Highly connected monochromatic subgraphs of multicolored graphs

Highly ConnectedMonochromatic SubgraphsOf Multicolored Graphs________________ Henry Liu,1 Robert Morris2 and Noah Prince3

1DEPARTMENT OF MATHEMATICS

UNIVERSITY COLLEGE LONDON

GOWER STREET

LONDON WC1E 6BT, UK

E-mail: [email protected]

2NEW HALL

THE UNIVERSITY OF CAMBRIDGE

CAMBRIDGE CB3 0DF, UK


3DEPARTMENT OF MATHEMATICS

UNIVERSITY OF ILLINOIS

409 W. GREEN STREET

URBANA, IL 61801, USA


Received June 6, 2005; Revised June 11, 2008

Published online 9 January 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jgt.20365

Abstract: We consider the following question of Bollobas: given an r-

coloring of E(Kn), how large a k-connected subgraph can we find using at

most s colors? We provide a partial solution to this problem when s 5 1

(and n is not too small), showing that when r 5 2 the answer is n�2k12,

when r 5 3 the answer is b(n�k)/2c11 or d(n�k)/2e11, and when r�1 is a

prime power then the answer lies between n/(r�1)�11(k2�k)r and (n�k1

1)/(r�1)1r. The case sZ2 is considered in a subsequent paper (Liu et

The work reported in this paper was done when the authors were at theUniversity of Memphis.Contract grant sponsor: Van Vleet Memorial Doctoral Fellowships.

Journal of Graph Theory

& 2009 Wiley Periodicals, Inc.

22

al.[6]), where we also discuss some of the more glaring open problems

relating to this question. & 2009 Wiley Periodicals, Inc. J. Graph Theory 61: 22–44, 2009

Keywords: Edge graph coloring; k-connected graph

1. INTRODUCTION

A graph G on n � k þ 1 vertices is said to be k-connected if whenever at mostk�1 vertices are removed from G, the remaining vertices are still connectedby edges of G. It is easy to see that given any graph G, either G or �G (thecomplementary graph) is connected. A substantial generalization of thisobservation, due to Bollobas, asks the following question: When we color theedges of the complete graph Kn with at most r colors, how large a k-connectedsubgraph are we guaranteed to find using only at most s of the colors? In thispaper we shall provide a partial answer to this question in the case s5 1, andin a subsequent paper [6] we shall consider the case s � 2, and in particularthe cases s ¼ 2, 2s ¼ r and s ¼ Yð

ffiffirpÞ, where a jump occurs. The majority of

the problem, however, remains wide open.Bollobas and Gyarfas [2] observed the following example in the case r5 2

and s5 1. First partition the vertices of Kn into five classes, four of order k�1(call these A1, A2, A3 and A4) and the fifth containing the remaining n�

4k þ 4 vertices (call it B). Color the edges between Ai and B red if i5 1 or 2,and blue if i5 3 or 4, and color the edges between Ai and Aj red if fi; jg 2ff1; 2g; f1; 3g; f2; 4gg and blue otherwise (i 6¼ j). Color the edges inside theblocks arbitrarily. The construction is pictured (Figure 1) with only the rededges drawn.

How large a k-connected subgraph using edges of only one color does thiscoloring contain? Suppose such a subgraph H uses more than n� 2k þ 2 of thevertices, and assume (by the symmetry between the given red and blue edges)that the edges of H are colored red. H must use some vertex v of A3 [ A4;suppose v 2 A3. But now if we remove the vertices of A1 \ V ðHÞ from H (to getH 0 say) then v is no longer in the same component of H 0 as any vertex ofðA2 [ A4 [ BÞ \ VðH 0Þ, and if n � 4k � 3 then such a vertex must exist.

Since jA1 \ V ðHÞj � jA1j ¼ k � 1, we have shown that if n � 4k � 3,we cannot guarantee finding a monochromatic k-connected subgraph onmore than n� 2k þ 2 vertices. Bollobas and Gyarfas conjectured that thisexample is extremal, i.e., that if n � 4k � 3 we can guarantee finding amonochromatic k-connected subgraph on at least n� 2k þ 2 vertices. (Notethat when n ¼ 4k � 4 the example above (with A1 and A2 blue cliques, and A3

and A4 red cliques) contains no monochromatic k-connected subgraph at all,so the conjecture really is the strongest possible.) They also gave a short proofof a somewhat weaker statement [2]. Using the ideas from their proof, we areable to show that the conjecture holds when n � 13k � 15. To state this resultwe shall need a little notation.

JOURNAL OF GRAPH THEORY 23

Journal of Graph Theory DOI 10.1002/jgt

Suppose we are given n; r; s; k 2 N, and a function f : EðKnÞ ! ½r�, i.e., anr-coloring of the edges of Kn. We assume always that n � 2. Given a subgraphH of Kn, write cf ðHÞ for the order of the image of EðHÞ under f, i.e.,cf ðHÞ ¼ jf ðEðHÞÞj, the number of different colors with which f colors H.Now, define

Mðf ; n; r; s; kÞ :¼ maxfjVðHÞj : H � Kn; cf ðHÞ � sg;

the order of the largest k-connected subgraph of Kn using at most s colorsfrom ½r�. Finally, define mðn; r; s; kÞ ¼ minf fMðf ; n; r; s; kÞg. Thus, the questionof Bollobas asks for the determination of mðn; r; s; kÞ for all values of theparameters. We shall state all our main results in terms of mðn; r; s; kÞ.

Our first result is the following; it is exactly the conjecture of Bollobas andGyarfas in the case n � 13k � 15.

Theorem 1. Let n; k 2 N, with n � 13k � 15. Then

mðn; 2; 1; kÞ ¼ n� 2k þ 2:

Unfortunately our method breaks down when n is much smaller than 13k,and an analysis of the situation for small values of k suggests that a com-pletely different approach may be necessary in this case.

For r42 the situation becomes a little more complicated. Almost 30 yearsago, while studying a different problem (on hypergraph covering), Furedi [3]and Gyarfas [4] showed independently that n=ðr� 1Þ � mðn; r; 1; 1Þ �n=ðr� 1Þ þ r whenever r�1 is a prime power, with equality in the lowerbound when ðr� 1Þ2 divides n. In Section 3 we shall give a short proof of thisresult. It is easy to modify the upper bound construction of Furedi andGyarfas to give mðn; r; 1; kÞ � ðn� k þ 1Þ=ðr� 1Þ þ cn;k;r when r�1 is a primepower, where c ¼ cn;k;r � r and c ¼ 0 when ðr� 1Þ2 divides ðn� rðk � 1ÞÞ (seeSection 3). The next result shows that this upper bound is essentially bestpossible for these values of r.

Theorem 2. Let n; k; r 2 N, with r � 3 and r�1 a prime power. Then,

n

r� 1� 11ðk2 � kÞr � mðn; r; 1; kÞ �

n� k þ 1

r� 1þ r:

Moreover, the lower bound holds for all 3 � r 2 N.

FIGURE 1.

24 MONOCHROMATIC SUBGRAPHS


Theorem 2 shows that mðn; r; 1; kÞ ¼ n=ðr� 1Þ þOð1Þ when r�1 is a primepower, for every k 2 N. Finally, we shall determine the function exactly whenr5 3.

Theorem 3. Let n; k 2 N, with n � 480k. Then

n� k þ 1

2� mðn; 3; 1; kÞ �

n� k þ 1

2þ 1:

We remark that moreover, equality holds in the lower bound of Theorem 3 ifand only if nþ k � 1 ðmod4Þ (see Corollary 16).

The rest of the paper is organized as follows. In Section 2 we shall proveTheorem 1, and in Section 3 we shall prove Theorems 2 and 3.

