Zero sum partition theorems for graphs

Sci. VOL. 17 NO. 4 (1994) 697-702

697

ZERO-SUM PARTITION

THEOREMS

FOR

GRAPHS

Y. CARO

Department

ofMathematics HaihUniversity

Oranim, Israel

I.KRASIKOV Y. RODITI’Y

School ofMathematics Tel-AvivUniversity

RamatAviv, Israel

Department

ofMathematics Beit-Berl College

Kfar-Saba,

Israel

(Received

November 13, 1992 andin revisedformFebruary 2,

1993)

ABSTP,.ACT. Letq

pn

beapowerofanodd prime p. Weshow that the t(q)

verticesofeverygraphGcanbe partitionedinto t(q)classes V(G)=

13

V such that the number of

t=l

edges in any inducedsubgraph <

Vi>

is divisible byq,where

t(q)<_

(q-l)

-(2(q-1)-1)4

ifq 2

n,

thent(q) 2q-1.

In

particular,is isshownthat t(3) 3and4_<t(5) <5.

KEY WORDS AND PHILSES. Zero-sum,

partition, clique-number. 1991 AMS SUBJECT

CLASSIFICATION CODES.

05C55.

1.

INTRODUCTION.

Long

ago, Gallaiprovedthefollowingpartition theorem:

TItEOREM A. Let Gbeagraph. Thenthere existsapartition AOB V(G), AnB such thatinthesubgraphsinducedon <A> and <B> all thedegreesare even

(see

[8]

problem5.17 o

[6]).

Recently, this theoremreceivedsome attention

([4],

[5]).

A

partial sampleof theproblems

evolved from Gallai’s theoremis:

PROBLEM

1: What is themaximumorder ofaninduced subgraph// ofagraph G, with n non-isolatedvertices, such thatin//allthedegrees areeven, respectively odd.

In

particularisit

true that everytreeon n>2 vertices contains aninduced subgraph onat least 2n-₃ 2 vertices in whichallthedegreesareodd?

For

someresults concerningthisproblem,see

[4]

and

[5].

PROBLEM2: Letk>2beaninteger. Does thereexistaconstantc c(k),depending only

onk,such thateverygraphcanbe partitionedintoc(k) vertex-disjointclasses o V: V(G)in

t=l

such away that inall the inducedsubgraphs <V >, < < c(k), all the degrees are divisibleby ?

Clearly,Theorem

A

statesthat c(2) 2. Theexistenceof c(k) fork>3is aninterestingopen problem.

Motivated by Theorem

A

and the above problems, we consider the following related question:

Y. CARO, I. KRASIKOV AND Y. RODITTY

graphG canbepartitionedinto t(k)vertex-disjoint

classes,

=w

V

V(G) in such away that in each of the inducedsubgraphs <

v,

>, < < thenumberofedgesisdivisiblebyk?

Obviously an affirmative answer to Problem 2

(at

least for /

odd)

_implies an affirmative answer to Problem 3 with t(k)< c(k). Hence Problem 3 is weaker, and in fact more tractable, thanProblem 2.

It is convenientforlater purpose tointroduce thefollowingnotations. Denoteby (a,:) the

smallest integer msuch that thehypergraph

a

can be partitionedintornclasses V(G)= V ina way that e(<V >)= O(modk), <i<m, where e( <V > is the number ofedges in theinduced subhypergraph V

i. Observe that for every

:

>_2 and every hypergraph G,(G,k)< X(G), where x(a) is the usual chromatic number of G, namely, the smallest number of colors such that each

edgehasverticesof at least twocolors.

Hence,

withthosenotationsProblem3is: doessup (G,k) t(k)<x _.fora_fixedk> 2?

G

Our main goal is to prove the existence of theconstant t(k) introduced in Problem 3. We

shall do it explicitly for k being a prime power, using the Baker-Schmidt Theorem

[3],

and

indicate extensions to non-prime power moduli aswell as to r-uniform hypergraphs.

We

shall

then consider in detailsomeparticular casesin which we were able tofurtherimprove uponour

general upper bounds.

In

order tokeep this introduction short andconcise weshallonlyremark thatournotationsfollow

[2]

andothers will bepresentedwhen needed.

2.

PARTITION THEOREMS AND PROOFS.

Westartwithanexact formulationsof themainresults.

TttEOREM

1. Letr, q>2be positive integers.

(1)

There exists aconstant t(q,r), dependingonlyon q and r, such that (H,q)< t(q,r),for

everyr-uniformhypergraph H.

(2)

Ifr 2, q 2kthent(q,2)=2q-1.

THEOREM2. Ifr 2,q io

n,

iois anodd prime, then

t(q,2)

<

(q-1)

-(2(q-

₄1)- 1)2

-t-.

9

THEOREM3. t(3,2) 3and4<t(5,2)<5.

