EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 7078
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Super Hop Roman Domination in Graphs
Leomarich F. Casinillo1, 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. LetG = (V (G), E(G)) be a simple undirected graph. A function f : V (G) → {0, 1, 2} is
a super hop Roman dominating function (SHRDF) on G if for every v ∈ V (G) with f(v) = 0, there
exist w, u ∈ V (G) with f(w) = 2 and f(u) ̸= 0 such that dG(v, w) = 2, and N2

G(u) ∩ {x ∈ V (G) :
f(x) = 0} = {v}. The weight of SHRDF f , denoted ωshR

G (f), is given by ωshR
G (f) =

∑
y∈V (G) f(y).

The super hop Roman domination number of a graph G, denoted γshR(G), is the minimum weight
of an SHRDF on G, that is, γshR(G) = min{ωshR

G (f) : f is an SHRDF on G}. In this paper, we
make an initial investigation of this newly defined variation of hop Roman domination in graphs.
Some bounds and exact values of the parameter are obtained and some characterizations on some
classes of graphs are given.
2020 Mathematics Subject Classifications: 05C69
Key Words and Phrases: Super domination, hop Roman domination, super hop Roman domi-
nation

1. Introduction

Domination is one of the major concepts in graph theory that is rigorously studied
by several discrete mathematicians due to its interesting theoretic structures [1], [2], [3],
[4], [5], [6], [7], [8], [9]. Roman dominating function is one of the topics in the theory
of domination that remains intriguing and have been a center of mathematics research.
Roman domination was pioneered by Cockayne et al. [5] in 2004 which is based on the
defence strategy of Roman Emperor Constantine the great around the fourth century A.D.
Currently, there are now several variations of Roman domination that has been published
in the literature of graph theory and can be found in [10], [11], [12]. In the year 2017,
Shabani et al. [13] formally introduced the hop Roman domination in graphs which is
extensively studied recently. In addition, super dominating sets in graphs initiated by

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.7078
Email addresses: leomarich.casinillo@g.msuiit.edu.ph (L. F. Casinillo),
sergio.canoy@g.msuiit.edu.ph (S. R. Canoy Jr.)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 2 of 15

Lemańska et al. [14] also captures the attention of many graph theorists. Motivated by
hop Roman domination and super domination, the author introduced a new parameter
called super hop Roman domination and investigated its mathematical properties.

Let G = (V (G), E(G)) be a simple, undirected and finite graph where V (G) is the
vertex set and E(G) is the edge set of G. The cardinality of V (G) denoted by |V (G)|
is called the order of G and the cardinality of E(G) denoted by |E(G)| is called the
size of G. The complement of a graph G denoted by G is the graph that satisfies the
following conditions: (i) V (G) = V (G); and (ii) uv ∈ E(G) if and only if uv /∈ E(G).
All needed basic concepts and terminologies used in this study which are not define are
found in [15], [16], [6]. Let x ∈ V (G). Then the open neighborhood of x in G is the set
NG(x) = {y ∈ V (G) : xy ∈ E(G)} and the closed neighborhood of a vertex x ∈ V (G) is
the set NG[x] = NG(x)∪ {x}. Let O ⊆ V (G). Then, the set NG(O) = N(O) =

⋃
v∈O

NG(v)

is called the open neighborhood of O and the set NG[O] = N [O] = N(X) ∪ X is called
the closed neighborhood of O. Let u and v be two distinct vertices in graph G. Then, the
distance between u and v denoted by dG(u, v) is the length of the shortest walk between
u and v in G. If there is no such walk between u and v in G, then we define the distance
as dG(u, v) = ∞. Now, let v ∈ V (G). Then, the set N2

G(v) = {u ∈ V (G) : degG(u, v) = 2}
is called the open hop-neighborhood and each element of N2

G(v) is called hop-neigbor of
vertex v. Moreover, for H ⊆ V (G), N2

G(H) =
⋃
v∈H

N2
G(v) and N2

G[H] = N2
G(H) ∪H.

A subset D of vertex set V (G) is a dominating set of G if for every v ∈ V (G) \ D,
there exists u ∈ D such that uv is an edge of G [6]. In that case, N [D] = V (G). The
domination number denoted by γ(G) is the minimum cardinality of a dominating set D in
G. If D is a dominating set with |D| = γ(G), then we call D a minimum dominating set
of G or a γ-set in G. A dominating set S ⊆ V (G) is called a super dominating set of G if
for every vertex u ∈ V (G)\S, there exists v ∈ S such that N(v)∩ (V (G)\S) = {u} [8]. In
that case, v is a private neighbor of u with respect to V (G) \ S. The smallest cardinality
of a super dominating set of G is called the super domination number denoted by γsp(G).
A super dominating set of cardinality γsp(G) is called γsp-set in G.

A set S ⊆ V (G) is called a hop dominating set of G if for every vertex in v ∈ V (G)\S,
there exists u ∈ S such that dG(u, v) = 2 [17]. The smallest cardinality of a hop dominating
set in G, denoted γh(G), is called the hop domination number of G. A hop dominating
set of cardinality γh(G) is called a γh-set in G. A hop dominating set S ⊆ V (G) is called
super hop dominating if for every vertex v ∈ V (G) \ S, there exists u ∈ D such that
N2

G(u) ∩ (V (G) \ S) = {v} [18]. The smallest cardinality of a super hop dominating set
in G, denoted γsh(G), is called the super hop domination number of G. A super hop
dominating set of cardinality γsh(G) is called a γsh-set in G. Hop domination and some
of its variants have beed studied previously in [18], [19], [20], [21], [22], and [23].