2. THE CASE r 5 2

Our first task is to prove Theorem 1. Given any 2-coloring f of EðKnÞ, wewrite R for the graph on VðKnÞ with edge set f �1ð1Þ, and B for the graph withedge set f �1ð2Þ, so EðRÞ [ EðBÞ ¼ EðKnÞ. We shall always refer to the colorsas ‘red’ and ‘blue,’ respectively.

The set of neighbors of a vertex x in a graph G will be denoted by GGðxÞ, orjust GðxÞ when it is clear to which graph we refer, and similarly the degree of xwill be denoted dGðxÞ, or simply dðxÞ. We shall write G½A� for the subgraph ofG induced by a set A � V ðGÞ, and G � A for the graph G½V ðGÞ n A�. IfC;D � V ðGÞ and C \D ¼ ;, then G½C;D� will denote the bipartite graph,with parts C and D, induced by G. For any undefined terms, see [1].

The following simple lemma appeared in [2]. We give the proof for the sakeof completeness.

Lemma 4. In any 2-coloring of EðKnÞ with dRðvÞ � 2k � 2 for everyv 2 VðKnÞ, either R is k-connected or B contains a k-connected subgraph on atleast n�k11 vertices.

Proof. If R is not k-connected, then B must contain a complete bipartitegraph H on n�k11 vertices. Let the part sizes be i and j. If 1 � i � k � 1,then j � n� 2k þ 2, and any vertex v in the i-set has dRðvÞ � 2k � 3, a con-tradiction. Hence i � k, and similarly j � k, so H is k-connected. &

We shall also need the following easy lemma, which will be usefulthroughout the entire paper.

Lemma 5. Let G be a bipartite graph with partite sets M and N such thatdðxÞ � k for every x 2M, and jGðyÞ \ GðzÞj � k for every pair y; z 2 N. Then Gis k-connected.

Proof. Let G be such a bipartite graph, and let C be any subset of V ðGÞ ofsize at most k�1. We wish to show that G0 ¼ G � C is connected. But this is



clear, since any two vertices x; y 2 V ðG0Þ \N have a common neighborin the graph G0 (since jGGðxÞ \ GGðyÞj � k and jCj � k � 1), and any vertexz 2M has a neighbor in V ðG0Þ \N, since dGðzÞ � k. The lemma follows. &

Finally, we make a trivial observation.

Observation 1. Let G be a graph, and v 2 V ðGÞ. If G � v is k-connected anddðvÞ � k, then G is also k-connected.

We are now ready to prove Theorem 1.

Proof of Theorem 1. Let k 2 N, with n � 13k � 15. The upper bound,mðn; 2; 1; kÞ � n� 2k þ 2, follows from the construction described in Section 1.To prove the matching lower bound, let f be a 2-coloring of EðKnÞ. We shall finda monochromatic k-connected subgraph of Kn on at least n� 2k þ 2 vertices.

By Lemma 4, we may assume that there exist vertices x1; y1 2 V ¼ V ðKnÞ

with dRðx1Þ � 2k � 3 and dBðy1Þ � 2k � 3, as otherwise the lemma gives us amonochromatic k-connected subgraph on at least n�k11 vertices. We con-struct (by choosing vertices one by one) maximal subsets X ¼ fx1; . . . ; xpg andY ¼ fy1; . . . ; yqg of V such that

(a) for each i 2 ½p�, dðxiÞ � 2k � 3 in the graph R� fx1; . . . ;xi�1g, and(b) for each i 2 ½q�, dðyiÞ � 2k � 3 in the graph B� fy1; . . . ; yi�1g.

Claim 1. minðp; qÞ � 8k � 11.

Proof. Let u ¼ jX n Y j, v ¼ jY n X j and r ¼ jX \ Y j, so p ¼ uþ r

and q ¼ vþ r. Let eRðX Þ be the number of red edges in H ¼ Kn½X [ Y �with an endpoint in X, and eBðY Þ be the number of blue edges in H with an

endpoint in Y. Now, there are uvþ urþ vrþr

2

� �edges in H with an

endpoint in X and an endpoint in Y. Each such edge contributes at least one to

eRðX Þ or eBðY Þ, so eRðX Þ þ eBðY Þ � uvþ urþ vrþr

2

� �. Also, by the defi-

nition of X, the number of edges in R with an endpoint in X is at mostð2k � 3ÞjX j, so eRðX Þ � ð2k � 3Þp, and similarly eBðY Þ � ð2k � 3Þq. Hence,

pq ¼ uvþ ruþ rvþ r2 � eRðX Þ þ eBðY Þ þ r2 �r

2

� �

� ð2k � 3Þpþ ð2k � 3Þqþ1

2r2 þ

1

2r

� ð2k � 3Þðpþ qÞ þ1

2pqþ

1

4ðpþ qÞ

since r � p and r � q. It follows that

1

2pq � 2k �

11

4

� �ðpþ qÞ;



and so dividing by pq=2, we get

1 �1

pþ

1

q

� �4k �

11

2

� �:

Therefore p or q is at most 8k � 11. &

Assume then, without loss of generality, that jX j � 8k � 11. Note that Xwas chosen to be maximal, so dR�X ðvÞ � 2k � 2 for every vertex v 2 V n X .Therefore, by Lemma 4, either R� X is k-connected, or B� X contains a k-connected subgraphH on at least n� jX j � k þ 1 vertices. Suppose the latter.By the definition of X, any vertex x 2 X sends at most 2k � 3 red edges intoH, and so x must send at least

jHj � 2k þ 3 � n� jX j � 3k þ 4 � n� 11k þ 154k

blue edges into H. So by Observation 1, B½VðHÞ [ X � is k-connected, and hasn�k11 vertices. Hence we may assume that R� X is k-connected.

Now choose a set M0 containing V n X by repeatedly moving fromX to M0 those vertices that send at least k red edges to M0. To be precise,set X0 ¼ X and M0 ¼ V n X , and at time t 2 N form Xt and Mt bychoosing a vertex v 2 Xt�1 with jGRðvÞ \Mt�1j � k if one exists, and settingXt ¼ Xt�1 n fvg and Mt ¼Mt�1 [ fvg. If no such vertex exists then stop theprocess, and set N ¼ Xt�1 and M 0 ¼Mt�1. Notice that every vertex in Nsends at most k�1 red edges into M0, and that R½M 0� is k-connected byObservation 1.

If jNj � 2k � 2 then R½M 0� is our desired subgraph, so assume thatjNj � 2k � 1. We wish to apply Lemma 5 to the bipartite graphG0 ¼ B½M 0;N�, but we may have some ‘bad’ vertices v 2M 0 withdG0 ðvÞ � k � 1. We must therefore first remove these vertices from M0.

Let U denote the set of bad vertices in M0, so

U ¼ fv 2M 0 : dG0 ðvÞ � k � 1g:

Since each vertex of N sends at most k�1 red edges into M0, R½M 0;N� has atmost jNjðk � 1Þ edges. But each vertex of U sends at least jNj � k þ 1 rededges into N. Thus, we have

jU jðjNj � k þ 1Þ � jNjðk � 1Þ;

and hence

jU j �jNjðk � 1Þ

jNj � k þ 1�ð2k � 1Þðk � 1Þ

k� 2k � 2;

since the function xðk � 1Þ=ðx� k þ 1Þ is decreasing for x4k � 1, andjNj � 2k � 1.

