International Journal of Computer Applications (0975 – 8887)
Volume 83 – No3, December 2013
Strong Dominating Sets of Some Arithmetic Graphs
M.Manjuri
B.Maheswari
Dept. of Applied Mathematics, S.P.Women’s
University, Tirupati-517502, Andhra Pradesh, India
Dept. of Applied Mathematics, S.P.Women’s
University, Tirupati-517502, Andhra Pradesh, India
ABSTRACT
In this graph vertex 1 becomes an isolated vertex. Hence we
consider Arithmetic
graph without vertex 1 as the
contribution of this isolated vertex is nothing when
domination parameters are studied.
Graph Theory has been realized as one of the most flourishing
branches of modern Mathematics finding widest applications
in all most all branches of Sciences, Social Sciences,
Engineering, Computer Science, etc. Number Theory is one of
the oldest branches of Mathematics, which inherited rich
contributions from almost all greatest
mathematicians,
ancient and modern. Using the number theoretic function
Euler totient function an Euler totient Cayley graph is defined
and in this paper we study the Strong domination parameters
of Euler totient Cayley graphs and Arithmetic V n Graphs.
Keywords
Euler totient Cayley graphs, Arithmetic V n Graphs, Strong
domination, Cayley Graph
Subject Classification: 68R10
1. INTRODUCTION
Nathanson [1] was the pioneer in introducing the concepts of
Number Theory, particularly, the ‘Theory of Congruences’
in Graph Theory, thus paved the way for the emergence of a
new class of graphs, namely “Arithmetic Graphs”. Inspired by
the interplay between Number Theory and Graph Theory
several researchers in recent times are carrying out extensive
studies on various Arithmetic graphs in which adjacency
between vertices is defined through various arithmetic
functions.
Cayley Graphs are another class of graphs associated with
elements of a group. If this group is associated with some
Arithmetic function then the Cayley graph becomes an
Arithmetic graph. The Cayley graph associated with Euler
totient function is called an Euler totient Cayley graph .
2. EULER TOTIENT CAYLEY GRAPH
For each positive integer , let
be the additive group of
integers modulo and be the set of all numbers less than
and relatively prime to . The Euler Totient Cayley
graph
) is defined as the graph whose vertex set V is
given by
and the edge set is given
by
.
Some properties of Euler totient Cayley graphs can be found
in [2] and the domination parameters of these graphs are
presented in [3].
The Euler Totient Cayley graph
) is a complete graph
if is a prime and it is
regular.
3. ARITHMETIC
PROPERTIES
Let
GRAPH AND ITS
be a positive integer such that
. Then the Arithmetic
graph is defined as
the graph whose vertex set consists of the divisors of
and
two vertices
are adjacent in
graph if and only if
GCD
for some prime divisor
of
Clearly,
graph is a connected graph. If
is a prime, then
graph consists of a single vertex. Hence it is connected. In
other cases, by the definition of adjacency in
there exist
edges between prime numbers, their prime powers and also
to their prime products. Therefore each vertex of
is
connected to some vertex in
It is observed that the domination parameters these graphs are
functions of , where is the core of
that is the number
of distinct prime divisors of [ 4 ].
Let
denote the
graph throughout this paper.
Strong domination
Let
be a graph and
Then,
strongly
dominates
if (i)
and (ii)
A set
is called a strong-dominating set (sd-set) of if every
vertex in
is strongly dominated by at least one vertex
in
The strong domination number
of G is the minimum
cardinality of a strong dominating set.
Some results on strong domination can be seen in [5].
In the following sections we find minimum strong
dominating sets of
and
graphs and obtain
their strong domination number in various cases.
4. STRONG DOMINATING SETS OF
EULER TOTIENT CAYLEY GRAPH
Theorem 4.1: If is a prime, then the strong domination
number of
is 1.
Proof: Let be a prime. Then
is a complete graph
and hence every vertex is of degree
So, any single
vertex dominates all other vertices in
Hence every
vertex in
is adjacent with a vertex in
where
is any vertex in
Obviously the degree of every vertex in
of
is equal to the degree of the vertex
in
Thus
is a strong dominating set with
minimum cardinality
Hence
1.
Theorem 4.2: The strong domination number of
is 2, if
where is an odd prime.
Proof: Let us consider
for
,
is an odd
prime. Then the vertex set is given by
.
Let
be decomposed into the following disjoint subsets.
1.
2.
3.
The set of odd integers which are less than
relatively prime to
But this set is nothing but since
The set of non-zero even integers,
The set of integers 0 and
and
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International Journal of Computer Applications (0975 – 8887)
Volume 83 – No3, December 2013
We now show that
is a dominating set of
By the definition of edges in
, it is clear
that the vertices in are adjacent with the vertex 0 and hence
they are dominated by the vertex 0.
