EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 3, 2024, 2084-2091
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Study on Even Sum Domination Number of Some
Graphs

Sejal H Karkar1,∗, D D Pandya2, S G Sonchhatra2, P DMaheta3, H M Rathod3,
S D Bhanderi3

1 Department of Humanities and Science, Lukhdhirji Engineering College, Morbi, Gujarat,
India.
2 Department of Humanities and Science, Government Engineering College, Rajkot, Gujarat,
India.
3 Department of Computer Engineering, Government Engineering College, Rajkot, Gujarat,
India.

Abstract. Let G be a connected graph and uv ∈ E(G). We say, the vertex v even sum dominates
u (u even sum dominates v) if deg(v)+deg(u) is an even number. A set S is an even sum dominating
set (ESDS) if every vertex v ∈ V is either in S or even sum dominated by a vertex in S. An even
sum dominating set S is a mimimal even sum dominating set if no proper subset S

′ ⊂ S is an
even sum dominating set. The even sum domination number γes(G) of a graph G is the minimum
cardinality of an even sum dominating set of G. In this paper, we discuss some properties and
bounds for this concept. We also derive even sum domination number for some standard graphs.

2020 Mathematics Subject Classifications: 05C45, 05C69

Key Words and Phrases: Degree of a vertex, Domination number, Even sum domination
number

1. Introduction

We consider simple, finite, connected and undirected graph G with vertex set V (G) and
edge set E(G). We follow West [9] for all standard terminology and notations while the
terms related to the theory of domination in graphs are used in the sense of Haynes et al.
[4]. We shall give brief summary of definitions which are useful for the present investi-
gations. The domination number is a well studied parameter as observed by Hedetniemi
and Laskar [5]. A set S ⊆ V (G) of vertices in a graph G = (V (G), E(G)) is called a
dominating set if every vertex v ∈ V (G) is either an element of S or is adjacent to an

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5313

Email addresses: sdpansuria@gmail.com (Sejal H. Karkar), ddpandya@gecrajkot.ac.in (D D Pandya),
sgsonchhatra@gecrajkot.ac.in (S G Sonchhatra), pdmaheta@gecrajkot.ac.in (P D Maheta),
hmrathod@gecrajkot.ac.in (H M Rathod), sdbhanderi@gecrajkot.ac.in (S D Bhanderi)

https://www.ejpam.com 2084 © 2024 EJPAM All rights reserved.



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2085

element of S. A dominating set S is a minimal dominating set if no proper subset S
′ ⊂ S

is a dominating set. The domination number γ(G) of a graph G is the minimum cardi-
nality of a dominating set in graph G. This concept is explored and reformed in various
fields to solved many real life problems. Furthermore, a number of broad formulations of
this idea have emerged recently; these definitions, however, rely on certain requirements
that can be applied to the dominating set, outside of it, or both[1, 2, 8]. This paper is
worked on one of the recently introduced parameter known as even sum domination [7].
This definition is dependent on the real-world situations when it is feasible to divide the
set’s elements into two partitions, with each partition being dominated by a same type
of that partition.
The degree of a vertex v in graph G, denoted as d(v) or deg(v), is the number of edges
incident to v, counting each loop twice. For any connected graph G and uv ∈ E(G), the
vertex v even sum dominates u (u even sum dominates v) if deg(v) + deg(u) is an even
number. Precisely, two adjacent vertices of odd degree as well as two adjacent vertices of
even degree can even sum dominate each other. A set S is called even sum dominating
set(ESDS) if every vertex v ∈ V is either an element of S or it is even sum dominated
by some vertex of S. An even sum dominating set S is a mimimal even sum dominating
set if no proper subset S

′ ⊂ S is an even sum dominating set. The even sum domination
number γes(G) of a graph G is the minimum cardinality of an even sum dominating set
of G.

v
1

v
2 v7

v
3

v4
v

6

v5

Figure 1 G

For the graph G given in Figure 1, the even sum dominating set S = {v1, v3, v4, v6}.