We complete the proof of Theorem 1 by setting M ¼M 0 nU , and applyingLemma 5 to the graph G ¼ B½M;N�. By the definition of U, dGðxÞ � k forevery vertex x 2M. Also,

jMj � n� jX j � jU j � n� 10k þ 13 � 3k � 2;



since jX j � 8k � 11, jU j � 2k � 2 and n � 13k � 15, and as observed earlier,dGðyÞ � jMj � k þ 1 for every y 2 N. Therefore,

jGGðyÞ \ GGðzÞj � jMj � 2k þ 2 � k

for every pair y; z 2 N, so by Lemma 5, G is k-connected.Since M [N ¼ V nU and jU j � 2k � 2, G is the desired monochromatic

k-connected subgraph. &

Remark 1. We can in fact improve (for k � 18) the bound on n to n �ð9þ

ffiffiffiffiffi10pÞk as follows. First note that n� 11k þ 154k still holds, so it will

suffice to show that jMj � 3k � 2. Set a ¼ 4þffiffiffiffiffi10p

. We havejMj ¼ n� jNj � jU j, so if jNj � ak þ 4, then jMj � 3k � 2 if n � ðaþ 5Þk. IfjNj � ak þ 5 however, then

jU joðak þ 5Þk

ða� 1Þk þ 6o

ak

a� 1¼ ð

ffiffiffiffiffi10p� 2Þk;

so if n � ðaþ 5Þk � 13 then jMj � n� 8k þ 11� jU j � 3k � 2.

We have the following rather weak corollary to Theorem 1 (and Remark1), which would be improved by further reducing the bound on n.

Corollary 6. For every graph G on n vertices, G or G has a bn=ð9þffiffiffiffiffi10pÞc-

connected subgraph on at least n� 2bn=ð9þffiffiffiffiffi10pÞc þ 2 vertices.

What happens when n is much smaller? For n close to 4k � 3 the problemseems to become much more complicated, so we have been forced to restrictourselves to small values of k. It is not difficult to prove the Bollobas–GyarfasConjecture when k ¼ 1 or 2 (see [2]). We have extended this to the case k5 3.

Theorem 7. For n � 9, m(n,2,1,3)5 n�4.

The proof of this result involves a somewhat lengthy and delicate caseanalysis. We provide only a brief sketch, and refer the interested reader to [5]for a complete proof.

For i; j 2 N, define KGi;jði þ jÞ to be a complete bipartite graph Ki;j � G. We

simply write KGi;j if i and j are known. Notice if G is not k-connected, then

there exists a KGi;jðjGj � k þ 1Þ.

Proof. Let f be a 2-coloring of EðKnÞ, and suppose that there is no mono-chromatic 3-connected subgraph of Kn on at least n� 4 vertices. We shall showthat there is a vertex of high degree in R and in B. Since R is not 3-connected,there exists a KB

i;jðn� 2Þ. If I � 3, then this KBi;jðn� 2Þ is 3-connected. If i ¼ 2,

then j ¼ n� 4; let I and J be the partite sets of sizes i and j, respectively. If R½J�is 3-connected, then it is the desired subgraph. Otherwise, B½J� has a connectedsubgraph on n� 6 vertices, which, along with the vertices of I, form a3-connected subgraph on n� 4 vertices. Hence i5 1, so there is a vertex x withdBðxÞ � n� 3. Similarly, there is a vertex y with dRðyÞ � n� 3.



Assume without loss of generality that xy 2 EðRÞ, and let N � GBðxÞ withjNj ¼ n� 3. Writing dN

G ðvÞ for jGGðvÞ \Nj, we can assume that dNB ðvÞon� 4

for all v 2 N, since otherwise there is a KB2;n�4 and we are done as before.

The remainder of the proof is an analysis of the following cases: eitherthere is a v 2 N with dN

B ðvÞ ¼ n� 5, or dNR ðvÞ � 2 for all v 2 N. In the latter

case, we consider the three subcases corresponding to when the set S ¼ fv 2

N : dNR ðvÞ ¼ 2g has cardinality 0, 1, or at least 2. &

Bollobas and Gyarfas noted that it is not even clear that in any 2-coloringof EðK4k�3Þ, there is a monochromatic k-connected subgraph at all. A proofof this could probably be used to improve the bound n � minðð9þffiffiffiffiffi10pÞk; 13k � 15Þ in Theorem 1.

3. GENERAL r AND s 5 1

In this section we consider the case s5 1, but for general r. The question ofBollobas thus becomes, what is mðn; r; 1; kÞ? We can derive an upper bound byconsidering finite affine planes.

Lemma 8. Let n; r; k 2 N, with n � rðk � 1Þ and r�1 a prime power. Then

mðn; r; 1; kÞ �n� k þ 1

r� 1þ r;

and if ðr� 1Þ2 divides ðn� rðk � 1ÞÞ, then mðn; r; 1; kÞ � ðn� k þ 1Þ=ðr� 1Þ.

Moreover mðn; 3; 1; kÞ � ðn� k þ 3Þ=2 for every n; k 2 N, and ifn � 2rðk � 1Þ, then m(n,r,1,k)5 0.

Proof. Let n; r; k 2 N, with n � rðk � 1Þ and r�1 a prime power. We shalldescribe a coloring f of the edges of Kn in which there is no monochromatic k-connected subgraph on more than ðn� k þ 1Þ=ðr� 1Þ vertices.

Since r�1 is a prime power, there exists a finite affine plane AFr�1 of orderr�1. Let p1; . . . ; pðr�1Þ2 be the points and P1; . . . ;Pr be the parallel classes ofAFr�1. Let C1; . . . ;Cr be disjoint subsets of V ðKnÞ, each of size k�1, and letW ¼ V ðKnÞ n

Sri¼1 Ci. Now divide W into ðr� 1Þ2 classes V1; . . . ;Vðr�1Þ2 of

about equal size (i.e., jðjVij � jVjjÞj � 1 for every pair i; j).The coloring f is defined as follows. If x 2 Vi and y 2 Vj are vertices of Kn

and i 6¼ j, then let f ðxyÞ ¼ t if and only if pi and pj lie on the same line in theclass Pt. If i ¼ j then f ðxyÞ may be chosen arbitrarily. If x 2 Ci, and y 2W ,then let f ðxyÞ ¼ i. If x 2 Ci and y 2 Cj, then let f(xy)5min(i,j).

Let ‘ 2 ½r�, and let G be a monochromatic, k-connected subgraph of Kn,with all edges colored ‘ by f. Suppose G contains vertices from two differentlines of P‘. Then removing the vertices V ðGÞ \ C‘ from G disconnects G, andjV ðGÞ \ C‘j � k � 1, a contradiction. So G contains vertices from at mostr�1 of the sets Vi. Similarly, G may contain no vertex of the set Ci if i 6¼ ‘.



Hence

jGj � ðr� 1Þn� rðk � 1Þ

ðr� 1Þ2

� �þ k � 1 �

n� k þ 1

r� 1þ r: ð1Þ

Since ‘ and G were arbitrary, this completes the proof of the first inequality,and if ðr� 1Þ2 divides n� rðk � 1Þ, then we can remove the r term from theright-hand side of (1).

If r5 3, we split into two cases: n� 3k þ 3 � 1 ðmod4Þ andn� 3k þ 3 6�1 ðmod 4Þ. If n� 3k þ 3 ¼ 4qþ 1, then exactly one of the sets Vi

has order qþ 1, and so (1) becomes jGj � 2qþ 1þ ðk � 1Þ ¼ ðn� k þ 2Þ=2.If n� 3k þ 3 � 0; 2 or 3 ðmod 4Þ, then

n� 3k þ 3

4

� ��

n� 3k þ 5

4

so

jGj �n� 3k þ 5

2þ k � 1 ¼

n� k þ 3

2:

To prove the final part of the lemma, let n � 2rðk � 1Þ, and consider thefollowing coloring g of EðKnÞ. First, partition the vertices of Kn into 2r setsD1; . . . ;D2r, each of size at most k�1. It is well-known (and easy to prove, see[1] for example) that one can partition the edges of K2r into r edge-disjointHamilton paths of length 2r� 1, with each vertex an end-vertex of exactlyone path; let these paths be Q1; . . . ;Qr. If x 2 Di and y 2 Dj with i 6¼ j andij 2 Qt, then let gðxyÞ ¼ t; if i ¼ j, and i is an end-vertex of Qt0 , then letgðxyÞ ¼ t0. It is easy to check that the above coloring contains no k-connectedmonochromatic subgraph, so if n � 2rðk � 1Þ then mðn; r; 1; kÞ ¼ 0. &

The color 2 subgraph of the coloring described in Lemma 8 when r5 3 isshown (Figure 2).

