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Combinatorial Algorithms Involving Pattern Containing and Avoiding Permutations

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Title: Combinatorial Algorithms Involving Pattern Containing and Avoiding Permutations
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Copyright Date: 2008

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Title: Combinatorial Algorithms Involving Pattern Containing and Avoiding Permutations
Physical Description: Mixed Material
Copyright Date: 2008

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Source Institution: University of Florida
Holding Location: University of Florida
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Iwouldrstliketothankmyfamilyforbeingsosupportivethroughoutmygraduateschoolcareer.Iwouldalsowanttothankmyadvisor,MiklosBona,forhisguidance.Hissuggestiontolookatstacksortingwasdenitelytheturningpointinmyresearch.IamveryappreciativeofDavidDrake,AndrewVince,andNeilWhitefortheirhelpfulsuggestionsandquestionsduringmyseminarpresentationsaswellasforservingonmycommittee.Inaddition,IwouldliketothankMeeraSitharamforagreeingtobeonmycommitteeandbringingacomputersciencepointofviewtothealgorithmsconsideredinmythesis.IwouldalsoliketothankKevinKeatingandJonathanKingforconsistentlyattendingmyseminartalksandaskingquestionsIhadnotconsidered.Inparticular,Chapter3isthedirectresultofaquestionProf.KingaskedmewhenIwaspresentingmyworkfromChapter2.Finally,IamgratefultoBruceSaganforhissuggestionsregardingChapter5.Thankstohiscomments,theproofsinthischapteraresignicantlylesscomplicated. iv

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page ACKNOWLEDGMENTS ............................. iv LISTOFFIGURES ................................ vi ABSTRACT .................................... viii 1ANINTRODUCTIONTOSTACKSORTING ............... 1 1.1SortingPermutationsbyOneStack ................. 1 1.2SortingPermutationsbytheRight-greedyAlgorithm ....... 2 1.3SortingPermutationsbytheLeft-greedyAlgorithm ........ 11 2ALGORITHMSFORSORTINGWITHtSTACKSINSERIES ..... 16 2.1ExamplesofUsingtheLeft-greedyandRight-greedyAlgorithms 16 2.2ComparingtheLeft-greedyandRight-greedyAlgorithms ..... 17 2.3ObtainingMorePermutationswhichareSortablebytStacksinSeries ................................ 24 2.4MoreInformationabouttheLeft-greedyAlgorithmfort>2 ... 32 3STACKSINACIRCULARSERIES .................... 38 4PARITYANDSORTABLEPERMUTATIONSOFAGIVENLENGTH 44 4.1ParityoftheNumberofPermutationsofLengthnwhicharet-StackSortablefort=1or2 .......... 44 4.2ParityoftheNumberofPermutationsofLengthnwhichareSortablebyTwoStacksinSeries .............. 47 5PERMUTATIONRECONSTRUCTION .................. 51 5.1Introduction .............................. 51 5.2TheCasewhenk=1 ......................... 54 5.3TheCasewhenk=2 ......................... 56 5.4TheCasewhenk=3 ......................... 58 5.5FurtherResultsandDirections .................... 61 REFERENCES ................................... 64 BIOGRAPHICALSKETCH ............................ 66 v

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Figure page 1{1Sorting4132 ................................ 1 1{2Sorting4312usingtheright-greedyalgorithm ............. 3 1{3Arootednonseparableplanarmapontwelvevertices. ........ 6 1{4A(1;0)treeonnineteenvertices. ................... 7 1{5Sorting4312usingtheleft-greedyalgorithm .............. 12 1{6A(0;1)treeonnineteenvertices. ................... 13 2{1Sorting3241failsusingtheright-greedyalgorithm ........... 17 2{2Sorting3241usingtheleft-greedyalgorithm .............. 18 2{3Sorting4123untilthethirdcriticalmomentusingtheleft-greedyal-gorithmandthendoingthesamewiththeright-greedyalgorithm. 19 2{4Stacksusingtheright-greedyalgorithmimmediatelybeforepienters 20 2{5Stackswhentheleft-greedyalgorithmhasbeencarriedoutuntilthe(i+1)stcriticalmoment. ........................ 21 2{6Stacksusingtheright-greedyalgorithmimmediatelyafterpienters. 21 2{7Stackswhentheleft-greedyalgorithmhasbeencarriedoutuntilthe(i+1)stcriticalmoment. ........................ 22 2{8Stacksusingtheright-greedyalgorithmimmediatelybeforexistomoveintothe(k+1)ststack. ..................... 22 2{9Stacksusingtheright-greedyalgorithmimmediatelybeforexistomoveintothe(k+1)ststack. ..................... 23 2{104231and42351beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm. ............................ 26 2{112341and23541beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm. ............................ 30 2{122541673failingtobesortedbythreestacksinseriesusingtheleft-greedyalgorithm. ............................ 33 vi

