Vol 4, No 2 (2013)

(1)

Available online throug

ISSN 2229 – 5046

International Journal of Mathematical Archive- 4(2), Feb. – 2013 298

DOMINATING SETS OF SQUARE OF CENTIPEDES

A. Vijayan

1

& K. Lal Gipson

*

1

Associate Professor, Department of Mathematics, Nesamony Memorial Christian College,

Marthandam, Kanayakumari District, Tamil Nadu, South India

*

Assistant Professor, Department of Mathematics, Mar Ephraem College of Engineering and

Technology,Elavuvilai, Mathandam, Kanayakumari District, Tamil Nadu, South India

(Received on: 15-01-13; Revised & Accepted on: 15-02-13)

ABSTRACTS

L

et G = (V, E) be a simple graph. A set S ⊆ V is a dominating set of G, if every vertex in V-S is adjacent to atleast one vertex in S. Let

P

_n∗2 be the square of centipede corresponding to the path

P

_n and let

D P

(

_n∗2

, )

i

denote the family of all dominating sets of

P

_n∗2 with cardinality i. In this paper, some properties of the dominating sets of centipedes are

exhibited. Also, we characterized the number of dominating sets of

P

_n∗2and

P

_n∗2

−

{ }

2 n

of cardinality i.

Keywords: domination set, domination number.

1. INTRODUCTION

Let G = (V, E) be a simple graph of order |V| = n. For any vertex v∈V, the open neighbourhood of ν is the set N (v) = {u∈V/uv∈E} and the closed neighbourhood of ν is the set N[v] = N(v) ∪{v}. For a set S ⊆ V, the open neighbourhood of S is N(S) = Uν∈S N (v) and the closed neighbourhood of S is N[S] = N(S) ∪ S. A set S ⊆ V is a dominating set of G,

if N[S] = V, or equivalently, every vertex in V-S is adjacent to atleast one vertex in S. The domination number of a

graph G is defined as the minimum size of a dominating set of vertices in G and it is denoted as

γ

( )

G

.

A path is a connected graph in which end vertices have degree one and the remaining vertices have degree two, and is denoted by

n

P

.The centipede

P

_n∗ consists of a path

P

_n in which each vertex imbedded with a pendant edge and a pendant vertex.

Definition: 1.1: The 2nd power of a graph with the same set of vertices as G and an edge between two vertices if and only if there is a path of length at most 2 between them.

As usual we use

 

_{ }

x

for the largest integer less than or equal to x and

 

_{ }

x

for the smallest integer greater than or equal to x. Also, we denote the set {1, 2… n} by [n], throughout this paper.

2. DOMINATING SETS OF SQUARE OF CENTIPEDES

For the construction of the dominating sets of the square of centipede,

P

_n∗2, we need to investigate the dominating sets

of

P

n 2

{ }

2 n

∗

₋

.In this section we investigate dominating sets of

P

_n∗2. Let

D P

(

_n∗2

, )

i

be the family of dominating

sets of

P

_n∗2with cardinality i. We shall find recursive formula for

D P

(

_n∗2

, )

i

.We need the following lemmas to obtain the result of this section:

Lemma 2.1: γ (

P

_n2) =

5 n

 

 

By lemma 2.1 and the definition of domination number, one has the following lemma:

Corresponding author: K. Lal Gipson*

*_{Assistant Professor, Department of Mathematics, Mar Ephraem College of Engineering and}

(2)

Lemma 2.2: For every

n

∈

N

,

i)

(

2

)

3

n

P

γ

∗

 

=  

_{ }

ii)

(

2

{ }

2 )

3

n

P

n

γ

∗

₋

 

=  

_{ }

iii)

(

2

, )

2

3

n

D P

∗

i

= Φ

if and only if i

 

_{ }

or i

n

 





iv)

(

2

{ }

2 , )

2

1

3

n

D P

∗

−

n i

= Φ

if and only if i

 

_{ }

or i

n

−

 





Proof:

i) Clearly

{

3, 7,11,...2

n

−

1 }

is a minimum dominating set for

P

_n∗2.If n is even or odd it contains

3 n

 

 

elements.

