Inthispaper,weareinterestedinrandomgraphswithaxeddegree

Transcription

1 ACRITICALPOINTFORRANDOM GRAPHSWITHAGIVENDEGREE DepartmentofMathematics PittsburghPA15213,U.S.A. Carnegie-MellonUniversity SEQUENCE MichaelMolloy UniversitePierreetMarieCurie EquipeCombinatoire BruceReed August14,2000 Paris,France CNRS thenalmostsurelyallcomponentsinsuchgraphsaresmall.wecan sumto1,weconsiderrandomgraphshavingapproximatelyinverticesofdegreei.essentially,weshowthatifpi(i?2)i>0thensuch Givenasequenceofnon-negativerealnumbers0;1;:::which Abstract graphsalmostsurelyhaveagiantcomponent,whileifpi(i?2)i<0 applytheseresultstogn;p;gn;m,andotherwell-knownmodelsof randomgraphs.therearealsoapplicationsrelatedtothechromatic numberofsparserandomgraphs. 1

2 berofverticesandthenumberofcyclesinthelargestcomponent.ofcourse, fromwhichthegraphsarepicked.inonestandardmodelwepickarandom thebehaviouroftheseparametersdependsontheprobabilitydistribution Inthispaperweconsidertwoparametersofcertainrandomgraphs:thenum- graphgn;mwithnverticesandmedgeswhereeachsuchgraphisequally 1IntroductionandOverview ofnandletngotoinnity.thepointm=12nisreferredtoasthecritical likely.weareinterestedinwhathappenswhenwechoosemasafunction size(n2=3).ifm=cnforc>12thenthereareconstants;>0dependent probabilitytendingtooneasntendstoinnity)gn;mhasnocomponentof pointorthedouble-jumpthresholdbecauseofclassicalresultsduetoerd}os oncsuchthata.s.gn;mhasacomponentonatleastnverticeswithat sizegreaterthano(logn),andnocomponenthasmorethanonecycle.if M=12n+o(n),thenalmostsurely(a.s.)thelargestcomponentofGn;Mhas andrenyi[8]concerningthedramaticchangeswhichoccurtotheseparametersatthispoint.ifm=cn+o(n)forc<12thenalmostsurely(i.e.with leastncycles,andnoothercomponenthasmorethano(logn)verticesor see[3],[11],or[14]. probability.ofcourse,wehavetosaywhatwemeanbyadegreesequence. sequencewhereeachgraphwiththatdegreesequenceischosenwithequal Ifthenumberofverticesinourgraph,nisxed,thenadegreesequenceis ofgn;m.formorespecicsonthesetwoparametersatandaroundm=12n morethanonecycle.thiscomponentisreferredtoasthegiantcomponent simplyasequenceofnnumbers.however,weareconcernedherewithwhat quenceofsequences".thus,wegeneralizethedenitionofdegreesequence: happensasymptoticallyasntendstoinnity,sowehavetolookata\se- Inthispaper,weareinterestedinrandomgraphswithaxeddegree valuedfunctionsd=d0(n);d1(n);:::suchthat 2.Pi0di(n)=n. 1.di(n)=0forin; GivenanasymptoticdegreesequenceD,wesetDntobethedegree Denition:Anasymptoticdegreesequenceisasequenceofinteger- sequencefc1;c2;:::;cng,wherecjcj+1andjfj:cj=igj=di(n)foreach i0..denedntobethesetofallgraphswithvertexset[n]withdegree 2

3 sequencedn.arandomgraphonnverticeswithdegreesequencedisa degreesequenced,wewantthesequencesdntobeinsomesensesimilar. alln1. uniformlyrandommemberofdn. Denition:AnasymptoticdegreesequenceDisfeasibleifDn6=;for constantsisuchthatlimn!1di(n)=n=i. Wedothisbyinsistingthatforanyxedi,theproportionofverticesof degreeiisroughlythesameineachsequence. Becausewewishtodiscussasymptoticpropertiesofrandomgraphswith Inthispaper,wewillonlydiscussfeasibledegreesequences. andweonlyclaimthingstobetrueforsucientlylargen. beenrandomregulargraphs.perhapsthemostimportantrecentresultisby Denition:AnasymptoticdegreesequenceDissmoothifthereexist graphforanyconstantk3,thengisa.s.hamiltonian. RobinsonandWormald[20],[21],whoprovedthatifGisarandomk-regular Throughoutthispaper,allasymptoticswillbetakenasntendsto1 quencecomesfromtheanalysisofthechromaticnumberofsparserandom graphs.thisisbecauseaminimally(r+1)-chromaticgraphmusthavemini- Inthepast,themostcommonlystudiedrandomgraphsofthistypehave mumdegreeatleastr.inanattempttodeterminehowmanyedgeswerenec- iedtheexpectednumberofsubgraphsofminimumdegreethreeinrandom cessarytoforcearandomgraphtoa.s.benot3-colourable,chvatal[7]stud- graphswithalinearnumberofedges.heshowedthatforc<c=1:442:::, Anothermotivationforstudyingrandomgraphsonaxeddegreese- theprobabilitythatarandomgraphonnverticeswithminimumdegree authorsusedaspecialcaseofthemaintheoremofthispapertoshowthat threeandatmost1:793nedgesisminimally4-chromaticisexponentially ponentiallylarge.intheworkthatmotivatedtheresultsofthispaper,the whileforc>ctheexpectednumberofsuchsubgraphsingn;m=cnisex- small[18].weusedthistoshowthatforcalittlebitbiggerthanc,the theexpectednumberofsuchsubgraphsingn;m=cnisexponentiallysmall, nentiallysmall.thissuggeststhatdeterminingtheminimumvalueofcfor expectednumberofminimally4-chromaticsubgraphsofgn;m=cnisexpo- whicharandomgraphwithcnedgesisa.s.4-chromaticmayrequiremore thanastudyofthesubgraphswithminimumdegree3. Recently Luczak[14]showed(amongotherthings)thatifGisarandom 3