Let f : V (G) → {0, 1, 2} be a function on G. Let the sets V0, V1, V2 be given as follows:

V0 = {v ∈ V (G) : f(v) = 0};
V1 = {v ∈ V (G) : f(v) = 1}; and



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 3 of 15

V2 = {v ∈ V (G) : f(v) = 2}.

In this case, we may denote f by f = (V0, V1, V2). A function f = (V0, V1, V2) is a
hop Roman dominating function (HRDF) on G if for every v ∈ V0, there exists u ∈ V2
such that dG(u, v) = 2. The weight of f is given by ωhR

G (f) =
∑

v∈V (G) f(v). The hop
Roman domination number of G, denoted γhR(G), is the minimum weight of an HRDF
on G, that is, γhR(G) = min{ωhR

G (f) : f is an HRDF on G}. Any HRDF f on G with
ωhR
G (f) = γhR(G) is called a γhR-function on G.

A function f = (V0, V1, V2) is a super hop Roman dominating function (SHRDF) on G
if it satisfies the following conditions:

(SHR1) f is a hop Roman dominating function on G; and

(SHR2) for each v ∈ V0, there exists w ∈ V1 ∪ V2 such that N2
G(w) ∩ V0 = {v}.

The weight ωshR
G (f) of an SHRDF f is given by ωshR

G (f) =
∑

u∈V (G) f(u), that is,
ωshR
G (f) = |V1|+2|V2|. The super hop Roman domination number ofG, denoted γshR(G), is

the minimum weight of an SHRDF onG, that is, γshR(G) = min{ωshR
G (f) : f is an SHRDF on G}.

Any SHRDF f on G with ωshR
G (f) = γshR(G) is called a γshR-function on G.

Consider the graph G with |V (G)| = 10 in Figure 1 below. Let f = (V0, V1, V2) be
a function on G such that V0 = {v5, v6, v8}, V1 = {v1, v2, v4, v7, v9, v10}, and V2 = {v3}.
Observe that f is a γshR-function on G. Hence, γshR(G) = 8.

1
v1

1v2 2
v3

1
v4

0
v5

0
v6

1
v7

0
v8 1

v9

1
v10

Figure 1: A graph G with γshR(G) = 8.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 4 of 15

In this paper, we do an initial investigation of super hop Roman domination in graphs.

2. Known Results

We shall need the results obtained by Canoy et al. in [18].

Theorem 1. Let G be a graph of order n ≥ 1. Then ⌈n2 ⌉ ≤ γsh(G). In particular,
n ≤ 2γsh(G).

Theorem 2. Let G be a graph of order n ≥ 1. Then γsh(G) = n if and only if each
component G′ of G is a complete graph.

Corollary 1. Let G be a connected graph of order n. Then γsh(G) = n if and only if
G = Kn.

3. Results

This section explores the properties of the super hop Roman dominating function in
graphs.

Proposition 1. Let G be a graph of order n ≥ 1 and f = (V0, V1, V2) be an SHRDF on
G. Then V1 ∪ V2 is a super hop dominating set on G. Moreover, if f is a γshR-function
on G, then the following statements hold:

(i) V0 = ∅ if and only if V2 = ∅. In this case, γshR(G) = |V (G)| = n; and

(ii) If V1 = ∅, then V2 is a γsh-set and γshR(G) = 2γsh(G) = n.

Proof. Assume that f = (V0, V1, V2) is an SHRDF on G and let v ∈ V (G) \ (V1 ∪ V2).
Then v ∈ V0. Since f satisfies (SHR2), there exists w ∈ V1∪V2 such that N2

G(w)∩V0 = {v}.
This implies that V1 ∪ V2 is a super hop dominating set on G.

Now, suppose f is a γshR-function on G. Let V0 = ∅. Assume for a moment that
V2 ̸= ∅, say w ∈ V2. Let W0 = V0, W1 = V1 ∪ {w}, and W2 = V2 \ {w}. Then
g = (W0,W1,W2) is a super hop Roman dominating function on G. It follows that

ωshR
G (g) = |W1|+ 2|W2|

= (|V1|+ 1) + 2(|V2| − 1)

= |V1|+ 2|V2| − 1

< ωshR
G (f)

= γshR(G),

a contradiction to the assumption that f is a γshR-function on G. Thus, V2 = ∅.
Conversely, suppose that |V2| = 0. Since f satisfies (SHR1), the assumption that

|V2| = 0 forces |V0| = 0. This, in turn, implies that |V1| = n. Thus, γshR(G) = |V1|+2|V2| =
|V1| = |V (G)| = n, showing that (i) holds.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 5 of 15

Next, suppose V1 = ∅. Then V1 ∪ V2 = V2 is a super hop dominating set. Suppose
V2 is not a γsh-set in G. Let V ′

2 be a γsh-set in G. Then, we obtain |V ′
2 | < |V2|. Define a

function g = (W0,W1,W2) on G where W0 = V (G) \ V ′
2 , W1 = ∅ and W2 = V ′

2 . Then g
is an SHRDF on G and ωshR

G (g) = 2|W2| < 2|V2| = γshR(G), a contradiction. Therefore
V2 is a γsh-set on G and γshR(G) = 2|V2| = 2γsh(G). By Theorem 1, γshR(G) = n. This
shows that (ii) holds.

