International Journal of Mathematical Analysis
Vol. 10, 2016, no. 13, 623 - 637
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ijma.2016.6343
On Pairs of Disjoint Dominating
Sets in a Graph
Edward M. Kiunisala
Mathematics Department
College of Arts and Sciences
Cebu Normal University
Cebu City, Philippines
Ferdinand P. Jamil
Department of Mathematics and Statistics
College of Science and Mathematics
MSU-Iligan Institute of Technology
Iligan City, Philippines
c 2016 Edward M. Kiunisala and Ferdinand P. Jamil. This article is disCopyright tributed under the Creative Commons Attribution License, which permits unrestricted use,
distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
In this paper, we investigate pairs of disjoint dominating sets A and
B in a graph G, where B is either an independent or a total dominating
set in G.
Mathematics Subject Classification: 05C69
Keywords: dominating set, independent dominating set, total dominating
set, pair of disjoint dominating sets
1
Introduction
In this study, we consider graphs G = (V (G), E(G)) which are finite, simple
and undirected. The basic terminologies used here are adapted from [4]. For
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Edward M. Kiunisala and Ferdinand P. Jamil
S ⊆ V (G), the symbol |S| denotes the cardinality of S. In particular, |V (G)|
is called the order of G. The cardinality |E(G)| of the edge set E(G) of G is
the size of G. If |E(G)| = 0, then G is an empty graph. An empty graph of
order n is denoted by Kn .
Given two vertices u and v of a graph G, a u-v geodesic is a shortest path
in G joining u and v. For a non-empty S ⊆ V (G), hSi denotes that subgraph
H of G for which |E(H)| is the maximum size of a subgraph of G with vertex
set S.
Let G and H be any graphs. The (disjoint) union of G and H is the graph
G ∪ H with vertex set V (G ∪ H) = V (G) ∪ V (H) and edge set E(G ∪ H) =
E(G) ∪ E(H), where V (G) and V (H) are considered disjoint. The join of
G and H is the graph G + H with vertex set V (G) ∪ V (H) and edge set
E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}. The corona G ◦ H of G and H is
the graph obtained by taking one copy of G and |V (G)| copies of H, and then
joining the ith vertex of G to every vertex in the ith copy of H. We denote
by H v that copy of H whose vertices are adjoined with the vertex v of G.
In effect, G ◦ H is composed of the subgraphs H v + v = H v + h{v}i joined
together by the edges of G. For any v ∈ V (G), G − v is the resulting graph
after removing from G the vertex v and all edges of G incident to v.
For any vertex v of a graph G, the closed neighborhood NG [v] of v consists
of v and all vertices of G adjacent to v. We also define NG (v) = NG [v] \ {v}.
For S ⊆ V (G), we define NG [S] = ∪v∈S NG [v] and NG (S) = ∪v∈S NG (v). A
dominating set in G is any S ⊆ V (G) for which NG [S] = V (G). In this case,
we also say S dominates V (G). If S = {u} dominates V (G), we simply say
u dominates V (G). A dominating set S is an independent dominating set if
uv ∈
/ E(G) for all u, v ∈ S. A dominating set S is a total dominating set if for
each u ∈ S there is v ∈ S such that uv ∈ E(G). The minimum cardinality of a
dominating (resp. independent dominating, total dominating) set is called the
domination number (resp. independence domination number, total domination
number) of G, denoted by γ(G) (resp. γi (G), γt (G)). The symbols D(G),
I (G) and T (G) are used to denote the collection of all dominating sets, the
collection of all independent dominating sets, and the collection of all total
dominating sets in G, respectively. We also use the terminologies γ-set, γi -set
and γt -set to mean any set in D(G), in I (G) and in T (G), respectively, of
minimum cardinality.
Domination is one of the most well-studied concepts in graph theory. The
reader is referred to [1, 3, 5, 6, 7, 8, 10, 11, 13, 22, 24, 25] for the fundamental
concepts and recent developments of the domination theory, and to [3, 12, 16,
25] for its various applications.
In 1962, O. Ore gave the classical result which can be stated as follows:
Theorem 1.1 [24] If a graph G has no isolated vertices and S is a minimum
dominating set, then V (G) \ S is a dominating set in G.
On pairs of disjoint dominating sets in a graph
625
It has motivated the introduction of the concept of inverse domination (see [21])
as well as the concept of disjoint domination (see [14]). A subset S ⊆ V (G)
is an inverse dominating set in G if S is a dominating set in G and there
is a minimum dominating set D in G such that S ∩ D = ∅. The minimum
cardinality of an inverse dominating set in G is the inverse domination number
of G, which is denoted by γ 0 (G).
