Bounds on double domination in squares of graphs

IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
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Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org 454
BOUNDS ON DOUBLE DOMINATION IN SQUARES OF GRAPHS
M. H. Muddebihal1
, Srinivasa G2
1
Professor, Department of Mathematics, Gulbarga University, Karnataka, India, mhmuddebihal@yahoo.co.in
2
Assistant Professor, Department of Mathematics, B. N. M. I. T , Karnataka, India, gsgraphtheory@yahoo.com
Abstract
Let the square of a graph G , denoted by 2
G has same vertex set as in G and every two vertices u and v are joined in 2
G if and
only if they are joined in G by a path of length one or two. A subset D of vertices of 2
G is a double dominating set if every
vertex in 2
G is dominated by at least two vertices of D . The minimum cardinality double dominating set of 2
G is the double
domination number, and is denoted by ( )2
d Gγ . In this paper, many bounds on ( )2
d Gγ were obtained in terms of elements of
G . Also their relationship with other domination parameters were obtained.
Key words: Graph, Square graph, Double dominating set, Double domination number.
Subject Classification Number: AMS-05C69, 05C70.
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1. INTRODUCTION
In this paper, we follow the notations of [1]. All the graphs
considered here are simple, finite and connected. As usual
( )p V G= and ( )q E G= denote the number of vertices
and edges of G , respectively.
In general, we use X〈 〉 to denote the subgraph induced by
the set of vertices X and ( )N v and [ ]N v denote the open
and closed neighborhoods of a vertex v, respectively. The
notation ( ) ( )( )0 1G Gα α is the minimum number of
vertices(edges) is a vertex(edge) cover of G .
Also ( ) ( )( )0 1G Gβ β is the minimum number of vertices
(edges) is a maximal independent set of vertex (edge) of G .
Let ( )deg v is the degree of a vertex v and as usual
( ) ( )( )G Gδ ∆ denote the minimum (maximum) degree of
G . A vertex of degree one is called an end vertex and its
neighbor is called a support vertex. Suppose a support
vertex vis adjacent to at least two end vertices then it is
called a strong support vertex. A vertex v is called cut
vertex if removing it from G increases the number of
components of G .
The distance between two vertices u and v is the length of
the shortest u v - path in G . The maximum distance
between any two vertices in G is called the diameter,
denoted by ( )diam G .
The square of a graph G , denoted by 2
G has the same
vertex set as in G and the two vertices u and v are joined
in 2
G if and only if they are joined in G by a path of length
one or two (see [1], [2]).
We begin by recalling some standard definitions from
domination theory.
A set S V⊆ is said to be a double dominating set of G , if
every vertex of G is dominated by at least two vertices of
S . The double domination number of G is denoted by
( )d Gγ and is the minimum cardinality of a double
dominating set of G . This concept was introduced by F.
Harary and T. W. Haynes [3].
A dominating set ( )S V G⊆ is a restrained dominating set
of G , if every vertex not in S is adjacent to a vertex in S
and to a vertex in V S− . The restrained domination number
of G , denoted by ( )re Gγ is the minimum cardinality of a
restrained dominating set of G . This concept was
introduced by G. S. Domke et. al.,[4].
A dominating set ( )S V G⊆ is said to be connected
dominating set of G , if the subgraph S is not
disconnected. The minimum cardinality of vertices in such a
set is called the connected domination number of G and is
denoted by ( )c Gγ [5].
A subset ( )2
D V G⊆ is said to be a dominating set of 2
G ,
if every vertex not in D is adjacent to some vertex in D .
The domination number of 2
G , denoted by ( )2
Gγ , is the

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minimum cardinality of a dominating set of 2
G .
Similarly, a dominating set D of 2
G is said to be total
dominating set of 2
G , if for every vertex ( )2
v V G∈ , there
exists a vertex u D∈ , u v≠ , such that u is adjacent to v
or if the subgraph D has no isolated vertex. The total
domination number of 2
G , denoted by 2
( )t Gγ is the
minimum cardinality of total dominating set of 2
G .
Domination parameters in squares of graphs were
introduced by M. H. Muddebihal et. al.,(see [6] and [7]).
Analogously, a subset ( )2
D V G⊆ is said to be double
dominating set of 2
G , if every vertex in 2
G is dominated
by at least two vertices of D . The double domination
number of 2
G , denoted by ( )2
d Gγ , is the minimum
cardinality of a double dominating set of 2
G . In this paper,
many bounds on ( )2
d Gγ were obtained in terms of
elements of G . Also its relationship with other different
domination parameters were expressed.
2. RESULTS
Theorem 2.1:
a. For any cycle pC , with 3p ≥ vertices,
( )2
2, 3.
1, 0(mod3)
3
, .
3
d p
for p
p
C for p
p
otherwise
γ

