Complete subgraphs with large degree sums

Complete Subgraphs with Large Degree Sums

Ralph J. Faudree MEMPHIS STATE UNIVERSITY

MEMPHIS, TENNESSEE

ABSTRACT

Let t (n ,k ) denote the Tu rh number-the maximum number of edges in a graph on n vertices that does not contain a complete graph K,,,. It is shown that if G is a graph on n vertices with n > k2(k - 1)/4 and rn 2 t (n ,k ) edges, then G contains a complete subgraph Kk such that the sum of the degrees of the vertices is a t least 2krn/n. This result is sharp in an asymptotic sense in that the sum of the degrees of the vertices of Kk is not in general larger, and if the number of edges in G is at most t (n ,k ) - ~n (for an appropriate E ) , then the conclusion is not in general true. 0 1992 John Wiley & Sons, Inc.

1. INTRODUCTION

Let t ( n , k ) denote the Turin number, which is the maximum number of edges in a graph on n vertices that does not contain a complete subgraph K k + , , and let T(n , k ) denote the corresponding unique extremal graph. Thus, T ( n , k) is the complete k-partite graph with the number of vertices in the parts distributed as equally as possible, and t ( n , k ) is the number of edges in this complete k-partite graph. Each of the vertices in T(n , k) has degree [ ( ( k - l)n)/kl or L((k - l)n)/kJ. A survey of results involving Turin numbers introduced in [9] can be found in an article by Simonovits [8].

By Turin’s Theorem [9] any graph with n vertices and m > t(n, k - 1) edges will contain a complete subgraph K k . How large the sum of the degrees in such a complete subgraph can be has been investigated in several papers. With that problem in mind, we give the following definition.

Definition 1. For any positive integer k and any graph G, Ak(G) will denote the maximum of the sum of the degrees of the k vertices of a complete

Journal of Graph Theory, Vol. 16, No. 4, 327-334 (1992) 0 1992 John Wiley & Sons, Inc. CCC 0364-9024/92/040327-08$04.00

328 JOURNAL OF GRAPH THEORY

subgraph Kk of G. Also, Ak(n, m ) will denote the minimum of Ak(G) taken over all graphs G with n vertices and m edges.

Determination of Ak(G) in the triangle case ( k = 3 ) has received the most attention. Edwards [6] showed that if G is a graph with n vertices and m 5: n2/3 edges, then A3(G) 2 6m/n. In [3] and [4 ] , A3(G) was considered for m > n2/4 . All of these results were improved in a paper by Fan [7] where he showed that if G is a graph with n vertices and m > n2/4 edges, then A3(G) > 21m/4n, and if m 1 n 2 / 3 and A is the maximum degree in G, then A3(G) 2 3 6 - 2n + 4m/A 2 6m/n.

If a graph G with n vertices and m edges is regular (of degree 2m/n), then the only possible values for Ak(G) are 0 (if G contains no Kk) or 2km/n oth- erwise. Therefore the maximum possible value for Ak(n, m ) is 2kmln. This motivates the following definition:

Definition 2. If G is a graph with n vertices and m edges, then a balanced Kk is a complete subgraph K k such that the sum of the degrees of its vertices is at least 2kmln.

We will prove the following result:

Theorem 1. m 2 (k - l )n2 /2k edges contains a balanced Kk. Thus,

For any positive integer k , any graph G with n vertices and

Ak(G) 2 2 k m / n ,

and

Ak(n, m ) = r2km/nl.

The proof of Theorem 1 will show that the balanced Kk can be obtained in “greedy” way- that is, by successively selecting vertices of maximum degree in the appropriate neighborhoods. The structure of Turin type graphs suggest that greedy algorithms might be successful in determining subgraphs. This was observed in [ l ] , [2 ] , and [ S ] , where it was shown (in a slightly stronger form) that any graph G of order n with more than t(n, k ) edges will have a vertex u whose neighborhood N induces a subgraph of order n’ with more than t(n: k - 1) edges.

