EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 3, 2024, 1585-1601
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Differentiating Odd Dominating Sets in Graphs

Mary Ann A. Carbero1∗, Gina A. Malacas1,2, Sergio R. Canoy, Jr.1,2

1 Department of Mathematics and Statistics, College of Science and Mathematics,
MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines
2 Center for Mathematical and Theoretical Physical Sciences, Premier Research Institute
of Science and Mathematics, MSU-Iligan Institute of Technology, 9200 Iligan City,
Philippines

Abstract. Let G = (V (G), E(G)) be a simple and undirected graph. A dominating set
S ⊆ V (G) is called a differentiating odd dominating set if for every vertex v ∈ V (G),
|N [v] ∩ S| ≡ 1(mod 2) and NG[u] ∩ S ̸= NG[v] ∩ S for every two distinct vertices u and
v in V (G). The minimum cardinality of a differentiating odd dominating set of G, denoted
by γo

D(G), is called the differentiating odd domination number. In this paper, we discuss
differentiating odd dominating sets and give bounds or exact values of the differentiating odd
domination numbers of some graphs. We give necessary and sufficient conditions for some graphs
to admit a differentiating odd dominating set. Moreover, we characterize the differentiating odd
dominating sets in graphs resulting from join, corona, and lexicographic product of some graphs
and determine the differentiating odd domination numbers of these graphs.

2020 Mathematics Subject Classifications: 05C69

Key Words and Phrases: Differentiating, domination, odd dominating, differentiating-dominating,
differentiating odd dominating

1. Introduction

Domination is one of the most explored areas in Graph Theory. Indeed, numerous
variations of domination have been introduced and investigated from various perspectives
and approaches (see [1], [2], [5], [7], [10], [13], [14], and [18]). One prominent area of
research in this domain is the investigation of differentiating-dominating sets in graphs,
which are alternatively referred to as identifying codes in certain contexts. This research
has roots dating back to 1998 when Karpovsky, Chakrabarty, and Levitin introduced
identifying codes (see [4]) and have been investigated further by Frick et al in 2008
(see [6]). Furthermore, in the study of Canoy and Malacas [12], they characterized
the differentiating-dominating sets in the join, corona, and lexicographic product of

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5213

Email addresses: maryann.carbero@g.msuiit.edu.ph (M. Carbero),
gina.malacas@g.msuiit.edu.ph (G. Malacas), sergio.canoy@g.msuiit.edu.ph (S. Canoy)

https://www.ejpam.com 1585 © 2024 EJPAM All rights reserved.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1586

graphs and determined the bounds or the exact differentiating-domination numbers of
the aforementioned graphs. Other studies related to the topic can be found in [8], [9],
[11], [15], [16], and [17].

In 1989, Sutner introduced the concept of odd dominating set under the name
“odd-parity cover” (see [19]). Specifically, he showed that every graph contains odd
dominating set in the context of cellular automata (see [19]). However, this parameter
has been studied very little. Previous studies on parity domination mainly focused on
algorithmic problems and even dominating sets [3]. In this paper, we introduce the
concept of differentiating odd dominating sets in graphs. Note that the concept of
differentiating-dominating set may be used to model problems which involve protection
in a given network where the goal is to specifically determine the exact location of an
intruder (e.g. burglar or fire). When used in this case as a protection strategy, an
element of a differentiating-dominating set may refer to a monitoring device or location
(vertex) where a monitoring device is positioned or placed. When, in addition, the num-
ber of these locations or monitors adjacent to a location (with or with no monitoring
device) is required to be odd for every location, then the concept of odd dominating set
is also imposed.

2. Terminologies and Notation

Let G = (V (G), E(G)) be a simple and undirected graph. The open neighborhood of
a vertex v of G is the set NG(v) = {u ∈ V (G) : uv ∈ E(G)} and its closed neighborhood
is the set NG[v] = NG(v) ∪ {v}. The open neighborhood of a subset S of V (G) is the
set NG(S) = ∪v∈SNG(v) and its closed neighborhood is the set NG[S] = NG(S) ∪ S.
Vertex v is a leaf if degG(v) = 1 and the vertex u ∈ (V (G) ∩NG(v)) is called a support
vertex. L(G) and S(G) denote the sets consisting of all leaves and support vertices in
G, respectively. A graph G of order n ≥ 3 is point distinguishing if for any two distinct
vertices u and v of G, NG[u] ̸= NG[v]. It is totally point determining if for any two
distinct vertices u and v of G, NG(u) ̸= NG(v) and NG[u] ̸= NG[v].

A set S ⊆ V (G) is a dominating set (respectively, total dominating set) in G if
NG[S] = V (G) (respectively, NG(S) = V (G)). The smallest cardinality of a dominating
set in G, denoted by γ(G), is called the domination number in G. A dominating set
in G with cardinality γ(G) is called a γ-set of G. A set of vertices S is called an
odd dominating set (respectively, even dominating set) if for every vertex v ∈ V (G),
|NG[v]∩S| ≡ 1(mod 2) (respectively, |NG[v]∩S| ≡ 0(mod 2)). The minimum cardinality
of an odd dominating set is called the odd domination number in G (respectively, even
domination number), denoted by γodd(G) (respectively, γeven(G)). Any odd dominating
set with cardinality γodd(G) is called a γodd-set.

