International J.Math. Combin. Vol.3(2017), 81-89 Minimum Equitable Dominating Randic Energy of a Graph P.Siva Kota Reddy (DepartmentofMathematics,SiddagangaInstituteofTechnology,B.H.Road,Tumkur-572103,India) K. N. Prakasha (DepartmentofMathematics,VidyavardhakaCollegeofEngineering,Mysuru-570002,India) GavirangaiahK (DepartmentofMathematics,GovernmentFirstGradeCollge,Tumkur-562102,India) E-mail: [email protected],[email protected],[email protected] Abstract: Inthispaper,weintroducetheminimumequitabledominatingRandicenergyof a graph and computed the minimum dominating Randic energy of graph. Also, established theupperandlowerboundsfortheminimumequitabledominatingRandicenergyofagraph. KeyWords: Minimumequitabledominatingset,Smarandachelyequitabledominatingset, minimum equitabledominating Randiceigenvalues, minimum equitabledominating 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)=(R ) is given by [1-3]. ij n n × 1 if v v , Rij = √didi i ∼ j 0 otherwise  We can see lower and upper bounds 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 graphof ordern with vertex setV(G)= v ,v ,v , ,v andedge set E. 1 2 3 n { ··· } A subset U of V(G) is an equitable dominating set, if for every v V(G) U there exists a ∈ − 1ReceivedDecember19,2016,Accepted August22,2017. 82 P.SivaKotaReddy,K.N.PrakashaandGavirangaiahK 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) | − | ≥ denotesthedegreeofvertexxinV(G). Anyequitabledominatingsetwithminimumcardinality is called a minimum equitable dominating set. Let E be a minimum equitable dominating set of a graph G. The minimum equitable dominating Randic matrix RE(G) = (RE) is given ij n n × by 1 if v v , √didi i ∼ j RE = 1 if i=j and v E, ij i  0 otherwise ∈ The characteristicpolynomialofRE(G) is denotedby φE(G,λ)=det(λI RE(G)). Since R − the minimum equitable dominating Randic Matrix is real and symmetric, its eigenvalues are real numbers and we label them in non-increasing order λ > λ > λ . The minimum 1 2 n ··· equitable dominating Randic Energy is given by n RE (G)= λ . (1) E i | | i=1 X Definition 2.1 The spectrum of a graph G is the list of distinct eigenvalues λ >λ > λ , 1 2 r ··· with their multiplicities m ,m ,...,m , and we write it as 1 2 r λ λ λ 1 2 r Spec(G)= ··· . m m m  1 2 r ···   This paper is organized as follows. In the Section 3, we get some basic properties of minimumequitabledominatingRandicenergyofagraph. IntheSection4,minimumequitable 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 1 P = , d d i j i<j X where d d is the product of degrees of two vertices which are adjacent. i j Proposition 3.1 The first three coefficients of φE(G,λ) are given as follows: R (i) a =1; 0 (ii) a = E ; 1 −| | (iii) a = E C P. 2 2 | | − MinimumEquitableDominatingRandicEnergyofaGraph 83 Proof (i) From the definition ΦE(G,λ)=det[λI RE(G)], we get a =1. R − 0 (ii) The sum of determinants of all 1 1 principal submatrices of RE(G) is equal to the × trace of RE(G) a =( 1)1 trace of [RE(G)]= E . 