4 thenallthecomponentsofsucharandomgrapharea.s.quitesmall.note giantcomponent.ourmaintheoremalsogeneralizesthisresult. graphonaxeddegreesequence1,withnoverticesofdegreelessthan2, howcloselythisparallelsthephenomenoninthemorestandardmodelgn;m. andatleast(n)verticesofdegreegreaterthan2,thenga.s.hasaunique foranymodelofrandomgraphsaslongas:(i)wecandeterminethedegree sequenceofgraphsinthemodelwithreasonableaccuracy,and(ii)oncethe graphwithdegreesequenceda.s.hasagiantcomponent,whileifq(d)<0 Notefurtherthattheseresultsallowustodetermineasimilarthreshold WesetQ(D)=Pi1i(i?2)i.EssentiallyifQ(D)>0thenarandom degreesequenceisdetermined,everygraphonthatdegreesequenceisequally likely.gn;pissuchamodel,andthus(asweseelater),ourresultscanbe Dnhas(i+o(1))nverticesofdegreeiforeachi0.Pickarandomvertex usedtoverifythepreviouslyknownthresholdforgn;p. inourgraphandexposethecomponentinwhichitliesusingabranching itsneighbours,repeatinguntiltheentirecomponentisexposed.nowwhen ofwhyitdetermineswhetherornotagiantcomponentexists.supposethat process.inotherwords,exposeitsneighbours,andthentheneighboursof avertexofdegreeiisexposed,thenthenumberof\unknown"neighbours Beforedeningtheparameterprecisely,wegiveanintuitiveexplanation increasesbyi?2.theprobabilitythatacertainvertexisselectedasa neighbourisproportionaltoitsdegree.thereforetheexpectedincreasein thenumberofunknownneighboursis(roughly)pi1i(i?2)i.thisis,of course,q(d). quickly.however,ifitispositivethenthenumberofunknownneighbours, formally. mainthrustofourarguments.wewillnowbegintostateallofthismore andthusthesizeofthecomponent,mightgrowquitelarge.thisgivesthe Thus,ifQ(D)isnegativethenthecomponentwilla.s.beexposedvery translate. insistthattheasymptoticdegreesequencesweconsiderarewell-behaved. Inparticular,whenthemaximumdegreeinourdegreesequencegrowswith n,wecanrunintosomeproblemsifthingsdonotconvergeuniformly.for example,ifd1(n)=n?dn:9e;di(n)=dn:9eifi=dpne,anddi(n)=0 1Hedidn'tusetheasymptoticdegreesequenceintroducedhere,buttheresults Thereareafewcaveats,soinorderforourresultstohold,wemust 4

5 However,thisisdeceivingasthereareenoughverticesofdegreepntoensure otherwise,then1=1,andi=0fori>1,andwegetq(d)=?1. thatagiantcomponentcontainingn?o(n)verticesa.s.exists. 1.Disfeasibleandsmooth. 2.i(i?2)di(n)=ntendsuniformlytoi(i?2)i;i.e.forall>0there Denition:AnasymptoticdegreesequenceDiswell-behavedif: 3. existsnsuchthatforalln>nandforalli0: L(D)=lim ji(i?2)di(n) n n!1xi1i(i?2)di(n)=n?i(i?2)ij< exists,andthesumapproachesthelimituniformly,i.e.: (a)ifl(d)isnitethenforall>0,thereexistsi;nsuchthatfor (b)ifl(d)isinnitethenforallt>0,thereexistsi;nsuchthat alln>n: foralln>n: jixi=1i(i?2)di(n)=n?l(d)j< WenotethatitisaneasyexercisetoshowthatifDiswell-behavedthen: L(D)=Q(D) ixi=1i(i?2)di(n)=n>t assparse: numberofedgesinourdegreesequence.wedenesuchadegreesequence Itisnotsurprisingthatthethresholdoccurswhentherearealinear 5

