Download 17. A GENERAL ITERATIVE METHOD FOR SOLVING MIXED EQUILIBRIUM PROBLEMS, COCOERCIVE QUASIVARIATIONAL INCLUSIONS PROBLEMS AND FIXED POINT PROBLEMS OF AN INFINITE FAMILY OF STRICTLY PSEUDOCONTRACTIVE MAPPINGS Author: P. Tianchai

INTERNATIONAL JOURNAL OF MATHEMATICAL SCIENCES AND APPLICATIONS
Volume 5 • Number 1 • January-June 2015
TOTAL DOMINATION NUMBER OF PRODUCT OF CYCLES
M. Thiagarajan1 and R. Bhaskaran2
1,2
School of Mathematics, Madurai Kamaraj University, Madurai-625021, Tamil Nadu, India
E-mail: [email protected] , [email protected]
Abstract: Let G = (V, E) be a graph. A subset of vertices S � V is called a total dominating set of G if every vertex of
is adjacent to some vertex in S. The total domination number �t(G) of a graph G is the cardinality of a minimum total
dominating set. In this paper, we deal with the problem of finding the total domination number for some cartesian
product of cycles �t(Ck × Cn) by enumeration process.
Keywords: Cycle, Cartesian Product, Total domination number.
1. INTRODUCTION
In this paper, we consider only the graphs that are undirected, without loops or multiple edge. We use mainly
Bondy and Murty’s book [1] for terminologies and notations. Let G be a graph with vertex set V(G) and edge set
E(G). If u, v � V then the edge joining u and v is denoted by uv. Any two vertices u,v � V(G) are said to be
adjacent if uv � E(G). For a vertex v � V (G), the set of vertices that are adjacent to v is said to be the
neighborhood of v, denoted by N(v).
For any non-empty subset A of V(G), we define the neighborhood of A as N ( A) = �N (v ) . The closed
v� A
neighborhood of A denoted by N[A] is A � N ( A) . A set D of vertices in a graph G is called a dominating set of G
if every vertex v � V is either an element of D or is adjacent to an element of D. The domination number �(G) of
a graph G is the minimum of the cardinalities of dominating sets of G. Inspired by the famous five queen problem
Cockayne, Dawes and Hedetniemi [2] introduced the concept of Total domination in graphs. A total dominating
set D is a subset of V such that N(D) = V(G). The total domination number �t(G) of a graph G is the minimum of
the cardinalities of total dominating sets.
The Cartesian product of two graphs G1 and G2 is defined as the graph G1 × G2 with vertex set V(G1) ×
V(G2) and (u1, u2) (v1, v2) � E(G1 × G2) if either u1 = v1 and u2v2 � E(G2) or u2 = v2 and u1v1 � E(G1).
Problems on various domination parameters of cartesian product of graphs are of great interest ever since Vizing
[10] conjectured that � (G � H ) � � (G ) � ( H ) . This conjecture is still open.
Domination number of cartesian product of cycles were intensively investigated by Sandi Klavzar and Norbert
Seifter [6], by Mohamed H. El-Zahar and Ramy S. Shaheen [4]. Total domination number of complete grid graphs
Pm � Pn were calculated for m = 1, 2, 3, 4 by Sylvain Gravier [7] and for m = 5, 6 by A. Klobucar [3]. For further
details regarding domination theory on graphs, we refer to the extensive survey book on various domination
parameters by Teresa W. Haynes et al. in [8].
2. NOTATIONS AND TERMINOLOGIES
Throughout this paper we denote the vertices of a path Pn or a cycle Cn by 1, 2, ...n.. This notation is convenient
to formulate the proofs. Thus the vertices of Ck × Cl are denoted by (i, j), where i = 1, 2, .... k and j = 1, 2, ... l.