FIGURE 2.



Lemma 8 gives us the upper bounds in Theorems 2 and 3. We shall nowshow that mðn; r; 1; 1Þ � n=ðr� 1Þ for every n and r. Furedi [3] and Gyarfas [4]discovered this while studying a hypergraph covering problem, namely, if onehas r partitions of ½n� such that every x; y 2 ½n� lie in a common block of atleast one of them, then how small can the largest block be? This is obviouslyequivalent to our problem, since the monochromatic components define rpartitions of V ðKnÞ, and if an edge is colored i then its endpoints lie in thesame block of the ith partition.

We present a short, simple proof of this result, the ideas of which will beextended to give the lower bound in Theorem 2.

Lemma 9. Let m; n 2 N and c 2 ½0; 1�. If G is a bipartite graph with part-sizesm and n, and eðGÞ � cmn, then G has a component of order at least c(m1n).

Proof. If c ¼ 0 the result is trivial, so assume c40. Let M and N be thepartite sets of sizes m and n, respectively, and let xy 2 EðGÞ. The order of thecomponent of G containing xy is at least dðxÞ þ dðyÞ. SinceX

xy2EðGÞ

ðdðxÞ þ dðyÞÞ ¼X

v2V ðGÞ

dðvÞ2 ¼Xv2M

dðvÞ2 þXv2N

dðvÞ2

�eðGÞ

m

� �2

mþeðGÞ

n

� �2

n ¼eðGÞ2ðmþ nÞ

mn;

there must be an edge xy with dðxÞ þ dðyÞ � eðGÞðmþ nÞ=mn � cðmþ nÞ. Theorder of the component of G containing xy is therefore at least cðmþ nÞ. &

Corollary 10. The order of the largest monochromatic component of an r-coloring of EðKm;nÞ is at least (m1n)/r.

This result is best possible, since if the partite sets are M and N, and jMjand jNj are both divisible by r, then we may partition M into partsM1; . . . ;Mr and N into parts N1; . . . ;Nr of equal size, and color all edgesbetween Mi and Nj with color i � j ðmod rÞ. The largest monochromaticcomponent in this coloring has order (m1n)/r.

Theorem 11. Let n; r 2 N. Then mðn; r; 1; 1Þ � n=ðr� 1Þ.

Proof. Let n; r 2 N, let f be an r-coloring of EðKnÞ, and let C be a mono-chromatic component of Kn. If C spans the whole of VðKnÞ, thenMðf ; n; r; 1; kÞ ¼ n, and we are done. Otherwise, the edges of Kn½C;V ðKnÞ n C�

are ðr� 1Þ-colored by f, since C is a (maximal) component. Thus, by Cor-ollary 10, Kn contains a monochromatic component of order at leastn=ðr� 1Þ. &

We now return to the situation for general k. The strategy we shall use toprove the lower bound in Theorem 2 is analogous to that used in the proof ofTheorem 11. First, in Lemma 13, we shall derive an (asymptotically tight)upper bound on the number of edges in a bipartite graph with no large



k-connected subgraph (as we did in Lemma 9). From there we simply de-termine how large a k-connected subgraph this ensures.

We shall use the following simple observation in the proof of Lemma 13.

Lemma 12. If a; b; c; d40, then

ab

aþ bþ

cd

cþ d�ðaþ cÞðbþ dÞ

aþ bþ cþ d:

Proof. Expanding the inequality shows it is equivalent toðad � bcÞ2 � 0. &

The next lemma is the key step in the proof of Theorem 2. It is the ana-logue of Lemma 9 for general k.

Lemma 13. Let q; ‘;m; n 2 N with m; n � ‘ and mþ n � 2‘ þ 1. Let G be abipartite graph with parts M and N of size m and n, respectively. If G has noð‘ þ 1Þ-connected subgraph on at least q vertices, then

eðGÞ �qðn� ‘Þðm� ‘Þ

mþ n� 2‘þ ð‘2 þ ‘Þðmþ n� 2‘Þ: ð2Þ

Proof. We prove this by induction on mþ n. To prove the base case,suppose that m ¼ ‘. The inequality

qðn� ‘Þðm� ‘Þ

mþ n� 2‘þ ð‘2 þ ‘Þðmþ n� 2‘Þ � mn;

reduces to ð‘2 þ ‘Þðn� ‘Þ � ‘n, which holds if n � ‘ þ 1. Similarly this in-equality is true if n ¼ ‘ and m � ‘ þ 1. Since eðGÞ � eðKm;nÞ ¼ mn, inequality(2) holds when mþ n ¼ 2‘ þ 1.

So let q; ‘;m; n 2 N, m; n � ‘ þ 1, and assume that the statement ofthe lemma holds if jMj þ jNj � mþ n� 1. Let G be a bipartite graph,with parts M and N of size m and n, respectively, and with noð‘ þ 1Þ-connected subgraph on at least q vertices. Suppose first thatq � mþ nþ 1. Then

qðn� ‘Þðm� ‘Þ

mþ n� 2‘þ ð‘2 þ ‘Þðmþ n� 2‘Þ

4ðn� ‘Þðm� ‘Þ þ ð‘2 þ ‘Þðmþ n� 2‘Þ

¼ mnþ ‘2ðmþ nþ 1� 2‘ � 2Þ � mn � eðGÞ;

and so inequality (2) holds in this case.Next suppose that q � mþ n. Since G contains no ð‘ þ 1Þ-connected sub-

graph on at least q � jGj vertices, G itself cannot be ð‘ þ 1Þ-connected, sothere exists a cutset C of size at most ‘. Let x 2M and y 2 N be disconnectedby C (i.e., they are in different components of G � C). Since m; n � ‘ þ 1, wecan choose a set C0 � C, x; y=2C0, with jC0 \Mj ¼ jC0 \Nj ¼ ‘. Since x and y



were in different components of G � C, they must be indifferent componentsof its subgraph G � C0, so G � C0 is disconnected.

Let G1 be a component of G � C0 and let G2 ¼ G � ðV ðG1Þ [ C0Þ. Fori ¼ 1; 2, let Hi be the subgraph induced by V ðGiÞ [ C0, and letmi ¼ jV ðHiÞ \Mj and ni ¼ jV ðHiÞ \Nj. Note that since jC0 \Nj ¼jC0 \Mj ¼ ‘, we have mi; ni � ‘, and 2‘ þ 1 � mi þ ni � mþ n� 1, sinceV ðG1Þ and V ðG2Þ are nonempty. Hence we can apply the induction hypoth-esis to the graphs H1 and H2, since if Hi contains an ð‘ þ 1Þ-connected sub-graph on at least q vertices then so does G.