6 K+o(1)forsomeconstantK. Denition:AnasymptoticdegreesequenceDissparseifPi0idi(n)=n= chosenuniformlyatrandomfromamongstallsuchgraphs.then: di(n)=0.letgbeagraphwithnvertices,di(n)ofwhichhavedegreei, nitethendissparse. degreesequenceforwhichthereexists>0suchthatforallnandi>n14?, Themainresultinthispaperisthefollowing: Notethatforawell-behavedasymptoticdegreesequenceD,ifQ(D)is Theorem1LetD=d0(n);d1(n);:::beawell-behavedsparseasymptotic (a)ifq(d)>0thenthereexistconstants1;2>0dependentond (b)ifq(d)<0andforsomefunction0!(n)n18?,di(n)=0 foralli!(n),thenforsomeconstantrdependentonq(d),g dependentond. suchthatga.s.hasacomponentwithatleast1nverticesand hasfewerthan2r!(n)2logncycles.also,a.s.nocomponentof a.s.hasnocomponentwithatleastr!(n)2lognvertices,anda.s. onecomponentofsizegreaterthanlognforsomeconstant 2ncycles.Furthermore,ifQ(D)isnitethenGa.s.hasexactly analogoustothecasec=1inthemodelgn;p=c=n,andwouldbeinteresting Theorem1(a)agiantcomponent. NotealsothatTheorem1failstocoverthecasewhereQ(D)=0.Thisis NotethatifQ(D)<0thenQ(D)isnite. ConsistentwiththemodelGn;M,wecallthecomponentreferedtoin Ghasmorethanonecycle. toanalyze. onaxedwell-behaveddegreesequencewithcn+o(n)edgesforanyc>1 isthatitisdiculttogeneratesuchgraphsdirectly.insteadishasbecome standardtostudyrandomcongurationsonaxeddegreesequence,and Pi1ii>2;Pi1i(i?2)i<0;Pi1i=1;0i1. thenga.s.hasagiantcomponent,asthereisnosolutionto OneimmediateapplicationofTheorem1isthatifGisarandomgraph usesomelemmaswhichallowustotranslateresultsfromonemodeltothe andrenedbybollobas[3]andalsowormald[22]. other.thecongurationmodelwasintroducedbybenderandcaneld[2] Amajordicultyinthestudyofrandomgraphsonxeddegreesequences 6

7 degreesequence,wedothefollowing: 1.FormasetLcontainingdeg(v)distinctcopiesofeachvertexv. 2.ChoosearandommatchingoftheelementsofL. Inordertogeneratearandomcongurationwithnverticesandaxed denedbythepairsinthematching.wesaythatacongurationhasa graphicalpropertypifitsunderlyingmultigraphdoes. ofarandomcongurationonadegreesequencemeetingtheconditionsof Theorem1issimplewithprobabilitytendingtoe?(D),forsome (D)<O(n1=2?).Theconditiondi(n)=0foralli>n1=4?isneededto Eachcongurationrepresentsanunderlyingmultigraphwhoseedgesare applythisresult.ifq(d)isnitethen(d)tendstoaconstant. urations,whichisclearlyequalforallgraphsonthesamedegreesequence Usingthemainresultin[17]itfollowsthattheunderlyingmultigraph andthesamenumberofvertices. propertypwithprobabilityatleast1?znforsomeconstantz<1,thena meetingtheconditionsoftheorem1(withq(d)possiblyunbounded)hasa Thisgivesusthefollowingveryusefullemmas: Also,anysimplegraphGcanberepresentedbyQv2V(G)deg(v)!cong- thenarandomgraphwiththesamedegreesequencea.s.hasp. meetingtheconditionsoftheorem1a.s.hasapropertyp,andifq(d)<1, Lemma2IfarandomcongurationwithagivendegreesequenceD Lemma1IfarandomcongurationwithagivendegreesequenceD conguration. opedindependentlybybollobasandfrieze[6],flajolet,knuthandpittel graphsonagivendegreesequence. [10],andChvatal[7].Bothmodelsareveryusefulwhenworkingwithrandom Thecongurationmodelisverysimilartothepseudographmodeldevel- UsingtheseLemmas,itwillbeenoughtoproveTheorem1forarandom inwhichwearestudyingthem,wecannowgiveamoreformaloverviewof theproof.theremainderofthissectionisdevotedtothisoverview.inthe seesomeapplicationsoftheorem1:theaforementionedworkconcerningthe followingtwosectionswegiveallthedetailsoftheproof.insection4,we Havingdenedthepreciseobjectsthatweareinterestedin,andthemodel 7