For any graph G which has no isolated points, Theorem 1.1 guarantees the
existence of a pair (S, D) of dominating sets in G with S ∩ D = ∅. Any such
pair is called a dd-pair . We denote by DD(G) the collection of all dd-pairs in
G. The minimum sum |S| + |D| among all dd-pairs in DD(G) is the disjoint
domination number of G, which is denoted by γγ(G). That is,
γγ(G) = min{|S| + |D| : (S, D) ∈ DD(G)}.
A dd-pair (S, D) with |S| + |D| = γγ(G) is called a γγ-pair.
Inverse domination is studied further in [9, 19, 20]. Disjoint domination
is also investigated in [16, 17, 19, 23]. In [26], an application of the concept
in information retrieval system is cited: In an Information Retrieval System,
we always have a set of primary nodes to pass on the information. In case,
the system fails, we have another set of secondary nodes to do the job in
the complement. Thus, the dominating sets and the elements in the inverse
dominating sets can stand together to facilitate the communication process.
In what follows, we investigate disjoint domination which involves either a
total dominating or an independent dominating set.
2
A pair of disjoint dominating and total dominating sets
The authors in [15, 17, 18] have determined certain conditions under which
a graph G has a vertex set V (G) having a pair of subsets consisting of a
dominating set and a total dominating set which are disjoint.
Theorem 2.1 [17] If G is a graph of minimum degree at least 2 such that no
component of G is a chordless cycle of length 5, then V (G) can be partitioned
into a dominating set D and a total dominating set T .
Theorem 2.2 [18] If G is a graph of minimum degree at least 3 with at least
one component different from the Petersen graph, then G contains a dominating set D and a total dominating set T which are disjoint and satisfy
|D| + |T | < |V (G)|.
A graph G is a (D, T )-graph if V (G) contains a pair of disjoint subsets
where one is a dominating set and the other a total dominating set. If G is
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Edward M. Kiunisala and Ferdinand P. Jamil
a nontrivial connected graph, then for any graph H, G + H and G ◦ H are
(D, T )-graphs.
A dt-pair in G is any pair (S, T ) such that S is a dominating set and T is
a total dominating set in G with S ∩ T = ∅. The symbol DT (G) denotes the
collection of all dt-pairs in G, and
γγt (G) = min{|S| + |T | : (S, T ) ∈ DT (G)}.
A dt-pair (S, T ) is called a γγt -set if |S| + |T | = γγt (G).
For all (D, T )-graphs G,
γ(G) + γt (G) ≤ γγt (G) ≤ |V (G)|.
(1)
If G = K3 ◦ K1 , then γt (G) = 3 = γ(G). For this graph, γ(G) + γt (G) =
γγt (G) = |V (G)| = 6.
Theorem 2.3 Let G be a (D, T )-graph. Then
(i) γγt (G) = 3 if and only if G = K1 + H, where H is connected and
γt (H) = 2.
(ii) γγt (G) = 4 if and only if either
(a) G = K1 + H, where H is a connected graph with γt (H) = 3, or
(b) G has a pair of disjoint γ-sets {u, v} and {x, y} such that xy ∈
E(G).
Proof : Statement (i) is clear.
(ii) Suppose that γγt (G) = 4, and let (S, T ) be a γγt -pair in G. Then
either |S| = 1 and |T | = 3 or |S| = 2 = |T |. Suppose that |S| = 1 and
|T | = 3, and put S = {v}, T = {a, b, c} and H = G − v. Then G = K1 + H
and T ⊆ V (H). Moreover, T is a total dominating set in H so that H is
connected and γt (H) ≤ 3. In view of Theorem 2.3(i), γt (H) = 3. Suppose
that G = K1 + H with H connected and γt (H) = 3, and let T be a γt -set in
H. Then (V (K1 ), T ) is a dt-pair in G and 3 ≤ γγt (G) ≤ 1 + 3 = 4. In view of
Theorem 2.3(i), γγt (G) = 4.
Suppose that |S| = |T | = 2. Clearly, hT i = K2 . In view of Statement (i), S
and T are γ-sets of G. Conversely, if γ(G) = 2, then γγt (G) > 3 by Statement
(i). Thus γγt (G) = 4.
The next theorem shows that γγt (G) − γ(G) − γt (G) and |V (G)| − γγt (G)
can each be made arbitrarily large.
Theorem 2.4 For each positive integer n,
627
On pairs of disjoint dominating sets in a graph
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Figure 1:
(i) there exists a connected graph G such that γγt (G) − γ(G) − γt (G) = n;
(ii) there exists a connected graph G such that |V (G)| − γγt (G) = n.
