Maximizing the sum of the squares of the degrees ... - Semantic Scholar

Discrete Mathematics 285 (2004) 57 – 66

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Maximizing the sum of the squares of the degrees of a graph Kinkar Ch. Das Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, West Bengal, India Received 13 February 2003; received in revised form 26 March 2004; accepted 20 April 2004

Abstract Let G = (V; E) be a simple graph with n vertices, e edges, and vertex degrees d1 ; d2 ; : : : ; dn . Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G and mi be the average of the degrees of the vertices adjacent to vertex vi ∈ V . It is proved that 2e +n−2 max{dj + mj : vj ∈ V } 6 n−1 with equality if and only if G is an Sn graph (K1; n−1 ⊆ Sn ⊆ Kn ) or a complete graph of order n − 1 with one isolated vertex. Using the above result we establish the following upper bound for the sum of the squares of the degrees of a graph G:    n
2e d1 n−2 d2i 6 e + d1 + (d1 − dn ) 1 − n−1 n−1 n−1 i=1 with equality if and only if G is a star graph or a regular  graph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. A comparison is made to another upper bound on ni=1 d2i , due to de Caen (Discrete Math. 185 (1998) 245). We also present several upper bounds for ni=1 d2i and determine the extremal graphs which achieve the bounds. c 2004 Elsevier B.V. All rights reserved.  Keywords: Graph; Degree sequence; Upper bound

1. Introduction All graphs considered here are nonempty, >nite, undirected, and have no loops or multiple edges. Let G = (V; E) be a simple graph with n vertices and e edges. To avoid trivialities we always assume that n ¿ 2. Also assume that the vertices are ordered such that d1 ¿ d2 ¿ · · · ¿ dn , where di is the degree of vi for i = 1; 2; : : : ; n and the average of the degrees of the vertices adjacent to vi is denoted by mi , respectively.  consider the following problem: >nd a function f(n; e; d1 ; dn ) with the property that ni=1 d2i 6 f(n; e; d1 ; dn ), where We n 2 i=1 di is the sum of the squares of the degrees of a graph G having n nodes, e edges, and d1 , d n being the highest and the lowest degree of vertices. We would like the bound to be sharp, i.e., we would like to have ni=1 d2i = f(n; e; d1 ; dn ) for some graphs.  We recall some known upper bounds for ni=1 d2i . 1. SzAekely et al. [9]: 2  n n

 2 di 6 di : i=1

i=1

E-mail addresses: [email protected], [email protected] (K.C. Das). c 2004 Elsevier B.V. All rights reserved. 0012-365X/$ - see front matter
doi:10.1016/j.disc.2004.04.007

(1)

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K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

2. D. de Caen [1]:   n
2e d2i 6 e +n−2 : n−1 i=1

(2)