Now consider the elements of
By the definition of
it
contains non-zero even numbers. So if
then
is an
odd number, because
is an odd prime. Hence
or
That is the vertex is adjacent with the vertex
This implies that the vertices of
are dominated by the
vertex
Thus every vertex of
is dominated by the
vertices of
Therefore
is a dominating set of
Now we show that
is a strong dominating set of
.
respectively. Hence the vertices in
are dominated by the
vertices
In a similar way, we can show that all vertices of each
collection in
are dominated by the vertices
Thus the set
becomes a dominating set
in
as it dominates every vertex in
Now we
show that is a strong dominating set of
.
We know that the graph
is
regular.
So, the degree of every vertex in
of
is equal
to the degree of every vertex in
Thus every vertex in
is strongly dominated by at least one vertex in
Therefore
is a strong dominating set of
By the properties of
we know that the graph
is
regular. So for
,
is
regular. That is every vertex of
is of
degree
We claim that
is minimal. Suppose we delete from
and let
Then
cannot be a dominating
set of
, since the vertices of
are dominated only
by the vertex 0 and 0
.
That is the degree of the vertices
of
of vertices in
in
Suppose
we
delete
any
other
vertex, say
from
. Let
.
Suppose
is a dominating set of
. Consider the
vertex
in
, then the vertex
is dominated by some
vertex in
and
This implies that
Now
are in consecutive order implies that
. So
for all
and
such that
and
For
, we get
, a contradiction because GCD
Therefore it follows that
cannot be a dominating set of
and hence
is minimal.
is equal to the degree
So the vertices in
are strongly dominated by the vertex
and the vertices in
are strongly dominated by the
vertex
Thus every vertex in
one vertex in
is strongly dominated by at least
Therefore is a strong dominating set of
Since the
graph
is
regular, a single vertex cannot
dominate the rest of
vertices. Therefore
becomes a minimal strong dominating set with cardinality 2.
Thus
Theorem 4.3: Suppose
is neither a prime nor
Let
, where
,
, … are primes
and
are integers ≥ 1. Then the strong
domination number of
is given by
where is the length of the longest
stretch of consecutive integers in , each of which shares a
prime factor with
Proof: Suppose
and
prime nor 2p. Let us consider the vertex set
given by
. Then the set
following disjoint subsets.
is neither a
of
falls into the
1.
2.
The set of integers relatively prime to ,
The set
where
is a collection of
consecutive integers in
such that for every
in
, GCD
3. The singleton set
By the definition of adjacency in
, it is
obvious that the vertices of are dominated by the vertex
Let be a subset in with maximum cardinality Suppose
where GCD
for
Since
are in consecutive order, we
have
Therefore it follows that
So the vertices
are adjacent with the vertices
If a minimal dominating set is formed in any other way, then
the cardinality of such a set is not smaller than that of
This
follows from the properties of the prime divisors of a number.
Hence becomes a strong dominating set of
minimum cardinality.
with
Therefore
5. STRONG DOMINATING SETS OF
ARITHMETIC
GRAPH
Theorem 5.1: If
primes and
domination number of
, where , ,…., are
are integers ≥ 1, then the strong
is given by
where is the core of .
Proof: Suppose
. Consider
with
vertex set
Now we have the following possibilities.
Case 1: Suppose
for all . Now we show that the
set
becomes a dominating set of
.
By the definition of
graph, the vertices in
are primes
, their powers and their
products.
All the vertices
, for which GCD
are
adjacent with the vertex
in
for all
Since
every vertex in
has atleast one prime factor viz.,
( as they are divisors of
, every vertex in
is adjacent with at least one vertex in
Thus
becomes a dominating set of
.
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International Journal of Computer Applications (0975 – 8887)
Volume 83 – No3, December 2013
Now we show that
.
By the definition of edges in
the degree of
of
the
degree
the degree of
is a strong dominating set of
, we see that the degree of
and the degree
of
the degree of
All the vertices
, for which
are adjacent with the vertex
vertices
are adjacent with the vertices
and
and
and
respectively.
Therefore
is a strong
dominating set of
We now prove that
is minimal. Suppose we
remove any from Then the vertices of the form ,
are not dominated by any other vertex
in D as
for
Therefore every ,
must be included into
If we form a minimum dominating set in any other
manner, then the order of such a set is not smaller than that of
This follows from the properties of prime divisors of a
number. Hence
and so on
are adjacent with the vertex
for only one
That is,
with exponent 1. Then
is the
Let
The vertex
is adjacent only with the vertex
in
for any , as
Now the degree of
the degree of
.Hence it follows as in Case 1 that the vertices in
strongly dominate the vertices of
.
Therefore
is a strong dominating set of
and
Similarly we can show that every vertex in
shares a common factor or or ……or with some
vertex in
Thus every vertex in
one vertex in
Therefore
becomes a dominating set of
is adjacent with the vertex
Thus every vertex in
by at least one vertex in
Therefore
strong dominating set of
is strongly dominated
is a
.