1 ⇒ − −| | (iii) a a ( 1)2a = ii ij − 2 1 i<j n(cid:12)(cid:12)aji ajj(cid:12)(cid:12) ≤X≤ (cid:12) (cid:12) (cid:12) (cid:12) = (cid:12)a a a(cid:12) a (cid:12)ii jj − (cid:12)ji ij 1 i<j n ≤X≤ = a a a a ii jj ji ij − 1 i<j n 1 i<j n ≤X≤ ≤X≤ = E C P. 2 | | − 2 Proposition 3.2 If λ ,λ ,...,λ are the minimum equitable dominating Randic eigenvalues 1 2 n of RE(G), then n λ 2 = E +2P. i | | i=1 X Proof We know that n n n λ2 = a a i ij ji i=1 i=1j=1 X XX n = 2 (a )2+ (a )2 ij ii i<j i=1 X X = 2 (a )2+ E ij | | i<j X = E +2P. | | 2 Theorem 3.3 Let G be a graph with n vertices and Then REE(G) n(E +2[P]) ≤ | | p where 1 P = d d i j i<j X for which d d is the product of degrees of two vertices which are adjacent. i j Proof Letλ ,λ , ,λ betheeigenvaluesofRE(G). NowbyCauchy-Schwartzinequal- 1 2 n ··· ity we have n 2 n n a b a 2 b 2 . i i i i ! ≤ ! ! i=1 i=1 i=1 X X X 84 P.SivaKotaReddy,K.N.PrakashaandGavirangaiahK Let a =1 , b = λ . Then i i i | | n 2 n n λ 1 λ 2 i i | |! ≤ ! | | ! i=1 i=1 i=1 X X X Thus, [REE]2 n(E +2P), ≤ | | which implies that [REE] n(E +2P), ≤ | | p i.e., the upper bound. 2 Theorem 3.4 Let G be a graph with n vertices. If R= det RE(G), then REE(G) (E +2P)+n(n 1)Rn2. ≥ | | − q Proof By definition, n 2 REE(G) 2 = λ i | |! i=1 (cid:0) (cid:1) X n n = λ λ i j | | | | i=1 j=1 X X n = λ 2 + λ λ . i i j | | ! | || | i=1 i=j X X6 Using arithmetic mean and geometric mean inequality, we have 1 n(n−1) 1 λ λ λ λ . i j i j n(n 1) | || | ≥  | || | − i=j i=j X6 Y6   Therefore, 1 n n(n−1) [REE(G)]2 λ 2 +n(n 1) λ λ i i j ≥ | | −  | || | i=1 i=j X Y6   1 n n n(n−1) λi 2 +n(n 1) λi 2(n−1) ≥ | | − | | ! i=1 i=1 X Y n = λi 2 +n(n 1)Rn2 | | − i=1 X 2 = (E +2P)+n(n 1)Rn. | | − MinimumEquitableDominatingRandicEnergyofaGraph 85 Thus, REE(G) (E +2P)+n(n 1)Rn2. ≥ | | − 2 q §4. Minimum Equitable Dominating Randic Energy of Some Standard Graphs Theorem 4.1 The minimum equitable dominating Randic energy of a complete graph K is n REE(K )= 3n 5. n n−1 − Proof Let K be the complete graph with vertex set V = v ,v , ,v . The minimum n 1 2 n { ··· } equitable dominating set =E = v . The minimum equitable dominating Randic matrix is 1 { } 1 1 1 ... 1 1 n 1 n 1 n 1 n 1 − − − −  1 0 1 ... 1 1  n 1 n 1 n 1 n 1 − − − −  1 1 0 ... 1 1  RE(Kn)=n−...1 n−...1 ... ... n−...1 n−...1.      1 1 ... 1 0 1  n 1 n 1 n 1 n 1  − − − −   1 1 ... 1 1 0  n 1 n 1 n 1 n 1   − − − −  The characteristic equation is n 2 λ+ 1 − λ2 2n−3λ+ n−3 =0 n 1 − n 1 n 1 (cid:18) − (cid:19) (cid:18) − − (cid:19) (2n 3)+√4n 3 (2n 3) √4n 3 1 and the spectrum is SpecER(Kn)= −2(n1−1) − −2(n−1−1) − nn−−12 . −   3n 5 Therefore, REE(K )= − . n n 1 2 − Theorem 4.2 The minimum equitable dominating Randic energy of star graph K is 1,n 1 − REE(K )=√5. 1,n 1 − Proof Let K be the star graph with vertex set V = v ,v , ,v . Here v be 1,n 1 0 1 n 1 0 − { ··· − } the center. The minimum equitable dominating set = E = V(G). The minimum equitable 86 P.SivaKotaReddy,K.N.PrakashaandGavirangaiahK dominating Randic matrix is 1 1 1 ... 1 1 √n 1 √n 1 √n 1 √n 1 − − − −  1 1 0 ... 0 0  √n 1 −  1 0 1 ... 0 0  RE(K1,n−1)=√n...−1 ... ... ... ... ... .    1 0 0 ... 1 0  √n 1   1− 0 0 ... 0 1  √n−1    The characteristic equation is λ(λ 1)n 2[λ 2]=0 − − − 2 1 0 spectrum is SpecE(K )= . R 1,n−1  1 n 2 1  −   Therefore, REE(K )=n. 1,n 1 − 2 Theorem 4.3 The minimum equitable dominating Randic energy of Crown graph S0 is n (4n 7)+√4n2 8n+5 REE(S0)= − − . n n 1 − Proof Let S0 be a crown graphof order 2n with vertex set u ,u , ,u ,v ,v , ,v n { 1 2 ··· n 1 2 ··· n} and minimum dominating set = E = u ,v . The minimum equitable dominating Randic 1 1 { } matrix is 1 0 0 ... 