A set S ⊆ V (G) is a differentiating set in a graph G if for every two distinct
vertices u and v in G, NG[u] ∩ S ̸= NG[v] ∩ S. It is a strictly differentiating set if it is
differentiating and NG[u] ∩ S ̸= S for all u ∈ V (G). A differentiating (respectively,
strictly differentiating) subset S of V (G) which is also dominating is called a



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1587

differentiating-dominating (respectively, strictly differentiating-dominating) set in a graph
G. The minimum cardinality of a differentiating-dominating (respectively, strictly
differentiating-dominating) set in G, denoted by γD(G) (respectively, γSD(G)), is called
the differentiating-domination (respectively, strictly differentiating-domination) number
in G. Any differentiating-dominating (respectively, strictly differentiating-dominating)
set with cardinality γD(G) (respectively, γSD(G)) is called a γD-set (respectively, γSD-
set). A set S ⊆ V (G) is a differentiating odd dominating set (respectively, differentiating
even dominating set) if it is both differentiating and odd dominating (respectively, both
differentiating and even dominating). A set S ⊆ V (G) is a strictly differentiating odd
dominating set (respectively, strictly differentiating even dominating set) if it is both
strictly differentiating and odd dominating (respectively, both strictly differentiating
and even dominating). The sets DOD(G) and DED(G) is the set of all differentiating
odd dominating sets and the set of all differentiating even dominating sets, respectively,
in G. The sets SDOD(G) and SDED(G) is the set of all strictly differentiating odd
dominating sets and the set of all strictly differentiating even dominating sets,
respectively, in G. The minimum cardinality of a differentiating odd dominating
(respectively, differentiating even dominating) set in G, denoted by γoD(G) (respectively,
γeD(G)), is called the differentiating odd domination number (respectively,
differentiating even domination number in G. The minimum cardinality of a strictly
differentiating odd dominating (respectively, strictly differentiating even dominating)
set in G, denoted by γoSD(G) (respectively, γeSD(G)), is called the strictly differentiating
odd domination number (respectively, strictly differentiating even domination number)
in G. Any differentiating odd dominating (respectively, strictly differentiating odd
dominating) set with cardinality γoD(G) (respectively, γoSD(G)) is called a γoD-set
(respectively, γoSD-set).

3. Results

Remark 1. Every differentiating odd dominating set in a connected graph G is an odd
dominating set.

Remark 2. Every differentiating odd dominating set in a connected graph G is a
differentiating-dominating set.

Theorem 1. Let G be a graph. Then G admits a differentiating-dominating set if and
only if it is point distinguishing.

Proof. Suppose G admits a differentiating set, say S. Suppose G is not point
distinguishing. Then there exist distinct vertices a, b ∈ V (G) such that NG[a] = NG[b].
This implies that NG[a]∩S = NG[b]∩S, a contradiction. Thus, G is point distinguishing.

For the converse, suppose that G is point distinguishing. Then S = V (G) is a
differentiating-dominating set, showing that G has a differentiating-dominating set.

Lemma 1. Let G be a connected graph of order m and let S be a differentiating
odd dominating set in G. Then m ≤ 2|S|−1. In particular, m ≤ 2γ

o
D(G)−1, i.e.,

γoD(G) ≥ ln(m)+ln(2)
ln(2) .



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1588

Proof. Let S be a differentiating odd dominating set in G and let k = |S|. Let
D = {Q ⊆ S : |Q| is odd}. Then |D| = 2k−1. Since S is differentiating odd dominating,
m ≤ 2k−1. If S is a γoD-set, then m ≤ 2γ

o
D(G)−1. This proves the assertion.

Theorem 2. Let G be a point distinguishing connected graph.

(i) If G has a support vertex v with |NG(v)| = 2, then G does not admit a
differentiating odd dominating set.

(ii) If S is a differentiating odd dominating set in G and v ∈ V (G) with
|NG(v) ∩ L(G)| ≥ 2, then NG(v) ∩ L(G) ⊆ S and v /∈ S.

Proof. Let w ∈ L(G)∩NG(v) and let z ∈ NG(v)\{w}. SupposeG has a differentiating
odd dominating set S. If w ∈ S, then v /∈ S because S is an odd dominating set. Since
S is a differentiating set, NG[w] ∩ S = {w} ̸= NG[v] ∩ S. This forces z ∈ S. However,
the assumption would imply that NG[v] ∩ S = {w, z}, contradicting the fact that S is
an odd dominating set. Thus, w /∈ S. Consequently, v ∈ S. Since S is odd dominating,
z /∈ S. It follows that NG[w] ∩ S = {v} = NG[v] ∩ S, contrary to the assumption that S
is a differentiating set. Therefore, G has no differentiating odd dominating set, showing
that (i) holds.

Next, suppose that S is a differentiating odd dominating set and v ∈ V (G) with
|NG(v)∩L(G)| ≥ 2. Suppose v ∈ S. Since S is odd dominating, (NG(v)∩L(G))∩S = ∅.
Let x, y ∈ NG(v) ∩ L(G) where x ̸= y. Then NG[x] ∩ S = NG[y] ∩ S = {v}, contrary to
the assumption that S is a differentiating set. Therefore, v /∈ S. Since S is a dominating
set, NG(v) ∩ L(G) ⊆ S. Thus, (ii) holds.

The next results follow from the preceding ones.

Corollary 1. For n ≥ 2, Kn and Pn do not admit a differentiating odd dominating set.

Corollary 2. Let Sn = K1,n be a star of order n + 1 where n ≥ 3. Then Sn admits a
differentiating odd dominating set if and only if n is odd. Moreover, if n is odd, then
S = V (Sn) \ {v0} where degG(v0) = n, is the only differentiating odd dominating set in
Sn. In particular, γoD(Sn) = n.

Proof. Let V (Sn) = {v0, v1, · · · , vn}, where degG(v0) = n. Suppose Sn admits a
differentiating odd dominating set, say S. By Theorem 2(ii), S = V (Sn) \ {v0}. Since
S is odd dominating, |NSn [v] ∩ S| = |S| = n is odd.