8 bemorespecicregardingtheorderinwhichweexposethepairsofthe randommatching. nishthissectionandthenskipaheadtothelastone. readerwhoisnotinterestedinthedetailsoftheproofmightwanttojust chromaticnumberofsparserandomgraphs,andanewproofofaclassical double-jumptheorem,showingthatthisworkgeneralizesthatresult.a whichhavedegreeiasfollows: exposed.avertexsomebutnotallofwhosecopiesareinexposedpairsis GivenD,wewillexposearandomcongurationFonnvertices,di(n)of Inordertoexaminethecomponentsofourrandomconguration,wewill partiallyexposed.allotherverticesareunexposed.thecopiesofpartially exposedverticeswhicharenotinexposedpairsareopen. 1.FormasetLconsistingofidistinctcopiesofeachofthedi(n)vertices Ateachstep,avertexallofwhosecopiesareinexposedpairsisentirely 2.RepeatuntilLisempty: whichhavedegreei. (a)exposeapairoffbyrstchoosinganymemberofl,andthen Allrandomchoicesaremadeuniformly. (b)repeatuntiltherearenopartiallyexposedvertices: Chooseanopencopyofapartiallyexposedvertex,andpairit withanotherrandomlychosenmemberofl.removethemboth choosingitspartneratrandom.removethemfroml. atime.whenanycomponentiscompletelyexposed,wemoveontoanew one;i.e.werepeatstep2(a). Essentiallyweareexposingtherandomcongurationonecomponentat ofthisfreedom,butinmostcaseswewillpickitrandomlyinthesamemanner 2(a).Inafewplacesinthispaper,itwillbeimportantthatwetakeadvantage withthesameprobabilityunderthisprocedure,andhencethisisavalidway inwhichwepickalltheothervertex-copies,i.e.unlesswestateotherwise, tochoosearandomconguration. Itisclearthateverypossiblematchingamongstthevertex-copiesoccurs wewillalwaysjustpickauniformlyrandommemberofl. NotethatwehavecompletefreedomastowhichvertexwepickinStep 8

9 isexposed.iftheneighbourofvchoseninstep2(b)isofdegreed,then Xigoesupbyd?2.Eachtimeacomponentiscompletelyexposedand werepeatstep2(a)thenifthepairexposedinstep2(a)involvesverticesof ofdegreedisrd,thentheprobabilitythatwepickacopyofavertexof degreed1andd2thenxiissettoavalueofd1+d2?2. Now,letXirepresentthenumberofopenvertex-copiesaftertheithpair degreedinstep3(b)isrd=pi1ri.thereforeinitiallytheexpectedchange inxiisapproximatelypi1i(i?2)di(n) Notethatifthenumberofvertex-copiesinLwhicharecopiesofvertices Walk",withanexpectedchangeofQ(D) Markovprocessverycloseindistributiontothewell-studied\Drunkard's standardresultofrandomwalktheory(seeforexample[9])impliesthatif Therefore,atleastinitially,ifthisvalueispositivethenXifollowsa Pj1jdj(n)=Q(D) Q(D)>0,thenafter(n)steps,Xiisa.s.oforder(n). K.SinceXi+1Xi?1always,a K: quickly,andthiswillgiveustheotherpartoftheorem1,asthesizesof seethatifq(d)isbounded,thenthisgiantcomponentisa.s.unique. thecomponentsoffareboundedabovebythedistancesbetweenvaluesof cyclesinit,andthiswillgiveustherstpartoftheorem1.wewillalso on(n)vertices.wewillseethatsuchacomponenta.s.hasatleast(n) Ontheotherhand,ifQ(D)<0,thenXia.s.returnstozerofairly Itfollowsthatourrandomcongurationa.s.hasatleastonecomponent isuchthatxi=0. Therearethreemajorcomplications: 1.Apurerandomwalkcandropbelow0.WheneverXireaches0,it OfcoursetherandomwalkfollowedbyXiisnotreallyassimpleasthis. 3.Asmoreandmoreverticesareexposed,theratioofthemembersofL 2.Weneglectedtoconsiderthatthesecondvertex-copychoseninStep Wewillcallsuchapairofvertex-copiesabackedge. resetsitselftoapositivenumber. ofxichanges. whicharecopiesofverticesofdegreedshifts,andtheexpectedincrease 2(b)mightbeanopenvertex-copyinwhichcaseXidecreasesby2. 9

10 1.ThiswillincreasetheprobabilityofXigrowinglarge,andsothisonly Thesecomplicationsarehandledasfollows: posesapotentialprobleminprovingpart(b).inthiscase,wewill 3.Inprovingpart(a),welookatourcomponentatatimewhentheexpectedincreaseinXiisstillatleast12itsoriginalvalue.Wewillsee thatthecomponentbeingexposedatthispointisa.s.agiantcomponent.inprovingpart(b),itisenoughtoconsidertheconguration 2.Wewillseethatthisa.s.doesn'thappenoftenenoughtoposeaserious problem,unlessthepartiallyexposedcomponentisalreadyofsize(n). o(n?1),andhenceevenifwe\tryagain"ntimes,thiswilla.s.never happen. showthattheprobabilityofacomponentgrowingtoobigisoforder 2GraphsWithNoLargeComponents twosections. Thisisaroughoutlineoftheproof.Wewillllinthedetailsinthenext signicantly. aftero(n)steps.atthispoint,theexpectedincreasehasn'tchanged theconditionsgivenintheorem1(b),thenfa.s.doesnothaveanylarge graphs.wewillrstprovethatiffisarandomcongurationmeeting components. InthissectionwewillprovethattheanalogueofTheorem1(b)holdsfor randomcongurations.lemma1willthenimplythatitholdsforrandom somefunction0!(n)n18?fhasnoverticesofdegreegreaterthan!(n),thenfa.s.hasnocomponentswithmorethan=dr!(n)2logne vertices. sequencednmeetingtheconditionsoftheorem1.ifq(d)<0andiffor GivenQ(D)<0set=?Q(D)=KandsetR=150 Lemma3LetFbearandomcongurationwithnverticesanddegree ThefollowingtheoremofAzumawillplayanimportantrole: 2. Azuma'sInequality[1]Let0=X0;:::;Xnbeamartingalewith jxi+1?xij1 10