2. Main Results

Let G = (V,E) be a connected graph and u, v ∈ V , then we say that u is even sum
neighbor of v if uv ∈ E and deg(u) + deg(v) is an even number. The even sum open
neighborhood Nes(v) of the vertex v is the set of vertices which are even sum neighbors
of v, that is Nes(v)={ u ∈ V (G):uv ∈ E and (deg(u) + deg(v)) is an even number}. We
say a vertex v ∈ V is an even sum isolate of S if Nes(v) ⊆ V − S. The vertex v is said to
be an even sum isolate of V if v has no even sum neighbors in V , that is, there does not
exist the vertex u ∈ V which is adjacent to v in such a way that deg(u) + deg(v) will be



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2086

an even number. For the graph G given in Figure 1, the vertices v1, v3 and v6 are even
sum isolates of S while the vertex v4 is an even sum isolate of V .
Theorem 1. An even sum dominating set S is a minimal even sum dominating set if
and only if for every u ∈ S, one of the following condition holds.
(a) No vertex in S even sum dominates u.
(b) There exists a vertex v ∈ V − S for which Nes(v) ∩ S = {u}.
Proof. Assume that S is a minimal even sum dominating set of G. Then for every vertex
u ∈ S, S − {u} is not an even sum dominating set. This means there is a vertex v in
V − S ∪ {u} is not even sum dominated by any vertex of S − {u}. Now either u = v, in
which case u is not even sum dominated by any vertex of S other than u, or v ∈ V − S.
If v is not even sum dominated by any vertex of S −{u} but even sum dominated by any
vertex of S, then vertex v has only one even sum neighbor u in S, that is Nes(v)∩S = {u}.
Conversely, suppose that S is an even sum dominating set and for every u ∈ S, one of
the two axioms holds. We will try to prove that S is an minimal even sum dominating
set. Suppose that S is not a minimal even sum dominating set, that is, there is at least
one vertex u ∈ S such that S − {u} is an even sum dominating set. Hence, u is even sum
dominated by at least one vertex in S−{u}, that is, there exists a vertex w ∈ S−{u} which
even sum dominates u. Hence, axiom (a) does not hold. Also if S − {u} is a dominating
set then every vertex in V − S is even sum dominated by some vertex of S − {u}, that is,
there exists an another vertex x in S such that x ∈ Nes(v) for some v ∈ V −S and x ̸= u.
Hence, axiom (b) does not hold. So neither axiom (a) nor (b) holds that contradicts our
assumption that one of these two axioms holds.
Theorem 2. If there exists any isolate of V in G then it must be in every minimal even
sum dominating set of G.
Proof. As per the definition an even sum isolate of V , the vertex v is not even sum
dominated by any vertex of V other than itself. Hence, it must be in every minimal even
sum dominating set of G.
Theorem 3.([6]). Every connected graph G of order n ≥ 2 has a dominating set S whose
complement V − S is also a dominating set.

For even sum dominating set there may not exist an even sum dominating set S whose
complement V − S is an even sum dominating set. For graph G, given in Figure 1,
S = {v1, v3, v4, v6} while V − S = {v2, v5, v7} is not an even sum dominating set.
Theorem 4. Let G be a connected graph with even sum dominating set S. If G has no
isolate of S and V then V − S of every minimal even sum dominating set S is an even
sum dominating set.
Proof. As S being any minimal even sum dominating set of G, every vertex v ∈ V − S
is even sum dominated by atleast one vertex of S. So, by the definition of even sum
domination, ∀v ∈ V − S can also even sum dominate some u ∈ S. Assume that V − S is
not an even sum dominating set. Therefore, there exists at least one vertex w ∈ S which
is not even sum dominated by any vertex of V − S. So, w is even sum dominated by only
some vertex of S that is, S − {w} is also an even sum dominating set. In this case S is
not minimal even sum dominating set of G which contradicts to our assumption that S
is a minimal even sum dominating set of G. Hence, V − S of every minimal even sum



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2087

dominating set S is an even sum dominating set.
Theorem 5. 1 ≤ γes(G) ≤ n
Proof. Let G be a simple graph and |G| is an odd number. As per the definition of an
even sum dominating set if G has atleast one vertex of n− 1 degree then γes(G) = 1 and
if the graph has all the vertices of odd degree than γes(G) = n.

