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Four-Cycles in GraphsWithout a GivenEven Cycle
Daniela Kuhn1 and Deryk Osthus2
1FREIE UNIVERSITAT BERLIN
INSTITUT FUR MATHEMATIK
ARNIMALLEE 2–6
D-14195 BERLIN, GERMANY
E-mail: [email protected]
2INSTITUT FUR INFORMATIK
HUMBOLDT-UNIVERSITAT ZU BERLIN
UNTER DEN LINDEN 6
D-10099 BERLIN, GERMANY
E-mail: [email protected]
Received July 29, 2002; Revised May 17, 2004
Published online in Wiley InterScience(www.interscience.wiley.com).
DOI 10.1002/jgt.20048
Abstract: We prove that every bipartite C2‘-free graph G contains a C4-free subgraph H with e(H)� e(G)=(‘� 1). The factor 1=(‘� 1) is bestpossible. This implies that ex(n;C2‘)� 2(‘� 1)ex(n; fC4;C2‘g), whichsettles a special case of a conjecture of Erdos and Simonovits. � 2004 Wiley
Periodicals, Inc. J Graph Theory 48: 147–156, 2005
Keywords: cycles; extremal problems
1. INTRODUCTION
Given a finite family F of graphs and an integer n, the Turan number exðn;FÞ of
F is the largest integer m such that there exists a graph on n vertices with m edges
� 2004 Wiley Periodicals, Inc.
147
which contains no member of F as a subgraph. Clearly exðn;FÞ � exðn;FÞ for
every F 2 F. Erdos and Simonovits [4] (see also Section 3.4 in [3]) asked whether
there always exists an F 2 F for which up to constants the converse also holds:
Conjecture A. For every finite family F of graphs, there exists an F 2 F such
that exðn;FÞ ¼ Oðexðn;FÞÞ.For the case when F consists of even cycles, this would mean that (up to
constants) the Turan number of F is given by that of the longest cycle in F .
Verstraete (personal communication) conjectured something stronger:
Conjecture B. For all integers k < ‘, there exists a positive c ¼ cð‘Þ such that
every C2‘-free graph G has a C2k-free subgraph H with eðHÞ � eðGÞ=c.
This conjecture was motivated by a result of Gyori [6] who showed that every
bipartite C6-free graph G has a C4-free subgraph which contains at least half of
the edges of G (the case where equality holds was characterized by Furedi, Naor,
and Verstraete [5]). Our main result generalizes that of Gyori.
Theorem 1. Let ‘ � 3 be an integer and G a C2‘-free bipartite graph. Then G
contains a C4-free subgraph H with eðHÞ � eðGÞ=ð‘� 1Þ.The complete bipartite graphs K‘�1;m show that we cannot replace 1=ð‘� 1Þ by
anything larger (see Proposition 5). Moreover, as every graph G has a bipartite
subgraph containing at least half of the edges of G, Theorem 1 carries over to
arbitrary graphs with an extra factor 1=2. It would be interesting to know whether
this factor is really necessary.
Theorem 1 immediately confirms Conjecture A in the following special case:
Corollary 2. Let ‘ � 3 be an integer. Then exðn; fC4;C2‘gÞ � exðn;C2‘Þ2ð‘�1Þ .
Obviously Corollary 2 remains true if we replace the C4 by any graph containing
a C4 as subgraph.
Conjecture B would follow if every C2‘-free graph G which can be obtained by
pasting together C2k’s has bounded average degree. (A graph G is pasted together
from C2k’s if G can be obtained from a C2k by successively adding new C2k’s
which have at least one edge in common with the previous ones.) Indeed, we may
assume that every edge of our given C2‘-free graph G lies in a C2k. But every C2k
is contained in a maximal subgraph D � G pasted together from C2k’s and
distinct such subgraphs D are edge-disjoint. If all these graphs D have bounded
average degree, then the union of a spanning tree of each of them is a C2k-free
subgraph of G as desired in Conjecture B.
Problem C. Given integers k < ‘, does there always exist a number d ¼ dð‘Þsuch that every C2‘-free graph which is pasted together from C2k’s has average
degree at most d?