Hence

(

2

)

3

n

P

γ

∗

 

=  

_{ }

.

ii) Clearly

{

3, 7,11,...2

n

−

3 }

is a minimum dominating set for

P

_n∗2

−

{ }

2 n

. If n is even or odd it contains

3 n

 

 

elements.

Hence

(

2

{ }

2 )

3

n

P

n

γ

∗

₋

 

=  

_{ }

iii) It follows from (i) and the definition of dominating set. iv) It follows from (ii) and the definition of dominating set.

For the construction of

D P

(

_n∗2

, )

i

we consider

{ }

{

}

2 2 2 2

1 1 3

(

_n

2 ,

1),

(

_n

,

1),

(

_n

2 2 ,

1),

(

_n

,

1).

D P

∗

−

n i

−

D P

∗₋

i

−

D P

∗₋

−

n

−

i

−

a n d D

P

∗₋

i

−

The families of these dominating sets can be empty or otherwise. Thus, we have eight combinations, whether these five families are empty or not. Two of these combinations are not possible (Lemma 2.3 (i), (ii) & (iii)).

Also, the combinations

D P

(

_n∗2

−

{ }

2 n i

,

− =

1)

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

do not need to be considered because it implies that

D P

(

_n∗2

, )

i

=

Φ

(See lemma 2.2 (iii)). Thus we only need to consider four combinations or cases. We consider those cases in Lemma (2.3).

Lemma 2.3:

i) If

D P

(

n2

{ }

2 n i

,

1)

D P

(

n 2₁

,

i

1)

D P

(

n ₁2

{

2 n

2 ,

}

i

1)

∗ ∗ ∗

− −

−

− =

−

− = Φ

2

3

(

_n

,

1)

then D P

∗₋

i

− = Φ

.

ii) If

D P

(

_n∗2

−

{ }

2 n i

,

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

then D P

(

_n∗₋₁2

,

i

− = Φ

1)

. iii) If

D P

(

_n∗2

−

{ }

2 n i

,

− =

1)

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

,

2

(

n

, )

then D P

∗

i

= Φ

.

iv) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

, 2

(

_n

, )

then D P

∗

i

≠ Φ

v) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

, ,

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− = Φ

1)

,then

D P

(

n 2₁

,

i

1)

∗

−

− ≠ Φ

.

Proof:

i) If

D P

(

_n∗2

−

{ }

2 n i

,

− =

1)

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− = Φ

1)

, by lemma 2.2(iii),(iv)

1

3 n

i

−

 

_{ }

 



or

i

−

1 

2 n

−

1

,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

2

,

1

3

3 n

i

−



_

−



_





(3)

Therefore,

1

2

1

3 n

i

−



_

−



_

or i

−

n

−









Suppose,

D P

(

_n∗₋2₃

,

i

− ≠ Φ

1)

then

3

1 2(

3)

3 n

i

n

−



_{ ≤ − ≤ −}









.

We have

1

3

1

3

3 n

n

i

−



 

_≤



_{≤ −}



 





 



.But

1

3 n

i

−



_

−



_







,which is a contradiction.

Therefore,

D P

(

_n∗₋2₃

,

i

− = Φ

1)

.

ii) If

D P

(

n2

{ }

2 n i

,

1)

D P

(

n 2₁

{

2 n

2 ,

}

i

1)

D P

(

n 2₃

,

i

1)

∗ ∗ ∗

− −

−

− =

−

− =

− = Φ

, by lemma 2.2(iii),(iv)

1

3 n

i

−

 

_{ }

 



or

i

−

1 

2 n

−

1

,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-3,

1

3

3 n

i

−



_

−



_







or

i

−

1 

2 n

-6.

There fore

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

1

.Therefore

1

3

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

2

holds.

