Table Of ContentNonamemanuscriptNo.
(willbeinsertedbytheeditor)
On equivalence between maximal and maximum antichains
over range-restricted vectors with integral coordinates and
the number of such maximal antichains
Shen-FuTsai
7
1
Received:2017/01/24
0
2
n Abstract Consider a strict partially ordered set P consisting of all d-dimensional
a vectors with integral coordinates restricted in a certain range. We found that any
J
maximalantichainisalsomaximum,andthemaximumsizehasasimpleexpression
4
intermsoftherange.Propertiesofthenumberofmaximalantichainsgiventherange
2
are explored. We present our proof, the application on other areas, and some open
] questions.
O
Keywords OrderedSet·(0−1)Matrix·ExtremalProblems
C
.
h
t 1 Introduction
a
m
ConsiderastrictpartiallyorderedsetP consistingofalld-dimensionalvectorswith
[
integral coordinates restricted in a certain range. The binary relation < is defined
1 as y=(y ,...,y )<x=(x ,...,x ) if y <x for every i∈[1,d]. Without loss of
1 d 1 d i i
v
generality,supposetherangeisgivenbyanpositiveintegervectorw=(w ,...,w ),
0 1 d
meaningthatforeachdimesioni,thecoordinateofeveryvectorinP isrestrictedto
5
7 [1,wi].ThenumberofvectorsinP is(cid:213) di=1wi.AnantichainisasubsetofP such
6 thatforanytwoelementsitcontains,neither<oritsinverse>holds.Amaximalan-
0 tichainisanantichainthatisnotapropersubsetofanyotherantichain.Amaximum
.
1 antichainisanantichainthathascardinalityatleastaslargeaseveryotherantichain.
0
7 IfinPwereplacethebinaryrelation<by≤,i.e.y=(y ,...,y )≤x=(x ,...,x )
1 d 1 d
1
if y ≤x for every i∈[1,d], then the size of maximal antichain varies. For exam-
: i i
v ple for w=(2,2), {(1,1)} is a maximal antichain, and so is {(1,2),(2,1)}. Their
i
X sizes differ by 1. On the other hand, under < the only two maximal antichains are
{(1,1),(2,1),(1,2)}and{(2,2),(2,1),(1,2)},bothwithcardinality3.Themainre-
r
a sultofthisworkisthatunder<,everymaximalantichainisalsomaximum,andthe
Shen-FuTsai
7319143rdAve.NE,Redmond,WA,USA
E-mail:parity@gmail.com
2 Shen-FuTsai
maximalsizeis(cid:213) d w −(cid:213) d (w −1).
i=1 i i=1 i
Becausewediscoveredtheresultwhileworkingonthefieldofmatrixexclusion1,
we will present and proveour result using its terminology.The rest of the paper is
organized as follows: in Section 2 we introduce matrix exclusion terminology and
reviewsomeliteratures.InSection3weprovethatmaximalandmaximumantichains
areequivalentinoursetting.WeexplorethenumberofmaximalantichainsinSection
4.InSection5weconcludewithsomeopenquestions.
2 Matrixexclusion
A matrixis binaryif eachofits entriesiseither 0or 1.Allthe matricesmentioned
inthispaperarebinary.Anentrywithvalue1isanoneentry;otherwiseitisazero
entry.AbinarymatrixAcontainsanotherbinarymatrixPifwecanobtainabinary
matrix A′ with the same dimensionality as P by deleting 0 or more columnsand 0
or morerowsfromA, and A′ ≥P forall i and j. Thedefinitionextendsnaturally
ij ij
tod-dimensionalmatrix– ineachdimensionwe delete0ormoreindicesandform
amatrixwiththesamedimensionalityasP.AavoidsPifAdoesnotcontainP.We
callPtheforbiddenmatrix.InthisworkweonlycareaboutP=I ,the2×...×2
2,d
d-dimensionalidentitymatrix.
Definition1 Let P be a d-dimensional matrix. ex(w,P) is the maximum number
of one entries a w ×w ×...×w matrix can have while avoiding P. ex(n,P)≡
1 2 d
ex((n,n),P),ex(n,P,d)≡ex((n,...,n),P).