Now we show that is minimum. Suppose we delete the
vertex from then the vertex
is not adjacent
with any other vertex in because
;
;….....;
for more than one
for all
Thus
a minimum strong dominating set of
That is
Suppose
. It is
is a strong dominating set of
,
,
and
Therefore
is a strong dominating set of
cardinality 2.
with
Now we show that the set
In this case the vertices of
and their products.
is a dominating
are primes
is
.
.
Sub case (ii): Suppose
Suppose
Then the vertex set of
is
Obviously
is a dominating set of
and this set is also a strong dominating set with
cardinality 1 as
,
set of
, we see that
;
,…,
Now the vertex
only.
Similar is the case with the deletion of any vertex
from , for
;
Here we have the following possibilities.
Suppose
.
By the definition of edges in
Hence
Then
clear that
as
is adjacent with at least
.
with cardinality
We have the following sub cases.
Sub case (i): Suppose
, the vertices
so on are adjacent with the vertex
GCD
Hence
Case 3: Suppose
the
The vertices
in
So, the degree of
is greater than the degree
of all other vertices in
Thus every vertex in
is
strongly dominated by at least one vertex in
Case 2: Suppose
only prime divisor of
in
That is
Now we show that the set
becomes a dominating set of
The vertices
as
are adjacent only with the vertex
in
;
Now
,
, Hence it follows as in Case
1 that the vertices in strongly dominate the vertices of
.
Therefore
set of
is a strong dominating
with cardinality
Hence
Sub case (iii): Suppose
That is
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International Journal of Computer Applications (0975 – 8887)
Volume 83 – No3, December 2013
Let
Subcase (ii) that
cardinality
. Then we can show as in
is a strong dominating set of
with
Hence
2
.
5
2
2
2 2
2 X5
Sub case (iv): Suppose
for
for
and
2
5
2
2x 5
Then
2
2X5
We can prove as in Case 1 and Sub case (i) of Case 3 that the
set
forms a minimum strong dominating set of
.
Hence
2x5
Fig.4
G (V100)
Strong Dominating Set = {2, 5}.
.
6. CONCLUSION
Using Number theory, it is interesting to study the strong
dominating sets of these Arithmetic graphs. This work gives
the scope for the study of strong dominating sets of product
graphs of these graphs and the authors have also studied this
aspect.
2
2 2
3
2x3x5
2
2x3x5
5
2
2x3x5
2
2
2x3x5
2
ILLUSTRATIONS
3
2 2
2x3
Euler totient Cayley Graphs
2x3
2
3x5
0
1
2x5
2
2x3
6
3x5
2
2x5
2
2
2x3
5
Fig.5
G (V180)
Strong Dominating Set = {2,3,15 }
3
4
Fig.1
Strong Dominating Set = {0}.
13
2
0
5
2x5
1
12
2
3
11
Fig.6
10
4
5
9
8
Strong Dominating Set = {10}.
6
7
Fig.2
Strong Dominating Set = {0,7}.
2
3
2x3x5
1
0
5
2
7
3x5
3
6
2x3
2x5
Fig.7
4
5
Fig.3
Strong Dominating Set = {2, 15}.
Strong Dominating Set = {0,1}.
Arithmetic Vn Graphs
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International Journal of Computer Applications (0975 – 8887)
Volume 83 – No3, December 2013
2
2
2
2x3x5x7
3
2x3x5x7
3
2x3x5x7
5
5
3x5x7
3x5x7
7
5x7
7
2x3
2x3x7
3x7
2x5
2x3
2x5x7
3x5
2x5
2x3x5
2x7
2
2x5x7
2
2x7
5x7
2
2x3x7
3x7
2
2
2x3
3x5
2
2x3x5
Fig.8
2
2x5
2x5x7
2
2x3x7
Strong Dominating Set = {2,6,35}.
2x3x5
2x7
Fig.10
2
Strong Dominating Set = {2,3,35 }.
3
2
2x3x5
7. REFERENCES
5
2x3x5
2
2
2
2x5
2x3
2
2x3
2x5
3x5
Fig.9
Strong Dominating Set = {2, 15}.
[1] Nathanson and B.Melvyn, -Connected components of
arithmetic graphs, Monat.fur.Math, 29, (1980), 219 –
220.
[2] L.Madhavi, Studies on domination parameters and enum
eration of cycles in some Arithmetic Graphs,Ph.D. Thesi
s
submitted to S.V.University, Tirupati,
India, (2002).
[3] S.Uma Maheswari, and B.Maheswari, – Domination
parameters of Euler Totient Cayley Graphs, Rev.Bull.Cal
Math.Soc., 19 (2), (2011), 207-214.
[4] S.Uma Maheswari, and B.Maheswari, – Some
Domination parameters of Arithmetic Graph Vn, IOSR,J
ournal of Mathematics, Volume 2, Issue 6, (Sep Oct 201
2), 14 -18.
[5] E.Sampathkumar, L. Pushpa Latha – Strong weak
domination and domination balance in graph,
Discrete Mathematics, 161(1996), 235-242.
IJCATM : www.ijcaonline.org
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