Hence,

D P

(

_n∗₋2₁

,

i

− = Φ

1)

.

iii) If

D P

(

n 2

{ }

2 n i

,

1)

D P

(

n ₁2

,

i

1)

D P

(

n ₁2

{

2 n

2 ,

}

i

1)

D P

(

n 2₃

,

i

1)

∗ ∗ ∗ ∗

− − −

−

− =

−

− =

− = Φ

. by lemma 2.2(iii),(iv),

1

3 n

i

−

 

_{ }

 



or

i

−

1 

2 n

−

1

,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

2

,

1

3 n

i

−



_

−



_







or

1

2 i

−



n

-3,

1

3

3 n

i

−



_

−



_







or

i

−

1 

2 n

-6.

Therefore,

1

3

3 n

i

−



_

−



_







or

i

−

1 

2 n

-1.

That is,

3

1

3

3 n

n

i



_

−



_

+ ≤

 

_{ }





 



or

i



2 n

.

Therefore,

3 n

i

 

_{ }

 



or

i



2 n

Hence,

D P

(

_n∗2

, )

i

= Φ

.

iv) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

,

then

1

2

1

3 n

i

n

  ≤ − ≤ −

 

 

,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

2

,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-3 and

1

3

3 n

i

−



_

−



_







or

i

−

1 

2 n

-6.

Therefore, we have

2 n

−

2 

i

− ≤

1

2 n

−

1

.

Therefore,

2 n

− ≤ − ≤

1 i

1 2

n

−

1

.

Therefore, i-1=2n-1

Therefore, i=2n

Therefore

D P

(

n 2

, )

i

∗

_{≠ Φ}

(4)

v) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− = Φ

1)

,

th en D P

(

_n∗₋₁2

,

i

− ≠ Φ

1)

We have

1

2

1

3 n

i

n

  ≤ − ≤ −

 

 

and

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-3.

Therefore,

2 n

− ≤ − ≤

2 i

1

2 n

−

1

.

Therefore,

2 n

− ≤ ≤

1 i

2 n

.

Therefore, i=2n-1 or i=2n.

Therefore, in either case, we have

D P

(

_n∗₋2₁

,

i

− ≠ Φ

1)

.

Theorem 2.4: For every

n

≥

7

i)

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

and

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

,if and only if i=2n.

ii)

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋₁2

,

i

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− = Φ

1)

,

D P

(

_n∗₋2₃

,

i

− = Φ

1)

,

if and only if i=2n-1

iii)

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋₁2

,

i

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− ≠ Φ

1)

,

and D P

(

_n∗₋₃2

,

i

− = Φ

1)

if and only if i=2n-2 or 2n-3 or 2n-4.

Proof:

i) (⇒) Since

D P

(

_n∗₋2₁

,

i

− =

1)

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

− =

1)

D P

(

_n∗₋2₃

,

i

− = Φ

1)

,we have

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-2,

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-3,

1

3

3 n

i

−



_

−



_







or

i

−

1 

2(

n

−

3)

.

Also,

1

2

1.

3 n

i

n

  ≤ − ≤ −

 

 

Therefore,

1

3 n

i

−



_

−



_







does not hold.

Therefore,

i

−

1 

2 n

−

2

.

Therefore

i

− ≥

1

2 n

−

1

.

Therefore,

i

≥

2 n

Together, we have i=2n.

(⇐) It follows from Lemma 2.2(iii), (iv).

ii) (⇒)Since

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

,

i

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− = Φ

1)

, by Lemma

2.2(iii),(iv), we have

1

2

1

3 n

i

n

  ≤ − ≤ −

 

 

,

1

2

3 n

i

n

−



_{ ≤ − ≤ −}









,and

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

-3.Also

1

3 n

i

−



_

−



_







Then

2 n

−

3 

i

− ≤

1

2 n

−

2

.