Now EðGÞ ¼ EðH1Þ [ EðH2Þ, so eðGÞ � eðH1Þ þ eðH2Þ, and by the induc-tion hypothesis we have

eðH1Þ þ eðH2Þ � qðn1 � ‘Þðm1 � ‘Þ

m1 þ n1 � 2‘þðn2 � ‘Þðm2 � ‘Þ

m2 þ n2 � 2‘

� �

þ ð‘2 þ ‘Þðm1 þm2 þ n1 þ n2 � 4‘Þ:

Applying Lemma 12 with a ¼ n1 � ‘, b ¼ m1 � ‘, c ¼ n2 � ‘ and d ¼ m2 � ‘,and using the identities m1 þm2 ¼ mþ ‘ and n1 þ n2 ¼ nþ ‘, we have

ðn1 � ‘Þðm1 � ‘Þ

m1 þ n1 � 2‘þðn2 � ‘Þðm2 � ‘Þ

m2 þ n2 � 2‘

�ðn1 þ n2 � 2‘Þðm1 þm2 � 2‘Þ

m1 þm2 þ n1 þ n2 � 4‘¼ðn� ‘Þðm� ‘Þ

mþ n� 2‘;

and hence

eðGÞ � eðH1Þ þ eðH2Þ � qðn� ‘Þðm� ‘Þ

mþ n� 2‘þ ð‘2 þ ‘Þðmþ n� 2‘Þ;

so the induction step is complete. The lemma follows immediately. &

The lower bound in Theorem 2 now follows from Lemma 13 and thefollowing well-known theorem of Mader [7].

Mader’s Theorem. Let a 2 R, and G be a graph with average degree a. Then Ghas an a=4-connected subgraph.

Note that since, in any r-coloring of Kn, some color occurs at leastnðn� 1Þ=2r times, Mader’s Theorem implies the existence of a monochro-matic ðn� 1Þ=4r-connected subgraph. This subgraph is k-connected ifn � 4krþ 1, and has at least ðn� 1Þ=4rþ 1 vertices. It is this weak bound thatwe shall need to prove the lower bound in Theorem 2.

Proof of Theorem 2. Let n; k; r 2 N with k � 2, r � 3 and r�1 a primepower. The upper bound on mðn; r; 1; kÞ follows from Lemma 8, so only thelower bound remains to be shown. If n � 11ðk2 � kÞðr2 � rÞ then the resultholds vacuously, so assume n411ðk2 � kÞðr2 � rÞ. Let f be an r-coloring ofEðKnÞ, and for 1 � i � r let GðiÞ denote the graph on V ðKnÞ with edge set



f �1ðiÞ. We shall find, for some i 2 ½r�, a k-connected subgraph of GðiÞ on atleast n=ðr� 1Þ � 11ðk2 � kÞr vertices.

Let H be a monochromatic k-connected subgraph of Kn of maximumorder, and suppose without loss that H has color 1. Let C ¼ V ðHÞ, jCj ¼ c,D ¼ V ðKnÞ n C and jDj ¼ d. By Mader’s Theorem, c � ðn� 1Þ=4rþ 14n=4r,and we may assume that con=ðr� 1Þ, since otherwise H is the desiredmonochromatic subgraph. Thus c; d4k, since r � 3 and n44kr. We shallapply Lemma 13 to the bipartite graph GðiÞ½C;D�, where i 2 ½2; r� is chosen tomaximize the number of edges in this graph.

Since H is maximal, no vertex of D sends more than k�1 edges of color 1into C ¼ VðHÞ, so by the pigeonhole principle, for some i 2 ½2; r� there are atleast dðc� k þ 1Þ=ðr� 1Þ edges between C and D of color i. Fix this i, let‘ ¼ k � 1 and let G ¼ GðiÞ½C;D�. By Lemma 13, if q 2 N satisfies

qðd � ‘Þðc� ‘Þ

ðcþ d � 2‘Þþ ð‘2 þ ‘Þðcþ d � 2‘Þo

dðc� ‘Þ

r� 1� eðGÞ;

or, equivalently,

qodðcþ d � 2‘Þ

ðd � ‘Þðr� 1Þ�ð‘2 þ ‘Þðcþ d � 2‘Þ2

ðc� ‘Þðd � ‘Þ; ð3Þ

then G contains a k-connected subgraph on at least q vertices.The theorem will now follow if we can show that the right-hand side

of (3) is greater than n=ðr� 1Þ � 11ðk2 � kÞr, by setting q equal to thisvalue. Since d4d � ‘, cþ d ¼ n and ‘2 þ ‘ ¼ ðk � 1Þ2 þ ðk � 1Þ ¼ k2 � k, wehave

dðcþ d � 2‘Þ

ðd � ‘Þðr� 1Þ�ð‘2 þ ‘Þðcþ d � 2‘Þ2

ðc� ‘Þðd � ‘Þ4

n� 2‘

r� 1�ðk2 � kÞðn� 2‘Þ2

ðc� ‘Þðd � ‘Þ:

It therefore only remains to bound ðc� ‘Þðd � ‘Þ from below. Sincecþ d ¼ n, ðc� ‘Þðd � ‘Þ is increasing with c for con=2, so since con=ðr� 1Þand r � 3, the minimum is achieved by taking c to be as small as possible.Hence, by setting c ¼ n=4r, we get

ðc� ‘Þðd � ‘Þ4n

4r� ‘

� � 4r� 1

4r

� �n� ‘

� �

4n

8r

� �ð8r� 3Þn

8r

� �4

n2

10r;

since n411ðk2 � kÞðr2 � rÞ48‘r and r � 3. We have therefore shown that if

q ¼n� 2‘

r� 1�ðk2 � kÞn2

ðn2=10rÞ

4

n

r� 1� 11ðk2 � kÞr;



then by Lemma 13 there exists a monochromatic k-connected subgraph on atleast q vertices. This completes the proof. &

Having proved Theorem 2, we can now use it (in place of Mader’s The-orem) to give the following slight improvement for sufficiently large valuesof n.

Theorem 14. Let n; k; r 2 N and E40 satisfy r � 3 and n � ð11ð2þ EÞ=EÞk2r2.Then

mðn; r; 1; kÞ �n

r� 1� 1þ

1

rðr� 2Þþ E

� �k2r:

In particular, if n � 44k2r2, then mðn; r; 1; kÞ � n=ðr� 1Þ � 2k2r.

Proof. The proof follows exactly as the proof of Theorem 2, but we cannow give the following improved bound on ðc� ‘Þðd � ‘Þ, since we knowc4n=ðr� 1Þ � 11k2r.

ðc� ‘Þðd � ‘Þ4n

r� 1� 11k2r� ‘

� �ðr� 2Þn

r� 1þ 11k2r� ‘

� �

4r� 2

ðr� 1Þ2

� �n2 � 11k2 r2 � 3r

r� 1

� �þ ‘

� �n� 121k4r2

4r� 2

ðr� 1Þ2

� �n2 � 11k2ðr� 1Þn

since n � 13k2r2 if E � 11. Let d ¼ Eðr� 2Þð1þ 1=rðr� 2Þ þ EÞ�1. Thend4Eðr� 2Þ=ð2þ EÞ, so

dn411k2r2ðr� 2Þ411k2ðr� 1Þ3;

since n � ð11ð2þ EÞk2r2Þ=E and r � 3. Thus

ðc� ‘Þðd � ‘Þ4r� 2� d

ðr� 1Þ2

� �n2;

so if

q ¼n� 2‘

r� 1�ðk2 � kÞðr� 1Þ2

r� 2� d4

n

r� 1�

k2ðr� 1Þ2

r� 2� d;

then there exists a monochromatic k-connected subgraph on at least q ver-tices. Now simply observe that we chose d so that

ðr� 1Þ2

r� 2� d¼ 1þ

1

rðr� 2Þþ E

� �r;

and the theorem follows. The final implication is attained by setting E ¼ 23and

recalling that r � 3. &



It would be interesting to know where in the ranges given by Theorems 2and 14 the truth lies. We strongly suspect that the upper bound from Lemma8 gives the correct answer.

Conjecture 1. Let n; k; r 2 N with r � 3, n � 2rðk � 1Þ þ 1, r�1 a prime powerand n� rðk � 1Þ divisible by ðr� 1Þ2. Then

mðn; r; 1; kÞ ¼n� k þ 1

r� 1:

Remark 2. By Lemma 8, mðn; r; 1; kÞ ¼ 0 if n � 2rðk � 1Þ. Hence the lowerbound on n in the conjecture cannot be weakened any further.

