Judicious partitions of uniform hypergraphs

0.12 MB·

by  John Haslegrave

Judicious partitions of uniform hypergraphs
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1
John Haslegrave ∗
0
2
January 23, 2017
n
a
J

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Abstract
2
The vertices of any graph with m edges may be partitioned into two

] partssothateachpartmeetsat least 2m edges. Bollob´as andThomason
O 3
conjectured that the vertices of any r-uniform hypergraph with m edges
C may likewise be partitioned into r classes such that each part meets at
. least r m edges. In this paper we prove the weaker statement that,
h 2r−1
for each r ≥ 4, a partition into r classes may be found in which each
t
a class meetsat least r m edges, a substantial improvementon previous
3r−4
m bounds.
[

1 1 Introduction
v
5
The vertices of any graph(indeed, any multigraph)may be partitionedinto
5
two parts, each of which meets at most two thirds of the edges ([4]; it also
8
5 appears as a problem in [2]). An equivalent statement in this case is that
0 each part spans at most one third of the edges. These two statements give
1. rise to different generalisations when a partition into more than two parts is
0 considered. In this paper we shall only address the problem of meeting many
7 edges; the problem of spanning few edges is addressed in [6] for the graph case
1 and [5] for the hypergraphcase.
:
v A particularly interesting case occurs when we partition the vertices of an
i r-uniform hypergraph into r classes. Bollob´as and Thomason (see [3], [7]) con-
X
jectured that every r-uniform hypergraph with m edges has an r-partition in
r
a which each class meets at least r m edges. The author [10] recently proved
2r−1
the conjecture for the case r =3.
The previous best known bound for each r >3 was provedby Bollob´as and
Scott [7], who in fact obtained a constant independent of r. The method they
usedwastoprogressivelyrefine a partitionby repartitioningthe verticesintwo
or three parts to increase the number of parts which met cm edges; in doing so
they showed that r disjoint parts meeting at least cm edges may be found for
c=0.27.
Our strategywill be to combine those ideas with the methods usedto prove
theconjecturedboundinthecaser =3[10]. Weshallobtainvaluesofc = r
r 3r−4
∗ResearchsupportedbytheEngineeringandPhysicalSciences ResearchCouncil
1
for r > 3, and so c → 1 as r → ∞. While this is a significant improvement
r 3
onpreviousbounds, itisstillsomedistancefromthe conjecturedbounds which
approach 1 in the limit.
2
All results obtained in this paper apply to hypergraphs in which repeated
edgesarepermitted, andinfactwe shallneed this extrageneralityinaninduc-
tion step.
2 New parts from old
In this section we give two lemmas which show that if we have a partition
inwhichsome partsmeetmany moreedges thanrequiredwe maylocallyrefine
thatpartitionandincreasethenumberofpartswhichmeettherequirednumber
of edges. The first of these lemmas was proved in [7]; the second is a new
result in the same spirit. The setting for eachlemma is the same. G is a multi-
hypergraph,butnotnecessarilyuniform;edgesmayhaveanynumberofvertices,
andedgesofanysizemayberepeated. Ghasmedges,butthemaximumdegree
of a vertex is less than cm where c>0 is an arbitrary constant. For A,B ⊂V
we write d(A) for the number of edges meeting A and d(A,B) for the number
meeting both A and B (we shall only use the latter notation where A and B
are disjoint).
Lemma 1 ([7]). Let c > 0 be a constant and let G be a multi-hypergraph on
vertexsetV with medges suchthat∆(G)<cm. IfAandB aredisjoint subsets
of V, each of which meet at least 2cm edges, then there is a partition of A∪B
into three parts, each of which meets at least cm edges.
Lemma 1 was proved in [7]; we provide their proof for completeness.
Proof. Wemayreplaceeachedgebyasubedgeifnecessarysothatnoedgemeets
eitherAorBinmorethanonevertex. WemaythenpartitionAintothreeparts,
A1,A2,A3, such that the union of any two meets at leastcm edges: we may do
this by taking A1 to be a maximalpartmeeting fewer than cm edges,and then
dividing A\A1 into two non-empty parts, since any set which strictly contains
A1 must meet at least cm edges and A2∪A3 = A\A1 meets d(A)−d(A1) >
2cm−cm = cm edges. We find a similar partition B = B1 ∪B2 ∪B3. It is
sufficient to find i,j such that d(A ∪B ) ≥ cm; then A ∪B ,A\A ,B \B
i j i j i j
will be a suitable partition. We shall show that this is always possible.