In contrast to our result, the field of matrix exclusion has mostly focused on
asymtotic behavior of ex(w,P), especially in two-dimensional matrices. Bienstock
and Gyo¨ni[1] showed ex(n,P)=Q (na (n)) for some trapezoidalpatternswith four
oneentries,wherea (n)istheextremelyslowgrowinginverseAckermannfunction.
Fu¨redi and Hajnal[2] raised the question of what matrices have extremal functions
linear in n? Subsequently an array of works emerged to respond to the quest, for
example [3,4,5,6]. There are much fewer works on higher dimensional forbidden
matrices.KlazarandMarcus[7]showedthatex(n,P,d)=O(nd−1)foranyk×...×k
permutation matrix P. Geneson and Tian[8] extended the result to all d-dimension
tuplepermutationmatrices.
3 Equivalencebetweenmaximalandmaximumantichains
For notational simplicity we define 2-contain same as contain, except when d =1
amatrix2-containsI aslongasithasmultipleoneentries.Likewise2-avoidand
2,d
ex2(w,P)followsnaturallyfrom2-containthusdefined.Theonlyreasontohave2-
containdefinedistocovertheweirdbuttrivialedgecased=1inthemathematical
1 Afterconjecturingandprovingtheresult,theauthorpostthemontheforumofCrowdMathProject[9]
underusernameparityhome,astheauthorwasoriginallyworkingonamatrixexclusiongameproposed
ontheforumbymoderatorJesseGeneson.
TitleSuppressedDuetoExcessiveLength 3
inductionusedbytheproof.
Theorem1 ex2(w,I )=(cid:213) d w −(cid:213) d (w −1). Moreover, every maximal w ×
2,d i=1 i i=1 i 1
...×w matrix that 2-avoids I has exactly ex2(w,I ) one entries. A maximal
d 2,d 2,d
matrixisonesuchthatflippinganyzerotoonemakesit2-containI .
2,d
Notethat(cid:213) d w −(cid:213) d (w −1)isthenumberofentrieswithatleastonecoordi-
i=1 i i=1 i
natebeing1,andalsothenumberofentrieswithatleastoneindexisuchthatx =w.
i i
TheresulthasimplicationonamatrixexclusiongameproposedbyGeneson[9].
Playedbym>1playersnumberedfrom0tom−1,thegamestartswithamatrixA
filledwithzeroentries.TheplayerstaketurntoflipazeroinAtoone,andaplayer
losesthegameifhisorhermovemakesAcontainapredefinedbinarymatrixP.Our
resultimpliesthatifP=I ,player[((cid:213) d w −(cid:213) d (w −1)) modm]alwaysloses
2,d i=1 i i=1 i
nomatterwhattheplayers’strategiesare.
Proof Thetheoremistrivialwhend=1orw =1forsomeiin[1,d].
i
Definition2 A d-row x=(x ,...,x ) of a d-dimensionalmatrix M is the set of
1 d−1
entrieswiththefirstd−1coordinatesbeingx ,...,x .Thereare(cid:213) d−1w d-rows,
1 d−1 i=1 i
eachwithw entries.
d
Definition3 The x -th d-cross section is the set of entries with the last coordinate
d
beingx .Therearew d-crosssections,eachwith(cid:213) d−1w entries.
d d i=1 i
LetMbeamaximalw ×...×w matrixthat2-avoidsI .Wewillbeprimarily
1 d 2,d
dealingwiththed-rowsofMthroughouttheproof.
Definition4 Ad-rowx=(x ,...,x )isaancestorofd-rowy=(y ,...,y )if
1 d−1 1 d−1
x<y. We say y is a descendantof x. A(x) and D(x) are the sets of ancestors and
descendantsofx,respectively.
Ancestoranddescendantrelationsaretransitive,andthegraphsofthesetwore-
lationsaredirectedacyclic.DefineM(x,x )=M .
d x1,...,xd
For a d-row x define l(x) and h(x) as the minimum and maximum y such that
M(x,y)=1,respectively.Theyarewelldefinedifthereisanoneentryinx.Foraset
of d-rowsZ, l(Z) and h(Z) are the smallest and largestd-coordinateof one entries
inZ,respectively,withl(0/)=¥ andh(0/)=−¥ .ForZ6=0/,theyarewelldefinedif
thereexistsz∈Zsuchthattheyarewelldefinedforz.Wefirstestablishthath(·)and
l(·)arewelldefinedeverywhere,i.e.everyd-rowofMhassomeoneentry.