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........ 34 2{14PartIIof2541673beingsortedbythreestacksinseries. ....... 35 2{15PartIof26351784beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm. .......................... 36 2{16PartIIof26351784beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm. .......................... 37 3{1Stacksinacircularseries ......................... 38 3{2Stacksatthemomentoffailure. ..................... 39 3{3PartIofsorting23451bythreestacksinacircularseries. ....... 42 3{4PartIIofsorting23451bythreestacksinacircularseries. ...... 43 4{1Thethree(0;1)treesonthreevertices. ................ 48 vii

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viii

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ix

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1.1 SortingPermutationsbyOneStack Figure1{1: Sorting4132 TheideasofstacksortingandsortingbystacksinserieswereintroducedbyKnuth[ 11 ].Inthisbook,heprovedthefollowingtheorem. 1

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Stacksortinghasalsoledtootherstudiesofpermutations.Duetothenatureofthetypesofpermutationsthatcouldbesortedbystacks,theconceptofpatternavoidancewasintroduced. 13 ]. 1.2 SortingPermutationsbytheRight-greedyAlgorithm 22 ],[ 23 ]introducedaright-greedyalgorithmtosortpermutationsbytstacksinseries.Theright-greedyalgorithmalwaystakestherightmostlegalmove.

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Anexampleofhowtheright-greedyalgorithmcanbeusedtosortapermutationwhenwehavetwostacksinseriesisshowninFigure 1{2 Figure1{2: Sorting4312usingtheright-greedyalgorithm

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Thekeytothisdierenceisthatthe5of35241forcesthe3outrstsoasnottoformthe231patternthatoccursif2entersthestacksbeforethe3leaves.Infact,itturnsoutthat2-stacksortablepermutationscanbecategorizedasfollows: 1. Thepermutationmustavoid2341and 2. any3241patterncontainedwithinthepermutationmustbepartofa35241pattern.West[ 22 ],[ 23 ]conjecturedthatthenumberof2-stacksortablepermutationsoflengthnwasW2(n)=2 (n+1)(2n+1)3nn:ThisconjecturewasshowntobetruebyZeilberger[ 24 ].Inhisproof,Zeil-bergerconsideredthelargestdecreasingsubsequencen(n1)(n2):::(ni+1)ofpermutationsoflengthn,appliedabijectionandweightenumerationtothepermutations,andthenusedaPolya-Schutzenberger-Tuttetransform.Allofthisresultedinaninthdegreefunctionalequationthatrequiredtheuseofacomputertosolve.Itturnsoutthatthesepermutationscanberelatedtocertaintypesofgraphsandplanarmaps. 1{3

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Figure1{3: Arootednonseparableplanarmapontwelvevertices. 1. Foranyleafvertexl,wehave(l)=a. 2. Ifvisavertexthatisneitheraleafnortheroot,lookatitschildrenv1;v2;v3;:::;vk.Thena(v)b+kXi=1(vi). 3. Finallyifristherootvertexanditschildrenarer1;r2;r3;:::;rk,then(r)=kXi=1(ri). 1{4 .Sincethen,Dulucq,Gire,andWest[ 8 ]andGouldenandWest[ 10 ]haveexhibitedbijectionsbetweennonseparablerootedplanarmapsonn+1edgesand2-stacksortablepermutationsoflengthn.Thenonseparablerootedplanarmapswithn+1edgeshadbeenenumeratedbyTutte[ 20 ].Inaddition,abijectionbetweenthesenonseparablerootedplanarmapsonn+1edgesand(0;1)treesonn+1verticeswasshownbyCori,Jacquard,andSchaeer[ 7 ].Thisisespecially