 =


= + ≡

 
 
 
b. For any complete graph pK , with 2p ≥ vertices,
( )2
2d pKγ = .
c. For any star 1,nK , with 2n ≥ vertices,
( )2
1, 2d nKγ = .
d. For any wheel pW , with 4p ≥ vertices,
( )2
2d pWγ = .
e. For any complete bipartite graph 1 2,p pK , with
1 2p p p+ = vertices,
( ),1 2
2
2p pd Kγ = .
Theorem 2.2: For any connected graph G with 3p ≥
vertices, ( )2
1
2
d
p
Gγ
 
≤ + 
 
.
Proof: For 2p ≤ , ( )2
2
d
p
Gγ
 
≤/  
 
. For 3p ≥ , we prove
the result by induction process. Suppose 3p V= ≤ in G ,
then ( )2
2
d
p
Gγ
 
=  
 
. Assume that the result is true for any
graph with p -vertices. Let G be a graph with 1p +
vertices. Then by induction hypothesis, it follows that
( )2 1
2
d
p
Gγ
+ 
≤  
 
. Hence the result is true for all graphs
with 3p ≥ vertices by induction process.
Theorem 2.3: For any connected graph G with 3p ≥
vertices, ( ) ( )2 2
d G G pγ γ+ ≤ . Equality holds if and only if
3 3,G C P≅ .
Proof: Let { }1 2, ,..., kS v v v= be the minimal set of vertices
which covers all the vertices in 2
G . Clearly, S forms a
dominating set of 2
G . Further, if there exists a vertex set
( )2
1V G S V− = in 2
G . Then '
S V D∪ = , where '
1V V⊆ in
2
G , be the set of vertices such that ( )2
v V G∀ ∈ , there
exists two vertices in '
S V D∪ = . Further, since every
vertex of 2
G are adjacent to at least two vertices of 2
G ,
clearly D forms a double dominating set of 2
G . Therefore,
it follows that D S p∪ ≤ . Hence ( ) ( )2 2
d G G pγ γ+ ≤ .
Suppose, 3 3,G C P≅/ . Then either 2 S D≠ or D S p∪ < ,
which gives a contradiction in both cases.
Suppose, 3 3,G C P≅ . Then in this case, 2 2 1 2D S= = ⋅ = .
Clearly, 3D S p∪ = = . Therefore, ( ) ( )2 2
d G G pγ γ+ = .
Theorem 2.4: For any connected ( ),p q - graph G ,
( ) ( )2 2
2 2dG Gγ γ≤ + .
Proof: Suppose { } ( )2
1 2, ,..., nS v v v V G= ⊆ be the minimal
set of vertices which covers all the vertices, such that
( ), 3dist u v ≥ for all { },u v S∈ . Then S forms a minimal
dominating set of 2
G . Further, if for every ( )2
v V G∈ ,
there exists at least two vertices { },u w S∈ such that ,u v∀ ,
( )N v and ( )N u belongs to ( )2
V G S− . Then S itself is a
double dominating set of 2
G . Otherwise, there exists at
least one vertex ( )x N S∈ such that { }S x D∪ = forms a
G . Since for any