Since each edge meets A in at most one place then d(A) = d(A ) (and
Pi i
likewise d(B) = d(B )). Similarly, if an edge meets both A and B then we
Pi i
may find unique i,j for which it meets A and B . Therefore,
i j
d(A,B)=Xd(Ai,Bj).
i,j
Also,
d(A ∪B )=d(A )+d(B )−d(A ,B ),
i j i j i j
2
so
Xd(Ai∪Bj) = X(d(Ai)+d(Bj)−d(Ai,Bj))
i,j i,j
= 3d(A)+3d(B)−Xd(Ai,Bj)
i,j
= 3d(A)+3d(B)−d(A,B)
≥ 3d(A)+2d(B)≥10cm.
Since there are nine terms in the LHS, at least one must exceed cm.
Thisproofshowsmorethanrequired: wecanalwaysfindapartitionofA∪B
into three parts, two meeting at least cm edges and the third meeting at least
10cm. However,we cannotalwaysfind apartitionintothree parts,twomeeting
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morethancm+1edgesandthethirdmeetingatleastcm,asseenbyconsidering
the case where each of A and B have three vertices, two meeting cm−1 edges
and one meeting 2 edges.
A part which does not meet the desired number of edges can also be useful
providedwehaveanotherpartmeetingsufficientlymanyedgestocombinewith
it.
Lemma 2. Let c > 0 be a constant and let G be a multi-hypergraph on vertex
set V with m edges such that ∆(G)<cm. If A and B are disjoint subsets of V,
with d(A) ≥ 2cm and d(A)+2d(B) ≥ 3cm, then there is a partition of A∪B
into two parts, each of which meets at least cm edges.
Proof. Asbefore,wemayreplaceeachedgebyasubedgeifnecessarysothatno
edgemeetseitherAorB inmorethanonevertex. Wemaythenagainpartition
A into three parts A1,A2,A3 such that the union of any two meets at least cm
edges. Itisnowsufficienttofindisuchthatd(A ∪B)≥cm;thenA ∪B,A\A
i i i
will be a suitable partition. We claim this is always possible.
Again,
d(A,B)=Xd(Ai,B)
i
and
d(A ∪B)=d(A )+d(B)−d(A ,B),
i i i
so
Xd(Ai∪B) = X(d(Ai)+d(B)−d(Ai,B))
i i
= d(A)+3d(B)−Xd(Ai,B)
i
= d(A)+3d(B)−d(A,B)
≥ d(A)+2d(B)≥3cm.
Since there are three terms in the LHS, at least one must be at least cm.
3
Inparticular,wemayfindsuchapartitionwhend(A)≥2cmandd(B)≥ cm;
2
this is the only case in which we shall apply Lemma 2.
3 The bounds
Throughout this section G is an r-uniform multi-hypergraph on vertex set
V with m edges. Our goalis to prove that there is a partitionof V into r parts
such that each part meets at least c m edges for some suitable constant c . In
r r
order to use Lemmas 2 and 1 we need to reduce to the case ∆(G) < c m. To
r
thatendwe apply inductiononr. If somevertexmeets atleastc m edges then
r
we may removethat vertex,replacing eachedge of G by a subedge of size r−1
not containing that vertex, and then use the result for r−1 to partition the
remaining vertices into r−1 parts, each meeting at least cr−1m edges; this is
sufficient so long as cr−1 ≥cr, which will be the case. We shall take c2 = 32, so
the case r =2 is known.
Our plan will be to look at partitions for which d(V ) is large, and show
Pi i
that if some parts meet too few edges then there are enough parts meeting at
least 2c m edges to allow us to construct a good partition by combining parts
r
as above.
Certainlyanypartitionwhichisoptimalinthesenseofmaximising d(V )
Pi i
alsosatisfiesthelocaloptimalityconditionthat d(V )cannotbeincreasedby
Pi i
moving a single vertex. We shallconsider particularlythe yet weakercondition
that d(V ) cannot be increased by moving a single vertex into V . We begin
Pi i r
by establishing bounds on partitions satisfying this condition.
