Vol 2, No 4 (2011)

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TOTAL NUMBER OF SPECIAL KINDS OF NEIGHBOURHOOD SETS OF GRAPHS

*M.P Sumathi and N. D. Soner

Department of Studies in Mathematics, University of Mysore, Mysore 570006, India

E-mail: [email protected], [email protected]

! " # $ ! " #

---

ABSTRACT

T

his paper is concerned with special kinds of neighbourhood sets of graphs

P

_n and

C

_n . The neighbourhood set, the split neighbourhood sets of a graph are considered using a recurrence relation, we give the number of all neighbourhood sets mentioned for the graphs

P

_n and

C

_n.

---

INTRODUCTION:

In this paper, we consider the path

P

_n and cycle

C

_n graphs of n vertices as graphs with

V

(

P

_n

)

=

{

x

₁

,

x

₂

...

x

_n

}

,

}

,...

,

{

)

(

P

_n

x

₁

x

₂

x

₂

x

₃

x

_n1

x

_n

E

=

₋ for

n

≥

3

and

V

(

C

_n

)

=

{

x

₁

,

x

₂

...

x

_n

}

,

E

(

C

_n

)

=

{

x

₁

x

₂

,

x

₂

x

₃

,...

x

_n−1

x

_n

,

x

_n

x

₁

}

for

n

3

. For

v

∈

V

, the closed neighbourhood of

V

is

N

[

v

]

=

{

u

∈

v

:

uv

∈

E

(

G

)}

{

v

}

.

A subset

S

of

V

(

G

)

is a neighbourhood set of G if

G

=

v∈S

N

(

v

)

. Where

N

(

v

)

is the subgraph of

G

induced by

N

[

v

]

. The neighbourhood number

η

(

G

)

of

G

is the minimum cardinality of a neighbourhood set

of

G

.

Neighbourhood sets of

P

_n and

C

_n:

The results considered in this section will be used to determine the total number of split neighbourhood sets of

P

_n and

n

C

.To determine the number of all neighbourhood sets of a path of

n

vertices, for all

n

≥

1

, we introduce the

following notations.

D

G

V

D

P

_n

)

{

(

)

:

(

=

⊆

N

is a neighbourhood set of

P

_n

}

.

}

:

)

(

{

)

(

1

P

n

=

D

∈

N

P

n

X

n

∈

D

N

.

)

(

)

(

)

(

_{1 1}

2

P

n

N

P

n

N

P

N

=

−

Denoting the cardinalities of the families

N

(

P

_n

),

N

₁

(

P

_n

)

and

N

₂

(

P

_n

)

respectively by

N

′

(

P

_n

)

,

N

₁′

(

P

_n

)

,

)

(

2

P

n ′

N

. We obtain the following equality,

1

),

(

)

(

)

(

=

₁′

+

₂′

≥

′

n

P

P

P

_n

_N

_n

_N

_n

N

(1)

of course

_N

′

(

P

_n

)

is the total number of neighbourhood sets of a path

P

_n on

n

vertices. It is easy to see that

1

)

(

₁

=

′

P

N

,

N

′

(

P

₂

)

=

3

,

N

′

(

P

₃

)

=

5

. These inequalities are the initial conditions for the recurrence relation

given in the following theorem.

Page: 433-437

& ' ( ( !

Theorem 1: For

n

≥

3

,

)

(

)

(

)

(

+₁ −₁

′ ′

′

+

=

_{n n}

n

P

P

P

N

N

N

Proof: Assume the

_N

is a neighborhood set of

P

_n+1 for

n

≥

3

. Then we have three possibilities:

(1) If

X

_n

,

X

_n+1

∈

D

, then

D

−

{

X

n+1

}

∈

N

1

(

P

n

)

.

(2) If

X

_n

∈

D

and

X

_n+1

∉

D

,then

D

∈

N

1

(

P

n

)

.

(3) If

X

_n

∉

D

and

X

_n−1

,

X

_n+1

∈

D

, then

D

−

{

X

_n+1

}

∈

_N

₂

(

P

_n

)

.

Thus, the number

(

+₁

)

′ n

P

N

of all neighbourhood sets of

P

n+₁ is described by the following equality.

