Discrete Splittings of the Necklace

(1)

Frédéri Meunier

∗

Deember 1,2007

Abstrat

ThispaperdealswithdiretproofsandombinatorialproofsofthefamousNeklae

theoremofAlon,GoldbergandWest. Thenewresultsareadiretprooffortheaseof

twothievesandthreetypesofbeads,andaneientonstrutiveproofforthegeneral

asewithtwothieves. Thislastproof usesatheoremofKyFanwhihisaversionof

Tuker'slemmaonerningubialomplexesinsteadofsimpliialomplexes.

KeyWords: onstrutiveproof;ubialomplex;splittingneklaes.

1 Introdution

A lassialappliation oftheBorsuk-Ulamtheoremis theSplitting Neklaetheorem (see

the book of Matou²ek [10℄). Provedrst by Goldberg andWest in 1985 [9℄, this theorem

got asimplerproof in 1986,foundbyAlon andWest andusing theBorsuk-Ulamtheorem

[4℄.

Twothieveshavestolenapreiousneklae,whihhas

n

^beads. ^These ^beads ^belong ^to

t

^dierent^types. Âssume ^that^thereîsânêven^numberôf^beadsôfêah^type,^say

2a i

^beads

of type

i

^, ^for ^eah

i ∈ {1, 2, . . . , t}

^, ^where

a i

îs â ^nonzero înteger. ^Remark ^that ^we ^have

2 P t

i=1 a i = n

^.

Thebeads are xed onan openhainmade of gold. An exampleisgivenin Figure 1.

Thedierenttypesareenodedbythenumbers1,2,3.

Aswedonotknowtheexatvalueofeahtypeofbeads,afairdivisionoftheneklae

onsistsofgivingthesamenumberofbeadsofeahtypetoeahthief. Thenumberofbeads

of eah type is even, hene suh adivision is alwayspossible: utthe hain at the

n − 1

possiblepositions. But thehain is made ofgold! It isa shamethat wedamageit. Isn't

it possibletodo thedivision withlessuts? Theansweris yes, asshownby thefollowing

theorem provedbyGoldbergandWest:

Theorem 1 Afairdivisionof theneklaewith

t

^types^of^beads ^between^two^thieves^an^be

done with nomorethan

t

^uts.

In 1987, Alon proved the following generalization, for aneklaehaving

qa i

^beads ^for

eahtype

i

^,

a i

înteger,ûsing âgeneralizationoftheBorsuk-Ulamtheorem [1℄:

Theorem 2 A fair division of the neklae with

t

^types ^of ^beads ^between

q

^thieves ^an ^be

done with nomorethan

t(q − 1)

^uts.

∗

LVMT,EoleNationaledesPontsetChaussées,6-8avenueBlaisePasal,CitéDesartesChamps-sur-

Marne,77455Marne-la-Valléeedex2,Frane. E-mail:frederi.meunierenp.fr

(2)

Figure1: Anopenneklae,with12beads,of3dierenttypes,tobedividedbetweenAlie

and Bob.

2 1 2 1 3 1 3 1 3 2 3 2

.

. .

. . . . . . . . .

.

N

^ ?

Alie Bob Alie Bob

Figure2: Afairdivision betweenAlieandBob.

For

q = 2

^,^this^last^theorem^oinides^with^Theorem^1. Furthermore,thisboundistight, evenwhen

a 1 = a 2 = . . . = a t = 1

^: ônsider^the^neklae^for^whihâll^the^beadsôf^the^rst

typeome rst,then allbeads oftheseond type,thenallbeads ofthethird type,andso

on.

Exept for partiular ases, all known proofs of this theorem are topologial and use

ontinuous tools. Setion 2is devoted to diret and non-topologial proofs for partiular

ases ofthese theorems: oneonthem isthe rstproof ofthis typeforthe ase

q = 2

^and

t = 3

^.

In Setion 3, we give apurely ombinatorial 1

and onstrutive proof for the omplete

ase

q = 2

^(although^the^topologialinspirationremainsnotable).

It uses a ubial version of Tuker's lemma found by Ky Fan (Theorem 3), involving

ubial omplexes instead of simpliial omplexes. We give a onstrutive proof of this

theorem (suh aproof wasnot known before thepresent paper; nevertheless, ourproof is

inspiredbytheonstrutiveproofgivenbyPresottandSuforaombinatorialgeneralization

of Tuker'slemmafound byKyFan, [13℄)and showthat it suitsto theneklaeproblem.

It givesaneientalgorithmforthedivisionoftheneklaebetweentwothieves: ontrary

to the known algorithm, our result needs no approximation or rounding method and no

arbitrarilynetriangulationoftheross-polytope. Furthermore,the

d

^-ubesôfôurûbial

omplexhaveasimpleinterpretation: eah

d

^-ube^orresponds^to^the^hoie^of

d

^beads. ^We

will also see that our method gives also an almost-fair division of the neklae between

two thieves whenthe numberof beads in any typeis not neessarilyeven(2-splittings in

theterminologyofAlon,Moshkovitzand Safra[3℄).

2 Diret proofs

2.1 A diret proof for

t ≤ 2

and any

q

This proofwasrstgivenin thepaperbyEpping,HohstättlerandOertel[7℄. This paper

deals with dierent aspets of a paint shop problem: one of the ases orresponds to the

Neklae theorem presented as aonjeture (Conjeture 10) by the authors who did not

1

AordingtoZiegler[15℄,a ombinatorialproofinatopologialontextisaproof usingno simpliial

approximation,nohomology,noontinuousmap.

(3)

theneklae,see [11℄.

If

t = 1

^,î.e.,îf^the^beadsâreônlyôf ône^type,^the^solutionîsstraightforward: splitthe neklaein

q

^parts^of^the^same^size,^whih^needs

q − 1

^uts^(and^has^omplexity

O(q)

^).

If

t = 2

^: ^the ^proof^works^by ^indution ^on

q

^. ^F^or

q = 1

^, ^there^is ^nothing^to ^prove. ^Let

usnowsupposethat

q ≥ 2

^. ^We(virtually)splittheneklaeinto

q

^disjointsub-neklaesof

a 1 + a 2

ônseutive^beads;^the^neklaeîsôpen ândhorizontal;thebeadsgetthenumbers 1to

n

^,^from ^left^to^right.

If oneofthese

q

sub-neklaesontains

a 1

^beads ^of ^type ¹^and

a 2

^beads ^of ^type^2, ^we

removethissub-neklae,andgiveitto oneofthethieves,andbyindution,weknowthat

thereisafairsplittingoftheremainingin

2(q −2)

^uts;^all^together,^this^gives

2 + 2(q −2) = 2(q − 1)

^uts.

Hene, we an assume that there is no suh sub-neklae among the

q

sub-neklaes.

Thus, thereisatleastonesub-neklaewhihontainsstritlymorethan

a 1

^beads^of^type

1,andanotherwhihhasstritlylessthan

a 1

^beads^of^type^1. ^Let's^all^the^rst^sub-neklae

M ⁺

^and^the^seond ^one

M ⁻

^.

Ifweslideawindowofsize

a 1 + a 2

^over^the^neklae,^starting^at

M ⁻

^until^we^reah

M ⁺

using movesofexatlyonebead,thewindowwillontain

a 1

^beadsôf ^type¹ât ^leastône.

