Averaging Sequences Deranged Mappings and a Problem of - PDF document

AveragingSequences,DerangedMappings,andaProblemofLampertandSlaterNeilJ.CalkinDepartmentofMathematicalSciencesClemsonUniversityClemson,SC29634-1907calkin@math.clemson.eduE.RodneyCaneldDepartmentofComputerScienceTheUniversityofGeorgiaAthens,GA30602,USAHerbertS.WilfDepartmentofMathematicsUniversityofPennsylvaniaPhiladelphia,PA19104-6395wilf@math.upenn.eduDedicatedtothememoryofGian-CarloRotaAbstractWeansweraquestionposedbyLampertandSlater[7].Considerasequenceofrealnumbersintheinterval[01]denedby=0,=1,and,forequalsanaverageofprecedingtermsinthesequence.Theweightsusedintheaverageareprovidedbyatriangulararrayn;kofprobabilitieswhoserowsumsare1.Whatisthelimitingbehaviorofasequencesodened?FortheLampert-Slatersequencetheweightn;kistheprobabilitythatarandomlychosenxed-pointfreemappingof[+1]omitsexactlyelementsfromitsimage.Togainsomeinsightintothisaveragingprocess,werstanalyzewhathappenswithasimplerarrayofweightsn;kdenedintermsofbinomialcoecients.Oneofourtheoremsstatesthatiftheweightsn;karecloselyconcentratedandthesequenceexhibitsoscillatorybehavioruptoacertaincomputablepoint,thenitwillexhibitoscillatorybehaviorfromthenon.WecarryoutthecomputationsnecessarytoverifythattheLampert-Slatersequencesatisesthehypothesesofthelattertheorem.Aresultonmartingales[1]isusedtoprovethecloseconcentrationoftheweightsn;kAMS-MOSSubjectClassication(1990):05A16,60C05,60G46 1IntroductionIn[7],thefollowingquestionisraised.Beginwithplayers,andrepeatthefollowing\knockout"procedurewhilethereremaintwoormoreplayers.Eachremainingplayerchoosesanotherplayeruniformlyatrandom;thesetofplayerssochosendropoutofthegame;thatis,weknockoutthechosenplayers.Thegameterminateswhenthereisasingleplayer,ornoplayer.LampertandSlaterintheirpaperconsidermoregeneralknockoutprocessesonagraphduringasingleroundofwhicheachvertexchoosesatrandomaneighborforknockingout.Theprocessdescribedabovecorrespondstobeingthecompletegraphonvertices.Thequestionis,asafunctionof=thenumberofplayerswhobeginthegame,whatistheexpectednumberofplayersremainingattheendoftheknockoutprocess?Clearly,thenumbersallliebetween0and1.Ourinitialcomputationsthrough=170revealedagentleoscillationbetweenmaximaintheneighborhoodof053andminimaintheneighborhoodof047,withanapparentslighttendencytowardsconvergence.Onemightconjectureatthatpointthatthelimitis12,andthattheconvergenceisslow.Tothecontrary,however,wehaveproventhatforappropriateconstantsa;b;andalllargecos(2:Thisimpliesthatthesequencehasnolimit.(AlllogarithmsinthispaperaretothebaseWhatistheconnectionbetweentheLampert-Slatersequenceandderangedmappings?Aderangedmappingoftheset[;:::nisafunctionnn]![n]suchthatforall.Thenumberofsuchfunctionsis(.Letbetheprobabilitythatarandomlychosenderangedmappingof[+1]omitsexactlyelementsfromitsimage.ThentheLampert-SlatersequenceisgivenbyThereadermaynotethattheprobabilitiessodenedarenonzeroonlyforintherange1,becauseanymappingof[+1]whichomitsexactlyelementsfromitsimageisconstant,andhencenotderanged.Nevertheless,wewritethebasicrecursion(2)asabove,withthesummationovertherange0,becausewewillwishtoconsiderotherunderlyingarrays.InSection2,inparticular,weanalyzeasequencedenedby(2)withadierentunderlyingtriangulararray.InSection3wegatherthenecessaryresultsonderangedmappings.InSection4,weprove(Theorem3)thatifthegivenprobabilitiessatisfycertainproperties(foremostofwhichiscloseconcentration),andifasucientlylongsegmentofthesequenceexhibitsatypeofoscillatorybehavior,then,infact,thesequenceoscillatesindenitely.Finally,inSection6,wedescribethecomputationsusedtoverifythattheprobabilitiesandsequenceposedbyLampertandSlaterdoindeedsatisfythehypothesesofTheorem3.Weconcludebyproving(1).Therolesofandchangefromsectiontosection.Thedenitionofisgivenatthestartofeachsection,andalwaysisdeterminedby(2). 2TheCoinFlippingGameInthissectionweconsiderthefollowingproblem:initiallycoinsareallheadsuponatabletop.Repeatthefollowingprocessuntilonlyoneornoneofthecoinsisheadsup: ipexactlyonceallofthecoinsthatarestillshowingheads.Whatistheprobabilitythatweterminatewithexactlyonehead?ThisquestionwasraisedasamonthlyProblemin1991andasolutionwasgivenin1994[9].Theanswerwasgiveninjustsucientdetailtoanswerthequestionpreciselyasasked,whichwastodecideifthelimitexistsornot(itdoesn't).Herewegiveamoredetailedsolutiontogivesomeadditionalinsightintothesortofbehaviorthatsuchproblemsexhibit,inacontextthatissimplerthantheknockoutsproblemthatwillbediscussedbelow,andisthemainobjectofthispaper.Itisclearthatthesequenceisgivenbytherecursion(2)withtheunderlyingprobabilities:=(2Weshallcomputetheordinarygeneratingfunctionforthissequence.Summingover2,weobtain 2=�x 2+2 2�xXn0qnx 2�xn;sof(x 2+2 2�xfx andnow,byiteratingthis,weobtain 2+x x)2+2x 3x)2+4x 8�7x+8x )2+x 21Xnn Withthisgeneratingfunctionwearenowreadytoprovethenonexistenceofthelimit.Infactwewillgivethreedierentproofsofthis.Let'sstateitasatheorem. Theorem1Theprobabilityofterminatingwithexactlyoneheadinan-headgamedoesn'ttendtoalimit.2.1FirstproofofTheorem1Nowwecangetanexplicit,niteformulaforthecoecients.Indeedwehave 2Xm02�m1 2�m)x)2=x +1)(1+1) +1) +1) x`j `j!(�1 Thecoecientofgivesus 2Xj n�1j!(� 1�2�j�1=n 28:n;1+Xj n�1j!(� anite,explicitformulaforourcoecients.Sumsofthistypehaveappearedelsewhere.Preciselythissequenceistreatedintheanalysisofaprobabilisticmodelinnumbertheory[8].In[4]wendadiscussionofthequantities Intermsofthem,our'saregivenby Nowaformulaforoursequencecanbereadofromthatofthe's,asgiveninthelatterpaper.Wehave 2log2 2n;1+Hn�1 2log2miH log2 mi log2mi log2 mi log2 wherethe'saretheharmonicnumbers.As,theaboveformulais 2log2mi log2milogNotethatisaperiodicfunctionoflog,toaccuracy(1).Asimilarformulaappearsin[5,Section5.2.2,equation(47)],whereitarisesintheasymptoticanalysisofasortingalgorithm.2.2SecondproofofTheorem1Thistimeweusethefactthatifexists,thenHence 21Xkk ��=) 21Xkk (1+(2Itiseasytoseethatthebehaviourofthiswhenissmallisclosetothebehaviourof 2Xk2k (1+2Write)=2(1+,and2.BytheEuler-Maclaurinsumformula, 2Z1(t)Z1(t)u0(t)=1 2log2 2Z1(t)u0(t)1 2log2 2Z10 )1�t (1+Intheintegral,ifwecutinhalf,weadd1totheargumentof,whichdoesn'tchangeitsvalue.Sotheintegralisinvariantunderhalvingof.Tolearnmore,wereplacebyitsFourierseriessin2nt n5 