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graph G with 2p ≥ , ( )2
2d Gγ ≥ . Therefore, it follows that
2
2
D
S
+
≤ . Clearly, ( ) ( )2 2
2 2dG Gγ γ≤ + .
Theorem 2.5: For any connected ( ),p q -graph G ,
with 3p ≥ vertices, ( ) ( )2
d G G pγ γ+ ≤ . Equality holds for
3 4 3 4 5 7, , , , ,C C P P P P .
Proof: Let { }1 1 2, ,..., mF v v v= be the set of all non end
vertices in G . Suppose { }1 2 1, ,..., ,kS v v v F k m= ⊆ ≤ , be
the minimal set of vertices which are at distance three
covers all the vertices of G . Then S itself forms a minimal
γ -set of G . Otherwise, there exists at least one vertex
[ ]v N S∈ such that { }S v∪ forms a minimal dominating set
of G . Now in 2
G , since ( ) ( )2
V G V G= , let
{ }1 2, ,..., iI u u u= be the set of all strong support vertices.
Suppose '
1D I F= ∪ , where '
1 1F F I⊆ − be the minimum
set of vertices which covers all the vertices in 2
G , such that
for every vertex ( )2
v V G∈ , there exists at least two
vertices { },u w D∈ where iv I∀ ∈ and
{ }' 2
1 [ ]j kv F v V G D∀ ∈ ∃ ∈ − has at least two neighbors
which are either iv or jv . Then D forms a minimal double
dominating set of 2
G . Therefore, it follows that
{ }D S v p∪ ∪ ≤ . Hence ( ) ( )2
d G G pγ γ+ ≤ .
Suppose 3 4 3 4, , ,G C C P P≅ . Then in this case, 2D = and
2S p= − . Clearly, it follows that D S p∪ = . Therefore,
( ) ( )2
d G G pγ γ+ = .
Suppose 5 7,G P P≅ . Then in this case,
2
p
D
 
=  
 
and
2
p
S
 
=  
 
. Clearly, it follows that
2 2
p p
D S
   
∪ = +   
   
. Therefore, ( ) ( )2
d G G pγ γ+ = .
Theorem 2.6: For any connected ( ),p q - graph G with
3p ≥ vertices, ( ) ( )2
2
d
diam G
G pγ
 
≤ −  
 
.
Proof: For 2p = , ( ) ( )2
2
d
diam G
G pγ
 
≤ −/  
 
. Hence
consider 3p ≥ . Suppose there exists two vertices
( ),u v V G∈ , which constitutes the longest path in G . Then
( ) ( ),dist u v diam G= . Since ( ) ( )2
V G V G= , there exists a
vertex set { }1 2, ,..., iD v v v= such that for every vertex
, 1jv D j i∈ ≤ ≤ , there exists at least one vertex
, 1kv D k i∈ ≤ ≤ . Also every vertex in 2
G is adjacent to at
least two vertices of D in 2
G . Then D forms a minimal
G . Since 2D ≥ and the
diametral path includes at least two vertices. It follows that,
( )2 2D p diam G≤ − . Clearly, ( ) ( )2
2
d
diam G
G pγ
 
≤ −  
 
.
Theorem 2.7: For any nontrivial tree with 3p ≥ vertices
and m cut vertices, then ( )2
1d T mγ ≤ + , equality holds
if 1,nT K≅ , 2n ≥ .
Proof: Let { }1 2, ,..., iB v v v= be the set of all cut vertices in
T with B m= . Suppose { }1 2, ,..., jA v v v= , 1 j i≤ ≤ be the
set of cut vertices which are at a distance two from the end
vertices of T and A B⊂ . Now in 2
T , all the end vertices
are adjacent with jv A∀ ∈ and { } { }B A− . Now in 2
T ,
since ( ) ( )2
V T V T= , for every vertex ( )2
v V T∈ , there
exists at least two vertices{ },u v B D∈ = in 2
T . Further,
since D covers all the vertices in 2
T , D itself forms a
minimal double dominating set of 2
T . Since every tree
T contains at least one cut vertex, it follows that
1D m≤ + . Hence ( )2
1d T mγ ≤ + .
Suppose T is isomorphic to a star 1,nK . Then in this case,
2D = and 1m = . Therefore, it follows that
( )2
1d T mγ = + .
Theorem 2.8: For any connected ( , )p q -graph G ,
( ) ( ) ( )2 2
d tG G Gγ γ≤ + ∆ .
Proof: For 2p = , the result follows immediately. Hence,
let 3p ≥ . Suppose { } ( )1 1 2, ,..., nV v v v V G= ⊆ be the set of
all vertices with ( )deg 2, 1iv i n≥ ≤ ≤ . Then there exists at
least one vertex 1v V∈ of maximum degree ( )G∆ . Now in
2
G , since ( ) ( )2
V G V G= , let { }1 1 2, ,..., kD v v v= 1V⊆ in
2
G . Suppose 1D covers all the vertices in 2
G and if the
subgraph 1D has no isolated vertex, then 1D itself is a
minimal total dominating set of 2
G . Otherwise, there exists
a set 1D H∪ , where ( )2
1H V G D⊆ − , forms a minimal
total dominating set of 2
G . Now let { }1 2 1, ,..., jD v v v V= ⊆
in 2
G be the minimal set of vertices, which covers all the
vertices in 2
G . Suppose ( )2
v V G∀ ∈ , there exists at least