The

a 1 + a 2

^beadsôf^the^windowân^be^given^toôneôf^the^thieves,ând^then^byîndution,

asbefore,

2 + 2(q − 2) = 2(q − 1)

^uts^are^suient.

The time omplexity an be omputed as follows: eah bead reeives the number of

beadsofeahofthetwotypespreedingitontheneklae(thebeaditselfinluded),whih

anbedonein

O(n)

operations: so,whenweonsiderasub-neklae,weknowimmediately howmanybeadsofeahtypeareinthissub-neklae;adihotomyallowsustondasub-

neklaehaving

a 1

^beads^of^type¹^and

a 2

^beads^of^type²ⁱⁿ

O(log n)

^. ^The^total^omplexity

is thus:

O(n + q log n)

^.

2.2 A diret proof for the ase

a ₁ = a ₂ = . . . = a t = 1

, any

t

^and ^any

q

Weonsider the beads oneafter theother; from left to right. Wealwayshavea urrent

thief. Whenonsideringthe

k

^th^bead,^we^hekîf^theûrrent^thief^hasâ^beadôf^this^type

ornot.

Ifhedoes: wehangetheurrentthief andtakeathiefnothavingabeadof thistype,

wegivethebeadtothisnewurrentthief andgotothenextbead.

Ifhedoesnot: wegivethebeadto theurrentthief andgotothenextbead.

Thisproedureensuresadivisionin atmost

t(q − 1)

^uts: ^indeed, ^we^never^hange^the

urrent thief when wemeet the rst bead of a given type. A hange of the urrentthief

is preisely aut ofthe neklae. Thenumberof utsis thus less thanorequalto

n

^, ^the

number of all beads, minus

t

^, ^the ^number ôf ^rst ^meetings ôf â ^type ôf ^beads: ⁱⁿ ^total:

n − t = qt − t = t(q − 1)

^.

Moreover,thisalgorithmisgreedyandhasomplexity

O(n)

^.

2.3 A diret proof for

t = 3

and

q = 2

2.3.1 Main steps

The neklae is identied with the interval

]0, n[⊂ R

^. ^The ^beads ^are ^numbered ^with ^the

integers

1, 2, . . . , n

^,^from^left^to^right. ^The

k

^th^beadôupiesûniformly^theînterval

]k−1, k[

^.

For

S ⊆]0, n[

^, ^we^denote ^by

φ i (S)

^the^quantity ^of^beads ^of^type

i

ⁱⁿ ^positions ^inluded

in

S

^. ^More^preisely,^we^denote ^by

1 ⁱ (u)

^the^mapping^whihîndiates îf^thereîs â^beadôf

(4)

type

i

^at ^point

u

^of

]0, n[

^:

1 ⁱ (u) =

1

^if^the

⌈u⌉

^th^bead^is^of^type

i 0

^if^not.

Wehavethen

φ i (S) :=

Z

S

1 ⁱ (u)du.

If there is a window of

n/2

^beads înduing â ^fair ^division, ^there îs ^nothing ^to ^prove.

Thus, we anassumethat there is no

x

^suh ^that

φ i (]x, x + n/2[) = a i

^for ^all

i = 1, 2, 3

^.

Moreover,weanalsoassumethat

φ 1 (]0, n/2[) > a 1

^. ^These^twoassumptionsarealledthe starting assumptions.

First, we unite the types 2 and 3 in a single type, alled type 2'. Hene, we dene

φ 2 ^′ := φ 2 + φ 3

^.

Then,webuildagraph

G

^,^whose^verties^are^triples

(c 1 , c 2 , c 3 )

^,^suh^that

• c 1

^,

c 2

^and

c 3

^are^integers,

• 0 ≤ c 1 ≤ c 2 ≤ c 3 ≤ n

^,

• φ 1 (]0, c 1 [∪]c 2 , c 3 [) = a 1

^and

φ 2 ^′ (]0, c 1 [∪]c 2 , c 3 [) = a 2 + a 3

^.

Distint triples

(c 1 , c 2 , c 3 )

^and

(d 1 , d 2 , d 3 )

^are ^neighbors ⁱⁿ

G

^if

|c i − d i | ≤ 1

^for ^all

i = 1, 2, 3

^.

Vertiesof

G

ôrrespond^thus^to ûtsôf^the ^neklae^giving^the^same^numberôf ^beads

to eah thief,and dividingfairly thebeads oftype1. Withoutlossofgenerality,wedeide

that Aliegetsthepart

]0, c 1 [∪]c 2 , c 3 [

^and^Bob ^the^remaining.

Remarkthatthestartingassumptionimpliesthatthereisnovertexoftheform

(0, c 2 , n)

^.

Inthenextparagraph,thefollowinglemmaisproved:

Lemma 2.1 Thefollowing holdsunder thestarting assumptions:

(i) Theverties

(c 1 , c 2 , c 3 )

ôf ôdd^degree ^haveêither

c 1 = 0

^or

c 3 = n

^.

(ii) Thenumberofvertiesofodddegreesuhthat

c 1 = 0

îdôddândêquals^the ^numberôf

vertiesofodd degreesuhthat

c 3 = n

^.

Theendoftheproofisthenstraightforward. Wesplitthesetofvertiesofodddegrees

of

G

^into ^two^disjoint ^subsets

A

^and

B

^:

A

îs ^the ^set ôf ^vertiesôf ôdd^degrees ^suh ^that

φ 3 (]0, c 1 [∪]c 2 , c 3 [) < a 3

^and

B

^the^setôf^vertiesôfôdd^degrees^suh^that

φ 3 (]0, c 1 [∪]c 2 , c 3 [) >

a 3

^(aording ^to^the^startingassumptions,there isnovertex

(c 1 , c 2 , c 3 )

^of ^odd^degree^suh

that

φ 3 (]0, c 1 [∪]c 2 , c 3 [) = a 3

^). ^Hene

|A| = |B|

ând ^this ^number îsôdd. ^This împlies^that

there isa pathin

G

^that ^links ^a ^vertex ⁱⁿ

A

^to ^a^vertexⁱⁿ

B

^.

φ 1

^and

φ 2 ^′ = φ 2 + φ 3

^are

onstanton

G

^. ^Moreover,^the^value^of

φ 3

^for^twoâdjaent^verties^diers^byât ^mostône.

Hene,weneessarilyhaveavertexonthis pathsuhthat

φ 1 = a 1

^,

φ 2 = a 2

^and

φ 3 = a 3

^.

2.3.2 Proof ofLemma2.1 (

t = 3

^and

q = 2

⁾

Intheproof, wedistinguishtheverties

(c 1 , c 2 , c 3 )

âording^to^the^typeôf^beadsâdjaent

to thepoint where the neklaeis ut. Below, therst

|

^represents

c 1

^, ^the^seond ^one

c 2

andthethird one

c 3

^. ^The^numbers^on^both^sides^of^the

j

^th^line

|

^show^the^type^of^the^bead

whihisontheleftsideofpoint

c j

ând^the^typeôn^the^right^sideôf^point

c j

^. ^F^or^instane:

. . . 1

|{z}

Alie

| 2 ^′ . . . 2 ^′

| {z }

Bob

| 1 . . . 1

| {z }

Alie

| 2 ^′ . . .

| {z }

Bob

.