Lemma 1. Let G be a graph of order n and let f = (V0, V1, V2) be a γshR-function on G.
Then |V2| ≤ |V0| and |V0| ≤ |V1|+ |V2|. Moreover, each of the following statements hold:

(i) If |V0| = |V2|, then γshR(G) = n.

(ii) If γshR(G) < n, then 1 ≤ |V1| < n.

Proof. If |V0| = 0, then |V2| = 0 by Proposition 1. Hence, |V0| = |V2|. So sup-
pose |V0| ̸= 0. Since f satisfies (SHR1), it follows that for each v ∈ V0, there exists
zv ∈ V2 ∩ N2

G(v). Hence, the assignment v → zv defines a function ψ from V0 into V2.
Since f is a γshR-function on G, ψ must be onto. Thus, |V2| ≤ |V0|.

Now, by Proposition 1(i), V1 ∪ V2 is a super hop dominating set. Hence, for every
v ∈ V0, there exists wv ∈ V1 ∪ V2 such that N2

G(wv) ∩ V0 = {v}. Define the function
h : V0 → {wv : v ∈ V0} by h(v) = wv for each v ∈ V0. Then h is a one-one and onto
function. Thus, |V0| = |{wv : v ∈ V0}|. Since {wv : v ∈ V0} ⊆ V1 ∪ V2, it follows that
|V0| ≤ |V1 ∪ V2|.

Next, if |V0| = |V2|, then we have γshR(G) = |V1|+2|V2| = |V1|+ |V2|+ |V0| = |V (G)| =
n. This shows that (i) holds.

Finally, suppose that γshR(G) < n. Then |V0| ̸= |V2| by (the contrapostive of) (i).
Assume that |V1| = 0. Then V2 is a γsh-set on G and γshR(G) = 2γsh(G) = n by
Proposition 1. This contradicts the assumption that γshR(G) < n. Therefore |V1| ≥ 1,
showing that (ii) holds. This proves the assertion.

Proposition 2. Let G be a graph of order n and let f = (V0, V1, V2) be a γshR-function
on G. Then each of the following statements holds:

(i) γshR(G) < n if and only if 1 ≤ |V2| < |V0|.

(ii) γshR(G) = n if and only if |V0| = |V2|.

Proof. (i) Suppose γshR(G) < n. By Lemma 1, and property (SHR1), we have
1 ≤ |V2| < |V0|.

For the converse, suppose that 1 ≤ |V2| < |V0|. Then

γshR(G) = ωshR
G (f) = |V1|+ 2|V2|

< |V1|+ |V2|+ |V0|
= n.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 6 of 15

(ii) Suppose γshR(G) = n. Assume for a moment that |V0| ̸= |V2|. By Proposition
1(i), |V0| ̸= 0 and |V2| ̸= 0. Lemma 1 would now imply that 1 ≤ |V2| < |V0|. This implies
that γshR(G) < n by (i), a contradiction to our assumption. Therefore, |V0| = |V2|.

The converse follows from Lemma 1(i).

Theorem 3. Let G be a graph of order n and let f = (V0, V1, V2) be an SHRDF on G.
Then, V1∪V2 is a minimal super hop dominating set of G if and only if each u ∈ V2, there
exists a vertex v ∈ V0 such that N2

G(v)∩V2 = {u} or dG(v, w) ̸= 2 for all w ∈ (V1∪V2)\{u}.

Proof. Let f = (V0, V1, V2) be an SHRDF on G of order n. Then by Proposition 1,
V1 ∪ V2 is a super hop dominating set on G.
(⇒) Assume that V1 ∪ V2 is a minimal super hop dominating set on G. Then for every
u ∈ V1 ∪V2, (V1 ∪V2) \ {u} is not a super hop dominating set of G. This means that there
exists v ∈ V (G) \ ((V1 ∪ V2) \ {u}) such that dG(v, w) ̸= 2 for all w ∈ (V1 ∪ V2) \ {u}.
Suppose that v ̸= u. It is worth noting that V1 ∪ V2 is a super hop dominating set, hence,
v must be super hop dominated by V1 ∪ V2. So, it follows that dG(u, v) = 2 which implies
that N2

G(v)∩V2 = {u}. Now, suppose that v = u. Then it simply follows that dG(v, w) ≠ 2
for every w ∈ (V1 ∪ V2) \ {u}.
(⇐) As for the converse, we assume that for every u ∈ V2, there exists v ∈ V0 such that
N2

G(v) ∩ V2 = {u}. Then it follows that for every v ∈ V0 is not hop dominated by the set
(V1 ∪ V2) \ {u}. On the other hand, assume that for every u ∈ V2, we have dG(u,w) ̸= 2
for all w ∈ (V1 ∪ V2) \ {u}. This implies that u can not be super hop dominated by any
vertex x ∈ (V1 ∪ V2) \ {u} and hence, (V1 ∪ V2) \ {u} is not a super hop dominating set of
G. Therefore, it is concluded that V1∪V2 is a minimal super hop dominating set of G.

The next result gives some bounds on the super hop Roman domination number of a
graph.

Theorem 4. Let G be a connected graph of order n ≥ 1. Then,

max{γsh(G), γR(G)} ≤ γshR(G) ≤ n.

