pick w ∈ I. By assumption, there exists vw ∈ V (G) \ S such that

[
S \ {w}

]
∪

{vw} = J is an ln(2,2)-set of G. Hence, J = D ∪ T, where T =
[
I \ {w}

]
∪ {vw}, is an

ln(2,2)-set containing D, a contradiction. Hence, S is the only forcing subset for S.
Therefore, fln(2,2)(G) = ln(2,2)(G).

Remark 9. A (2,2)-locating set in G does not exist for some graph G. In particular, if
γ(G) = 1, then G has no (2,2)-locating set.

Proposition 8. For any path Pn with n ≥ 4 vertices,

fln(2,2)(Pn) =


0, if n = 4 and n ≥ 7 is odd,

2, if n ≥ 8 is even,

3, if n = 6,

4, if n = 5.

Proof: Suppose that Pn = [v1, v2, . . . , vn]. If n = 4, then V (P4) is the only ln(2,2)-set
of P4. Thus, by Remark 8 (i), fln(2,2)(P4) = 0. Suppose that n = 5. By Example 2,
ln(2,2)(P5) = 4. Then Q1 = {v1, v2, v3, v4}, Q2 = {v1, v2, v3, v5}, Q3 = {v1, v2, v4, v5},



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1076

Q4 = {v1, v3, v4, v5} and Q5 = {v2, v3, v4, v5} are the ln(2,2)-sets of P5. Clearly, for
every vi ∈ Qj there exists vk ∈ V (P5) \ Qj such that [Qj \ {vi}] ∪ {vk} where
i, j, k ∈ {1, 2, 3, 4, 5} is an ln(2, 2)-set of P5. Thus, by Theorem 7, fln(2,2)(P5) = 4.
Suppose that n = 6. By Example 2, ln(2,2)(P6) = 4. Then the ln(2,2)-sets of P6

are R1 = {v1, v2, v4, v5}, R2 = {v1, v2, v4, v6}, R3 = {v1, v3, v4, v5}, R4 = {v1, v3, v4, v6},
R5 = {v1, v3, v5, v6}, R6 = {v2, v3, v4, v5} and R7 = {v2, v3, v4, v6}. Note that {v3, v5, v6}
is the forcing subset of R5 and the minimum forcing subset of P6. Thus, fln(2,2)(P6) = 3.

Now, suppose that n ≥ 7 and n is odd. By Example 2, ln(2,2)(Pn) =
n+ 1

2
. Then

S = {v1, v3, v5, . . . , vn−2, vn} is the only ln(2, 2)-set of Pn. Thus, by Remark 8 (i),
fln(2,2)(S) = 0 = fln(2,2)(Pn).

Next, suppose that n ≥ 8 and n is even. Then

T1 = {v1, v2, v4, . . . , vn−2, vn},
T2 = {v1, v3, v4, . . . , vn−2, vn},
T3 = {v1, v3, v5, . . . , vn−3, vn−2, vn−1},
T4 = {v1, v3, v5, . . . , vn−3, vn−2, vn},
T5 = {v1, v3, v5, . . . , vn−1, vn} and

T6 = {v2, v3, v4, . . . , vn−2, vn}

are the ln(2, 2)-sets of Pn. Hence, no vertex of Pn is contained in a unique ln(2, 2)-set.
Thus, fln(2,2)(Pn) ≥ 2. It can be seen that {v1, v2}, {vn−2, vn−1}, {vn−1, vn} and {v2, v3}
are uniquely contained in T1, T3, T4 and T5, respectively. Therefore,

fln(2,2)(T1) = fln(2,2)(T3) = fln(2,2)(T4) = fln(2,2)(T5) = 2 = fln(2,2)(Pn).

Proposition 9. For any cycle Cn with n ≥ 4 vertices,

fln(2,2)(Cn) =


0, if n = 4,

1, if n ≥ 8 is even,

2, if n ≥ 7 is odd,

4, if n = 5 and 6.

