Journal of Combinatorial Theory, Series B 88 (2003) 29–43
http://www.elsevier.com/locate/jctb
Partitions of graphs with high minimum degree
or connectivity
Daniela Kühna and Deryk Osthusb,1
b
a
Mathematisches Seminar, Universität Hamburg, BundesstraX e 55, 20146 Hamburg, Germany
Institut f ür Informatik, Humboldt-Universität zu Berlin, Unter den Linden 6, 10099 Berlin, Germany
Received 2 May 2001
Abstract
We prove that there exists a function f ðcÞ such that the vertex set of every f ðcÞ-connected
graph G can be partitioned into sets S and T such that each vertex in S has at least c
neighbours in T and both G½S and G½T are c-connected. This implies that there exists a
function gðc; HÞ such that every gðc; HÞ-connected graph contains a subdivision TH of H so
that G V ðTHÞ is c-connected. We also prove an analogue with connectivity replaced by
minimum degree. Furthermore, we show that there exists a function hðcÞ such that the vertex
set of every graph G of minimum degree at least hðcÞ can be partitioned into sets S and T such
that both G½S and G½T have minimum degree at least c and the bipartite subgraph between S
and T has average degree at least c:
r 2003 Elsevier Science (USA). All rights reserved.
Keywords: Graph partitions; Minimum degree; Connectivity; Topological minors
1. Introduction
It is well known that the vertex set of every graph G of minimum degree at least
2c 1 can be partitioned into sets S and T such that the bipartite subgraph ðS; TÞG
of G between S and T has minimum degree at least c: Moreover, Hajnal [4] and
Thomassen [11] showed that for every c there exists k ¼ kðcÞ such that the vertex set
of every graph of minimum degree at least k can be partitioned into sets S and T
E-mail addresses: [email protected] (D. Kuhn),
.
[email protected]
(D. Osthus).
1
Graduate school ‘‘Combinatorics, Geometry and Computation’’, supported by Deutsche Forschungsgemeinschaft Grant GRK/588-1.
0095-8956/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved.
doi:10.1016/S0095-8956(03)00028-5
30
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
such that the graphs G½S and G½T induced by these sets both have minimum degree
at least c: Later, Stiebitz [9] proved that k ¼ 2c þ 1 suffices. The complete graph on
2c þ 1 vertices shows that this is best possible. Hajnal [4] and Thomassen [11] also
proved an analogue of their result where minimum degree is replaced by
connectivity.
On the other hand, it is easily seen that we cannot simultaneously require that
G½S; G½T and ðS; TÞG have large minimum degree or connectivity if G has large
minimum degree or connectivity (see Proposition 6). In Section 2, we show that we
can nevertheless strengthen the results of Hajnal and Thomassen by requiring that
each vertex in S has many neighbours in T:
Theorem 1. For every cAN there exists k ¼ kðcÞAN such that the vertex set of every
graph G of minimum degree at least k can be partitioned into non-empty sets S and T
such that both G½S and G½T have minimum degree at least c and every vertex in S has
at least c neighbours in T:
Theorem 2. For every cAN there exists k ¼ kðcÞAN such that the vertex set of every
k-connected graph G can be partitioned into non-empty sets S and T such that both
G½S and G½T are c-connected and every vertex in S has at least c neighbours in T:
Theorem 2 can be applied to show the existence of non-separating structures in
highly connected graphs. A well-known result in this area is the theorem of
Thomassen [10] which states that every ðc þ 3Þ-connected graph G contains an
induced cycle C such that G V ðCÞ is c-connected. In our case, if H is a structure
whose existence is guaranteed by high connectivity, then Theorem 2 implies that in
every highly connected graph G there exists a copy of H such that G V ðHÞ is still
highly connected. Indeed, if S; T is a partition as in Theorem 2, then any copy of H
in G½S will do. For example, an application of the result of Mader that every graph
of sufficiently large average degree contains a subdivision of a given graph H (see e.g.
[3, Theorem 3.6.1]) immediately yields the following.
Corollary 3. For every cAN and every graph H there exists k ¼ kðc; HÞAN such that
every k-connected graph G contains a subdivision TH of H such that G V ðTHÞ is cconnected.
Similarly, an ‘induced analogue’ of Corollary 3 follows from the result of [6] that
for every sAN and every graph H there exists d ¼ dðs; HÞ such that every Ks;s -free
graph G with average degree at least d contains an induced subdivision of H:
Corollary 4. For all c; sAN and every graph H there exists k ¼ kðc; s; HÞAN such
that every k-connected Ks;s -free graph G contains an induced subdivision TH of H such
that G V ðTHÞ is c-connected.
Analogous results for graphs of large minimum degree can be deduced from
Theorem 1.
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
31
Strengthening the theorem of Mader, Larman and Mani [7] showed that every
sufficiently highly connected graph G contains a subdivision of a given graph H with
prescribed branch vertices. Thus the following conjecture of Thomassen, which
greatly strengthens Theorem 2, would even imply the existence of a non-separating
such subdivision.