andintegratetermwise(whichHardyisfondofnotingcannotbejustiedbyabsoluteconvergencebutitcanbyboundedconvergence)toobtain, 2log2 2Z10 )1�t (1+ 2log2 2Xn11 102 1�t (1+ 2log2 2Xn11 sin(2)cos(2 (1+ 2Xn11 cos(2)sin(2 (1+ 2log2 2Xn11 cos(2 (log2)sinh(2log2) 2log2 (log2) sinh(2log2)cos(2showingthat)isperiodicinlog(andnotconstant),andhencethatdoesn'texist.2.3ThirdproofofTheorem1 whereisthefractionalpartofthebase-2logarithmofThisimpliesthatdoesn'ttendtoalimit.Toprovetheassertedasymptoticformula,weextractthecoecientof),tond 21Xk�k1�2�kn�1:6 Hence, 21Xkt+�k 1�2t+�k n!n�1=1 21Xk=�t2�k 1�2�k Now,if3,then n!n�1 1�2+t 3 n!n�1= 1�n1=3 andso Now,if3,then n!n�1=e�2�k 2�k�2k andso 2Xk2�ke�2�k 2�k�2k no1 n=1 21Xk=�ke�2�k+O1 n=1 21Xk=+ke�2+k+O1 asclaimed.Bytakingmorecarewiththeexpansionof ,weobtain ns2()�1 2s3()+1 n2s3()�5 6s4( 8s5()+O1 n3;7 inwhich Thiscan,ofcourse,beexpandedtoanyxednumberofterms.Bywayofexample,actualcomputationgives


,and(0)+ ns21 2s3+1 n2s35 (0)+ 7213524304938


3PropertiesofDerangedMappingsThroughoutthissectionequalstheprobabilitythatarandomlychosenderangedmappingofofn+1]omitsexactlypointsfromitsimage,0.InSection5wegiveanalgorithmforcomputingtheseprobabilities.Theliteratureonrandommappingsisvast,seeforexample[6],althoughthexed-point-freepropertydoesnotappearoften.In[2]theasymptoticdistributionformanystatisticsonmappingsisderived,includingthenormalityoftheimagesize.Ofcourse,thenumberofelementsomittedfromtherangediersfromthelatteronlybyaconstant.Itturnsoutwedonotneedinformationonthedistributionofthisstatisticsomuchasweneedaboundondeviationfromaveragebehavior;thatis,acloseconcentrationresult.ThenecessarytheoremappearsinChapter7(Martingales)ofthebook[1]byAlon,Spencer,andErd}os.Thetotalnumberofderangedmappingsof[+1]is,sinceinconstructingafunctionforeachintegerinthedomaintherearechoicesfor).Assigningeachofthesemappingstheprobabilityisaninstanceofthefollowinggeneralsituation:lettherebegivenadomainarange,andanjjmatrixofprobabilitieswhoserowsumsareall1.Theentryappearingatrowandcolumnequalstheprobabilitythatafunctionsatises,thelattereventsbeingindependentoverdierent.Thegivenmatrixdeterminesaprobabilitydistributiononthesetofallfunctions.Theuniformprobabilityspacewhichweareconsideringforderangedmappingsisthecaseinwhichbothandhavesize+1,alldiagonalentriesofthematrixare0,andallodiagonalentriesare1Continuingtofollow[1,p.89],letbeafunctional(weshallbeinterestedinbeingthenumberofpointsomittedfromtheimageof),andlet


bean-gradationoftherange.Thesequence;:::Xdenedonfunctions)forallisamartingale.(See[1]forundenedterms.)Notethat)istheconstant),independently,and)is).Weneedthefollowingconcentrationresult: Theorem2[1,p.90].Letaprobabilitymeasureonanitefunctionspacebegivenasabovebyanmatrixofnonnegativenumberswhoserowsumsare,andletbeafunctionalsatisfyingtheLipschitzconditionf;fdieronlyonwithrespecttoagivengradation(3).Then,forandallwehaveLetusremarkthatwithhi],0+1,and)=thenumberofpointsomittedfromtherangeof,theLipschitzconditionissatised.Thus,wehaveacloseconcentrationresultfortheprobabilitiesWecompletethissectionbycomputingthemeanandstandarddeviationofthedistribution:Ofcoursethemeanissimply+1timestheprobabilitythataparticularelement,say1,isomittedfromtherange:+1)(1Turningto,westartwith+1)whichisobtainedbycountingtriples(i;j;f),withaxed-point-freemapping,i;jtwodistinctelementsnotintherangeof,andthendividingbythetotalnumberofxed-point-freemappings.Adding,andsubtracting,wend+1)+1)(1+1)Somecarefulcalculationshowsthatdiersfrom(+1))bylessthan1.Werecordforfutureuse:for+1)1for 