For the converse, suppose that n is odd. Then S = V (Sn) \ {v0} is a differentiating
odd dominating set in Sn.

Note that if n is odd, then S = V (Sn)\{v0} is the only differentiating odd dominating
set in Sn by Theorem 2(ii). Hence, γoD(Sn) = n.

Corollary 3. Let G be any non-trivial connected graph of order n and let m be a positive
odd integer with m ≥ 3. Then there exists a connected graph H obtained from G such
that γoD(H) = mn. Moreover, if every vertex in G has even degree, then γodd(H) = n.
In particular, the difference γoD(G)− γodd(G) can be made arbitrarily large.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1589

Proof. Let V (G) = {v1, v2, · · · , vn} and let V (Km) = {x1, x2, · · · , xm}. Let H be
the graph obtained from G by adding the edges vixj for each i ∈ {1, 2, · · · , n} and for
each j ∈ {1, 2, · · · ,m}. By Theorem 2(ii), S = V (H) \ V (G) is a γoD-set in H. Thus,
γoD(H) = mn. Cleary, γodd(H) = n.

Suppose now that |NG(v)| is even for every v ∈ V (G). Since V (G) is a minimum
dominating set in H and |NH [x] ∩ V (G)| is odd for every x ∈ V (H), V (G) is a γodd-set
in H. Thus, γodd(H) = n.

Suppose γoD(G) = 1, say S = {v} is a γoD-set in G. If there exists w ∈ V (G) \ {v},
then NG[v] ∩ S = NG[w] ∩ S = {v}, contrary to the fact that S is a differentiating set.
Thus, G = K1. We state this formally.

Remark 3. Let G be a graph. Then γoD(G) = 1 if and only if G = K1.

Remark 4. There exists no connected graph G with γoD(G) = 2.

To see this, suppose that such a connected graphG exists. Then |V (G)| = 2 according
to Lemma 1. Hence, G = K2. This, however, is not possible by Corollary 1.

Theorem 3. Let G be a connected graph of order n ≥ 4. If G admits a differentiating
odd dominating set, then max{γD(G), γodd(G), 3} ≤ γoD(G) ≤ n − |S(G)|. Moreover,
γoD(G) = 3 if and only if G = K1,3.

Proof. By Remarks 1, 2, 3, and 4, max{γD(G), γodd(G), 3} ≤ γoD(G). Next, let S be
a γoD-set in G. Let v ∈ S(G) and let xv ∈ L(G) ∩ NG(v) be fixed. Since S is an odd
dominating set, v ∈ S or xv ∈ S but not both. Let

DG = {w ∈ V (G) \ S : w = v ∈ S(G) or w = xv}.

Then |DG| = |S(G)| and S ⊆ V (G) \DG. It follows that γ
o
D(G) = |S| ≤ n− |S(G)|.

For the second part, suppose that γoD(G) = 3. From Lemma 1, it follows that n = 4.
It can easily be verified that among the connected graphs of order 4, only K1,3 satisfies
the given property. Thus, G = K1,3.

The converse is easy.

Let Sp = K1,p and Sq = K1,q be stars with central vertices (support vertices) v0 and
w0, respectively. Then the double star Sp,q is the graph obtained from Sp and Sq by
adding the edge v0w0.

Corollary 4. Let Tn be a tree of n ≥ 4. If Tn has a differentiating odd dominating set,
then γoD(Tn) ≤ n− |S(Tn)| with equality holding if |NTn(v)∩L(Tn)| is odd and at least 3
for every v ∈ S(Tn). In particular, if Tn = Sp,q (a double star), where p ≥ 3 and q ≥ 3
and are odd, then γoD(Tn) = n− 2 = p+ q.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1590

Proof. Suppose S is a γoD-set in Tn. By Theorem 3, γoD(Tn) ≤ n − |S(Tn)|. Next,
suppose that |NTn(v) ∩ L(Tn)| is odd and at least 3 for every v ∈ S(Tn). By theorem
2(ii), it follows that NTn(v) ∩ L(Tn) ⊆ S and v /∈ S for every v ∈ S(Tn). Therefore,
S = NTn(v)∩L(Tn) and γoD(Tn) = n−|S(Tn)|. From this, it follows that γoD(Tn) = p+q
when Tn = Sp,q.

Theorem 4. γoD(Cn) = n for n ≥ 4.

Proof. Let Cn = [v1, v2, ..., vn, v1] and let S be a γoD-set in Cn. Suppose S ̸= V (Cn).
Then there exists v ∈ V (Cn) \ S. Without loss of generality, we may assume that
v = v1. Since S is odd dominating, v2 ∈ S or vn ∈ S but not both. Assume that
v2 ∈ S. Then vn /∈ S. Since |NCn [v2] ∩ S| must be odd, v3 /∈ S. This implies that
NCn [v1]∩S = NCn [v2]∩S = {v2}, contrary to the assumption that S is a differentiating
set. Therefore, S = V (Cn) and γoD(Cn) = n.

Theorem 5. [6] Let Cn be the cycle on n vertices. For n ≥ 3, γD(C2n) = n.

Corollary 5. Let n be a positive integer such that n ≥ 3. Then there exists a connected
graph G such that γoD(G) − γD(G) = n. In other words, the difference γoD − γD can be
made arbitrarily large.

Proof. Let G = C2n. By Theorem 5, γD(C2n) = n and by Theorem 4,
γoD(C2n) = 2n. Therefore, γoD(G)− γD(G) = n.

Theorem 6. Let G = Kn1,n2,...,nk
be the complete k-partite graph with

2 ≤ n1 ≤ n2 ≤ · · · ≤ nk, where k ≥ 2. Then G admits a differentiating odd
dominating set if and only if

∑
j ̸=t nj is even for every t ∈ {1, 2, . . . , k}. Moreover,

in this case, γoD(G) =
∑k

j=1 nj.

