Degree sequence and independence in K(4)-free graphs

We investigate whether K,-free graphs with few repetitions in the degree sequence may have independence number o(n). We settle the cases r = 3 and r >/5, and give partial results for the very interesting case r=4 . In an earl ier art icle I-4] it is shown that t r iangle-free graphs in which no term of the degree sequence occurs more than twice must be bipar t i te , but that there exist t r iangle-free graphs with a rb i t ra r i ly large ch roma t i c number in which no degree occurs more than three times. It is easy to see that the large chromat ic number in such graphs is not due to small independence number , since the ne ighbors of a vertex are independent , and the restr ic t ion on repeated degrees forces there to be vertices of large degree. If one excludes K 4 ra ther than K3, the s i tua t ion is cons iderab ly more myster ious, F o r a g raph G we define f ( G ) to be the m a x i m u m number of occurrences of an e lement of the degree sequence of G. Tha t is, i f f ( G ) = k , then some set of k vertices have the same degree but no set of k + 1 vertices have the same degree. Note that f ( G ) >~ 2 if G has at least two vertices. T h r o u g h o u t the paper we will assume that our graphs have no isola ted vertices so that 6 ~> 1. The number of vertices of G will be deno ted by n or v(G). The number of edges of G will be deno ted by e(G). We shall use /3 to denote the independence number of a graph. We address the fol lowing quest ion in this paper . * Corresponding author. Z This research was partially supported by NSA grant MDA-89-H-2036. 2 This research was partially supported by NSA grant MSPPS-054-91. 0012-365X/95/$09.50 © 1995-Elsevier Science B.V. All rights reserved SSDI 0 0 1 2 3 6 5 X ( 9 3 ) E 0 2 2 6 T 286 P. Erd6s et al. / Discrete Mathematics 141 (1995) 285-290 Question 1. Suppose G is a graph with no Kr and with f(G)<~ k. Is it possible that ~=o(n)? The answer to Quest ion 1 is no if r = 3 or r = 4 and k ~< 3. The answer to Quest ion 1 is yes if r ~> 5. These answers are provided by Theorems 1, 2, 6, and 7. For K4-free graphs w i t h f ( G ) ~> 4 we do not know whether fl = o (n) is possible. This is an interesting open question. Theorem 1. If G contains no K 3 and f(G) <<, k, then fl>~n/k, so fl v~ o(n). Proof. There exists a vertex of degree at least n/k. The neighbors of this vertex form an independent set. [] This settles the case r = 3 of Quest ion 1 with a negative answer. Theorem 1 also raises the following interesting question. Question 2. Suppose G contains n o g 3 a n d f ( G ) ~< k. Is the lower bound fl >~ n/k best possible? We will show next that the lower bound on fl given in Theorem 1 is sharp for k = 2 and asymptot ica l ly sharp for k = 4 before cont inuing our considerat ion of Question 1. Invest igat ing Quest ion 2 is an interesting open p rob lem for k other than 2 or 4. For each positive integer n, let H , be the graph on 4n vertices with vertex classes A={ai}7=l, B={bl}7=l, C={ci}7=1, and D={di}7=l and with edge set defined as follows. Edges {a, bj: i<~j}, {cid/ i>~j}, {a,di: l <~i<~n}, {b,ci: l <~i<<,n}, {bidi+l: ivan}, and {ciai+l: ivan} (see Fig. 1). The subgraph of H . induced by A • B is a bipart i te half-graph, as is the subgraph induced by C w D. If G is a bipart i te half g raph on 2n vertices, then it is easy to see that f ( G ) = 2 and fl=n. Thus the lower bound given in Theorem 1 can be exact for k = 2 . We now consider the case when k = 4. The degrees of the vertices of H . are as follows: d(al)=d(dl)=n+3-i if i~: 1, d(al)=d(dl)=n-~ 1, d(bl)=d(cl)=i+2 if i 4: n, and d(b.)=d(c.)=n+ 1. Hence every degree between 3 and n is repeated 4 times, while degree n + 1 is repeated 8 times, so f (H. ) = 8. The 4 vertices of H . with subscript k are said to be level k of H . and denoted by Lk for each k f rom 1 to n. We note that the vertices in levels 1 through k induce Hk, which will be useful in the induct ion a rgument which follows. We claim that f l (H.)=n+l. This can be checked for n = l and n = 2 . The set A w {c. } is an independent set of n + 1 vertices in H. , so we concentra te on showing f l (H.) ~< n + 1. Suppose this is true for n ~< r and suppose S is an independent set of r + 3 vertices in Hr+ 1. If IS n L,+ 11 ~< 1, then, since the first r levels of H,+ 1 induce H,, we have by induction that ISI ~< (r+ 1)+ 1 < r + 3 ; a contradict ion. Thus S contains 2 elements in L,+ 1. Each level is a 4 cycle, so there are only two possibilities. We assume without loss of generali ty that S contains a,+ 1 and c,+ 1. N o w c,+ 1 is adjacent P. ErdlJs et a l / Discrete Mathematics 141 (1995) 285-290 287