Proof : (i) Write K2 = [x, y] and obtain G as in Figure 1 by adding to K2 the
pendant edges v1 x, v2 x and uj y, j = 1, 2, . . . , n. Then {x, y} is a γ-set and at
the same time a γt -set in G. On the other hand, there is a unique γγt -pair
in G, namely S = {v1 , v2 , u1 , u2 , . . . , un } and T = {x, y}. For this graph G,
γγt (G) − γt (G) − γ(G) = [(n + 2) + 2] − 2 − 2 = n.
(ii) Consider the graph G = K2 + K1,n = K1 + (K1 + K1,n ). Since γt (K1 +
K1,n ) = 2, γγt (G) = 3 by Theorem 2.3. Thus, |V (G)| − γγt (G) = (n + 3) − 3 =
n.
3
A pair of disjoint dominating and independent dominating sets
Recall that by an independent set we mean any S ⊆ V (G) such that uv ∈
/ E(G)
for all u, v ∈ S, and the symbol β(G) denotes the maximum cardinality of an
independent set in G. An independent set S with |S| = β(G) is called a β-set.
Lemma 3.1 let G be any graph with no isolated vertices. If S ⊆ V (G) is an
independent set in G, then V (G) \ S is a dominating set in G.
Let G be a graph without an isolated vertex. A di-pair in G is any pair
(S, I) of subsets of V (G), where S is a dominating set and I is an independent
dominating set in G such that S ∩I = ∅. We use the symbol DI (G) to denote
the collection of all di-pairs in G, and define
γγi (G) = min{|S| + |I| : (S, I) ∈ DI (G)}.
A di-pair (S, I) with |S| + |I| = γγi (G) is called a γγi -pair.
It is worth noting that Lemma 3.1 guarantees existence of di-pairs (S, I),
where I is a γi -set in G. An inverse independent dominating set in G is a
dominating set S such that S ⊆ V (G)\I for some γi -set I in G. The minimum
cardinality of an inverse independent dominating set is denoted by γi0 (G). Any
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Edward M. Kiunisala and Ferdinand P. Jamil
inverse independent dominating set with caridanilty γi0 (G) is called a γi0 -set in
G. Apparently,
γ(G) ≤ γi0 (G) ≤ |V (G)| − γi (G).
(2)
If G is the complete graph Km , where m ≥ 2, then γ(G) = 1 = γi0 (G). If G
is the complete bipartite graph Km,n , where 1 ≤ m ≤ n, then γi (G) = m and
γi0 (G) = n so that γi0 (G) = n = |V (G)| − γi (G). Hence, the above estimates
are sharp.
The parameters γi0 (G) and γi (G) are not directly related. Note, in particularly, that γi (P3 ) < γi0 (P3 ). On the other hand, γi0 (G1 ) < γi (G1 ), where G1
is the graph as in Figure 2. Verify that γi (G1 ) = 3 and γi0 (G1 ) = 2, the latter
being determined by the blackened vertices.
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Figure 2:
The parameters γi0 (G) and γ 0 (G) are not also directly related. Note that
γ 0 (P2 ◦ K2 ) = 4 > 3 = γi0 (P2 ◦ K2 ). On the other hand, we have γ 0 (G2 ) =
2 < 3 = γi0 (G2 ), where G2 is the graph as in Figure 2. The blackened vertices
comprise a γi0 -set in G2 . In fact, γ 0 (G) − γi0 (G) and γi0 (G) − γ 0 (G) can each be
made arbitrarily large. Let n be a positive integer. If G is the graph G3 as
in Figure 2, then the sets {xk , yj : k = 1, 2, . . . , n + 1; j = 1, 2, . . . , n + 2} and
{u, yj : j = 1, 2, . . . , n + 2} are a γ 0 -set and a γi0 -set in G, respectively. In this
case, γ 0 (G)−γi0 (G) = (2n+3)−(n+3) = n. On the other hand, if G = K2,n+2 ,
then γ 0 (G) = 2 and γi0 (G) = n + 2 so that γi0 (G) − γ 0 (G) = n.
Theorem 3.2 Let G be a connected graph of order n. Then
(i) γi0 (G) = 1 if and only if either G = K2 or G = K2 + G∗ , for some graph
G∗ of order n − 2.
(ii) For n ≥ 2, γi0 (G) = n − 1 if and only if G = K1 ,n−1 ;
(iii) For n ≥ 4, γi0 (G) = n − 2 if and only if either G = K1 + (Kn−3 ∪ K2 )
or there exists u, v ∈ V (G) such that dG (u, v) = 2 and hV (G) \ {u, v}i=
Kn−2 .
Proof : Statement (i) is clear.