We also have the following conjecture for the bipartite version of thequestion. It says that the order of the largest k-connected subgraph equals theupper bound given in Corollary 10 (and so does not depend on k), as long asthe partite sets are large.

Conjecture 2. Let m; n; k; r 2 N, with r � 3 and m; n � rk. Any r-coloring of theedges of Km;n contains a monochromatic k-connected subgraph on at leastðmþ nÞ=r vertices.

Although we have been unable to prove Conjectures 1 and 2, Theorem 3shows that Conjecture 1 holds in the case r5 3. We shall next prove thisresult. We begin with an easy lemma.

Lemma 15. Let k; p; q 2 N satisfy 3p � q � p � 24k, and let P and Q be setswith jPj ¼ p and jQj ¼ q. Let Kp;q be the complete bipartite graph with parts Pand Q. Suppose the edges of Kp;q are 3-colored in such a way that each vertex inP sends at most k edges of color 3 into Q, and each vertex in Q sends at most kedges of color 2 into P.

Then the subgraph induced by edges of color 1 contains a k-connected sub-graph G with jP n VðGÞj � 16k, and jQ n V ðGÞj � 8k. In particular,jV ðGÞj � pþ q� 24k.

Proof. Let k; p; q 2 N satisfy 3p � q � p � 24k, and let f be a 3-coloring ofEðKp;qÞ satisfying the conditions of the lemma. Let

SP ¼ fv 2 P : v sends at most 3q=4 edges of color 1 into Qg

and

SQ ¼ fv 2 Q : v sends at most 3p=4 edges of color 1 into Pg

be sets of ‘bad’ vertices. We shall remove the bad sets and applyLemma 5.

We need to bound jSPj and jSQj from above. Since each vertex of Q has atmost k incident edges of color 2, we have jf �1ð2Þj � kq, and similarlyjf �1ð3Þj � kp. Also, since each vertex of SP has at least q=4 incident edges ofcolor 2 or 3, we have jf �1ð2Þj þ jf �1ð3Þj � jSPjðq=4Þ, and similarly



jf �1ð2Þj þ jf �1ð3Þj � jSQjðp=4Þ. Thus,

jSPj �4

qðjf �1ð2Þj þ jf �1ð3ÞjÞ �

4kðpþ qÞ

q� 8k

and

jSQj �4

pðjf �1ð2Þj þ jf �1ð3ÞjÞ �

4kðpþ qÞ

p� 16k:

Now, let P0 ¼ P n SP and Q0 ¼ Q n SQ, and let G be the bipartite graph withvertex set P0 [Q0, and edge set f �1ð1Þ. If x 2 P0, then x sends at least 3q=4edges of color 1 into Q, so

dGðxÞ � 3q=4� jSQj � 18k � 16k4k;

and similarly if y; z 2 Q0, then

jGGðyÞ \ GGðzÞj � 3p=4þ 3p=4� p� jSPj

¼ p=2� jSPj � 12k � 8k > k;

so the conditions of Lemma 5 are satisfied. Thus by Lemma 5, G is k-con-nected. Since also jP n V ðGÞj ¼ jSPj � 8k and jQ n V ðGÞj ¼ jSQj � 16k, G isthe desired subgraph. &

Given graphs G and H, define G [H to be the graph with vertex setV ðGÞ [ V ðHÞ and edge set EðGÞ [ EðHÞ. We shall also use the followingtrivial observation.

Observation 2. Let k 2 N. If G and H are k-connected graphs, andjV ðGÞ \ V ðHÞj � k, then the graph G [H is also k-connected.

We are now ready to prove Theorem 3.

Proof of Theorem 3. Let n; k 2 N with n � 480k. The upper bound onmðn; 3; 1; kÞ follows from Lemma 8, so only the lower bound remains to beshown.

Let f be a 3-coloring of the edges of Kn, and let V ¼ V ðKnÞ. For eachi 2 f1; 2; 3g, let Gi be the subgraph of Kn with vertex set V and edge set off �1ðiÞ, and assume that Gi has no k-connected subgraph on more than ðn�kÞ=2 vertices. We begin by covering V with monochromatic k-connectedsubgraphs.

Claim 1. There exist (not necessarily disjoint) subsets A1, A2, and A3 of V suchthat Gi½Ai� is k-connected, and A1 [ A2 [ A3 ¼ V .

Proof. Assume, without loss of generality, that eðG1Þ � eðG2Þ � eðG3Þ. ByMader’s Theorem (and since n412k), there exists a maximal set A1 � V , withjA1j � n=12, such that G1½A1� is k-connected. If jA1j4ðn� kÞ=2, then G1½A1�

is a monochromatic k-connected subgraph on more than ðn� kÞ=2 vertices,contradicting our assumption, so (writing Ac

1 for V n A1) we havejAc

1j4n=24jA1j.



For i ¼ 2; 3, let Hi ¼ Gi½A1;Ac1� be the bipartite graph induced by the edges

of color i and the sets A1 and Ac1. Since A1 is maximal, each vertex of Ac

1 sendsat most k�1 edges of color 1 into A1 (by Observation 1), and so has degree atleast jA1j � k þ 1 in H2 [H3. Hence

eðH2Þ þ eðH3Þ � jAc1jðjA1j � k þ 1Þ �

11n

12

n

12� k þ 1

� �4

n2

15;

the second inequality holding because the function �x2 þ ðn� k þ 1Þx isincreasing for xoðn� k þ 1Þ=2, and the third holding because n495k.

Since eðH2Þ � eðH3Þ, we obtain eðH2Þ4n2=30, so the average degree in H2

is at least n=15. Applying Mader’s Theorem again, we deduce that H2 con-tains an ðn=60Þ-connected subgraph H 02. Since H2 is bipartite, H 02 mustcontain at least n=60 vertices of each class of H2; in particular, it must containat least 8k vertices of A1 (since n � 480k).

Let A2 be a maximal set containing V ðH 02Þ such that G2½A2� is k-connected.We have now found sets A1 and A2, with Gi½Ai� k-connected for i ¼ 1; 2. Wecomplete the proof by using Lemma 5 to find a k-connected graph in G3

containing ðA1 [ A2Þc.

Let X ¼ A1 \ A2 and Y ¼ ðA1 [ A2Þc. Notice that jA2j � ðn� kÞ=2, since

otherwise we would have a monochromatic k-connected subgraph on morethan ðn� kÞ=2 vertices, contradicting our assumption. Since VðH 02Þ � A2 and,as observed above, H 02 contains at least 8k vertices of A1, we have jX j � 8k.Hence also

jY j ¼ n� jA1j � jA2j þ jA1 \ A2j

� n� ðn� kÞ þ jX j ¼ jX j þ k � 9k;

since jA1j; jA2j � ðn� kÞ=2.We want to apply Lemma 5 to the bipartite graph G3½X ;Y �, but first we

must remove the vertices of degree at most k�1 from X, as in the proof ofTheorem 1. As in that proof, let

U ¼ fv 2 X : jGG3ðvÞ \ Y j � k � 1g:

Since G½A1� and G½A2� are maximal monochromatic k-connected subgraphs,each vertex v 2 Y can send only at most k�1 edges of color 1 into A1, andk�1 edges of color 2 into A2. Therefore, v must send at least jX j � 2k þ 2edges of color 3 into X ¼ A1 \ A2. This is true for every v 2 Y , so G3½X ;Y �has at most jY jð2k � 2Þ nonedges. But each vertex of U sends at least jY j �k þ 1 nonedges into Y. Thus, we have

jU jðjY j � k þ 1Þ � jY jð2k � 2Þ;

and hence

jU j �2jY jðk � 1Þ

jY j � k þ 1�

18kðk � 1Þ

8k þ 1o3k;



since the function 2xðk � 1Þ=ðx� k þ 1Þ is decreasing for x4k � 1, andjY j � 9k.