4ATheoremAboutOscillatoryBehaviorThroughoutthissectiondenotesagenericarrayofprobabilitiesaboutwhichweshallassumecertainhypothesesandprovecertainresults.Weshalluse)todenotethefunctioncos(theconstantsbeinggiven.Withnolossonemaytaketheconstantspositive,andwedoso.Anarray,determiningasequence,andafunction)understoodtohavebeengiven,forintegersandwedeneI;J=maxIkthemaximumabsolutedierencebetweenandtheapproximation)overthehalfopeninterval(I;JWeintroduceoneadditionalconvenientnotationinvolvingthesymbol.Wheneverappearsinanequation,itstandsforarealnumberwhosevalueisintheinterval[+1].Itis,ofcourse,notnecessarilythesamevalueateachappearance;moreover,inagivenequation,itsvaluemaydependonthefreevariablesfoundintheequation.Ifwewanttosay,forexample,thattworealvaluedfunctions)and)dierinabsolutevaluebynomorethan10forall,wewouldwrite)+10Precisely,thissaysthat,denedon10,neverexceeds1inabsolutevalue.Twofurtherexamplesofthisnewnotation,bothusedinthesequel,arelog(1+3(6)bTherstisprovenbynotingthat(log(1+isanincreasingfunctionof1.ThesecondfollowsfromTaylor'sformulawithremainder.Thenexttheoremgivesconditionsontoquantifyandprovethenotionthatifapproximateswellonasucientlylargeinterval(I;J],thenthereisasubstantiallylargerinterval(I;K]wheretheapproximationisonlyslightlylessgood.Theorem3Given:atriangulararray,ofprobabilitieswhoserowsumsequal,andsixpositiveconstants.Letbedenedbytherecursion(2),cos(I;Jbedenedby(5),)bethemeanofthe-throw,)bethevarianceofthe-throw,andbethesetofsuchthat +1).Assumethefollowingconditions:+1)(1+=n+1)+1Thenthereexistconstants,andsuchthatforeverypairofintegersIJandthereexistsaninteger(1+suchthatJ;KI;JProof.Webeginbytellinghowtochoose,and.LetChoosesolargeandpositivebutsosmallthatmax=;and(1+)(1+)(1+Then,chooseandbytheformulasc=2(1+=max(1LetIJbetwointegerssatisfyingcondition(8).Weclaimthatmaybetakenas(1+Toseethis,let+1beanintegerinthehalfopeninterval(J;K].(Inwhatfollows,wesometimeswithoutexplicitmention.)WehaveLetting,weseethatthesecondsumontherightofthepreviousequationequals.Toboundtherstsum,werstcheckthatI;J].Byassumption,+1) +1);sincethelatterisincreasingfor+1)+1)+1)(1+1)and).Intheotherdirection,byassumption,+1)(1+);sincethelatterisanincreasingfunctionof+1)+1)(1+)+(+1)(1+)(1++1)(1+)(1+(1+)(1+(1+)(1+)(1+and].Hence,asasserted,I;J],andI;JForthedurationoftheproof,letn;k))beimplicitlydenedby+1)Using(for +1) +1) +1) andtheassumptionthat,wehave(again,for+1) (1+ Thus,(6)isapplicable,andlog()=log(+1) +1) forUsing(7)wecalculate,for+1) +1) Tocontinue,wecomputemax(;n +1)aswellas max( =max(1;n=1),wehavealtogether I;J(1+Usingtheassumptionsabout,and,wehave(log(+1))+bnwhichinconjunctionwith(13)implies(9).ThiscompletestheproofofTheorem3.Nowwemaystateacorollarywhich,inconjunctionwithcomputation,willpermitustoprovethatsequencesofthetypewearestudyingdonothavelimits.LetsatisfythehypothesesofTheorem3,letbedenedby(12),andassumethatsatisfyconditions(10)and(11).Letbethelargestintegerpossible,subjecttoThen,foreveryI;N(1+ 1�e�=3=6:13 Inparticular,ifthelatterislessthan,thenthesequencehasnolimit.Proof.Bythedenitionof