Caen has pointed out that (1) and (2) are incomparable. He has also mentioned that the bound (2) is perhaps a bit more useful than (1), since it depends only on  n and e rather than the full-degree sequence. The kth moment of the degree sequence of a graph G is k (G) = 1n ni=1 dki . In particular, n2 (G) is the sum of the squares of the degrees of a graph G. FIuredi and KIundgen [2] gave asymptotically sharp bounds for k (G) when G is in a monotone family. The object of this paper is to establish the following bound:    n
d1 2e n−2 d2i 6 e + d1 + (d1 − dn ) 1 − n−1 n−1 n−1 i=1 which is sharp for a star graph or a regulargraph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. We present two other upper bounds for ni=1 d2i in terms of n, e, d1 and dn . Also we determine the extremal graphs which achieve the upper bound. 2. Lemmas and results Lemma 2.1. Let G be a connected graph and Dij = {di + dj : vi vj ∈ E}. Then all Dij ’s are equal if and only if G is a regular graph or a bipartite semiregular graph. Proof. It is well known that G is a regular graph or a bipartite semiregular graph if and only if the line graph of G is regular. Using this we get the required result. Pan [7] has proved the following result in Theorem 2.4. Lemma 2.2 (Pan [7]). Let G be a connected bipartite graph. Then d1 + m1 = d2 + m2 = · · · = dn + mn holds if and only if G is a bipartite regular graph or a bipartite semiregular graph. Lemma 2.3. Let G be a connected graph. For vi ∈ V , the degree of vi and the average of the degrees of the vertices adjacent to vi are denoted by di and mi , respectively. Then d1 + m1 = d2 + m2 = · · · = dn + mn holds if and only if G is a regular graph or a bipartite semiregular graph. Proof. If G is a regular graph or a bipartite semiregular graph then d1 + m1 = d2 + m2 = · · · = dn + mn holds. Conversely, let d1 + m1 = d2 + m2 = · · · = dn + mn holds. Now we have to prove that G is a regular graph or a bipartite semiregular graph. Let us take a vertex vn with the lowest degree, say r. Further let vertex vn be adjacent to the vertices vi1 ; vi2 ; : : : ; vir . If possible let all the vertices vi1 ; vi2 ; : : : ; vir be not of equal degrees and among these vertices let d be the highest degree corresponding to the vertex vi1 . For vertex vn , r + mn ¡ r + d and for vertex vi1 , d + mi1 ¿ d + r. From these two relations, we get r + mn ¡ d + mi1 . But this is not possible since r + mn = d + mi1 . Therefore we can say that vi1 ; vi2 ; : : : ; vir have equal degrees s (say). Let vertex vi1 be adjacent to the vertices vn ; vir+1 ; vir+2 ; : : : ; vir+s−1 . Since r is the lowest degree, then the degrees of the vertices vn ; vir+1 ; vir+2 ; : : : ; vir+s−1 are all greater than or equal to r. If possible let one of them be greater than r. Then for vertex vi1 , s + mi1 ¿s + r. Therefore dn + mn = r + s ¡ s + mi1 . Again this is not possible. Therefore vi1 is adjacent to all r degree vertices. Similarly vi2 ; vi3 ; : : : ; vir are all adjacent to the r degree vertices. Continuing the procedure, we can show that r degree vertices are adjacent to the s degree vertices and s degree vertices are adjacent to the r degree vertices. First we assume that G is not bipartite. Then G has an odd cycle vj vj1 vj2 : : : vjk−1 vjk vjk vjk−1 : : : vj2 vj1 vj . Therefore the degree of the vertex vj is either r or s. Let r be the degree of the vertex vj . Since r degree vertices are adjacent to s degree vertices and s degree vertices are adjacent to r degree vertices, therefore vj1 and vj1 both are s degree vertices. Similarly we can say that vjk and vjk both are r degree vertices if k is even, or s degree vertices if k is odd. But both vik and vik are adjacent vertices. Therefore r and s must be equal. Again let s be the degree of the vertex vj . Similarly we can show that r and s must be equal. Hence the graph G is r-regular graph.

K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

59

Next we assume that G is bipartite. It follows from Lemma 2.2 that G is a bipartite regular graph or a bipartite semiregular graph. Lemma 2.4. Let G be a simple graph with n vertices, e edges and vertex degrees d1 ; d2 ; : : : ; dn . Then n
dj2 6 e max{dj + mj : vj ∈ V }

(3)

j=1

with equality if and only if max{dj + mj : vj ∈ V } = di + mi for each nonisolated vertex vi . Proof. We have n n
n
n n n



dj mj = {dk : vj vk ∈ E} = {dk : vj vk ∈ E} = dk2 : j=1

j=1 k=1

k=1 j=1

k=1

Therefore n n

dj2 = dj (dj + mj ) 2 j=1

j=1

6 max{dj + mj : vj ∈ V }

n


dj

j=1

= 2e max{dj + mj : vj ∈ V }: Equality holds in (3) if and only if n
dj (k − (dj + mj )) = 0; where k = max{dj + mj : vj ∈ V }:

