Minimum Equitable Dominating Randic Energy of a Graph

2018·0.17 MB·

by  Rajendra P.

International J.Math. Combin. Vol.1(2018), 97-108
Minimum Equitable Dominating Randic´ Energy of a Graph
Rajendra P.
(BharathiCollege,PGandResearchCentre,Bharathinagara,571422,India)
R.Rangarajan
(DOSinMathematics,UniversityofMysore,Mysuru,570006,India)
[email protected],[email protected]
Abstract: Let G be a graph with vertex set V(G), edge set E(G) and di is the degree
of its i-th vertex vi, then the Randic´ matrix R(G) of G is the square matrix of order n,
whose(i,j)-entryisequalto √d1idj ifthei-thvertexvi andj-thvertexvj ofGareadjacent,
and zero otherwise. The Randic energy [3] RE(G) of the graph G is defined as the sum of
the absolute values of the eigenvalues of the Randic´matrix R(G). A subset ED of V(G) is
called an equitable dominating set [11], if for every vi V(G) ED there exists a vertex
∈ −
vj EDsuchthatvivj E(G)and di(vi) dj(vj) 1. Inthecontrast,suchadominating
∈ ∈ | − |≤
set ED is Smarandachely if di(vi) dj(vj) 1. Recently, Adiga, et.al. introduced, the
| − | ≥
minimum covering energy Ec(G) of a graph [1] and S. Burcu Bozkurt, et.al. introduced,
Randic´ Matrix and Randic´ Energy of a graph [3]. Motivated by these papers, Minimum
equitable dominating Randic´energy of a graph REED(G) of some graphs are worked out
and boundson REED(G) are obtained.
Key Words: Randic´matrix anditsenergy,minimum equitabledominatingset and mini-
mum equitable dominating Randic´energy of a graph.
AMS(2010): 05C50.
§1. Introduction
Let G be a graph with vertex set V(G) and edge set E(G). The adjacency matrix A(G) of
the graph G is a square matrix, whose (i,j)-entry is equal to 1 if the vertices v and v are
i j
adjacent,otherwisezero[12]. SinceA(G)issymmetric,itseigenvaluesareallreal. Denotethem
by λ ,λ ,...,λ , and as a whole, they are called the spectrum of G and denoted by Spec(G).
1 2 n
The energy of graph [12] G is
ε(G)= n λ .
i=1| i|
P
The literature on energy of a graph and its bounds can refer [4,8,9,10,12]. The Randic´
matrix R(G)=(r ) of G is the square matrix of order n, where
ij
1ReceivedJuly28,2017,AcceptedMarch1,2018.
98 RajendraP.andR.Rangarajan
1 , if v and v are adjacent vertices in G;
(rij)= √didj i j
0, otherwise.

TheRandicenergy[3]RE(G)ofthegraphGisdefinedasthesumoftheabsolutevaluesof
the eigenvaluesof the Randic´matrix R(G). Let ρ ,ρ , ,ρ be the eigenvalues ofthe Randic´
1 2 n
···
matrix R(G). Since R(G) is symmetric matrix, these eigenvalues are real numbers and their
sum is zero. Randic´energy [3] can be defined as
RE(G)= n ρ
i=1| i|
For details of Randic´energy and its bounds, canPrefer [2, 3, 5, 6, 7].
Let G be a graph with vertex set V(G) = v ,v , ,v and ED is minimum equitable
1 2 n
{ ··· }
dominating set of G. Minimum equitable dominating Randic´ matrix of G is n n matrix
×
R (G)=(r ), where
ED ij
1 , if v and v are adjacent vertices in G;
√didj i j
(rij)= 1, if i=j and vi ∈ED;
0, otherwise.
ThecharacteristicpolynomialofRED(G)isdenotedbydet(ρI RED(G))= ρI RED(G).
− | − |
Since R (G) is symmetric, its eigenvalues are real numbers. If the distinct eigenvalues of
ED
R (G) are ρ >ρ > >ρ with their multiplicities are m ,m , ,m then spectrum of
ED 1 2 r 1 2 r
··· ···
R (G) is denoted by
ED
ρ ρ ρ
1 2 r
SpecR (G)= ··· .
ED
m m m 
1 2 r
···
 
