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.
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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 .
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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)
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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
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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
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