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'~ DISCRETE

MATHEMATICS

ELSEVIER Discrete Mathematics 150 (1996) 411 414

Note

The Erd6s-S6s conjecture for graphs of girth 5

Stephan Brandt"'*, Edward Dobson b

a FB Mathematik, Freie Universitht Berlin, Graduiertenkolleg 'Alg. Diskr. Mathematik', Arnimallee 2-6.

14195 Berlin, Germany

b Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA

Received 12 October 1993; revised 17 October 1994

Abstract

We prove that every graph of girth at least 5 with minimum degree 6 >~ k/2 contains every

tree with k edges, whose maximum degree does not exceed the maximum degree of the graph.

An immediate consequence is that the famous Erd6s-S6s Conjecture, saying that every graph of

order n with more than n(k - 1)/2 edges contains every tree with k edges, is true for graphs of

girth at least 5.

AMS Subject Classifications (1991): Primary 05C35, Secondary 05C05

Keywords: Erd6s-S6s conjecture; Tree; Girth

One of the most challenging problems in extremal graph theory is the Erd6s-S6s

Conjecture for trees from 1963 (see [4]):

Conjecture 1. Every graph of order n with more than n(k - 1)/2 edges contains every

tree with k edges as a subgraph.

The related Erd6s-S6s Conjecture for forests (see [4]) was verified by the first

author in [2]. Since a complete solution of Conjecture 1 seems to be out of reach,

partial solutions might be interesting. We prove that Conjecture 1 is true for graphs

without cycles of length 3 and 4.

Theorem 1. Every graph of order n and girth at least 5 with more than n(k - 1)/2 edges

contains every tree with k edges as a subgraph.

* Corresponding author. E-mail: brandt(a math.fu-berlin.de.

0012-365X/96/$15.00 © 1996--Elsevier Science B.V. All rights reserved

SSD1 0012-365X[95)00207-3

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S. Brandt, E. Dobson/Discrete Mathematics 150 (1996) 411-414

We derive this as a direct consequence of the following sufficient degree condition

for a graph of girth at least 5 to contain every tree with k edges. Let 6 and A denote the

minimum and maximum degree, respectively.

Theorem 2. Let G be a 9raph with 9irth at least 5 and T be a tree with k edoes. If

6(G) >>. k/2 and A(G) >t A(T) then G contains T as a subgraph.

For a subset S ~ V(G) let N(S) be the set of vertices with at least one neighbor in S.

The central tool in the proof of Theorem 2 is the following simple lemma.

Lemma 1. Let G be a 9raph of order n without isolated vertices. If S is a subset of V(G)

where every pair of vertices in S has distance at least 3 then I S I <~ n/2.

Proof. Since S is an independent set where no two vertices have a common neighbor

we get with 6(G) >>. 1,

n/> IS w N(S)I = ISI + IN(S)I >i 21SI,

hence[Sly< n/2. []

Before performing the proof we need to fix some terminology. For undefined basic

concepts we refer the reader to introductory graph theoretical literature, e.g. [1].

For a graph G and a vertex v ~ V(G) let G - v denote the subgraph induced by

V(G)\{v} and similarly G- S is the subgraph induced by V(G)\S for a subset

S ~_ V(G). An embeddin9 of a graph H in G is an injection a: V(H)~ V(G) where

vw E E(H) implies a(v)tr(w)E E(G). A vertex of degree 1 in a tree is a leaf and

a penultimate vertex is a leaf in the subtree of T which is obtained by deleting all leaves

of T(and the incident edges). Call a sequence (Ti), 1 ~< i ~< p, of subtrees a resolution of

T if Tl is a star, Tp = T, and T/_ 1 is obtained from T~ by deleting all leaf neighbors of

a penultimate vertex of minimum degree in T~. We will prove that we can extend an

embedding of Ti- ~ in G to an embedding of T~ in G in a greedy fashion, which is the

induction step in the proof of Theorem 2.