Note that to settle Conjecture B, it would be sufficient to obtain an affirmative
answer to the question in Problem C for bipartite graphs G. The next result deals
with the simplest case of Problem C for bipartite graphs.
148 JOURNAL OF GRAPH THEORY
Theorem 3. Let ‘ � 3 be an integer and suppose that G is a bipartite C2‘-free
graph which is obtained by pasting together C4’s. Then the average degree of G is
at most 16‘.
Clearly, Conjecture B would imply that every C2‘-free graph can be made into a
graph of girth larger than 2‘ by deleting at most a constant fraction of its edges.
The following simple result shows that at least we can achieve any desired girth
in this way by forbidding a longer even cycle of suitable length.
Theorem 4. For every even integer g � 4, there are infinitely many integers ‘such that every C2‘-free graph G contains a subgraph H whose girth is greater
than g and such that eðHÞ � eðGÞ=2ð4‘Þðg�2Þ=2.
A related conjecture of Thomassen [8] states that for all integers g and k there
exists an integer d such that every graph G of average degree at least d contains a
subgraph of average degree at least k and girth greater than g. Unlike what is
expected in the case when C2‘ 6� G for some ‘ � g=2, d clearly cannot depend
linearly on k here (when g is fixed and even). For regular graphs, the truth of
Thomassen’s conjecture was observed by Alon (see [7] for the argument) and the
case g � 4 was proved in [7].
This paper is organized as follows. In Section 2 we prove Theorem 1 together
with tighter bounds for graphs with few edges compared to ‘ (Lemma 6) and
many edges (Corollary 7). Theorem 3 is then proved in Section 3 and Theorem 4
in the last section.
2. ELIMINATING 4-CYCLES IN GRAPHS WITHOUT A GIVENEVEN CYCLE
Let us first introduce some notation. We denote the number of edges of a graph G
by eðGÞ and its number of vertices by jGj. We write dðGÞ :¼ 2eðGÞ=jGj for the
average degree of G. The girth of G is the length of its shortest cycle. A k-cycle is
a cycle of length k, which we abbreviate by Ck. The successor of a vertex x on a
directed cycle ~CC is the neighbor y of x on C for which the edge xy is directed from
x to y. The predecessor of x is the neighbor y of x on C for which xy is directed
from y to x. Given distinct vertices x and y on ~CC, we write x~CCy for the directed
subpath of ~CC joining x to y. The set of neighbors of a vertex x 2 G is denoted by
NGðxÞ. Given graphs F and G, we say that G is F-free if it does not contain F as a
subgraph. We often denote a bipartite graph with vertex classes A and B by ðA;BÞ.The following proposition implies that in Theorem 1 we cannot replace the
factor 1=ð‘� 1Þ by anything larger.
Proposition 5. Given integers ‘ � 3 and m, there exists a C2‘-free bipartite
graph G with at least m edges for which the maximum number of edges in a C4-
free subgraph is exactly eðGÞ=ð‘� 1Þ þ ‘�12
� �.
FOUR-CYCLES IN GRAPHS 149
Proof. Consider the complete bipartite graph G ¼ ðA;BÞ with jAj ¼ ‘� 1
and jBj ¼ m and let H be any C4-free subgraph of G. Let X be a maximal set of
edges of H whose endvertices in B are distinct. (So jXj � jBj.) Given an edge
e ¼ ab 2 H � X with a 2 A and b 2 B, let Ae be the (unordered) pair of vertices
which consists of a and the endvertex in A of the unique edge in X incident
with b. As H is C4-free, the pairs Ae are distinct for different edges e 2 H � X.
Thus
eðHÞ � jXj þ ‘� 1
2
� �� jBj þ ‘� 1
2
� �¼ eðGÞ
‘� 1þ ‘� 1
2
� �:
To see that this bound can be attained, suppose that m � ‘�12
� �and consider the
C4-free subgraph H of G which is obtained as follows. For each pair a 6¼ a0 of
vertices in A, choose a vertex b 2 B and join b to both a and a0 but to none of the
other vertices in A. Join every remaining vertex in B to exactly one (arbitrary)
vertex in A. &
Given a graph G and a cycle C of length at least 4 in G, we say that C is
augmented if there is an edge xy on C such that the two neighbors of x; y in
C � xy are adjacent in G. Every such edge xy is called a special edge of C. So a
special edge lies on a 4-cycle in G. Moreover, every 4-cycle C of a graph G is
augmented and each edge on C is special.
