Cmjv03i03p0298

J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

Total Edge Fixed Geodomination Polynomial of Cycles and Wheels AYYAKUTTY VIJAYAN1 and THANKARAJ BINU SELIN2 1

Assistant Professor, Department of Mathematics, Nesamony Memorial Christian College Marthandam-629165, Kanyakumari District, Tamilnadu, INDIA. 2 Department of Mathemetics, Marthandam College of Engineering and Technology, Kuttakuzhi, Kanyakumari District, Tamilnadu, INDIA. (Received on: May 13, 2012) ABSTRACT We introduce a total edge fixed geodomination polynomial of cycles and wheels. Let G = (V, E) be a simple graph. In10, the concept of total edge fixed geodomination polynomial of G is defined as Gt(G, x) =

∑

ek∈E(G)

n −2

Gek(G, x) where Gek(G, x) =

∑

gek(G, i)xi, gek(G, i) is the

i=ge (G)

k

number of edge fixed geodominating sets of graph G with cardinality i, where ek is a fixed edge of G and gek (G) is the ek geodomination number of G. In this paper, we obtain the total edge fixed geodomination polynomial of cycles and wheels. Also we study some properties of total edge fixed geodominating sets and its coefficients.

Keywords: Edge fixed geodominating set, Total edge fixed geodomination polynomial.

1. INTRODUCTION Let G = (V,E) be a simple graph of order n. An edge of a graph is said to be pendant if one of its vertices is a pendant vertex. A vertex of a graph is said to be

pendant if its neighbour contains exactly one vertex. The distance d (u, v) between two vertex u and v in a connected graph G is the length of the shortest u–v path. A u–v path of length d (u, v) consists of all vertices lying on some u–v geodesic of G, while for S ⊆ V, I

Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

299 [S] =

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

U

I [u, v]. A set S of vertices is a

u,v ∈ S

geodetic set if I [S] = V, and the minimum cardinality of a geodetic set is the geodetic number g (G). The concept of edge fixed geodomination was introduced in8. Let e = x y be any edge of a connected graph G of order atleast 3. A set S of vertices of G is an e-geodominating set if every vertex of G lies on either an x-u geodesic or y-u geodesic in G for some element u in S. The minimum cardinality of an e-geodominating set of G is defined as the e-geodomination number of G and is denoted by ge(G) or gx y(G).

A wheel Wn is a graph with n vertices x1, x2,…,xn with x1 having degree n – 1 and all the other vertices having degree 3. The vertex x1 is adjacent to all the other vertices, and for i = 2, 3,…..,n, xi is adjacent to xi+1 and xn is adjacent to x2.

A cycle is a graph with an equal number of vertices and edges whose vertices can be placed around a circle so that two vertices are adjacent if and only if they appear consecutively along the circle. A cycle with n vertices is denoted by Cn.

n −2

∑

Gek (Cn, x) =

gek (Cn, i) xi.

i=ge (Cn )

k

Where gek (Cn, i) is the number of edge fixed geodominating sets of Cn with cardinality i. where ek is a fixed edge.

The total edge fixed geodomination polynomial of Cn is defined as

∑

Gt (Cn, x) =

ek ∈E(Cn )

If gt (Cn) =

Gek(Cn , x)

min {g

ek ∈E(Cn )

ek (Cn)}, then we can write

n −2

Gt(Cn, x) =

∑

gt(Cn, i) xi where gt (Cn, l) is

i=g (Cn ) t

the total number of edge geodominating sets of cardinality i.

fixed

Theorem: 2.2

2. TOTAL EDGE FIXED GEODOMINATION POLYNOMIAL OF CYCLES

The total edge fixed polynomical of cycle Cn is

In this section we introduce the total edge fixed geodomination polynomial of cycle. Let Cn, n ≥ 3 be the cycle with n vertices. Let V (Cn) = [n] and E [Cn] = {{1, 2}, {2, 3},……, {n - 1, n}} .

n −3 Gt (Cn, x) = nx(1 + x) n −4 nx(1 + x) ( x + 2)

Definition: 2.1

Let Cn be the cycle with n vertices v1, v2,…,vn. Let the edges of Cn be e1,e2,…en.