Slightly stronger results follow from the proof of Theorem 1, and one such corollary of this type is the following result.

Corollary 1. n > k2(k - 1)/4 and m 2 t ( n , k ) edges contains a balanced K k . Thus,

For any positive integer k , any graph G on n vertices with

&(G) 2 2km/n ,

and

Ak(n,m) = r2km/nl.

SUBGRAPHS WITH LARGE DEGREE SUMS 329

To exhibit the asymptotic sharpness of Theorem 1, examples will be described to verify the following result.

Theorem 2. For any positive integer k 2 3 there is an E > 0 such that for any m < ( k - l)n2/2k - En, there is a graph G with n vertices and m edges that does not contain a balanced K k .

Of course, the inequality m < (k - l)n2/2k - En in Theorem 2 can be replaced by the inequality m < t(n, k ) - En and the conclusion would still be valid, since ( k - l)n2/(2k) - t (n , k ) 5 k / 2 .

2. MAIN RESULTS

We start with some trivial observations concerning Ak(G). Let G be a graph with n vertices and m edges, and let { u I , u 2 , . . . , un} be the vertices that give the degree sequence (d1 ,d2 , . . . ,dn) with d l 2 d2 2 ... 2 d,. Thus, Z:=Idi = 2m. The “average” degree in G is 2m/n, and so

A , ( G ) = d , 2 2m/n.

The “average edge degree” also gives a lower bound for A2(G). This is a result of the following inequalities that are a consequence of a convexity argument.

1 1 “ A2(G) L - 2 2 (d, + d,) = - E d ; 2

2m 1 = ~ ,,,+G m 1=1 m

This balanced edge cannot be selected in a greedy way. In general, the selection of a largest degree vertex uI of a graph, and then the selection of a vertex u2 in the neighborhood of u1 of largest degree will not always give an edge such that the sum of the degrees of the endvertices is at least 4m/n when m < n2/4. Such a graph will be described in the next section. How- ever, if the number of edges in a graph G with n vertices is m 2 n2/4, then the balanced edge can be selected in a greedy way by selecting a vertex of maximum degree and then selecting a vertex of maximum degree that is in the neighborhood of the first vertex.

Lemma 1. If G is a graph with n vertices and m L t(n,2) = C(n2 - 1)/41 edges, then A2(G) 5 4m/n. Also, a balanced edge can be selected in a greedy way.

ProoJ: Select a vertex u1 of maximal degree d, in G , and let N I be the neighborhood of ul. Then, select a vertex u2 in NI that has maximal degree, say d2 , relative to G . If d , + dz 1 4m/n, then we are done, so suppose not.


By the selection of u1 and u2, any vertex in N1 has degree at most d l , and any vertex in N 1 has degree at most d 2 . Therefore, adding the degrees of the vertices in G gives the following inequality.

d l d 2 + (n - d l ) d l = INlld2 + (n - INl l )d l 2 2m. (1)

Since, by assumption, d z < (4m/n) - d l , we have

This i s equivalent to the inequality

However, Eq. ( 3 ) gives a contradiction, since d l 2 r2m/nl 2 4 2 , so the first factor in ( 3 ) is nonnegative and the second factor is nonpositive. This completes the proof of Lemma 1. I

The same greedy approach used in the proof of Lemma 1 can be used to prove the following, which is the main result:

Theorem 1. Let k be a positive integer. If G i s a graph with n vertices and m 2 ( k - 1)n2/2k edges, then A,(G) 2 2km/n. Also, a balanced Kk can be selected in a greedy way.

Prooj The proof will be by induction on k . It is trivial for k = 1, and Lemma 1 implies the result fork = 2. Therefore we can assume that k 2 3 . Consider a fixed k 2 3 and assume the theorem is true for all smaller values. We will proceed as in the proof of Lemma 1 to show it true for k .