The strategy of the proof of Theorem 1 is to successively mark edges of our
given C2‘-free graph G and to ensure that no marked edge lies in a C4 by deleting
a bounded number of edges (at most ‘� 2) for each marked one. This strategy
will also be used in the proof of Theorem 4, but the argument there will be much
simpler.
Proof of Theorem 1. We will need the following claim.
Suppose that G is a C2‘-free bipartite graph and D is an even augmented
cycle in G whose length 2k is maximal under the condition that k � ‘.Let xy be a special edge of D. Then there exists a set X of at most ‘� 2
edges of G � xy such that, firstly, every edge in X lies on a 4-cycle of
G and, secondly, every 4-cycle of G containing the edge xy has an edge
in X.
ð*Þ
Before we prove ð*Þ, let us show how it implies the theorem. Given our graph
G, choose an augmented even cycle D0 in G as in ð*Þ. (As a 4-cycle is augmented,
we may assume that such a D0 exists.) Let x0y0 be a special edge of D0 and apply
ð*Þ to obtain a set X0 � EðG � x0y0Þ. Mark x0y0 and delete all edges in X0 from G
to obtain a new graph G1. If G1 is not C4-free, choose a maximal augmented even
cycle D1 as above and a special edge x1y1 of D1. Apply ð*Þ to obtain a set
X1 � EðG1 � x1y1Þ. As x0y0 does not lie in a 4-cycle of G1 but x1y1 as well as
150 JOURNAL OF GRAPH THEORY
every edge in X1 does, the edges x0y0 and x1y1 are distinct and x0y0 =2 X1. Mark
x1y1 and delete all the edges in X1 to obtain a new graph G2. We continue in this
fashion until we obtain a C4-free graph H. As in the process, no edge was marked
more than once and no marked edges were deleted, it follows that eðHÞ �eðGÞ=ð‘� 1Þ, as desired.
To prove ð*Þ, first note that k < ‘ since G is C2‘-free. Let C be the set of all 4-
cycles in G which contain xy. Then each C 2 C has a vertex in D � fx; yg(otherwise ðD [ CÞ � xy would be an augmented cycle of length 2k þ 2 � 2‘,contradicting the choice of D). Denote by Ax the set of all those neighbors a of x
in VðD � yÞ for which xa lies in a 4-cycle (which may contain y). Let Bx be the
set of all those neighbors b of y in VðD � xÞ for which there exists a vertex
outside D which is joined to both x and b. Then Ax and Bx are disjoint, since G is
bipartite and all edges joining x to Ax or y to Bx lie on a 4-cycle. Since every
C 2 C contains an edge joining x to Ax or an edge joining y to Bx, we are done
if jAxj þ jBxj � ‘� 2, as in that case we can set X :¼ fxa : a 2 Axg[fyb : b 2 Bxg. Thus, if we define sets Ay � NGðyÞ and By � NGðxÞ similarly,
we may assume that both jAxj þ jBxj � ‘� 1 and jAyj þ jByj � ‘� 1.
Let Vevenx be the set of all vertices of D � x whose distance from x on D is even
and let Voddx be the set of all vertices of D � y with odd distance from x. Thus
Ax;By � Voddx and Bx;Ay � Veven
x since G is bipartite. Fix the orientation of D in
which x is the successor of y. As jVevenx j ¼ jVodd
x j ¼ k � 1, but by our assumption
jAxj þ jBxj � ‘� 1 � k, there are vertices ax 2 Ax and bx 2 Bx such that bx is the
successor of ax on x~DDy. Indeed, if there are no such vertices then the successors of
vertices in Ax cannot lie in Bx and thus jBxj � jVevenx j � jAxj ¼ k � 1 � jAxj, a
contradiction. Similarly, there are vertices ay 2 Ay and by 2 By such that ay is the
successor of by on x~DDy. Furthermore, if k < ‘� 1, then we can additionally
require that ay 6¼ bx.