Here, Star graph K1, n achieves lower bound for any odd number n while it achieves
the upper bounds for even number n. The graph containing the property given in below
theorem also achieves the upper bound.
Theorem 6. Let G be the graph in which every vertex of even degree is adjacent to the
vertices of odd degree only and every vertex of odd degree is adjacent to the vertices of
even degree only then γes(G) = |V (G)|.
Proof. Here sum of degree of any two adjacent vertices of G will be an odd number. So any
vertex of G can not even sum dominate any other vertex of G. Therefore to construct an
even sum dominating set of minimum cardinality we include all the vertices of G. Hence,
γes(G) = |V (G)|.
Theorem 7 ([3]). A connected graph G is Euler if and only if the degree of each vertex
is even.
Theorem 8. If G is an Euler graph then γes(G) = γ(G).
Proof. Let G is an Euler graph then by Theorem 7, the degree of each vertex is even.
So, any two vertices of G can even sum dominate each other if they are adjacent to each
other. Hence, γes(G) = γ(G).

Corollary 1. γes(Cn) =
⌈n
3

⌉
.

Proof. As the cycle Cn is an Euler graph, according to Theorem 8, γes(Cn) = γ(Cn) =
⌈n
3

⌉
.

3. Even sum domination number of some standard graphs

Theorem 1. γes(Pn) = 2 +

⌈
n− 2

3

⌉
.

Proof. Let v1, v2, v3, . . . , vn be the n vertices of Pn where v1 and vn are pendant vertices.
So degree of v1 and vn is 1 while degree of v2,...,vn−1 is 2. Therefore v1 and vn can even
sum dominate themselves only and they are not even sum dominated by their neighbors.
So, they must be in even sum dominating set S. Now as remaining vertices v2,...,vn−1

having even degree they even sum dominate their selves and their neighbors. So to even

sum dominate remaining n − 2 vertices we need

⌈
n− 2

3

⌉
vertices. Therefore, we require

2 +

⌈
n− 2

3

⌉
vertices to even sum dominate all the vertices of Pn.

Theorem 2. γes(Km,n) =


2, if m+ n is an even number ,

where m, n > 1 .

|V (Km,n)|, if m+ n is an odd number.



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2088

Proof. Let V1 and V2 be two subsets of Km,n such that V1 ∪ V2 = V (Km,n) where |V1| = m
and |V2| = n and v1,v2,...,vm are vertices of V1 and u1,u2,...,un are vertices of V2. Now
according to the values of m and n we consider two cases as below.
Case 1: m+ n is an even number.
That is either m and n both are even number or both are odd number. According to that
we consider two subcases as given below.
Subcase 1.1:If m and n both are even number.
If m and n both are even number then v1,v2,...,vm,u1,u2,...,un are vertices of even degree.
According to the definition of even sum domination every vi, i = 1, 2, ...,m can even sum
dominate itself and all the vertices of V2. Similarly every uj , j = 1, 2, ..., n can even sum
dominate itself and all the vertices of V1. Therefore, it is enough to consider one vertex
from V1 and one vertex from V2 to even sum dominate all the vertices of Km,n.
Subcase 1.2:If m and n both are odd number.
If m and n both are odd number then deg(vi)+deg(uj) is even number for 1 ≤ i, j ≤ m,n.
So from the definition of even sum domination every vi, i = 1, 2, ...,m can even sum
dominate itself and all the vertices of V2. Similarly every uj , j = 1, 2, ..., n can even sum
dominate itself and all the vertices of V1. Therefore, it is enough to consider one vertex
from V1 and one vertex from V2 to even sum dominate all the vertices of Km,n. Hence,
from the above both subcases γes(Km,n) = 2.
Case 2: m+ n is an odd number.
If m+n is an odd number then either m is an odd number and n is an even number or n is
an odd number andm is an even number. Therefore, in both the cases deg(vi)+deg(uj) will
be an odd number for 1 ≤ i, j ≤ m,n. So from the definition of even sum domination every
vi, i = 1, 2, ...,m can even sum dominate itself only and similarly for every uj , j = 1, 2, ..., n
can even sum dominate itself only. Therefore to even sum dominate all the vertices ofKm,n