_______________________________________________________________________________________
two vertices { },u w D∈ which are adjacent to at least one
vertex of D and at least two vertices of ( )2
V G D− . Then
D forms a dγ - set of 2
G . Otherwise D I∪ , where
( )2
I V G D⊆ − , forms a minimal double dominating set of
2
G . Clearly, it follows that ( )1D I D H G∪ ≤ ∪ + ∆ .
Therefore, ( ) ( ) ( )2 2
d tG G Gγ γ≤ + ∆ .
( ) ( )2
dG Gγ γ≤ . Equality holds if and only if ( ) 2Gγ =
with ( ) 3diam G ≤ .
Proof: If { } ( )1 1 2, ,..., nV v v v V G= ⊆ be the set of vertices
with ( )deg 2, 1iv i n≥ ≤ ≤ . Then { }1 2 1, ,..., kS v v v V= ⊆
forms a minimal dominating set of G . Now without loss of
generality in 2
G , since ( ) ( )2
V G V G= . If
{ }2 1 2, ,..., kV v v v= be the set of vertices with ( )deg 2kv < .
If 2 ( )V V G∈ , then the vertices which are at a distance at
least two are adjacent to each vertex of 2V in 2
G . Hence
1 2S V D∪ = where 1S S⊆ forms a minimal double
dominating set of 2
G . If 2V φ= , then 3S V D∪ = where
3 1V V⊆ forms a minimal double dominating set of 2
G .
Further, since every vertex in 2
G is adjacent to atleast two
vertices of D , it follows that S D≤ . Hence,
( ) ( )2
dG Gγ γ≤ .
Suppose ( ) 2Gγ ≠ with ( ) 3diam G ≤ . Then in this case
( ) 1diam G = and hence, 1S = . Clearly, S D< .
Therefore ( ) ( )2
dG Gγ γ< , a contradiction.
Further, if ( ) 2Gγ = with ( ) 3diam G ≤/ . Then in this case,
( ) 4diam G ≥ . Clearly, D S> . Therefore,
( ) ( )2
dG Gγ γ< , again a contradiction.
Hence ( ) ( )2
dG Gγ γ= if and only if ( ) 2Gγ = with
( ) 3diam G ≤ .
( ) ( )2
0 1d G p Gγ α≤ − + . Equality holds for pK .
Proof: Let { } ( )1 2, ,..., nA v v v V G= ⊆ , where
( )deg 2, 1iv i n≥ ≤ ≤ , be the minimum set of vertices which
covers all the edges of G , such that ( )0A Gα= . Now in
2
G since ( ) ( )2
V G V G= , let { }1 2, ,..., kD v v v A= ⊆ be the
set of vertices such that for every vertex ( )2
v V G∈ , there
exists at least two vertices { },u w D∈ in 2
G . Further, if
D covers all the vertices in 2
G , then D itself is a double
dominating set of G . Clearly, it follows that 1D p A≤ − +
and hence ( ) ( )2
0 1d G p Gγ α≤ − + .
Suppose pG K≅ . Then in this case, 1A p= − and 2D = .
Clearly, it follows that 1D p A= − + and
hence ( ) ( )2
0 1d G p Gγ α= − + .
( ) ( )2
d tG Gγ γ≤ .
Proof: Let { } ( )1 2, ,..., nK u u u V G= ⊆ be the set of vertices
such that [ ]i jN u N u φ ∩ =  , where 1 i n≤ ≤ , 1 j n≤ ≤ .
Suppose there exists a minimal set
{ } ( )1 1 2, ,..., kK u u u N K= ∈ , such that the subgraph
1K K∪ has no isolated vertex. Further, if 1K K∪ covers
all the vertices in G , then 1K K∪ forms a minimal total
dominating set of G . Since ( ) ( )2
V G V G= , there exists a
vertex set { }1 2 1, ,..., mD v v v K K= ⊆ ∪ in 2
G , which covers
all the vertices in 2
G and for every vertex ( )2
v V G∈ , there
exists at least two vertices { },u w D∈ . Clearly, D forms a
minimal double dominating set of 2
G . Therefore, it follows
that 1D K K≤ ∪ . Hence ( ) ( )2
d tG Gγ γ≤ .
Theorem 2.12: For any connected ( ),p q -graph G ,
( ) ( )2
0 1d G Gγ β≤ + . Equality holds for pK .
Proof: For 2p = , the result is obvious. Hence let 3p ≥ .
Suppose { } ( )1 2, ,..., mF u u u V G= ⊆ be the set of all vertices
with ( )deg 1, 1iv i m= ≤ ≤ . Then '
F F∪ , where
( ) [ ]' '
,F V G F F N F⊆ − ∉ forms a maximal independent
set of vertices, such that ( )'
0F F Gβ∪ = . Since
( ) ( )2
V G V G= , let { } ( )2
1 1 2, ,..., nD v v v V G F= ⊆ − and
( )1D N F∈ . Suppose ( )2
2 1D V G D⊆ − such that
1 2D D D∪ = forms a minimal set of vertices which covers
all the vertices in 2
G . Further, if for every vertex
( )2
v V G∈ , there exists at least two vertices { },u w D∈ .
Then D forms a minimal double dominating set of 2
G .
Since every graph G contains at least one independent
vertex, it follows that '
1D F F≤ ∪ + . Therefore,
( ) ( )2
0 1d G Gγ β≤ + .