By definition of Bx 3 bx, there is a vertex v outside D which is joined to both
x and bx. Similarly, there is a vertex w outside D which is joined to both y and
by. Then v and w are distinct since G is bipartite. We will now distinguish two
cases.
Case 1. ay ¼ bx:
By construction, in this case we must have k ¼ ‘� 1. But then the graph D0
obtained from D by deleting both xy and axbx ¼ byay and adding the two paths
xvbx and ywby is a cycle of length 2k þ 2 ¼ 2‘ (Fig. 1), a contradiction.
Case 2. ay 6¼ bx.
Let x0 be the successor of x on ~DD and y0 the predecessor of y. Thus x0y0 2 G since
xy was a special edge of D. Denote by D0 the cycle obtained from D by removing
all edges of the path x0xyy0 as well as the edges axbx and ayby and by adding the
edge x0y0 as well as the paths axxvbx and ayywby (Fig. 1). Then D0 is an
augmented cycle of length 2k þ 2 � 2‘, contradicting the choice of D. (Indeed,
xv is a special edge of D0.) &
FOUR-CYCLES IN GRAPHS 151
The following simple probabilistic argument improves the bound in Theorem 1
in the case when eðGÞ < ð3ð‘� 1Þ=4Þ3.
Lemma 6. Every bipartite graph G has a C4-free subgraph H with eðHÞ �34
eðGÞ2=3.
Proof. Set m :¼ eðGÞ. Consider the random subgraph G0 of G which is
obtained by including each edge of G with probability p :¼ m�1=3 independently
of all other edges of G. Denote the number of edges of G0 by X and the number of
its 4-cycles by Y. Since G is bipartite, every 4-cycle in G is determined by two of
its independent edges (actually, we only need that G is K4-free for this). As there
are two possibilities for the choice of these edges in each 4-cycle, the number of
4-cycles in G is at most m2=4. Thus
EðX � YÞ � pm � p4m2=4 ¼ 3m2=3=4:
So there exists an outcome G0 for which X � Y � 3m2=3=4. But this means that
the (C4-free) graph H obtained from G0 by deleting one edge on every C4 of G0
has at least 3m2=3=4 edges, as desired. &
The complete bipartite graph Kr; r2ð Þ shows that the bound in Lemma 6 is best
possible up to the value of the constant. Indeed, as we saw in the proof of
Proposition 5, the maximum number of edges of a C4-free subgraph of Kr; r
2ð Þ is
precisely 2 r2
� �. On the other hand, it is well known that the complete graph Kr has a
C4-free subgraph with c1r3=2 ¼ c2eðKrÞ3=4edges (see e.g., [1, Ch. VI, Theorem
2.8]). Thus at first sight, one might think that perhaps every graph G should have
a C4-free subgraph with at least c3eðGÞ3=4edges, but the above remark shows that
this is not the case.
FIGURE 1. Illustrating Case 1 and Case 2 in the proof of Theorem 1.
152 JOURNAL OF GRAPH THEORY
Combining the deletion process of the proof of Theorem 1 with the probabi-
listic argument above, we can come closer to the upper bound in Proposition 5:
we obtain the correct order of magnitude for the additive term.