Lemma 3. Let V1,V2,...,Vr be a partition for which Pid(Vi) cannot be in-
creased by moving a vertex into V . Then
r
r
Xd(Vi)≥(r+1)m−rd(Vr).
i=1
Proof. For eachi<r and eachv ∈V , since movingv into V does notincrease
i r
the sum, the number of edges e such that e∩V = {v} must be at least the
i
number of edges e containing v which do not meet V : the first quantity is the
r
decrease in d(V ) effected by moving v and the second is the increase in d(V ).
i r
Thus
X X |{e:e∩Vi =v}|≤X X |{e∋v :e∩Vr =∅}|.
i6=rv∈Vi i6=rv∈Vi
Since, for v 6=w, {e:e∩V =v} and {e:e∩V =w} are disjoint,
i i
X X |{e:e∩Vi =v}| = X| [ {e:e∩Vi =v}|
i6=rv∈Vi i6=r v∈Vi
= X|{e:|e∩Vi|=1}|.
i6=r
4
For 1 ≤ i < r, let E = {e : |e∩V | = 1}. For each edge e write f(e) for the
i i
number of parts (including V ) which meet e. If f(e) < r then |e∩V | > 1 for
r i
some i, and so e is in at most f(e)−1 of the E ; trivially if f(e)= r then e is
i
in at most r−1=f(e)−1 of the E . Thus
i
X|{e:|e∩Vi|=1}| ≤ X(f(e)−1)
i6=r e
r
= Xd(Vi)−m.
i=1
Also,
X X |{e∋v :e∩Vr =∅}|=r(m−d(Vr)),
i6=rv∈Vi
since each edge not meeting V is counted exactly r times in the sum, once for
r
each vertex it contains. Combining the above relations, we see that
r
Xd(Vi)−m≤r(m−d(Vr)),
i=1
as required.
The reason for considering the condition that d(V ) cannot be increased
Pi i
by moving a single vertex into V is that, as we shall see, it is preserved by
r
moving vertices into V . If we are able to start from a partition which satisfies
r
the condition and in which V1,...,Vr−1 are “good” (in the sense of meeting at
leastacertainproportionofedges)then we cantry to improvethe partitionby
moving vertices into V while keeping the other parts good. In this way we will
r
either obtain a partition into r good parts or we will be forced to stop because
V1,...,Vr−1 areallminimalgoodsets. Weformalisetheseideasinthefollowing
lemma.
Lemma 4. Let V1,V2,...,Vr be a partition for which Pid(Vi) is as large as
possible, ordered such that d(V1) ≥ d(V2) ≥ ··· ≥ d(Vr), and let c > 0 be a
constant. If d(Vr−1) ≥ cm then either there exists a partition W1,W2,...,Wr,
with W ⊆V for 1≤i≤r−1 (and so W ⊇V ), such that each part meets at
i i r r
least cm edges, or there exists a partition W1,W2,...,Wr, again with Wi ⊆ Vi
for 1≤i≤r−1, such that, for each i6=r, d(W )≥cm but d(W \{w})<cm
i i
for any w ∈W , and also
i
r−1
Xd(Wi)>(r+1)(m−cm).
i=1
Proof. Suppose U1,U2,...,Ur is a partition with Ui ⊆ Vi for each i < r and
so U ⊇ V . For any v ∈/ U , say v ∈ U , let U′ = U \{v}, U′ = U ∪{v},
r r r i i i r r
5
V′ =V \{v} and V′ =V ∪{v}. Then
i i r r
d(U )−d(U′) = |{e:e∩U ={v}}|
i i i
≥ |{e:e∩V ={v}}|
i
= d(V )−d(V′)
i i
and
d(U′)−d(U ) = |{e∋v :e∩U =∅}|
r r r
≤ |{e∋v :e∩V =∅}|
r
= d(V′)−d(V )
r r
so
d(U′)+d(U′) ≤ (d(U )+d(V′)−d(V ))+(d(V′)−d(V )+d(U ))
i r i i i r r r
= d(U )+d(U )+(d(V′)+d(V′)−d(V )−d(V ))
i r i r i r
≤ d(U )+d(U ),
i r
meaningthatU1,U2,...,Ur alsosatisfiestheconditionthatPid(Ui)cannotbe
increased by moving a vertex into U .
r
Foreachi<r,then,letW beaminimalsubsetofV satisfyingd(W )≥cm,
i i i
and let W =V \ W . If d(W )≥cm then this is a suitable partition with
r Si<r i r
each part meeting at least cm edges; if not then Lemma 3 ensures that
r−1
Xd(Wi) ≥ (r+1)m−rd(Wr)−d(Wr)
i=1
> (r+1)(m−cm),
as required.