11 forall0i<n.let>0bearbitrary.then eachi:maxje(f()j1;2;:::i+1)?e(f()j1;2;:::i)jci f()=f(1;2;:::;n)bearandomvariabledenedbythesei.iffor Thisyieldsthefollowingveryusefulstandardcorollary. CorollaryLet=1;2;:::;nbeasequenceofrandomevents.Let Pr[jXnj>pn]<e?2=2 jf?e(f)j>tisatmost:2exp wheree(f)denotestheexpectedvalueoff,thentheprobabilitythat argumentsseeeither[16]or[5]. Formoredetailsonthiscorollaryandanexcellentdiscussionofmartingale 2Pc2i! insection1. InordertoproveLemma3,wewillanalyzetheMarkovprocessdescribed?t2 sumofdeg(v)?2overallverticesvcompletelyorpartiallyexposedduring therstisteps.wenotethatwi=xi+2yi?2ci. backedgesformed,andcibethenumberofcomponentsthathavebeenat leastpartiallyexposedduringtherstisteps.wealsodenewitobethe congurationhavebeenexposed.similarly,weletyibethenumberof NowWi\stalls"wheneverabackedgeisformed,andonlychangeswheneveranewvertexiscompletelyorpartiallyexposed.Forthisreason,itis easiertoanalyzewiwhenitisindexednotbythenumberofpairsexposed, rstjnewvertices(partiallyorcompletely)exposed. ablewhichdoesexactlythis.weletzjbethesumofdeg(v)?2overthe butbythenumberofnewverticesexposed.thusweintroduceanothervari- expectedincreaseasxi,butbehavesmuchmorenicely.inparticular,it Section1.Specically,ifaftertherstjverticeshavebeencompletelyor isn'taectedbytherstandsecondcomplicationsdiscussedattheendof ThereasonthatweareintroducingZjisthatithasthesameinitial RecallthatXiisthenumberofopenvertex-copiesafteripairsofour 11

12 partiallyexposed,thereareexactlyri(j)unexposedverticesofdegreei,then Zj+1=Zj+(i?2)withprobabilityiri(j)=Piri(j). thann?2. thejthvertexispartiallyexposed;i.e.wij=zj. vinf,theprobabilitythatvliesonacomponentofsizeatleastisless randomvariableijwhichisthenumberofpairsexposedbythetimethat Recallthat=R!(n)2logn. Lemma4SupposethatFisasdescribedinLemma3.Givenanyvertex NowinordertodiscussXiandZjatthesametime,wewillintroducethe thatxi>0forall1i.thus,wewillconsidertheprobabilityofthe latter. canalsogetzixi?2.thisisbecauseateachiterationweeitherhavea theprobabilitythatvliesonacomponentthatlargeisatmosttheprobability Notethatforanyi,ifCi=1thenWi=Xi+2Yi?2Xi?2.Infact,we Proof: backedgeorexposeanewvertex.thusiniterationi,wehaveexposedi?yi HerewewillinsistthatvistherstvertexchoseninStep2(a).Therefore step;thereforezizi?yi?yiwi?yixi+yi?2xi?2. newvertices,thereforewi=zi?yi.nowzidecreasesbyatmostoneateach thatxi>0forall1iisatmosttheprobabilitythatz>?2.we willconcentrateonthisprobability,aszibehavesmuchmorepredictably thanxi. NowifXi>0forall1i,thenC=1.Therefore,theprobability rstjvertex-copieschosenwereallcopiesofverticesofdegree1.ifthiswere thecasethentheexpectedincreaseinzjwouldbe:?2ṫhisistruebecausetheexpectedincreaseofzjwouldbehighestifthe?+o(1).weclaimthatforj,theexpectedincreaseinzjislessthan InitiallytheexpectedincreaseinZjisPi1i(i?2)di(n)=Pi1idi(n)= forsucientlylargen,asj=o(n)andidi(n)!iuniformly. Therefore,theexpectedvalueofZislessthan?2+deg(v)<?3.?(d1(n)?j)+Pi2i(i?2)di(n) (d1(n)?j)+pi2idi(n)+o(1)=?+o(1) 12?2