Corollary 7. Let ‘ � 3 be an integer and let G be a C2‘-free bipartite graph
with eðGÞ � ð3ð‘� 1Þ=8Þ3. Then G contains a C4-free subgraph H with
eðHÞ � eðGÞ‘� 1
þ 33ð‘� 1Þ2
83� 1:
Proof. Let x0 :¼ ð3ð‘� 1Þ=8Þ3. Run the deletion process described after ð*Þ
in the proof of Theorem 1 until it returns a graph Gi whose number of unmarked
edges lies in the interval ½x0; x0 þ ‘� 1�. This can be done since in each step of
the process precisely one edge will be marked and at most ‘� 2 edges will be
deleted. (The edge ei is not counted as a marked edge of Gi.) Let x be the number
of unmarked edges of Gi and let G0 be the graph consisting of these edges. Apply
Lemma 6 to obtain a C4-free subgraph H0 of G0 with eðH0Þ � 3x2=3=4. As no
marked edge of Gi lies in a C4 of Gi, the graph H consisting of H0 together with
the set M of all marked edges of Gi is C4-free. Moreover,
eðHÞ � jMj þ eðH0Þ � eðGÞ � x
‘� 1þ 3x2=3
4
� eðGÞ‘� 1
� x0 þ ‘� 1
‘� 1þ 3x
2=30
4¼ eðGÞ
‘� 1þ 33ð‘� 1Þ2
83� 1;
as required. &
We now outline how the proof of Corollary 7 can be modified to find a C4-free
subgraph H with eðHÞ � eðGÞ=ð‘� 1Þ þ ‘� 2 in every C2‘-free bipartite graph
G with eðGÞ > ð‘� 1Þð‘� 2Þ. This improves the bound in Corollary 7 when ‘ is
small; and for ‘ ¼ 3 it matches the bound in Proposition 5 and is already due to
Gyori [6]. This time we stop the deletion process if the number x of unmarked
edges of Gi lies in the interval ½ð‘� 1Þð‘� 2Þ þ 1; ð‘� 1Þ2�. We then choose a
spanning forest F of the graph consisting of the unmarked edges of Gi. As is
easily seen (e.g., by induction on the number of edges), such a forest must have at
least ð‘� 1Þ þ ð‘� 2Þ edges. So the C4-free subgraph of G consisting of all the
marked edges of Gi together with the edges in the chosen spanning forest F has at
least
eðGÞ � x
‘� 1þ ð‘� 1Þ þ ð‘� 2Þ � eðGÞ
‘� 1þ ‘� 2
edges.
FOUR-CYCLES IN GRAPHS 153
3. GRAPHS PASTED TOGETHER FROM 4-CYCLES
Before proving Theorem 3, let us show that it is best possible in the sense that we
cannot replace the condition that the graph G is pasted together from 4-cycles by
the weaker assumption that every edge of G lies in many 4-cycles.
Proposition 8. For all integers ‘ � 3 and k � 1, there exists a C2‘-free bipartite
graph G with minimum degree at least k such that each edge lies in at least
ð‘� 2Þðk � 1Þ many 4-cycles.
Proof. Let G0 be a bipartite graph of minimum degree at least k and girth
greater than 2‘. Denote the vertex classes of G0 by A and B. Let G be the bipartite
graph obtained from G0 by replacing every vertex b 2 B by a set Xb of ‘� 1 new
vertices, each of which is joined to precisely the vertices in NG0 ðbÞ. Then the
minimum degree of G is at least k and each edge of G lies in at least
ð‘� 2Þðk � 1Þ many 4-cycles. Moreover, it is easily checked that for every cycle
C of length at most 2‘ in G, there exists a vertex b 2 B such that C lies within the
complete bipartite subgraph of G whose vertex classes are Xb and NG0 ðbÞ. Thus G
is C2‘-free. &
For the proof of Theorem 3, we need the following result which was shown by
Bondy and Simonovits in their proof of [2, Lemma 2].
Lemma 9. Let ‘ � 2 be an integer. Then every C2‘-free bipartite graph of radius
at most ‘ has average degree at most 16‘.
Proof of Theorem 3. By Lemma 9 it suffices to show that the radius of G is
at most ‘� 1. Suppose that G is obtained by successively pasting together the 4-
cycles C1;C2; . . . ;Cm in that order. Fix a vertex x 2 C1. We will show that every
other vertex y 2 G has distance at most ‘� 1 from x. Since x 2 C1, there exists a
sequence C1 ¼: D1;D2; . . . ;Dk of distinct 4-cycles in G such that y 2 Dk and
such that Di and Diþ1 share at least one edge for all 1 � i < k. Let ei be an edge
of Di that lies on Diþ1 and let ek be any edge of Dk incident with y. We now
inductively define even cycles D01; . . . ;D0
k such that x; ei 2 D0i and jD0
ij �jD0
i�1j þ 2. Set D01 :¼ D1 and suppose that we have already defined D0
1; . . . ;D0i for
some i � 1. Using that G is bipartite, it is easily seen that D0i [ Diþ1 contains a
cycle D0iþ1 with the required properties. Note that, since G is C2‘-free and
jD0ij � jD0
i�1j þ 2, we must have jD0ij � 2‘� 2 for all i � k. But D0
k contains both
x; y and thus the distance between x and y is at most ‘� 1, as desired. &
4. LONGER EVEN CYCLES
Theorem 4 follows immediately from repeated applications of the following
lemma. The basic argument in the proof of this lemma was already used to prove
Theorem 1. It relies on the fact that if ‘ satisfies the condition in the statement,
154 JOURNAL OF GRAPH THEORY
then every graph obtained from (sufficiently many) Cg’s by pasting them together
in a chain-like fashion (where consecutive Cg’s share exactly one edge) must
contain a C2‘.