theinequalityholdsforIn.Letand(1+ForIn,wehavebyTheorem3,I;NLet(1+;againbyTheorem3,forIn,wehaveI;NByinduction,with(1+,wendthatforallI;NSince(1+,wehave(1+and,since(1+(1+i=i= TheCorollaryfollows.5AwarmupexerciseToillustratetheuseoftheCorollary,wegiveanotherproofthatthelongtermbehaviorofthecoin ippingproblemisoscillatory.Theninthefollowingsectionwewillapplyittotheknockoutsproblem.TousetheCorollaryrequiressomewhatlengthycalculations.Wewilldescribetheprocessindetailforthisexample,andbemorebriefwiththesecondexample.ThersttaskistondalloftheconstantsinTheorem3.Ourtriangularmatrixofprobabilitieshereis1).Itiseasytoverifythatthetailestimate whichisahypothesisofTheorem3,holds,bystandardestimates(e.g.,AppendixAof[1])ofthetailsofthebinomialdistribution.Itisalsonothardtodeterminethat2andBecause2,wetakelog(2).From(12)wehave000557393854568Todetermine,tentatively,ourapproximatingfunction),wecomputetherstonethousandvaluesof,andsolvefor,andtominimizethesumofsquarescos(log(sin(log(Togetthemorefamiliarformof),wesolveforandsuchthatcos(cos(Wendthatisabout7,rathersmall.Knowing,weguessinsistingthatthelasttwotermsontherightsideof(15)sumtolessthan6.Inthisway,wedecidetotake=20000.From(14)wetake261.Nowwerepeattheleastsquarest,usingthe\ocial"limits9000insteadofthetentativelimits400000in(16).Thisyieldsthevaluesforthefunction),andgivesalsoI;N]=maxIk(log(Thecomputedvalueforagreescloselytothetheoreticalvalue1(2log(2))giveninSection2.Finally,wecomputetherightsideof(15)tobe0


,whichislessthan


,andwehaveconrmedthathasnolimit.6ComputationsfortheKnockoutsProblemThroughoutthissectionagainequalstheprobabilitythatarandomlychosenderangedmappingontheset[+1]omitsexactlypointsfromitsimage.Denen;k)tobethenumberofwaysto partitiontheset[]intoanorderedcollectionofblocks,suchthatfor1elementdoesnotbelongtothe-thblock.Suchanorderedpartitioncorrespondsinanaturalwaytoaxed-point-freemappingof[]whoseimageisexactlytheset.Hence,andAmappingwhichomitsnoelementfromitsimageisaxed-point-freepermutation,alsoknownasaderangement.Therecursionforcountingderangementsiswellknown[3],andwehave0)=Nowletusconsider)whenisatleast1.Theorderedpartitionscountedbyareoftwovarieties.Intherstvariety,wehavethosepartitionsinwhichelement+1isasingletonblock;inthesecondvarietywehavethosepartitionsinwhichelement+1belongstoablockofsizetwoorgreater.Tocreateanorderedpartitionofthersttype,weproceedinthreesteps:(1)chooseaninteger,intherange1+1;(2)chooseanorderedpartitionof[]intoblockssuchthatisnotinthe-thblockfor1ijandsuchthat+1isnotintheblockfor;(3)insertasthe-thblock.Wedenedn;k)ascountingorderedpartitionsof[]intoblockssuchthatelementisforbiddenfromthe-theblock.However,amoment'sre