Therefore, i-1=2n-2

Thus, i=2n-1

(5)

iii) (⇒) Since

{ }

{

}

2 2 2 2

1 1 3

(

_n

2 ,

1)

,

(

_n

,

1)

,

(

_n

2 2 ,

1)

,

(

_n

,

1)

D P

∗

−

n i

− ≠ Φ

D P

∗₋

i

− ≠ Φ

D P

∗₋

−

n

−

i

− ≠ Φ

and D P

∗₋

i

− = Φ

, by

Lemma 2.2(iii),(iv), we have

1

2

1

3 n

i

n

  ≤ − ≤ −

 

 

,

1

2

3 n

i

n

−



_{ ≤ − ≤ −}









,

1

2

3

3 n

i

n

−



_{ ≤ − ≤ −}









.Also .

3

1

3 n

i

−



_

−



_







or

i

−

1 

2 n

−

6

.Also

i

− ≥

1

2 n

−

5

.

Therefore,

2 n

− ≤ − ≤

5 i

1

2 n

−

3

.Therefore,

2 n

− ≤ ≤

4 i

2 n

−

2

.

Hence i=2n-2,2n-3,2n-4

(⇐) It follows from Lemma 2.2(iii), (iv).

Now, we construct following Theorem.

Theorem 2.5:

i) If

D P

(

n 2

{ }

2 n i

,

1)

∗

₋

_{− ≠ Φ}

,and

D P

(

n 2₁

,

i

1)

D P

(

n ₁2

{

2 n

2 ,

}

i

1)

D P

(

n 2₃

,

i

1)

∗ ∗ ∗

−

− =

−

− =

−

− = Φ

then

{

{ }

}

2 2

(

_n

, )

2 /

(

_n

2 ,

1)

D P

∗

i

=

X

∪

n

X

∈

D P

∗

−

n i

−

ii) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

,

i

− ≠ Φ

1)

,

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− ≠ Φ

1)

and

D P

(

n 2₃

,

i

1)

∗

−

− = Φ

, then

{

{ }

{

}

{

}

{

}

{

}

2 2 2

1 1 2 1

2

3 3 1

(

, )

2 /

(

2 ,

1),

2 1 /

(

,

1)

2 2, 2

3 /

(

2 2 ,

1)

n n n

n

D P

i

X

n

X

D P

n i

X

n

D P

i

X

n

X

D P

n

i

∗ ∗ ∗

−

∗ −

=

∪

∈

−

∪

−

∪

−

∈

−

iii) If

D P

(

_n∗2

−

{ }

2 n i

,

− ≠ Φ

1)

D P

(

_n∗₋₁2

,

i

− ≠ Φ

1)

D P

(

_n∗₋2₁

−

{

2 n

−

2 ,

}

i

− ≠ Φ

1)

and

D P

(

_n∗₋2₃

,

i

− ≠ Φ

1)

,then

{

{ }

{

}

{

}

{

}

{

}

{

}

2

1 1

2

2 2 1

2

3 3 1

2 2

4 3 3

4 ₂ ₂

4 3 3

2 /

(

2 ,

1),

2 1 /

(

,

1),

2 3 /

(

2 2 ,

1)

,

2 1 /

(

,

1),1

(

,

1)

2 3 /

(

,

1),1

(

,

1)

n

n n

X

n

X

D P

n i

X

n

X

D P

i

X

n

X

D P

n

i

n

X

D P

i

D P

i

X

n

X

D P

i

D P

i

∗

∗ − ∗

−

∗ ∗

− −

∗ ∗

− −

∪

∈

−

∪

−

∈

−

∪

−

∈

−



−

∈

−

∉

−



∪ 

−

∈

−

∈

−



Proof:

i) By theorem 2.1(i),i=2n.Since in this case

D P

(

_n∗2

, )

i

=

D P

(

_n∗2

, 2 )

n

=

{ }

[ ]

2 n

and

{ }

{

[

]

}

2 2

(

_n

2 ,

1)

(

_n

2 , 2

1)