13 veryclosetoitsexpectedvalue. andf()=z.weneedtobound WewillusethecorollaryofAzuma'sInequalitytoshowthatZisa.s. Letbethesetofunexposedverticesatthispoint.Thesizeofisn?i. iwillindicatethechoiceoftheithnewvertexexposed,i=1;:::;, istheeventthatxisthe(i+1)stnewvertexexposed. Supposethatwearechoosingthe(i+1)stvertextobepartiallyexposed. Foreachx2,deneEi+1(x)tobeE(Zj1;2;:::;i+1)wherei+1 Consideranytwoverticesu;v2.WewillboundjEi+1(u)?Ei+1(v)j. je(f()j1;2;:::i+1)?e(f()j1;2;:::i)j: nextvertex.now,z=zj?1+(px2sdeg(x)?2)+deg(y1)?2+deg(y2)?2, thedistributionofthisorderisunaectedbythepositionsofu;v. w.therefore,themostthatchoosingbetweenu;vcanaecttheconditional wherey1isthejthvertexexposed(eitheruorv)andy2iseitheru;v;or expectedvalueofzistwicethemaximumdegree,i.e.jei+1(u)?ei+1(v)j Considertheorderthattheverticesin?fu;vgareexposed.Notethat 2!(n). LetSbethesetoftherst?2verticesunderthisorder,andletwbethe wehavethat Since,E(f()j1;2;:::i)=Xx2PrfxischosengEi+1(x); Z>0isatmost:2exp ThereforebythecorollaryofAzuma'sInequality,theprobabilitythat je(f()j1;2;:::i+1)?e(f()j1;2;:::i)j2!(n): AndnowLemma3followsquiteeasily: ProofofLemma3:?(3R!(n)2logn)2 2P(2!(n))2!=2n?2 <n?2: 72R 13 2

14 ofsizeatleastiso(1).thereforea.s.noneexist. throughouttheexposureofourconguration. ByLemma4,theexpectednumberofverticeswhichlieoncomponents wouldnotbeabletoreach0withinr!(n)2lognsteps. WealsogetthefollowingCorollary: Corollary3UnderthesameconditionsasLemma3,a.s.Xi<2 Proof: BecauseXidropsbyatmost2ateachstep,ifitevergotthathigh,it seethatita.s.hasnomulticycliccomponents. asinlemma3.fa.s.hasnocomponentwithatleast2cycles. WewillnowshowthatFa.s.doesn'thavemanycycles.First,wewill Lemma5LetFbearandomcongurationmeetingthesameconditions Proof: Chooseanyvertexv.LetEvbetheeventthatvliesonacomponentof 2 ofthiscomponent,xi<2. sizeoftherstcomponentisatmostthenthesecondbackedgemustbe sizeatmostwithmorethanonecycle,andthatthroughouttheexposure chosenwithinatmost+2steps.thereforetheprobabilitythateholdsis lessthan WewillinsistthatvistherstvertexexaminedunderStep2(a).Ifthe sotheprobabilitythatevholdsforanyviso(1). as!(n)<n18?. ThereforebyLemma3andCorollary3a.s.nocomponentsofFhave ThereforetheexpectednumberofverticesforwhichEvholdsiso(1)and 2!2 M?2?32=on?1 morethanonecycle. asinlemma3.fa.s.haslessthan2logncycles. thatita.s.doesnothavemanycycliccomponents. WecannowshowthatFa.s.doesnothavemanycycles,byshowing Lemma6LetFbearandomcongurationmeetingthesameconditions Proof: Wewillshowthata.s.throughouttheexposureofF,atmost2logn 2 back-edgesareformed.therestwillthenfollow,sincebylemma5,a.s.no infisexactlythenumberofbackedges. componentcontainsmorethanonecycle,andsoa.s.thenunmberofcycles 14

15 i.clearlythenumberofbackedgesformedisatmostthenumberofsuccessfulti's,plusthenumberofbackedgesformedattimeswhenxi>2. NowbyCorollary3,weknowthattherearea.s.noneofthelattertypeof M?2i+12and1otherwise. Nowthenumberofvertex-copiestochoosefromisPj1jdj(n)?2i+ ThereforetheexpectedvalueofT,thenumberofsuccessfulTi'sis: E(T)=2+(M?2)=2 Xi=1M?2i+1=log(M)(1+o(1)) 2 M?2i+1,for addallofltobi.lettibetheeventthatamemberofbiischoseninstep tion,thenletbiconsistofany2ofthem.otherwise,letbiconsistofthe sizeofbiupto.ofcourseiflistoosmalltodothis,thenwewilljust openvertex-copiesandenougharbitrarilychosenmembersofltobringthe FirstwemustdeneasetBiofunmatchedvertex-copies: Foreachi,iftherearemorethan2openvertex-copiesattheithitera- 1=M?2i+1.ThereforetheprobabilityofTiholdingis2 backedges,sowewillconcentrateonthenumberoftheformertype. muchbiggerthane(t). NowwewilluseasecondmomentargumenttoshowthatTisa.s.not Therefore,byChebyshev'sinequality,theprobabilitythatT>1:5log(M) E(T2)=Xi6=j =(E(T)2+E(T))(1+o(1)) (M?2i+1)(M?2j+1)+E(T) 2 2logn,provingtheresult. isatmost1=(4e(t))(1+o(1))=o(1). ProofofTheorem1(b): ThisclearlyfollowsfromLemmas2,3,5,and6. AndnowwecanproveTheorem1(b). Therefore,a.s.thenumberofbackedgesformedislessthan1:5log(M)< 2 15