(5)

This vertexis ofdegree0,beausenomoveofa

|

ân^beômpensated^by ânother^moveⁱⁿ

order to maintainthe numberof beads of types1 and2' reeivedby eah thief. Another

exampleisthefollowing(where

c 1 = 0

^):

|1 . . . 1|1 . . . 1|2 ^′ . . . .

Thedegreeofsuhavertexequals3. Indeed,thereare3neighbors:

|1 . . . |11 . . . |12 ^′ . . . , 1| . . . 11| . . . 1|2 ^′ . . .

and

1| . . . 1|1 . . . |12 ^′ . . .

Inanyofthese threeases,themoveofautisompensatedbyanothermove.

Let us ompute the degree of a vertex

(c 1 , c 2 , c 3 )

ⁱⁿ ^general. ^Three ^ases ^have ^to ^be

distinguished

1.

c 1 = 0

^: ^the ^vertex^is ^of ^the^form:

|a . . . b|c . . . d|e . . .

^with

a, b, c, d, e ∈ {1, 2 ^′ }

^. ^F^our

subaseshavetobedistinguished,thehekingbeingeasy:

•

^If

b 6= e

^and

c = d

^,^the^degree^is¹^if

a 6= c

^and³^if^not.

•

^If

b 6= e

^and

c 6= d

^,^the^degree^is¹^if

b = c

^and³^if^not.

•

^If

b = e

^and

c = d

^, ^the ^degree^is ⁴^if

a = b = c

^, ² ^if

a 6= b

^, ^and ⁰^if

a = b

^and

b 6= c

^.

•

^If

b = e

^and

c 6= d

^,^the^degree^is^2.

2.

c 3 = n

^: ^thisâseân ^be^treatedêxatlyâs^theâse^before.

3.

c 1 > 0

^and

c 3 < n

^: ^we ênumerate ^subases ^while âvoiding ^subases ^whih ân ^be

dedued from theothers bysymmetries ortheexhange of 1and 2': wegiveonlya

fewsubases,forthehekingisalwaysstraightforward:

•

^Subase

. . . 1|1 . . . 1|1 . . . 1|1 . . .

^: ^the^degree^equals^6.

•

^Subase

. . . 1|1 . . . 1|1 . . . 1|2 ^′ . . .

•

^Subase

. . . 1|1 . . . 1|1 . . . 2 ^′ |1 . . .

•

^Subase

. . . 1|1 . . . 1|1 . . . 2 ^′ |2 ^′ . . .

^: ^the^degree êquals^2. Îndeed, ^there âreêxatly

twoneighbors:

. . . 11| . . . 11| . . . 2 ^′ |2 ^′ . . .

^and

. . . |11 . . . |11 . . . 2 ^′ |2 ^′ . . .

^.

•

^Subase

. . . 1|1 . . . 1|2 ^′ . . . 2 ^′ |1 . . .

•

^Subase

. . . 1|1 . . . 2 ^′ |1 . . . 1|2 ^′ . . .

•

^Subase

. . . 1|1 . . . 1|2 ^′ . . . 2 ^′ |2 ^′ . . .

^: ^the^degreeêquals^2. Îndeed,^there âreêxatly

twoneighbors:

. . . 1|1 . . . 12 ^′ | . . . 2 ^′ 2 ^′ | . . .

^et

. . . |11 . . . |12 ^′ . . . 2 ^′ |2 ^′ . . .

^.

•

^Subase

. . . 1|1 . . . 2 ^′ |1 . . . 2 ^′ |2 ^′ . . .

•

^Subase

. . . 1|2 ^′ . . . 2 ^′ |1 . . . 1|2 ^′ . . .

•

^Subase

. . . 1|2 ^′ . . . 1|2 ^′ . . . 2 ^′ |1 . . .

Alldegreesareeven.

(6)

It followsthat a vertex of odd degreeis neessarily suh that

c 1 = 0

^or

c 3 = n

^. ^This

provespoint(i) ofLemma 2.1. Notethat avertex hasan odd degreeifand only if

b 6= e

(withthenotationsdenedin therstaseabove).

Weprovenowthat: Thereisan odd numberof vertiesof odd degree suhthat

c 1 = 0

^.

Let

ǫ > 0

^be â^very^small ^real. ^We ^rst ^show ^that â ^vertex ^has ân ôdd ^degreeîf ^the

followingholds: oneof theintervals

]c 2 − 1, c 3 − 1 + ǫ[

^and

]c 2 , c 3 + ǫ[

^has^a^proportion ^of

beads of type1stritly smallerthan

a 1

a 1 +a 2 +a 3

and theother hasaproportionof beads of

type1stritlygreaterthan

a 1

a 1 +a 2 +a 3

. Formally,wearegoingto showthat theoddverties

(0, c 2 , c 3 )

^are^suh^that

φ 1 (]c 2 − 1, c 3 − 1 + ǫ[)

a 1 + a 2 + a 3 + ǫ − a 1

a 1 + a 2 + a 3

and

φ 1 (]c 2 , c 3 + ǫ[)

a 1 + a 2 + a 3 + ǫ − a 1

a 1 + a 2 + a 3

haveoppositesigns.

Thistwoquantitiesarerespetivelyequalto

a 1 + 1 {b=1} − (1 − ǫ) 1 {d=1}

a 1 + a 2 + a 3 + ǫ − a 1

a 1 + a 2 + a 3

and

a 1 + ǫ1 {e=1}

a 1 + a 2 + a 3 + ǫ − a 1

a 1 + a 2 + a 3

,

where

b

^,

d

^, ^and

e

âre ^beads ⁱⁿ ^positions^speiedⁱⁿ ^the^rstâseâbove,ând

1 ^P

^takes^the

value1(resp. 0)if

P

îs^true^(resp. ^false). Ît îsêasy ^to ^see^that ^these ^two^quantities ^have

oppositesignsifandonlyif

b 6= e

^. ^Hene ^they^haveôpposite^signsîfândônlyîf^the^vertex

hasodddegree.

We show now that there is an odd number of suh verties. With the same kind of

argumentsasin Subsetion2.1,i.e., withtheinterval

]c 2 , c 3 + ǫ[

^moving^from ^left^to^right,

wesee thatthereisanoddnumberofsuh situations,forthesignof

φ 1 (]u, u + a 1 + a 2 + a 3 + ǫ[)

a 1 + a 2 + a 3 + ǫ − a 1

a 1 + a 2 + a 3

hangesstritlywhentheinterval

]u, u + a 1 + a 2 + a 3 + ǫ[

^moves^from

]0, a 1 + a 2 + a 3 + ǫ[

to

]a 1 + a 2 + a 3 , n + ǫ[

^(this ^is^a^true^for

ǫ = 0

^, ^aording^to ^the^startingassumptions,and thus forall

ǫ > 0

^smallênough). ^Hene, ^thereîs ânôdd^numberôf ^verties ôfôdd^degree

suhthat

c 1 = 0

^.

By symmetry, there is also an odd number of verties of odd degree having

c 3 = n

^.