Select a vertex u1 of maximal degree d l in G, and let N 1 be the neighborhood of u I . Then, select a vertex u2 in N I that has maximal degree, say d 2 , relative to G, and let N2 be the neighborhood of u2. Continue in this way to select vertices u3, u4, . . . , Vk. For each i the vertex u, will be a vertex of maximal degree d , in N1 n NZ n . . . n N,-, . Denote the neighborhood of u, in G by N,. By the induction assumption N I n N2 n ... n N,-l # 0, and so each u, can be chosen.

If d l + d 2 + ... + dk 2 2km/n, then we are done, so suppose this is not true. We will show that this leads to a contradiction. If we denote the ver- t i c e s o f G b y N , , , t h e n f o r ( l S i % k - l ) , e a c h v e r t e x i n N o n N 1 n . . . n N,-, n N, has degree at most d , by the maximality of the degree d , . Also, each vertex in N 1 n N z n ... n Nk-l has degree at most d k . Therefore, the number of edges in G is at most


For each (1 I i I k - l ) , we have INo n N , n ... n N,-, rl m,l 5 (n - d,), and IN1 fl N2 f l ... rl Nk-,l 2 n - X!Zj(n - d,) = X!L,'d, - (k - 2)n. There- fore the sum of the degrees of G gives the following inequality for the number of edges in G.

k - 1 c (n - d,)d, + c d , - (k - 2)n)dk 3 2 m . , = I (:r : ( 4 )

If we let d = d , + d 2 + . . . + dk- , , and use the supposition that dk < n

2km d, we get the following strict inequality from (4) . _ -

By a convexity argument we have XfL,'dz 2 (k - 1)(&)*. Thus ( 5 ) implies the next inequality

(F - d ) (d - (k - 2)n) + dn - (k - 1 ) (k / (k - 1) )2 - 2m > 0 ,

which is equivalent to

By the induction assumption we know that the first factor in (6) is nonnegative and the second factor is nonpositive since d 2 2(k - l )m/n and m 2

(k - l )n2/2k. Thus, (6) gives a contradiction, which completes the proof of Theorem 1. I

Corollary 1. Let k be a positive integer. If G is a graph with n > k2(k - 1) /4 vertices and m > t(n, k ) edges, then Ak(G) > 2km/n. Also, a balanced Kk can be selected in a greedy way.

ProoJ: The proof of Corollary 1 is identical to the proof of Theorem 1 up to Eq. ( 6 ) , and the fact that the first factor in (6) is nonnegative. To verify that the second factor of (6) is nonpositive is equivalent to showing that d > (k - 1)2n/k. If n = kl + r with 0 5 r < k , then t ( n , k ) =

(k - l )n2 / (2k) - r(k - r)/(2k). Note that

d 2 1 2(k - 1 ) m] g p(kn- ' ) ("'.2 - w) 2k 2k

(k - 1)'n r(k - r ) (k - 1 ) ( k - 1)'n k '

- = [ k kn


The last inequality is a consequence of n > k2(k - 1)/4. Hence, (6) also gives a contradiction in this case as well, which completes the proof of Corollary 1. I

Note that if m > (k - 1)2n/2k - n/(2k(k - l)), then the same contradiction from Eq. (6) can be obtained. Thus m > ( k - 1)2n/2k - n/(2k(k - 1)) edges ensures a balanced K k .

3. EXAMPLES

For each k 2 3 and each m < (k - l)n2/2k - ~n (for an appropriate positive E ) , we want to construct a graph G with n vertices and m edges that does not contain a balanced Kk. For rn 5 t(n, k - 1) this can easily be done. Any subgraph of the Turin graph T(n , k - 1) has this property since it contains no K k . Therefore, we only consider the cases where m >