(4)

j=1

Now suppose that equality in (3) holds. Then (4) holds. Now each term of this summation is nonnegative and since sum is equal to zero, therefore for each vi either di = 0 or max{dj + mj : vj ∈ V } = k = di + mi . Conversely, let max{dj + mj : vj ∈ V } = di + mi for each nonisolated vertex vi . Therefore we can see easily that the equality holds in (3). Theorem 2.5. Let G be a connected graph with n vertices, e edges and vertex degrees d1 ; d2 ; : : : ; dn . Then n
dj2 6 e max{dj + mj : vj ∈ V } j=1

with equality if and only if G is a regular graph or a bipartite semiregular graph. Proof. Using above Lemmas 2.3 and 2.4, we get the required result. Lemma 2.6. Let G be a simple graph of order n with degree sequence d1 ; d2 ; : : : ; dn . Then n


d2i mi = 2 {di dj : vi vj ∈ E} = {di mi + dj mj : vi vj ∈ E}: i=1

16i; j6n

16i; j6n

For vi ∈ V , the degree of vi and the average of the degrees of the vertices adjacent to vi are denoted by di and mi , respectively.  Proof. We have di mi = j {dj : vi vj ∈ E}. Therefore n n



d2i mi = {di dj : vi vj ∈ E} = 2 {di dj : vi vj ∈ E} i=1

and

n
i=1

i=1

d2i mi =

n
i=1

j

di (di mi ) =

16i; j6n


16i; j6n

{di mi + dj mj : vi vj ∈ E}:

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K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

Lemma 2.7 (Papendieck and Recht [8]). Let q1 ; q2 ; : : : ; qn be n positive numbers, then: n pi pi pi min 6 i=1 6 max n i i qi q qi i=1 i for any real number p1 ; p2 ; : : : ; pn . Equality holds on either side if and only if all the ratios pi =qi are equal. Theorem 2.8. Let G be a connected graph with n vertices, e edges and vertex degrees d1 ; d2 ; : : : ; dn . Then  n
di (di + mi ) + dj (dj + mj ) d2i 6 e max : vi vj ∈ E di + dj i=1

(5)

with equality if and only if G is a regular graph or a bipartite semiregular graph. Proof. We have
{di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} 16i; j6n




=

16i; j6n




=




{d2i + d2j : vi vj ∈ E} +

{di mi + dj mj : vi vj ∈ E}

16i; j6n

{(di + dj )2 : vi vj ∈ E};

by Lemma 2:6:

(6)

16i; j6n

If a1 ; a2 ; : : : ; an are n positive numbers then we have n  n 2 2 i=1 ai i=1 ai ¿ n n with equality if and only if a1 = a2 = · · · = an . Using this result in (6), we get




{di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} ¿

16i; j6n

2 {di + dj : vi vj ∈ E}

16i; j6n

e

:

(7)

We have the well-known identity [5, Exercise 10.30], n
i=1

d2i =




{di + dj : vi vj ∈ E}:

(8)

16i; j6n

From (7) and (8), we get  n
16i; j6n {di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} 2  di 6 e : 16i; j6n {di + dj : vi vj ∈ E} i=1 Using Lemma 2.7, we get  n
di (di + mi ) + dj (dj + mj ) d2i 6 e max : vi vj ∈ E : di + dj i=1

(9)

Now suppose that equality in (5) holds. Then the equality holds in (7) and (9). In particular we have from (7), that di + dj for any edge vi vj ∈ E are equal. Since G is a connected graph, according to the Lemma 2.1, G is a regular graph or a bipartite semiregular graph. Also these graphs satisfy the equality (9). Conversely, it is easy to verify that equality in (5) holds for regular graph and bipartite semiregular graph. Theorem 2.9. Let G be a connected graph with n vertices, e edges and vertex degrees d1 ; d2 ; : : : ; dn . Then  n
Wi d2i 6 e max : vi ∈ V ; Di i=1

(10)

K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

61

  where Wi = j {di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} and Di = j {di + dj : vi vj ∈ E}. Moreover, the equality holds if and only if G is a regular graph or a bipartite semiregular graph. Proof. We have n

Wi = 2 {di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} i=1

16i; j6n

and n
i=1




Di = 2

{di + dj : vi vj ∈ E}:

16i; j6n

Therefore  n Wi 16i; j6n {di (di + mi ) + dj (dj + mj ) : vi vj ∈ E} i=1  = n i=1 Di 16i; j6n {di + dj : vi vj ∈ E}  ¿

16i; j6n

{di + dj : vi vj ∈ E} e

From (8) and (11), we get n n
Wi d2i 6 e i=1 n i=1 Di i=1 

6 e max

Wi : vi ∈ V Di

;

by (7):

(11)

:

(12)