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Figure1{4: A(1;0)treeonnineteenvertices. interestingbecausethereisalsoabijectionbetweenpermutationswhichcanbesortedbytwostacksinseriesandasimilarclassoflabeledtrees.However,noneoftheproofsoftheformulaforW2(n)directlyindicatewhyW2(n)<3nn.Inparticular,noinjectionfromthe2-stacksortablepermutationsintoasetofsize3nnhasbeenfound.Apossiblekeytothisproblemisthat2-stacksortablepermutationsoflengthncanberepresentedasuniquewordsoflength3nconsistingofthethreeletters,d,l,andu.Thedrepresentsamovefromtheinputtotherststack,thelrepresentsamovefromtherststacktothesecondstack,andtheurepresentsamovefromthesecondstacktotheoutput.Somenecessaryconditionsareclear,suchasthefactthattheithdmustappearbeforetheithlwhichinturnmustappearbeforetheithu.Itisalsonotdiculttoseethatanywordcorrespondingtoa2-stacksortablepermutationmustnothaveconsecutivelettersllorud.However,satisfyingtheseconditionsisnotenoughtoguaranteeawordthatrepresentsa2-stacksortablepermutation.Determiningsucientconditionsisanunsolvedproblem.Anotherinterestingconnectionto2-stacksortablepermutationsisasetLnoflatticepaths.Thesepathsstartandendat(0;0),neverleavetherstquadrant,andusesomecombinationof3nstepswhereeachofthestepsisone

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ofthefollowingtypes:(1;1),(0;1),and(1;0).Kreweras[ 12 ]provedamoregeneralresultwhichshowswhatthenumberofsuchpathsis. 19 ]andtheotherisacorollaryofaresultduetoStankovaandWest[ 17 ].

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Infact,everytimeanentryistobebroughtintothestacks,theentriesalreadyinthestacksmusteachbemovedleft.Thus2;3;4;:::t+1willeachbeintheirownstack.Asthereareonlytstacks,thesetentriestakeupallofthemandsot+2cannotenterthestacksunlessthe2leaves.Thuswecannotgettheidentitypermutationandsothispermutationisnott-stacksortable.3

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decidingwhichentrytoputineachposition,wehaveatmostt+1choicessinceassoonaswedecidetotakeanentryfromDk,wemusttakethelowestentrythathasnotbeenusedalready.Notethatthereareclearcasessuchastherstpositionwhenwehavestrictlylessthant+1choices.Hencethenumberofpermutationsoflengthnthatavoid(t+2)(t+1):::321islessthan(t+1)2n.3Alsointerestinginthestudyoft-stacksortablepermutationsisthefactthatthenumberof1-stacksortablepermutationsisW1(n)=2nn 2(n+1)(2n+1):ThiswouldleadonetobelievethatthatperhapsforlargervaluesoftwewouldhaveWt(n)=(t+1)nn 22 ]andlaterbyGuibertin2003(notpublished)seemstoindicatethatthisisnotthecasefort=3.ThisisbecauseW3(n)appearstohaveprimefactorsmuchlargerthan4n.Nonetheless,itwouldbeinterestingtoknowifthenumeratorofthisformulastillprovidesanupperboundasitdidinthecaseswhenn=1andn=2.