Cn:

The edge fixed geodomination polynomial of a cycle Cn is defined as

geodomination

if n is odd if n is even

Proof:

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300

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

and ge1(Cn, 2) = (n − 3) C1. By the similar argument ge1(Cn, 3) = (n − 3) C2.

Now we consider ge1 (Cn, i). Any set in V (G) with v i 

 2  +1

and any i – 1 vertices

from the n – 3 vertices v3, v4,…, v i 

 2  −1

v i  , v i 

 2  + 2

 2 

,

….., vn is a edge fixed

Case (i) : If n is odd.

geodominating set of Cn of cardinality i. That is ge1 (Cn, i) = (n − 3) Ci.

Fix e1. Then clearly ge1 (Cn) = 1.

Now ge1(Cn, 1) = 1.

Therefore

  Here S = v i   +1   2   is the only edge fixed geodominating set of Cn with cardinality 1.

Every edge fixed geodominating set contains

v i  .  2  +1

Thus

  v  i   +1   2  

Ge1 (Cn, x) = [x + (n − 3)C1 x2 + (n – 3) C2 x3 +……+ (n – 3) Cn−3 xn−2] = x [1 + (n − 3)C1 x + (n – 3) C2 x2 +……+ (n – 3) Cn−3 xn−3] = x (1 + x)n−3.

Since Cn has n edges,

together with

Get (Cn, x) = nx (1 + x) n – 3,

each of the remaining n – 3 vertices form the edge fixed geodominating set of cardinality 2. Thus there are (n − 3) C1 fixed edge geodominating set of cardinality 2 to corresponding to e1.

Case (ii) : If n is even

Consider ge1 (Cn, 2).

Now

Here the vertices v1 and v2 are fixed and they do not form the part of the edge fixed geodminating sets S. Thus S = 

   v3 , v i   ,  v4 , v i   , . . . + 1 + 1  2   2    

   vn , v i    + 1  2    

  

Gt (Cn, x) = n x (1 + x)n−3, if n is odd.

Fix e1. Then clearly ge1 (Cn) = 1. ge1(Cn,

1)

    v i +1  , v i + 2   2   2 

= are

2.

Here

the

edge

S

=

fixed

geominating set of Cn with cardinality i.

Thus, v i together with each of the 2

+1

remaining n–4

vertices other than v1, v2,

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301

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

v i and v i 2

+1

2

forms the edge fixed geodominating set of cardinality 2 corresponding to e1.

Consider ge1(Cn, 2)

together with each of the

+1

remaining n–4 vertices other than



v1,v2, v i 2

+1



  +2 2 

Here the vertices v1 and v2 are fixed and they do not form part of edge fixed geodominating set S. They are given by

and  v i ,v i



2

+1









 











 

S =  v i , v3  ,...,  v i , vn  ,  v i , v3  ,...,  v i , vn  ,  v i , v i   and +1 +1 +2 +2 +1 +2



2

2

 

2

 

2

2



2

ge1 (Cn, 2) = (n – 4) C1 + (n – 4) C1 + 1.

By the similar argument ge1 (Cn, 3) = (n – 4) C1 + (n – 4) C2 + (n – 4) C2.

Now we consider ge1 (Cn, i). Any set in V (G) with v i 2

vertices other than v i 2

+2

, {v3, v4,….., v i 2

the n−4 vertices other than v i 2

+1

+1

,vi 2

+3

and any i – 1 vertices from the n– 4

. . .vn} & v i 2

,{v3, v4,….., v i , v i 2

+1

2

+2

+2

and any i – 1 vertices from

, . . .vn} & { v i 2

+1

,vi 2

+2

} with i–2

vertices from the n–4 vertices forms edge fixed geo dominating set of cardinality i.