Let X 0 ¼ X nU , and consider the bipartite graph G3½X0;Y �. By the defi-

nition of U, each vertex in X0 has degree at least k in this graph. Also, asnoted above, each vertex of Y sends at most 2k � 2 edges of color 1 or 2 intoX0, so any two vertices in Y have at least

jX 0j � 4k þ 4 ¼ jX j � jU j � 4k þ 448k � 3k � 4k ¼ k

common neighbors in X0.So G3½X

0;Y � satisfies the conditions of Lemma 5, and therefore by thatlemma G3½X

0;Y � is k-connected. Let A3 be a maximal set such that G3½A3� isk-connected, and X 0 [ Y � A3. Since V n ðA1 [ A2Þ ¼ Y � A3, this completesthe proof of Claim 1. &

For the remainder of the proof, fi; j; ‘g will always be the set f1; 2; 3g,though the order will vary. Let A1;A2;A3 be the (maximal) sets given byClaim 1, and for each i (i.e., for each triple i; j; ‘ with fi; j; ‘g ¼ f1; 2; 3g), letai ¼ jAi n ðAj [ A‘Þj, bi ¼ jðAj \ A‘Þ n Aij, and c ¼ jA1 \ A2 \ A3j (seeFigure 3).

By Claim 1, Xi

ai þX

i

bi þ c ¼ n: ð4Þ

Our initial assumption says that jAij ¼ ai þ bj þ b‘ þ c � ðn� kÞ=2 for eachtriple i; j; ‘. Summing over i ¼ 1; 2; 3 and subtracting (4) givesX

i

bi þ 2c �n� 3k

2; ð5Þ

whilst summing pairwise and subtracting (4) gives

ai � bi þ cþ k ð6Þ

for each i 2 f1; 2; 3g.

FIGURE 3.



Now, observe that since Gj½Aj� is a maximal monochromatic k-connectedsubgraph, each vertex of Ai n Aj sends at most k�1 edges of color j intoAj n Ai, for each pair i; j. We wish to apply Lemma 15 to the pairof setsAi n Aj and Aj n Ai; the next claim (which we shall also prove using Lemma15) allows us to do so.

Claim 2. ai � n=6 for each i 2 f1; 2; 3g.

Proof. Let i 2 f1; 2; 3g and suppose aion=6. Note that by (6) we also havebi þ c � ai � kon=6. Assume, without loss of generality, thatjAj n A‘j � jA‘ n Ajj. We shall apply Lemma 15 with P ¼ Aj n A‘ andQ ¼ A‘ n Aj.

Let p ¼ jAj n A‘j and q ¼ jA‘ n Ajj. By assumption, q � p. Now observethat p � 24k, since otherwise

jAjj ¼ n� jA‘ n Ajj � jAi n ðAj [ A‘Þj ¼ n� p� ai45n

6� 24k4

n

2;

since aion=6 and n472k, which contradicts our assumption thatjAjj � ðn� kÞ=2. Also note that q � 3p, since

pþ q ¼ jAjDA‘j ¼ n� jAi n ðAj [ A‘Þj � jAj \ A‘j ¼ n� ðai þ bi þ cÞ;

so if q43p, then

jAjj ¼ jAj n A‘j þ jðAj \ A‘Þ n Aij þ jAj \ A‘ \ Aij

¼ qþ bi þ c43ðn� ai � bi � cÞ

4þ bi þ c

43ðn� aiÞ

44

5n

8;

which again contradicts our assumption that jAjj � ðn� kÞ=2.Hence k, p and q satisfy 3p � q � p � 24k. Also, as observed above, each

vertex of P ¼ Aj n A‘ sends at most k�1 edges of color ‘ into Q ¼ A‘ n Aj,and similarly each vertex of Q sends at most k�1 edges of color j into P, bymaximality of A‘ and Aj. So by Lemma 15, there must exist a monochromatick-connected subgraph in Gi½P;Q� on at least pþ q� 24k vertices. Since

pþ q� 24k ¼ n� ðai þ bi þ cÞ � 24k42n

3� 24k4

n

2

(because bi þ coaion=6 and n4144k), this contradicts our assumption thatthere is no monochromatic k-connected subgraph in Gi on more than ðn�kÞ=2 vertices. This final contradiction completes the proof of the claim. &

We shall now apply Lemma 15 to the sets Ai n Aj and Aj n Ai, for each pairi and j. Let i; j 2 f1; 2; 3g with i 6¼ j, and assume, without loss of generality,that jAi n Ajj � jAj n Aij. We shall apply Lemma 15 with P ¼ Ai n Aj andQ ¼ Aj n Ai.

Let p ¼ jAi n Ajj and q ¼ jAj n Aij. By assumption, q � p. Now,p � jAi n ðAj [ A‘Þj ¼ ai, so by Claim 2, p � n=6. Since n � 144k, it follows



that p � 24k, and since q � jAjjon=2, it also follows that 3p � q. As observedearlier, each vertex of P sends at most k�1 edges of color j into Q, and eachvertex of Q sends at most k�1 edges of color i into P, since Ai and Aj aremaximal.

Therefore, applying Lemma 15 to the sets P and Q, we obtain a k-con-nected subgraph of G‘ (where, as usual, ‘ ¼ f1; 2; 3g n fi; jg) omitting at most16k vertices of P and at most 8k vertices of Q. Let this subgraph be L‘.

We obtain in this way three k-connected subgraphs, L1, L2 and L3. Foreach ‘ 2 f1; 2; 3g, let M‘ be the vertex set of a maximal k-connected subgraphof G‘ containing L‘. Now, for each pair i 6¼ j, let Xij ¼ Ai n ðAj [M‘Þ be theset of vertices in Ai n Aj avoided by M‘, and let xij ¼ XijAi. Also, for each‘ 2 f1; 2; 3g, let Z‘ ¼M‘ n ðAi [ AjÞ, and let z‘ ¼ jZ‘j. We have, therefore, foreach triple i; j; ‘, that

ai þ aj þ bi þ bj � xij � xji þ z‘ � jM‘j �n� k

2; ð7Þ

by assumption, since M‘ is the vertex set of a monochromatic k-connectedsubgraph.

Although we have so far been approximating wildly, we must now beprecise. Summing the inequalities (7) over ‘ ¼ 1; 2; 3, we get

2X

i

ai þ 2X

i

bi �X

i;j

xij þX

i

zi �3ðn� kÞ

2;

which is equivalent to

nþ 3k

2� 2cþ

Xi;j

xij �X

i

zi;

sinceP

ai þP

bi þ c ¼ n. Combining this withP

bi þ 2c � ðn� 3kÞ=2, weobtain

0 �X

i

bi �n� 3k

2� 2c �

Xi;j

xij �X

i

zi � 3k; ð8Þ

soP

xij �P

zi � 3k, and by the pigeonhole principle, there exists an i 2f1; 2; 3g such that

xij þ xi‘ � zi � k: ð9Þ

We fix this i for the remainder of the proof. We shall show that inequality (9)implies that jXij [ Xi‘j � k, and deduce that Gi½Ai [Mi� is k-connected.

Indeed, let j 2 f1; 2; 3g n fig, and consider a vertex v 2 Xij. We shall showthat v 2Mi and therefore that Xij \ Xi‘ � Zi. Let ‘ 2 f1; 2; 3g n fi; jg, and re-call that Xij � Ai n ðAj [M‘Þ, so v=2Aj and v=2M‘. Since v=2Aj and Aj is max-imal, v sends at most k�1 edges of color j into Aj, and since v=2M‘ and M‘ ismaximal, v sends at most k�1 edges of color ‘ into M‘.

