Energy of a Hypercube and its Complement

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1 Iteratioal Joural of Algebra, Vol. 6, 01, o. 16, Eergy of a Hypercube ad its Complemet Xiaoge Che School of Iformatio Sciece ad Techology, Zhajiag Normal Uiversity Zhajiag Guagdog, P.R. Chia oamc@163.com Wawe Xie school of Mathematics ad Computatio Sciece, Shezhe Uiversity Shezhe , P.R. Chia xieww@16.com Abstract Let Q deote the complemet of -dimesioal hypercube Q, ad let EG ad LEG deote, respectively, the ordiary eergy ad Laplacia eergy of a graph G. We obtai LEQ =EQ = ad LE Q =E Q = +1 +, where 1. Mathematics Subject Classificatio: 05C50 Keywords: Graph,Complemet,Hypercube,Eigevalues,eergy of graph, Laplacia eergy of graph 1 Itroductio Let G be a graph assumed simple throughout with s vertices vertex set deoted by V G ad t edgesedge set deoted by eg, ad let AG =a ij be the adjacecy matrix for G. The characteristic polyomial of the adjacecy matrix AG, i.e. P G, λ =P AG,λ=detλI AG is called the characteristic polyomial of the graph G. The eigevalues of the matrix AG are called the eigevalues of graph G. Suppose the distict eigevalues of graph G are deoted by λ 1,λ,,λ r with correspodig multiplicities m 1,m,,m r, where m i is a oegative iteger i =1,,,r. The eergy of a graph G, deoted by EG, is defied as r EG = m i λ i. i=1

2 800 Xiaoge Che ad Wawe Xie The eergy of G was first defied by Gutma i This cocept arose i chemistry where certai umerical quatities, such as the heat of formatio of a hydrocarbo, are related to total π-electro eergy that ca be calculated as the eergy of a appropriate molecular graph[1]. Gutma ad Zhou [] defied the Laplacia eergy of a graph G usig its Laplacia matrix DG AG, where AG is the adjacecy matrix ad DG is the diagoal matrix of degrees: r LEG = m i μ i t i=1 s where μ 1,μ,,μ r are the Laplacia eigevalues of DG AG. t is the umber of edges of G, ad t/s is the average degree of a vertex of G. The -dimesioal hypercube Q is the simple graph whose vertices are the the -tuples with etries i {0,1} ad whose edges are the pairs of -tuples that differ i exactly oe positio, the umber of its vertices ad edges are ad 1 respectively. The complemet Ḡ of a simple graph G is the simple graph with vertex set V G defied by uv eḡ if ad oly if uv / eg. Let Q deote the complemet of -dimesioal hypercube Q, ad let BQ ad B Q deote, respectively, the adjacecy matrices of Q ad Q. I [3],we have B Q =J I BQ, BQ 1 = B Q 1 = ,BQ +1 =,B Q +1 = BQ I I BQ B Q J I J I B Q where 1,I is the idetity matrix, J is the matrix of all 1 s. There are a umber of results o the eergy of graphs [1,,4,5,6,7]. I recet years, it is foud that hypercubes are becomig a fudametal tool i computer sciece, codig ad cryptography, geetic algorithms ad discrete eural etwor etc[8]. I this paper, we have solved the problem for EQ,E Q,LEQ ad LE Q. The ordiary eergy of hypercubes Let P Q,λ=detλI BQ be the characteristic polyomial of Q. It is easy to verify that P Q 1,λ=detλI BQ 1 = λ + 1λ 1, ad P Q +1,λ=PQ,λ+1PQ,λ 1, where 1. We have the followig. Lemma.1 [3] Let Q be -dimesioal hypercube.the Q has +1 distict eigevalues. They are give by q = +, ad the eigevalue q has multiplicity, =0, 1,,,, where 1, is biomial coefficiet. Oe ca easily prove the followig.,,

3 Eergy of a hypercube ad its complemet 801 Lemma. Let be a oegative iteger. The 1 =0 = ; = 1 ; 3 = 1 ; 4 +1 = ; 5 j j=0 = Theorem.3 Let EQ be the ordiary eergy of -dimesioal hypercube Q. The EQ =, where 1, a is the ceilig of umber a. Proof By Lemma.1 ad Lemma., we have EQ = =0 q = =0 + = =0 + = +1 = =0 =0 + = +1 = +1 = =0 =0 + =0 =0 =0 =0 = =0 + =0 =0 = =0 + 1 = =0, where a is the floor of umber a. Case 1. If is eve, the =0 = 1 + 1, =0 = = 1. Hece EQ = =0 = Case. If is odd, the =0 = 1, =0 = Hece EQ = 1. =0 = +1. = = 1 = +1 Combiig above all cases we complete the proof. 3 The ordiary eergy of the complemet of hypercubes Let P Q,λ=detλI B Q be the characteristic polyomial of Q. Clearly, P Q 1,λ=detλI B Q 1 = λ. For P Q,λ,, we have the followig.