The minimum equitable dominating Randic´energy of G is defined as
n
RE (G)= ρ .
ED i
| |
i=1
X
Example 1.1 Let W be a wheel graph, with vertex set V(G) = v ,v ,v ,v ,v , and let
5 1 2 3 4 5
{ }
its minimum equitable dominating set be ED = v . Then minimum equitable dominating
1
{ }
Randic´matrix R (W ) is
ED 5
1 1 1 1 1
√12 √12 √12 √12
 1 0 1 0 1 
√12 3 3
RED(W5)=√112 13 0 13 0 
 
 1 0 1 0 1 
√12 3 3 
 
 1 1 0 1 0 
√12 3 3 
 
0.6666 0 0.2324 1.4342
SpecR (W )= − .
ED 5
 1 2 1 1 
 
MinimumEquitableDominatingRandic´EnergyofaGraph 99
The minimum equitable dominating Randic´energy of W is RE (W )=2.3332.
5 ED 5
§2. Bounds for the Minimum Equitable Dominating Randic´Energy of a Graph
Lemma 2.1 If ρ ,ρ , ,ρ are the eigenvalues of R (G). Then
1 2 n ED
···
n
ρ = ED
i
| |
i=1
X
and
n
1
ρ2 = ED +2 ,
i | | d d
i j
i=1 i<j
X X
where ED is minimum equitable dominating set.
Proof (i) The sum of eigenvalues of R (G) is
ED
n n
ρ = r = ED .
i ii
| |
i=1 i=1
X X
(ii) Consider, the sum of squares of ρ ,ρ ,...,ρ is,
1 2 n
n n n n
ρ2 = r r = (r )2+ r r
i ij ji ii ij ji
i=1 i=1j=1 i=1 i=j
X XX X X6
n
= (r )2+2 (r )2
ii ij
i=1 i<j
X X
n
1
ρ2 = ED +2 .
i | | d d 2
i j
i=1 i<j
X X
Upper and lower bounds for RE (G) is similar proof to McClelland’s inequalities [10],
ED
are given below
Theorem 2.2(UpperBound) Let G be a graph with ED is minimum equitable dominating set.
Then
1
REED(G) vn ED +2 .
≤ u | | didj
uu Xi<j
t  
Proof Let ρ ,ρ , ,ρ be the eigenvalues of R (G). By Cauchy-Schwartz inequality,
1 2 n ED
···
we have
n 2 n n
a b a2 b2 , (1)
i i! ≤ i! i!
i=1 i=1 i=1
X X X
where a and b are any real numbers.
100 RajendraP.andR.Rangarajan
If a =1, b = ρ in (1), we get
i i i
| |
n 2 n n
1
ρ 12 ρ 2 , [RE (G)]2 n ED +2
i i ED
| |! ≤ ! | | ! ≤ | | didj
i=1 i=1 i=1 i<j
X X X X
 
by Lemma 2.1,
1
RE (G) n ED +2 .
ED v
≤u | | didj
u Xi<j
u
t  
Theorem 2.3(Lower Bound) Let G be a graph with ED is minimum equitable dominating
| |
set and d is degree of v . Then
i i
1
REED(G) ED +2 +n(n 1)Dn2,
≥ | | d d −
s i<j i j
X
where D = n ρ .
i=1| i|
Q
Proof Consider
n 2 n
[RE (G)]2 = ρ = ρ 2+ ρ ρ . (2)
ED i i i j
" | |# | | | | | |
i=1 i=1 i=j
X X X6
By using arithmetic and geometric mean inequality, we have
1
n(n−1)
1
ρ ρ ρ ρ
i j i j
n(n 1) | | | | ≥  | | | |
− i=j i=j
X6 Y6
  1
n n(n−1)
ρ ρ n(n 1) ρ 2(n 1)
i j i −
| | | | ≥ − | | !
i=j i=1
X6 Y
n 2/n
ρ ρ n(n 1) ρ . (3)
i j i
| | | | ≥ − | |!
i=j i=1
X6 Y
Now using (3) in (2), we get
n n 2/n
[RE (G)]2 ρ 2+n(n 1) ρ ,
ED i i
≥ | | − | |!
i=1 i=1
X Y
n
1
[REED(G)]2 ED +2 +n(n 1)Dn2, where D = ρi ,
≥| | d d − | |
i j
i<j i=1
X Y
1
REED(G) ED +2 +n(n 1)Dn2.
≥ | | d d −
s i<j i j
X
MinimumEquitableDominatingRandic´EnergyofaGraph 101
§3. Bounds for Largest Eigenvalue of R (G) and its Energy
ED
Proposition 3.1 Let G be a graph and ρ (G) = max ρ be the largest eigenvalue of
1 1 i n i
≤≤ {| |}
R (G). Then
ED
1 1 1
v ED +2 ρ1(G) ED +2 .
uuun| | Xi<j didj≤ ≤s| | Xi<j didj
t  
Proof Consider,
n
1
ρ2(G) = max ρ 2 ρ 2 = ED +2
1 1≤i≤n {| i| }≤ | i| | | didj
i=1 i<j
X X
1
ρ (G) ED +2
1
≤ | | d d
s i<j i j
X
Next,
n
1
n ρ2(G) ρ2 = ED +2
1 ≥ i | | d d
i j
i=1 i<j
X X
1 1
ρ2(G) ED +2
1 ≥ n| | d d 
i j
i<j
X
 