Now suppose that equality in (10) holds. Then all inequalities in the above argument must be equalities. In particular, we have from (11) that di +dj for any edge vi vj ∈ E are equal. Since G is a connected graph, according to the Lemma 2.1, G is a regular graph or a bipartite semiregular graph. Also these graphs satisfy the equality (12). Conversely, it is easy to verify that equality in (10) holds for regular graph and bipartite semiregular graph. 3. An upper bound on max{di + mi : vi ∈ V } Theorem 3.1. Let G be a graph with n vertices, and e edges. Then 2e max{dj + mj : vj ∈ V } 6 +n−2 n−1

(13)

with equality if and only if G is an Sn graph (K1; n−1 ⊆ Sn ⊆ Kn ) or a complete graph of order n − 1 with one isolated vertex. Proof. Let vi be the vertex of the graph G where di + mi is the maximum. Therefore max{dj + mj : vj ∈ V } = di + mi = di + (T=di ), T = sum of the degrees of the adjacent vertices of the vertex vi . We have 2e = di + T + (n − di − 1)pi , where pi is the average degree of the vertices nonadjacent to vertex vi in the graph G. Now we have to prove that T di + T + (n − di − 1)pi di + 6 + n − 2; di n−1   T i:e:; (n − di − 1) n − 2 + pi − ¿ 0: (14) di If di = n − 1 then (14) holds, otherwise di 6 n − 2. Two cases arise (i) 0 6 pi ¡ 1, (ii) pi ¿ 1. Case (i): 0 6 pi ¡ 1. In this case there is at least one isolated vertex. So, T=di 6 n − 2. Therefore (14) holds as di 6 n − 2. Case (ii): pi ¿ 1. Again (14) holds as T=di 6 n − 1 and di 6 n − 2. This completes the proof of >rst part.

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K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

Now suppose that equality in (13) holds. Then   T (n − di − 1) n − 2 + pi − = 0: di

(15)

From (15), we get either di = n − 1 or (T=di ) − pi = n − 2. Case (i): di = n − 1. In this case graph G is Sn . Case (ii): T=di − pi = n − 2. Since d1 ¿ T=di , we have that d1 ¿ n − 2 + pi :

(16)

Since pi ¿ 0 and d1 6 n − 1, from (16) we get pi ∈ [0; 1]. If pi ∈ (0; 1], then d1 = n − 1 and hence graph G is Sn . Otherwise pi = 0, from (16) we get d1 = n − 1 or n − 2. Subcase (a): pi = 0 and d1 = n − 1. In this subcase graph G is Sn . Subcase (b): pi = 0 and d1 = n − 2. In this subcase let v1 be the vertex having degree n − 2. Therefore vertex v1 is nonadjacent to one vertex. Also pi = 0, therefore either vertex vi is adjacent to all the remaining vertices of the graph G or vi is nonadjacent to some vertices which are isolated vertices. Since d1 = n − 2, vi is nonadjacent to one isolated vertex and di = n − 2. Therefore T = (n − 2)(n − 2) and 2e = (n − 1)(n − 2). Hence the graph G is an Sn graph or a complete graph of order n − 1 with one isolated vertex. Conversely, let G be an Sn graph or a complete graph of order n − 1 with one isolated vertex. When G = Sn , at the central vertex vk , dk + mk = n − 1 +

2e − (n − 1) 2e = + n − 2: n−1 n−1

Therefore max{dj + mj : vj ∈ V } = dk + mk = 2e=(n − 1) + n − 2. When G is a complete graph of order n − 1 with one isolated vertex, max{dj + mj : vj ∈ V } = 2(n − 2) =

2e + n − 2: n−1

Hence the theorem. Lemma 3.2. Let G be a graph with n vertices, and e edges. Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G. Then the inequality   2e di n−2 d i + mi 6 + di + (d1 − dn ) 1 − (17) n−1 n−1 n−1 holds for each nonisolated vertex vi . Moreover, the equality holds if and only if di = n − 1 or vertex vi is adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. Proof. Let vi be any vertex of degree di . Two cases arise (i) di = n − 1, (ii) di ¡ n − 1. Case (i): di = n − 1. Therefore di + mi = n − 1 +