Proof: Suppose that Cn = [v1, v2, . . . , vn, v1]. If n = 4, then V (C4) is the only
ln(2, 2)-set of C4. Thus, by Remark 8 (i), fln(2,2)(C4) = 0. If n = 5, then the ln(2, 2)-sets
are B1 = {v1, v2, v3, v4}, B2 = {v1, v2, v3, v5}, B3 = {v1, v2, v4, v5}, B4 = {v1, v3, v4, v5}
and B5 = {v2, v3, v4, v5}. Clearly, for each vi ∈ Bj there exists vk ∈ V (C5) \ Bj such
that [Bj \ {vi}] ∪ {vk} where i, j, k ∈ {1, 2, 3, 4, 5} is an ln(2, 2)-set of C5. Thus, by
Theorem 7, fln(2,2)(C5) = 4. Suppose that n = 6. Then Ei,j = V (C6) \ {vi, vj} for all
i, j ∈ {1, 2, . . . , 6} are the ln(2,2)-sets of C6. Thus, for every vk ∈ Ei,j where k ̸= i, j there
exists vi ∈ V (C6) \Ei,j such that [Ei,j \ {vk}] ∪ {vi} = Ej,k is an ln(2,2)-set of C6. Hence,



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1077

by Theorem 7, fln(2,2)(C6) = 4.

Next, suppose that n ≥ 7 and n is odd. By Examples 1 and 2, ln2(Cn) =
n+ 1

2
= ln(2,2)(Cn) . By similar argument as in the proof of Proposition 7, fln(2,2)(Cn) = 2.

Now, suppose that n ≥ 8 and n is even. By Example 2, ln(2,2)(Cn) =
n

2
. Then

F1 = {v1, v3, v5, . . . , vn−3, vn−1} and F2 = {v2, v4, v6, . . . , vn−2, vn} are the only
ln(2, 2)-sets of Cn with v1 ∈ F1 and v1 /∈ F2. Thus, by Remark 8(ii), fln(2,2)(F1)
= 1 = fln(2,2)(Cn).

6. Forcing 2-Metric Dimension in the Join of Graphs

The join of two graphs G and H, denoted by G + H, is the graph with vertex-set
V (G + H) = V (G)∪̇V (H) and edge-set E(G + H) = E(G)∪̇E(H) ∪ {uv : u ∈ V (G),
v ∈ V (H)}.

In view of Theorem 2, we have the following theorem.

Theorem 8. Let G be a connected graph with |V (G)| ≥ 3 and let K1 = ⟨v⟩. Then a
proper subset S of V (K1 + G) is a 2-metric basis of K1 + G if and only if one of the
following holds:

(i) S is an ln(2,2)-set of G,
(ii) S = {v} ∪ T , where T is an ln(2,1)-set of G.

Theorem 9. Let K1 = ⟨v⟩ and G a connected graph with |V (G)| ≥ 3 and
ln(2,2)(G) = ln(2,1)(G). Then

fdim2(K1 +G) =

®
0, if G has a unique ln(2,2)-set ,

f ln(2,2)(G), if G has no unique ln(2,2)-set.

Proof: Suppose that G has a unique ln(2,2)-set, say T . Since ln(2,2)(G) = ln(2,1)(G), by
Theorem 8, T is a unique 2-metric basis for K1+G. By Remark 4 (i), fdim2(K1+G) = 0.
Now, suppose that G has at least two ln(2,2)-sets. Let A be an ln(2,2)-set of G and let F
be a forcing subset for A such that

fln(2,2)(G) = fln(2,2)(A) = |F |.

By Theorem 8, A is a 2-metric basis of K1 +G. Thus,

fdim2(K1 +G) ≤ fln(2,2)(A) = fln(2,2)(G).

Let S0 be a 2-metric basis for K1 + G such that fdim2(K1 + G) = fdim2(S0).
By Theorem 8, S0 is an ln(2,2)-set of G. Let F0 be a forcing subset for S0 with
|F0| = fdim2(S0). Hence,

fdim2(K1 +G) = fdim2(S0) = |F0| ≥ fln(2,2)(S0) ≥ fln(2,2)(G).

Therefore, fdim2(K1 +G) = fln(2,2)(G).



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1078

Example 3. (1.) For the fan Fn = K1 + Pn, where n ≥ 2,

fdim2(Fn) =


0, if n = 2, 3 and n ≥ 7 is odd,

1, if n = 5,

2, if n ≥ 8 is even,

3, if n = 4, 6.

(2.) For the wheel Wn = K1 + Cn, where n ≥ 3,

fdim2(Wn) =


0, if n = 3, 4,

1, if n ≥ 8 is even,

2, if n ≥ 7 is odd,

3, if n = 5, 6.