(ii) By Theorem 3.2(i), the result holds for n = 2. We proceed with
n ≥ 3. Suppose that γi0 (G) = n − 1, and let S ⊆ V (G) be a γi0 -set in G. Let
On pairs of disjoint dominating sets in a graph
629
v ∈ V (G) \ S. Then NG [v] = V (G), that is, vx ∈ E(G) for all x ∈ V (G) \ {v}.
We claim that xy ∈
/ E(G) for all x, y ∈ V (G) \ {v}. Suppose that there exist
x, y ∈ V (G) \ {v} such that xy ∈ E(G). Then v, y ∈ NG [x] ⊆ NG [S \ {y}].
Thus, S \ {y} is an inverse independent dominating set in G, a contradiction.
Therefore, G = K1 ,n−1 . The converse is obvious.
(iii) Suppose that n ≥ 4 and γi0 (G) = n − 2, and let S be a γi0 -set in G. Let
u, v ∈ V (G) \ S. We consider two cases.
Case 1: Suppose that NG [v] = V (G). First, we claim that hSi = Kn−2 .
Suppose that there exist x, y ∈ S such that xy ∈ E(hSi). Either xu, yu ∈
/ E(G)
or, say, ux ∈ E(G). Since u, y, v ∈ NG [S\{y}], S\{y} is an inverse independent
dominating set in G, a contradiction. This shows that hSi = Kn−2 . Next, we
claim that there is a unique x ∈ S such that xu ∈ E(G). For the existence
part, note that if xu ∈
/ E(G) for all x ∈ S, then G = K1 ,n−1 and, by Theorem
0
3.2(ii), γi (G) = n − 1, a contradiction. To show the uniqueness, suppose that
there exists y ∈ S \ {x} with uy ∈ E(G). Then x, y ∈ NG [u] ⊆ NG [S ∗ ],
where S ∗ = (S \ {x, y}) ∪ {u}. This means that S ∗ is an inverse independent
dominating set in G and γi0 (G) ≤ n − 3, a contradiction. The two claims above
together imply that G is the graph obtained by adding to K1,n−1 the edge xu.
That is, G = K1 + (Kn−3 ∪ K2 ).
Case 2: Suppose that {u, v} is a γi -set in G. Then uv ∈
/ E(G). Let P = [u =
x1 , x2 , . . . , xk = v] be a u-v geodesic in G. Since {u, v} is a γ-set in G, k ≤ 4.
We claim that if k = 4, then G = P = P4 . Suppose that k = 4 and G 6= P4 .
Let y ∈ V (G) \ V (P4 ). Then either uy ∈ E(G) or vy ∈ E(G). Assume that
uy ∈ E(G). Note that u, v, x2 ∈ NG [{y, x3 }] ⊆ NG [S \ {x2 }]]. This means that
S \ {x2 } is an inverse independent dominating set in G, a contradiction. Thus,
if k = 4, then G = P4 . In this case, dG (u, x3 ) = 2 and hV (G) \ {u, x3 }i = K2 .
Suppose that k < 4. Since uv ∈
/ E(G), k = 3 and P = [u = x1 , x2 , x3 = v].
This means that dG (u, v) = 2. Let x, y ∈ S such that xy ∈ E(G). Put
S ∗ = S \ {y} whenever x = x2 , and put S ∗ = S \ {x} whenever x 6= x2 . Then
V (G) = NG [S ∗ ]. Accordingly, S ∗ is an inverse independent dominating set in
G, a contradiction. Therefore, hSi = Kn−2 .
Now, we prove the converse. First, suppose that G = K1 + (Kn−3 ∪ K2 ).
Let V (K1 ) = {v}, u ∈ V (K2 ), and let S = V (G) \ {u, v}. Then {v} and
S are a γi -set and a γi0 -set, respectively, in G. Thus, γi0 (G) = |S| = n − 2.
Next, suppose that G has distinct vertices u and v such that dG (u, v) = 2 and
hSi = Kn−2 , where S = V (G) \ {u, v}. Since G is connected, for each x ∈ S,
xu ∈ E(G) or xv ∈ E(G). Thus {u, v} is an independent dominating set in
G. Since u ∈
/ NG [v], {v} is not a dominating set in G. Similarly, {u} is not
a dominating set in G. Since n ≥ 4, {x} is not a dominating set in G for all
x ∈ S. Thus {u, v} is a γi -set in G. It also follows that S is a γi0 -set in G, and
γi0 (G) = n − 2.
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Edward M. Kiunisala and Ferdinand P. Jamil
For all graphs G with no isolated vertices,
γi (G) + γ(G) ≤ γγi (G) ≤ γi (G) + γi0 (G).