Lemma 10. Let g � 4 be an even integer. For every integer ‘ > g=2 with
2‘ � 2 mod ðg � 2Þ, every C2‘-free graph G of girth at least g contains a sub-
graph H of girth greater than g and such that eðHÞ � eðGÞ=4ð‘� 1Þ.Proof. Let k 2 N be such that 2‘ ¼ g þ kðg � 2Þ. Let D0 be an even cycle in
G satisfying the following two properties:
* D0 contains an edge e0 ¼ xy that lies on a g-cycle in G,* the length r0 of D0 is maximal under the condition that r0 � 2‘ and
r0 � 2 mod ðg � 2Þ.
(As every g-cycle is a candidate for D0, we may assume that D0 exists.) Then
r0 < 2‘, as G is C2‘-free. Let C be a maximal family of g-cycles in G which
contain e0 but share no other edge incident to x or y with each other. We will show
that jCj � r0 � 2. Let C;C0 2 C. By the choice of D0, both C and C0 meet
VðD0Þnfx; yg. (If e.g., VðCÞ \ ðVðD0Þnfx; ygÞ ¼ ; then we would have chosen
the cycle ðC [ D0Þ � xy instead of D0.) Moreover, since G has girth g, it follows
that VðCÞ \ VðC0Þ ¼ fx; yg. So C and C0 meet VðD0Þnfx; yg in distinct vertices.
Thus jCj � r0 � 2.
Now we mark e0 and delete all those edges incident with x or y which lie on a
cycle in C except for e0. We say that each of the deleted edges is assigned to e0.
Let G1 denote the resulting graph. Then e0 does not lie on a g-cycle of G1. If G1 is
not Cg-free, consider again an even cycle D1 in G1 which contains an edge e1
lying on a g-cycle of G1 and whose length r1 is maximal under the condition that
r1 � 2‘ and r1 � 2 mod ðg � 2Þ. In a similar fashion as before, we mark e1 and
delete some edges from G1 which we assign to e1. Denote the graph thus obtained
by G2. As e0 does not lie in a g-cycle of G1 but e1 as well as every edge in
EðG1ÞnEðG2Þ does, we have e0 2 G2 and e0 6¼ e1. We continue in this fashion
until we obtain a Cg-free subgraph H of G. H contains all the marked edges and
no edge was marked twice. Moreover, to each marked edge ei we assigned at most
2ðri � 2Þ < 4ð‘� 1Þ deleted edges. Thus eðHÞ � eðGÞ=4ð‘� 1Þ, as required. &
Proof of Theorem 4. We will prove the following assertion.
For every even integer g � 4, there exists an integer k such that for each
‘ > k with ‘ � 1 mod k every C2‘-free graph G contains a subgraph H
whose girth is greater than g and such that eðHÞ � eðGÞ=2ð4‘Þðg�2Þ=2.
So suppose that we are given a C2‘-free graph G. First choose a bipartite
subgraph G0 of G with eðG0Þ � eðGÞ=2. Note that now we may successively apply
Lemma 10 to G0 to obtain a subgraph H of G with the required properties
provided that 2‘ > g and 2‘ � 2 mod ði � 2Þ for all even integers 4 � i � g. The
FOUR-CYCLES IN GRAPHS 155
latter conditions are satisfied if 2‘ > g and ‘ � 1 mod j for all integers 1 <j < g=2. But if we set k :¼
Qg=2j¼1 j then this holds for every ‘ > k with
‘ � 1 mod k. &
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156 JOURNAL OF GRAPH THEORY