Before proving the theorems, we shall need the Baker-Schmidt theorem and some recent result of

[1]

which isgiven hereinfulldetail inordertokeepthe paper self-contained.

THEOREM

B

(Baker-Schmidt

[3]).

Let q be a prime pow,er. If s>d(q-1)+l and

hl(Zl,

.,Zs),h2(Zl,

,Zs),

/.

,hl(Zl,

,Zs)

.

Z[Zl,

-,Xs] are polynomials satisfying

hl(0

hl(0

0, and

:

degh <_d, then there exists a binary vector a_#0 such that

hl(a

hl(a

=_O(rnodq).

Let

Zq

be the set ofintegersroodqconsideredasanadditive group.

THEOREM

C

(Alon

and

Caro

[1]).

Let q be a prime power, and let U=(V,E) by a

hypergraphof rankr

(the

sizeof thelargestedge)and let

f:EZq

bea

Zq-coloring

of E(H). Then thereisasubset Uof V, where U >

vI

r(q-1) such that < U> =O(modq). Thesameholds forqanon-primepowerprovided U >

IV

rq31og2q.

PROOF

OF

THEOREM

C.

For

q aprimepower, let mbe themaximuminteger such that thereexists a set M of m vertices ofH so that <M> _=_{O(modq). If} m>

IV

I-r(q-1) there is

nothing toprove. Otherwise, therearemorethan r(q-1) vertices in V\M. Define W V\M and let Fbe the set of all edgesofHthatcontain atleast onevertex ofW. Associateeach vertexw in

w

withavariable_{Zw and consider the}followingpolynomial equation

h=

y

f(e)

H

Zw O(mdq).

e.F w.e

699

is at most r, Theorem B (for t=1) implies that there is a nontrivial solution in which each

variable x w is either 0 or 1. Let W’ be the set of all the vertices for which

xw=

in this solution. Note that for those vertices, the polynomial h counts, mod q, the weight of the edges

added to <M>, when adding w’ to M. Thus, e< MUW’> O(rnod q), contradicting the maximality ofm. This proves theassertionof the theoremfor q aprime power. Forthe second partofthe theorem referto

[1].

13

PROOF

OF THEOREM1.

(1)

To prove thispart, weadopt agreedy strategy, namely, weshall trytodelete, ineach stage, the largest possibleinducedsubhypergraph whose number ofedges is divisiblebyq. Using the above result, if qis a prime power then H contains an induced subhypergraph G, such that [G _> H r(q- 1) and <//\G> <r(q-1). Thus, <H\G> can bedecomposed into subsets of order at most r-1, which contain no edges. Since H is r-uniform we conclude that in this

case, (q isaprimepower), t(q,r)<

+r(q-

_r-11)_--r-l"rq- Ifqisanon-primepower,thenthesame

rq31og2q

argumentyieldst(q,r)<

+

r-(2)

We already know by part

(1)

that t(q,2)<2q-1 for q=2

k.

It remains to present a graph that realizes the upper bound. Consider the complete graph

K2q

_1" Observe that for < <2q-1,

()

O(mod q), q 2

k.

Hencein adecomposition of

K2q_

each classconsists asingle vertexandweactuallyneed2q- classes.

Hence

t(2k,2)

2k

+

1. 13

Before proving Theorem2,weneedalemma.

A

vertez-coloring ofa graph G is a function g:V(G){I,2,...,k} in whichadjacent vertices

have different colors.

A

color-classisa class ofverticeshavingthesame color.

By

w w(G) we

denotethe clique-number ofG.

LEMMA. Suppose

G is agraph on n vertices. Let x be aminimum vertex coloring of G usingx(G) colors such that thenumber,

Xll,

of color classes ofsize one isas small aspossible. Then,

(i)

Eachvertex inany color classC,

ICI

>_3isadjacent to allverticesof

x

_1.

(ii)

At

leastonevertexin anycolor classC,

ICI

2isadjacenttoallverticesof X_1.

(iii)

IfG

n+

[XI[

n+w-1

(iv)

IfG_#Kntheneb(G,q)< ₂ < ₂

PROOF.

(i)

Suppose

therewerevertices C,

ICI

>3, and

w

x

suchthat (v,w). E(G). Thenwe cantake (v,w)as acolorclass, leaving(G)asbefore,but

xll

isreducedbyone, acontradiction tothe minimality of

(ii)

Let C {u,v}. If thereis avertex w X such that (w, u), (w, v) E(G) then thevertices

{w,u,v} formacolorclassreducingx(G).

Suppose

nowthat there areverticesw,z X such that (w,u),(z,v) E(G). Then wecantake {w,u} and {z,v}ascolorclasses,and again x(G)isreduced. This contradiction shows thatatleast oneofthevertices u,visadjacenttoall

x

_1.

