Minimum Equitable Dominating Randic Energy of a Graph

Page 1

International J.Math. Combin. Vol.3(2017), 81-89

Minimum Equitable Dominating Randic Energy of a Graph P.Siva Kota Reddy (Department of Mathematics, Siddaganga Institute of Technology, B.H.Road, Tumkur-572103, India)

K. N. Prakasha (Department of Mathematics, Vidyavardhaka College of Engineering, Mysuru- 570 002, India)

Gavirangaiah K (Department of Mathematics, Government First Grade Collge, Tumkur-562 102, India) E-mail: reddy math@yahoo.com, prakashamaths@gmail.com, gavirangayya@gmail.com

Abstract: In this paper, we introduce the minimum equitable dominating Randic energy of a graph and computed the minimum dominating Randic energy of graph. Also, established the upper and lower bounds for the minimum equitable dominating Randic energy of a graph.

Key Words: Minimum equitable dominating set, Smarandachely equitable dominating set, minimum equitable dominating Randic eigenvalues, minimum equitable dominating Randic energy.

AMS(2010): 05C50. §1. Introduction Let G be a simple, finite, undirected graph, The energy E(G) is defined as the sum of the absolute values of the eigenvalues of its adjacency matrix. For more details on energy of graph see [5, 6]. The Randic matrix R(G) = (Rij )n×n is given by [1-3].

Rij =

 

√1 di di

 0

if vi ∼ vj , otherwise

We can see lower and upper bounds on Randic energy in [1,2]. Some sharp upper bounds for Randic energy of graphs were obtain in [3].

§2. The Minimum Equitable Dominating Randic Energy of Graph Let G be a simple graph of order n with vertex set V (G) = {v1 , v2 , v3 , · · · , vn } and edge set E. A subset U of V (G) is an equitable dominating set, if for every v ∈ V (G) − U there exists a 1 Received

December 19, 2016, Accepted August 22, 2017.


82

P.Siva Kota Reddy, K. N. Prakasha and Gavirangaiah K

vertex u ∈ U such that uv ∈ E(G) and |deg(u) − deg(v)| ≤ 1, and a Smarandachely equitable dominating set is its contrary, i.e., |deg(u) − deg(v)| ≥ 1 for such an edge uv, where deg(x) denotes the degree of vertex x in V (G). Any equitable dominating set with minimum cardinality is called a minimum equitable dominating set. Let E be a minimum equitable dominating set E of a graph G. The minimum equitable dominating Randic matrix RE (G) = (Rij )n×n is given by  √1    di di if vi ∼ vj , E Rij = 1 if i = j and vi ∈ E,    0 otherwise E The characteristic polynomial of RE (G) is denoted by φE R (G, λ) = det(λI − R (G)). Since the minimum equitable dominating Randic Matrix is real and symmetric, its eigenvalues are real numbers and we label them in non-increasing order λ1 > λ2 > · · · λn . The minimum equitable dominating Randic Energy is given by

REE (G) =

n X i=1

|λi |.

(1)

Definition 2.1 The spectrum of a graph G is the list of distinct eigenvalues λ1 > λ2 > · · · λr , with their multiplicities m1 , m2 , . . . , mr , and we write it as 

λ1

Spec(G) =  m1

λ2

···

m2

· · · mr

λr

.

This paper is organized as follows. In the Section 3, we get some basic properties of minimum equitable dominating Randic energy of a graph. In the Section 4, minimum equitable dominating Randic energy of some standard graphs are obtained.

§3. Some Basic Properties of Minimum Equitable Dominating Randic Energy of a Graph Let us consider P =

X 1 , dd i<j i j

where di dj is the product of degrees of two vertices which are adjacent. Proposition 3.1 The first three coefficients of φE R (G, λ) are given as follows: (i) a0 = 1; (ii) a1 = −|E|;

(iii) a2 = |E|C2 − P .


83

Minimum Equitable Dominating Randic Energy of a Graph

E Proof (i) From the definition ΦE R (G, λ) = det[λI − R (G)], we get a0 = 1.