ectionwillrevealthatn;k)willalsocountcorrectlyschemeofforbiddinginwhichacertainelementisdeniedmembershipinthe-thblock,andtheelementssosingledoutarealldistinct.Hence,thenumberoforderedpartitionsinstep(2)aboveequalsn;k),andthetotalnumberofpartitionsoftherstvarietyis(+1)n;kTocreateanorderedpartitionofthesecondtype:(1)partition[]into+1blocks,keepingoutofthe-thblock;(2)chooseablockintowhich+1istheninserted.Notethatstep(1)isfeasible,andthatstep(2)doesnotcreateanyforbiddenmemberships,duetotheassumption1.Summarizing,wehavethenicerecursion:+1)(n;kn;k1))Letusremarkthatifwellthe=0columnofthearraywiththefactorials1insteadofthederangementnumbers1,andthenlltherestofthetable(where1)byexactlythesamerecursion(17),theresultingtablecontains(n;n),wheren;b)istheStirlingnumberofthesecondkind.Theprobabilitiesassociatedwith(n;ncorrespondtochoosingamappingof[+1]atrandom,withnorequirementthemappingbexed-point-free.ThisamountstoaLampert-Slaterknockoutgameinwhichself-eliminationispermitted. Altogether,then,wecancomputeinitialrowsofthen;k)arrayinanumberofarithmeticoperationswhichisproportionaltothenumberofvaluescomputed.Thissuggeststhattherstvaluesofthesequencecanbecomputedinquadratictime,butsuchaconclusionignoresafurthermultiplicativefactorofinthecomplexityduetothesizeoftheintegeroperandsinvolved.Itis,however,feasibletocomputetherst1776valuesofusingtheabovescheme,providedthecalculationisdonein oatingpoint,andnotexactly.Therearisesthequestionofroundingerror.Toconrmreliability,thecomputationshavebeencarriedoutintwodierentprecisions,rstwithDigits=22,thenwithDigits=32.(ForthoseunfamiliarwiththesymboliccomputationsystemMaple,\Digits"isaglobalvariablesetbytheuserwhichcontrolsthenumberofdigitskeptin oatingpointcomputations.)Thetworesultsagree,outto=1776,intherstsixteenplacesalways.Foranyonewishingtorepeatthecalculations,wereportthefollowingsamplevalues,obtainedwithDigits=32,4767534353123257282220563501866652829933875860791739826500429501FromSection3,weknowthatand).NotethatthefourassumptionsneededtoapplyTheorem3arefullled:theinequalitiesneededforandareimpliedby(4),andtheconcentrationresultforisgiveninTheorem2.Because,wehave.Thevalueschosenfor,andwereagaindeterminedbyaleastsquarest.Hereisthesummaryofallcomputations:=1776=505I;NOnenalactofarithmeticsrevealsI;N(1+ since00243issmallerthan,wehaveproven(1).References[1]NogaAlon,JoelH.Spencer,andPaulErd}os,TheProbabilisticMethod,JohnWiley&Sons,Inc.,NewYork,1992. [2]JimArneyandEdwardA.Bender,Randommappingswithconstraintsoncoalescenceandnumberoforigin,Pacic.J.Math.(1982),269{294.[3]LouisP.Comtet,AdvancedCombinatorics,D.Reidel,1974.[4]PeterKirschenhofer,ANoteonAlternatingSums,TheElectronicJournalofCombinatorics(2)(1996),#R7.[5]DonaldErvinKnuth,TheArtofComputerProgramming,VolumeIII:SortingandSearchingAddison-Wesley,Reading,Massachusetts,(1973).[6]ValentinKolchin,RandomMappings,OptimizationSoftware,NewYork(1986).[7]DouglasE.LampertandPeterJ.Slater,Parallelknockoutsinthecompletegraph,AmericanMath.Monthly(1998),556{558.[8]ShuguangLi,OnArtin'sConjectureforCompositeModuli,DoctoralDissertation,UniversityofGeorgia,1998[9]LennartRade,proposer,ProblemE3436,TheAmer.Math.Monthly(1991),p.366;solu-tionbyO.P.Lossers,ibid.(1994),p.78.