122
M. THIAGARAJAN AND R. BHASKARAN
Let x �V (Cm ) and y � V (Cn ) , then we define x × y as a subgraph of Cm � Cn induced by the vertices
{(1, 2),� , ( x, y )} . It is to be noted that x � y � Cm � Cn . For m �V (Cl ) , we define the set (Ck)m = Ck × {m}
and for n � V (Ck), n (Cl ) = {n} � Cl . We call (Ck ) m as the mth column of Ck � Cl and n (Cl ) the nth row of
m�n
Ck � Cl . The set B =
� (C ) , where n � 0, m � 1, m � n � l is called a k � (n � 1) block of
k i
consecutive columns
i= m
of Ck � Cl .
Figure 1: k × (n + 1) Block of Cn × Cx
In this paper, we determine the total domination number of Ck � Cn . Given any n, we can always write it as
n = tq � r , where 1 � t � n, q � 1 and r = 0,1,� , t � 1 . The idea of the proof is to split Ck � Cn into a primary
block Q1 and a secondary block Q2. Primary block Q1 contain q number of k × t blocks, where each k × t block
is called the base block. We choose the vertices of a total dominating set D of Cm × Cn by enumeration process
such that there is no overlapping (i.e. no vertex is totally dominated by more than one vertex). Another important
point to be considered is that when we repeat the pattern of above formed total dominating set in the following base
blocks we must retain the character of total domination, i.e. there is no isolated vertex in the induced subgraph
G[D].
3. MAIN RESULTS
We know that as it can be easily shown that for any cycle Cn,
�n
�� 2 � 1 ifn � 2(mod 4)
� t (Cn ) = �
�� n � ,
otherwise
� �
�� � 2 �
TOTAL DOMINATION NUMBER OF PRODUCT OF CYCLES
123
and consequently we have
Proposition 3.1: For any n � 3, we have
�n
�� 2 � 1, ifn � 2(mod 4)
� t ( K1 � Cn ) = �
��n�,
otherwise.
�� �� 2 ��
Proof: Noting that K1 � Cn � Cn , we have � t ( K1 � Cn ) = � t (Cn ) .
Theorem 3.2: [9] For any n � 3,
� � 2n �
� � 3 � � 1, if n � 4(mod 6)
�� �
� t ( P2 � Cn ) = �
� � 2n � ,
otherwise.
�� �� 3 ��
� 4n �
Theorem 3.3: For any n � 3 , � t (C3 � Cn ) = � � .
�5 �
Proof: Let S' be a total dominating set for the base block of C3 � Cn . If we start with the vertex (1,1) then
we can include one among {(1, 2), (2,1), (3,1), (1, t )} . When we include (2,1) or (3,1) in our total dominating
set, we can totally dominate 5 vertices but in the case of including (1,2) or (1,t) we can totally dominate one
more vertex as compared to the previous case. Among these two, we choose (1,t) as it aids to maintain
adjacency with the adjacent block. Now {(1,1), (1, t )} are in our total dominating set. So we can not include
{(2, 2), (3, 2), (2, t � 1), (3, t � 1)} to avoid over-lapping. Thus we include (2,3) and (3,3) in our total dominating
set.
Now {(1,1), (2,3), (3, 2), (1, t )} , is our total dominating set which implies t � 4 . If t = 4 then over-lapping
occurs and when t = 5 the set {(1,1), (2,3), (3,3), (1,5)} totally dominates C3 � C5 . Thus any C3 � Cn can be split
in to a primary block Q1 (q number of 3 × 5 base blocks) and if necessary a secondary block Q2 3 × r, where
r = 1, 2,3, 4,5) . In Q1 each base block is totally dominated by
� n �1�
S = {(1,1 � k ), (2,3 � k ), (3,3 � k ), (1,5 � k ) /k = 0,1,� , �
�}
� 5 �
Figure 2: Total Dominating Set for Q1
124
M. THIAGARAJAN AND R. BHASKARAN
If n � 0 (mod 5) then C3 � Cn can be split in to
n
number of 3 × 5 base blocks and each base block 3 × 5 is
5
totally dominated by four vertices as described above. Thu � t (C3 � Cn ) s =
� 4n �
4n
= � �.