(ii) The sum of determinants of all 1 × 1 principal submatrices of RE (G) is equal to the trace of RE (G) ⇒ a1 = (−1)1 trace of [RE (G)] = −|E|. (iii)

(−1)2 a2

aii aij

=

a a

ji jj 1≤i<j≤n X = aii ajj − aji aij X

1≤i<j≤n

X

=

1≤i<j≤n

aii ajj −

= |E|C2 − P.

X

aji aij

1≤i<j≤n

2

Proposition 3.2 If λ1 , λ2 , . . . , λn are the minimum equitable dominating Randic eigenvalues of RE (G), then n X λi 2 = |E| + 2P. i=1

Proof We know that n X

λ2i

=

i=1

n X n X

aij aji

i=1 j=1

n X X (aij )2 + (aii )2

=

2

=

X 2 (aij )2 + |E|

i<j

i=1

i<j

=

2

|E| + 2P.

Theorem 3.3 Let G be a graph with n vertices and Then RE E (G) ≤ where P =

p n(|E| + 2[P ])

X 1 dd i<j i j

for which di dj is the product of degrees of two vertices which are adjacent. Proof Let λ1 , λ2 , · · · , λn be the eigenvalues of RE (G). Now by Cauchy - Schwartz inequality we have !2 ! n ! n n X X X ai b i ≤ ai 2 bi 2 . i=1

i=1

i=1


84

P.Siva Kota Reddy, K. N. Prakasha and Gavirangaiah K

Let ai = 1 , bi =| λi |. Then n X i=1

!2

|λi |

n X

i=1

1

!

n X i=1

2

|λi |

!

Thus, [RE E ]2 ≤ n(|E| + 2P ), which implies that [RE E ] ≤ i.e., the upper bound.

p n(|E| + 2P ),

2

Theorem 3.4 Let G be a graph with n vertices. If R= det RE (G), then RE E (G) ≥

q 2 (|E| + 2P ) + n(n − 1)R n .

Proof By definition, 2 RE E (G)

= =

n X

i=1 n X i=1

=

!2

| λi |

| λi |

n X i=1

n X j=1 2

| λi |

!

| λj | +

X i6=j

| λi || λj | .

Using arithmetic mean and geometric mean inequality, we have X 1 | λi || λj | ≥ n(n − 1) i6=j

1   n(n−1) Y  | λi || λj | .

i6=j

Therefore,

[RE E (G)]2

n X

n X

=

i=1

i=1

n X i=1

1   n(n−1) Y | λi |2 +n(n − 1)  | λi || λj |

i6=j

| λi |2 +n(n − 1) | λi |2 +n(n − 1)R

n Y

i=1 2 n

2

= (|E| + 2P ) + n(n − 1)R n .

| λi |2(n−1)

1 ! n(n−1)


85

Minimum Equitable Dominating Randic Energy of a Graph

Thus, RE E (G) ≥

q 2 (|E| + 2P ) + n(n − 1)R n .

2

§4. Minimum Equitable Dominating Randic Energy of Some Standard Graphs

Theorem 4.1 The minimum equitable dominating Randic energy of a complete graph Kn is RE E (Kn ) = 3n−5 n−1 .

Proof Let Kn be the complete graph with vertex set V = {v1 , v2 , · · · , vn }. The minimum equitable dominating set = E = {v1 }. The minimum equitable dominating Randic matrix is 

1

  1  n−1  1   n−1 RE (Kn ) =  .  ..    1  n−1 1 n−1

1 n−1 1 n−1

...

1 n−1

0 .. .

... .. .

1 n−1 1 n−1

...

1 n−1 1 n−1

1 n−1

0 .. .

...

...

1 n−1 1 n−1 1 n−1

.. .

0 1 n−1

1 n−1  1  n−1   1  n−1 

. ..  .    1  n−1  0

The characteristic equation is λ+

1 n−1

n−2 2n − 3 n−3 λ2 − λ+ =0 n−1 n−1

 and the spectrum is SpecE R (Kn ) = Therefore, RE E (Kn ) =

√ (2n−3)+ 4n−3 2(n−1)

√ (2n−3)− 4n−3 2(n−1)

−1 n−1

1

1

n−2

.