Proof. Let Sn1 , Sn2 , . . . , Snk
be the partite sets in G. Suppose G admits a

differentiating odd dominating set S. Let j ∈ {1, 2, . . . , k} and let v ∈ Snj .
Suppose v /∈ S. Note that since S is odd dominating, |NG[v] ∩ S| = |NG(v) ∩ S| is
odd. Pick any w ∈ Snj \ {v}. Since S is differentiating and NG(w) ∩ S = NG(v) ∩ S, it
follows that w ∈ S and NG[w] ∩ S = {w} ∪ (NG(v) ∩ S). Since |NG(v) ∩ S| is odd,
|NG[w] ∩ S| is even, contrary to the assumption that S is an odd dominating set.
Therefore, Snj ⊆ S for each j ∈ {1, 2, . . . , k}, i.e., S = V (G). Now, let t ∈ {1, 2, . . . , k}
and let a ∈ Snt . Then NG[a]∩S = NG[a] = {a}∪ (∪j ̸=tSnj ). Since S is odd dominating,
|NG[a]| = 1 +

∑
j ̸=t |Snj | = 1 +

∑
j ̸=t nj is odd. This implies that

∑
j ̸=t nj is even.

For the converse, suppose that
∑

j ̸=t nj is even for every t ∈ {1, 2, . . . , k}. Let
D = V (G) and let x, y ∈ V (G) with x ̸= y. Suppose first that x, y ∈ Sr for some
r ∈ {1, 2, . . . , k}. Since y /∈ NG[x], NG[x] ∩ D = NG[x] ̸= NG[y] = NG[y] ∩ D.
Next, suppose that x ∈ Sp and y ∈ Sq for p ̸= q, where p, q ∈ {1, 2, . . . , k}. Since
V (Sq) \ {y} ⊆ NG[x] \ NG[y], NG[x] ∩ D = NG[x] ̸= NG[y] = NG[y] ∩ D. Hence, D
is a differentiating set. Next, let w ∈ V (G) and let w ∈ St. Then, by assumption,



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1591

|NG[w] ∩D| = |NG[w]| = 1 +
∑

j ̸=t nj is odd. Therefore, D = V (G) is a differentiating
odd dominating set in G.

Whenever the given property is satisfied, we find that S = V (G) is the only
differentiating odd dominating set in G. Thus, γoD(G) = |V (G)| =

∑k
j=1 nj .

The next result is immediate from Theorem 6.

Corollary 6. Let Km,n be a complete bipartite graph such that m ≥ 2 and n ≥ 2. Then
Km,n admits a differentiating odd dominating set if and only if m and n are both even.
Moreover, γoD(Km,n) = m+ n.

Theorem 7. Let G1, G2, · · · , Gk be the components of G. Then G admits a
differentiating odd dominating set if and only if Gj admits a differentiating odd
dominating set for each j ∈ {1, 2, . . . , k}. In this case,

γoD(G) =
k∑

j=1

γoD(Gj).

Proof. Suppose G admits a differentiating odd dominating set, say S. Let
Sj = S ∩ V (Gj) for each j ∈ {1, 2, . . . , k}. Since S is dominating, Sj is dominating
in Gj for each j ∈ {1, 2, . . . , k}. Next, let j ∈ {1, 2, . . . , k} and let u, v, w ∈ V (Gj),
where u ̸= v. Since S is differentiating odd dominating,

NGj [u] ∩ Sj = NG[u] ∩ S ̸= NG[v] ∩ S = NGj [v] ∩ Sj

and |NGj [w]∩Sj | is odd. This implies that Sj is a differentiating odd dominating set in
Gj .

For the converse, suppose that each component Gj admits a differentiating odd
dominating set, say Dj . Then, clearly, S

′ = ∪k
j=1Dj is a differentiating odd dominating

set in G.
Now, let S0 be a γoD-set in G. Then S′

j = S0 ∩ V (Gj) is a differentiating odd

dominating set in Gj for each j ∈ {1, 2, . . . , k} and S0 = ∪k
j=1S

′
j . Hence,

γoD(G) = |S0| =
k∑

j=1

|S′
j | ≥

k∑
j=1

γoD(Gj).

On the other hand, ifD′
j is a γ

o
D-set inGj for each j ∈ {1, 2, . . . , k}, then S′

0 = ∪k
j=1D

′
j

is a differentiating odd dominating set in G. It follows that

γoD(G) ≤ |S′
0| =

k∑
j=1

|D′
j | =

k∑
j=1

γoD(Gj).

This proves the desired equality.

Corollary 7. Let G be a graph. Then each of the following holds:



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1592

(i) If G = Kn, then γoD(G) = n.

(ii) γoD(G) = 2 if and only if G = K2.

(iii) γoD(G) = 3 if and only if G ∈ {K3,K1,3}.

The join G+H of two graphs G and H is the graph with V (G+H) = V (G)∪V (H)
(disjoint union) and E(G+H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G) and v ∈ V (H)}.

Theorem 8. Let G be a non-trivial point distinguishing graph and let K1 = ⟨v⟩. Then
S ⊆ V (K1 +G) is a differentiating odd dominating set in K1 +G if and only if one of
the following holds:

(i) S = {v} ∪ SG where SG = V (G) ∩ S ̸= ∅ satisfies the following:

(a) |SG| is even and SG is a strictly differentiating set in G

(b) |NG[u] ∩ SG| is even for all u ∈ V (G)

(ii) S ⊆ V (G), |S| is odd, and S is a strictly differentiating odd dominating set in G.

