ErdEis, P., S. Fajtlowicz and W. Staton, Degree sequences in triangle-free graphs, ... c dk-. ~dk~2n2+21i+i; k=2n+l k=l. (tO12-365X/91/$03.50 @ 1991- Elsevier ...
Discrete Mathematics 92 (1991) 85-88 North-Holland
85
Degree sequences in triangle-fr graphs Paul Erdh, Siemion Fajtlowicz and William Staton Mathematical
Institute, Hungarian
Academy
of Sciences, Reriitanoda u. 13-15 j:-1053
Budapest, Hungary
Received 4 January 1989 Revised 19 April 1989 Dedicated to Professor R.G. Stanton on the occasion of his 68tb biihday.
Abstract
ErdEis, P., S. Fajtlowicz and W. Staton, Degree sequences in triangle-free graphs, Discrete Mathematics 92 (1991) 85-88.
The computer program Graffiti has conjectured that the chromatic number of a triangle-free graph does not exceed the maximum number of occurrences of an element of the degree sequence [I]. We prove that indeed if no degree occurs more than twice, the graph is bipartite. On the other hand, we prove that every triangle-free graph is an induced subgraph of a triangle-free graph in which no degree occurs more than three times. In the two lemmas which follow, we assume that {di: 1 s i S 4n + 3) is a nondecreasing sequence of positive integers and that no three terms are equal. We assume also that all subscripts are appropriately bounded. Lemma 1. (a) dk+z (b) A+t - A ar.
dk 3
1;
Proof. (a) is obvious since no three terms are equal. (b) is an easy induction beginning with (a). El Lemma 2. (4
5
d,-zd 2n
4n+l (W
c k=2n+l
k 3 2n2; k=l
k=2n+l
dk-
~dk~2n2+21i+i; k=l
(tO12-365X/91/$03.50 @ 1991-
Elsevier Science Publishers B.V. All rights reserved
P. Erd6.s et al.
86 dn+2
(4
c
2?I+t
d*-
c b-i-1
4n+3
(4
d+2n2+2n;
k=l
k=2n+2
dk-
2
c
d,a2?12+4n+2.
A=1
k=2n+2
Proof.
6) (b)
2
dk-~dk=$(d~+a-d&n=2n2.
k=2n+l
4n+1 c, k=2n+l
k=l
k=l
k=l
dk - 2 dk=db+l+
2
dk-
-$ dk k=l
k=2n+l
k=l
~d4n+l+2n2 32n+1+2n2. 4n+2 2 it=&+2
Zn+1
2?l+1 dk -
2 k=l
dk=&a+2-&et
+
2 k=l
(&+k
-dk)
2n+1 ~&m+2-&ta+2+
c k=l
(4
Z=n+(2n+l)n=2n2+2n. 4n+3
2n+1
2 k=2n+2
dk-
2 k=l
dk=&+3+
k=2n+2
32n+2+(2n2+2n)=2n2+4n+2.
0
Note that for equality to hold in any of the statements of Lemma 2, the sequence must be such that dk+* - dk = 1 for all appropriately bounded k. This condition implies that there are no ‘gaps’ in the sequence and that every term appears exactly twice with the possible exceptions of the first and last terms, which may appear either once or twice. Sequences satisfying dk+* - dk = 1 for all k will be called compact. Note that compact sequences are determined by their first two terms. For a graph G, we define f =f(G) to be the maximum number of occurrences of any term of the degree sequence of G. As usual, S = S(G) is the smallest element of the degree sequence of G. In order to avoid trivialities, we assume our graphs have no isolated vertices. Theorem 1.Let G be a triangle-free graph with f(G) = 2. Then G is bipartite, b(G) = 1, and the degree sequence of G is compact. Proof. Suppose G has 4n vertices, vl, . . . , v4,_ with degree sequence d I,-.., ddn in nondecreasing order. Let H be the subgraph induced by V&f!: - - . , v+ The number of edges in H is at least ;(
k
=!$
21i+l
dk-
2 d,),
k=l
Degree sequencesin triangle-freegraphs
87
with equality oniy in case the vertices not in H send all their edges into N. By Lemma 2(a), this is at least n* edges, with equality only in case the degree sequence of G is compact. But H is triangle-free with 2n vertices, so Turan’s theorem forces the conclusion that H is Km,, and that the degree sequence of G is indeed compact. Consider the vertices uI, . . . , v2” not in H. If (X, Y) is the bipartition of H, then, since G is triangle-free, no ui (1 c i s 2n) sends edges into both X and Y. Hence each vi may be piaced either with X or with Y to form a bipartition of G. Since the vertices Ui (1 z i s 2~) have degrees not exceeding n, it follows that 6 = 1. This concludes the proof in case G has 4n vertices. In case G has 4n + 2 or 4n + 3 vertices, the proof is nearly identical, using Lemma 2(c) and (d). In case G has 4n + 1 vertices, we let H be the subgraph induced by u~,,+~ and note that by Lemma 2(b), H has at least n* + n + 4 edges. %+I,..., Turan’s theorem forbids this. It follows that no triangle-free graph with 4n + 1 vertices may have f(G) = 2. El We have seen that triangle-free graphs G with f(G) = 2 must have 6 = 1. It is natural to ask whether an upper bound on f(G) imposes an upper bound on S. Lemma 3. For every positive integer n there is a bipartite graph G with 8n vertices, 6=n+l,
andf(G)=3.
Proof. Let G have vertices Xi and yj for 1 < i d 4n. Join Xi to yj with an edge if oue of these conditions holds: (i) i+j>4n+l; (ii) lsi