16 3GraphsWithGiantComponents InthissectionwewillprovetheanalogueofTheorem1(a)forrandomcon- gurations.lemmas1and2willthenimplythattheorem1(a)holds. existconstants1;2>0dependentondsuchthatfa.s.hasacomponent sequencednmeetingtheconditionsoftheorem1.ifq(d)>0thenthere withatleast1nverticesand2ncycles.moreover,theprobabilityofthe converseisatmostzn,forsomexed0<z<1. Lemma7LetFbearandomcongurationwithnverticesanddegree Firstwewillshowthatagiantcomponentexistswithhighprobability: 0<<a.s.Zdne>n.Moreover,theprobabilityoftheconverseisat hold. discussedintheprevioussection.again,thekeywillbetoconcentrateon therandomvariablezj. AsinSection2,wewillproveLemma7byanalyzingtheMarkovprocess ThroughoutthissectionwewillassumethattheconditionsofLemma7 partnerispi(n)=idi(n)=pj1jdj(n)=ii=k+o(1). most(z1)n,forsomexed0<z1<1. Proof: Forsimplicity,wewillassumethatnisaninteger. Initially,theprobabilitythatavertex-copyofdegreeiischosenasa Lemma8Thereexists0<<1;0<<min(14;K4)suchthatforall attheendofsection2,i.e.thatfactthattheratiosofunexposedverticesof dierentdegreesareshifting. (n)steps.thuswehavetoworryaboutthethirdcomplicationdescribed UnlikeinSection2,wehavetoconsiderthebehaviourofourwalkafter i>iischosen,wesubtractonefromzjinsteadofaddingi?2toit,then thatifwechangezjslightlybysayingthateverytimeavertexofdegree ofhighdegree.sowhatwewilldoisshowthatwecanndavaluei,such wewillstillhavepositiveexpectedincrease. one,suchthatforeach2ii,iisalittlelessthantheinitialprobability Wewillthenshowthatwecanndasequence1;:::;isummingto Itturnsoutthatthisproblemismuchlessseriousifwecanignorevertices wewouldstillhaveapositiveexpectedincrease. ofavertexofdegreeibeingchosen.however,ifweweretoadjustzjalittle furtherbyselectingavertexofdegreeiwithprobabilityiateachstep,then Wewillcallthis\adjustedZj"Zj.Clearly,ifwendsomeJsuchthat 16

17 asbigastheprobabilitythatzj>r.wewillconcentrateonthesecond afterjsteps,theprobabilityofchoosingavertexofdegreeiisstillatleast ifor2ii,thentheprobabilitythatzj>rforanyrisatleast probabilityaszjismuchsimplertoanalyze. 1.Pi=1; 2.0<i<ii=K,for2ii,unless0=i=ii=K; 3.Pi1i(i?2)i>0. Moreformally,whatwewishdoischooseasequence1:::;isuchthat Notethat:Xi1(i?2)pi(n)=Xi1(i?1)pi(n)?Xi1pi(n) ii,pi>i>0unlesspi=i=0,1=1?2?3?:::?i,and PiSinceDiswell-behavedandQ(D)>0,thereexistsisuchthat i=2(i?1)pi>1+0,forsome0>0andsucientlylargen. Therefore,wecanchooseasequence1;:::;isuchthatforall2 Setpi=ii=K. =Xi2(i?1)pi(n)?1: walk: i=2(i?1)i=1+0 ConsidertherandomvariableZjwhichfollowsthefollowingrandom Zj+1=Zj+(i?2)withprobabilityi,1ii. Z0=0 2.ItfollowsthatPi1(i?2)i=0 Fori=2;:::;i,chooseanyi>0suchthatii?i 2. =minf2;:::;i;k4g.clearly,afteratmostiterations,theprobability ofchoosingacopyofavertexofdegreei2isatleasti.therefore,for 0jn,therandomvariableZjmajorisesZj;i.e.foranyR: NowtheexpectedincreaseinZjatanystepis0 Pr[Zj>R]Pr[Zj>R:] 2.Thusthelemma K<i,andset followsbyletting=04,asitiswell-known(seeforexample[9])thatzn 17

18 asclaimed. isa.s.concentratedarounditsexpectedvaluewhichis2n,andthatthe probabilityofdeviatingfromtheexpectedvaluebymorethan(n)isaslow analyzexj.inordertodothis,recallthattherandomvariableijisdened tobethenumberofpairsexposedbythetimethatthejthvertexispartially exposed;i.e.wij=zj. a.s.existssome1iidne.suchthatxi>n,where=min(2;14). WehavejustshownthatZja.s.growslarge.However,wereallywantto Lemma9Thereexists0<0<suchthatforany0<0there 2 Moreover,theprobabilityoftheconverseisatmost(z2)nforsome0<z2<1, dependenton. chosenisxi a.s.w<2nfor0. orinpairshavebeenexposed.weclaimthatwecanchoose0suchthat Proof: Forsimplicity,wewillassumethatnisaninteger. Atanystepi;1iIn,theprobabilitythatanopenvertex-copyis WewillcountW,thenumberofbackedgesformedbeforeeitherXi>n 0ifXj>nandatmostp= NowIjj+YIjj+Zj Therefore,ateachstep,theprobabilitythatsuchabackedgeisformedis Kn?2i+o(1),regardlessofthechoicesmadeprevioustothatstep. ifxjn. 2(+)n. variablebin(p;in). Thusthenumberofsuchcopieschosenismajorisedbythebinomial ThereforethelemmafollowssolongaspInp(+)n<2n,whichis K?2?2 12 thatthereisa.s.agiantcomponent: equivalentto4+4<k,yielding0. atstepi=id0newilla.s.haveatleast1nverticesand2ncycles.moreover, atleast NowifXinforall1i<InthenWisequaltoYIn. theprobabilityoftheconverseisatmost(z3)n,forsomexed0<z3<1. NowthatweknowthatXia.s.getstobeaslargeas(n),wecanshow Lemma10Thereexists1;2>0suchthatthecomponentbeingexposed ThereforeXIn=Zn?2YInwhichwithprobabilityatleast1?(z1)nis 2n,whichyieldsourresult. 2 18