Proof. Let S be a differentiating odd dominating set in K1 + G. Let V (K1) = {v}
and let SG = V (G) ∩ S. Consider the following cases:

Case 1: v ∈ S.
Then S = {v} ∪ SG. Since S is differentiating in K1 + G, SG ̸= ∅. Since S is odd
dominating in K1+G, |NK1+G[v]∩S| = |{v}|+ |SG| is odd. It follows that |SG| is even.
Suppose SG is not differentiating in G. Then there exist x, y ∈ V (G) and x ̸= y such
that NG[x] ∩ SG = NG[y] ∩ SG. It follows that

NK1+G[x] ∩ S = {v} ∪ (NG[x] ∩ SG) = NK1+G[y] ∩ S,

a contradiction to the fact that S is differentiating in K1 + G. Therefore, SG is
differentiating in G. Furthermore, suppose SG is not strictly differentiating in G. Then
there exists w ∈ V (G) such that NG[w] ∩ SG = SG. It follows that

NK1+G[w] ∩ S = {v} ∪ SG = NK1+G[v] ∩ S,

a contradiction to the fact that S is differentiating in K1 + G. Thus, SG is strictly
differentiating in G. This proves that (a) holds.

Now, let u ∈ V (G). Since S is odd dominating, |NK1+G[u]∩S| = |{v}|+ |NG[u]∩SG|
is odd. Thus, |NG[u] ∩ SG| is even. This shows that (b) holds. Hence, (i) holds.

Case 2: v ̸∈ S.
Then S ⊆ V (G). Since S is odd dominating, |NK1+G[v] ∩ S| = |S| is odd. Since S is
odd dominating and v /∈ S, it follows that S is odd dominating in G. Moreover, as in
Case 1, S is strictly differentiating in G. Therefore, (ii) holds.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1593

Conversely, suppose (i) holds. Let x, y be distinct vertices in V (K1 + G). If
x, y ∈ V (G), then NG[x] ∩ SG ̸= N [y] ∩ SG by (a). It follows that

NK1+G[x] ∩ S = {v} ∪ [NG[x] ∩ SG] ̸= {v} ∪ [NG[y] ∩ SG] = NK1+G[y] ∩ S.

Suppose x = v. Then NG[y] ∩ SG ̸= SG because SG is strictly differentiating. Since
NK1+G[x] ∩ S = SG ∪ {v}, then NK1+G[x] ∩ S ̸= NK1+G[y] ∩ S. Since |SG| is even and
(b) holds, it follows that |NK1+G[z] ∩ S| is odd for all z ∈ V (K1 +G). Therefore, S is a
differentiating odd dominating set in K1 +G.

Next, suppose (ii) holds. Since S is strictly differentiating-dominating
set in G, S is differentiating-dominating in K1 + G. Let w ∈ V (K1 + G).
If w = v, then |NK1+G[w] ∩ S| = |S| is odd, by assumption. If w ∈ V (G), then
|NK1+G[w]∩S| = |NG[w]∩S| is odd because S is odd dominating in G. Therefore, S is
a differentiating odd dominating set in K1 +G.

The graphs G and K1 + G in Figure 1 illustrate the graphs described in Thereom
8(i).

z

d

h

c

g

x

w

u

b

f

y

a

e

G : z

d

h

c

g

x

w

u

b

f

y

a

e

v
K1 +G :

Figure 1: Graphs G and K1 +G illustrating Theorem 8 (i)

In the next results, we use the following parameters for any graph G′ admitting the
given set:

γeoD (G′) = min{|S| : |S| is even and S ∈ DOD(G′)}

γeoSD(G
′) = min{|S| : |S| is even and S ∈ SDOD(G′)}

γooSD(G
′) = min{|S| : |S| is odd and S ∈ SDOD(G′)}

γeeSD(G
′) = min{|S| : |S| is even and S ∈ SDED(G′)}

Corollary 8. Let G be a point distinguishing graph.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1594

(i) If G admits a strictly differentiating odd dominating set with odd cardinality, then
γoD(K1+G) ≤ γooSD(G) and equality holds if G does not admit a differentiating even
dominating set.

(ii) If G admits a strictly differentiating even dominating set with even cardinality, then
γoD(K1 +G) ≤ γeeSD(G) + 1 and equality holds if G does not admit a differentiating
odd dominating set with odd cardinality.

Corollary 9. All fans Fn = K1+Pn of order n+1 have no differentiating odd dominating
set for all n.

Proof. Clearly, F1 and F2 do not admit a differentiating odd dominating set.
Suppose now that n ≥ 3. Let K1 = ⟨v⟩ and G = Pn = [v1, v2, . . . , vn]. Suppose Fn

has a differentiating odd dominating set, say S. Suppose S = {v} ∪ SG, where SG

satisfies (a) and (b) in Theorem 8. If v1 ∈ SG, then v2 ∈ SG by property (b). Hence, by
the same property (b), v3 /∈ SG. This implies that NG[v1]∩SG = NG[v2]∩SG = {v1, v2},
contradicting property (a). This forces v1 /∈ SG. By property (b), v2, v3 /∈ SG. Thus,
NG[v1] ∩ SG = NG[v2] ∩ SG = ∅, contradicting (a). Therefore, S ⊆ V (Pn) and
satisfies (ii) in Theorem 8. If v1 ∈ S, then v2, v3 /∈ S because S is odd dominating
in G. This would imply that SG is not differentiating in G, contradicting (ii). Hence,
v1 /∈ S. Since S is odd dominating in G, v2 ∈ S and v3 /∈ S. This implies that
NG[v1] ∩ S = NG[v2] ∩ S = {v2}, contrary to the assumption that S is differentiating
in G. Therefore, such S does not exist, i.e., Fn does not admit a differentiating odd
dominating set.