(3.) For the star Sn = K1 +Kn of order n+ 1,

fdim2(Sn) = 0.

As a consequence of Theorem 3, we have the following results.

Theorem 10. Let G and H be nontrivial connected graphs such that
ln(2,2)(G) = ln(2,1)(G) and ln(2,2)(H) = ln(2,1)(H). A proper subset S of V (G + H) is
a 2-metric basis for G + H if and only if S = SG ∪ SH where SG = V (G) ∩ S and
SH = V (H)∩S are ln2-sets of G and H, respectively such that SG or SH is an ln(2,2)-set.
In particular,

dim2(G+H) = min{ln(2,2)(G) + ln2(H), ln2(G) + ln(2,2)(H)}.

Theorem 11. Let G and H be nontrivial graphs such that (2, 2)-locating set of G and
H do not exist. Then S ⊆ V (G + H) is a 2-metric basis for G + H if and only if
S = SG ∪ SH where SG and SH are ln(2,1)-sets of G and H, respectively. In particular,
dim2(G+H) = ln(2,1)(G) + ln(2,1)(H).

Theorem 12. Let G and H be nontrivial graphs such that ln(2,2)(G) = ln(2,1)(G) and
ln(2,2)(H) = ln(2,1)(H). Then

fdim2(G+H) = min{fln(2,2)(G) + fln2(H), f ln2(G) + fln(2,2)(H)}.

Proof: Suppose that ln(2,2)(G) + ln2(H) < ln2(G) + ln(2,2)(H). By Theorem 10,
dim2(G + H) = ln(2,2)(G) + ln2(H). Let SG be an ln(2,2)-set of G with
fln(2,2)(G) = fln(2,2)(SG) and FG be a forcing subset of SG with
fln(2,2)(SG) = |FG|. Let SH be an ln2-set of H such that fln2(H) = fln2(SH) and
FH be a forcing subset of SH with fln2(SH) = |FH |. By Theorem 10, S = SG ∪ SH is a
2-metric basis of G+H. We claim that FG ∪ FH is a forcing subset of S. Suppose there
exists a 2-metric basis S′ ̸= S of G+H with FG∪FH ⊆ S′. By Theorem 10, S′ = S′

G∪S′
H



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1079

where S′
G is an ln(2,2)-set of G and S′

H is an ln2-set of H. Since FG∪FH ⊆ S′, FG∩FH = ∅
and S′

G ∩ S′
H = ∅, hence, FG is not a forcing subset of SG or FH is not a forcing subset

of SH , a contradiction. Thus, FG ∪ FH is a forcing subset of S. This implies that

fdim2(G+H) ≤ fdim2(S)

≤ |FG ∪ FH |
= |FG|+ |FH |
= fln(2,2)(SG) + fln2(SH)

= fln(2,2)(G) + fln2(H).

Now, let W be a 2-metric basis for G+H with fdim2(G+H) = fdim2(W ) and let F
be a forcing subset of W such that fdim2(W ) = |F |. Then by Theorem 10, W = WG∪WH

where WG is an ln(2,2)-set of G and WH is an ln2-set of H or WG is an ln2-set of G and
WH is an ln(2,2)-set of H. Since F ⊆ W = WG ∪WH , consider the following cases:
Case 1. F ⊆ WG

Then WH is a unique ln2-set of H and F is a forcing subset of WG. For if there exists
ln(2,2)-set W ′

G ̸= WG of G and F ⊆ W ′
G, then F ⊆ W ′ = W ′

G ∪ WH . By Theorem 10,
W ′ ̸= W is a 2-metric basis for G+H, a contradiction.
Case 2. F ⊆ WH

Then WG is a unique ln(2,2)-set of G. By similar argument as in Case 1, F is a forcing
subset of WH .
Case 3. F = A ∪B where A ⊆ WG and B ⊆ WH .

Then A is a forcing subset of WG since if there exists an ln(2,2)-set
W ′

G ̸= WG of G and A ⊆ W ′
G, then W ′ = W ′

G ∪ WH is a 2-metric basis of G + H
with F ⊆ W ′ ̸= W , a contradiction. Similary, B is a forcing subset of WH . Therefore
in any case F = FWG

∪ FWH
where FWG

and FWH
are forcing subsets of WG and WH ,

respectively where it may happen that FWG
or FWH

is empty. Hence,

fdim2(G+H) = fdim2(W )

= |F |
= |FWG

∪ FWH
|

= |FWG
|+ |FWH

|
≥ fln(2,2)(WG) + fln2(WH)

≥ fln(2,2)(G) + fln2(H).