1 1
ρ (G) ED +2
1 v
≥ un| | didj
u Xi<j
u
t  
Therefore,
1 1 1
ED +2 ρ (G) ED +2 .
v 1
uun| | Xi<j didj≤ ≤s| | Xi<j didj
u
t  
Proposition 3.2 If G is a graph and n ED +2 1 , then
≤| | i<j didj
P
REED(G)≤ |ED|+2nPi<j di1dj +vu(n−1)|ED|+2 di1dj − |ED|+2nPi<j di1dj !2.
uu Xi<j
t  
Proof We know that,
n n
1 1
ρ2 = ED +2 , ρ2 = ED +2 ρ2 (4)
i | | d d i | | d d − 1
i j i j
i=1 i<j i=2 i<j
X X X X
102 RajendraP.andR.Rangarajan
By Cauchy-Schwarz inequality, we have
n 2 n n
a b a2 b2 .
i i! ≤ i! i!
i=2 i=2 i=2
X X X
If a =1 and b = ρ , we have
i i i
| |
n 2 n
ρ (n 1) ρ 2
i i
| |! ≤ − | |
i=2 i=2
X X
1
[RE (G) ρ ]2 (n 1) ED +2 ρ2
ED −| 1| ≤ − | | d d − 1
i j
i<j
X
 
1
REED(G) ≤ ρ1+vu(n−1)|ED|+2 didj −ρ21 (5)
uu Xi<j
t  
Consider the function,
1
F(x)=x+ (n 1) ED +2 x2 .
v
u − | | didj − 
uu Xi<j
t  
Then,
x (n 1)
F′(x)=1 − .
− ED +p2 1 x2
| | i<j didj −
q
P
Here F(x) is decreasing in
1 1 1
ED +2 , ED +2 .
n| | d d  | | d d 
i j i j
i<j i<j
X X
   