(3)
Theorem 3.3 Let G be a nontrivial connected graph of order n ≥ 2. Then
(i) γγi (G) = 2 if and only if G has two distinct vertices each of which
dominates V (G).
(ii) γγi (G) = 3 if and only if G = K1 + H, where γ(H) = 2.
(iii) γγi (G) = 4 if and only if either G = K1 + H where γ(H) = 3 or
γi (G) = 2 = γi0 (G).
Proof : (i) Clearly, γγi (G) = 2 if and only if γ(G) = 1 and γ 0 (G) = 1. The
latter holds if and only if G has two distinct vertices each of which dominates
V (G).
(ii) Suppose that γγi (G) = 3 and (D, I) is a γγi -pair in G. Either |D| = 1
and |I| = 2 or |I| = 1 and |D| = 2. In either case, G has a vertex v such that
NG [v] = V (G). Thus, G = K1 + H for some subgraph H of G. By Theorem
3.3(i), it is necessary that γ(H) = 2. Conversely, suppose that G = K1 + H,
where γ(H) = 2. Then γi (G) = 1 and γi0 (G) = 2. Inequality 3 implies that
γγi (G) ≤ 3. In view of Theorem 3.3(i), γγi (G) = 3.
(iii) Suppose that γγi (G) = 4, and let (D, I) be a γγi -pair in G. Then
exactly one of the following holds: (1) |D| = 1 and |I| = 3; (2) |I| = 1 and
|D| = 3; (3) |D| = 2 = |I|. If (1) or (2) holds, then G has a vertex that
dominates V (G) so that G = K1 + H for some subgraph H of G. Moreover, by
Theorem 3.3(i) and Theorem 3.3(ii), γ(H) = 3. Suppose that |I| = 2 = |D|,
and let I = {u, v}. Since I is an independent dominating set in G, γi (G) ≤ 2.
Suppose that γi (G) = 1, and let w ∈ V (G) for which NG [w] = V (G). Since
uv ∈
/ E(G), w 6= u and w 6= v. Thus {u, v} and {w} constitute a di-pair in G
so that γγi (G) ≤ 3, a contradiction. This means that γi (G) = 2 and {u, v} is
a γi -set in G. Consequently, D is an inverse independent dominating set in G.
Hence, γi0 (G) ≤ |D| = 2. If γi0 (G) = 1, then, by Theorem 3.2(i), either G = K2
or G = K2 + G∗ for some subgraph G∗ of G. In either case, γγi (G) = 2, a
contradiction. Therefore, γi0 (G) = 2.
Conversely, suppose that G = K1 + H where γ(H) = 3. Then γγi (G) ≤ 4.
In view of the preceeding results, γγi (G) ≥ 4. Finally suppose that γi (G) =
2 = γi0 (G). Then 3 ≤ γγi (G) ≤ 4. Theorem 3.3(ii) above implies that
γγi (G) = 4.
4
Join of graphs
Proposition 4.1 Let G and H be graphs with γi (G) < γi (H).
631
On pairs of disjoint dominating sets in a graph
(i) If G is an empty graph, then γi0 (G + H) = γ(H).
(ii) If G is a nonempty graph, then
(
1, if G = K2 , K2 + G∗ f or some G∗ < G,
γi0 (G + H) =
2, otherwise.
Proof : Let S ⊆ V (G + H) be a γi0 -set in G + H, and let I ⊆ V (G + H) be a
γi -set in G + H such that S ⊆ V (G + H) \ I. Since γi (G) < γi (H), I ⊆ V (G).
If G is an empty graph, then I = V (G) and, consequently, S ⊆ V (H) and
S is a γ-set in H. That is, γi0 (G + H) = |S| = γ(H). This proves (i).
Suppose that G is a nonempty graph. Then V (G) 6= I. Take u ∈ V (G) \ I
and v ∈ V (H). Then S ∗ = {u, v} is a dominating set in G + H and S ∗ ⊆
V (G + H) \ I. Thus, γi0 (G + H) ≤ |S ∗ | = 2. Consequently, |S| ≤ |S ∗ |. Suppose
that |S| = 1. Then I and S and are both singleton subsets of V (G). Thus
G has two distinct vertices u and v such that both u and v dominate V (G).
Accordingly, either G = K2 or G = K2 + G∗ for some subgraph G∗ of G. On
the other hand, it can be readily verified that if G = K2 or G = K2 + G∗ for
any graph G∗ , then γ(G + H) = 1 = γi0 (G + H). This proves (ii).
Proposition 4.2 Let G and H be graphs with γi (G) = γi (H).
(i) If G is an empty graph, then
(
|V (G)|, if γ(H) = 1 or H is an empty graph,
γi0 (G + H) =
2,
otherwise.