Consequently, fdim2(G+H) = fln(2,2)(G) + fln2(H). Similarly, if
ln2(G) + ln(2,2)(H) < ln(2,2)(G) + ln2(H), then,

fdim2(G+H) = fln2(G) + fln(2,2)(H).

Example 4. Consider the join of two graphs C4 and P7. Note that ln(2,2)(C4) = 4
= ln(2,1)(C4) and ln(2,2)(P7) = 4 = ln(2,1)(P7). Since C4 and P7 both having unique
ln(2,2)-sets, fdim2(C4 + P7) = fln(2,2)(C4) = fln(2,2)(P7) = 0.



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1080

Example 5. Consider the join of two graphs C7 and P6. Then ln(2,2)(C7) = 4
= ln(2,1)(C7) and ln(2,2)(P6) = 3 = ln(2,1)(P6). Since C7 and P6 have no unique
ln(2,2)-sets, fdim2(C7 + P6) = min{fln(2,2)(C7) + fln2(P6), f ln2(C7) + fln(2,2)(P6)}
= min{2 + 2, 2 + 3} = min{4, 5} = 4.

Theorem 13. Let G and H be nontrivial connected graphs where both G and H have
no (2, 2)-locating sets. Then

fdim2(G+H) =


0, if both G and H have

unique ln(2,1)-sets,

f ln(2,1)(G) + fln(2,1)(H), otherwise.

Proof: Suppose that both G and H have unique ln(2,1)-sets. Let SG and SH be the
unique ln(2,1)-sets of G and H, respectively. By Theorem 11, S = SG ∪ SH is the unique
2-metric basis for G + H. By Remark 4 (i), fdim2(G + H) = 0. Next, suppose that
G has no unique ln(2,1)-set. Let SG and SH be ln(2,1)-sets of G and H, respectively.
Thus, by Theorem 11, S = SG ∪ SH is a 2-metric basis for G + H. Let FG and FH be
forcing subsets of SG and SH , respectively such that fln(2,1)(G) = fln(2,1)(SG) = |FG|
and fln(2,1)(H) = fln(2,1)(SH) = |FH |. We claim that FG ∪ FH is a forcing subset of
S = SG ∪ SH . Clearly, FG ∪ FH ⊆ SG ∪ SH = S. Suppose there exists 2-metric basis
S′ ̸= S with FG ∪ FH ⊆ S′ for G + H. By Theorem 11, S′ = S′

G ∪ S′
H where S′

G and
S′
H are ln(2,1)-sets of G and H, respectively. Since S′ ̸= S, SG ̸= S′

G or SH ̸= S′
H . Also,

since FG ∪ FH ⊆ S′, FH ⊆ S′
H and FG ⊆ S′

G. This is a contradiction since FG and FH

are forcing subsets of SG and SH , respectively. Hence, FG ∪ FH is a forcing subset of S.
Thus,

fdim2(G+H) ≤ fdim2(S)

≤ |FG ∪ FH |
= |FG|+ |FH |
= fln(2,1)(G) + fln(2,1)(H).

Next, suppose that S0 is a 2-metric basis for G + H such that
fdim2(G + H) = fdim2(S0). By Theorem 11, S0 = S0

G ∪ S0
H where S0

G and S0
H are

ln(2,1)-sets of G and H, respectively. Let F0 = F 0
G∪F 0

H , where F 0
G ⊆ S0

G and F 0
H ⊆ S0

H , be
a forcing subset of S0 such that fdim2(S0) = |F0|. Suppose F 0

G ⊆ D0
G for some ln(2,1)-set

D0
G of G with D0

G ̸= S0
G. Then S′

0 = D0
G ∪ S0

H is a 2-metric basis for G+H, S′
0 ̸= S0, and

F0 ⊆ S′
0. This contradicts the assumption that F0 is a forcing subset of S0. Thus, F 0

G is
a forcing subset of S0

G. Similarly, F 0
H is a forcing subset of S0

H . Hence,

fdim2(G+H) = fdim2(S0) = |F0| = |F 0
G ∪ F 0

H |
= |F 0

G|+ |F 0
H |

≥ fln(2,1)(S
0
G) + fln(2,1)(S

0
H)

≥ fln(2,1)(G) + fln(2,1)(H).