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1.3 SortingPermutationsbytheLeft-greedyAlgorithmIn2000,Atkinson,Murphy,andRuskuc[ 1 ]usedaleft-greedyalgorithmforsortingpermutationsusingtwostacksinseries.Theleft-greedyalgorithmalwaystakestheleftmostlegalmove.Hereweconsideralegalmovetobemovinganentrytoastackwhereitissmallerthantheentrybelowitormovinganentrytotheoutputifitisthesmallestentrynotalreadythere.Anexampleoftheleft-greedyalgorithmbeingappliedtosortapermutationbytwostacksinseriesisshowninFigure 1{5

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Figure1{5: Sorting4312usingtheleft-greedyalgorithm

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sucientconditionsismoredicult.WordssatisfyingtheseconditionsarecalledGM-words.ThisclassicationofwhatmakesawordcorrespondtoapermutationallowedAtkinson,Murphy,andRuskuctocountthenumberofsuchpermutationsofagivenlengthn.FirstanirreducibleGM-wordwasdenedtobeaGM-wordWsuchthatitcouldnotbebrokendowntogetW=W1W2wherebothW1andW2wereGM-wordsthemselves.TheythensetupabijectionbetweenirreducibleGM-wordsandaclassofdescriptiontreescalled(0;1)treesonnvertices,previouslydenedinDenition 1.17 .ThesetreeswereenumeratedbyCori,Jacquard,andSchaeer[ 7 ]byconstructingabijectionbetweenthemand2-colorablecubicplanarmapswith3nedgeswhichwereenumeratedbyTutte[ 20 ]. 1{6 Figure1{6: A(0;1)treeonnineteenvertices. ThisbijectionallowedAtkinson,Murphy,andRuskuc[ 1 ]tothenenumeratethenumberofpermutationsoflengthnwhicharesortablebytwostacksinseries.TheyconstructedabijectionbetweenthesetofallGM-wordsoflength3nandplaneforestsof(0;1)treesonnverticestodeducethefollowingtheorem.

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2:Anexplicitformulaforznisgivenby7n23n2 2(1)n1+3nXi=22i+1(2i4)! 3 ].Thisistherstresultwherethenumberofpermutationsofagivenlengththatavoidonebasissetofpermutationsisthesameasthenumberofpermutationsofthesamelengththatavoidabasissetofadierentcardinality.Despitethenumericalcorrelationbetweenpermutationsthatavoidallthepermutationsintheinnitebasisdescribedaboveandthepermutationsthatavoidonlythepermutation1342,thereisnodirectbijectionlinkingthetwo.However,Bona'sproofoftheenumerationofthepermutationsoflengthnthatavoid1342hadalsoshownabijectionbetween(0;1)treesandasubclassofpermutationsthatavoid1342.

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1.20 referringtopermutationswhicharesortablebytstacksinserieswhenusingtheright-greedyalgorithmalsoappliestothesepermutations.Thisissincethemonotonicityrequirementswehaveplacedonthestackswillnotallow234:::(t+1)(t+2)1tobesortedbytstacksinseries.Unfortunatelyaswewillshow,theleft-greedyalgorithmceasestobeoptimalorprovideaclosedclassassoonastherearemorethantwostacksinseries.

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2.1 ExamplesofUsingtheLeft-greedyandRight-greedyAlgorithmsInthenextsection,wewillcomparethetwoalgorithms.Itwillbeshownthattheuseoftheleft-greedyalgorithmwillallowustosortanypermutationthatcouldhavebeensortedbytheright-greedyalgorithmwhensortingbytstacksinseries.Inaddition,whenwehavetwoormorestacksinseries,therearepermutationsthatwillbesortedbyonlytheleft-greedyalgorithmwheneachalgorithmisapplied.Wenowpresentanexampleofapermutationwhichcanonlybesortedbytheleft-greedyalgorithmwhenwetryapplyingeachalgorithmwithtwostacksinseries.Thisalsogivesinsightintowhytheright-greedyalgorithmdoesnotyieldaclosedclassofpermutations,sinceinsertingalargeelementintoanappropriateplacemightforcetheright-greedyalgorithmtomakeamoreoptimalmovethanitwouldwithoutthelargeelement.Note:Forthepurposesofthischapter,wewillsaythatanalgorithmfailswhenwecannotmoveanyentrytoastackwithoutviolatingtheincreasingordercondi-tiononthestacksandwecannotmovethenextentryoftheidentitypermutationintotheoutput. 2{1 ,noticethatthe4cannotentertherststackbecauseitislargerthan3.Likewisethe3cannotmovelefttothesecondstackbecauseitislargerthan2.Finallythe2cannotexitthestacksbecausethe1isnotalreadyintheoutput. 16