Conjecture (Thomassen [12]). For every cAN there exists k ¼ kðcÞAN such
that if G is a k-connected graph and X DV ðGÞ consists of at most c vertices then
the vertex set of G can be partitioned into non-empty sets S and T such that X DS;
each vertex in S has at least c neighbours in T and both G½S and G½T are cconnected.
If true, this would also imply a conjecture of Lovász (see [12]) that there exists a
function f ðcÞ such that, for any two vertices x and y in an f ðcÞ-connected graph G;
there exists an induced x–y path P such that G V ðPÞ is c-connected. Kriesell [5]
observed that Tutte’s theorem on 3-connected graphs can be used to show that
f ð1Þ ¼ 3 and proved that f ð2Þ ¼ 5:
Finally, for graphs of high minimum degree we can strengthen the result of Hajnal
and Thomassen in a different direction. It turns out that instead of asking for high
minimum degree on one side of the bipartite graph ðS; TÞG as in Theorem 1, one can
require ðS; TÞG to have high average degree:
Theorem 5. The vertex set of every graph G of minimum degree at least 232 c can be
partitioned into non-empty sets S and T such that both G½S and G½T have minimum
degree at least c; ðS; TÞG has average degree at least c and jSj; jTjXjGj=218 :
In Section 4, we present some results about partitions of graphs of high average
degree or chromatic number.
Let us now introduce some notation. Given a graph G; we write eðGÞ for the
number of its edges, dðGÞ :¼ 2eðGÞ=jGj for its average degree, dðGÞ for its minimum
degree and wðGÞ for its chromatic number. If S and T are disjoint sets of vertices of
G; then we denote by G½S the subgraph of G induced by S and write ðS; TÞG for the
bipartite subgraph of G whose vertex classes are S and T and whose edges are all the
S–T edges in G: We denote by eG ðS; TÞ or, if this is unambiguous, by eðS; TÞ the
number of these edges and call ðS; TÞG the bipartite subgraph of G between S and T.
We write NðxÞ for the neighbourhood of a vertex xAG and dG ðxÞ or dðxÞ for the
degree of x in G:
2. Proof of Theorems 1 and 2
Before proving Theorems 1 and 2, let us first observe that these results are best
possible in the sense that one cannot additionally require that each vertex in T has
many neighbours in S; i.e. that ðS; TÞG has large minimum degree.
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D. Kuhn,
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.
Proposition 6. For every kAN there is a k-connected graph G whose vertex set cannot
be partitioned into non-empty sets S and T such that each vertex of G has a neighbour
in both S and T:
Proof. Let Gkn be the bipartite graph whose vertex classes are X :¼ f1; y; ng and the
set X ðkÞ consisting of all k-element subsets of X ; and in which xAX is joined to
Y AX ðkÞ if xAY : It is easy to see that Gkn is k-connected if n is sufficiently large
compared to k: Consider any partition of V ðGkn Þ into non-empty sets S and T: As we
may assume that nX2k 1; one of S; T contains at least k vertices from X : Suppose
that this is true for S and let Y AX ðkÞ be a vertex of Gkn corresponding to a set of k
vertices in X -S: Then Y has no neighbour in T; which proves the proposition. &
In the proof of Theorems 1 and 2 we will apply the following quantitative versions
(see [4, Remark 3 and Theorem 4.3] or [9]) of the results mentioned in Section 1.
Theorem 7. The vertex set of every graph G of minimum degree at least 3c can be
partitioned into non-empty sets S and T such that both G½S and G½T have minimum
degree at least c:
Theorem 8. The vertex set of every 8c-connected graph G can be partitioned into nonempty sets S and T such that both G½S and G½T are c-connected.
In the proof of Theorem 1 we will also use the following simple proposition (see
e.g. [3, Proposition 1.2.2]) and a lemma which is an easy consequence of Theorem 7.
Proposition 9. Every graph G contains a subgraph of minimum degree at least dðGÞ=2:
Lemma 10. The vertex set of every graph G of minimum degree at least 3c can be
partitioned into non-empty sets S and T such that both G½S and G½T have minimum
degree at least c and every subgraph of G½S has average degree less than 6c:
Proof. Theorem 7 implies that V ðGÞ can be partitioned into non-empty sets S and T
such that both dðG½SÞXc and dðG½TÞXc: Choose S and T satisfying these
properties such that S is minimal. Then S; T are as desired. Indeed, if G½S had a
subgraph of average degree at least 6c; then by Proposition 9 it would also contain a
subgraph H of minimum degree at least 3c: Apply Theorem 7 to obtain a partition
S 0 ; T 0 of V ðHÞ: Let X be a maximal subset of S\T 0 such that S0 DX and dðG½X ÞXc:
Then each vertex in S\ðT 0 ,X Þ has at least 3c c ¼ 2c neighbours in G outside X :
Hence X ; V ðGÞ\X is a partition contradicting the minimality of S. &
The following lemma shows that if S; T is a partition as provided by Lemma 10,
then we can alter it by successively moving vertices from S to T to obtain a partition
S 0 ; T 0 satisfying Theorem 1 except that G½S 0 is only required to have large average
degree. Theorem 1 will then immediately follow since G½S0 ; having large average
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
33
degree, contains a subgraph of large minimum degree. The ‘moreover’ part of
Lemma 11 will only be used in the proof of Theorem 2.