(ii) If G is a nonempty graph, then
(
1,
γi0 (G + H) =
2,
if γ(G) = 1,
otherwise.
Proof : Suppose that G is an empty graph. Then |V (G)| = γi (H) ≤ |V (H)|. If
|V (G)| = |V (H)|, then H is an empty graph, and γi0 (G+H) = |V (G)|. Suppose
that |V (H)| > |V (G)|, and I ⊆ V (H) is a γi -set in H. Then I 6= V (H). Pick
u ∈ V (H) \ I. For any v ∈ V (G), {u, v} is an inverse independent dominating
set in G + H. This implies that γi0 (G + H) ≤ 2. Let S be a γi0 -set in G + H.
Then, in particular, |S| = 1 if and only if |V (G)| = 1 and γ(H) = 1. This
proves (i).
Suppose that G is a nonempty graph. Then γi (G) < |V (G)|. Let I ⊆ V (G)
be a γi -set in G. Choose u ∈ V (G) \ I and v ∈ V (H). Then {u, v} is an
inverse independent dominating set in G + H. Thus γi0 (G + H) ≤ 2. Clearly,
γi0 (G + H) = 1 if and only if γ(G) = 1. This proves (ii).
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Edward M. Kiunisala and Ferdinand P. Jamil
Corollary 4.3 Let G be a graph without isolated vertices. Then γi0 (G) = 1 if
and only if G = K2 + H for some graph H of order |V (G)| − 2.
Proof : Suppose that γi0 (G) = 1. Then there exists distinct(vertices) u and
v ∈ V (G) such that {u} and {v} are γ-sets in G. Since uv ∈ E(G), G =
h{u, v}i + H = K2 + H.
The converse immediately follows from Proposition 4.1 and Proposition
4.2.
Proposition 4.4 Let G and H be graphs with γi (G) ≤ γi (H).
(i) If G and H are nonempty, then either γγi (G + H) = 2 or γγi (G + H) =
2 + γi (G). Further, γγi (G + H) = 2 if and only if γ(G) = γ(H) = 1 or
G contains two vertices each of which dominates V (G).
(ii) If G is an empty graph, then
(
2 + |V (G)|,
γγi (G + H) =
|V (G)| + γ(H),
if γi (G) = γi (H) < |V (H)|,
otherwise.
Proof : To prove (i), let G and H be nonempty graphs. Let (S, I) be a γγi pair in G + H. Then either I ⊆ V (G) or I ⊆ V (H). Note that since G
is nonempty, if I ⊆ V (G), then I 6= V (G). Similarly, if I ⊆ V (H), then
I 6= V (H). In any case, we may choose u ∈ V (G) \ I and v ∈ V (H) \ I. Since
{u, v} is a dominating set in G + H, the definition of S implies that |S| ≤ 2.
Suppose that |S| = 1. If S ⊆ V (H), then γi (H) = 1 so that γi (G) = 1. This
means that γ(G) = γ(H) = 1 and γγi (G + H) = 2. Suppose that S ⊆ V (G).
Pick x ∈ V (G) \ S and y ∈ V (H). Then {x, y} and S constitute a di-pair
in G + H. It follows that 2 ≤ γγi (G + H) ≤ 3. If γγi (G + H) 6= 2, then
γγi (G + H) = 3 = 2 + |S| = 2 + γi (G). On the other hand, γγi (G + H) = 2
if and only if |I| = |S| = 1. Note that |S| = |I| = 1 if and only if either
γ(G) = γ(H) = 1 or G contains two vertices each of which dominates V (G).
Finally, suppose that |S| = 2. Since γi (G + H) = γi (G) and (S, I) is a γγi -pair,
γγi (G + H) = 2 + γi (G).
Now, we prove (ii). If G is the trivial graph, then γγi (G + H) = 1 + γ(H).
Let G be an empty graph. Then γ(G) ≥ 2. Suppose that γi (G) = γi (H) <
|V (H)|. Then γi (G + H) = γi (H). Take a γi -set I in H and take u ∈ V (G)
and v ∈ V (H) \ I. Then I and {u, v} constitute a γγi -pair in G + H so that
γγi (G + H) = 2 +|V (G)|. Next, if H is empty, then G + H = Km,n , where m =
|V (G)| and n = |V (H)|. In this case, γγi (G + H) = m + n = |V (G)| + γ(H).
Finally, suppose that H is nonempty but γ(G) < γ(H). Then V (G) and any
γ-set in H constitute a γγi -pair in G + H. Thus, γγi (G + H) = |V (G)| + γ(H).
On pairs of disjoint dominating sets in a graph
633
Proposition 4.5 Let G be a graph with no isolated vertex.