Lemma1 Everyd-rowofMhassomeoneentry.
4 Shen-FuTsai
Proof Givenad-rowx,ifA(x)isempty,thenM(x,w )=1.IfD(x)isempty,then
d
M(x,1)=1.Ifneithersetisempty,since(1,1,...,1)∈A(x),l(A(x))iswelldefined,
and so is h(D(x)) as (w ,...,w )∈D(x). We then have l(A(x))≥h(D(x)), be-
1 d−1
causeotherwiseM2-containsI .Foranyy∈[h(D(x)),l(A(x))],havingM(x,y)=1
2,d
doesnotmakeM2-containI . ⊓⊔
2,d
Lemma2 Givenanyd-rowx,theset{y: M(x,y)=1}isacontiguoussegment.
Proof If M(x,y )=M(x,y )=1 for y ≤y , then setting M(x,y)=1 for any y∈
1 2 1 2
[y ,y ]doesnotmakeM2-containI . ⊓⊔
1 2 2,d
w(M),thenumberofoneentriesinM,isthen(cid:229) h(x)−l(x)+1.
x
Lemma3
h(x)=min(w ,l(A(x))) (1)
d
Proof It is trivial when A(x)=0/, so we only discuss when A(x) is not empty. If
h(x)>l(z)forsomez∈A(x),thenM 2-containsI .Ontheotherhand,ifh(x)<
2,d
l(A(x)),settingM(x,l(A(x)))=1doesnotmakeM2-containI . ⊓⊔
2,d
Lemma4
l(x)=max(1,h(D(x))) (2)
Proof It is trivial when D(x)=0/, so we only discuss when D(x) is not empty. If
l(x)<h(z)forsomez∈D(x),thenM 2-containsI .Ontheotherhand,ifl(x)>
2,d
h(D(x)),settingM(x,h(D(x))=1doesnotmakeM2-containI . ⊓⊔
2,d
Thetwolemmasaboveimplyl(x)≤l(z)ifD(x)⊂D(z)andsimilary,h(x)≤h(z)
ifA(z)⊂A(x).
We are now going to prove the theorem by mathematical induction on d and
w throughthe followinglemmas. Assume the theoremholdsfor 1,2,...,d−1 di-
d
mensions,andforw=(w ,...,w ,1),(w ,...,w ,2),...,(w ,...,w ,w −1)
1 d−1 1 d−1 1 d−1 d
withd dimensions.
Lemma5 GivenamatrixM,ifh(·)andl(·)arewelldefinedandsatisfyEquation1
andEquation2foreveryd-rowx,thenM2-avoidsI andismaximal.
2,d
Proof Clearly, if there is an one entry in some d-row x with d-coordinate y larger
thanmin(w ,l(A(x))),thenyeithergoesbeyondw ,whichisimpossible,ory>l(z)
d d
forsomez∈A(x),andM 2-containsI .Similarlywe couldnothaveoneentryin
2,d
anyd-rowxwithd-coordinatelessthanmax(1,h(D(x))).
Ontheotherhand,ifallentriesinsomed-rowxhaved-coordinateissmallerthan
min(w ,l(A(x))), setting M(x,min(w ,l(A(x)))) to 1 does not make M 2-contain
d d
I .Thesimilarargumentworksforl(x). ⊓⊔
2,d
TitleSuppressedDuetoExcessiveLength 5
We know that h(x)=w if A(x)=0/. In fact, if the converse is true things are
d
mucheasier.Let’sdenote{x: h(x)=w ,A(x)6=0/}byX(M),amoredifficultsetof
d
d-rowsthatwewilldealwithlater.