Lemma 11. Let c; k; c; rAN be such that cXr and kX26 ccr: Let G be a graph of
minimum degree at least k and let S; T be a partition of V ðGÞ such that
dðG½SÞXc; dðG½TÞXc and every subgraph of G½S has average degree less than
cc: Then there exists S0 DS such that, writing T 0 :¼ V ðGÞ\S0 ; every vertex in S 0 has at
least r neighbours in T 0 ; dðG½S0 ÞXc=8 and dðG½T 0 ÞXr: Moreover, T 0 can be obtained
from T by successively adding vertices having at least r neighbours in the superset of T
already constructed.
Proof. Let ADS be the set of all vertices in S having at least 4cc neighbours in S;
and let C :¼ S\A: Thus 4ccjAjp2eðG½SÞ; and so, as every subgraph of G½S has
average degree less than cc;
ccjAj eðG½SÞ
p
:
eðG½AÞp
2
4
ð1Þ
Define B to be the set of all those vertices in S which have less than r neighbours in
T: (So if B ¼ |; then S0 :¼ S and T 0 :¼ T would be a partition as required in the
lemma.) Since dðGÞXkX4cc þ r; we have BDA: Also
kjBj
pðk rÞjBjp2eðG½SÞ:
2
ð2Þ
Let B 0 be the set of all those vertices in B which have at least r S
neighbours in C: For
every vertex xAB 0 choose a set Nx of r neighbours in C: Let eS ð xAB
S 0 Nx Þ denote the
number of all edges in G½S which are incident to some
vertex
in
xAB 0 Nx : (So here
S
we also count those edges with both endvertices in xAB 0 Nx :Þ Then
eS
[
xAB 0
!
ð2Þ
Nx pjB 0 j4ccr p
16ccr eðG½SÞ eðG½SÞ
p
:
k
4
ð3Þ
S
Thus, if we add xAB 0 Nx to T then we can ensure that every vertex in B 0 has at least
r neighbours in the resulting superset of T while maintaining high average degree in
the resulting subgraph of G½S: We now have to deal with the vertices in B\B 0 : As
some of these vertices may have (nearly) all their neighbours in B\B 0 ; but we
nevertheless want to ensure that every vertex in S 0 has many neighbours in T 0 ; we
will move all of B\B 0 into T 0 : For each vertex of B\B 0 which has most of its
neighbours outside B\B 0 (and thus in A\ðB\B 0 Þ), we will also move some of these
neighbours into T 0 in order to ensure that G½T 0 has large minimum degree. The
main difficulty is that the removal of these neighbours should not decrease the
average degree of the resulting subgraph of G½S too much.
So let A1 :¼ B\B 0 ; A2 :¼ A\A1 and let A01 DA1 be S
the set of all those vertices which
have at least k=2 neighbours in A2 : Put C 0 :¼ C \ xAB 0 Nx (Fig. 1). We will show
34
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
Fig. 1. The set-up of the proof of Lemma 11.
that we can add A1 ,
S
xAB 0
Nx to T together with a set A02 DA2 such that
(i) each vertex in A01 has at least r neighbours in A02 ;
(ii) eðA02 ; C 0 ÞpeðA2 ; C 0 Þ=2:
As we shall see, the partition thus obtained is as required in the lemma. However,
let us first show that there exists a set A02 satisfying (i) and (ii). Let a1 ; a2 ; y be any
enumeration of the vertices in A2 such that their degrees in ðA2 ; C 0 ÞG form a nonincreasing sequence. For every xAA01 let Rx be the set consisting of the r rightmost
neighbours of x in a1 ; a2 ; y : Note that, writing ½ai for the set fa1 ; y; ai g;
If aARx then x has at least k=2 r neighbours in ½a:
ð*Þ
S
Let A02 :¼ xAA0 Rx : Then clearly A02 satisfies condition (i). To show that A02 also
1
satisfies (ii), we first prove the following
Claim. For all aAA2 at most half of the vertices in ½a are contained in A02 :
Suppose not and let a be a vertex in A2 such that more than half of the vertices in
½a are contained in A02 : Consider the bipartite subgraph H of G½S between ½a and
the set An1 consisting of all those vertices xAA01 for which ½a-Rx a|: Since ½a-A02
contains at most r vertices lying in a common set Rx ; it follows that jAn1 jXj½aj=ð2rÞ:
Moreover, ð * Þ implies that each vertex in An1 has degree at least k=2 rXk=4 in H:
So if jAn1 jXj½aj; then
eðHÞX
kjAn1 j kjHj ccjHj
X
X
;
4
8
2
while if jAn1 jpj½aj; we have
eðHÞX
kjAn1 j kj½aj kjHj ccjHj
X
X
X
:
4
8r
16r
2
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
35
Thus, in each case H is a subgraph of G½S of average degree at least cc; contradicting our assumption. This proves the claim.