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Suppose pG K≅ . Then in this case, G contains exactly one
independent vertex and by Theorem 2.1(b), it follows that
( ) ( )2
0 1d G Gγ β= + .
Theorem 2.13: For any non-trivial tree T ,
( ) ( )2
1d reT Tγ γ≤ + .
Proof: Let { } ( )1 2, ,..., nF v v v V T= ⊆ be the set of vertices
with ( ) { }deg 1, , 1i iv v F i n= ∀ ∈ ≤ ≤ . Suppose for every
vertex ( )v V T F∈ − , there exists a vertex u F∈ and also a
vertex ( )x V T F∈ − . Then F itself is a restrained
dominating set of T . Otherwise, there exists at least one
vertex ( )w V G F∈ − , such that { }'
D F w= ∪ forms a
minimal restrained dominating set of T . Let
{ }1 2, ,..., kD u u u V F= ⊆ − in 2
T be the minimal set of
vertices which are chosen such that ( )2
v V T∀ ∈ , there
exists at least two vertices { },y z D∈ . Further, since
D covers all the vertices in 2
T , clearly D forms a minimal
T . Therefore, it follows that
'
1D D≤ + due to the distance between vertices of T is
one. Hence ( ) ( )2
1d reT Tγ γ≤ + .
Theorem 2.14: For any connected graph G ,
( ) ( )2
1d cG Gγ γ≤ + .
Proof: Suppose { } ( )1 2, ,..., nC v v v V G= ⊆ be the set of all
cut vertices in G . Further, if C I∪ , where ( )I N C∈ with
( ) { }deg 2,i iv v I≥ ∀ ∈ be the minimal set of vertices which
covers all the vertices in G and if the sub graph C I∪ is
connected. Then C I∪ forms a minimal connected
dominating set of G . Let { }1 2, ,..., kD v v v= be the minimal
set of vertices which covers all the vertices in 2
G . Suppose
for every vertex ( )2
v V G∈ , there exists at least two
vertices{ },u w D∈ . Then D itself forms a minimal double
dominating set of 2
G . Therefore, it follows that
1D C I≤ ∪ + and hence ( ) ( )2
1d cG Gγ γ≤ + .
REFERENCES
[1]. F. Harary, Graph Theory, Adison-Wesley, Reading,
Mass., 1972.
[2]. F. Harary and I. C. Ross, The square of a tree, Bell
System Tech. J. 39, 641-647, 1960.
[3]. F. Harary and T. W. Haynes, Double domination in
graphs, Ars Combinatorica 55, 201-213, 2000.
[4]. G. S. Domke, J. H. Hattingh, S. T. Hedetniemi, R. C.
Laskar and L. R. Markus, Restrained Domination in
Graphs, Discrete Mathematics 203, 61–69, 1999.
[5]. E. Sampathkumar and H. B. Walikar, The Connected
Domination Number of Graphs, J.Math.Phy.Sci. 13,
607-613, 1979.
[6]. M. H. Muddebihal, G. Srinivasa and A. R. Sedamkar,
Domination in Squares of Graphs, Ultra Scientist,
23(3)A, 795–800, 2011.
[7] M. H. Muddebihal and G. Srinivasa, Bounds on Total
Domination in Squares of Graphs, International Journal
of Advanced Computer and Mathematical Sciences,
4(1), 67–74, 2013

Bounds on double domination in squares of graphs

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