Corollary 10. The wheel Wn = K1 +Cn admits a differentiating odd dominating set if
and only if n is odd and n ̸= 3. Moreover, if n is odd and n ≥ 5, then γoD(Wn) = n.

Proof. Suppose Wn admits a differentiating odd dominating set, say S. Since
W3 = K4 is not point distinguishing, n ̸= 3. LetK1 = ⟨v⟩ andG = Cn = [v1, v2, . . . , vn, v1].
Suppose S = {v} ∪ SG, where SG satisfies (a) and (b) in Theorem 8(i). If v1 ∈ SG, then
v2 ∈ SG or vn ∈ SG but not both by (b). We may assume that v2 ∈ SG. Then vn /∈ SG.
Again, by property (b), v3 /∈ SG. It follows that NG[v1] ∩ SG = NG[v2] ∩ SG = {v1, v2},
contradicting property (a). Consequently, v1 /∈ SG. By property (b), v2, v3, vn ∈ SG

and v4 /∈ SG (hence, n ̸= 4). This implies that NG[v2] ∩ SG = NG[v3] ∩ SG = {v2, v3},
contradicting property (a). Therefore, v /∈ S. Thus, S ⊆ V (G) and satisfies (ii) of
Theorem 8. Suppose S ̸= V (G). We may assume that v1 ∈ V (G) \ S. Since S is odd
dominating, v2 ∈ S or vn ∈ S but not both. Assume that v2 ∈ S. Then vn /∈ S. Again,
since S is odd dominating, v3 /∈ S. Therefore, NG[v1]∩S = NG[v2]∩S = {v2}, contrary
to the fact that S is differentiating in G. Thus, S = V (G). Since S is odd dominating
and v /∈ S, |NWn [v] ∩ S| = |V (Cn)| = n is odd.

For the converse suppose that n ≥ 5 and is odd. By Theorem 8, S0 = V (Cn) is a
differentiating odd dominating set in Wn.

As seen earlier, if n is odd and n ≥ 5, then V (Cn) is the only differentiating odd
dominating set in Wn. Accordingly, γ

o
D(Wn) = n.



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1595

The next result is due to Canoy and Malacas [12].

Theorem 9. [12] Let G and H be non-trivial graphs of orders m ≥ 2 and n ≥ 2,
respectively. Then S ⊆ V (G + H) is a differentiating-dominating set in G + H if and
only if SG = V (G) ∩ S and SH = V (H) ∩ S are differentiating sets in G and H,
respectively, and either SG or SH is strictly differentiating.

Theorem 10. Let G and H be non-complete graphs of order m ≥ 4 and n ≥ 4,
respectively. Then S ⊆ V (G + H) is a differentiating odd dominating set in G + H
if and only if S = SG ∪ SH , where SG and SH are differentiating sets in G and H,
respectively, either SG or SH is strictly differentiating, and one of the following
statements holds:

(i) |SG| and |SH | are even, and SG and SH are both odd dominating sets in G and H,
respectively.

(ii) |SG| and |SH | are odd, and SG and SH are both even dominating sets in G and H,
respectively.

(iii) |SG| is odd, |SH | is even, SG is odd dominating in G, SH is even dominating set
in H.

(iv) |SG| is even, |SH | is odd, SG is even dominating in G, and SH is odd dominating
sets in H.

Proof. Let S ⊆ V (G + H) be a differentiating odd dominating set in G + H. Let
SG = V (G) ∩ S and SH = V (H) ∩ S. By Theorem 9, SG and SH are differentiating-
dominating sets in G and H, respectively, and either SG or SH is stricly differentiating.
Now, since S is odd dominating in G + H, |NG+H [x] ∩ S| = |NG[x] ∩ SG| + |SH | and
|NG+H [y]∩S| = |NH [y]∩SH |+ |SG| are odd for every x ∈ V (G) and for every y ∈ V (H).
Hence, if |SG| is even (or |SH | is even), then |NH [y] ∩ SH | is odd (resp. |NG[x] ∩ SG|
is odd). This implies that SH is odd dominating (resp. SG is odd dominating). If |SG|
is odd (or |SH | is odd), then |NH [y] ∩ SH | is even (resp. |NG[x] ∩ SG| is even). Hence,
SH is even dominating (resp. SG is even dominating). Therefore, (i), or (ii), or (iii), or
(iv) holds.

For the converse, suppose that S = SG ∪ SH , where SG and SH are differentiating-
dominating sets in G and H, respectively, and either SG or SH is strictly
differentiating. Then S is a differentiating set in G+H. Now let p ∈ V (G+H). Then
|NG+H [p]∩S| = |NG[p]∩SG|+ |SH | if p ∈ V (G) and |NG+H [p]∩S| = |NH [p]∩SH |+ |SG|
if p ∈ V (H). Hence, if one of (i), (ii), (iii), and (iv) holds, then S is an odd dominating
set in G+H.

Corollary 11. Let G and H be a non-complete graphs of order m ≥ 4 and n ≥ 4,
respectively, such that G +H admits differentiating odd dominating set. If both G and
H do not have an even dominating set and both G and H admit a strictly differentiating
set then

γoD(G+H) = min{γeoD (G) + γeoSD(H), γeoD (H) + γeoSD(G)},



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1596

where we set γeoSD(G
′) = +∞ whenever SDOD(G′) = ∅, where G′ ∈ {G,H}.

Example 1. Let G = Sp,q and H = Sr,t (double stars), where p, q, r, and t are odd
numbers greater than 2. By Corollary 4, γoD(G) = γeoD (G) = γeoSD(G) = p + q and
γoD(H) = γeoD (H) = γeoSD(H) = r + t. By Corollary 11, γoD(G+H) = p+ q + r + t.