2

1 D P

∗

−

n i

− =

D P

∗

−

n

− =

n

−

,

then we have the result.

ii) Let

{

{ }

2

{ }

}

1 1

2 /

1

(

n

2 ,

1)

Y

=

X

∪

n

X

∈

D P

∗

−

n i

−

{

}

2

}

2 2

2 1 /

2

(

n 1

,

1)

Y

=

X

∪

n

−

X

∈

D P

∗₋

i

−

and

{

}

2

{

}

3 3

2 2, 2

3 /

3

(

n 1

2 2 ,

1),

Y

=

X

∪

n

−

n

−

X

∈

D P

∗₋

−

n

−

i

−

. Obviously,

Y

₁

Y

₂

Y

₃

D P

(

n2

, )

i

∗

∪ ∪ ⊆

(2.1) Now let

Y

D P

(

n 2

, )

i

∗

∈

.If

2 n

∈

Y

, then at least one of the vertives labeled 2n-1 or 2n-2 is in Y.In either cases,

{ }

2

{ }

1

2

1

(

n

2 ,

1)

Y

=

X

∪

n forsome X

∈

D P

∗

−

n i

−

,that is

Y

∈

Y

₁.

If 2n

∉

Y

and 2n-1

∈

Y

, then 2n-2

∈

Y

, So

Y

=

X

₂

∪

{

2 n

−

1 }

forsome D P

(

_n∗₋₁2

,

i

−

1)

,that is

Y

∈

Y

₂.Now suppose that2n-2

∈

Y

,2n

∉

Y

and 2n-1

∉

Y

,then at least one of the vertices labeled

2n-3,2n-4 is in Y. If 2n-4

∈

Y

, then

{

}

2

{

}

3

2 2, 2

3

(

n 1

2 2 ,

1)

(6)

Thus, we have proved that

D P

(

_n∗2

, )

i

⊆ ∪ ∪

Y

₁

Y

₂

Y

₃. (2.2) From (2.1) and (2.2), We have

{

{ }

{

}

{

}

{

}

2 2 2

1 1 2 1

2

3 3 1

(

, )

2 /

(

2 ,

1),

2 1 /

(

,

1)

2 2, 2

3 /

(

2 2 ,

1),

n n n

n

D P

i

X

n

X

D P

n i

X

n

D P

i

X

n

X

D P

n

i

∗ ∗ ∗

− ∗

−

=

∪

∈

−

∪

−

∪

−

∈

−

iii) Let

{ }

2

{ }

1 1

2 /

1

(

n

2 ,

1),

Y

=

X

∪

n

X

∈

D P

∗

−

n i

−

{

}

2

2 2

2 1 /

2

(

n 1

,

1),

Y

=

X

∪

n

−

X

∈

D P

∗₋

i

−

{

}

2

{

}

3 3

2 3 /

3

(

n 1

2 2 ,

1)

,

Y

=

X

∪

n

−

X

∈

D P

∗₋

−

n

−

i

−

and

{

}

{

}

2 2

4 3 3

4 4 ₂ ₂

4 3 3

2 1 /

(

,

1),1

(

,

1)

2 3 /

(

,

1),1

(

,

1)

n n

n

X

D P

i

D P

i

Y

X

n

X

D P

i

D P

i

∗ ∗

− −

∗ ∗

− −



−

∈

−

∉

−

_



=

_{∪ }

_

−

∈

−

∈

−





.

Obviously,

Y

₁

∪ ∪ ∪ ⊆

Y

₂

Y

₃

Y

₄

D P

(

_n∗2

, )

i

(2.3)

Now, let

Y

D P

(

n2

, )

i

∗

∈

. If

2 n

∈

Y

,then at least one of the vertices labeled 2n-1 or 2n-2 is in Y. In either cases,

{ }

2

{ }

1

2

1

(

n

2 ,

1)

Y

=

X

∪

n forsome X

∈

D P

∗

−

n i

−

; that is

Y

∈

Y

₁.