),

(

)

(

)

(

)

(

+_{1 1 1 2} −₁

′ ′

′ ′

+

+

=

_{n n n}

n

P

P

P

P

_N

_N

_N

N

(2)

it follows that for

n

≥

3

, we have

)

(

)

(

)

(

_{1 1 2 1}

1

P

n

=

N

P

n−

+

N

P

n−

N

and

)

(

)

(

_{1 1}

2

P

n

=

N

P

n−

N

(3)

Applying these equalities in equation (3) to the first of the sum in equation (2), we obtain

)

(

)

(

)

(

)

(

)

(

P

_{n 1}

_N

₁

P

_{n 1}

_N

₂

P

_n1

_N

₁

P

_n

_N

₂

P

_n

N

+

=

−

+

−

+

+

′

Hence

N

′

(

P

_n+₁

)

=

N

′

(

P

_n−₁

)

+

N

′

(

P

_n

)

.

The following results concern the total number of neighbourhood sets of the cycle

C

_n on n vertices for

n

≥

4

.First

we introduce the following notations:

N

N

(

C

_n

)

=

{

D

subseteqV

(

G

)

:

is a neighbourhood set of

C

_n }.

D

X

Cn

D

c

_n

=

∈

₁

∈

01

(

)

{

N

(

)

:

(

N

and

X

_n

∉

D

)

or

(

X

₁

∉

D

and

X

_n

∈

D

)}

}

,

:

)

(

{

)

(

₁

11

c

n

=

D

∈

N

Cn

X

X

n

∈

D

N

By

_N

′

(

C

_n

)

,

_N

₀₁′

(

C

_n

)

and

_N

₁₁′

(

C

_n

)

we mean the cardinalities of families

_N

(

C

_n

)

,

_N

₀₁

(

c

_n

)

and

_N

₁₁

(

c

_n

)

respectively.Using these numbers, we obtain the following equality.

N'(Cn) = N'_{01(Cn)+ N}'_11(Cn)

_n

_≥

₄

_,

₍₄₎

It is easy to check that

_N

′

(

C

₄

)

=

7

and

_N

′

(

C

₅

)

=

11

.

Theorem 2: For

n

≥

5

,

)

(

)

(

)

(

C

_n1 ′

C

_n1 ′

C

_n ′

+

=

₋

+

N

N

N

Proof:Let

n

be a neighbourhood set of

C

_n for

n

≥

5

. Consider the following cases:

(i) If

X

1

∉

D

and

X

2

,

X

n

∈

D

, then

D

∈

N

11

(

H

1

)

where

)

(

)

(

)

(

H

₁

V

C

X

₁

Page: 433-437

& ' ( ( ! )

Hence

H

₁

≈

C

_n−1

(ii)

IfX

_n

∉

D

and

X

₁

,

X

_n−1

∈

D

then

D

∈

_N

₁₁

(

H

₂

)

, where

)

(

)

(

)

(

H

₂

V

C

_n

X

_n

V

=

−

and

E

(

H

₂

)

=

{

E

(

C

_n

)

−

(

X

_n−1

X

_n

,

X

_n

X

₁

)}

(

X

_n−1

X

₁

)

Hence

H

₂

≈

C

_n−1

(iii)

IfX

₁

,

X

_n

∈

D

and

X

₂

∉

D

then

D

∈

N

₀₁

(

H

₁

)

(iv)

IfX

₁

,

X

₂

,

X

_n

∈

D

then

D

∈

_N

₁₁

(

H

₁

)

Hence from case (i),(ii),(iii) and (iv) we have that

)

(

)

(

)

(

_{11 1 11 2}

01

C

n

H

H

′ ′

′

+

=

N

N

N

)

(

2

)

(

_{11 1}

01 −

′ ′

=

_n

n

C

C

_N

N

)

(

2

/

)

(

)

(

_{01 1 11 1}

11 − −

′ ′

′

+

=

_{n n}

n

C

C

C

_N

_N

N

)

(

)

(

)

(

C

_{n 01}′

C

_{n 11}′

C

_n ′

+

=

N

N

N

=

2

_N

₁₁′

(

C

_n−1

)

+

_N

₀₁′

(

C

_n−1

)

/

2

+

_N

₁₁′

(

C

_n−1

)

2

(

₀₁

(

−₂

)