2e − n + 1 2e = + n − 2: n−1 n−1

Hence (17) holds. Case (ii): di ¡ n − 1. Suppose all the vertices which were nonadjacent to the vertex vi are now adjacent to the vertex vi . Suppose this new graph is G1 and max{dj + mj : vj ∈ V } occurs at the vertex vi and let it be di + mi . Since G1 is an Sn graph, di + mi =

2(e + n − di − 1) + n − 2: n−1

(18)

K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

63

Suppose that T is the sum of the degrees of the adjacent vertices of the vertex vi in the graph G. Therefore T d i + m i = di + : di Now, we have T + (n − di − 1)(pi + 1) T − ; (di + mi ) − (di + mi ) = n − di − 1 + n−1 di where pi is the average degree of the vertices nonadjacent to vertex vi in the graph G, 2e 2(n − di − 1) i:e:; di + mi = +n−2+ − (n − di − 1) n−1 n−1 T T + (n − di − 1)(pi + 1) by (18) + ; n−1 di   T (n − di − 1) 2e +n−2− n − 2 + pi − = n−1 n−1 di −

=

   di T n−2 2e : + di − 1 − pi − n−1 n−1 n−1 di

(19)

We have pi − dn ¿ (T=di ) − d1 . Using this result in (19), we get   2e di n−2 + di + 1 − (d1 − dn ): d i + mi 6 n−1 n−1 n−1 Now suppose that equality in (17) holds. Then di = n − 1 or −(pi − (T=di )) = d1 − dn , that is, di = n − 1 or (d1 di − T )=di = ((n − di − 1)dn − k)=(n − di − 1), where k is the sum of the degrees of the nonadjacent vertices of the vertex vi in G. From (d1 di − T )=di = ((n − di − 1)dn − k)=(n − di − 1), we get that (n − di − 1)dn − k ¿ 0; as T 6 d1 di ; n − di − 1 i:e:;

k 6 (n − di − 1)dn ;

i:e:;

k = (n − di − 1)dn :

Therefore all the vertices those are nonadjacent to the vertex vi are of degree dn . Using this fact in (d1 di − T )=di = ((n − di − 1)dn − k)=(n − di − 1), we get T = d1 di . Therefore all the vertices those are adjacent to the vertex vi are of degree d1 . Conversely, let di = n − 1 or vertex vi be adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. Then we can easily see that the equality holds in (17). Corollary 3.3. Let G be a graph with n vertices, and e edges. Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G. Then   2e d1 n−2 di + m i 6 + d1 + (d1 − dn ) 1 − (20) n−1 n−1 n−1 holds for each nonisolated vertex vi . Moreover, the equality holds if and only if di = n − 1 or vertex vi (degree is d1 ) is adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. Proof. Since (17) is true for all nonisolated vertices vi and di 6 d1 , therefore we get (20). Using Lemma 3.2 we conclude that the equality holds if and only if di = n − 1 or vertex vi (degree is d1 ) is adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. Theorem 3.4. Let G be a connected graph with n vertices, and e edges. Then the best upper bound of max{di +mi : vi ∈ V } in terms of n and e is 2e=(n − 1) + n − 2. Proof. Since G is a connected graph, we can construct an Sn graph (K1; n−1 ⊆ Sn ⊆ Kn ) with the same number of vertices and edges as G. We know that for Sn graph, 2e max{di + mi : vi ∈ V } = + n − 2: n−1

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K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

Using Theorem 3.1 we conclude that the best upper bound of max{di +mi : vi ∈ V } in terms of n and e is 2e=(n−1)+n−2. Hence theorem.

4. Main results In this section, we will give three upper bounds for the sum of the squares of the degrees of a graph and determine the extremal graphs which achieve the bounds. Theorem 4.1. Let G be a graph with n vertices and e edges. Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G. Then n


d2i

 6e

i=1

  2e d1 n−2 + d1 + (d1 − dn ) 1 − n−1 n−1 n−1

(21)