Definition5 X(M)≡{x: h(x)=w ,A(x)6=0/}
d
Lemma6 IfX(M)=0/,thenw(M)=(cid:213) d w −(cid:213) d (w −1).
i=1 i i=1 i
Proof TransformMtoM′ byremovingthew -thd-crosssection.Thensinceh(x)=
d
w ifandonlyifA(x)=0/,wehave
d
w(M′)=w(M)−|{x: A(x)=0/}|
WeclaimthatM′ isstillmaximal.First,weshowthattherearestilloneentryfor
everyd-rowin{x: A(x)=0/}.IfD(x)=0/,thenM′(x,1)=1.IfD(x)6=0/,thensince
h(z)<w foreachz∈D(x),l(x)<w inM,meaningM′stillhasoneentryind-row
d d
x.Sol(·)andh(·)arestillwelldefinedinM′.Second,sincetheonlychangeinh(·)
and l(·) is the decrease of h(·) by 1 for {x: A(x)=0/}, Equation1 and Equation2
stillholdinM′,andsobyLemma5,M′ isstillmaximal.Byassumption
w(M′)=w w ...w (w −1)−(w −1)(w −1)...(w −1)(w −2)
1 2 d−1 d 1 2 d−1 d
A(x)=0/ ifandonlyifatleastoneofthecoordinatesx ,...,x is1.So
1 d−1
|{x: A(x)=0/}|=w w ...w −(w −1)(w −1)...(w −1)
1 2 d−1 1 2 d−1
Addingthemtogether,wehave
d d
(cid:213) (cid:213)
w(M)= w − (w −1)
i i
i=1 i=1
⊓⊔
Tocopewithd-rowsinX(M),wewillshowthatwecanalwaysconvertMtoM′
whichstill2-avoidsI maximally,with|X(M′)|<|X(M)|,andw(M)=w(M′).The
2,d
firststep,inLemma7,istofindapairofd-rowsxandx′whichwecouldmanipulate
justalittlebittoreduce|X(·)|whilekeepingmaximalityandw(·)intact.
Wesaythata d-rowx=(x ,...,x )isasemi-ancestorofd-rowif y≤xand
1 d−1
x6=y.
Lemma7 IfX(M)6=0/,thenthereared-rowsxandx′,suchthatx=(x′+1,...,x′ +
1 d−1
1),h(x)=w >l(x),andx′ hasnootherdescendantzwithh(z)=w .
d d
Proof Inthefirststagewefindad-rowxwith h(x)=w >l(x)andA(x)6=0/.We
d
startfromanarbitraryd-rowx∈X(M).Ifl(x)=w ,thenthereisdescendantzofx
d
withh(z)=w becausetheonlyotherpossibilityisD(x)=0/,whichleadstol(x)=1,
d
a contradiction.We set x to z and see if l(x)<w . We repeatthe process untilwe
d
findaxwithh(x)=w >l(x),whichisguaranteedtohavenon-emptyancestorset
d
6 Shen-FuTsai
A(x).Suchxexistsbecausethereareonlyfinitenumberofd-rows,andwheneverwe
hitonewithoutdescendants,l(x)=1<w .Wethensetx′=(x −1,...,x −1).
d 1 d−1
In the second stage, if x′ has no other descendant z with h(z)= w then we
d
are done. If it does, then since D(z)⊂D(x), l(z)≤l(x)<w , i.e. z qualifies the
d
requirement for x too. Moreover x is a semi-ancestor of z, so we set x to z and
x′ =(x −1,...,x −1). We repeat the process until x′ has no other descendant
1 d−1
z with h(z)=w . This process terminates because in the sequence of x we found,
d
eachelementisasemi-ancestorofthesubsequentelement,andthissequencecould
notbeinfinitelylong. ⊓⊔
Lemma8 If X(M)6=0/, we can convert M to another maximal matrix M′ that 2-
avoidsI with|X(M′)|<|X(M)|andw(M′)=w(M).
2,d
Proof By Lemma 7, we find d-rows x and x′ such that x=(x′ +1,...,x′ +1),
1 d−1
h(x)=w >l(x), and x′ has no other descendant z with h(z)=w . Construct M′
d d
identicaltoMexcepth′(x)=w −1andl′(x′)=w −1.WehaveX(M′)=X(M)−x
d d
andw(M′)=w(M).