By the claim there exists an injection which sends every vertex aAA02 to a vertex in
A2 \A02 preceding a in the sequence a1 ; a2 ; y : The choice of the enumeration
a1 ; a2 ; y now implies (ii). S
Put T 0 :¼ T,A1 ,A02 , xAB 0 Nx and S0 :¼ V ðGÞ\T 0 : Let us first show that G½S0 has large average degree. Clearly,
ðiiÞ
eðG½S0 ÞXeðA2 \A02 ; C 0 Þ X eðA2 ; C 0 Þ=2:
Thus if eðA2 ; C 0 ÞXeðG½SÞ=4 then eðG½S 0 ÞXeðG½SÞ=8; as required. To verify the
other case, note that
!
[
eðG½S0 ÞX eðG½SÞ eðG½AÞ eS
Nx eðA1 ; C 0 Þ eðA02 ; C 0 Þ
xAB 0
ð1Þ;ð3Þ
eðG½SÞ eðG½SÞ
jB\B 0 jr eðA02 ; C 0 Þ
4
4
ð2Þ;ðiiÞ eðG½SÞ
eðA2 ; C 0 Þ
:
X
4
2
X eðG½SÞ Thus eðG½S0 ÞXeðG½SÞ=8 also holds in the case when eðA2 ; C 0 ÞpeðG½SÞ=4: This
proves that dðG½S 0 ÞXc=8: S
Furthermore, every vertex in S0 DS\A1 ¼ S\ðB\B 0 Þ has at
least r neighbours in ðT, xAB 0 Nx ÞDT 0 :
Let us now verify that T 0 can be obtained from T by successively adding vertices
having
S at least r neighbours inSthe superset of T already constructed. Clearly, as
B- xAB 0 Nx ¼ |; T1 :¼ T, xAB 0 Nx can be constructed in this way. As
A02 DðA\BÞ,B 0 ; every vertex in A02 has at least r neighbours in T1 ; and thus also
T1 ,A02 can be constructed. Since by (i) every vertex in A01 has at least r neighbours in
A02 ; we can now construct T1 ,A02 ,A01 : Finally, to construct T; we now successively
move vertices from A1 \A01 to the other side of the partition as long as they have at
least r neighbours there. Suppose that we are not able to exhaust all of A1 \A01 in this
way, but are left with some non-empty set X : As every vertex in X DA1 \A01 DB\B 0
has less than r neighbours in C; less than k=2 neighbours in A2 ; less than r
neighbours in T 0 \X and as X ,C,A2 ,ðT 0 \X Þ ¼ V ðGÞ; it follows that the graph
G½X has minimum degree at least k k=2 2rXcc; contradicting our assumption
on G½S: Thus T 0 can be constructed in the required way. In particular as
dðG½TÞXcXr this implies that dðG½T 0 ÞXr; so S 0 and T 0 are as desired. &
Proof of Theorem 1. We will prove the theorem for kX210 6c2 ¼ 211 3c2 : Apply
Lemma 10 to obtain a partition S 00 ; T 00 with dðG½S00 ÞX16c; dðG½T 00 ÞX16c and such
that every subgraph of G½S00 has average degree less than 6 16c: Then apply Lemma
11 to find a partition S 0 ; T 0 such that dðG½S 0 ÞX2c; dðG½T 0 ÞXc and each vertex in
S 0 has at least c neighbours in T 0 : By Proposition 9, the graph G½S0 contains a
subgraph H of minimum degree at least c: Then S :¼ V ðHÞ; T :¼ V ðGÞ\S is a
partition as required. &
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D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
To prove Theorem 2, we need the following weak version of a theorem of Mader
(see e.g. [3, Theorem 1.4.2]) and an analogue to Lemma 10.
Theorem 12. Every graph G has a IdðGÞ=4m-connected subgraph.
Lemma 13. The vertex set of every 9c-connected graph G can be partitioned into nonempty sets S and T such that G½T is c-connected, the average degree of G½S is at least
c and every subgraph of G½S has average degree at most 32c:
Proof. Theorem 8 implies that V ðGÞ can be partitioned into non-empty sets S and T
such that dðG½SÞXc; G½T is c-connected and jTjX2c: Choose S and T satisfying
these properties such that S is minimal. We will show that S and T are as desired.