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Figure2{1: Sorting3241failsusingtheright-greedyalgorithm 2{2 ,noticethatmovingthe3tothesecondstackbeforethe2trapsitintherststackiswhatgivestheleft-greedyalgorithmtheabilitytosortthispermutation.Theresultsshownintheremainderofthischapterarereproducedwithpermissionfromtheauthor'spaper[ 18 ]. 2.2 ComparingtheLeft-greedyandRight-greedyAlgorithmsThefactthatanypermutationthatcanbesortedbytheright-greedyalgo-rithmwillalsobesortedbytheleft-greedyalgorithmontstacksinserieswillfollowfromthesubsequentlemma.Itwasalreadyknownforthecasewhent=1,sincebothalgorithmsarethesameandwhent=2sincetheleft-greedyalgorithmisknowntobeoptimal. 2{3 weshowthesortingof4123usingtwostacksinseriesbyeachalgorithmcarriedoutuntilthethirdcriticalmoment.

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Figure2{2: Sorting3241usingtheleft-greedyalgorithm

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Figure2{3: Sorting4123untilthethirdcriticalmomentusingtheleft-greedyalgo-rithmandthendoingthesamewiththeright-greedyalgorithm.

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wellbytheinductionhypothesis.Thereforetheright-greedyalgorithmfailsbythispointalso,sothelemmatriviallyholdsforallsubsequentpi'sincludingpi+1.Thuswemayassumethatpicanenterthestacksbytheleft-greedyalgorithm.Also,sincewearestilltriviallydoneiftheright-greedyalgorithmfailsbeforepientersthestacks,wemayassumepicanenterthestacksbytheright-greedyalgorithmaswell.Hencewhenusingtheinductionhypothesis,wemayassumethatthestacksus-ingtheright-greedyalgorithmareasinFigure 2{4 attheithcriticalmoment,withallentriesbeingatleastasfarleftinthestacksusingtheleft-greedyalgorithmasthoseusingtheright-greedyalgorithm. Figure2{4: Stacksusingtheright-greedyalgorithmimmediatelybeforepienters Whenpienterstherststackbyeachalgorithm,westillhavetheconditionthateachentryisatleastasfarleftbytheleft-greedyalgorithmasbytheright-greedyalgorithmsincenoentriesmoveleftexceptforpimovingintotherststackinbothalgorithms.Theconditioncontinuestoholdifwecarryouttheleft-greedyalgorithm(whilenotcontinuingtocarryouttheright-greedyalgorithmontheotherseriesofstacks)untilitistimeforpi+1toenterthestacksorthealgorithmfailssinceweareonlymovingentriesleftinthestacksusingtheleft-greedyalgorithm.Wenowhavethestacksthatusetheleft-greedyalgorithmbeingasshowninFigure 2{5 andthestacksthatusetheright-greedyalgorithmasshowninFigure 2{6 .Nowifwhenwecarryouttheright-greedyalgorithmuntilpi+1shouldenterorthealgorithmfails,anditresultswitheachentrybeingnofurtherleftthanwhere