we must include all the vertices of Km,n. Hence, γes(Km,n) = m+ n = |V (Km,n)|.
Definition 1. The middle graph M(G) of a graph G is the graph whose vertex set is
V (G) ∪ E(G) and in which two vertices are adjacent whenever either they are adjacent
edges of G or one is a vertex of G and the other is an edge incident with it.

Theorem 3. γes(M(Pn)) = 4 +

⌈
n− 6

2

⌉
.

Proof. Let v1, v2, . . . , vn be the vertices and e1, e2, . . . , en−1 be the edges of path Pn.
Then V (M(Pn)) = {v1, v2, . . . , vn, e1, e2, . . . , en−1} and let V (M(Pn)) = V1 ∪ V2 where
V1 = {v1, v2, . . . , vn} and V2 = {e1, e2, . . . , en−1}. Here, d(v1) = d(vn) = 1, d(vi) = 2 for
i = 2, 3, . . . , n − 1, d(e1) = 2 for n = 2, d(e1) = d(e2) = 3 for n = 3, d(e1) = d(en−1) = 3
for i = 3, 4, . . . , n and d(ei) = 4 for i = 2, 3, . . . , n − 2. As per the definition of even sum
domination the vertex v1 and the vertex e1 can even sum dominate each other and the
vertex vn and the vertex en−1 can even sum dominate each other. So, let’s consider e1 and
en−1 in even sum dominating set S. Now v2 and vn−1 can even sum dominate e2 and en−2

respectively other than themselves while e2 and en−2 can even sum dominate their three
neighbor vertices other than themselves. Therefore e2, en−2 ∈ S. So after considering
e1, en−1, e2 and en−2 in S, in total six vertices v1, v2, v3, vn, vn−1, vn−2 from V1 and six
vertices e1, e2, e3, en−1, en−2, en−3 from V2 will be even sum dominated. Now to even sum



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2089

dominate remaining n − 6 vertices from V1 and n − 7 vertices from V2 it is enough to

consider

⌈
n− 6

2

⌉
from V2. Thus, e1, en−1, e2, en−2 and

⌈
n− 6

2

⌉
vertices from V2 even

sum dominate all the vertices of M(Pn).

Theorem 4. γes(M(Cn)) =


n

2
, if n is an even number,

n+ 1

2
, if n is an odd number.

Proof. Let v1, v2, v3, . . . , vn be the vertices and e1, e2, e3, . . . , en be the edges of cycle Cn.
Then V (M(Cn)) = {v1, v2, . . . , vn, e1, e2, . . . , en}. Here, d(vi) = 4 for 1 ≤ i ≤ n, d(ei) = 2
for 1 ≤ i ≤ n. Therefore, every vi even sum dominates four vertices other than itself while
every ei even sum dominates two vertices other than itself. If n is an even number then in
order to form an even sum dominating set of minimum cardinality it is enough to consider

either
n

2
vertices, either v1, v3, v5, ..., vn−1 or v2, v4, v6, ..., vn from M(Cn). Now if n is an

odd number then to form an even sum dominating set of minimum cardinality it is enough

to consider
n+ 1

2
vertices, which may be v1, v3, ..., vn or v2, v4, ..., vn−1, vn from M(Cn).

Definition 2. Let G be a graph with V (G) = S1∪S2∪ . . . St∪T where Si is the set having
at least two vertices of same degree and T = V (G)−∪Si where i = 1, 2, . . . , t. The degree
splitting graph DS(G) is obtained from G by adding vertices w1, w2, . . . , wt and joining
wi to each vertex of Si for 1 ≤ i ≤ t.

Theorem 5. γes(DS(Pn)) =


2, if n is an odd number ,

2 +

⌈
n− 2

3

⌉
, if n is an even number.

