A RELATIONSHIP BETWEEN BOUNDS ON THE SUM OF SQUARES OF DEGREES OF A GRAPH

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1 J. Appl. Math. & Computig Vol. 21(2006), No. 1-2, pp Website: A RELATIONSHIP BETWEEN BOUNDS ON THE SUM OF SQUARES OF DEGREES OF A GRAPH YEON SOO YOON AND JU KYUNG KIM Abstract. Let G =(V,E) be a simple graph with vertices ad e edges. Let d i be the degree of the ith vertex v i V ad m i the average of the degrees of the vertices adjacet to vertex v i V. It is kow by Cae [1] ( ad Das [2] that 4e2 d d2 e max{d j + m j v j V } e 1 + 2). I geeral, the equalities do ot hold i above iequality. It is show that a graph G is regular if ad oly if 4e2 = d d2.i fact, it is show a little bit more strog result that a graph G is regular if ad oly if 4e2 = d d2 = e max{d j + m j v j V }. For a graph G with >2 vertices, it is show that G is a complete graph ( K if ad oly if 4e2 = d d2 = e max{d j + m j v j V } = e 1 + 2). AMS Mathematics Subject Classificatio : 05C99,68R10. Key words ad phrases : Graph, degree square, upper boud. 1. Itroductio A simple graph is a graph with o loops or multiple edges. For ay vertex v V, the degree of v is the umber of edges meetig at v. A graph G is regular of degree r if all the vertices of G have the same degree r. Acomplete graph is a graph i which every two distict vertices are joied by exactly oe edge. The complete graph with vertices is deoted by K. I fact, the complete graph K is regular of degree 1. A bipartite graph G =(V,E) is a graph whose vertex set V ca be split ito sets V 1 ad V 2 i such a way that each edge of the graph jois a vertex i V 1 to a vertex i V 2. It is easy to distiguish the vertices i V 1 from those i V 2 by drawig the former i black ad the latter i white. A complete bipartite graph G =(V,E) is a bipartite graph i which each black vertex i V 1 is joied to each white vertex i V 2 by exactly oe edge, that is, all vertices i V 1 have the same degree s =#V 2 ad all vertices i V 2 have the Received Jue 14, Correspodig author. c 2006 Korea Society for Computatioal & Applied Mathematics ad Korea SIGCAM. 233

2 234 Yeo Soo Yoo ad Ju Kyug Kim same degree r =#V 1. The complete bipartite graph with r black vertices ad s white vertices is deoted by K r,s. A complete bipartite graph of the form K 1,s is called a star graph. Throughout this paper, let G =(V,E) be a simple graph with vertices ad e edges. Let d i be the degree of the ith vertex v i V ad m i the average of the degrees of the vertices adjacet to vertex v i V. We recall some kow bouds for d d 2. The Cauchy-Schwarz iequality yields a lower boud = 1 (d d ) 2 d d 2. It is kow by D. de Cae [1] that d d2 e( 1 + 2). Also, it is kow by K. Das [2] that d d 2 e max{d j + m j v j V } e Thus we have the followig iequality; 4e2 V } e( d d 2 e max{d j + m j v j + 2). I 2004, K. Das [2] obtaied equivalet coditios to be d d 2 = e max{d j + m j v j V } ad e max{d j + m j v j V } = e respectively. The object of this paper is to fid equivalet coditios to be = d d2 ad 4e2 = e( 1 + 2) respectively. It is show that a graph G is regular if ad oly if 4e2 = d d2. I fact, it is show a little bit more strog result that a graph G is regular if ad oly if = d d2 = e max{d j + m j v j V }. For a graph G with >2 vertices, it is show that G is a complete graph K if ad oly if = d d2 = e max{d j + m j v j V } = e Bouds o the sum of squares of degrees of a graph Let G =(V,E) be a simple graph with vertices ad e edges. Let d i be the degree of the ith vertex v i V ad m i the average of the degrees of the vertices adjacet to vertex v i V. The followig lemma is well kow result called by the hadshakig lemma. Lemma 1. (The Hadshakig Lemma) d d =.

3 Bouds o the sum of squares of degrees of a graph 235 From the Cauchy-Schwarz Iequality a i b i a i 2 b i 2, takig ( ) 2 ( a i = d i ad b i = 1, we have that d i d 2 i ). Thus we have, from the above result ad the hadshakig lemma, a lower boud of the sum of squares of degrees of a graph as follows; Propositio 2. 4e2 = 1 (d d ) 2 d d 2. I 1998, D. de Cae [1] obtaied a upper boud of the sum of squares of degrees of a graph as follows; Theorem 3. [1] d d2 e( 1 + 2). I 2004, K. C. Das [2] obtaied a upper boud of the sum of squares of degrees of a graph, which is less the equal to Cae s upper boud as follows; Theorem 4. [2] (1) d d 2 e max{d j + m j v j V }. (2) max{d j + m j v j V } From the above three results, we have the followig remark. Remark 5. 4e2 d d2 e max{d j + m j v j V } e( 1 + 2). The followig example shows that the equalities do ot hold i above iequality. Example 6. Cosider the followig graph; The 4e2 = < ( 2 =44< = 136 e max{d j + m j v j V } < =8( ) = e 1 ) = O the other had, K. C. Das [2] obtaied a equivalet coditio to be his upper boud equal to the sum of squares of degrees of a graph as follows;