Therefore, fdim2(G+H) = fln(2,1)(G) + fln(2,1)(H).



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1081

Example 6. Consider the join of two graphs G and H, where (2,2)-locating sets do not
exist for both graphs.

a

b

c

d

f

g

h

i

j
k

G:

H:

e

The join G + H

Note that G has ln(2,1)-sets W1 = {a, c, d, uf}, W2 = {b, c, d, e}, W3 = {b, c, d, f},
W4 = {b, d, e, f} and W5 = {c, d, e, f}. On the other hand, H have ln(2,1)-sets
Z1 = {g, h, j, k}, Z2 = {g, h, i, k} and Z3 = {g, i, j, k}. Clearly, a ∈ W1 and a /∈ W2,
W3,W4,W5. Also, {h, j} ⊆ Z1 and {h, j} ⊈ Z2, Z3. Thus, fln(2,1)(G) = 1 and
fln(2,1)(H) = 2. Hence,

fdim2(G+H) = fln(2,1)(G) + fln(2,1)(H) = 1 + 2 = 3.

7. Forcing 2-Metric Dimension in the Corona of Graphs

The corona of two graphs G and H, denoted by G◦H, is the graph obtained by taking
one copy of G of order n and n copies of H, and then joining every vertex of the ith copy
of H to the ith vertex of G. For v ∈ V (G), denote by Hv the copy of H whose vertices
are attached one by one to the vertex v. Subsequently, denote by v+Hv the subgraph of
the corona G ◦H corresponding to the join ⟨{v}⟩+Hv, v ∈ V (G).

Let S =
⋃

v∈V (G)

Sv ⊆ V (G ◦H) be a 2-resolving set of G ◦H and p, q ∈ V (Hv) where

p ̸= q for v ∈ V (G). Then rG◦H(p/S) and rG◦H(q/S) differ in at least 2-positions. By
Remark 2, rHv(p/Sv) and rHv(q/Sv) must differ in at least two distinct positions. By
definition of G ◦ H, if p, q ∈ V (Hv) \ Sv, then there exist at least two distinct vertices
r, s ∈ V (Hv) ∩ Sv such that either r, s ∈ NHv(p) \ NHv(q) or r, s ∈ NHv(q) \ NHv(p) or
r ∈ NHv(p) \ NHv(q) and s ∈ NHv(q) \ NHv(p). Similarly, if p ∈ V (Hv) \ Sv and
q ∈ Sv, there exists a vertex w ∈ V (Hv) ∩ Sv such that w ∈ NHv(p) \ NHv(q) or
w ∈ NHv(q) \ NHv(p). Hence, Sv is a 2-locating set of Hv. Thus, Theorem 4 is
modified in the next result.

Theorem 14. Let G and H be nontrivial connected graphs. Then S ⊆ V (G ◦ H) is a
2-metric basis for G ◦H if and only if S =

⋃
v∈V (G)

Sv where Sv is an ln2-set of H
v for all

v ∈ V (G). In particular,
dim2(G ◦H) = |V (G)|ln2(H).



D. Managbanag, H. Rara / Eur. J. Pure Appl. Math, 16 (2) (2023), 1068-1083 1082

Theorem 15. Let G and H be nontrivial connected graphs. Then

fdim2(G ◦H) =

®
0, if H has a unique ln2-set ,

|V (G)|fln2(H), if H has no unique ln2-set.

Proof: Suppose H has a unique ln2-set. For each v ∈ V (G), let Mv ⊆ V (Hv) be
the unique ln2-set of H

v. By Theorem 15, S =
⋃

v∈V (G)

Mv is the unique 2-metric basis for

G◦H. Thus, by Remark 4 (i), fdim2(G◦H) = 0. Suppose H does not have unique ln2-set.
For each v ∈ V (G), let Qv ⊆ V (Hv) be an ln2-set of H

v such that fln2(H
v) = fln2(Qv).