(i) If γ(G) = 1, then γγt (G + K1 ) = 3.
(ii) If γ(G) > 1, then γγt (G + K1 ) = min{1 + γt (G), 2 + γ(G)}.
Proof : (i) If γ(G) = 1, then G is connected and γt (G) = 2. By Theorem 2.3,
γγt (G + K1 ) = 3.
(ii) Suppose that γ(G) > 1, and let D = V (K1 ) = {v}. Let T ⊆ V (G) be a
γt -set in G. Then (D, T ) is a dt-pair in G+K1 . Thus, γγt (G+K1 ) ≤ 1+γt (G).
Let D∗ ⊆ V (G) be a γ-set in G. Since G has no isolated vertex, D∗ 6= V (G).
Let u ∈ V (G) \ D∗ , and put T ∗ = {u, v}. Then T ∗ is a γt -set in G + K1 and
(D∗ , T ∗ ) is a dt-pair in G+K1 . Hence, γγt (G+K1 ) ≤ 2+γ(G). Combining with
the above result, γγt (G) ≤ min{1 + γt (G), 2 + γ(G)}. Now, let (S, T ) be a dtpair in G + K1 . Either v ∈ S or v ∈ T . Suppose that v ∈ S. Then T ⊆ V (G),
and is a total dominating set in G. Thus, |S| + |T | ≥ 1 + γt (G). Suppose that
v ∈ T . Then T ∩V (G) 6= ∅ and S ⊆ V (G) is a dominating set in G. This yields
|S| + |T | ≥ 2 + γ(G). This means that |S| + |T | ≥ min{1 + γt (G), 2 + γ(G)}.
Since (S, T ) is arbitrary, γγt (G + K1 ) ≥ min{1 + γt (G), 2 + γ(G)}, and the
conclusion follows.
Proposition 4.6 Let G and H be nontrivial graphs. Then
(
3, if γ(G) = 1 or γ(H) = 1,
γγt (G + H) =
4, otherwise.
Proof : Clearly 3 ≤ γγt (G + H) ≤ 4. Suppose that γγt (G + H) = 3. By
Theorem 2.3, G + H = K1 + G∗ , where G∗ is connected and γt (G∗ ) = 2. Let
V (K1 ) = {v}. If v ∈ G, then, in particular, vx ∈ E(G) for all x ∈ V (G) \ {v}
so that γ(G) = 1. Similarly, if v ∈ V (H), then γ(H) = 1. Conversely,
suppose that γ(G) = 1 and let v ∈ V (G) such that NG [v] = V (G). Then
NG+H [v] = V (G + H). Since G is nontrivial, we can pick u ∈ V (G) \ {v}. For
any w ∈ V (H), {u, w} is a total dominating set in G∗ = (G + H) − v. Thus
γt (G∗ ) = 2 and, by Theorem 2.3, γγt (G + H) = 3.
5
Corona of graphs
Canoy et al. provided the following four results for the corona of graphs.
Theorem 5.1 [11] Let G be a connected graph and H any graph. Then C ⊆
V (G ◦ H) is a dominating set in G ◦ H if and only if C ∩ V (H v + v) is a
dominating set in H v + v for every v ∈ V (G).
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Edward M. Kiunisala and Ferdinand P. Jamil
Theorem 5.2 [5] Let G be a connected graph and H any graph. Then C ⊆
V (G ◦ H) is an independent dominating set in G ◦ H if and only if C ∩ V (G)
is an independent set in G and C ∩ V (H v + v) is an independent dominating
set in H v + v for every v ∈ V (G).
Theorem 5.3 [5] Let G be a connected graph of order n and H any graph
with γ(H) 6= 1. If C ⊆ V (G ◦ H) is a γi -set in G ◦ H, then C ∩ V (G) is a
β-set in G.
Theorem 5.4 [11] Let G be a connected graph and let H be any graph. Then
C ⊆ V (G ◦ H) is a total dominating set in G ◦ H if and only if for every
v ∈ V (G), either
(i) V (v + H v ) ∩ C is a total dominating set in H v + v; or
(ii) v ∈ C and NG (v) ∩ C 6= ∅.
In particular, Theorem 5.2 and Theorem 5.3 yield the following corollary.
Corollary 5.5 Let G and H be connected graphs with γ(H) > 1, and let
S ⊆ V (G ◦ H). Then S is a γi -set in G ◦ H if and only if S ∩ V (G) is a β-set
in G and S ∩ V (H v ) is a γi -set in H v for all v ∈ V (G) \ S.
Proposition 5.6 Let G and H be any graphs. Then
(i) γi0 (G ◦ H) = |V (G)| + (γ(H) − 1)β(G); and
(ii) γγi (G ◦ H) = |V (G)|(1 + γi (H)) − (γi (H) − γ(H))β(G).