/

2

₁₁

(

−₂

))

₀₁

(

−₁

)

/

2

₁₁

(

−₁

)

′ ′

′ ′

+

+

+

=

_N

C

_n

_N

C

_n

_N

C

_n

_N

C

_n

=

_N

′₀₁

(

C

_n−2

)

+

2

_N

₁₁′

(

C

_n−2

)

+

1

/

2

(

2

_N

₁₁′

(

C

_n−2

))

+

_N

₁₁′

(

C

_n−1

)

=

_N

′₀₁

(

C

_n−2

)

+

2

_N

₁₁′

(

C

_n−2

)

+

_N

₁₁′

(

C

_n−2

)

+

_N

₁₁′

(

C

_n−1

)

₀₁

(

−₂

)

₀₁

(

−₁

)

₁₁

(

−₂

)

₁₁

(

−₁

)

′ ′

′ ′

+

+

+

=

N

C

_n

N

C

_n

N

C

_n

N

C

_n

From equation (4) the above equation reduces to

)

(

)

(

)

(

)

(

)

(

₀₁′ _{−2 11}′ _{−2 01}′ _{−1 11}′ ₋₁ ′

+

+

+

=

_{n n n n}

n

C

C

C

C

C

_N

_N

_N

_N

N

)

(

)

(

)

(

−₂ −₁

′ ′

′

+

=

_{n n}

n

C

C

C

N

N

N

)

(

)

(

)

(

C

_n1 ′

C

_n1 ′

C

_n ′

+

=

₋

+

N

N

N

Split neighbourhood set of

P

_n and

C

_n:

Using the numbers

N

(

P

_n

)

and

N

(

C

_n

)

we determine the number of split neighbourhood sets of path and the cycle

of graphs on n vertices. First we give supportive theorem that characterize the split neighbourhood set of

P

_n and

n

C

.

Lemma 3: Any neighbourhood set

D

of

P

_n

n

≥

3

, is a split neighbourhood set of

P

_n if and only if

,

V

(

P

D

)

_{( )}

K

₁

n P

n

−

≠

.

Proof: Let S be a split neighbourhood set D of

P

_n .According to the definition of

S

, it follows that

V

(

P

_n

)

−

D

is disconnected. Thus

V

(

P

)

D

_{( )}

not

K

₁

n P

n

−

≈

, proving necessity.

For sufficiency, let

D

be a neighbourhood set of

P

_n ,

n

≥

3

, and suppose ,

1 ) (

)

(

P

D

K

V

n P

n

−

≠

. Since

H

=

V

(

P

n

−

D

)

(Pn) is an induced subgraph of

P

n , then any connected

Page: 433-437

& ' ( ( ! #

component of the subgraph

H

. We show that

H

1 is not a unique connected component of

H

. First, observe that

H

1 contains atmost two vertices. Otherwise there would exist a vertex of ₁

(

)

_{( )}

n P

n

D

P

V

H

H

⊂

=

−

not

neighbour of

D

in

P

_n . Consequently,

H

₁

≈

P

₁ or

H

₁

≈

P

₂ . Hence

H

has at least two connected

components, because

H

₁

≠

P

₁ or

H

₁

≠

P

₂ by premise. This shows that

(

)

_{( )}

n P

n

D

P

V

H

=

−

is disconnected.

Moreover, since

D

is also a neighbourhood set of

P

_n , it is a split neighbourhood set of

P

_n , completes the proof of

the theorem.

Similar to the case of

P

_n , we have a result concerning the split neighbourhood sets of

C

_n .

Lemma 4: Any neighbourhood set

D

of

C

_n ,

n

≥

4

is a split neighbourhood set of

C

_n if and only if

,

V

(

C

)

D

_{( )}

K

₁ n C

n

−

≠

.

Additionally, observe that there is only one neighbourhood set

D

of

P

_n such that , and there

are exactly

n

neighbourhood sets

D

of

P

n such that

V

(

P

)

D

_{( )}

K

₁ n P

n

−

≈

.

In special cases of

C

_n , we have one Neighbouhood set D such that and

n

for

1 ) (

)

(

C

D

K

V

n C

n

−

≈

.