Let M(Qv) ⊆ Qv be a forcing subset for Qv with fln2(Qv) = |M(Qv)|. Then by Theorem
15, SQ =

⋃
v∈V (G)

Qv is a 2-metric basis of G ◦ H. Let C =
⋃

v∈V (G)

M(Qv). Then C is a

forcing subset for SQ. Thus,

fdim2(G ◦H) ≤ fdim2(SQ)

≤ |C|
= |V (G)||M(Qv)|
= |V (G)|fln2(Qv)

= |V (G)|fln2(H).

Next, let S′ be a 2-metric basis for G ◦ H such that
fdim2(G ◦H) = fdim2(S

′). Then by Theorem 15, S′ =
⋃

v∈V (G)

Rv where Rv is an ln2-set

of Hv for each v ∈ V (G). Let C ′ be a forcing subset of S′ such that fdim2(S
′) = |C ′|. We

claim that C ′ ∩Rv = Cv is a forcing subset for Rv for all v ∈ V (G). Suppose there exists
w ∈ V (G) such that C ′ ∩ Rw = Cw is not a forcing subset for Rw. Let R′

w be an ln2-set
of Hw with Cw ⊆ R′

w and R′
w ̸= Rw. Then

S′′ =

Ñ ⋃
v∈V (G)\{w}

Rv

é
∪R′

w

is a 2-metric basis for G ◦H with S′ ̸= S′′ and C ′ ⊆ S′′, a contradiction. Thus, Cv is a
forcing subset for Rv for each v ∈ V (G). Let C ′ =

⋃
v∈V (G)

Cv. Then

fdim2(G ◦H) = |C ′| =
∑

v∈V (G)

|Cv| ≥
∑

v∈V (G)

fln2(H
v) = |V (G)|fln2(H).

Therefore, fdim2(G ◦H) = |V (G)|fln2(H).

Example 7. Consider the corona of two graphs P3 and C4. Since C4 has a unique ln2-set,
fdim2(P3 ◦ C4) = |V (P3)|fln2(C4) = 3 · 0 = 0.

Example 8. Consider the corona of two graphs K3 and C5. Since C5 has no unique
ln2-set, fdim2(K3 ◦ C5) = |V (K3)|fln2(C5) = 3 · 3 = 9.



REFERENCES 1083

Acknowledgements

This research is funded by the Department of Science and Technology - Accelerated Sci-
ence and Technology Human Resource Development Program (DOST-ASTHRDP), Philip-
pines.

References

[1] R.F. Bailey and I. Yero. Error-Correcting Codes from k- Resolving Sets. Discussiones
Mathematicae, Graph Theory, 39:341–355, 2019.

[2] S. Bau and A.F. Beardon. The Metric Dimension of Metric Spaces. Comput. Methods
Funct. Theory, 13:295–305, 2013.

[3] L.M. Blumenthal. Theory and Applications of Distance Geometry. Clarendon Press,
Oxford, 1953.

[4] J. Cabaro and H. Rara. On 2- Resolving Sets in Join and Corona of Graphs. European
Journal of Pure and Applied Mathematics, 14(3):773–782, 2021.

[5] F. Harary and R.A. Melter. On the Metric Dimension of a Graph. Ars Combin.,
2:191–195, 1976.

[6] T.P. Zivkovic, F. Harary and D.J. Klein. Graphical Properties of Polyhexes: Perfect
Matching Vector and Forcing. J. Math. Chem., 6:295–306, 1991.

[7] M. Heydarpour and S. Maghsoudi. The Metric Dimension of Geometric Spaces.
Topology Appl., 178:230–235, 2014.

[8] D.J. Klein and M. Randic. Innate Degree of Freedom of a Graph. J. Comput. Chem.,
8:516–521, 1987.

[9] A. Estrada-Moreno, J. Rodrıguez-Velazquez and I. Yero. The k- Metric Dimension
of aGraph. Applied Mathematics and Information Sciences, 9:2829–2840, 2015.

[10] V. Saenpholphat and P. Zhang. On Connected Resolvability of Graphs. Australian
Journal of Combinatorics, 28:25–37, 2003.

[11] P.J. Slater. Leaves of Trees. Congress. Numer., 14:549–559, 1975.

[12] P.J. Slater. Dominating and Reference Sets in Graphs. J. Math. Phys. Sci.,
22:445–455, 1988.

[13] G. Chartrand, P. Zhang and Kalamazoo. The Forcing Dimension of a Graph. Math-
ematica Bohemica, 126(4):711–720, 2001.