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Figure2{5: Stackswhentheleft-greedyalgorithmhasbeencarriedoutuntilthe(i+1)stcriticalmoment. Figure2{6: Stacksusingtheright-greedyalgorithmimmediatelyafterpienters. itwasbytheleft-greedyalgorithm,thenwearedone.Supposethereisanentryxsuchthatxwasinthekthstackbytheleft-greedyalgorithmatthispoint,butxmovesfurtherleftbytheright-greedyalgorithmbeforepi+1enters.Choosextobetherstentryforwhichthishappens(intheorderofthestepsoftheright-greedyalgorithm)andlookatwhatishappeningwhenxmovesfromthekthstacktothe(k+1)ststack.Werstnotethatifsuchanentryexists,the(k+1)ststackisactuallyastackandnottheoutput.Thisisbecausexwouldnotbeinthelaststackbytheleft-greedyalgorithmsinceanyentrythathasalreadyhadallentriessmallerthanitenterthestacks,leavesbytheleft-greedyalgorithmbeforeanewentryenters.Thusnoentrycanleavethestacksbytheright-greedyalgorithmthathasnotalsoleftbytheleft-greedyalgorithm.Sincexdidnotmovetothe(k+1)ststackbytheleft-greedyalgorithm,theremustbeanentryy0
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2{7 andthestackswheretheright-greedyalgorithmwasappliedmustbeasinFigure 2{8 atthispoint. Figure2{7: Stackswhentheleft-greedyalgorithmhasbeencarriedoutuntilthe(i+1)stcriticalmoment. Figure2{8: Stacksusingtheright-greedyalgorithmimmediatelybeforexistomoveintothe(k+1)ststack. Becausey0isinthestacksandfurtherrightthanx,ify0couldhavemoved,itwouldhavepriorityoverxbytheright-greedyalgorithm.Ify0wasnotthetopentryofitsstackatthispoint,thenthesmallerentryatthetopofthestackwouldhavepriorityoverx.Thustheremustbeay1
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Figure2{9: Stacksusingtheright-greedyalgorithmimmediatelybeforexistomoveintothe(k+1)ststack. Thusthereisnoentrythatmovesfurtherleftbytheright-greedyalgorithmthanbytheleft-greedyalgorithmatthe(i+1)stcriticalmoment.ThereforetheLemmaisproved.3Nowwearereadytoprovethemainresultofthischapter.

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Toseethattheleft-greedyalgorithmsortsmorepermutationsthantheright-greedyalgorithmwhent>1,wesimplyneedanexampleofapermutationthatisnotsortablebytstacksinseriesusingtheright-greedyalgorithm,butissortableontstacksusingtheleft-greedyalgorithm. 2{1 .InExample2.2itwasshownthattheleft-greedyalgorithmcouldbeusedtosuccessfullysort3241bytwostacksinseriesinFigure 2{2 .Hencewehaveprovedthefollowingproposition. ObtainingMorePermutationswhichareSortablebytStacksinSeriesInthefollowinglemmaweshowonewaytoobtainpermutationsoflengthnthataresortablebytstacksinserieswhentheleft-greedyalgorithmisappliedfrompermutationsoflengthn1whichcanbesortedusingtheleft-greedyalgorithmont1stacksinseries.Itwillalsobeshownthatifapermutationoflengthn1issortablebyt1stacksinseries,thenapermutationoflengthnthatissortablebytstacksinseriescanbeconstructedfromitthesameway.

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However,thiswillnotholdtruewhenwerestrictourselvestousingtheright-greedyalgorithm. 2{10 weshow4231and42351beingsortedbythreestacksinseries.Here4231isshownbeingsortedbythreestacksinseriesinsteadoftwoinordertodemonstratetheprocessdescribedintheproofofLemma 2.8 .Proof:Sincepissortableononlyt1stacksinseriesbytheleft-greedyalgorithm,weknowthatifwetrytosortouraugmentedpermutationp0withtheleft-greedyalgorithmontstacks,aseachentryinp0isabouttoenterthestacksuntil(possibly)afternenters,therightmoststackwillbeempty.Thisisbecausetheleft-greedyalgorithmwouldalwaysmovetheentriesinthestacksleftbeforeallowinganewentrytoenter.Henceifwecouldnotgetoneoftheoriginalentriesofptomoveoutthisrststack(beforenhasthepossibilityofinterfering),pwouldnotbesortablebytheleft-greedyalgorithmont1stacks.Becauseofthis,itisclearthatallentriesuptoandincludingnofp0willenterthestacksbytheleft-greedyalgorithm.Claim:Ifweignorethemovesmadebyn,theleft-greedyalgorithmonp0proceedsexactlythesamewayasitdoesonpwhenwetrytosortbothbytstacks.AnexampleofthisisclaimbeingtruehasalreadybeenshowninFigure 2{10 .ProofofClaim:Thisisevidentuntiltheentrynentersthestackswhensortingp0.Ifthereissomediscrepancy,letmovingxfromthekthstacktothe(k+1)ststackbetherstmovemadeonp0thatdoesnotmatchthatonp.Notethatx6=n.Say

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Figure2{10: 4231and42351beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm.