Therefore ge1 (Cn, i) = (n – 4) Ci−1 + (n – 4) Ci – 1 + (n − 4) Ci −2. Therefore Ge1 (Cn, x) = 2x + [(n – 4) C0 + (n−4)C1 + (n−4)C1] x2 + [(n−4) C1+(n−4) C2 + (n−4) C2] x3 + [(n−4)C2 + (n−4)C3 + (n−4)C3] x4 +…. + [(n−4)Cn−5 + (n−4)Cn–4 + (n−4)Cn−4 ] x n – 3 + x n – 2 = 2x + [ (n−4) C0 x2 + (n−4) C1 x3 +…+ (n – 4)Cn−5 xn−3 ] + [ (n−4) C1 x2 + (n – 4) C2 x3 + … + (n – 4) Cn – 4 xn – 3] + [ (n – 4) C1 x2 + (n – 4 ) C2 x3 + … + (n – 4) Cn – 4 x n–3 ] + (n – 4) Cn – 4 x n – 2] = 2x+[(n−4) C0 x2 + (n − 4) C1 x3 + … + (n − 4)Cn –5 x3 + (n−4) Cn–4 xn −2] + 2 [(n−4) C1 x2 + (n – 4) C2 x3 + … + (n−4)Cn−4 xn−3] = 2x +[1 + (n −4) C1 x + (n – 4) C1 x2 + ….. + (n – 4) Cn – 4 x n – 4 ]+ x2 [1 + (n – 4) C1 x + (n – 4) C2 x2 +… (n – 4) Cn – 4 x n – 4] =2x (1 + x) n – 4 + x2 (1 + x) n – 4 Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

302

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

=x (1 + x) n – 4 (x + 2) Since Cn has n edges Gt (Cn, x) = nx (1 + x) n – 4 ( x + 2) if n is even.

Table 1 gt (Cn, j), the number of total edge fixed geodominating set of Cn of cardinality j.

j

1

2

3

4

5

6

7

8

9

10

11

12

13

n 3

3

4

8

4

5

5

10

5

6

12

30

24

6

7

7

28

42

28

7

8

16

72

128

128

48

8

9

9

54

135

180

135

54

9

10

20

130

360

550

500

270

80

10

11

11

88

308

616

710

616

318

88

11

12

24

204

768

1680

2352

2184

1344

528

120

12

13

13

130

585

1560

2730

3276

2730

1560

585

130

13

14

28

294

1400

3990

7560

9996

9408

6300

2940

910

168

14

15

15

180

990

3300

7425

11880

13860

11880

7425

3300

990

180

gt (C2n+1, 1) = 2n+1 ∀ n ≥ 1

Thorem 2.3

(ii)

The following properties hold for the the coefficients of Gt (Cn x).

(iii) gt (C2n+1, i) =

(i)

gt (C2n, 1) = 4n ∀ n ≥ 2

15

(2n+1)(2n − 2)(2n − 3)...(2n − i) (i − 1)!

∀ n ≥ 2

Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

303 (iv)

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

gt (C2n, i) = 2n(2n −4)(2n −5)...(2n − i −1)(4n − i − 5) ∀n ≥ 2 (i −1)!

(v) gt (C2, n−2) = n ∀ n ≥ 3 (vi) S2n = 3n22n−3 ∀ n ≥ 2 (vii) S2n+1 = 22n−2 (2n + 1) ∀ n ≥ 1 Where S2n = Total number of edge fixed geodominating sets in C2n.

S2n + 1 = Total number of edge fixed geodominating sets in C2n + 1.

Proof:

(ii)

by theorem 2.2, we have ge1 (C2n+1, 1) = 1, ∀ n ≥ 1 Now gt (C2n+1, 1) = ge1 (C2n+1, 1) + ge2 (C2n+1, 1) + ….+ gen (C2n+1, 1) = (1+ 1+…+1) (2n+1 times) = 2n + 1. (iii) by theorem 2.2. gt (Cn, i) = n (n – 3 ) Ci −1 if n is odd. gt (C2n+1, i) = (2n + 1) (2n – 2) Ci−1 = (2n+1)(2n − 2)...(2n − i) ∀ n ≥ 2 (i − 1)!

(i)

by theorem 2.2, we have ge1 (C2n, 1) = 2 ∀ n ≥ 2 Now gt (C2n, 1) = ge1 (C2n, 1) + ge2 (C2n, 1) + ….+ gen (C2n, 1) = 2 + 2+…+2 (2n times) = 4n .