First note that G½S cannot be 8c-connected. For otherwise, by Theorem 8, S can be
partitioned into two non-empty sets S1 and S2 such that each G½Si is c-connected.
As G was 2c-connected and jSj; jTjX2c; there are at least 2c independent S–T edges
in G: So some Si ; S2 say, must contain endvertices of at least c of these edges. But
then G½T,S2 is c-connected, and so S1 ; T,S2 is a partition contradicting the
minimality of S:
Let us now show that G½S does not even contain an 8c-connected subgraph (and
hence, by Theorem 12, no subgraph of average degree at least 32c). Suppose that H
is an 8c-connected subgraph of G½S: As G½S is not 8c-connected, it has a cut-set X
with jX jo8c: Let C be the unique component of G½S X such that
V ðHÞDV ðCÞ,X : Let S0 :¼ V ðCÞ,X and T 0 :¼ V ðGÞ\S 0 : Then G½T 0 is cconnected. Indeed, if it has a cut-set Y with jY joc; then, as G½T is c-connected,
there is a component D of G½T 0 Y such that TDV ðDÞ,Y : Since there is no path
from F :¼ T 0 \ðV ðDÞ,Y Þ to C within G½S which avoids X and no path in G from F
to T which avoids S0 ,Y ¼ V ðCÞ,X ,Y ; it follows that X ,Y separates C from F
in G; contradicting the fact that G is 9c-connected (Fig. 2). We may now successively
move vertices from S0 to T 0 if they have less than 8c neighbours in the subset of S 0
already constructed. (Thus each of these vertices has at least c neighbours in the
superset of T 0 and therefore adding it preserves c-connectedness.) This process must
terminate as H is a subgraph of G½S 0 with minimum degree at least 8c: The partition
obtained in this way contradicts the minimality of S: &
Fig. 2. X ,Y separates C from F in G:
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
37
Proof of Theorem 2. We will prove the theorem for kX216 c2 : First, apply Lemma 13
to obtain a partition S 00 ; T 00 such that dðG½S00 ÞX25 c; G½T 00 is 25 c-connected and
such that every subgraph of G½S00 has average degree less than 32 25 c: Now apply
Lemma 11 to find a partition S 0 and T 0 such that each vertex in S 0 has at least c
neighbours in T 0 ; dðG½S 0 ÞX4c; dðG½T 0 ÞXc and T 0 can be obtained from T 00 by
successively adding vertices having at least c neighbours in the superset of T 00 already
constructed. Thus G½T 0 is c-connected, since G½T 00 is. By Theorem 12, G½S0 contains an c-connected subgraph H: As each vertex in S 0 \V ðHÞ has c neighbours in
T 0 ; G½T 0 ,ðS 0 \V ðHÞÞ ¼ G V ðHÞ is c-connected. Thus S :¼ V ðHÞ; T :¼ V ðGÞ\S
is a partition as required. &
It would be interesting to know whether the quadratic bounds on k in Theorems 1
and 2 can be replaced by linear ones.
3. Proof of Theorem 5
Similar to Section 2, before proving the theorem, let us first show that it is best
possible in the sense that we cannot always partition the vertex set of a graph of large
minimum degree into non-empty sets S and T satisfying Theorems 1 and 5
simultaneously, i.e. we cannot always find S; T such that both G½S and G½T have
large minimum degree, ðS; TÞG has large average degree and every vertex in S has
many neighbours in T:
Proposition 14. For every kAN there is a k-connected graph G whose vertex set cannot
be partitioned into non-empty sets S and T such that dðG½SÞX1;
dðG½TÞX1; dððS; TÞG ÞÞX1 and each vertex in S has at least one neighbour in T:
Note that Proposition 6 is a special case of Proposition 14, but has the advantage
that its proof is less technical. Moreover, we will refer to the proof of Proposition 6
again in Section 4 to deduce a result about partitions of graphs of large chromatic
number and large minimum degree.
Proof. Let Gkn ; X and X ðkÞ be as defined in the proof of Proposition 6 and suppose
that S; T is a partition of V ðGkn Þ contradicting Proposition 14. Then 0ojX -Sjok:
(Indeed, if jX -SjXk and Y is a set consisting of any k vertices in X -S; then all
neighbours of the vertex Y AX ðkÞ are contained in S; contradicting our assumptions
on S and T:) Moreover, as any vertex Y in X ðkÞ -S must contain at least one of the
at most k 1 points in X -S (when Y is viewed as a k-element subset of X ), we have
jX
ðkÞ
-Sjp
k1
X
k1
i¼1
i
!
n
ki
!
p2k nk1 :
38
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
If n is sufficiently large, this implies
!
n
eðS; TÞp jX -Sj
þ jX ðkÞ -Sjk
k1
!
jGn j
k k1 1 n
p kð1 þ 2 Þn o
p k:
2 k
2
Thus dððS; TÞGn Þo1; a contradiction.