Let m = yn2, where ( k - 2)/(2(k - 1)) 5 y < ( k - 1)/2k. We will now describe a k-partite graph with n vertices and m = yn2 edges that does not have a balanced K k . For appropriate positive numbers a and b with 0 < 2b I a < 1, consider any bipartite graph Ba,h that has an vertices and is regular of degree bn. Now, consider the complete ( k - 1)-partite graph with an vertices in one part and (1 - a)n/(k - 2) vertices in each of the remaining k - 2 parts. Let G(a, b , k ) be the graph obtained from this complete ( k - 1)-partite graph by placing the bipartite graph Bo,b in the part with an vertices. Thus G,(a, b , k ) is a k-partite graph on n vertices with ay1 vertices having degree (1 - a + b)n and the remaining vertices having degree ( (k - 3 + a ) / ( k - 2))n. Thus, G , ( a , b , k ) has (a(1 - a + b ) + (1 - a ) ( ( k - 3 + a) / (k - 2) ) )n2 /2 edges. If we make the following selec- tions for a and b we will get a graph, which we will denote by G,(y), that has n vertices and yn2 edges:

t ( n , k - 1).

a = Jy~, 4(k - 2)y - 2k + 6

and

4(k - 2)y - 2k + 6 2 = (E) J--- - k - 2 '

Note that if y = (k - 1)/2k, then graph G,(y) is just the complete k- partite graph with n/k vertices in each part, since a = 2/k and b = l / k . For y < (k - 1)/2k, each complete subgraph Kk has precisely one vertex in


each of the k-parts, and so

2(1 - u + b) + ( k - 2 )

= ( k - 1 - a + 2b)n,

which reduces to the following for our choice of a and b:

'(',(y)) = ( k ( k k - - 2 3 ) I 2k k + - 2 4 d v ) n .

Showing that A(G,(y)) < 2kyn is equivalent to showing that

(7) k ( k - 3 ) I k + 2 J 4 ( k - 2 ) y - 2k + 6

< 2kY 9 k - 2 2 k - 4 k

when y 2 ( k - 2) / (2(k - 1 ) ) . Inequality (7) can be verified with straight- forward algebraic manipulations; in fact, y ? (k - 3) /2k is sufficient for this purpose. Hence G,(y) does not contain any balanced Kk.

Of course an, bn, and yn2 must be integers for the graphs G,(y) to have any meaning. This can only be ensured for yn2 < ( k - 1)n2/ (2k) - En for appropriate E . However, subgraphs of the examples G,(y) along with the subgraphs of the Turin graph T(n , k - 1 ) gives the following theorem:

Theorem 2. For any positive integer k 2 3 there is an E > 0 such that for any m < ( k - 1)n2/2k - En there is a graph G with n vertices and m edges that does not contain a balanced K k .

It was mentioned earlier that the greedy algorithm does not always pro- duce a balanced edge in a graph G with n vertices and m < n 2 / 4 edges. Consider for example for n even and r < (n - 4 ) / 2 , the graph G(n, r ) that is the vertex disjoint union of a Kni2 and a vertex u of degree n/2 - 1 and its neighborhood that induces a subgraph that is regular of degree r. The graph G(n, r ) has n vertices and m = (n - 2 ) (n + 2r + 4 ) / 8 edges. If the greedy algorithm first selects the vertex u, which is a vertex of maximal degree n/2 - 1 , then the resulting edge selected by the algorithm will have its degree sum equal to n / 2 + r. For r < ( n - 4 ) / 2 we have n / 2 + r < 4rn/n. There are other similar graphs constructed from vertex disjoint complete graphs (more than one) and a star with a neighborhood that induces a regular subgraphs that illustrates that the greedy algorithm does not work for rn 5 n 2 / 8 as well.

4. QUESTIONS

In many results in Turin type extremal theory a graph of chromatic number k can replace the complete graph K k . Thus, it is natural to ask if in Theo-


rem 1 the balanced complete graph Kk can be replaced by an arbitrary balanced graph of chromatic number k when n is sufficiently large. If so, what other results in Turan type extremal theory have an analogue for “balanced subgr a ph s”?

ACKNOWLEDGMENT

This research was partially supported by ONR research grant N000014-88- K-0070 and NAS Exchange grant.

The author would like to thank the Computer and Automation Institute and the Mathematical Institute of the Hungarian Academy of Sciences for their hospitality during the time this research was completed. Also, the author would like to thank the referee for pointing out some mistakes in the initial version of this article.

References

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Complete subgraphs with large degree sums