Furthermore,wehavethe followingequivalene, sinethe ondition

b 6= e

^is ^the ^same^for

theverties

(0, c 2 , c 3 )

^and

(c 2 , c 3 , n)

^:

deg(0, c 2 , c 3 )

^odd

⇔ deg(c 2 , c 3 , n)

^odd

.

Thelemmaisproved.

2.3.3 Final omments

Let'sgiveanillustrationofthisprooffortheexampleofFigure1.

Theorrespondinggraph

G

^isillustratedin Figure3. Thereisapath(atually,many) linking vertex

(0, 4, 10)

^, ^whih ^gives ^more ^beads ^of ^type ³ ^to ^Alie, ^to ^vertex

(4, 10, 12)

^,

(7)

(1, 4, 9) (0, 4, 10) (0, 5, 11) (1, 5, 10)(2, 6, 10) (3, 7, 10) (4, 8, 10) (5, 9, 10) (4, 9, 11)(4, 10, 12) (5, 10, 11)

(5, 11, 12) (5, 8, 9)

(3, 6, 9) (2, 7, 11) (1, 2, 7)

(5, 6, 7)

(3, 4, 7)

Figure3: Graph

G

orrespondingto theexample(with

t = 3

^and

q = 2

⁾^of^Figure^1.

whih givesmore beads of type3 to Bob. The verties

(1, 5, 10)

^and

(2, 6, 10)

^lie ^on ^this

pathandprovidefairdivisions.

On the other hand, it is not lear whether it is possible to derive from this proof an

eientalgorithmthatwouldhaveabetteromplexitythan

O(n ³ )

⁽

G

^might^be^not^expliitly

omputed).

Furthermore,themethod istedious. Is itpossibletondsomesimpliation?

3 Ky Fan's ubial theorem and onstrutive proof for

two thieves

3.1 Preliminaries

Wegivenow aonstrutiveand ombinatorial proof of Theorem 1. We ould think that

suh aproofould be easilydedued from Tuker'slemma. Indeed,Theorem 1isadiret

appliation of Borsuk-Ulam's theorem, whih has a onstrutive proof through Tuker's

lemma. ThisisdoneforinstanebySimmonsandSuin[14℄,buttheresultisnotompletely

satisfatorybeausethismethodrequiresatriangulation

T

ofthe

(t + 1)

-dimensionalross- polytopesuh thatthediameter ofanysimplexin

T

is a

O(1/(nt))

^, ând^then ît^requiresâ

rounding method.

Ourmethoddoesnotpresentthosediulties. Itbasesitselfonatheorem(Theorem3)

foundbyKyFanin 1960whihisaubialversionofTuker'slemma[8℄. Wegivebelowa

onstrutiveproofofthistheorem. Theubialomplexweneedisompletely natural.

3.2 Notation and denitions

Let

C

^be^a^ube,

∂C

^the^boundary^of

C

^, ^and

V (C)

^the^set^of^verties^of

C

^.Two^faets^of^a

d

^-ubeâre^said^to ^beâdjaent îf^theirintersetion isa

(d − 2)

^-ube. ^For^a^given^faet

σ

^of

C

^,^there îsônlyônenon-adjaentfaet. Twonon-adjaentfaetsareopposite.

Aubialomplex isaolletion

C

ofubesembeddedinaneulidianspaesuh that

•

^if

τ ∈ C

andif

σ

îsâ^faeôf

τ

^then

σ ∈ C

and

•

^if

σ

^and

σ ^′

^areⁱⁿ

C

,then

σ ∩ σ ^′

îsâ^fae ôf^both

σ

^and

σ ^′

^.

Aubialomplexissaidto bepureifallmaximalubeswithrespettoinlusionhave

samedimension.

(8)

V ( C )

îs^the^setôf^vertiesôf^theûbialômplex

C

.

A

d

-dimensional ubial pseudo-manifold

M

is apure

d

-dimensional ubialomplexin whiheah

(d − 1)

-dimensionalubeisontainedin atmosttwo

d

-dimensionalubes. The boundaryof

M

,denotedby

∂ M

,isthesetofall

(d−1)

^-ubes

σ

^suh^that

σ

^belongs^to^exatly

one

d

-dimensionalubeof

M

.

Letus denoteby

C ^d

theubialomplexwhih istheunionofthe ube

^d := [−1, 1] ^d

and allitsfaes(seeFigure6).

Let

k ≥ 0

^be ^an ^integer. ^Given

k

^distint ^integers

i 1 , i 2 , . . . , i k

^, ^eah ^of ^them ⁱⁿ

{1, 2, . . . , d}

^,^and

k

^numbers

ǫ 1 , ǫ 2 , . . . , ǫ k

^,^eah^of^them^belonging^to

{−1, +1}

^,^we^denote

F

i 1 i 2 . . . i k

ǫ 1 ǫ 2 . . . ǫ k

:= {(x 1 , x 2 , . . . , x d ) ∈ ^d : x i _j = ǫ j

^for

j = 1, 2, . . . , k},

whihisa

(d − k)

^-fae^of

^d

^. ^For

k = 0

^,^this^fae^is

^d

^itself.

We endow

∂ ^d

^with hemispheres. These hemispheres are denoted

+H 0

^,

−H 0

^,

+H 1

^,

−H 1

^,

. . .

^,

+H d−1

^,

−H d−1

^.

+H i

^(resp.

−H i

⁾^is^the^set^of^points

(x 1 , x 2 , . . . , x d )

^of

∂ ^d

^suh

that

x i+1 ≥ 0

^(resp.

x i+1 ≤ 0

^),^and ^suh ^that, ^if

i ≤ d − 2

^, ^then

x j = 0

^for

j > i + 1

⁽ⁱⁿ

Figure 4,thehemispheresof

∂ ³

^arehighlighted).

Thearrier hemisphere ofaubeis thehemisphereofsmallestpossibledimensionon-

tainingit.

Observation2.1 For

i ∈ {0, 1, . . . , d − 1}

^,

H i

^is homeomorphi tothe ball

B ⁱ

^,

(+H i ) ∪ (−H i )

^is homeomorphi to the

i

-dimensional sphere

S ⁱ

^, ^if

i ≥ 1

^,

∂(+H i ) = ∂(−H i ) = (+H i−1 ) ∪ (−H i−1 )

^and

H d−1 ∪ −H d−1 = ∂ ^d

^.

Let

C

and

D

be ubial omplexes, and let

λ : V ( C ) → V ( D )

^be^a ^map.

λ

îs âlled â

ubial mapif

λ

^satises^the^following^onditions:

1. forevery

σ ∈ C

,thereis a

τ ∈ D

suhthat

λ(V (σ)) ⊆ V (τ)

2.

λ

^takesâdjaent^verties^toâdjaent^vertiesôr^both^to ^the^same^vertex.

Byaslightabuseof notation,wewrite

λ : C → D

. Weemphasizethefatthat

λ(V (σ))

ⁱⁿ

thedenition aboveisnotneessarilythevertexsetofaube.