&
k
By Proposition 9, every graph G of large average degree contains a subgraph H of
large minimum degree. The following lemma implies that if G has many vertices of
large degree, then H can be chosen to contain a constant fraction of the vertices in G:
Lemma 15. Let cAN and 0oap1: Let G be a graph and let X DV ðGÞ be a set of at
least ajGj vertices having degree at least c in G: Then G has a subgraph of minimum
degree at least ac=4 which contains at least jX j=4 of the vertices in X :
Proof. By Proposition 9, G contains a subgraph of minimum degree at least ac=2:
Let HDG be maximal with minimum degree at least ac=4 and suppose that it
contains less than jX j=4 of the vertices in X : By the choice of H; every vertex of
G V ðHÞ has less than ac=4 neighbours in H: Thus G V ðHÞ still has at least
3jX j=4X3ajGj=4 vertices having degree at least c ac=4 ¼ ð1 a=4Þc in G V ðHÞ:
Thus
ð3ajGj=4Þð1 a=4Þc 3a 3c ac
dðG V ðHÞÞX
X
X :
jGj
4 4 2
By Proposition 9, G V ðHÞ contains a subgraph H 0 of minimum degree at least
ac=4: But H,H 0 contradicts the maximality of H: &
We will need the following special case of Chernoff ’s inequality (see [1, Theorems
A.11 and A.13]).
Lemma 16. Let X1 ; y; Xn be independent 0-1 random variables with PðXi ¼ 1Þ ¼ p
P
for all ipn; and let X :¼ ni¼1 Xi : Then PðjX EX jXEX =2Þp2eEX =16 and PðjX EX jXEX =4Þp2eEX =64 :
Proof of Theorem 5. Let k :¼ 232 c and let ðA; BÞG be a spanning bipartite subgraph
of G of minimum degree at least k=2: Suppose that jAjXjBj: Delete edges if necessary
to obtain a bipartite graph H in which each vertex in A has precisely k=2 neighbours
in B: (Then B may contain some vertices of degree less than k=2:) Consider a random
partition of B into sets B1 ; B2 where each vertex of B is included in B1 with
probability 1=2 independently of all other vertices in B: Call a vertex aAA good if it
has at least k=8 neighbours in each Bi ; and bad otherwise. Given aAA; write Na for
the number of those neighbours of a in H that lie in B1 : Lemma 16 together with the
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
39
fact that EðNa Þ ¼ k=4 implies that
Pða is badÞpPðjNa EðNa ÞjXEðNa Þ=2Þp2eEðNa Þ=16 p1=4:
Hence the expected number of bad vertices in A is at most jAj=4: Thus, there exists a
partition B1 ; B2 of B such that the set A0 of good vertices in A satisfies
3jAj
:
ð4Þ
jA0 jX
4
Let H 0 :¼ H ðA\A0 Þ: Now consider a random partition of A0 into A01 and A02 where
again each vertex of A0 is included in A01 with probability 1=2 independently of all
other vertices in A0 : Then, by Lemma 16,
PðjjA01 j jA0 j=2jXjA0 j=8Þ ¼ PðjjA01 j EðjA01 jÞjXEðjA01 jÞ=4Þ
0
p 2eEðjA1 jÞ=64 p1=3:
ð5Þ
Let us say that a vertex bAB is good if dH 0 ðbÞX3k=217 and if each A0i contains at least
one-quarter of the neighbours of b in H 0 ; otherwise call b bad. Given a vertex bAB;
let Nb0 denote the number of those neighbours of b in H 0 that lie in A01 : Using Lemma
16 again, if dH 0 ðbÞX3k=217 we have
0
Pðb is badÞpPðjNb0 EðNb0 ÞjXEðNb0 Þ=2Þp2eEðNb Þ=16 p1=214 :
Call an edge of H 0 good if it is incident to a good vertex of B; and bad otherwise. Let
Y denote the number of bad edges of H 0 : Let E 0 be the set of all those edges in H 0
whose endvertex in B has degree less than 3k=217 in H 0 : As
jE 0 jp
3kjBj 3kjAj ð4Þ kjA0 j eðH 0 Þ
p 17 p 15 ¼ 14 ;
217
2
2
2
it follows that
EðY Þ ¼
X
Pðthe endvertex of e in B is badÞ
eAEðH 0 Þ
p jE 0 j þ
X
Pðthe endvertex of e in B is badÞ
eAEðH 0 Þ\E 0
p jE 0 j þ eðH 0 Þ=214 peðH 0 Þ=213 :
Now Markov’s inequality implies that
PðY XeðH 0 Þ=212 ÞpPðY X2EY Þp1=2:
Together with (5) this implies that there exists a partition A01 ; A02 of A0 such that at