(iv)

by theorem 2.2

gt(Cn, i) = n [ (n – 4) Ci – 2+ 2(n – 4) Ci – 1] if n is even Therefore gt(C2n, i) = 2n [ (2n – 4) Ci – 2 + 2(2n – 4) Ci – 1]

= 2n  (2n − 4)(2n − 5)...((2n − 4) − (i − 3)) + 2(2n − 4)(2n − 5)...(2n − 4) − (i − 2)    (i − 2)! (i − 1)! 



= 2n(2n − 4)(2n − 5)...(2n − i −1) 1+ 2(2n − i − 2)  (i − 2)!

 

i −1

 

= 2n(2n − 4)(2n − 5)...(2n − i − 1) x(4n − i − 5) (i − 1)!

(v) It is enough to prove gt (C2n, 2n – 2) = 2n and gt (C2n + 1, (2n + 1) – 2) = 2n + 1 Using (iv) gt(C2n, 2n – 2) =

2 n (2 n − 4 )(2 n − 5 )...(2 n − 2 n + 2 − 1 ) x (4 n − 2 n + 2 − 5 ) (2 n − 2 − 1 )!

= 2n.

( 2 n − 4 ) ( 2 n − 5 ) . . .1 . ( 2 n − 3 ) (2 n − 4 )!

= 2n Using (iii) gt(C2n+1, 2n+1 – 2) = (2n+1)(2n − 2)...(2n − 2n+1) (2n+1 − 2 − 1)!

= (2n + 1) Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

304

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

Therefore gt (Cn, n – 2) = n

∀ n ≥ 3.

(vi) by theorem 2.2 Sum of the coefficients of Gt (C2n, x) = 2n {2[1 + (2n–4)C1 + (2n–4)C2 + . . . . + (2n–4) C2n–4] + [1 + (2n–4)C1 + S2n (2n–4)C2 + . . . + (2n–4)C2n–4]} = 2n { 2(1+1) 2n – 4 + (1+ 1) 2n – 4} = 2n x 3 x 2 2n – 4 = 3n 2 2n – 3. (vii)

by theorem 2.2

Sum of the coefficients of Gt (C2n+1, x) S 2n + 1 = (2n + 1) [1 + (2n + 1 – 3)C1 + (2n + 1 – 3)C2 + . . .+ (2n + 1 – 3)C2n + 1 – 3 ] = (2n + 1) [1 + (2n – 2)C1 + (2n − 2)C2 + . . . + (2n – 2) C 2n – 2] = (2n + 1) [ 1+ 1]2n – 2 = (2n + 1) 2 2n – 2.

3. TOTAL EDGE FIXED GEODOMINATION POLYNOMIAL OF WHEELS

The total edge fixed geodomination polynomial of Wn is defined as

In this section we introduce the total edge fixed geodomination polynomial of a wheel. Let Wn, n ≥ 4 be the wheel with n vertices. Let V (Wn) = [n] and E[(Wn] = {{1, 2}, {1, 3}, ……., {1, n}, {2, 3}, {3, 4},……., {n – 1, n}, {n, 2}}.

Gt (Wn, x) =

∑

ek ∈E(Wn )

If gt(Wn) =

min

ek ∈E(Wn )

Gek(Wn, x) [gek(Wn)] n −2

∑

then we can write Gt (Wn, x) = i=g

Definition 3.1

t(Wn )

i

The edge fixed geodomination polynomial of a wheel Wn is defined as

gt(Wn, i)x , where gt (Wn, i) is the total number of edge fixed geodominating sets of cardinality i.

n −2

Gek(Wn, x) =

∑

gek(Wn, i)xi

i=ge (G)

k

where gek(Wn, i) is the number of edge fixed geodominating sets of Wn with cardinality i where ek is a fixed edge.

Theorem 3.2: The total edge fixed geodomination polynomial of wheel Wn is given by Gt (Wn, x) = (n–1) [ xn – 5+ 4x n – 4 + 5x n – 3 + 2xn – 2], n ≥ 6

Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

305

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

By a theorem 2.13 in [8]. For the wheel Wn n ≥ 6, = K1+ Cn−1 ,

there are 3C1, edge fixed geodominating set of cardinality n–4 coressponding to ek. The vertices vk and vk+1 are fixed and they do not form the edge fixed geodominating set.

gek (Wn) =  n − 5

Therefore gek (Wn, n–4) = 3C1

Proof:

 n − 4

if e k is a edge of C n −1 otherwise

By the Similar argument gek(Wn, n−3) = 3C2, gek (Wn, n–2) = 3C3. Therefore Gek (Wn, x) = [x n – 5 + 3C1xn – 4 + 3C2xn + 3C3 x n – 2] Case (ii) : Let ej be an edge which has one end at k1.