We know that F (x) 0,
′
≤
x (n 1)
1 − 0.
− ED +p2 1 x2 ≤
| | i<j didj −
q P
We have
ED +2 1
x | | i<j didj .
≥s n
P
Since,
n ED +2 1 and |ED|+2 i<j di1dj ρ ,
1
≤| | d d n ≤
i j P
i<j
X
MinimumEquitableDominatingRandic´EnergyofaGraph 103
we have
|ED|+2 i<j di1dj |ED|+2 i<j di1dj ρ ED +2 1 .
s n ≤ n ≤ 1 ≤| | d d
P P i j
i<j
X
Then, equation (5) become
REED(G) |ED|+2 i<j di1dj +v(n 1) ED +2 1 |ED|+2 i<j di1dj 2 .
≤ nP u − | | didj − nP ! 
uu Xi<j
t  
2
§4. Minimum Equitable Dominating Randic´Energy of Some Graphs
Theorem 4.1 If K is complete graph with n vertices, then minimum equitable dominating
n
Randic´energy of K is
n
3n 5
RE (K )= − .
ED n
n 1
−
Proof LetK bethecompletegraphwithvertexsetV(G)= v ,v ,...,v andminimum
n 1 2 n
{ }
equitable dominating set is ED = v , we have characteristic polynomial of R (K ) is
1 ED n
{ }
ρ 1 1 1 1 1 1
− n−1 n−1 ··· n−1 n−1 n−1
− − − − −
(cid:12) 1 ρ 1 1 1 1 (cid:12)
(cid:12) n−1 n−1 ··· n−1 n−1 n−1(cid:12)
(cid:12) − − − − − (cid:12)
(cid:12)(cid:12) n−11 n−11 ρ ··· n−11 n−11 n−11(cid:12)(cid:12)
|ρI −RED(Kn)|=(cid:12)(cid:12)(cid:12) −... −... ... ... −... −... −... (cid:12)(cid:12)(cid:12) .
(cid:12) (cid:12)
(cid:12) 1 1 1 ρ 1 1 (cid:12)
(cid:12) n−1 n−1 n−1 ··· n−1 n−1(cid:12)
(cid:12) − − − − − (cid:12)
(cid:12) 1 1 1 1 ρ 1 (cid:12)
(cid:12) n−1 n−1 n−1 ··· n−1 n−1(cid:12)
(cid:12) − − − − − (cid:12)
(cid:12)(cid:12)(cid:12) n−−11 n−−11 n−−11 ··· n−−11 n−−11 ρ (cid:12)(cid:12)(cid:12)n×n
(cid:12) (cid:12)
(cid:12) (cid:12)
R = R R , k = 3,4, ,n 1,n. Then, we get ρ+ 1 common from R to R
k′ k − 2 ··· − n 1 3 n
−
and we have (cid:16) (cid:17)
ρ 1 1 1 1 1 1
− n−1 n−1 ··· n−1 n−1 n−1
− − − − −
(cid:12) 1 ρ 1 1 1 1 (cid:12)
(cid:12) n−1 n−1 ··· n−1 n−1 n−1(cid:12)
(cid:12) − − − − − (cid:12)
(cid:12) 0 1 1 0 0 0 (cid:12)
n 2(cid:12) − ··· (cid:12)
|ρI−RED(Kn)|= ρ+ n 1 1 − (cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12) .
(cid:18) − (cid:19) (cid:12) (cid:12)
(cid:12) 0 1 0 1 0 0 (cid:12)
(cid:12) − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 1 0 0 1 0 (cid:12)
(cid:12) − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 1 0 0 0 1 (cid:12)
(cid:12) − ··· (cid:12)n n
(cid:12) (cid:12) ×
(cid:12) (cid:12)
(cid:12) (cid:12)
104 RajendraP.andR.Rangarajan
C =C +C + +C , we get the characteristic polynomial
2′ 2 3 ··· n
n 2
1 − 2n 3 n 3
ρI R (K ) = ρ+ ρ2 − ρ+ − ,
ED n
| − | n 1 − n 1 n 1
(cid:18) − (cid:19) (cid:20) (cid:18) − (cid:19) (cid:18) − (cid:19)(cid:21)
1 (2n 3) √4n 3 (2n 3)+√4n 3
SpecRED(Kn) = n−−1 −2(n−−1) − −2(n−1) − .
n 2 1 1 
−
 