most eðH 0 Þ=212 of the edges of H 0 are bad and such that jjA01 j jA0 j=2jpjA0 j=8: Thus
3jA0 j
ð6Þ
jA0i jX
8
for i ¼ 1; 2: We will now show that there exist sets Ani pA0i and Bni DBi such that each
ðAni ; Bni ÞH 0 has large minimum degree and ðAn2 ; Bn1 ÞH 0 contains a constant fraction of
the edges of H 0 : Once we have obtained these sets, we will suitably extend An1 ,Bn1 to
S and An2 ,Bn2 to T: To find An1 and Bn1 ; consider the graph H1 :¼ ðA01 ; B1 ÞH 0 : As every
40
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
vertex in A01 has at least k=8 neighbours in B1 and
ð6Þ
jA01 j X
3jA0 j ð4Þ jAj jGj jH1 j
X X X
;
8
4
8
8
ð7Þ
we may apply Lemma 15 with X :¼ A01 and a :¼ 1=8 to obtain sets An1 DA01 and
Bn1 DB1 such that the minimum degree of ðAn1 ; Bn1 ÞH 0 is at least k=28 and
jA0 j ð7Þ jGj
jAn1 jX 1 X 5 :
4
2
ð8Þ
Thus
eH 0 ðAn1 ; Bn1 ÞX
kjAn1 j ð8Þ kjA01 j ð6Þ 3kjA0 j
X 10 X
:
28
2
213
But as at most eðH 0 Þ=212 ¼ kjA0 j=213 edges in H 0 +ðAn1 ; Bn1 ÞH 0 ; are bad it follows that
at least kjA0 j=212 of the edges from ðAn1 ; Bn1 ÞH 0 are good in H 0 : In other words,
X
dH 0 ðbÞXkjA0 j=212 ;
bABn1 ; b is good
and therefore there are also at least kjA0 j=214 edges in H 0 joining (the good vertices
in) Bn1 to A02 :
0
Now consider the graph ðA02 ; Bn1 ÞH 0 and let A}
2 be the set of all vertices in A2 which
15
n
have at least k=2 neighbours in B1 : Then
kjA}
kjA0 j
kjA0 j
2 j
þ 15 XeH 0 ðA02 ; Bn1 ÞX 14 ;
2
2
2
and therefore
jA}
2 jX
jA0 j ð4Þ jAj jGj
X 15 X 16 :
214
2
2
ð9Þ
}
Now consider the graph ðA}
2 ; B2 ÞH 0 : As in this graph the degree of every vertex in A2
}
16
is at least k=8; we may apply Lemma 15 with X :¼ A2 and a :¼ 1=2 to obtain sets
n
n
n
21
An2 DA}
and
2 and B2 DB2 such that ðA2 ; B2 ÞH 0 has minimum degree at least k=2
jA} j ð9Þ jGj
jAn2 jX 2 X 18 :
2
4
n
15
Since every vertex in A}
neighbours in Bn1 ; it follows that
2 +A2 has at least k=2
eH 0 ðAn2 ; Bn1 ÞX
Let
kjAn2 j kjGj
X 33 :
215
2
S 0 :¼ An1 ,Bn1
and
T 0 :¼ An2 ,Bn2 :
Then
ð8Þ
jS0 j X jGj=25 ;
jT 0 jXjGj=218 ;
dðG½S 0 ÞXk=28 Xc; dðG½T 0 ÞXk=221 Xc and ðS 0 ; T 0 ÞG has at least kjGj=233 edges.
Choose S+S 0 and T+T 0 such that S and T are disjoint, both G½S and G½T have
minimum degree at least k=221 and so that S,T is maximal. Then S,T ¼ V ðGÞ:
Indeed, otherwise every vertex in G ðS,TÞ has at least k 2k=221 neighbours
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
41
outside S,T; i.e. dðG ðS,TÞÞXk=2: But then S; V ðGÞ\S is a partition contradicting the choice of S; T: Note that eG ðS; TÞXeG ðS 0 ; T 0 ÞXeH 0 ðAn2 ; Bn1 ) and so
dððS; TÞG ÞXk=232 ¼ c; as desired. &
It would be interesting to know whether a result analogous to Theorem 5 holds for
graphs of high connectivity.
4. Partitions with constraints on the average degree or the chromatic number
We now derive a partition result for graphs of large average degree: the vertex set
of every such graph can be partitioned into S and T such that each of G½S; G½T and
ðS; TÞG still has large average degree. This follows straightforwardly from the next
theorem, which is a consequence of a result of Porter [8]. (The bounds in Porter’s
result were improved to best-possible ones by Bollobás and Scott [2].)
Theorem 17. The vertex set of every graph G can be partitioned into non-empty sets S
and T such that ðS; TÞG has at least eðGÞ=2 edges while both G½S and G½T have at
most eðGÞ=4 þ jGj edges.
Corollary 18. The vertex set of every graph G can be partitioned into non-empty sets S
and T such that ðS; TÞG has at least eðGÞ=2 edges while both G½S and G½T have at
least eðGÞ=4 3jGj edges.