We are now ready to prove the main result. We shall show that if we start
from a partition maximising d(V ) then either we have enough elbow room
Pi i
to obtain a good partition by repeated application of Lemmas 1 and 2 or we
may use Lemma 4 to obtain a good partition.
Theorem 5. Let c2 = 32, c3 = 59 and cr = 3rr−4 for r >3. If G is an r-uniform
multi-hypergraph with m edges there is a partition of the vertex set into r parts
with each part meeting at least c m edges.
r
Proof. Weuseinductiononr;thecaser =2isknown. Forr >2,ifsomevertex
v meets atleastc m edgesthen we may,by replacingeachedge witha subedge
r
of size r−1 not containing v and applying the r−1 case, find a partition of
the other vertices into r−1 parts each meeting at least cr−1m > crm edges.
Together with {v}, this is a suitable r-partition. Thus we may assume that no
vertex meets c m edges.
r
NowletV1,V2,...,Vr beapartitionforwhichPid(Vi)isaslargeaspossible,
ordered such that d(V1) ≥ d(V2) ≥ ··· ≥ d(Vr). If d(Vr) ≥ crm then we are
6
done, so we may assume d(V ) < c m. We consider three cases based on the
r r
values of d(Vr−1) and d(Vr).
Case 1. d(Vr−1)≥crm.
By Lemma 4, either we have a good partition or we may find a partition
W1,W2,...,Wr such that, for each i < r, d(Wi) ≥ crm but Wi is minimal
for this property, and further that r−1d(W ) ≥ (r+1)(m−c m). Suppose
Pi=1 i r
the partition found is of the latter type. For each i < r, since ∆(G) < c m,
r
|W | > 1. Let v,w be two vertices in W ; by minimality of W the number of
i i i
edges meeting W only at v is more than d(W )−c m, and so is the number
i i r
meeting W only at w. These sets of edges are disjoint from each other and
i
from the set of edges meeting Wi in more than one vertex; thus, writing d2(X)
for the number of edges meeting X in more than one vertex,
d(Wi)>2(d(Wi)−crm)+d2(Wi),
i.e.
d(Wi)+d2(Wi)<2crm.
However,eachedge not meeting W meets atleastone other partatmore than
r
one vertex, so
r−1
Xd2(Wi)>m−crm.
i=1
r−1
Combining this with the bound on d(W ) gives
Pi=1 i
r−1
X(d(Wi)+d2(Wi))>(r+2)(m−crm),
i=1
so for some i<r
r+2
d(Wi)+d2(Wi)> (m−crm).
r−1
Consequently, using our upper bound on d(Wi)+d2(Wi),
r+2
(m−c m)<2c m
r r
r−1
⇒ (r+2)(1−c )<2c (r−1)
r r
⇒ r+2<3rc
r
and so c > r+2.
r 3r
For r =3, c = 5 = r+2; for r ≥4,
r 9 3r
r
c =
r
3r−4
r+2 2−8r
= −
3r 3r(3r−4)
r+2
≤ ;
3r
7
in either case a contradictionis obtained(from our assumptionthat we did not
get a good partition from Lemma 4). (End of Case 1.)
Note that, using Lemma 3, if d(V )<c m then
r r
r r−2
d(Vr−1)−crm = Xd(Vi)−Xd(Vi)−d(Vr)−crm
i=1 i=1
r
> Xd(Vi)−(r−2)m−2crm
i=1
≥ (r+1)m−rd(V )−(r−2)m−2c m
r r
> (r+1)m−rc m−(r−2)m−2c m
r r
= 3m−(r+2)c m,
r
sofor r =3, since c < 3 , Case1 is the only possible case. Forthe remaining
r r+2
cases, then, we assume r≥4.
Case 2. crm>d(Vr−1)≥d(Vr)≥ cr2m.