WeclaimthatM′isstillamaximalmatrixthat2-avoidsI .Theonlychangewe
2,d
make is to decrease h(x) and l(x′) by 1. Clearly Equation 1 is not violated by M′,
becausex′isanancestorofx.TheonlypossibilitythatM′violatesEquation2isthat
x′ has another descendant z with h(z)=w . But this is impossible because of the
d
choicewemakewithx′. ⊓⊔
Finally,givenanymatrixMthatmaximally2-avoidsI ,ifnecessarybyapplying
2,d
Lemma8afinitenumberoftimeswecanconvertittoanothermaximalM′ withthe
samenumberofoneentries,still2-avoidsI ,andwithX(M′)=0/.ThenbyLemma
2,d
6wehavew(M)=w(M′)=(cid:213) d w −(cid:213) d (w −1).Thisconcludesourproof. ⊓⊔
i=1 i i=1 i
4 Numberofmaximalantichains
Afterprovingtheequivalenceofsuchmaximalandmaximumantichains,itisnatural
toaskhowmanymaximalantichainstherearegivenrangevectorw.Inthissection
weexplorethisquantity.
Definition6 R(w)issetofmaximalantichainsoverrange-restrictedvectorswithin-
tegralcoordinatesboundedbyw.Theorem1saysthateachelementofR(w)hassize
(cid:213) d w −(cid:213) d (w −1).
i=1 i i=1 i
Forwwithlength2,|R(w)|canbeexpressedinbinomialcoefficient.
Lemma9 |R([w1,w2])|=(cid:0)w1w+1w−21−2(cid:1)
Wefoundapropertyof|R(w)|whenwcontainsa2.
Lemma10 |R([w,2])|=R|(w)|
TitleSuppressedDuetoExcessiveLength 7
Proof Wewillshowthatthereisbijectivemapping f :R(w)→R([w,2]).Denotethe
lengthofvector[w,2]byd.ForamatrixMwithdimensionw ×...×w ×2,de-
1 d−1
fineS asthesetofitsd-rowsxwithM(x,y)=1ifandonlyifi≤y≤ j.
i,j
ForamatrixM∈R([w,2]),byLemma1andLemma2itsd-rowsarepartitioned
intodisjointsetsS for1≤i≤ j≤2.WeclaimthatS ∈R(w).Thisisbecause
i,j 1,2
d−1
(cid:213)
w(M)=2|S |+|S |+|S |= w +|S |
1,2 1,1 2,2 i 1,2
i=1
Maximizationofw(M) is equivalentto maximizationof|S |, whichbydefini-
1,2
tionleadstoS ∈R(w).
1,2
Thefollowingdescribesaninjective f.ForamatrixN∈R(w),weformamatrix
M by setting S =N and S =S =0/. Next, whenever there is a d-row x∈/ N
1,2 1,1 2,2
not in S nor S , we place it in either of them so that M still 2-avoidsI . Note
1,1 2,2 2,d
thatx couldnotbeboththeancestoranddescendantofsomed-rowsin N,because
otherwise N 2-containsI . Also, x must be either an ancestor or descendantof
2,d−1
some d-row in N, as otherwise N+x is a largerset that 2-avoidsI . Hence we
2,d−1
can always determine whether a d-row x∈/ N should be in S or S . Apparently
1,1 2,2
whenwearedoneMisamaximalmatrixthat2-avoidsI .
2,d
f is surjective because we have established that given a matrix M ∈R([w,2]),
S ∈R(w). ⊓⊔
1,2
Acorollaryfollowsimmediately.
Corollary1 Ifmax w ≤2,then|R(w)|=min w.
i i i i
5 Openquestions
Giventhesimpleexpression(cid:213) d w −(cid:213) d (w −1),itisnaturaltoaskifourresult
i=1 i i=1 i
canbeprovedwithsimplermethod,especiallywithoutthetechnicalandtrickyma-
neuvers in Lemma 7 and 8. Also, is there a better and perhaps more intuitive way
thanEquation1andEquation2tocharacterizeamaximalmatrixthat2-avoidsI ?
2,d
One of the motivations of pursuing an alternative characterization is to derive the
expressionmoregeneralthanSection4forthenumberofmaximalmatricesM that
2-avoidsI intermsofrangew=(w ,...,w ).
2,d 1 d
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