Proof. Since every super hop Roman dominating function is Roman dominating,
it follows that γR(G) ≤ γshR(G). Let f = (V0, V1, V2) be a γshR-function on G. By
Proposition 1, V1 ∪ V2 is a super hop dominating set. This implies that γsh(G) ≤
|V1| + |V2| ≤ |V1| + 2|V2| = γshR(G). Therefore, max{γsh(G), γR(G)} ≤ γshR(G). Since
h = (∅, V (G),∅) is an SHRDF on G, we have γshR(G) ≤ ωshR

G (h) = |V (G)| = n.
We note that for any graph G, it is easy to show that γshR(G) ≤ 2γsh(G). Indeed, if

S be a γsh-set on G, then g = (V ′
0 , V

′
1 , V

′
2), where V ′

0 = V (G) \ S, V ′
1 = ∅, and V ′

2 = S,
is a SHRDF on G. Thus, we obtain γshR(G) ≤ ωshR

G (g) = |V ′
1 | + 2|V ′

2 | = 2|S| = 2γsh(G).
However, 2γsh(G) is not always the best upper bound because n ≤ 2γsh(G) by Theorem
1.

The next result simply says that the super hop Roman domination number of a graph
is the sum of the super Roman domination numbers of its components. For completeness,
we show its proof here.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 7 of 15

Theorem 5. Let G1, G2, . . . , Gk be the components of graph G of order n. Then f is an
SHRDF on G if and only if the restriction function f |Gj is an SHRDF on Gj for each
j ∈ {1, 2, ...k}. Moreover, γshR(G) =

∑k
j=1 γshR(Gj).

Proof. Suppose f = (V0, V1, V2) is an SHRDF on G. For each j ∈ {1, 2, ..., k}, let
V j
0 = V0∩V (Gj), V j

1 = V1∩V (Gj) and V j
2 = V2∩V (Gj). Then f |Gj = (V j

0 , V
j
1 , V

j
2 ) for all

j ∈ {1, 2, ..., k}. Let j ∈ {1, 2, ..., k} and let v ∈ V j
0 . Then v ∈ V0. Since f is a hop Roman

dominating function on G, it follows that there exists w ∈ V2 such that dG(w, v) = 2. This
implies that w ∈ V j

2 and dGj (w, v) = 2, showing that f |Gj is a hop Roman dominating
function on Gj . Moreover, since f satisfies (SHR2), there exists z ∈ V1 ∪ V2 such that
N2

G(z) ∩ V0 = {v}. This means that z ∈ V j
1 ∪ V j

2 and N2
Gj

(z) ∩ V0 = {v}. Therefore, f |Gj

is an SHRDF on Gj . If, in particular, f is a γshR-function on G, then

γshR(G) = ωshR
G (f) = |V1|+ 2|V2|

=

k∑
j=1

∣∣V j
1 |+ 2

k∑
j=1

∣∣V j
2 |

=

k∑
j=1

(∣∣V j
1 |+ 2

∣∣V j
2 |
)

≥
k∑

j=1

γshR(Gj).

Next, suppose that f |Gj = (W j
0 ,W

j
1 ,W

j
2 ) is an SHRDF on Gj for each j ∈ {1, 2, ...k}.

Then V0 =
⋃k

j=1W
j
0 , V1 =

⋃k
j=1W

j
1 and V2 =

⋃k
j=1W

j
2 . Let v ∈ V0. Then v ∈ V j

0 for
some j ∈ {1, 2, ...k}. Since f |Gj is a hop Roman dominating function on G, it follows that
there exists w ∈ V j

2 such that dGj (w, v) = 2. It follows that w ∈ V2. Also, since f |Gj

satisfies (SHR2), there exists z ∈ V j
1 ∪ V j

2 such that N2
G(z) ∩ V

j
0 = {v}. This implies that

z ∈ V1 ∪ V2 and N2
G(z) ∩ V0 = {v}. Accordingly, f is an SHRDF on G. If, in particular,

f |Gj is a γshR-function on Gj for all j ∈ {1, 2, ...k}, then

k∑
j=1

γshR(Gj) =

k∑
j=1

ωshR
G (f |Gj ) =

k∑
j=1

(∣∣W j
1 |+ 2

∣∣W j
2 |
)

=

k∑
j=1

∣∣W j
1 |+ 2

k∑
i=1

∣∣W j
2 |

= |V1|+ 2|V2|
≥ γshR(G).

This proves the assertion.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 8 of 15

Corollary 2. Let G1, G2, . . . , Gk be the components of a graph G of order n. If Gj

is complete for every j ∈ {1, 2, · · · , k}, then γshR(G) = n. In particular, γshR(Kn) =
γshR(Kn) = n.

Proof. This follows from Theorem 2, Theorem 4, and Theorem 5.

Proposition 3. Let G be a graph of order n ≥ 1. Then

(i) γshR(G) = 1 if and only if G = K1;

(ii) γshR(G) = 2 if and only if G ∈ {K2,K2}; and

(iii) γshR(G) = 3 if and only if G ∈ {P3,K3,K3,K2 ∪K1}.

Proof. Let f = (V0, V1, V2) be a γshR-function on G.

(i) Suppose γshR(G) = 1. Then |V1| + 2|V2| = 1. This implies that |V2| = 0. By
Proposition 1(i), |V0| = 0. Hence, |V1| = |V (G)| = 1, i.e., G = K1.
The converse is clear.

(ii) Suppose γshR(G) = |V1| + 2|V2| = 2. Then |V2| ≤ 1. Suppose |V2| = 1. Then
|V1| = 0 and |V0| ̸= 0. Let V2 = {v} and let w ∈ V0. Then dG(v, w) = 2. Let
x ∈ NG(w) ∩ NG(v). Since |V1| = 0, this implies that V2 is not a hop dominating
set in G, a contradiction. Thus, |V2| = 0. This implies that |V0| = 0 and γshR(G) =
|V1| = |V (G)| = 2. Therefore, G ∈ {K2,K2}.
Conversely, suppose that G ∈ {K2,K2}. Then clearly, γshR(G) = 2.