3n − 5 . n−1

2

Theorem 4.2 The minimum equitable dominating Randic energy of star graph K1,n−1 is RE E (K1,n−1 ) =

√ 5.

Proof Let K1,n−1 be the star graph with vertex set V = {v0 , v1 , · · · , vn−1 }. Here v0 be the center. The minimum equitable dominating set = E = V (G). The minimum equitable


86

P.Siva Kota Reddy, K. N. Prakasha and Gavirangaiah K

dominating Randic matrix is 

1

 √ 1  n−1  √ 1  n−1 E R (K1,n−1 ) =   ..  .   1 √  n−1

√1 n−1

√1 n−1

...

√1 n−1

1

0

...

0

0 .. .

1 .. .

... .. .

0 .. .

0

0

...

1

0

0

...

0

√1 n−1

√1 n−1 

 0    0   ..  . .    0   1

The characteristic equation is

 spectrum is SpecE R (K1,n−1 ) =

λ(λ − 1)n−2 [λ − 2] = 0 2

1

0

1

n−2

1

.

2

Therefore, RE E (K1,n−1 ) = n.

Theorem 4.3 The minimum equitable dominating Randic energy of Crown graph Sn0 is RE

E

(Sn0 )

√ (4n − 7) + 4n2 − 8n + 5 = . n−1

Proof Let Sn0 be a crown graph of order 2n with vertex set {u1 , u2 , · · · , un , v1 , v2 , · · · , vn } and minimum dominating set = E = {u1 , v1 }. The minimum equitable dominating Randic matrix is 

1

  0    0   .  ..    0  E 0 R (Sn ) =   0   1   n−1  .  ..    1  n−1 1 n−1

0

0

...

0

0

1 n−1

...

0

0

...

0

0

...

0 .. .

0 .. .

... .. .

0 .. .

1 n−1 1 n−1

.. .

... .. .

1 n−1

0

0

...

0

1 n−1

...

1 n−1

1 n−1

1 n−1

1 n−1 1 n−1

...

1

0

...

0

.. .

... .. .

1 n−1 1 n−1

.. .

0 .. .

0 .. .

... .. .

0 .. .

0

...

1 n−1

0

0

...

0

1 n−1

...

0

0

0

...

0

0 .. . 1 n−1 1 n−1

..

.

1 n−1 1 n−1

0 .. .

1 n−1  1  n−1   1  n−1 

..  .    0   . 0    0   ..  .    0   0


87

Minimum Equitable Dominating Randic Energy of a Graph

The characteristic equation is λ+

1 n−1

n−2 λ−

0 spectrum is SpecE R (Sn )  √ √

=

1 n−1

n−2 λ2 −

1 λ−1 n−1

2n − 3 n−3 2 λ − λ+ =0 n−1 n−1

(2n−3)+ 4n−3 2(n−1)

1+ 4n2 −8n+5 2(n−1)

√ (2n−3)− 4n−3 2(n−1)

1 n−1

−1 n−1

√ 1− 4n2 −8n+5 2(n−1)

1

1

1

n−2

n−2

1

Therefore, RE

E

(Sn0 )

√ (4n − 7) + 4n2 − 8n + 5 . = n−1

.

2

Theorem 4.4 The minimum equitable dominating Randic energy of complete bipartite graph Kn,n of order 2n with vertex set {u1 , u2 , · · · , un , v1 , v2 , · · · , vn } is √ 2 n−1 √ RE (Kn,n ) = + 2. n E

Proof Let Kn,n be the complete bipartite graph of order 2n with vertex set {u1 , u2 , · · · , un , v1 , v2 , · · · , vn }. The minimum equitable dominating set = E = {u1 , v1 } with a minimum equitable dominating Randic matrix 

1

 0   0   0  . . RE (Kn,n ) =  .  1 n  1 n 1  n 1 n

1 n 1 n 1 n 1 n

1 n 1 n 1 n 1 n

1 n 1 n 1 n 1 n

...

1

0

0

...

0

0

0

...

0

0

0

...

0

0

0

0

0

0

...

0

0

0

...

0

0

0

...

0 .. .

0 .. .