Then gej(Wn ) = n–4.

Here ej = (Vn, Vn+j- 1).

Let Wn be a wheel with n vertices v1, v2, ….vn. Let the edges of Wn be e1, e2,.., e2n – 2.

There are n – 1 edges lies in Cn – 1 and n – 1 edges has one end in K1.

Case (i): Let ek be an edge which lies in Cn – 1. Then gek (Wn) = n–5. ek = { vk + 1, vk} Here S = {v1,v2,…, vk – 2, vk + 3, …, vn – 1 } is the only edge fixed geodominating set of Wn with cardinality n – 5. Then gek (Wn, n–5) = 1. Considers gek (Wn, n–4). Every edge fixed geodominating set contains S. Thus S together with each of the remaining 3 vertices forms the edge fixed geodominating set of cardinality n – 4. Thus

Here every edge fixed geodominating set contains vertices other than vn, vn+j-1, Vn+j-2, Vn+j. Also S = { v1, v2,…., vn+j-3, vn+j+1, vn } is the only edge fixed geodominating set of Wn with cardinality 1. Thus gej (Wn, n–4) = 1 Consider gej (Wn, n – 3).

Here the vertices Vn, Vn+j – 1 are fixed and they do not form the any edge fixed geodominating set. Thus S together with each of the remaining 2C1 vertices forms the edge fixed geodominating sets of cardinality n – 3. Thus gej (Wn, n–3) = 2C1.

By the similar argument gej (Wn, n–2).= 2C2.

Therefore Gej (Wn, x) = xn – 4 + 2C1x n – 3 + 2C2 xn – 2.

Since Wn has (n – 1) edges in Cn – 1 and (n – 1) edges has one end at k1, we have Gt (Wn, x) = (n–1) [x n – 5 + 3C1x n – 4 + 3C2xn – 3 + 3C3x n – 2 + x n – 4 + 2C1xn – 3 + 2C2 xn – 2 ] = (n–1) [ xn – 5 + 4xn – 4 +5xn – 3 +2xn – 2]

Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

306

Ayyakutty Vijayan, et al., J. Comp. & Math. Sci. Vol.3 (3), 298-307 (2012)

Remark: The total edge fixed geodomination polynomial of W4 and W5 is given by.

(i) Gt (W4, x) = 6x2 (ii) Gt (W5, x) = 12x + 20x2

Table 2 gek (Wn, j) , the number of total edge fixed geodominating set of Wn of cardinality j, n ≥ 6 j

1

2

3

4

5

6

7

8

9

10

11

12

6

5

20

25

10

7

0

6

24

30

12

8

0

0

7

28

35

14

9

0

0

0

8

32

40

16

10

0

0

0

0

9

36

45

18

11

0

0

0

0

0

10

40

50

20

12

0

0

0

0

0

0

11

44

55

22

13

0

0

0

0

0

0

0

12

48

60

24

14

0

0

0

0

0

0

0

0

13

52

65

26

15

0

0

0

0

0

0

0

0

0

14

46

70

13

14

15

n

Theorem 3.3 The following properties hold for the coefficients of Gt (Wn, x), for all n ≥ 6. (i) (ii)

gt (Wn, n–2) = 2n–2 gt (Wn, n–3) = gt (Wn, n–4) + gt (Wn , n–5) (iii) gt (Wn, n–3) = 5 (n–1) (iv) gt (Wn, n–4) = 4 (n–1) (v) gt (Wn, n–5) = n–1 (vi) Sn = 12 (n–1)

28

Where Sn = Total number of edge fixed geodominating sets in Wn. Proof: Proof is obvious by using theorem 3.2.

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Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)

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9. Vijayan. A and Binu Selin. T, An introduction to geodetic Polynomial of a graph, Submitted.

10. Vijayan. A and Binu Selin. T, On Total edge fixed geodominating sets and polynomials of a graph, submitted.

Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)