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

Theorem 11. [12] Let G (not necessarily point distinguishing) and let H be non-trivial
connected graphs. Then C ⊆ V (G ◦ H) is a differentiating-dominating set in G ◦ H if
and only if for every v ∈ V (G), one of the following is true:

(i) v ∈ C, NG(v) ∩ C ̸= ∅, and C ∩ V (Hv) is a differentiating set in Hv;

(ii) v ∈ C, NG(v) ∩ C = ∅, and C ∩ V (Hv) is a strictly differentiating set in Hv;

(iii) v /∈ C, NG(v) ∩ C ̸= ∅, and C1 = V (Hv) ∩ C is a differentiating-dominating set
in Hv; or

(iv) v /∈ C, NG(v)∩C = ∅, and C1 = V (Hv)∩C is a strictly differentiating-dominating
set in Hv.

Theorem 12. Let G be a non-trivial connected graph and let H be any non-trivial graph
such that G ◦H admits a differentiating odd dominating set. Then S ⊆ V (G ◦H) is a
differentiating odd dominating set in G ◦H if and only if S = SG ∪ [∪v∈V (G)Sv], where
SG ⊆ V (G) and Sv ⊆ V (Hv) for each v ∈ V (G), and satisfies the following conditions:

(i) For each v ∈ SG with |NG(v) ∩ SG| ̸= 0, Sv is differentiating even dominating in
Hv, and either

(a) |NG(v) ∩ SG| and |Sv| are odd or

(b) |NG(v) ∩ SG| and |Sv| are even.

(ii) For each w ∈ V (G)\SG with |NG(w)∩SG| ≠ 0, Sw is differentiating odd dominating
in Hw, and either

(c) |NG(w) ∩ SG| is even and |Sw| is odd or

(d) |NG(w) ∩ SG| is odd and |Sw| is even.

(iii) For each v ∈ V (G) with |NG(v) ∩ SG| = 0, it holds that

(e) Sv is a strictly differentiating even dominating set in Hv and |Sv| is even if
v ∈ SG and



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1597

(f) Sv is strictly differentiating odd dominating and |Sv| is odd if v ∈ V (G) \SG.

Proof. Suppose S is a differentiating odd dominating set in G ◦ H. Let
SG = S ∩V (G) and let Sv = S ∩V (Hv) for each v ∈ V (G). Then S = SG∪ [∪v∈V (G)Sv].
Let v ∈ SG such that |NG(v) ∩ SG| ≠ 0. By Theorem 11, Sv is a differentiating set in
Hv. Suppose first that |NG(v) ∩ SG| is odd. Since S is odd dominating in G ◦ H,
|NG◦H [p] ∩ S| = |NHv [p] ∩ Sv| + |{v}| is odd for every p ∈ V (Hv). This implies
that |NHv [p] ∩ Sv| is even for every p ∈ V (Hv). Thus, Sv is even dominating in Hv.
Moreover, because |NG◦H [v]∩S| = |NG(v)∩SG|+ (|{v}|+ |Sv|) is also odd, |Sv| is odd.
This shows that (i)(a) holds. Similarly, (i)(b) also holds. Therefore, (i) holds.

Next, let w ∈ V (G) \ SG with |NG(w) ∩ SG| ≠ 0. Again, by Theorem 11, Sw is a
differentiating set in Hw. Suppose |NG(w) ∩ SG| is even. Since S is odd dominating in
G ◦H, |NG◦H [q] ∩ S| = |NHw [q] ∩ Sw| is odd for every q ∈ V (Hw). It follows that Sw is
differentiating odd dominating in Hv. Since |NG◦H [w]∩S| = |NG(w)∩SG|+ |Sw| is odd
and |NG(w)∩SG| is even, |Sw| is odd. This shows that (ii)(c) holds. Similar arguments
will show that (ii)(d) holds. Thus, (ii) holds.

Finally, let v ∈ V (G) such that |NG(v) ∩ SG| = 0. Suppose first that v ∈ SG. Then,
by Theorem 8(i), |Sv| is even and Sv is a strictly differentiating even dominating set in
Hv. If v ∈ V (G) \ SG, then Sv is odd and Sv is strictly differentiating odd dominating
in Hv by Theorem 8(ii). Thus, (iii) holds.

For the converse, suppose that S satisfies (i), (ii), and (iii). By Theorem 11, S is a
differentiating-dominating set in G ◦ H. Let x ∈ V (G ◦ H) \ S and v ∈ V (G) be such
that x ∈ V (v +Hv). Suppose |NG(x) ∩ SG| ≠ 0. If v ∈ SG, then

|NG◦H [x] ∩ S| = |NG(x) ∩ SG|+ (|Sv|+ 1) if x = v;

otherwise, |NG◦H [x] ∩ S| = |NHv [x] ∩ Sv| + 1. By (a) and (b), |NG◦H [x] ∩ S|
is odd. Suppose v ∈ V (G) \ SG. Then |NG◦H [x] ∩ S| = |Sv| if x = v; otherwise,
|NG◦H [x] ∩ S| = |NHv [x] ∩ Sv|. By (c) and (d), |NG◦H [x] ∩ S| is odd. Lastly, suppose
that |NG(x) ∩ SG| = 0. Then parts of (e) and (f) would imply that |NG◦H [x] ∩ S| is
odd. Therefore, S is a differentiating odd dominating set in G ◦H.

Corollary 12. Let G be a non-trivial connected graph of order n and let H be a graph
that admits a strictly differentiating odd dominating set with odd cardinality. Then
γoD(G ◦ H) ≤ |V (G)|γooSD(H) and equality holds if H = Kp, where p is odd and at
least 3.