AordingtoLemma18ofthepaperofEhrenborgandHetyei[6℄,if

λ : V ( ⁿ ) → V ( ⁿ )

is anon-injetiveubialmap andif

ρ

îsâ^faetôf

ⁿ

^,^then ^there^are^0,²^or⁴^faets

σ

^of

ⁿ

^suh^that

λ(V (σ)) = V (ρ)

^. ^It^implies^the^following^lemma:

Lemma 2.2 Let

λ : V ( ^d ) → V ( ^m )

^be ^a ^ubial ^map ^(from

C ^d

into

C ^m

), with

m ≥ d

^,

and let

ρ

^be ^a

(d − 1)

^-fae ^of

^m

^. ^If

λ

îs ^not înjetive, ^there âre

0

^,

2

^or

4

^faets

σ

^of

^d

suhthat

λ(V (σ)) = V (ρ)

^. ^Moreover,^if^there^are

4

^suh^faets^and^if

σ

îsôneôf^them,^then

the faetopposite to

σ

îsâlsoône ôf^these

4

^faets.

Proof: Let

λ : V ( ^d ) → V ( ^m )

^be^anon-injetiveubialmapandlet

ρ

^be^a

(d−1)

^-fae

of

^m

^. ^If^there ^is ^no^faet

σ

^suh ^that

λ(V (σ)) = V (ρ)

^, ^there ^is^nothing ^to ^prove. ^So^let

us assumethat there is afaet

σ

^of

^d

^suh ^that

λ(V (σ)) = V (ρ)

^. ^We^want ^to^apply ^the

result ofLemma 18of [6℄, whihis validif

λ : V ( ^d ) → V ( ^d )

^. ^Let ^us^prove^that ^we^an

makethisassumption,thatis,that thereisa

d

^-fae^of

^m

^whihôntains^theîmageôf

λ

^.

•

^If

λ(V ( ^d )) = V (ρ)

^,^any

d

^-fae^ontaining

ρ

ôntains^theîmageôf

λ

^.

•

^If ^not, ^there ^is

v ∈ V ( ^d ) \ V (σ)

^suh ^that

λ(v) ∈ / V (ρ)

^. ^Let

w

^be ^the^vertex ^of

σ

whihisadjaentto

v

^,^and^let

w ^′

^be^the^single^vertex^of

σ

^suh^that

d(w, w ^′ ) = d − 1

^,

where

d(·, ·)

îs^the^distaneⁱⁿ^the^1-skeletonôfâûbe,ôuntedⁱⁿ^numberôfêdges. Ît

(9)

isstraightforwardtohekthat

d(λ(w ^′ ), λ(v)) = d

^(take^the^minimal^fae ^ontaining

both

ρ

^and

λ(v)

^). ^Let

v ^′

^be^any^vertex^of

^d

^,^we^have^the^following^hain^ofinequalities:

d = d(v, w ^′ ) = d(v ^′ , v)+d(v ^′ , w ^′ ) ≥ d(λ(v ^′ ), λ(v))+d(λ(v ^′ ), λ(w ^′ )) ≥ d(λ(v), λ(w ^′ )) = d.

Hene,

d(v ^′ , v) = d(λ(v ^′ ), λ(v))

^and

d(v ^′ , w ^′ ) = d(λ(v ^′ ), λ(w ^′ ))

^. ^It ^implies^that

λ(v ^′ )

^,

forany

v ^′

ⁱⁿ

^d

^,^isⁱⁿ ^the

d

^-ube^of

^m

^ontaining

ρ

^and

λ(v)

^.

Weanthus assumethat

λ : V ( ^d ) → V ( ^d )

^.

Ifthere are4faetssuhthat

λ(V (σ)) = V (ρ)

⁽ⁱⁿ^this ^ase^one^has^neessarily

d ≥ 2

^),

at least two of them, say

σ

^and

σ ^′

^, ^are ^adjaent. ^Let

σ ^′′

^be ^another ^faet ^adjaent ^to

σ

and

σ ^′

^. ^Let

x ∈ V (σ ^′ ∩ σ ^′′ ) \ V (σ)

^, ^and ^let

y

^be ^the ^vertex ^adjaent^to

x

ⁱⁿ

σ

^. ^We^have

y ∈ σ ∩ σ ^′ ∩ σ ^′′

^, ^and

λ(x)

^adjaent ^to

λ(y)

^. ^Moreover,^let

z

^be^the ^vertex^of

σ

^suh ^that

λ(z) = λ(x)

^(whih^exists^byinjetivity).

z

^is^adjaent^to

y

ⁱⁿ

σ

^and^annot^beⁱⁿ

σ ^′

^. ^Hene

z

îs^theônly^vertexôf

σ

^whihîsât^distane¹ôf

y

^and^notⁱⁿ

σ ^′

^. ^It^implies^that

z

^isⁱⁿ

σ ^′′

^.

As

λ(z) = λ(x)

^and^as

z

^and

x

^are^two^dierent^verties^of

σ ^′′

^,

λ(V (σ ^′′ )) 6= V (ρ)

^. ^A ^faet

σ ^′′

^adjaent^to

σ

^and^to

σ ^′

^annot^satisfy

λ(V (σ ^′′ )) = V (ρ)

^.

3.3 Ky Fan's ubial formula

Supposethat wehaveaubialmap

λ

^of

M

,a

t

-dimensional ubialpseudo-manifold, into

C ^m

, theubialomplexmadeby

^m

^and^its^faes.

Wedenoteby

α λ

i 1 i 2 . . . i m−t

ǫ 1 ǫ 2 . . . ǫ m−t

thenumberofthose

t

^-ubes

σ

^of

M

suhthat

the verties of

σ

^are ^mapped ^under

λ

ⁱⁿ â ône-to-one ^way ônto ^the ^verties ôf ^the

t

^-fae

F

i 1 i 2 . . . i m−t

ǫ 1 ǫ 2 . . . ǫ m−t

(if

m = t

^,^this^fae^is

^m

^itself).

Wewilllatermakeuseofthefollowingtheorem(forthease

m = t

^),^whose^onstrutive

proofisgivenbelow.

Theorem 3 Let

M

betheubialpseudo-manifoldwithoutboundaryobtainedbysubdividing

∂ C ^t+1

bymeansof thehyperplanes

x i = _n ^j

^where

i ∈ {1, 2, . . . , t + 1}

^and

j ∈ {−n + 1, −n + 2, . . . , −1, 0, 1, . . . , n − 2, n − 1}

^. ^Let

λ

^be^a^ubial ^map ^from

M

into

C ^m

,with

m ≥ t

^. ^If

λ

is antipodal(i.e. if

λ(−x) = −λ(x)

^),^then ^we^have

•

^if

m > t

^,

X

ǫ 2 ,...,ǫ _m−t =−1,+1

α λ

t + 1 t + 2 . . . m +1 ǫ 2 . . . ǫ m−t

= 1

^mod

2.

•

^if

m = t

^,^there^is^a

t

^-ube

σ

^of

M

suhthat

λ(V (σ)) = V ( C ^m )

^.

For

t = 2

^and

n = 2

^,

M

isillustratedinFigure4.

3.4 Construtive proof of Theorem 3

Wefollow roughlythe method found by Presott and Su [13℄ for provinga similar result

onerningsimpliialomplexes.