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movingyfromthelthstacktothe(l+1)ststackisthecorrespondingmoveonp.Becausethepreviousmovesallcorrespondedbeforethispoint,eachentrypi6=nisinthesamepositionforbothpermutationsuntilwemakethismove.Notethatneitherxnorycanbeleavingthestacksfortheoutputatthistime.Thisisbecausebothpermutationshavethesameoutputandifonepermutationhasanentryotherthannatthetopoftheleftmoststack,theotherhasthesameentrythere.Thusiftheentryshouldbethenextout,itistheleftmostmoveforboth.Case1:k>lInthiscase,movingxfromthekthstacktothe(k+1)ststackisthefurtherleftmove.Butthenxshouldhavebeenabletomovewhentheleft-greedyalgorithmwasappliedtopaswellsincetherecanbenoadditionalentriesinthe(k+1)ststackforpthatarenotthereforp0.Case2:l>kThesameargumentasinCase1holdsbecausetheonlyextraentrythatcouldbeinthe(l+1)ststackforp0isnandsincealltheotherentriesaresmaller,ncannotpreventanotherentryfrommovingleft.Thiscompletestheproofoftheclaim.Sinceweknowncanenterthetstacksforthepermutationp0usingtheleft-greedyalgorithmandallentries1;2;3;:::;(n1)canenterthetstacksforthepermutationpbytheleft-greedyalgorithm,bytheclaimweknowalltheentriesofp0canenterthetstackswhenusingtheleft-greedyalgorithm.Hencep0issortablebytstacksinseriesusingtheleft-greedyalgorithm.3Itiseveneasiertoshowtheresultistrueifweallowanyalgorithmtobeusedtosortthepermutation.

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2{10 .(Allotherentriesthatenterthestacksafternmayeasilymovepastnsincetheyaresmallerthann.)3Asstatedbefore,thismethodofconstructingpermutationsthataresortablebytstacksinserieswillnotworkifweonlywanttoallowtheright-greedyalgo-rithmtosortthepermutation.

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2{11 .Proof:Thisresultisclearforwhennisputinthelastposition.ThisisshowninFigure 2{11 .Supposenwasinsertedintooneofthersttpositions.Itisclearthatnmayenterthestacksatitsturnsincethereareatmostt1entriesaheadofnandtherearetstacks.TheremainderoftheprooffollowsexactlyastheproofforLemma3.1.3Thepreviouspropositionallowsustogivethefollowinglowerboundonthenumberofpermutationsoflengthnwhicharesortablebytstacksinseries.

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Figure2{11: 2341and23541beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm.

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Thepreviouspropositionwillalsoholdforsortingpermutationsontstacksinseriesusinganyalgorithmthatworks.Werstneedthefollowinglemma.

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Considerinstanceswhenweinsertnintooneofthersttpositionsofapermutationoflengthn1thatissortablebytheright-greedyalgorithmontstacksinseries.Inthiscase,theresultisnotalwaysapermutationthatisstillsortablebytheright-greedyalgorithmontstacksinseries. 2.4 MoreInformationabouttheLeft-greedyAlgorithmfort>2Althoughtheleft-greedyalgorithmsortsanypermutationthatcanbesortedbytstacksinseriesfort=1andt=2,theleft-greedyalgorithmisnotoptimalwhent>2. 2{12 weshowhowtheleft-greedyalgorithmfailstosortthispermutationbythreestacksinseries.InFigures 2{13 and 2{14 weshowhowthispermutationcanbesortedbythreestacksinserieswhenoneiscarefulwithhowtheentriesmovethroughthestacks.Noticethatbeingpatientwithmovingthe4leftiswhatmakesthedierence.Also,thepermutationssortedbytheleft-greedyalgorithmarenotaclosedclasswhent>2.(Sinceitisknown[ 11 ]thatthepermutationssortedbytstacksinseriesisaclosedclass,thisisactuallyenoughtoshowthattheleft-greedyalgorithmisnotoptimalfort>2.)

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Figure2{12: 2541673failingtobesortedbythreestacksinseriesusingtheleft-greedyalgorithm.