Proof. Let Sv ⊆ V (Hv) be a strictly differentiating odd dominating set with odd
cardinality such that |Sv| = γooSD(H

v) for each v ∈ V (G). Set S = ∪v∈V (G)Sv. By
Theorem 12, S is a differentiating odd dominating set in G ◦ H. Thus,
γoD(G ◦ H) ≤ |S| = nγooSD(H). If H = Kp, then γoSD(H) = p. Desired equality
follows now from Theorem 2(ii).

Corollary 13. Let G be a non-trivial connected graph of order n and let H be a graph
that admits a differentiating even dominating set with even cardinality. If G is r-regular,
where r is a positive even integer, then γoD(G ◦H) ≤ n+ nγeeD (H).



M. Carbero, G. Malacas, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 17 (3) (2024), 1585-1601 1598

The lexicographic product G[H] also has V (G[H]) = V (G) × V (H) as its vertex
set, and u = (u1, u2) is adjacent with v = (v1, v2) whenever u1v1 ∈ E(G) or u1 = v1 and
u2v2 ∈ E(H).

Observe that any subset C of V (G) × V (H) (infact, any set of ordered-pairs) can
be written as C = ∪v∈A({v} × Bv), where S ⊆ V (G) and Bv ⊆ V (H) for each v ∈ S.
Henceforth, we shall use this form to denote any subset C of V (G)V (H).

Theorem 13. [12] Let G (not necessarily point distinguishing) and H be non-trivial
connected graphs. Then C = ∪x∈S({x}×Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each
x ∈ S, is a differentiating-dominating set in G[H] if and only if

(i) S = V (G);

(ii) Tx is a differentiating set in H for every x ∈ V (G);

(iii) Tx or Ty is strictly differentiating in H whenever x and y are adjacent vertices of
G with NG[x] = NG[y]; and

(iv) Tx or Ty is (differentiating) dominating in H whenever x and y are distinct
non-adjacent vertices of G with NG(x) = NG(y).

Theorem 14. Let G be a non-trivial connected graph and let H be a non-trivial point
distinguishing connected graph. Then S = ∪v∈A({v} × Bv), where A ⊆ V (G) and
Bv ⊆ V (H) for each v ∈ A, is a differentiating odd dominating set in G[H] if and
only if

(i) A = V (G);

(ii) Bv is a differentiating set in H for every v ∈ V (G);

(iii) Bv or Bu is strictly differentiating in H whenever u and v are adjacent vertices of
G with NG[u] = NG[v];

(iv) Bv or Bu is differentiating-dominating in H whenever u and v are distinct
non-adjacent vertices of G with NG(u) = NG(v); and

(v) For every v ∈ V (G) and for every p ∈ V (H), |NH [p]∩Bv|+
∑

w∈NG(u) |Bw| is odd.

Proof. Suppose S is a differentiating odd dominating set in G[H]. Then (i), (ii),
(iii) and (iv) hold by Theorem 13. Let v ∈ V (G) and p ∈ V (H). Then

NG[H][(v, p)] = [{v} × (NH [p] ∩Bv)] ∪ [∪w∈NG(v)({w} ×Bw)].

Since S is odd dominating in G[H], |NG[H][(v, p)] ∩ S| = |NH [p] ∩Bv|+
∑

w∈NG(v) |Bw|
is odd, showing that (v) holds.

For the converse, suppose that S satifies the five conditions. Since (i), (ii), (iii),
and (iv) hold, S is a differentiating-dominating set in G[H] by Theorem 13. By (v), it
follows that S is odd dominating.



REFERENCES 1599

Corollary 14. Let G be non-trivial totally point determining graph and let H be a point
distinguishing connected graph. If H admits a differentiating odd dominating set with
even cardinality, then γoD(G[H]) ≤ |V (G)|γeoD (H).

Proof. For each v ∈ V (G), let Bv be a differentiating odd dominating set with
even cardinality such that |Bv| = γeoD (H). Then S = ∪v∈V (G)({v} × Bv) satisfies the
first four properties in Theorem 14. Now let v ∈ V (G) and p ∈ V (H). Since Bv is
differentiating odd dominating in H, |NH [p] ∩ Bv| is odd. Moreover, since |Bw| is
even for every w ∈ V (G), it follows that

∑
w∈NG(v) |Bw| is even. Therefore,

|NH [p] ∩ Bv| +
∑

w∈NG(v) |Bw| is odd, showing that property (v) in Theorem 14 is
also satisfied. Accordingly, S is a differentiating odd dominating set in G[H] and
γoD(G[H]) ≤ |S| = |V (G)|γeoD (H).

Conclusion

The concept of differentiating odd dominating set has been introduced and initially
investigated in this study. The differentiating odd domination number of a graph on
at least four vertices is at least equal to the maximum of the odd domination number,
the differentiating-domination number of the graph and, 3, and at most equal to the
difference of the order of the graph and the number of its support vertices. As shown
in this study, some graphs do not admit this kind of dominating set. The newly defined
concept and parameter have been investigated for the join, corona, and lexicographic
products of some classes of graphs. It may be interesting and worthwhile to find necessary
and sufficient conditions for a graph to admit a differentiating odd dominating set, study
the complexity of the decision problem involving the parameter, and investigate the
parameter for some other families of graphs.

Acknowledgements

The authors would like to express their gratitude to the referees for their
insightful comments and suggestions. Special thanks must go to the Department of
Science and Technology - Accelerated Science and Technology Human Resource
Development Program (DOST-ASTHRDP)-Philippines and MSU-Iligan Institute of
Technology for providing funding for this research.

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