(iii) Suppose γshR(G) = |V1| + 2|V2| = 3. Then |V2| ≤ 1. If |V2| = 0, then |V0| = 0 and
|V1| = |V (G)| = 3. Hence, G ∈ {P3,K3,K3, P2 ∪K1}. Next, suppose that |V2| = 1.
Then |V1| = 1 and |V0| ≥ 1. Let v ∈ V2 and u ∈ V0. Then dG(u, v) = 2. Now, let
x ∈ NG(v)∩NG(u). Since f satisfies (SHR1) and x ∈ NG(v), it follows that x ∈ V1.
Suppose now that |V (G)| ≥ 4. Let w ∈ V (G) \ {u, v, x}. Since |V1| = |V2| = 1, it
follows that w ∈ V0 and dG(w, v) = 2. Let y ∈ NG(v) ∩ NG(w). Again, this will
imply that y ∈ V1. Hence, x = y. However, since u ̸= w, it follows that f does not
satisfy (SHR2), a contradiction. Thus, |V (G)| = 3 and G = P3.
The converse is clear.

Theorem 6. Let G be a graph. Then γshR(G) = 4 if and only if it satisfies one of the
following:

(i) |V (G)| = 4; or

(ii) V (G) = {x, y, p, q, v} such that NG(v) = {p, q}, N2
G(v) = {x, y}, x ∈ N2

G(p) \N2
G(q)

and y ∈ N2
G(q) \N2

G(p).



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 9 of 15

Proof. Let f = (V0, V1, V2) be a γshR-function on G. Assume that γshR(G) = 4. Then
|V1| + 2|V2| = 4 and so, |V2| ≤ 2. First, suppose that |V2| = 0. Then |V0| = 0 and
|V1| = |V (G)| = 4. Next, suppose |V2| = 1. Then |V1| = 1 and |V0| ≥ 1. Suppose |V0| ≥ 3.
Let a, b, c ∈ V0. Since f satisfies (SHR1), we have N2

G(v) ∩ V0 = {a, b, c}. This forces
|NG(p)∩V0| = 1 for every p ∈ V1 because f satisfies (SHR2). However, this is not possible
because |V1| = 2 and |V0| ≥ 3. Therefore, |V0| ≤ 2. Consequently, 4 ≤ |V (G)| ≤ 5.
If |V0| = 1, then |V (G)| = 4. Suppose |V0| = 2. Then |V (G)| = 5. Let V0 = {x, y},
V1 = {p, q} and V2 = {v}. Then NG(v) ∩ V0 = V0 = {x, y}. Since f satisfies (SHR2), we
may assume that N2

G(p)∩V0 = {x} and N2
G(q)∩V0 = {y}. Let z ∈ NG(x)∩NG(v). Since

p ∈ N2
G(x) and y ∈ N2

G(v), it follows that z /∈ {p, y}. This forces z = q. Hence, q ∈ NG(v).
Similarly, p ∈ NG(v). This shows that (ii) holds. Finally, suppose |V2| = 2. Then |V1| = 0
and, by Lemma 1, |V0| = 2. It follows that |V (G)| = 4.

For the converse, suppose first that |V (G)| = 4. By Theorem 4 and Proposition 3,
we have γshR(G) = 4. Next, suppose that (ii) holds. Let V0 = {x, y}, V1 = {p, q}, and
V2 = {v}. By assumption, h = (V0, V1, V2) is an SHRDF on G. Hence, Proposition 3
would imply that γshR(G) = ωshR

G (h) = 4.

Proposition 4. Let G = Pn with n ≥ 1. Then

γshR(G) =

{
n, if n ≤ 8,

n− k, if n ≥ 9,

where k = ⌊n+1
10 ⌋.

Proof. Assume that G = Pn = [v1, v2, ..., vn] with n ≥ 1. Let f = (V0, V1, V2) be a
γshR-function on G and let n ≤ 8. If V1 = V (G), then we are done. Assume for a moment
that γshR(G) < n. Then V1 ̸= V (G) and by Proposition 2, it follows that |V0| > |V2|. This
implies that there exists u ∈ V2 such that |N2

G(u)∩V0| = 2. Let x, y ∈ N2
G(u)∩V0. Then by

definition of super dominating set, there exists a, b ∈ V1 ∪ V2 such that N2
G(a) ∩ V0 = {x}

and N2
G(b) ∩ V0 = {y}. Clearly, u /∈ {a, b}. Hence, n ≥ 9, a contradiction since n ≤ 8.

Thus, γshR(G) = n whenever n ≤ 8. Now, let n ≥ 9. Then, consider the following cases:
Case 1: n ≡ 0 (mod 10)

Let n = 10k where k ∈ N. Then k = ⌊n+1
10 ⌋ = n

10 . Now, let Si = {v10i−9, v10i−1, v10i}

and Di = {v10i−6, v10i−5, v10i−4} for each i ∈ {1, 2, ..., k}. Set W1 =
k⋃

i=1

Si, W2 =
k⋃

i=1

Di

and W0 = V (G) \ (W1 ∪W2). Then, g = (W0,W1,W2) is an HRDF on G. Then, it is easy
to see that for every v = vj ∈ W0, there exists u = vt ∈ W2 such that dG(u, v) = 2 and
there exists w = vl ∈ (W1 ∪W2) such that N2

G(w)∩W0 = {v}. By construction, it follows
that g is a γshR-function on G. Thus, we get

γshR(G) = ωshR
G (g) = |W1|+ 2|W2|

=
k∑

i=1

|Si|+ 2
k∑

i=1

|Di|



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 10 of 15

= 3k + 2(3k)

= 9k

= 10k − k

= n− k

where k = n
10 .