5
� 5 �
Figure 3: Total Dominating Set for Q2
� � n � 1� �
If n � 1, 2,3, 4(mod 5) then n columns in can be split in to a primary block Q1 � �
� � ( number of 3 × 5
�� 5 ��
base blocks) totally dominated by S and a secondary block Q2 which is totally dominated by a set isomorphic to
� 4n �
R1, R2, R3, R4 respectively as shown above. In all other cases it can be easily shown that � t (C3 � Cn ) = � � .
�5 �
ifn � 0(mod 4)
� n,
�
� (C � Cn ) = � n � 1, ifn � 1,3(mod 4)
Theorem 3.4: For any n � 4 , t 4
�n � 2, ifn � 2(mod 4).
�
Proof: Here we start with (1, 1) in our total dominating set for C4 × Cn. Then we can include one among
{(1, 2), (2,1), (1, t ), (4,1)} . In our case, we choose {(1,1), (4,1)} . From {(1,1), (4,1)} we can totally dominate
the following vertices {(1, 2), (2,1), (1, t ), (3,1), (4, 2), (4, t )} . Since the inclusion of (2, 2), (3, 2) cause overlapping
we exclude them. If we include (2, 3), (3,3) in the total dominating set then from {(1,1), (2,3), (3,3), (4,1)} we can
totally dominate all the vertices of 4 � 4 . Thus we have found a total dominating set {(1,1), (4,1), (2,3), (3,3)} for
the base block 4 � 4 . If n � 0(mod 4) then C4 � Cn is totally dominated by the set
� n � 1� �
4n
= n.
� . Thus � t (C4 � Cn ) =| S |=
S = {(1,1 � k ), (4,1 � k ), (2,3 � k ), (3,3 � k ) k = 0,1,� , �
�
� 4 ��
4
Figure 4: Total Dominating Set for Q1
125
TOTAL DOMINATION NUMBER OF PRODUCT OF CYCLES
If n = 4q + r, where 1 � r � 3 then C4 � Cn can be split in to a primary region containing q number of 4 × 4
blocks which is totally dominated by S and secondary region containing a 4 × r, which is totally dominated by a set
isomorphic to R1, R2 and R3 respectively.
Figure 5: Total Dominating Set for Q2
When
n � 2( mod 4) , we have � t (C4 � Cn ) = � t (C4 � Cn� 2 ) � 4 = (n � 2) � 4 = n � 2. If n � 1( mod 4) ,
we have
� t (C4 � Cn ) = � t (C4 � Cn�1 ) � 2 = (n � 1) � 2 = n � 1. When n � 3( mod 4) , we have
� t (C4 � Cn ) = � t (C4 � Cn �3 ) � 4 = (n � 3) � 4 = n � 1.
Theorem 3.5: For any n � 5,
� 4n
,
ifn � 0,3( mod 6)
�
� 3
��� 4n �
� t (C5 � Cn ) = � � � , ifn � 1, 2,5(mod 6)
�� 3 �
� � 4n �
ifn � 4(mod 4).
�� � ,
�� 3 �
Proof: By enumeration process, we construct the base block 5 × t, (where t is to be determined) for C5 × Cn
as follows: First we start with (1,1) and as a consequence we have to include any one among {(2,1),(1,2),(5,1),(1, t )}
to maintain the characteristic of total domination. Here we include (5,1). From {(1,1),(5,1)} we can totally dominate
{(2,1),(1, 2),(1, t ),(5, t ), (4,1), (5, 2)} . To avoid overlapping we include (3, 2),(3,3) . Now with {(1,1),(5,1), (3, 2),(3,3)} we can
totally dominate
{(2,1),(1,2),(5,2),(4,1),(1, t ),(5, t ),(3,1),(2,2),
(4,2),(2,3),(3, 4),(4,3)}.