•

^A

d

^-ube^of

M

(with

0 ≤ d ≤ t

⁾îs^said^to^be^fullîfîtsîmageûnder

λ

^is ^of^the^form

F

d + 1 d + 2 . . . m ǫ 1 ǫ 2 . . . ǫ m−d

.

(10)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

. . .

. .

. . .

. .

.. .. . .. . .. .. . .. .. . .. . .. .. . .. . .. .. . .. . .. ..

.

. .

.

. . .. .. . .. . .. .. . .. . .. .. . .. . .. .. . .. .. . .. . .. .. . .. . .

.

. .. . .. .. . .. . .. .. . .. . .. .. . .. .. . .. . .. .. . .. . .. ..

.

.. . .. . .. .. . .. . .. .. . .. .. . .. . .. .. . .. . .. .. . .. .

. . .

. .

.

. . .

. .

. . .

. .

. . .

.

. . .

. .

. . .

. .

. . .

. .

. . .

. .

.

. .

.

.. .. . .. . .. .. . .. .. . .. . .. .. . .. . .. .. . .. . .. ..

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

. .

.

I

?

K

+H 0

−H 0

+H 1

−H 1

+H 2

−H 2

- 6

x 1

x 2

x 3

Figure4: Anillustrationof

M

for

n = 2

^and

t = 2

^.

(11)

•

^A

d

^-ube^of

M

issaidtobealmost-fullifitisnotfullbutoneofitsfaetsisfull.

Exeptintheasewhen

d = m = t

^,ôneân^dene^the^signôfâ^full

d

^-ube^as^the^value

of

ǫ 1

ⁱⁿ ^theexpressionsabove.

Lemma 3.1 For analmost-full

d

^-ube

τ

^,^there^are^only ^threepossibilities:

1. eitheritsvertiesaremappedunder

λ

ⁱⁿâône-to-one^wayônto^the^vertiesôfâ

d

^-fae

(withthe entries

j

^and

ǫ j−d+1

^deleted)

F

d . . . ˆ j . . . m

ǫ 1 . . . ˆ ǫ j−d+1 . . . ǫ m−d+1

,

2. or

λ

îs^not înjetive ôn ^the ^vertex^set ôf

τ

^and

τ

^has êxatly ^two^full ^faets, êah ôf

themhavingthe samesign,

3. or

λ

îs ^notînjetive ôn ^the ^vertex ^set ôf

τ

^and

τ

^has êxatly ^four ^full ^faets,êah ôf

them having the same sign. Moreover, in this ase, if

σ ⊆ τ

^is ^a ^full ^faet, ^then ^the

opposite of

σ

îsâlsoâ^full^faet.

Proof: Firstremarkthat if

λ

îsînjetive,^theîmage ôf

τ

^is ^a

d

^-fae ^(this^easy^assertion

anbeprovedbyindution). Sine

τ

^is^not^full, ^this^image ^has^neessarily^the^form ^given

in point1.

Hene,oneanassumethat

λ

îs^notînjetive. ^Take^two^full^faets ôf

τ

^, ^say

σ

^and

σ ^′

^.

Sinetheyarefull,

λ

îsînjetiveônêahôf^them.

λ(V (σ))

^and

λ(V (σ ^′ ))

^are^the^vertex^sets

of

(d − 1)

^-faes.^They^have^a^non-emptyintersetion,otherwise

λ

^would^have^been^injetive

on

τ

^. ^Moreover,âording^to^the^denitionôfâ^fullûbe,^theîmageôf^two^fullûbesôf^the

samedimensionare parallel. But,twoparallel faesof aubehavingsamedimensionand

a nonemptyintersetion areequal. Thus,

λ(V (σ)) = λ(V (σ ^′ ))

^, ând ^there îsâ

(d − 1)

^-fae

ρ

^suh ^that ^for^all ^full ^faets

σ

^of

τ

^, ^one^has

λ(V (σ)) = V (ρ)

^. ^This ^settles ^the^statement

aboutthesign. Lemma2.2allowsto onlude.

Wean then extendthe notionof sign to almost-full ubes: thesign of analmost-full

d

^-ubeîs^the^signôfânyôfîts^full^faets.

Aubeissaidto beagreeableifitssignmathesthesignofitsarrierhemisphere.

Wedenethefollowinggraph

G

^. ^A^ube

σ

^with^arrier^hemisphere

±H d

îsâ^nodeôf

G

ifitisoneofthefollowing:

(1) anagreeablefull

(d − 1)

^-ube,

(2) anagreeablealmost-full

d

^-ube,^or

(3) afull

d

^-ube.

Twonodes

σ

^and

τ

^are^adjaentⁱⁿ

G

^if^all^the^following^hold:

(a)

σ

îsâ^faetôf

τ

^,

(b)

σ

^is^full,

()

τ

^has^the^dimensionôfîtsârrierhemisphere, and

(d) thesignofthearrierhemisphereof

τ

^mathes^the^sign^of

σ

^.

Welaim that

G

îsâ ^graphⁱⁿ ^whih êvery^vertex ^has^degree^1, ^2,ôr ^4. Furthermore, a vertex hasdegree 1ifand only ifits arrierhemisphere is

±H 0

^, ôrîfît îs â^full

t

^-ube.

Using antipodality, we willthen show that thepoint

H 0

^is ^neessarily^linked^by ^a^pathⁱⁿ

G

^to ^a^full

t

^-ube.

Toseewhy, let'sonsiderthethreekindsofnodesin

G

^:

(1)An agreeablefull

(d − 1)

^-ube

σ

^,^with^arrier^hemisphere

±H d

^,îsâ^faetôfêxatly^two

d

^-ubes,êahôf^whih ^must^beâ^fullôrânâgreeablealmost-full ubein thesamearrier.

(12)

Thissatisestheadjaenyonditions(a)(d)with

σ

^,^hene

σ

^has^degree²ⁱⁿ

G

^. ^The^ube

σ

^annot^be^adjaentⁱⁿ

G

^toânyôfîts^faets,ôtherwiseît^wouldôntraditôndition^().

(2) An agreeablealmost-full

d

^-ube

σ

^, ^whose ^arrier^hemisphere ^is

±H d

^, ^is ^adjaent ⁱⁿ

G

to its twoor four faets that are full

(d − 1)

^-ubes ^see ^Lemma ^3.1. ^A ^full ^faet ^of

σ

^is

automatiallyanodeof

G

^,^sineîfîtîs^notâgreeable,^thenîtîsârried^by

∓H d−1

^the^sign

ofthearrierofthisfullfaetisthentheoppositeofthesignof

σ

^. ^(Adjaeny^ondition^(d)

is satised beause

σ

îsâgreeable ând ^beause ânâgreeable almost-full

d

^-ube ^has^always

thesamesignasitsfullfaets).

(3)Exeptif

d = 0

^,^a^full

d

^-ube

σ

^with^arrier

±H d

^has^exatly^one^faet

τ

^that^is^full^and

whose signisthesameasthearrierhemisphereof

σ

^. ^Hene

σ

^is^adjaent^to

τ

ⁱⁿ

G

^.