4 80 Xiaoge Che ad Wawe Xie Lemma 3.1 Let P Q,λ=detλI B Q be the characteristic polyomial of Q, the P Q,λ= =0 P BQ 1,λ+ P 1 1J,λ+ 1 where, J is matrix of all 1 s. Proof It is easy to chec that If A ad C are square matrices of the same order,ad AC = CA, the λi A C det = detλi A CdetλI A + C, C λi A where I is the idetity matrix. Hece P Q,λ=detλI B Q λi 0 B Q 1 J I = det 0 λi J I B Q 1 λi B Q 1 J + I = det J + I λi B Q 1 = detλi B Q 1 +J IdetλI B Q 1 J + I = detλi + BQ 1 detλ +1I J + B Q 1. I a similar maer we arrive at detλ + I 1 1J + B Q = detλ + I + BQ 1 detλ + +1I +1 1J + B Q 1, =1,,,. If =, the detλ + +1I +1 1J + B Q 1 = detλ + +1I +1 1J + B Q 1 = detλ + 1I 1 1J + B Q 1 = detλ + 1I 1 1J = P 1 1J,λ+ 1. Thus we arrive at the coclusio that P Q,λ= =0 P BQ 1,λ+ P 1 1J,λ+ 1, where. Let E Q is the ordiary eergy of Q. Obviously, E Q 1 = 0, ad E Q = 4, For >, we have Theorem 3. Let Q be the complemet of -dimesioal hypercube Q. The E Q = +1 +, where. Proof Let Q j be j-dimesioal hypercube, ad q,j be the eigevalue of BQ j,where j =1,,, 1, =0, 1,,,j. By Lemma.1, q,j = j + is the eigevalue of BQ j too, ad the eigevalue q,j of BQ j has multiplicity j, =0, 1,,,j. The eigevalues of 1 1J are 0 ad. By Lemma. ad Lemma3.1, The eigevalues of Q are 0 1, 1, j+ 1 j, j =1,,, 1, =0, 1,,,j.

5 Eergy of a hypercube ad its complemet 803 i.e. +1, 1, +1+, j =1,,, 1, =0, 1,,,j. Hece E Q = 1 j j=1 =0 +1+ j = 1 = j j=1 + = 1 = = 1 = = 1 = = = = = = = +1 = = +1 = Case 1. If is eve, By Lemma.1,3, we have +1 =0 = 1 + 1, +1 = =. Hece E Q +1 = +1 = = = +1 + Case. If is odd, By Lemma.1,4, we have +1 =0 = , +1 =0 = Hece E Q = = Combiig above all cases we complete the proof. Clearly, if = 1, the the theorem holds too. 4 The Laplacia eergy of hypercubes ad its complemet The simple graph G is -regular if its each vertex degree is. Let G be a -regular, ad let s ad t deote, respectively, the umber of its vertices ad edges. It is to show that if λ 1,λ,,λ s are the eigevalue of -regular graph G, the λ 1, λ,, λ s are the Laplacia eigevalues of DG AG. Hece LEG = s j=1 λ j t = s s j=1 λ j = s j=1 λ j = EG. We have

6 804 Xiaoge Che ad Wawe Xie Lemma 4.1 Let G be a -regular graph. The LEG = EG. Theorem 4. Let be a iteger, 1. The 1 LEQ =EQ = ; LE Q =E Q = +1 + ; 3 EQ +E Q =LEQ +LE Q = Proof Obviously, Q ad Q are -regular ad 1-regular respectively. From Theorem.3, Theorem 3. ad Lemma 4.1, it is evidet to see that the theorem holds. Remar 4.3 A graph is called itegral if its spectrum cosists etirely of itegers. By Lemma.1 ad Lemma 3.1, we have that Q ad Q are itegral. Remar 4.4 Suppose G is a graph of order. G is called hypereergetic if the eergy EG ofg satisfies EG > 1. It is easy to prove that EQ = > 1, where 7, ad E Q = +1 + > 1, where 5. Clearly, Q ad Q are hypereergetic. ACKNOWLEDGEMENTS. This wor is supported by the Natioal Sciece Foudatio of Guagdog No: Refereces [1] I. Gutma, The eergy of a graph: old ad ew results, i Algebraic Combiatorics ad Applicatios, A. Bette, A. Koher, R. Laue, ad A.Wasserma, eds., Spriger, Berli, 001, [] I. Gutma ad B. Zhou, Laplacia eergy of a graph, Liear Algebra Appl., , [3] Ji Xu ad Ruibi Qu, The Spectra of Hypercubes, Joural of Egieerig Mathematics, , 1 5. [4] Balarisha, The eergy of a graph, Liear Algebra Appl [5] I. Shparlisi, O the eergy of some circulat graphs, Liear Algebra Appl., , [6] J. Koole, V. Moulto, Maximal eergy graphs, Adv. Appl. Math C 5. [7] W. Ya, L. Ye, O the miimal eergy of trees with a give diameter, Appl. Math. Lett

7 Eergy of a hypercube ad its complemet 805 [8] BethayM Y Cha Fracis,Chi Y L ad Poo C, Optimal simulatio of full biary trees o faultyhypercubes. IEEE Trasactios o Parallels ad Distributed Systems, Received: April, 01