Proof. Put n :¼ jGj and m :¼ eðGÞ: Let S; T be a partition satisfying the conditions
in Theorem 17 such that eðS; TÞ is minimal. Let us show that S; T is as required in
the corollary. Suppose not. Then eðG½SÞom=4 3n (say). Thus
X
X
dðS;TÞG ðtÞ ¼ eðS; TÞ4m=2 þ 2nX2eðG½TÞ ¼
dG½T ðtÞ;
tAT
tAT
and hence there exists a vertex tAT which has more neighbours in S than in T: But
then S0 :¼ S,ftg and T 0 :¼ T\ftg contradicts the choice of S; T since
eðG½S 0 ÞpeðG½SÞ þ nom=4 2n; eðG½T 0 ÞpeðG½TÞ
and
eðS; TÞ4eðS 0 ; T 0 ÞX
eðS; TÞ n4m=2 þ n: &
Note that Corollary 18 implies that for every e40 the vertex set of every graph G
with m edges and average degree at least 6=e can be partitioned into S and T such
that both G½S and G½T contain at least mð1=4 eÞ edges and ðS; TÞG contains at
least m=2 edges.
For a bipartite graph G ¼ ðA; BÞ this (with m=2 replaced by mð1=2 eÞ) can also
be proved using a straightforward probabilistic argument which we outline below.
Consider a random partition of A into A1 and A2 : Then with non-zero probability
most of the edges of G will have the property that their endvertex b in B is good in the
sense that about half of the neighbours of b lies in A1 and the other half lies in A2 :
Now consider any enumeration of the good vertices in B such that their degrees in G
42
D. Kuhn,
D. Osthus / Journal of Combinatorial Theory, Series B 88 (2003) 29–43
.
form a non-increasing sequence, and let B1 be the set of all good vertices with odd
indices. It is easily seen that A1 ,B1 ; A2 ,ðB\B1 Þ is a partition as desired. Since every
graph G contains a spanning bipartite subgraph G 0 with eðG 0 ÞXeðGÞ=2; this
argument also gives Corollary 18 with much weaker bounds.
To conclude this section, let us prove a simple result about partitions of graphs of
large chromatic number and large minimum degree: the vertex set of every such
graph can be partitioned into sets S and T such that both G½S and G½T have large
chromatic number and ðS; TÞG has large minimum degree. Note that we cannot
additionally require G½S and G½T to have large minimum degree. In fact, there need
not even exist a partition S; T such that each of G½S; G½T and ðS; TÞG has minimum
degree at least one. (Let Hkn be the graph obtained from the graph Gkn defined in the
proof of Proposition 6 by making Gkn ½X complete. Then wðHkn ÞXn and dðHkn Þ ¼ k;
but the proof of Proposition 6 shows that there is no partition S; T having the
properties in question.)
Proposition 19. The vertex set of every graph G can be partitioned into non-empty sets
S and T such that wðG½SÞ ¼ JwðGÞ=2n; wðG½TÞ ¼ IwðGÞ=2m and every vertex xAG
has at least d n ðxÞ :¼ minfIwðGÞ=2m; dðxÞ=2g neighbours in the partition set not
containing x:
Proof. Let S; T be a partition with wðG½SÞ ¼ JwðGÞ=2n and wðG½TÞ ¼ IwðGÞ=2m
and such that eðS; TÞ is maximal under these conditions. (By partitioning V ðGÞ into
the vertices in the first JwðGÞ=2n colour classes and the remaining IwðGÞ=2m colour
classes, it is clear that such a partition exists.) Let x be a vertex in S and suppose that
x has less than d n ðxÞpwðG½TÞ neighbours in T: Then the chromatic number of
G½T,fxg is still wðG½TÞ: As wðGÞpwðG½S\fxgÞ þ wðG½T,fxgÞ; this implies that
wðG½S\fxgÞ ¼ wðG½SÞ: But as d n ðxÞpdðxÞ=2; we have eðS\fxg; T,fxgÞ4eðS; TÞ;
contradicting the choice of S; T: Similarly, it can be shown that each vertex xAT has
at least d n ðxÞ neighbours in S: Thus S; T is a partition as required. &
It is not hard to show that the vertex set of every graph G of large chromatic
number and large average degree can be partitioned into S; T such that both G½S
and G½T have large chromatic number and each of G½S; G½T and ðS; TÞG has large
average degree. The idea is to use an averaging argument to find a set X DV ðGÞ
(consisting of some of the colour classes of G) such that G½X still has large
chromatic number but only a small fraction of the edges of G is incident to vertices in
X : Then one can apply Corollary 18 to G X to obtain a partition S 0 ; T 0 : Adding
one-half of the colour classes of G½X to each of S 0 and T 0 gives a partition of V ðGÞ
as desired.
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