Case 2: n ≡ r (mod 10) where 1 ≤ r ≤ 8
Let n = 10k + r where k ∈ N and 1 ≤ r ≤ 8. Then k = ⌊n+1

10 ⌋ = n−r
10 where r ∈

{1, 2, ..., 8}. In view of Case 1, we let S′
i = {v10i−9, v10i−1, v10i} andD′

i = {v10i−6, v10i−5, v10i−4}

for each i ∈ {1, 2, ..., k}. Again, set W ′
1 =

( k⋃
i=1

S′
i

)
∪ {vn−r+1, ..., vn} where 1 ≤ r ≤ 8,

W ′
2 =

k⋃
i=1

D′
i and W ′

0 = V (G) \ (W ′
1 ∪W ′

2). So, g′ = (W ′
0,W

′
1,W

′
2) is an HRDF on G. Note

that for every v′ = vj ∈ W ′
0, there exists u′ = vt ∈ W ′

2 such that dG(u′, v′) = 2 and there
exists w′ = vl ∈ (W ′

1 ∪W ′
2) such that N2

G(w
′) ∩W ′

0 = {v′}. By construction, it implies
that g′ is a γshR-function on G. Hence, we have

ωshR(g
′) = |W ′

1|+ 2|W ′
2|

=
( k∑
i=1

|Si|+ r
)
+ 2

k∑
i=1

|Di|

= 3k + r + 2(3k)

= 9k + r

= (10k + r)− k

= n− k

where k = n−r
10 for all r ∈ {1, 2, ..., 8}.

Case 3: n ≡ 9 (mod 10)
Let n = 10p+9 where p ∈ N∪{0}. Then k = ⌊n+1

10 ⌋ = p+1 = n+1
10 where p ∈ N∪{0}.

Again, by Case 1, we let S′′
i = {v10i−9, v10i−1, v10i} and D′′

i = {v10i−6, v10i−5, v10i−4}

for each i ∈ {1, 2, ..., k}. Now, set W ′′
1 =

( k⋃
i=1

S′
i

)
∪ {vn−8, vn}, W ′′

2 =
( k⋃
i=1

D′′
i

)
∪

{vn−5, vn−4, vn−3} and W ′′
0 = V (G) \ (W ′′

1 ∪W ′′
2 ). Thus, g′′ = (W ′′

0 ,W
′′
1 ,W

′′
2 ) is an HRDF

on G. Now, for every v′′ = vj ∈ W ′′
0 , there exists u′′ = vt ∈ W ′′

2 such that dG(u′′, v′′) = 2
and there exists w′′ = vl ∈ (W ′′

1 ∪W ′′
2 ) such that N2

G(w
′′) ∩W ′′

0 = {v′′}. By construction,
it follows that g′′ is a γshR-function on G. Hence, we have

γshR(G) = ωshR
G (g′′) = |W ′′

1 |+ 2|W ′′
2 |

=
( p∑
i=1

|S′′
i |+ 2

)
+ 2

( p∑
i=1

|D′′
i |+ 3

)



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 11 of 15

= 3p+ 2 + 2(3p+ 3)

= 9p+ 8

= (10p+ 9)− (p+ 1)

= n− k

where k = n+1
10 .

This proves the assertion.

The proof of the next result is similar to that of Proposition 4.

Proposition 5. Let G = Cn with n ≥ 3. Then,

γshR(G) =


n, if n ∈ {3, 4, 6, 7, 8, 9}
4, if n = 5

n− k, if n ≥ 10,

where k = ⌊ n
10⌋.

Theorem 7. Let G be a graph of order n. Then γsh(G) = γshR(G) if and only if each
component of G is complete. In this case, γsh(G) = γshR(G) = n.

Proof. Suppose γsh(G) = γshR(G) and let f = (V0, V1, V2) be a γshR-function on G. By
Proposition 1, V1 ∪ V2 is a super hop dominating set on G. Hence, γsh(G) ≤ |V1|+ |V2| ≤
|V1|+ 2|V2| = γshR(G). Since γsh(G) = γshR(G), it follows that |V2| = 0. By Proposition
1(i), we have |V0| = 0. This implies that |V1| = n = |V (G)|. Thus, γsh(G) = γshR(G) = n.
By Theorem 2, we find that each component of G is complete.

For the converse, suppose that every component of G is complete. By Theorem 2 and
Corollary 2, we have γsh(G) = γshR(G) = n.

Theorem 8. Let G be a graph of order n such that γsh(G) < γshR(G). Then γshR(G) =
γsh(G) + 1 if and only if there exist a set S ⊆ V (G) and vertex v such that S ∪ {v} is a
γsh-set of G and V (G) \ (S ∪ {v}) ⊆ N2

G(v).