Proof of Theorem 2. Consider a resolution (Ti) of T. Clearly we can embed the star

T~ in G by mapping its center on a vertex of maximum degree in G. For 2 ~< i ~< p

assume we have an embedding tr of Ti_ 1 in G and we want to extend it to an

embedding of T~. Let v be the penultimate vertex of minimum degree in Ti whose leaf

neighbors w~, w2 ..... wr are not in T~_ 1. Let x be another penultimate vertex of T~

which must exist since T~ is not a star. Note that by the minimality requirement on the

degree of v the vertex x has at least r leaf neighbors. Let G' be the subgraph of

G induced by the vertices of a(Ti_ ~ - v) and split G' into two graphs G~, G~ where

G~ is induced by a(x) and the images of the leaf neighbors of x and G~ = G' - V(G'I).

Observe that G~ is a star by the girth requirement and G~ is connected since it

contains a spanning tree.

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S. Brandt, E. Dobson/Discrete Mathematics 150 (1996) 411-414

413

Now we estimate the number dG,(tr(v)) of neighbors of a(v) in G'. Since G has girth

at least 5, every pair of neighbors of a(v) in G' has distance at least 3 in G'. So a(v) has

at most one neighbor in G~, and at most [G~ I/2 neighbors in G~ by Lemma 1 unless

I G~I = 1. But in the latter case a(v) is adjacent to the single vertex in G~ and therefore

to no vertex in G~. Hence we get

dG,(,r(v)) <<. L IGI I/2/+ 1 ~< L(k +1 - 2(r + 1))/2 J + 1 = [k/2-] - r,

so a(v) has at least r neighbors in G- V(G'), on which we map Wl, w2 ..... w,. So

G contains T~, and, by induction, also Tp = T. []

An earlier, considerably longer proof of Theorem 2 by the second author can be

found in [3].

Proof of Theorem 1. Take an induced subgraph H of G of smallest order which

satisfies e(H)>lHl(k-1)/2.

Clearly, A(H)>~k and since e(H-v)<~([Hl-l)

(k - 1)/2 for every vertex v we have 6(H) >1 k/2. So H satisfies the requirements of

Theorem 2, hence H, and therefore G, contains every tree with k edges. []

With a little more effort it can be derived that the only graphs of girth at least 5 with

more than L n(k -1)/2.] edges which do not contain every tree with k edges have

maximum degree k - 1 and they only miss the star with k edges (see [31).

Final remarks. Even if we replace the condition A(G) >~ A(T) of Theorem 2 by the

stronger requirement fi(G) ~> A (T), the conclusion is best possible for some values ofk.

Define the tree Tk+l with k + 1 edges by taking two stars K~.Fk/2 q and K1.Lk/2j and

adding an edge between two leaves. If, for even k, a (k/2)-regular graph with girth

5 and diameter 2 exists then this graph cannot contain Tk + 1, since the images of the

two centers of the stars of Tk + 1 are joined by a path of length 3 in G, hence they must

have a common neighbor outside the path. It is well-known (see e.g. [1, p. 161]), that

such graphs exist for k = 4, 6,14 -- the 5-cycle, the Petersen graph and the Hoffman-

Singleton graph, respectively -- and possibly for k = 114, but for no other values of k.

Nevertheless, it might be possible to extend Theorem 2 with the requirement

6(G) >~ A(T) to graphs of larger girth.

Conjecture 2 (Dobson [3]). Let G be a graph with girth g/> 2t + 1 and T be a tree

with k edges. If 6(G) t> k/t and 6(G) >~ A(T) then G contains T as a subgraph.

Note that this is a well-known fact for t =1 and for t =2 it follows from

Theorem 2.

Note added in proof. Very recently, Sacl6 and Wo~niak generalized Theorem 1

to graphs without 4-cycles and answered the case t =3 of Conjecture 2 in the

affirmative.

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S. Brandt, E. Dobson/Discrete Mathematics 150 (1996) 4ll-414

Acknowledgements

This research was performed while the first author was at the University of Twente,

The Netherlands, and the second author was at the University of Cambridge,

England.

References

[1] B. Bollobas, Graph Theory (Springer, New York 1979).

I-2] S. Brandt, Subtrees and subforests of graphs, J. Combin. Theory Ser. B 61 (1994) 63--70.

[3] E. Dobson, Some problems in extremal and algebraic graph theory, Ph.D. Dissertation, Louisiana

State University, Baton Rouge, 1995.

[4] P. Erd6s, Extremal problems in graph theory, in: M. Fiedler, ed., Theory of Graphs and its

Applications (Academic Press, New York, 1965) 29-36.