Proof. The path Pn has two pendant vertices and the remaining n − 2 vertices are of
degree 2. Thus, V (Pn) = {vi; 1 ≤ i ≤ n} = S1 ∪ S2 where S1 = {v1, vn} and S2 = {vi; 2 ≤
i ≤ n− 1}. To obtain DS(Pn) from Pn, add two vertices w1 and w2 corresponding to S1

and S2 respectively. Thus, V (DS(Pn)) = V (Pn)∪{w1, w2} and |V (DS(Pn))| = n+2. We
shall consider two cases according to values of n.
Case 1: n is an odd number.
According to the definition of even sum domination the vertex w1 even sum dominates
both the vertices v1, vn and itself also. Therefore, by considering w1 in S, all the vertices
of S1 will be even sum dominated. So, w1 ∈ S. Now w2 can even sum dominate every
vi from S2 while every vi from S2 can even sum dominate only its three neighbors other
than itself. Therefore, we also consider w2 in S. Hence S = {w1, w2} will be an even sum
dominating set of minimum cardinality. Therefore, γes(DS(Pn)) = 2.
Case 2: n is an even number.
According to the definition of even sum domination the vertex w1 even sum dominates
both the vertices v1, vn and itself also. Therefore, if we consider w1 in S then all the
vertices of S1 are even sum dominated. In S2, deg(vi) = 3 which is odd number while
deg(w2) = n − 2 which is even number w2 even sum dominates itself only and every vi
can even sum dominates only its neighbor from S2 and itself. So, from n − 2 vertices of



S. H Karkar et al. / Eur. J. Pure Appl. Math, 17 (3) (2024), 2084-2091 2090

S2 it is enough to consider

⌈
n− 2

3

⌉
vertices from S2. Therefore, S forms an even sum

dominating set of minimum cardinality. Hence, γes(DS(Pn)) = 2 +

⌈
n− 2

3

⌉
.

Definition 3. The wheel Wn is defined to be the join K1 + Cn. The vertex correspond-
ing to K1 is known as the apex and the vertices corresponding to cycle are known as rim
vertices while the edges corresponding to cycle are known as rim edges.

Theorem 6. γes(Wn) =


1, if n is an odd number ,

γes(Cn) + 1, if n is an even number.
Proof. Consider the wheel Wn with the vertices v1, v2, . . . , vn as its rim vertices and v0 as
its apex vertex. According to value of n we consider following two cases.
Case 1: n is an odd number.
In this case, deg(vi) = 3, 1 ≤ i ≤ n and deg(v0) = n, which is an odd number. So,
deg(vi) + deg(v0) will be an even number. Therefore, the vertex v0 even sum dominates
every rim vertex. Thus it is enough to consider an apex vertex v0 in even sum dominating
set S. Hence, γes(Wn) = 1.
Case 2: n is an even number.
In this case, deg(vi) = 3, 1 ≤ i ≤ n and deg(v0) = n, which is an even number. The apex
vertex v0 does not even sum dominate any rim vertex and it is not even sum dominated
by any rim vertex also. So the vertex v0 must be in even sum dominating set S while each

rim vertex can even sum dominate its both neighbors likewise cycle. Thus we need
⌈n
3

⌉
vertices to even sum dominate all the rim vertices. Therefore, it is enough to consider⌈n
3

⌉
+1 vertices to construct the even sum dominating set of minimum cardinality. Hence,

γes(Wn) = γes(Cn) + 1.

4. Concluding Remarks

We have derived some basic results on even sum domination number. In different sport
centers of any sport academy even sum domination can be used to schedule match with-
out wasting time of players and using minimum resources for some particular game like
Carrom, Tennis, Badminton in which we need even number of players. We have discussed
some properties and boundaries for this concept. We have also discussed even sum domi-
nation number for some standard graphs. In future research, It may be fascinating to find
even sum domination number of some graphs obtain by binary operations and compare
various domination models with even sum domination.

Acknowledgements

The authors thank the reviewers and editors of European Journal of Pure and Applied
Mathematics, for reviewing the paper and for the comments and suggestions they provided.



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