Suppose that k parts meet at least 2c m edges, l meet fewer than c m, and
r r
the remainingr−k−l meetatleastc mbutfewerthan2c m. UsingLemma2
r r
wemaycombineapartmeeting atleast2c medgesandapartmeetingatleast
r
crm to producetwo partsmeeting atleastc m; we know,since V meets fewest
2 r r
edges, that eachpart meets at least crm and so we can obtain a good partition
2
provided k ≥l. We shall show that this must be so.
Since r ≥4, 2cr ≤1, and since crm>d(Vr−1), l≥2. So
r
Xd(Vi) < 2crm(r−k−l)+crml+mk
i=1
= 2c mr−c lm+(1−2c )km.
r r r
However,by Lemma 3,
r
Xd(Vi)>(r+1)m−rcrm,
i=1
so
2rc −lc +k(1−2c )>(r+1)−rc ;
r r r r
however,if k <l then
2rc −lc +k(1−2c ) ≤ 2rc −lc +(l−1)(1−2c )
r r r r r r
= (2r−1)c +(l−1)(1−3c ).
r r
Since l≥2 and 1−3c <0,
r
(2r−1)c +(l−1)(1−3c ) ≤ (2r−4)c +1
r r r
= r+1−rc
r
8
(the final equality follows since c = r ), and a contradiction is obtained.
r 3r−4
Case 3. crm>d(Vr−1) and cr2m >d(Vr).
Supposethatj partsmeetatleast2c medges,kmeetatleast crm butfewer
r 2
than c m, l meet fewer than crm, and the remaining r−j−k−l meet at least
r 2
c m but fewer than 2c m. Using Lemma 2 k times and Lemma 1 l times, we
r r
mayfindr disjointpartswhichmeetatleastc medgesprovidedthatj ≥k+2l
r
(each application of Lemma 2 requires one part which meets 2c m edges and
r
eachapplicationofLemma1requirestwo). We shallshowthatthis mustbe so.
From Lemma 3, since d(V )< crm and r ≥4,
r 2
r
1 d(V ) r+1 c
i r
X > −
r m r 2
i=1
r+1 5c
r
= 2c + −
r
r 2
r2−2r−8
= 2c +
r
r(6r−8)
≥ 2c ;
r
similarly, since c ≤ 1,
r 2
r
1 d(V ) r+1 c
i r
X > −
r m r 2
i=1
3
>
4
2 c
r
≥ + ;
3 6
so, since c > 1 for every r, Lemma 6, following, ensures that j ≥k+2l.
r 3
We now give the result needed to fill in the gap.
Lemma 6. Let 1 ≤c≤ 1. If wehave afinite, non-emptycollection of numbers
3 2
in [0,1], whose arithmetic mean a is at least max 2c,2 + c , of which j are at
(cid:0) 3 6(cid:1)
least 2c, k are at least c but less than c, l are less than c, and the rest are at
2 2
least c but less than 2c, then j ≥k+2l.
Proof. We proceed by induction on k+l; if k = l = 0 then there is nothing
to prove. If k ≥ 1 then one of our numbers is less than the mean, so another
must exceed the mean; since the mean of those two numbers is at most 1+c ≤
2
2c ≤ a, either there are no other numbers (in which case we are done) or the
mean of the remaining numbers is at least a and we are done by induction. If
l ≥ 1 then one number is less than c; if also j ≤ 1 then the total is less than
2
1+ c +(n−2)2c < 2cn, so there must be two numbers which are at least 2c.
2
Since the mean of these three numbers is at most 2 + c ≤ a, either there are
3 6
no other numbers (in which case we are done) or the mean of the remaining
numbers is at least a and again we are done by induction.
9
The value of c which we use is tight only in the second case of Theorem 5
r
(for r > 4); if we were to use instead some c′ > c it would be feasible for two
r r
of our parts to meet just under c′ edges while the remaining parts each met
r
just under 2c′ edges, giving us no immediate way to use Lemmas 1 and 2. To
r
improve further on these bounds using a similar method, then, we might seek
to proveanaloguesof Lemmas 1 and2 for parts whichmeet morethan βcm for
some 1 < β < 2. To do this, however, we would need a way to impose some
stronger assumption on the degrees of vertices than simply ∆(G)<cm.
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