Proof. Suppose γshR(G) = γsh(G) + 1 and let f = (V0, V1, V2) be a γshR-function on
G. If γsh(G) = |V1|+ |V2|, then the assumption implies that |V1|+ |V2|+ 1 = |V1|+ 2|V2|.
Hence, |V2| = 1 and |V1| = γsh(G) − 1. This implies that V1 ∪ V2 is a γsh-set in G.
Let V2 = {v} and S = V1. Then V0 = V (G) \ (S ∪ {v}). Since f satisfies (SHR1),
we have V (G) \ (S ∪ {v}) ⊆ N2

G(v). Next, suppose that γsh(G) < |V1| + |V2|. Then
γsh(G) + 1 ≤ |V1| + |V2|. Since γshR(G) = γsh(G) + 1, we have |V1| + 2|V2| ≤ |V1| + |V2|.
Hence, it implies that |V2| = 0 and so, |V0| = 0 and |V1| = n. Consequently, γsh(G) = n−1.
Let Q = V (G) \ {x} be a γsh-set on G. Then there exists v ∈ Q such that x ∈ N2

G(v). Let
S = Q \ {v}. Then S ∪ {v} = Q is γsh-set in G and V (G) \ (S ∪ {v}) = {x} ⊆ N2

G(v).
For the converse, suppose there exist a set S ⊆ V (G) and vertex v such that S ∪ {v}

is a γsh-set of G and V (G) \ (S ∪ {v}) ⊆ N2
G(v). Let V0 = V (G) \ (S ∪ {v}), V1 = S, and



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 12 of 15

V2 = {v}. Then g = (V0, V1, V2) is an SHRDF on G. Hence,

γshR(G) ≤ ωshR
G (g) = |V1|+ 2|V2| = |S|+ 2 = (γsh(G)− 1) + 2 = γsh(G) + 1.

Since γsh(G) < γshR(G), it follows that γshR(G) = γsh(G) + 1.

The join of graphsG andH is the graphG+H with vertex set V (G+H) = V (G)∪V (H)
and edge set E(G+H) = E(G) ∪ E(H) ∈ {uv : u ∈ V (G) and v ∈ V (H)}.

Theorem 9. Let H be a non-complete graph. Then f = (V0, V1, V2) is an SHRDF on a
graph G = Kn +H if and only if the following conditions are satisfied:

(i) V (Kn) ⊆ V1 ∪ V2; and

(ii) f |H is an SHRDF on H.

Proof. Suppose that f = (V0, V1, V2) is an SHRDF on G. Let x ∈ V (Kn). Since
xy ∈ E(G) for all y ∈ V (G) \ {x} and f is an HRDF on G, it follows that x /∈ V0.
Hence, x ∈ V1 ∪ V2. This shows that (i) holds. This implies that V0 ⊆ V (H). Note that
f |H = (V H

0 , V H
1 , V H

2 ), where V H
0 = V0, V H

1 = V1 ∩ V (H), and V H
2 = V2 ∩ V (H). Let

v ∈ V H
0 . Since f is an HRDF on G, there exists w ∈ V2∩N2

G(v). Hence, w ∈ V H
2 , showing

that f |H is an HRDF on H. Since V1 ∪ V2 is a super dominating set in G, there exists
z ∈ V1 ∪ V2 such that N2

G(z) ∩ V0 = {v}. This implies that z ∈ V H
1 ∪ V H

2 . Thus, f |H is a
SHRDF on H. This shows that (ii) also holds.

For the converse, suppose (i) and (ii) hold. Let V H
1 = V1 ∩ V (H), V H

2 = V2 ∩ V (H),
V n
1 = V1 ∩ V (Kn), and V n

2 = V2 ∩ V (Kn). From (i), it follows that V0 = V H
0 ⊆ V (H).

If V0 = ∅, then f = (∅, V H
1 ∪ V n

1 , V
H
2 ∪ V n

2 ) is an SHRDF on G. Suppose V0 ̸= ∅. Let
x ∈ V0. Since f |H = (V0, V

H
1 , V H

2 ) is an HRDF on H, there exists y ∈ V H
2 ∩N2

H(x). Hence,
y ∈ V2 ∩N2

G(x), showing that f = (V0, V1, V2) is an HRDF on G. Also, since f |H satisfies
(SHR2) on H, it follows that there exists z ∈ V H

1 ∪V H
2 such that N2

H(z)∩V0 = {x}. Since
V H
1 ∪V H

2 ⊆ V1∪V2, we find that z ∈ V1∪V2. Consequently, f = (V0, V1, V2) is an SHRDF
on a graph G.

The following corollaries below are direct consequence of Theorem 9.

Corollary 3. Let G be any graph. Then γshR(Kn +G) = n+ γshR(G).

Proof. Let D = V (Kn) and let g = (V0, V1, V2) be a γshR-function on G. Let V ′
0 = V0,

V ′
1 = D ∪ V1, and V ′

2 = V2. Then f = (V ′
0 , V

′
1 , V

′
2) is an SHRDF on Kn + G by Theorem

9. Hence, we have

γshR(Kn +G) ≤ ωshR
Kn+G(f) = |V ′

1 |+ 2|V ′
2 |

= (|D ∪ V1|) + 2|V2|
= (|D|+ |V1|) + 2|V2|
= |D|+ (|V1|+ 2|V2|)
= n+ γshR(G).



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 13 of 15

On the other hand, let h = (W0,W1,W2) be a γshR-function on Kn + G. By Theorem
9, we have V (Kn) ⊆ W1 ∪W2 and h|G is an SHRDF on G. Since h is a γshR-function,
we have V (Kn) ⊆ W1, i.e., W1 = V (Kn) ∪ (W1 ∩ V (G)) and h|G = (W0,W1 ∩ V (G),W2)
Thus,

γshR(Kn +G) = ωshR
Kn+G(h) = |W1|+ 2|W2|

= (|V (Kn)|+ |W1 ∩ V (G)|) + 2|W2|
= n+ |W1 ∩ V (G)|+ 2|W2|
= n+ ωshR

G (h|G)
≥ n+ γshR(G).