In the next step we include (1, 4) and (5, 4). At this stage we can totally dominate 5 � 4 but we cannot consider
it as a base block since overlapping occurs. If we add (3, 5) and (3, 6) then the collection totally dominates 5 × 6
and it is a minimum one. Thus we fix 5 × 6 as our base block having
S = {(1,1),(5,1),(3, 2),(3,3),(1, 4),(5, 4),
(3,5),(3,6)}
dominating set. Hence given any C5 × Cn can be split into a primary region Q1 containing
blocks and if necessary a secondary region Q2 � 5 � r , where r = 1,2,3,4,5 . Suppose
is totally dominated by the set S = {(1,1 � 6k ), (5,1 � 6k ),(3,2 � 6 k ),(3,3 � 6k ),(1, 4 � k ),
� n �1� �
(5,4 � 6 k ),(3,5 � 6k ), (3,6 � 6 k )/k = 0,1,�, �
��
� 6 ��
� n � 1�
� 6 �
�
�
n � 0( mod 6)
as its total
number of 5 × 6
then given
C5 � Cn
126
M. THIAGARAJAN AND R. BHASKARAN
Figure 6: Total Dominating Set for Q1
4n
Thus we have � t (C5 � Cn ) =| S |= 3 . If r > 0 where r � n( mod 6) then primary region of C5 � Cn is totally dominated
by S and its secondary region is done by a set isomorphic to Rr as shown below
Figure 7: Secondary Block Q2 of C5 × Cn
In the case of
�n�
n � 1( mod 6) given C5 × Cn is divided into a primary region Q1 which consist of � �
�6�
× 6 block and secondary region
Q2 � R1 . Each 5 × 6 block is totally dominated by eight vertices and Q2 by a couple
of vertices thus we have � t (C5 � Cn ) =
of
n�2
6
number of
5�6
4(n � 1)
4 n � 2 � 4n �
�2=
= � � . If n � 2( mod 6)
3
3
�3�
then primary block Q1 of C5 � Cn consist
blocks and a 5 × 2 block thus we have.
� t (C5 � Cn ) =
Similarly when
number of 5
n � 3, 4,5( mod 6)
4(n � 2)
4n � 1
�3=
3
3
we can calculate its � t (C5 � Cn ) as
4n � 4n �
,
3 �� 3 ��
and
� 4n �
� 3�
� �
respectively. Hence we get
the desired result.
REFERENCES
[1]
J.A. Bondy and U.S.R. Murty, “Graph Theory with Applications”, Macmillan Press, London (1976).
[2]
E.J. Cockayne, R.M. Dawes and S.T. Hedetniemi, “Total Domination in Graphs”, Networks 10(1980) 211219.
TOTAL DOMINATION NUMBER OF PRODUCT OF CYCLES
127
[3] A. Klobuár, “Total Domination Numbers of Cartesian Products”, Mathematical Communications
9(2004) 35-44.
[4]
Mohamed H. El-Zahar and Ramy S.Shaheen, “On the Domination Number of the Product of Two Cycles”,
ARS Combinatoria, 84(2007) 51-64.
[5]
Nasrin Soltankhah, “Results on Total Domination and Total Restrained Domination in Grid Graphs”,
International Mathematical Forum, 5(2010)no. 7 319-332.
[6]
Sandi Klavzar and Norbert Seifter, “Dominating Cartesian products of Cycles”, Discrete Applied Mathematics,
59(1995)129-136.
[7] Sylvain Gravier, “Total Domination Numbers of Grid Graphs”, Discrete Applied Mathematics,
121 (2002)119-128.
[8]
Teresa W. Haynes, Stephen T. Hedetniemi and Peter J. Slater, “Fundamentals of Domination in Graphs”,
Marcel Dekker, New York (1998).
[9]
M. Thiagarajan and R. Bhaskaran, “Total Domination Numbers of Cylindrical Grid Graphs”, Communicated.
[10] V.G. Vizing, “The Cartesian Product of Graphs”, Vychisl. Sistemy, 9(1963) 30-43.