(iii)

Clearly,

x

forms acomplete subgraph. IfG_#Knthen by

(i)

or

(ii),

we can always addsomevertex to

x

toform alarger complete subgraph.

Hence,

Ix

+

_<

n-

Xll

andforanyqq(G,q)<X(G) weobtainthe required result using

(iv)

Asx<

Xll-I

₂

(iii).

13

In

theproofof the

Lemma

wehaveused the trivial bound (G)<

+

Hence,

weraisethefollowing problem:

CARO, I. KRASIKOV AND Y. RODITTY

Therelevantconstant forourproblemsisabout half.

PROOF OF THEOREM2. Observe first that if qis anodd integer, then

()--O(nodq)

and

(q+l

₂

=-

O(modq). Consider a graph G and estimate (G,q). Supposewe have deleted the lgest subgraph sured by Theorem C. Then weareleft with acollection A of2q-l 2q-2 vertices. We may assume w.l.o.g, that

AI=

2q-2 by adding isolated vertices. If < A> contains a completegraph

Kt,

qthen wemaydeletea

Kq

from <A> and take all the remaining vertices as singles. We have at most 2+(2q-2)-q=q closes which is smaller than the stated bound.

Hence we may assume that thelgest cliquein <A> is of order at most q-1. Put a=q-1,

Suppose

thatwe havein <

a

>, vertex disjoint independent sets of order 3. Delete those sets

fom <A>, the resultinggraphis denoted A_1. Consideraminimum vertex coloring ofA in

theLemma. Callavertex vin

AleX

strongifit isadjacent toallverticesof

x

and let Ybe the

setof

MI

strongvertices. Put z

IX I.

Thus, (a

I)

+

w( <Y > ).

By

thelemma and the fact that A 2a-3k,wefind:

(1)

(G,q)k+2a-3k+l=2(a-k)+l.

Also,wehave thetrivialboundon

(2)

(G’q)<l+k+2a-3k+-

₂

=l+a+r-k2

Let

0(Y)

be thecardinalityof the mimal independent setin <Y>. Then,

O0(Y)

52since

independent sets of size 3 have been removed.

We

shM1 use a lower bound for

0(F)

obtned from Turk’s Theorem

(see

[2])

and the fact that < > istriglefr,nely,

(3)

O(Y

m,d}

Y

wheredis theaveragedegrin <Y>.

Hence,

w(<Y> Y

.

By (ii)

ofthe lemma,

YI

>2a-3k-z

Hence,

2 Rewrite

(4)

follows:

(5)

4,2

2,(4.

+

)

+

42

+ +

0. Then,either8a-24k-3<0,or

4a

+

+

(8a-24k 3)

(6)

₄

or _4a

+

_(Sa

24k

3)

(7)

,x

5

Observe first that if 8a-24k-3<0, that is

a<3+

then by

(2)

we obtn that

wch is obviously better th the und of the threm. Otherwise, since r₅a

(6),

must be ignored.

Now

toobtn the required result substitute

(7)

with 0 the

st

choice,into

(1).

MA

1.

(i)

It isey to

show,

by simpledeletionof independent

prs,

that

O(G,q)53(q

.1),

for q

d prime. Sothe result of Threm 2 isbetter th this triviMboundonlyforq>11.

(ii)

The

ave

undsc slightly improved usingaresult of Sheer

[9]

concerning the

indendent

number ofatrigle-frgraph. The computations aresimple but tedious.

PROOF

OF

THEOM

3. Observefirst that wemay sume before that

Mter

deleting thelgest subaph guested by Threm

C,

we eleK withexactly 2q-2 vertices. Sothat for q 3 we have 4 vertices that

remn.

A

direct checking of 1graphs on 4-vertices

reveMs

that M1of them e ptitionable into at most two inducedsubgraphs in which the number of

ZERO-SUM PARTITION THEOREMS FOR GRAPHS 701

The wheel

Ws:

shows thatindeed t(3,2) 3.

Thecase q 5 ismore complicatedsince wehavemore than 12,000

graphs

on 8 verticesand

hand-checking is tedious. Our goal is to show that for every graph G on 8 vertices (G,5)<4. Thiswould imply (5,2)_<5. Wesplit the proofintoseveralcases:

(i)

w>5.

In

this case,wearedonesince

e(KS)

10,

G\Ks]

<_3and (G,5) <4.

(ii)

<4.

Thenby

Lemma

1,

IX

l[

_<3.

(iia)

If

[Xl[

0by

(iv)

ofthe

Lemma

we aredone.

(iib)

If

[Xl[

then we have at most 7 vertices in at most 3 color classes with sizes at least2,sothat againwe aredone.