Ontheotherhand,

σ

îs^the^faet ôfêxatly^two

(d + 1)

^-ubesⁱⁿ

+H d+1 ∪ −H d+1

^: ^one

in

+H d+1

^and^oneⁱⁿ

−H d+1

^, ^but

σ

^is ^adjaentⁱⁿ

G

^to êxatlyôneôf ^them: ^whih ône îs

determined bythesignof

σ

^,^sine^adjaeny^ondition^(d)^must^be^satised.

Finally,

σ

îs ôf ^degree²ôr⁴ⁱⁿ

G

^unless

d = 0

^or

σ

^is^a^full

t

^-ube: ^if

d = 0

^,

σ

^is ^the

point

±H 0

^and^it^has^no^fae,^hene

σ

^has^degree^1;^and^if

σ

^is ^a^full

t

^-ube,^then

σ

^is^not

thefaetofanyotherube,andisthereforeofdegree1.

Everynodein

G

^has^degree²^or^4,^exept^for^the^points

±H 0

^and^for^the^full

t

^-ubes.

Wetransform

G

^into^a^new ^graph

G ^′

^: ^we^replae^eah ^node

v

^of^degree⁴ⁱⁿ

G

^by ^two

nodes

v ^′

^and

v ^′′

^,ând^we^replae^the^fourêdges,^by^four^newêdges: ^two^linking

v ^′

^and^a^pair

of oppositefullfaets,andtwolinking

v ^′′

ând^theôther^pairôfôpposite^full^faets.

In

G ^′

^,âll^vertiesâreôf^degree¹ôr^2,^hene

G ^′

îs^theûnionôf^disjoint^pathsând^yles.

Notethattheantipodeofapathin

G ^′

îsâlsoâ^pathⁱⁿ

G ^′

^. ^No^path^an^have^antipodal

endpoints, otherwise it should be its own antipode and have a node oran edge whih is

antipodal to itself. Hene the number of endpoints is a multiple of four. Sine exatly

twoendpointsare

H 0

^and

−H 0

^,^thereâre^twieânôdd^numberôfêndpoints^whih âre^full

t

^-ubes. ^If

m > t

^,^half ^of^them^have^a^positive^sign^(if

m = t

^,^a^full

t

^-ube^has^no^sign).

A path of

G ^′

^starting ⁱⁿ

H 0

^an ^not ^terminate ⁱⁿ

−H 0

^, ^but ⁱⁿ ^a^full

t

^-ube. ^Thus ^we

haveanalgorithmwhih allowstondafull

t

^-ube. ^It ^isillustratedinFigure 5.

3.5 Splitting the neklae between Alie and Bob

We an easily dedue Theorem 1 from Theorem 3. Indeed let

m = t

^, ^and ^let

x :=

(x 1 , x 2 , . . . , x t+1 )

^be ^a^vertex ^of

M

. We order the

x i

ⁱⁿ ^the înreasing ôrder ôf ^their âb-

solute value. When

|x i | = |x i ′ |

^, ^we^put

x i

^before

x i ′

^if

i < i ^′

^. ^This^gives^apermutation

π

suhthat

|x π(1) | ≤ |x π(2) | ≤ . . . ≤ |x π(t+1) | = 1.

Then, wedene

λ = (λ 1 , λ 2 , . . . , λ t ) : M → C ^t

asfollows: weuttheneklaeat points

n|x i |

^(at^most

t

^are^dierent^from ^1). ^By^onvention,

x π(0) := 0

^; ^we^dene:

λ i (x) :=

+1

^if

P t

j=0

^sign

(x ǫ(j) )φ i

n|x π(j) |, n|x π(j+1) |

> 0

−1

^if^not,

for

i = 1, 2, . . . , t

^, ^where

ǫ(j)

^is ^the^smallest^integer

k

^suh ^that

|x k | ≥ |x _π(j+1) |

^, ^and ^with

theonventionsign

(0) := 0

^.

This formula anbe understood as follows:

λ i

^equals

+1

^(resp.

−1

⁾ ^if ^the ^rst ^thief

gets stritlymore(resp. stritlyless) beads oftype

i

^than^the^seondône,^when^weûtât

points

n|x j |

ôf^the^neklae,ând^when^the^signôf

x k

^shows^the^thief^who^gets^the

(j + 1)

^th

sub-neklae,where

k

^is^the^smallest^integer^suh^that

n|x k |

îs^greater^thanôrêqual^to^the

oordinateofanypointofthesub-neklae.

(13)

(+1, +1, +1)

(+1, −1, +1) (−1, −1, +1)

(+1, −1, −1) (−1, −1, −1)

(+1, +1, −1) (−1, +1, −1)

Figure 5: Illustration of the onstrutive proof of Theorem 3. The triples indiate the

orrespondinglabelontheubeimage oftheFigure 6.

.

. .

.

. . .. .. . .. .. . .

6 +

-

x 1

x 2

x 3

(−1, −1, −1)

k s ?

-

6 (−1, −1, +1)

(−1, +1, −1) (+1, +1, +1) (−1, +1, +1)

(+1, +1, −1) (+1, −1, −1)

(+1, −1, +1)

Figure6: An illustrationof

C ^m

for

m = 3

^.

(14)

Theonlyproblemwiththisdenitionisthat

λ

^may^not^be^antipodal: ^if

X t

j=0

sign

(x ǫ(j) )φ i

n|x π(j) |, n|x π(j+1) |

= 0,

then

λ i (x) = λ i (−x) = −1

^. ^This ân^be âvoided ^by â ^slight ^generi perturbation of the valueofallthebeads: adivisionusingutsbetweenbeadsanthennotbefairwithrespet

to anytypeofbead 2

.

λ

îsâûbial^map. Îndeed,^the^rstôndition^deningûbial^mapsîs^trivially^satised

andtheseondonefollowsfromthefattheimagesby

λ

ôf^twoâdjaent^vertiesôf

M

have at mostoneoordinatethatdiersevenintheasewhenthereis

j

^suh^that

ǫ(j)

^diers.

Theorem 3 showsthat there is a

t

^-ube

σ

^whose ^image^by

λ

^is

^t

^. Ôneôf ^the^vertiesôf

this

t

^-ubeôrresponds^to â^fair^division. Îndeed,^for âny^vertex

x

^of ^suh ^a

σ

^, ^moving ^to

eahofthe

t

^adjaent^verties^of

x

^hanges^the^valueôfêahôf^the

t

^omponents^of

λ(x)

^.

Theonstrutiveproof given abovefor Theorem3 suitsto theneklae problem. The

ubialomplexisompletelynaturalinthisontext: a

k

^-ube⁽

k ≤ t

⁾^orresponds^preisely

tothehoieof

k

^beads. ^Moreover,^it^isstraightforwardtohekifa

k

^-ube^is^fullⁱⁿ^terms

of beads. WehavethusaonstrutiveproofofTheorem1.

3.6 General disrete neklae splitting theorem

In[3℄,Alon,MoshkovitzandSafraprovethefollowingextensionofTheorem2(withows).