If 2n

∉

Y

and 2n-1

∈

Y

,then 2n-2

∈

Y

,so

Y

=

X

₂

∪

{

2 n

−

1 }

forsome D P

(

_n∗₋2₁

,

i

−

1)

that is

Y

∈

Y

₂.Now suppose that 2n-2

∈

Y

,2n

∉

Y

and 2n-1

∉

Y

,then at least one of the vertices labeled 2n-3,2n-4 is in Y.If 2n-4

,

Y

∈

then

Y

=

{

X

₃

∪

{

2 n

−

3 }

forsome X

₃

∈

D P

(

_n∗₋₁2

−

{

2 n

−

2 ,

}

i

−

1)

}

,

that is

Y

∈

Y

₃.Now suppose that 2n-3

∈

Y

,2n-2

∉

Y

,2n-1

∉

Y

,then at leat one of the vertices labeled 2n-4,2n-5 is in

Y.If 2n-5

∈

Y

,

then

{

}

{

}

2 2

4 3 3

4 ₂ ₂

4 3 3

2

1 (

,

1),1

(

,

1)

2

3 (

,

1),1

(

,

1)

n n

n

forsomeX

D P

i

D P

i

Y

X

n

forsomeX

D P

i

D P

i

∗ ∗

− −

∗ ∗

− −



−

∈

−

∉

−

_



=

_{∪ }

_

−

∈

−

∈

−





.

Thus we have proved that

D P

(

_n∗2

, )

i

⊆ ∪ ∪ ∪

Y

₁

Y

₂

Y

₃

Y

₄

.

(2.4)

From (2.3) and (2.4)

{

{ }

{

}

{

}

{

}

{

}

{

}

2

1 1

2

2 2 1

2

3 3 1

2 2

4 3 3

4 ₂ ₂

4 3 3

2 /

(

2 ,

1),

2 1 /

(

,

1),

2 3 /

(

2 2 ,

1)

,

2 1 /

(

,

1),1

(

,

1)

2 3 /

(

,

1),1

(

,

1)

n

n n

X

n

X

D P

n i

X

n

X

D P

i

X

n

X

D P

n

i

n

X

D P

i

D P

i

X

n

X

D P

i

D P

i

∗

∗ − ∗

−

∗ ∗

− −

∗ ∗

− −

∪

∈

−

∪

−

∈

−

∪

−

∈

−



−

∈

−

∉

−



∪ 

−

∈

−

∈

−



Theorem 2.6: For every n

≥

7,

{ }

{

}

2 2 2 2 2

1 1 3

(

_n

, )

(

_n

2 ,

1)

(

_n

,

1)

(

_n

2 2 ,

1)

(

_n

,

1)

D P

∗

i

=

D P

∗

−

n i

− +

D P

∗₋

i

− +

D P

∗₋

−

n

−

i

− +

D P

∗₋

i

−

Proof: We consider the three cases in Theorem 2.2

i) By Theorem 2.2( i),

D P

(

_n∗2

, )

i

=

{

X

∪

{ }

2 n

/

X

∈

D P

(

_n∗2

−

{ }

2 n i

,

−

1)

}

.Therefore we have the result in this case.

ii) ByTheorem2.2(ii), we have 2

(

_n

, )

D P

∗

i

=

A

₁

∪

A

₂

∪

A

₃Where,

{ }

(

2

)

2

(

2

{

}

)

1 n

2 ,

1 ,

2

(

n 1

,

1),

3 n 1

2 2 ,

1

(7)

1 1

,

2 1

X

∈

A X

∈

X

,and for every

X

₂

∈

A

₂,

2 n

∉

X

₂,so

A

₁

∩

A

₂

= Φ

.Also since for every

X

₂

∈

A

₂and

3 3

X

∈

A

,

2 n

− ∈

1 X and n

₂

,

2 − ∉

1 X

₃, we have

A

₂

∩

A

₃

= Φ

.For every

X

₃

∈

A

₃,

3

2 n

− ∉

1 X

, and 2n-3

∉

X

₃,but for every

X

₁

∈

A

₁,atleast one of 2n-3 or 2n-1 is in

X

₃;because

{ }

3 3

2 X

= ∪

Y

n

for some

Y

₃

∈

D P

(

_n∗2

−

{ }

2 n i

,

−

1

)