Proof : Verify that both claims are true for |V (G)| = 1 or γ(H) = 1. In what
follows, we assume that |V (G)| > 1 and γ(H) > 1. Let S ⊆ V (G ◦ H) be a
γi0 -set in G ◦ H, and let I ⊆ V (G ◦ H) \ S be a γi -set in G ◦ H. By Theorem
5.3, I ∩ V (G) is a β-set in G and for each v ∈ V (G ◦ H) \ I, I ∩ V (H v ) is
a γi -set in H v . For each v ∈ I ∩ V (G), choose a γ-set Sv in H v , and define
S ∗ = (∪v∈I∩V (G) Sv ) ∪ (V (G) \ I). By Theorem 5.1, S ∗ is a dominating set in
G ◦ H. Since S ∗ ∩ I = ∅, S ∗ is an inverse independent dominating set in G ◦ H.
Thus, |S| ≤ |S ∗ |. On the other hand, for each v ∈ I ∩ V (G), v ∈
/ S so that
S ∩ V (H v + v) = S ∩ V (H v ). By Theorem 5.2, S ∩ V (H v ) is a dominating set
in H v , and thus, |Sv | ≤ |S ∩ V (H v + v)| for all v ∈ I ∩ V (G). This yields
X
|S| ≥
|Sv | + |V (G) \ I| = |S ∗ |.
v∈I∩V (G)
Therefore, |S| = γ(H)β(G) + |V (G)| − β(G) = |V (G)| + (γ(H) − 1)β(G).
On pairs of disjoint dominating sets in a graph
635
Let (S, I) be a γγi -pair in G◦H, and for each v ∈ V (G), let Iv = I ∩V (H v +
v) and Sv = S ∩ V (H v + v). By Theorem 5.2, I ∩ V (G) is an independent
set in G and Iv = I ∩ V (H v ) is an independent dominating set in H v for each
v ∈ V (G) \ I. Also, by Theorem 5.1, Sv is a dominating set in H v + v for each
v ∈ V (G). By the choice of (S, I), |Sv |+|Iv | = γ(H)+1 for all v ∈ I∩V (G), and
|Sv | + |Iv | = 1 + γi (H) for all v ∈ V (G) \ I. Since γ(H) ≤ γi (H), |I ∩ V (G)| =
β(G). Thus, |S| + |I| = β(G)(1 + γ(H)) + (|V (G)| − β(G))(1 + γi (H)). Or,
equivalently, γγi (G ◦ H) = |V (G)|(1 + γi (H)) − β(G)(γi (H) − γ(H)).
Corollary 5.7 If γi (H) = γ(H), then γγi (G ◦ H) = γγ(G ◦ H) for any graph
G.
Theorem 5.8 Let G be a connected nontrivial grah and H any graph. Then
γγt (G ◦ H) = |V (G)|(1 + γ(H)).
Proof : For each v ∈ V (G), let Dv ⊆ V (H v ) be a γ-set in H v . Then D =
∪v∈V (G) Dv is a dominating set in G ◦ H. Thus, (D, V (G)) is a dt-pair in G ◦ H.
This means that
X
γγt (G ◦ H) ≤ |D| + |V (G)| = |V (G)| +
γ(H) = |V (G)|(1 + γ(H)).
v∈V (G)
Conversely, let (D, T ) be a γγt -pair in G ◦ H, and let v ∈ V (G). By Theorem
5.1, Dv = D ∩ V (H v + v) is a dominating set in H v + v. In view of Theorem
5.4, we consider two cases:
Case 1: Suppose that Tv = V (H v + v) ∩ T is a total dominating set in H v + v.
Since (D, T ) is a γγt -pair and {v} dominates V (H v +v), if v ∈
/ Tv , then |Dv | = 1
and Tv ⊆ V (H v ) and |Tv | = γt (H). In this case, |Dv | + |Tv | = 1 + γt (H). On
the other hand, if v ∈ Tv , then |Dv | + |Tv | = γ(H) + 2.
Case 2: Suppose that v ∈ T and NG (v)∩T 6= ∅. By the same reason, Tv = {v}
and Dv ⊆ V (H v ) with |Dv | = γ(H). Thus, |Dv | + |Tv | = 1 + γ(H).
Combining all the above cases, |Dv | + |Tv | ≥ 1 + γ(H). Therefore,
X
γγt (G ◦ H) ≥
(1 + γ(H)) = |V (G)|(1 + γ(H)).
v∈V (G)
This completes the proof of the theorem.
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Received: March 28, 2016; Published: May 10, 2016