with equality if and only if G is a star graph or a regular graph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. Proof. Using (3) and (20), we get the result (21). Equality holds in (21) if and only if equality holds in (3) and (20). Therefore the equality holds in (21) if and only if max{dj + mj : vj ∈ V } = di + mi = 2e=(n − 1) + (n − 2)=(n − 1)d1 + (d1 − dn )(1 − (d1 =(n − 1))), for each nonisolated vertex vi . Using Corollary 3.3 we conclude that the equality holds in (21) if and only if G is a star graph or a regular graph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. Remark. Now we shall see that the value obtained by applying (21) is better than the value obtained by applying (2). From (21) we have n


d2i 6 e

i=1



  2e d1 + n − 2 − {n − 2 − (d1 − dn )} 1 − : n−1 n−1

Since 1 − (d1 =(n − 1)) ¿ 0 and n − 2 − (d1 − dn ) ¿ 0 for connected or disconnected graphs, therefore the upper bound given by (21) is always lower than the bound given by (2). Theorem 4.2. Let G be a graph with n vertices, and e edges. Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G. Then n


d2i 6

i=1

2e(2e + (d1 − dn )(n − 1)) n + d 1 − dn

(22)

with equality if and only if G is a regular graph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. Proof. From (17), we get d i + mi 6

2e di + d1 − dn + (n − 2 − (d1 − dn )) ; n−1 n−1

for any nonisolated vertex vi :

Multiplying both sides by di and taking summation over i, we get 2

n


d2i =

i=1

i:e:;

n
i=1

n
i=1

d2i 6

di (di + mi ) 6

n d2i (2e)2 + 2e(d1 − dn ) + (n − 2 − (d1 − dn )) i=1 ; n−1 n−1

2e(2e + (d1 − dn )(n − 1)) : n + d 1 − dn

K.C. Das / Discrete Mathematics 285 (2004) 57 – 66

65

Now suppose that equality in (22) holds. Then the equality holds in (17) for each nonisolated vertex. Therefore each nonisolated vertex is adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. If isolated vertices exist in the graph G, then dn = 0. So each nonisolated vertices are adjacent to all d1 degree vertices, that is, G is a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. If isolated vertex is not present in the graph G, then each vertex is adjacent to the d1 degree vertices and nonadjacent to the dn degree vertices. Therefore all the vertices have same degree, that is, G is a regular graph. Conversely, let G be a regular graph or a complete graph Kd1 +1 with n − d1 − 1 isolated vertices. Therefore we can see easily that the equality holds in (22). Theorem 4.3. Let G be a graph with n vertices, and e edges. Also, let d1 , dn be, respectively, the highest degree and the lowest degree of G. Then n


d2i 6 2e(d1 + dn ) − nd1 dn

(23)

i=1

with equality if and only if graph G has only two type of degrees d1 and dn . Proof. We have n


d2i =

i=1

n


(di (di − dn ) + di dn )

i=1

6

n


(d1 (di − dn ) + di dn )

i=1

= 2e(d1 + dn ) − nd1 dn : Equality holds if and only if n


di (di − dn ) =

i=1

n


d1 (di − dn );

i=1

i:e:;

n


(d1 − di )(di − dn ) = 0:

(24)

i=1

Now suppose that equality in (23) holds. Then the equality holds in (24). Now each term of this summation is nonnegative and since sum is equal to zero, therefore either di = d1 or di = dn for i = 1; 2; : : : ; n: Therefore graph G has only two type of degrees d1 and dn . Conversely, if G has only two type of degrees d1 and dn , then n


(d1 − di )(di − dn ) = 0:

i=1

Hence the equality holds. 5. Application Now we give an application for upper bound on the sum of squares of degrees in a graph. Let t(G) denote the number of triangles in G. It was >rst observed by Goodman [3] that t(G) + t(G c ), where G c denotes the complement of G, is determined by the degree sequence:  2 1
(n − 1) n(n − 1)(n − 5) t(G) + t(G c ) = di − + : (25) 2 v ∈V 2 24 i

Goodman [4] raised the question of >nding a best possible upper bound of the form t(G) + t(G c ) 6 B(n; e), and he explicitly conjectured an expression for B(n; e). This conjecture was proved recently by Olpp [6]. Moreover, the expression of B(n; e) is rather complicated. We remark that three sharp upper bound on t(G) + t(G c ) follows from (25) and (21), (22), (23).

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Acknowledgements The author is grateful to the reviewers for their valuable comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9]

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