0 .. .

... .. .

1 n 1 n 1 n 1 n

1 n 1 n 1 n 1 n

1 n 1 n 1 n 1 n

.. .

.. .

.. .

1 n  1 n  1 n  1 n

..  . .  0   0   0  0

The characteristic equation is λ2n−4 (λ2 −

n−1 2 n−1 )[λ − 2λ + ]=0 n n

Hence, spectrum is 

 SpecE R (Kn,n ) =

1+

q

1 n

√ n−1 √ n

1

√ 2 n−1 √ Therefore, RE (Kn,n ) = + 2. n E

1

1−

q

1

1 n

0 2n − 4

√ n−1 √ n

1

.

2


88

P.Siva Kota Reddy, K. N. Prakasha and Gavirangaiah K

Theorem 4.5 The minimum equitable dominating Randic energy of cocktail party graph Kn×2 is 4n − 6 RE E (Kn×2 ) = . n−1 Proof Let Kn×2 be a Cocktail party graph of order 2n with vertex set {u1 , u2 , · · · , un , v1 , v2 , · · · , vn }. The minimum equitable dominating set = E = {u1 , v1 } with a minimum equitable dominating Randic matrix 

1 2n−2

1

  1  2n−2   1  2n−2  1   2n−2  . . RE (Kn×2 ) =   .   0    1  2n−2  1   2n−2

0 1 2n−2 1 2n−2

.. .

1 2n−2

0 1 2n−2 1 2n−2

1 2n−2

1 2n−2 1 2n−2

1 2n−2 1 2n−2 1 2n−2

...

0

1 2n−2

...

0

1 2n−2

0 .. .

... .. .

1 2n−2 1 2n−2 1 2n−2

1 2n−2 1 2n−2

1 2n−2 1 2n−2

...

1

1 2n−2

...

1 2n−2

0

...

1 2n−2 1 2n−2 1 2n−2

0

0

1 2n−2 1 2n−2 1 2n−2

0 .. .

...

.. .

...

.. .

1 2n−2 1 2n−2

1 2n−2 1 2n−2

0 1 2n−2

.. .

1 2n−2 1 2n−2

0 1 2n−2

1 2n−2  1  2n−2   1  2n−2 

    .   1  2n−2   1  2n−2   1  2n−2  0 0 .. .

The characteristic equation is λn−1 λ +

1 n−1

n−2

(λ − 1)[λ2 −

2n − 3 n−3 λ+ ]=0 n−1 n−1

Hence, spectrum is 

 SpecE R (Kn×2 ) = Therefore, RE E (Kn×2 ) =

√ 2n−3+ 4n−3 2(n−1)

1

√ 2n−3− 4n−3 2(n−1)

1

1

1

4n − 6 . n−1

0

−1 n−1

n−1 n−2

.

2

References [1] S.B. Bozkurt, A. D. Gungor, I. Gutman, A. S. Cevik, Randic matrix and Randic energy, MATCH Commum. Math. Comput. Chem. 64 (2010) 239-250. [2] S. B. Bozkurt, A. D. Gungor, I. Gutman,, Randic spectral radius and Randic energy, MATCH Commum. Math. Comput. Chem. 64 (2010) 321-334. [3] Serife Burcu Bozkurt, Durmus Bozkurt, Sharp Upper Bounds for Energy and Randic Energy, MATCH Commum. Math. Comput. Chem. 70 (2013) 669-680. [4] I. Gutman, B. Furtula, S B. Bozkurt, On Randic energy, Linear Algebra Appl., 442 (2014) 50-57. [5] I. Gutman, The energy of a graph, Ber. Math. Stat. Sekt. Forschungsz. Graz, 103(1978),


Minimum Equitable Dominating Randic Energy of a Graph

89

1-22. [6] I. Gutman, The energy of a graph: old and new results, Combinatorics and applications, A. Betten, A. Khoner, R. Laue and A. Wassermann, eds., Springer, Berlin, (2001), 196-211. [7] G. Indulal, I. Gutman, A. Vijayakumar, On distance energy of graphs, Match Commun. Math. Comput. Chem., 60(2008), 461-472.


Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.