Supposethattheneklaehas

n

^beads,êahôfâêrtain^type

i

^,^where

1 ≤ i ≤ t

^. ^Suppose

there are

A i

^beads^of^type

i

^,

1 ≤ i ≤ t

^,

P t

i=1 A i = n

^,^where

A i

^is^not^neessarily^a^multiple

of

q

^. ^A

q

^-splittingôf ^the^neklaeîsâ^partitionôfîtînto

q

^parts,êahônsistingôfâ^nite

number of non-overlapping sub-neklaes of beads whose unionaptures either

⌊A i /q⌋

^or

⌈A i /q⌉

^beads ^of^type

i

^,^for^every

1 ≤ i ≤ t

^.

Theorem 4 Everyneklaewith

A i

^beads ^of^type

i

^,

1 ≤ i ≤ t

^,^has ^a

q

^-splitting^requiring^at

most

t(q − 1)

^uts.

For

q = 2

^, ^the^proof^of ^the^preeding ^setion ^shows^something^more. ^The^proof^is ^still

valid,evenifthe

A i

^'sâre^notêven. Îf

A i

^is^odd,^the

(t − 1)

^-faet^of

^t

^whose

i

^th^oordinate

isequalto1(resp.

−1

^). ^orrespond^to^division^giving

⌈A i /2⌉

^beads^of^type

i

^to^Alie^(resp.

Bob)and

⌊A i /2⌋

^beads^of^type

i

^to^Bob^(resp. ^Alie). ^We^have^thus^the^following^theorem,

also withaonstrutiveproof:

Theorem 5 Every neklae with

A i

^beads ^of ^type

i

^,

1 ≤ i ≤ t

^, ^has ^a

2

^-splitting ^requiring

atmost

t

^uts. ^Moreover, ^for ^eah^type

i

^of ^beads^suh^that

A i

îsôdd,^we ân^hoose^whih

thief gets

⌈A i /2⌉

^beads^of ^type

i

^.

NotethatthistheoremanalsobederivedfromtheproofofAlon,MoshkovitzandSafra.

4 Open questions

There arealotof questionsthatremainopen.

1. Isitpossibleto extendthediretproofsofSetion 2.1to otherases?

2

Ifonedoesnotwanttouseaperturbation,thereisanotherwayforavoiding0:hooseapartiularbead

ofeahtypeanddene

λ _i ( x ) := +1

^(resp.

− 1

⁾^if^the^rst^thief^gets^stritly^more^(resp. ^stritly^less)^beads

oftype

i

thantheseondthief,oriftherstthiefandtheseondonegetthesamenumberofbeadsoftype

i

andtherstonegets(resp. doesnotget)thepartiularbeadoftype

i

.

(15)

forinstaneinthepaperofAlon[2℄.

3. Whatistheomplexityofthefollowingproblem?

INPUT: A neklae, with

t

^beads,

qa i

^beads ^of ^eah ^type ⁽

a i

îs ân înteger), ând

q

thieves,

TASK: Findafairdivisionoftheneklaeusingnomorethan

t(q − 1)

^uts,

This questionis still open(for moreaboutproblems of this type,see Papadimitriou

[12℄). Theorrespondingoptimizationproblem,namelyndingtheminimumnumber

ofutsprovidingafair division,isNP-hard,evenif

q = 2

^and

a 1 = a 2 = . . . = a t = 1

(see[5℄,[11℄).

4. IsthereananalogueofTheorem 5foranynumber

q

^of^thieves?

Referenes

[1℄ N. Alon,Splittingneklaes,Advanes inMath.,63(1987),247-253.

[2℄ N.Alon,Non-onstrutiveproofsinombinatoris,Pro. oftheInternational Congress

of Mathematiians,Kyoto 1990,Japan,Springer Verlag,Tokyo, pp.1421-1429,1991.

[3℄ N. Alon, D. Moshkovitz and S. Safra, Algorithmi Constrution of Sets for

k

^-

Restritions,toappearinThe ACM Transations onAlgorithms.

[4℄ N. Alon and D. West, The Borsuk-Ulam theorem and bisetion of neklaes, Pro.

Amer.Math. So., 98(1986),623-628.

[5℄ P. S. Bonsma, T. Epping and W. Hohstättler, Complexity results on restrited in-

stanesofapaintshopproblemforwords,DisreteAppliedMathematis,to appear.

[6℄ R.EhrenborgandG.Hetyei,GeneralizationsofBaxter'stheoremandubialhomology,

J. CombinatorialTheory,SeriesA, 69(1995),233-287.

[7℄ T.Epping,W.HohstättlerandP.Oertel,Complexityresultsonapaintshopproblem,

DisreteApplied Mathematis,136(2004),217-226.

[8℄ K.Fan,Combinatorialpropertiesofertain simpliialandubialvertexmaps,Arh.

Math., 11(1960),368-377.

[9℄ C.H. GoldbergandD. West,Bisetionofirleolorings,SIAMJ.Algebrai Disrete

Methods,6(1985),93-106.

[10℄ J.Matou²ek,UsingtheBorsuk-Ulamtheorem,Springer-Verlag,Berlin-Heidelberg-New

York,2003.

[11℄ F.MeunierandA.Sebö,Paintshop,oddyleandsplittingneklae,submitted.

[12℄ C.Papadimitriou,Ontheomplexityoftheparityargumentandotherineientproofs

ofexistene,J. ComputerandSystem Sienes,48(1994),498-532.

[13℄ T. Presottand F.E. Su,A onstrutiveproofof KyFan'sgeneralizationof Tuker's

lemma,J. Combin. Theory,Series A,111(2005),257-265.

[14℄ F. W. Simmons and F. E. Su, Consensus-halving via theorems of Borsuk-Ulam and

Tuker,Mathematial soialsienes,45 (2003),15-25.

[15℄ G.M.Ziegler,GeneralizedKneseroloringtheoremswithombinatorialproofs,Inven-

tionesMathematiae,147(2002),671-691.

Discrete Splittings of the Necklace

∗

n

t

2a i

i

i ∈ {1, 2, . . . , t}

a i

2 P t

i=1 a i = n

n − 1

t

t

qa i

i

a i

t

q

t(q − 1)

∗

N

^ ?

q = 2

a 1 = a 2 = . . . = a t = 1

q = 2

t = 3

q = 2

d

d

d

t ≤ 2

q

t = 1

q

q − 1

O(q)

t = 2

q

q = 1

q ≥ 2

q

a 1 + a 2

n

q

a 1

a 2

2(q −2)

2 + 2(q −2) = 2(q − 1)

q

a 1

a 1

M +

M −

a 1 + a 2

M −

M +

a 1

a 1 + a 2

2 + 2(q − 2) = 2(q − 1)

O(n)

a 1

a 2

O(log n)

O(n + q log n)

a 1 = a 2 = . . . = a t = 1

t

q

k

t(q − 1)

n

t

n − t = qt − t = t(q − 1)

O(n)

t = 3

q = 2

]0, n[⊂ R

1, 2, . . . , n

k

]k−1, k[

S ⊆]0, n[

M ⁺

M ⁻

M ⁻

M ⁺

a ₁ = a ₂ = . . . = a t = 1

1 ⁱ (u)

1 ⁱ (u) =

1 ⁱ (u)du.

φ 2 ^′ := φ 2 + φ 3

φ 2 ^′ (]0, c 1 [∪]c 2 , c 3 [) = a 2 + a 3

φ 2 ^′ = φ 2 + φ 3