,so atleast one of 2n-3 of 2n-1 must be in

Y

₃. Hence we have the result.

iii) By Theorem 2.2(ii)

D P

(

n 2

, )

i

∗ ₌

1 2 3 4

A

∪

A

∪

A

∪

A

,Where,

{ }

{

}

2 2 2

1

(

n

2 ,

1),

2

(

n 1

,

1),

3

(

n 1

2 2 ,

1)

A

=

D P

∗

−

n i

−

A

=

D P

∗₋

i

−

A

=

D P

∗₋

−

n

−

i

−

and

A

₄

=

D P

(

_n∗₋2₃

,

i

−

1)

Since for every

X

₁

∈

A X

₁

,

₂

∈

X

₁ ,and for every

X

₂

∈

A

₂,

2 n

∉

X

₂,so

A

₁

∩

A

₂

= Φ

.Also since for every

2 2

X

∈

A

and

X

₃

∈

A

₃,

2 n

− ∈

1 X and n

₂

,

2 − ∉

1 X

₃,we have

A

₂

∩

A

₃

= Φ

.For every

X

₃

∈

A

₃,

3

2 n

− ∉

1 X

, and 2n-3

∉

X

₃,but for every

X

₁

∈

A

₁, atleast one of 2n-3 or 2n-1 is in

X

₃,because

{ }

3 3

2 X

= ∪

Y

n

for some

Y

₃

∈

D P

(

_n∗2

−

{ }

2 n i

,

−

1

)

so atleast one of 2n-3 of 2n-1 must be in

Y

₃. For every

4 4

X

∈

A

,2n-1

∉

X

₄and2n-3

∉

X

₄ but for every

X

₁

∈

A

₁, atleast one of 2n-3 or 2n-1 is in,

X

₄because

{ }

4 4

2 X

= ∪

Y

n

for some

Y

₄

∈

D P

(

_n∗2

−

{ }

2 n i

,

−

1

)

, atleast one of 2n-3 of 2n-1 must be in

Y

₄.Hence we have the result.

3. Domination sets of

(

P

_n∗2

−

{ }

2 n

)

For the construction of

(

P

_n∗2

−

{ }

2 n

)

, we consider

{

}

{

}

{

}

2 2 2 2 2

1 1 2 2 3

(

_n

,

1), (

_n

2 2 ,

1), (

_n

,

1) , (

_n

2 4 ,

1)

(

_n

2 6 ,

1)

D P

∗₋

i

−

D P

∗₋

−

n

−

i

−

D P

∗₋

i

−

D P

∗₋

−

n

−

i

−

a n d

D P

∗₋

−

n

−

i

−

The families of these dominating sets can be empty or otherwise. Thus, we have eight combinations, whether these five families are empty or not. Two of these combinations are not possible (Lemma 2.3 (i), (ii) & (iii)).

Also, the combinations

{

}

{

}

{

}

2 2 2 2 2

1 1 2 2 3

(

_n

,

1)

(

_n

2 2 ,

1)

(

_n

,

1)

(

_n

2 4 ,

1)

(

_n

2 6 ,

1)

D P

∗₋

i

− =

D P

∗₋

−

n

−

i

− =

D P

∗₋

i

− =

D P

∗₋

−

n

−

i

− =

D P

∗₋

−

n

−

i

− = Φ

does not need to be considered because it implies that

D P

(

_n∗2

−

{ }

2 n i

,

−

1)

=

Φ

(See lemma 2.5 (iii)). Thus we only need to consider four combinations or cases. We consider those cases in theorem