Therefore, γshR(Kn +G) = n+ γshR(G). This establishes the desired equality.

Corollary 4. Let n be a positive integer greater than or equal to 3. Then each the following
holds:

(i) γshR(Sn) = γshR(K1,n) = n+ 1;

(ii) γshR(Fn) = γshR(K1 + Pn) =

{
n+ 1, if n ≤ 8

n− k + 1, if n ≥ 9,

where k = ⌊n+1
10 ⌋.

(iii) γshR(Wn) = γshR(K1 + Cn) =

{
n+ 1, if 3 ≤ n ≤ 9

n− k + 1, if n ≥ 10,

where k = ⌊ n
10⌋.

Theorem 10. Let G and H be any non-complete graphs. Then, f = (V0, V1, V2) is a
SHRDF on G+H if and only if f |G and f |H are SHRDF on G and H, respectively.

Proof. Let f = (V0, V1, V2) be an RDF on G + H, and let V G
i = Vi ∩ V (G) and

V H
i = Vi ∩ V (H) for i ∈ {0, 1, 2}. Then f |G = (V G

0 , V
G
1 , V

G
2 ) and f |H = (V H

0 , V H
1 , V H

2 ).

Suppose that f is an SHRDF on G + H. Let v ∈ V G
0 . Since f satisfies (SHR1) and

(SHR2) on G+H, there exists u ∈ V2 such that dG+H(u, v) = 2 and there exists w ∈ V1∪V2
such that N2

G+H(w) ∩ V0 = {v}. Since vx ∈ E(G +H) for all x ∈ V (H), it follows that
u ∈ V G

2 , w ∈ V G
1 ∪ V G

2 , dG(u, v) = 2, and N2
G(w)∩ V G

0 = {v}. Thus, f |G is an SHRDF on
G. Using similar argument, f |H is also an SHRDF on H.

Conversely, suppose that f |G and f |H are SHRDF on G and H, respectively. Let
v′ ∈ V0. Then, either v′ ∈ V G

0 or v′ ∈ V H
0 . Without loss of generality, suppose that

v′ ∈ V G
0 . Since f |G is an SHRDF on G, there exists u′ ∈ V G

2 such that dG(u′, v′) = 2
and there exists w′ ∈ V G

1 ∪ V G
2 such that N2

G(w
′) ∩ V G

0 = {v′}. Since V G
2 ⊂ V2 and

V G
1 ∪ V G

2 ⊆ V1 ∪ V2, it implies that u′ ∈ V2 and w′ ∈ V1 ∪ V2 for which dG+H(u′, v′) = 2
and N2

G+H(w′) ∩ V G
0 = {v′}. Thus, f is an SHRDF on G+H.



L. F. Casinillo, S. R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7078 14 of 15

The next result follows from Theorem 10.

Corollary 5. Let G and H be any two graphs. Then, γshR(G+H) = γshR(G)+γshR(H).

Proof. Let g = (V G
0 , V

G
1 , V

G
2 ) and h = (V H

0 , V H
1 , V H

2 ) be γshR-functions on G and H,
respectively. Let Vi = V G

i ∪ V H
i for each i ∈ {0, 1, 2}. Then g = f |G and g = f |H where

f = (V0, V1, V2). By Theorem 10, f is an SHRDF on G+H. Hence, we have

γshR(G+H) ≤ ωshR
G+H(f) = |V1|+ 2|V2|

= |V G
1 ∪ V H

1 |+ 2|V G
2 ∪ V H

2 |
= (|V G

1 |+ 2|V G
2 |) + (|V H

1 |+ 2|V H
2 |)

= γshR(G) + γshR(H).

Now, let f ′ = (V ′
0 , V

′
1 , V

′
2) be a γshR-function on G + H. Then it follows that f ′|G =

(V ′
0 ∩ V (G), V ′

1 ∩ V (G), V ′
2 ∩ V (G)) and f ′|H = (V ′

0 ∩ V (H), V ′
1 ∩ V (H), V ′

2 ∩ V (H)) are
SHRDF on G and H, respectively, by Theorem 10. Thus, we get

γshR(G+H) = ωshR
G+H(f ′) = |V ′

1 |+ 2|V ′
2 |

= |(V ′
1 ∩ V (G)) ∪ (V ′

1 ∩ V (H))|+ 2|(V ′
2 ∩ V (G)) ∪ (V ′

2 ∩ V (H))|
= (|V ′

1 ∩ V (G)|+ 2|V ′
2 ∩ V (G)|) + (|V ′

1 ∩ V (G)|+ 2|V ′
2 ∩ V (G)|)

= ωshR
G (f ′|G) + ωshR

H (f ′|H)

≥ γshR(G) + γshR(H).

Therefore, γshR(G+H) = γshR(G) + γshR(H). This establishes the desired equality.

The next result is a direct consequence of Corollary 2 and Corollary 5.

Corollary 6. If G = Km,n where m,n ≥ 1, then γshR(G) = m+ n.

4. Conclusion

This study introduced a new variation of hop Roman domination called super hop
Roman domination. Some bounds and exact values of the super hop Roman domination
number of some classes of graphs were determined. Necessary and sufficient conditions for
functions to be super hop Roman dominating in the join of some graphs were obtained.
The newly defined parameter can be studied for other classes of graphs and sharp and
tight bounds on the parameter may be obtained.

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