The minimum equitable dominating Randic´energy of K is
n
3n 5
RE (K )= − .
ED n
n 1 2
−
Theorem 4.2 If S ,(n 4) is star graph with n vertices, then minimum equitable dominating
n
≥
Randic´energy of S is
n
RE (S )=n.
ED n
Proof Let S ,(n 4) be the star graph with vertex set V(G) = v ,v , ,v and
n 1 2 n
≥ { ··· }
minimumequitabledominatingsetisED = v ,v , ,v ,wehavecharacteristicpolynomial
1 2 n
{ ··· }
of R (S ) is
ED n
ρ 1 1 1 1 1 1
− √−n 1 √−n 1 ··· √−n 1 √−n 1 √−n 1
− − − − −
(cid:12) 1 ρ 1 0 0 0 0 (cid:12)
(cid:12)(cid:12)√−n−1 − ··· (cid:12)(cid:12)
(cid:12) 1 0 ρ 1 0 0 0 (cid:12)
ρI RED(Sn) =(cid:12)(cid:12)(cid:12)(cid:12)√−n...−1 ... −... ·..·.· ... ... ... (cid:12)(cid:12)(cid:12)(cid:12) .
| − | (cid:12) (cid:12)
(cid:12)(cid:12)(cid:12)√−n11 0 0 ··· ρ−1 0 0 (cid:12)(cid:12)(cid:12)
(cid:12)(cid:12)(cid:12)√−n−−11 0 0 ··· 0 ρ−1 0 (cid:12)(cid:12)(cid:12)
(cid:12)(cid:12)(cid:12)(cid:12)√−n−11 0 0 ··· 0 0 ρ−1(cid:12)(cid:12)(cid:12)(cid:12)n×n
(cid:12) (cid:12)
R =R R , k =3,4,(cid:12) ,n. Then taking (ρ 1) common from R to R(cid:12) , we get
k′ k− 2 ··· − 3 n
ρ 1 1 1 1 1 1
− √−n 1 √−n 1 ··· √−n 1 √−n 1 √−n 1
− − − − −
(cid:12) 1 ρ 1 0 0 0 0 (cid:12)
(cid:12)√−n 1 − ··· (cid:12)
(cid:12) − (cid:12)
(cid:12) 0 1 1 0 0 0 (cid:12)
(cid:12) − ··· (cid:12)
|ρI −RED(Sn)|=(ρ−1)n−2(cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12) .
(cid:12) (cid:12)
(cid:12) 0 1 0 1 0 0 (cid:12)
(cid:12) − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 1 0 0 1 0 (cid:12)
(cid:12) − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 1 0 0 0 1 (cid:12)
(cid:12) − ··· (cid:12)n n
(cid:12) (cid:12) ×
(cid:12) (cid:12)
(cid:12) (cid:12)
C =C +C + +C . Then, the characteristic polynomial
2′ 2 3 ··· n
ρI R (S ) =ρ(ρ 1)n 2(ρ 2),
ED n −
| − | − −
MinimumEquitableDominatingRandic´EnergyofaGraph 105
0 1 2
SpecRED(Sn)= .
1 n 2 1
 − 
 
The minimum equitable dominating Randic´energy of S is RE (S )=n.
n ED n
2
Theorem 4.3 If K , where m< n and m n 2 is complete bipartite graph with m+n
m,n
| − |≥
vertices, then minimum equitable dominating Randic´energy of K is RE (K )=m+n.
m,n ED m,n
Proof Let K , where m < n and m n 2 be the complete bipartite graph with
m,n
| − | ≥
vertex set V(G) = v ,v ,...,v ,u ,u , ,u and minimum equitable dominating set is
1 2 m 1 2 n
{ ··· }
ED =V(G), we have characteristic polynomial of R (K ) is
ED m,n
ρ 1 0 ... 0 0 1 1 1 1
− √−mn √−mn ··· √−mn √−mn
(cid:12) 0 ρ 1 ... 0 0 1 1 1 1 (cid:12)
(cid:12)(cid:12)(cid:12)(cid:12) ... −... ... ... ... √−m...n √−m...n ·..·.· √−m...n √−m...n(cid:12)(cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12) 0 0 ··· ρ−1 0 √−m1n √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12)
ρI R (K ) =(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−m1n √−m1n ... √−m1n √−m1n(cid:12)(cid:12)(cid:12).
| − ED m,n | (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12)
(cid:12) 1 1 1 1 0 ρ 1 0 0 (cid:12)
(cid:12)(cid:12)√−mn √−mn ··· √−mn √−mn − ··· (cid:12)(cid:12)
(cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n 0 0 ··· ρ−1 0 (cid:12)(cid:12)(cid:12)
(cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n 0 0 ··· 0 ρ−1(cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
R =R R , k =(cid:12) 1,2,3, ,m 1andR =R R , d=m+2,m+3, ,m+(cid:12) n.
k′ k− m ··· − d′ d− m+1 ···
Then, taking (ρ 1) common from R to R and R to R , we get
1 m 1 m+2 m+n
− −
1 0 0 1 0 0 0 0
··· − ···
(cid:12) 0 1 0 1 0 0 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12) ... ... ·..·.· ... −... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 1 1 0 0 0 0 (cid:12)
(cid:12) ··· − ··· (cid:12)
(ρ 1)m+n−2(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−m1n √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12).
− (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12)
(cid:12) 0 0 0 0 1 1 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12) ... ... ·..·.· ... ... −... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 0 0 1 0 1 0 (cid:12)
(cid:12) ··· − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 0 0 1 0 0 1 (cid:12)
(cid:12) ··· − ··· (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
106 RajendraP.andR.Rangarajan
C =C +C + +C ,
m′ +1 m+1 m+2 ··· m+n
1 0 0 1 0 0 0 0
··· − ···
(cid:12) 0 1 0 1 0 0 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12) ... ... ·..·.· ... −... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 1 1 0 0 0 0 (cid:12)
(cid:12) ··· − ··· (cid:12)
(ρ 1)m+n−2(cid:12)(cid:12)(cid:12) 0 0 ··· 0 ρ−1 √−mnn √−m1n ··· √−m1n √−m1n(cid:12)(cid:12)(cid:12).
− (cid:12)(cid:12)(cid:12)√−m1n √−m1n ··· √−m1n √−m1n ρ−1 0 ··· 0 0 (cid:12)(cid:12)(cid:12)
(cid:12) 0 0 0 0 0 1 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)(cid:12) ... ... ·..·.· ... ... ... ... ·..·.· ... ... (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 0 0 0 0 1 0 (cid:12)
(cid:12) ··· ··· (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 0 0 0 0 0 1 (cid:12)
(cid:12) ··· ··· (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
The characterist(cid:12)ic polynomial ρI RED(Km,n) =ρ(ρ 1)m+n−2(ρ 2), (cid:12)
| − | − −
0 1 2
SpecR (K )= .
ED m,n
1 m+n 2 1
−
 