How many edges of color j or ‘ can v send into Aj n ðAi [ A‘Þ? Since M‘

contains V ðL‘Þ, we know that M‘ avoids at most 16k vertices of Aj n Ai, so by



the observations above, v sends at most 17k � 1 edges of color ‘ into Aj n Ai,and so at most 18k � 2 edges of color j or ‘ into Aj n Ai. Thus, v sends at most18k � 2 edges of color j or ‘ into Aj n ðAi [ A‘Þ � Aj n Ai.

Now, Mi avoids at most 16k vertices of the set Aj n A‘, and so at most 16kvertices of Aj n ðAi [ A‘Þ. Therefore, the number of edges of color i goingfrom v into Mi is at least

jAj n ðAi [ A‘Þj � 16k � ð18k � 2Þ ¼ ai � 34k þ 2

4n

6� 34k4k;

since ai � n=6 by Claim 2, and n4210k. But Mi was chosen to be amaximal monochromatic k-connected subgraph, so if v sends at least kedges of color i into Mi, it follows that v 2Mi.

Now, suppose that in fact v 2 Xij \ Xi‘. Since Xij � Ai n Aj

and Xi‘ � Ai n A‘, it follows that v 2 Ai n ðAj [ A‘Þ. Recall that Zi ¼

Mi n ðAj [ A‘Þ and it is clear that, in this case, v 2Mi implies v 2 Zi. HenceXij \ Xi‘ � Zi, and so zi � jXij \ Xi‘j.

It now follows immediately that

jXij [ Xi‘j ¼ jXijj þ jXi‘j � jXij \ Xi‘j � xij þ xi‘ � zi � k;

by inequality (9). The following claim now gives us the final contradiction.

Claim 3. Gi½Ai [Mi� is k-connected, and has order at least 3n/4.

Proof. Let i be as in inequality (9), and fi; j; ‘g ¼ f1; 2; 3g. We have shownthat jXij [ Xi‘j � k, and that Xij [ Xi‘ � Ai \Mi. By the definitions of Ai andMi, the graphs Gi½Ai� and Gi½Mi� are k-connected. Therefore, by Observation2, Gi½Ai [Mi� is k-connected.

Now, since Mi contains V ðLiÞ, we know that Mi avoids at most 24k ver-tices of AjDA‘, so Ai [Mi avoids at most 24k þ jðAj \ A‘Þ n Aij ¼ 24k þ bi

vertices of V. By inequality (8), we have (very weakly), thatbi �

Pm bm �

Pu;v xuv � 96k, since xuv � 16k for each u 6¼ v, u; v 2 f1; 2; 3g.

Thus,

jAi [Mij � n� ð24k þ biÞ � n� 120k �3n

4;

since n � 480k. This completes the proof of the claim. &

So Gi½Ai [Mi� is a monochromatic k-connected subgraph on more thanðn� kÞ=2 vertices, contradicting our assumption that no such subgraph exists.This contradiction proves the theorem. &

It is now easy to obtain the exact value of mðn; 3; 1; kÞ whenever n � 480k.



Corollary 16. Let n; k 2 N, with n � 480k. Then

mðn; 3; 1; kÞ ¼

ðn� k þ 1Þ=2 if nþ k � 1 ðmod4Þ;

ðn� k þ 2Þ=2 if nþ k � 0 or 2 ðmod4Þ;

ðn� k þ 3Þ=2 if nþ k � 3 ðmod 4Þ:

8><>:

Proof. Let n; k 2 N, with n � 480k. If nþ k � 1 ðmod4Þ, thenn� 3k þ 3 � 0 ðmod4Þ, so by Lemma 8 and Theorem 3 we havemðn; 3; 1; kÞ ¼ ðn� k þ 1Þ=2. If nþ k � 0 or 2 ðmod4Þ, then ðn� k þ 2Þ=2 isthe only integer in the range given by Theorem 3, so clearlymðn; 3; 1; kÞ ¼ ðn� k þ 2Þ=2. If nþ k � 3 ðmod4Þ, then we have mðn; 3; 1; kÞ �ðn� k þ 3Þ=2 by Lemma 8.

It remains to prove the lower bound in the case nþ k � 3 ðmod4Þ. Todo this, we follow the proof of Theorem 3, making a couple of small al-terations.

To be precise, let n; k 2 N, with n � 480k, and nþ k � 3 ðmod4Þ. Let f be a3-coloring of EðKnÞ, and assume that Gi contains no monochromatic k-connected subgraph on more than ðn� k þ 1Þ=2 vertices for i ¼ 1; 2; 3, whereGi is as defined above. Using this assumption, the proof goes through exactlyas above, except inequality (8) becomes

0 �X

i

bi �n� 3k þ 3

2� 2c �

Xi;j

xij �X

i

zi � 3k þ 3: ð10Þ

If xij þ xi‘ � zi � k for any triple fi; j; ‘g, then we would be done as in theproof of Theorem 3, so assume not. So inequality (10) is in fact an equality.But then

n� 3k þ 3

2¼ 2c

with c 2 N, which means that nþ k � 1 ðmod4Þ, a contradiction. This provesthe corollary. &

Remark 3. The bound n � 480k is, of course, likely to be far from best possible.By Lemma 8, we know that n ¼ 6k � 6 is not sufficient to guarantee the ex-istence of a monochromatic k-connected subgraph. We conjecture, along thelines of Bollobas and Gyarfas, that if n � 6k � 5 then mðn; 3; 1; kÞ �ðn� k þ 1Þ=2 (this is Conjecture 1 in the case r5 3).

We finish by stating the obvious question: what happens when r�1 is not aprime power? Our lower bound still holds in this case, so we have the fol-lowing easy corollary of (the proof of) Theorem 2.

Corollary 17. Let r; k 2 N, and n!1. Let r0 be the largest integer less than orequal to r such that r0�1 is a prime power. Then

n

r� 1þ oðnÞ � mðn; r; 1; kÞ �

n

r0 � 1þ oðnÞ:



In particular,

1

6þ oð1Þ

� �n � mðn; 7; 1; kÞ �

1

5þ oð1Þ

� �n:

Problem 1. Find a constant c ¼ cðrÞ (if one exists) such that

mðn; r; 1; kÞ ¼ ðcþ oð1ÞÞn

for those r 2 N which are not prime powers.

ACKNOWLEDGMENTS

The authors thank Bela Bollobas for suggesting the problem to them, and forhis ideas and encouragement. They would also like to thank ETH Zurich, andTrinity College, Cambridge, where part of this research was carried out. Thesecond and the third authors were partially supported during this research byVan Vleet Memorial Doctoral Fellowships.

References

[1] B. Bollobas, Modern Graph Theory, Springer, New York, 1998.

[2] B. Bollobas and A. Gyarfas, Highly connected monochromaticsubgraphs, Discrete Math 308 (2008), 1722–1725.

[3] Z. Furedi, Maximum degree and fractional matchings in uniformhypergraphs, Combinatorica 1 (1981), 155–162.

[4] A. Gyarfas, Partition coverings and blocking sets of hypergraphs (inHungarian), Comm Comp Automat Inst Hung Acad Sci 71 (1977), 62.

[5] H. Liu, R. Morris and N. Prince, Highly connected monochromaticsubgraphs of multicolored graphs: addendum (manuscript).

[6] H. Liu, R. Morris and N. Prince, Highly connected multicoloredsubgraphs of multicolored graphs, Discrete Math 308 (2008), 5096–5121.

[7] W. Mader, Existenz n– fach zusammenhangender Teilgraphen inGraphen genugend grosser Kantendichte, Abh Math Sem UnivHamburg 37 (1972), 86–97.



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Highly connected monochromatic subgraphs of multicolored graphs