(iic)

If

Ix

2and wehaveacolor class ofsize >3,than takeby

Lemma,

a

K4\e

(with

edges)

and the rest of the

(at

most)

four vertices contain at least one color class ofsize 2 and again we are done. Otherwise, all remaining color classes areofsize 2. Thenis is known from the

Lemma

1 that thereis atleastonevertex ineach color class thatisadjacent toallverticesof

X_1. Letuandvbe suchverticesofdistinctcolor classes. If (u,v). E(G) thenwehavea

K4\e

and we are done. Otherwise all such vertices are adjacent and togetherwith X create a complete subgraphofsize5andweareincase

(i).

(iid)

If

[Xl[

3, theneitherthere is asingle color class ofsize5 andwe aredoneorthere

istwocolor classes each ofsizeatleast two. Again,it isknown from theLemmathat thereis at least onevertex in each color class that is adjacent to all vertices of X_1.

Let

u and v be such vertices. If (u,v). E(G) thenwehave a

K4\e

and wearedone. Otherwise < X

1U{u,v}

> induces

acomplete subgraphofsize5andwe areincase

(i).

4

Now to show that t(5,2)>_4, consider the graph

Ksn

+

_4" Take

V(Ksn

+

4)

=/

lAi

where,

[AI[

=5n+l and

[Ai[

1,i-2,3,4, and the number ofedges of the induced subgraphs on the vertex sets A isdivisibleby5. Thisgives

k(Ksn

+

4,5)

_<4.

In

order to show that

(Ksn

+ 4,5)

_>4,

and the orders of the setsarez _z2,z3, respectively. Then,

suppose that

V(KSn + 4)

lAi’

(8)

z

+

z

2

+

z3 5n

+

4.

On

theother

hand,

each of the numbers _Zl,Z2,z3, should have been0or (rood 5).

So

that

(8)

is

impossible. I:i

REMARK

2.

We

could not find a

graph

such that (G,5)=5.

Its

existence can give a

completeanswerin thatcase.

THEOREM

4.

Let

qbeanodd prime power and let G beagraph such that allitsinduced

subgraphsoforderatmost 2q-2 axeperfect graphs. Then (G,q)<_q.

In

particularthisholds for

perfectgraphs.

PROOF. By

Theorem C, G hasa subgraph H,[HI _> [G[ -2q

+

2 such that e(U) O(modq). Clearly [G\H[ _<2q-2. If G\Hcontainsaclique

Kt,

>_qtakea

Kq

and all the remainingvertices as singles to obtain a partition into at most q classes. Otherwise by perfectness, (G\U,q) x(G\H) w(G\U) _<q- whichyields againapartitioninto atmostqclasses.

I. KRASIKOV AND Y. RODITTY

THEOREM

5. LetGbeagraph. Then,

(i)

O(G,2) if[_{e(G)=_O(mod}2).

(ii)

<b(G,2) 2iff e(G)=__{l(mod 2) and there is}avertexofodddegree.

(iii)

O(G,2) 3iff e(G)=__{l(rnod 2)}andalldegreesare even.

PROOF.

(i)

Trivial.

(ii)

Clearly, if e(G)=_ l(mod 2) then (G,2)>_2 and if deg u=l(mod 2) then the required partitionisG\{u},{u}=(G,2)=2.

(iii)

As e(G)= l(mod 2)we have (G,2)>2.

We

will show that (G,2)=2is impossibleand

byt(2,2) 3wemusthave (G,2) 3.

Indeedsuppose b(G,2)=2 and let A,B bethevertex classes.

As

e<A> e<B> 0(rnod 2) and e(G)= l(mod 2) wemusthaveodd number ofedgesbetween A andB.

But

thisispossibleiff

anodd numberofvertices ofAhave anodddegree in <A> andfor each of them thereisodd

number ofedges connectingit to < B>.

But

this isimpossible sincein <A> therewill bean

oddnumberofverticesofodd degree. El

REMARK

3.

(1)

By

imitationof theproofof Theorem 2we canshowthat ifqisprimepower and -1

thent(q,r)<_q

+

r(qr_

)"

(2)

A

deep theorem of Kierstead

[7]

states that if G is both

K1,3-free

and

Ks\e-free

thenw(G)_<x(G)< w(G)+1.

Hence

using the arguments ofTheorem4 we obtainthat for such a graphGandfor qanodd prime power(G,q)<q

+

1.

Finally,wepose thefollowingconjecture.

CONJECTURE.

Ifq#

2/

then t(q,2)<q(1+o(I)).

ACKNOWLEDGEMENT. We

would like tothankthereferee for hiscareful reading and helpful remarks.

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1.

ALON, N.

& CARO,

Y., On

threezero-sum

Raxnsey-type

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B.,

Eztremal

Graph

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Press,

New York,

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R.C.

&

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W.M.,

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o

and 1,

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CARO,

Y.,

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I.

&

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H.A.,

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