0 0 1 ... 1 1 n 1 n 1 n 1 − − −  0 0 0 ... 0 1 0 ... 1 1  n 1 n 1 n 1 − − −  0 0 0 ... 0 1 ... 1 0 1   n 1 n 1 n 1  ... ... ... ... ... −... ... .−.. ... −...       0 0 0 ... 0 1 ... 1 1 0  RE(Sn0)= n−1 n−1 n−1 .  0 1 1 ... 1 1 0 ... 0 0   n 1 n 1 n 1   − − −   1 0 1 ... 1 0 0 ... 0 0  n 1 n 1 n 1   −... ... −... ... −... ... ... ... ... ...       1 1 0 ... 1 0 0 ... 0 0  n 1 n 1 n 1   − − −   1 1 1 ... 0 0 0 ... 0 0  n 1 n 1 n 1   − − −  MinimumEquitableDominatingRandicEnergyofaGraph 87 The characteristic equation is n 2 n 2 λ+ 1 − λ 1 − λ2 1 λ 1 λ2 2n−3λ+ n−3 =0 n 1 − n 1 − n 1 − − n 1 n 1 (cid:18) − (cid:19) (cid:18) − (cid:19) (cid:18) − (cid:19)(cid:18) − − (cid:19) spectrum is SpecE(S0) R n (2n 3)+√4n 3 1+√4n2 8n+5 (2n 3) √4n 3 1 1 1 √4n2 8n+5 = −2(n−1) − 2(n−−1) −2(n−−1) − n−1 n−−1 − 2(n−−1) .  1 1 1 n 2 n 2 1  − −  (4n 7)+√4n2 8n+5  Therefore, REE(S0)= − − . n n 1 2 − Theorem 4.4 The minimum equitable dominating Randic energy of complete bipartite graph K of order 2n with vertex set u ,u , ,u ,v ,v , ,v is n,n 1 2 n 1 2 n { ··· ··· } 2√n 1 REE(K )= − +2. n,n √n Proof LetK bethecompletebipartitegraphoforder2nwithvertexset u ,u , ,u , n,n 1 2 n { ··· v ,v , ,v . The minimum equitable dominating set = E = u ,v with a minimum 1 2 n 1 1 ··· } { } equitable dominating Randic matrix 1 0 0 0 ... 1 1 1 1 n n n n 0 0 0 0 ... 1 1 1 1 n n n n 0 0 0 0 ... 1 1 1 1  n n n n   0 0 0 0 ... 1 1 1 1  n n n n RE(Kn,n)=... ... ... ... ... ... ... ... ....   1 1 1 1 ... 1 0 0 0 n n n n    1 1 1 1 ... 0 0 0 0 n n n n    1 1 1 1 ... 0 0 0 0 n n n n    1 1 1 1 ... 0 0 0 0 n n n n    The characteristic equation is n 1 n 1 λ2n−4(λ2 − )[λ2 2λ+ − ]=0 − n − n Hence, spectrum is SpecE(K )= 1+ n1 √√n−n1 1− n1 0 −√√n−n1 . R n,n  1q 1 1q 2n 4 1  −   2√n 1 Therefore, REE(K )= − +2. n,n √n 2 88 P.SivaKotaReddy,K.N.PrakashaandGavirangaiahK Theorem 4.5 The minimum equitable dominating Randic energy of cocktail party graph K n 2 × is 4n 6 REE(K )= − . n 2 × n 1 − Proof LetK beaCocktailpartygraphoforder2nwithvertexset u ,u , ,u ,v ,v , n 2 1 2 n 1 2 × { ··· ,v . The minimum equitable dominating set = E = u ,v with a minimum equitable n 1 1 ··· } { } dominating Randic matrix 1 1 1 1 ... 0 1 1 1 2n 2 2n 2 2n 2 2n 2 2n 2 2n 2 − − − − − −  1 0 1 1 ... 1 0 1 1  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2 − − − − − −  1 1 0 1 ... 1 1 0 1  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2  − − − − − −   1 1 1 0 ... 1 1 1 0  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2  RE(Kn×2)= ...− ...− ...− ... ... ...− ...− ...− ... .    0 1 1 1 ... 1 1 1 1   2n 2 2n 2 2n 2 2n 2 2n 2 2n 2  − − − − − −   1 0 1 1 ... 1 0 1 1  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2  − − − − − −   1 1 0 1 ... 1 1 0 1  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2  − − − − − −   1 1 1 0 ... 1 1 1 0  2n 2 2n 2 2n 2 2n 2 2n 2 2n 2   − − − − − −  The characteristic equation is λn−1 λ+ 1 n−2(λ 1)[λ2 2n−3λ+ n−3]=0 n 1 − − n 1 n 1 (cid:18) − (cid:19) − − Hence, spectrum is SpecER(Kn×2)= 2n−23(+n1−√14)n−3 11 2n−23(−n1−√14)n−3 n 0 1 nn−−112 . − −   4n 6 Therefore, REE(K )= − . n 2 × n 1 2 − References [1] S.B. 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