19 1nofthesewillbematchedwithmembersof,andatleast2nofthesewill beforethiscomponentisentirelyexposed.wewillshowthata.s.atleast vertices.formasetconsistingofexactlyonecopyofeachofthem. Proof: lemma. bematchedwithotheropenvertex-copiesfrom.clearlythiswillprovethe Notethatatthispointtherearea.s.atleastn?20n?n>n5unexposed ThereisasetofXIopenvertex-copieswhosepartnersmustbeexposed whereeachmatchingisequallyprobable.theexpectednumberofpairs containingonevertexfromeachof;isatleastn5xi cedureforexposingfsimplygeneratesarandommatchingamongstthem NowthereareM?2Ivertex-copiesavailabletobematched.Ourprobers,anditfollowsfromtheChernoboundsthatthesenumbersarea.s.at asclaimed.thereforethecomponenta.s.hasatleast1nverticesandat leasthalfoftheirexpectedvalueswiththeprobabilityoftheconverseaslow pectednumberofpairsofopenvertex-copieswhichformanedgeoffis M2?IXI Thepreviouslemmasgiveusalowerboundof21n;22nonthesenum- M?2I2. 2M?I,andtheex- least2ncycles. thenfa.s.hasexactlyonecomponentonmorethantlognvertices,for AndnowLemma7followsquiteeasily: ProofofLemma7: ThisisclearlyacorollaryofLemma10. WewillnowseethatFa.s.hasonlyonelargecomponent. Lemma11IfFisarandomcongurationasdescribedinLemma7, 2 ertyaifuandvlieoncomponentsofsizeatleast1nandtlognrespectively. WewillshowthatforanappropriatechoiceofT,theprobabilitythat(u;v) sizeatleast1n.wewillseeherethatnoothercomponentsoffarelarge. someconstanttdependentonthedegreesequence. haspropertyaiso(n?2),whichisenoughtoprovethelemma. Consideranyorderedpairofvertices(u;v).Wesaythat(u;v)hasprop- Proof: Recallthatwemaychooseanyvertex-copywewishtostarttheexposure WehavealreadyshownthatFa.s.hasatleastonegiantcomponentof with.wewillchooseu. ByLemma9,therea.s.existssomeIId1ne,suchthatXI>nwhere 19

20 thecontrary.denetobethesetofopenvertex-copiesafteristeps. exposedacopyofv,then(u;v)doesnothavepropertya,sowewillassume afteristeps,wearenotstillexposingtherstcomponent,c1,orifwehave =min(0 restofc1untillater.wewillseethatifc2getstoobig,thenitwilla.s. exposingv'scomponent,c2,immediately,andputotheexposureofthe Herewewillbreakfromthestandardmethodofexposure.Wewillstart 2;1 2;14),andsowecanassumethistobethecase.Notethatif v,andexposingitspartner.wecontinueexposingpairs,alwayschoosing includeamemberof. acopyofapartially-exposedvertexwhichisknowntobeinc2(ifoneis available),andexposingitspartner.wechecktoseeifthispartnerliesin.thiswouldimplythatvliesinc1.oncec2isentirelyexposed,ifitis disjointfromc1thenwereturntoexposingtherestofc1andcontinueto exposefinthenormalmanner.notethatthisisavalidwaytoexposef. WeexposeC2inthefollowingway.Westartbypickinganycopyof atleasttlognstepstoexposethiscomponent.therefore,theprobability thatvliesonacomponentofsizegreaterthantlognwhichisnotc1isat Also,ifvliesonacomponentofsizegreaterthanTlogn,thenitmusttake most: Ateachstep,theprobabilitythatamemberofischosenisatleastK. tozeroasn!1,soa.s.noneexist. forasuitablevalueoft. ThereforetheexpectednumberofpairsofverticeswithpropertyAtends 1?KTlogn=on?2 morethanonecycle. thenfa.s.hasnomulti-cycliccomponentonatmosttlognvertices,for anyconstantt. ItonlyremainstobeshownthatFa.s.hasnosmallcomponentswith Lemma12IfFisarandomcongurationasdescribedinLemma7, Proof: 2 ofstep2(a).nowifthiscomponentcontainsatmosttlognvertices,then itisentirelyexposedafteratmosto(n1=4)stepsasthemaximumdegreeis n1=4?. Considertheprobabilityofsomevertexvlyingonsuchacomponent. Wewillinsistthatweexposeanedgecontainingvintherstexecution 20