For

n

≥

3

, we introduce the notation

D

G

V

D

G

S

(

)

=

{

⊆

(

)

:

is a split neighbourhood set of

G

}

|

)

(

|

)

(

G

S

G

S

=

From the above, we have the following corollary, which will be used in proving theorem.

Corollary 5:

s

(

P

_n

)

=

d

(

P

_n

)

−

(

1

+

n

)

for

n

≥

3

and

s

(

C

_n

)

=

d

(

C

_n

)

−

(

1

+

n

)

It is easy to see that

s

(

P

₃

)

=

1

,

s

(

P

₄

)

=

3

and

s

(

P

₅

)

=

7

.

Theorem 6: For

n

≥

5

,

s

(

P

n+₁

)

=

s

(

P

n−₁

)

+

s

(

P

n

)

+

(

n

−

1

)

Proof: Let

n

≥

5

, according to corollary (3), for

P

_n+1 we have that

)

2

(

)

(

)

(

P

₁

d

P

₁

n

s

_n+

=

_n+

−

+

since

d

(

P

_n+1

)

=

d

(

P

_n−1

)

+

d

(

P

_n

)

by theorem (1), we obtain

)

2

(

)

(

)

(

P

₁

d

P

₁

n

s

_n+

=

_n+

−

+

=

d

(

P

_n−₁

)

+

d

(

P

_n

)

−

(

2

+

n

)

=

d

((

P

_n−₁

)

−

n

+

n

)

+

d

((

P

_n

)

−

(

n

+

1

)

+

(

n

+

1

))

−

(

2

+

n

)

=

d

((

P

_n−1

)

−

n

)

+

d

((

P

_n

)

−

(

n

+

1

))

−

(

2

+

n

+

n

+

(

n

+

1

))

Page: 433-437

& ' ( ( ! *

since

(

n

+

1

+

n

−

2

−

n

=

(

n

−

1

)

, it follows that

)

1

(

))

1

(

)

((

)

)

((

₁

−

+

−

+

+

−

=

d

P

n−

n

d

P

n

n

n

Finally applying corollary (3)to the expressions in brackets,

)

1

(

)

(

)

(

)

(

P

+₁

=

s

P

−₁

+

s

P

+

n

−

s

_{n n n} .

Theorem 7: For

n

≥

5

,

)

1

(

)

(

)

(

)

(

C

+₁

=

s

C

−₁

+

s

C

+

n

−

s

_{n n n}

Proof: Let

n

≥

5

, putting

n

+

1

in place of

n

in corollary 3, it follows

)

2

(

)

(

)

(

C

₁

d

C

₁

n

s

_n+

=

_n+

−

+

according to theorem (2),

)

(

)

(

)

(

C

_n1

d

C

_n1

d

C

_n

d

+

=

−

−

hence,

)

2

(

)

(

)

(

C

₁

d

C

₁

n

s

_n+

=

_n+

−

+

=

d

(

C

_n−₁

)

+

d

(

C

_n

)

−

(

2

+

n

)

=

d

((

C

_n−1

)

−

n

+

n

)

+

d

((

C

_n

)

−

(

n

+

1

)

+

(

n

+

1

))

−

(

2

+

n

)

=

d

((

C

_n−₁

)

−

n

)

+

d

((

C

_n

)

−

(

n

+

1

))

+

(

n

+

1

+

n

−

2

−

n

)

since

(

n

+

1

+

n

−

2

−

n

=

(

n

−

1

))

, it follows that

=

d

((

C

n−₁

)

−

n

)

+

d

((

C

n

)

−

(

n

+

1

))

+

(

n

−

1

)

Finally applying corollary (3) to the expressions in brackets,

)

1

(

)

(

)

(

)

(

C

+₁

=

s

C

−₁

+

s

C

+

n

−

s

_{n n n} .

REFERENCES:

[1] B.Janakiram; Split neighbourhood in graph, Proceedings of the Jangjeon Mathematical Society, 7 (2), (2004), 129-136.

[2] V.R.Kulli and B.Janakiram; The split domination number of a graph, Graph Theory Notes of New York, XXXII: 3, New York Academy of Sciences,(1997) 16-19

[3] M.Zwierzchowski,Total numbers of special kinds of dominating sets of graphs, Graph Theory Notes of New York XLV1, New York Academy of Sciences, (2004) 13-19.