4 236 Yeo Soo Yoo ad Ju Kyug Kim Theorem 7. [2] d d2 = e max{d j + m j v j V } if ad oly if G is a regular graph or a complete bipartite graph. Now, uder what coditios do 4e2 the above questio as follows; ad d d 2 equal? We ca aswer Theorem 8. A graph G is regular if ad oly if 4e2 = d d 2. Proof. Suppose that G is a regular of degree r. Sice e = 1 2 r ad d i = r, we kow that d d 2 = r 2. Moreover, we kow that 4e2 = r 2 = r 2. Thus we have d d2 = 4e2. O the other had, suppose that d d2 =. The we kow that 0 = d d2 4e2 = d d 2 1 (d d ) 2 = d d 2 1 (d d 2 +2(d 1 d d 1 d ) +2(d 2 d d 2 d )+ +2(d 1 d)) = 1 (( 1)(d d 2 ) 2((d 1 d d 1 d ) +(d 2 d d 2 d )+ +(d 1 d)) = 1 { ((d 1 d 2 ) 2 + +(d 1 d ) 2 )+((d 2 d 3 ) 2 + +(d 2 d ) 2 )+ +(d 1 d ) 2}. Sice each (d i d j ) 2 0, for each i, j, we kow that d i = d j ad G is a regular graph. I fact, we ca get, by Theorem 7 ad Theorem 8, a little bit more strog result as follows; Corollary 9. A graph G is regular if ad oly if 4e2 = d d2 = e max{d j+ m j v j V } Corollary 10. If G is a complete graph K, the 4e2 = d d 2 = e max{d j + m j v j V }. O the other had, K. C. Das [2] obtaied a equivalet coditio to be his upper boud equal to Cae s upper boud as follows;

5 Bouds o the sum of squares of degrees of a graph 237 Theorem 11. [2] max{d j + m j v j V } = if ad oly if G is a S graph (K 1, 1 S K ) or a complete graph of order 1 with oe isolated vertex. Now,uder what coditios do 4e2 the above questio as follows; ad e( 1 + 2) equal? We ca aswer Theorem 12. Let G be a graph with >2vertices. The G is a complete graph K if ad oly if 4e2 = d d2 = e max{d j + m j v j V } = e( 1 + 2). Proof. First we ote that G is a complete graph if ad oly if e = ( 1) 2. Moreover, we kow that e = ( 1) 2 iff ( 2) = ( 1)( 2) iff 4e 4e = + ( 1)( 2) iff 4e( 1) = ( +( 1)( 2)) iff 4e2 = e Thus we have that G is complete if ad oly if 4e2 = e( 1 + 2). Uder what coditios do d d2 ad e( 1 + 2) equal? We do ot have a complete aswer of this questio. But we have oly a partial aswer of this as follows; Theorem 13. Let G be a complete bipartite graph K r,s, where s ( > 1). The G is a star graph, that is G = K 1, 1 if ad oly if d d 2 = e 1 ). + 2 Proof. Suppose that G = K r,s, where s > 1. Sice there are r vertices of degree s ad s vertices of degree r, d d2 = rs2 + sr 2 = rs(r + s). Moreover, it follows, from the fact = r + s ad e = rs, that 2rs e = rs r + s 1 + r + s 2 ( r 2 + s 2 ) +4rs 3r 3s +2 = rs. r + s 1 Thus we kow that d d2 = e( 1 + 2) ( r 2 + s 2 ) +4rs 3r 3s +2 iff rs(r + s) =rs r + s 1

6 238 Yeo Soo Yoo ad Ju Kyug Kim iff rs r s +1=0 iff r(s 1) = s 1 iff r =1. So, we kow that G is a star graph, that is, G = K 1, 1 if ad oly if d d 2 = e( 1 + 2). Refereces 1. D. de. Cae, A upper boud o the sum of squares of degrees i a graph, Discrete Math., 185(1998), K. C. Das, Maximizig the sum of the squares of the degrees of a graph, Discrete Math., 285(2004), J. Harris ad J. Hirst ad M. Mossighoff, Combiatorics ad graph Theory, Spriger- Verlag New York, Ic Y. L. Pa, Sharp upper bouds for the Laplacia graph eigevalues, Liear Algebra Appl., 355(2002), L. A. Székely ad L. H. Clark ad R. C. Etriger, A iequality for degree sequeces, Discrete Math., 103(1992), R. J. Wilso ad J. J. Watkis, Graphs, A itroductory Approach, Joh Wiley & Sos, Ic., Yeo Soo Yoo received his BS from Haam Uiversity ad Ph. D at Korea Uiversity uder the directio of Moo Ha Woo. Sice 1985 he has worked for Haam Uiversity. Departmet of Mathematics, Haam Uiversity, Daejeo , Korea. yoo@haam.ac.kr Ju Kyug Kim received her BS from Mokwo Uiversity ad MS i Educatio at Haam Uiversity uder the directio of Yeo Soo Yoo.