The minimum equitable dominating Randic´energy of K is RE (K )=m+n.
m,n ED m,n
2
Theorem4.4 IfK ,(n 3)iscocktailpartygraphwith2nvertices, thenminimumequitable
n 2
× ≥
dominating Randic´energy of K is 4n 6.
n×2 n−−1
Proof LetK ,(n 3)bethecocktailpartygraphwithvertexsetV(G)= v ,v , ,v ,
n 2 1 2 n
× ≥ { ···
u ,u , ,u andminimumequitabledominatingsetisED = v ,u ,wehavecharacteristic
1 2 n 1 1
··· } { }
polynomial of R (K ) is
ED n 2
×
λ 1 0 −1 −1 −1 −1 −1 −1
− 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2
(cid:12) 0 λ 1 −1 −1 −1 −1 −1 −1 (cid:12)
(cid:12) − 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2(cid:12)
(cid:12) (cid:12)
(cid:12) −1 −1 λ −1 −1 0 −1 −1 (cid:12)
(cid:12)2n−2 2n−2 2n−2 ··· 2n−2 ··· 2n−2 2n−2(cid:12)
(cid:12) (cid:12)
(cid:12) −1 −1 −1 λ −1 −1 −1 −1 (cid:12)
(cid:12)2n−2 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2 2n−2(cid:12)
(cid:12)(cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12)(cid:12)
|ρI−RED(Kn×2)|=(cid:12)(cid:12)(cid:12)(cid:12)2n−−12 2n−−12 2n−−12 2n−−12 ··· λ 2n−−12 ··· 2n−−12 0 (cid:12)(cid:12)(cid:12)(cid:12).
(cid:12)(cid:12)(cid:12)2n−−12 2n−−12 0 2n−−12 ··· 2n−−12 λ ··· 2n−−12 2n−−12(cid:12)(cid:12)(cid:12)
(cid:12)(cid:12) ... ... ... ... ... ... ... ... ... ... (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) −1 −1 −1 −1 −1 −1 λ −1 (cid:12)
(cid:12)2n−2 2n−2 2n−2 2n−2 ··· 2n−2 2n−2 ··· 2n−2(cid:12)
(cid:12) (cid:12)
(cid:12) −1 −1 −1 −1 0 −1 −1 λ (cid:12)
(cid:12)2n−2 2n−2 2n−2 2n−2 ··· 2n−2 ··· 2n−2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)