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Figure2{13: PartIof2541673beingsortedbythreestacksinseries.

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Figure2{14: PartIIof2541673beingsortedbythreestacksinseries.

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2{15 and 2{16 Figure2{15: PartIof26351784beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm.

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Figure2{16: PartIIof26351784beingsortedbythreestacksinseriesusingtheleft-greedyalgorithm.

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3{1 Figure3{1: Stacksinacircularseries Howmucheasierdoeshavingthisextraallowancemakeittosortpermuta-tions?Thisquestionisansweredinthefollowingtwopropositions. 38

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Wemayassumethattheleft-greedyalgorithmwasappliedasitisoptimalinthecasewhenwehavejusttwostacks.Letpibetherstentryofpthatcannotlegallyenterthestacks.Theremustbeanentrycintherststackthatissmallerthanpiwhichkeepspifromenteringthestacks.Likewise,thereisanentrybinthesecondstackthatissmallerthancwhichkeepscfrommovingtothenextstack.Finallytheremustbeanentryasmallerthanbthathasnotyetenteredthestacksandthuspreventsbfromgoingtotheoutput.SeeFigure 3{2 Figure3{2: Stacksatthemomentoffailure. Inorderforthispermutationtobesortedbytwostacksinacircularseries,thereneedstobeawayforpitoenterthestacks.Forthistohappen,ccannotbeintherststack.Thusatsomepointweneedtomovecleft.Sincetheleft-greedyalgorithmwasappliedtotrytosortthepermutationbytwostacksinseries,weknowthatateverystagetheremusthavebeenasmallerentrytotheleftofcorabovec.Ourextraallowancemayallowustomovethesmallerentryfromtheleftofcbacktotherststack,butthentheentryisabovecandstillmustbemovedtotheleftofcbeforeccanmove.Hencethisdoesnothelpmovecleftinanyway.Iftheproblemwasthatthesmallerentrywasabovec,theleft-greedyalgorithmwouldhavemoveditleftwhenitwaspossibleandthismovedoesnotrequireourcircularallowance.Thusthereisnowaytomovecleftandthuswecannotbringpiinthestacksusingtwostacksinacircularseries.Hencenopermutationcanbesortedbytwostacksinacircularseriesthatcouldnothavebeensortedbytwostacksinseries.3

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Sinceaddinganotherpossiblemovedidnothelpatallwhensortingbytwostacksinseries,itmaybesurprisingtolearnthatoncetherearemorethantwostacksinseriestheadditionofthecircularmovemakesanimmensedierence.Infact,everypermutationcanbesortedbythreeormorestacksinacircularseries.Thedierencecanbeseeninpartbyobservingthattheclassofpermutationsthataresortablebytstacksinseriesisclosedandthereforecategorizedbypatternavoidance.Thusifthereisevenasinglepatternthatmustbeavoidedforapermutationtobesortable,bythepreviouslymentionedMarcus-TardosTheoremweknowthenumberofsuchpermutationsoflengthnisboundedbycnwherecisaconstantdependentonlyonthebasisofpatternstobeavoided.

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andtheycanmoveasiftherearenootherentriesinthestackssinceeachissmallerthananyoftheremainingentries.Withthesmallerelementsoutoftheway,pk+1canfreelyentertherststack.Onceagainthei1smallestentriescanmoveasthoughtherearenootherentriesinthestackandsobytheinductionhypothesis,wecanmovethemallbacktotherststack.Togetthek+1entriesinthesecondstack,theprocessisthesameasbefore.Beforelettingpk+1enterthestacks,movethekentriesalreadytheretothesecondstack.Thenmovethei1smallestentriestothethirdstack.Nowpk+1canentertherststackandmovetothesecondstack.Finallythei1smallestentriescanmovebacktothesecondstack.Ifwearetryingtogetthek+1entriesintothethirdstack,weneedaslightvariationonthismethod.Onceagain,beforelettingpk+1enterthestacks,putthekoriginalentriesinthethirdstack.Next,bringpk+1intotherststackandovertothesecondstack.Nowmovethei1smallestentriestotherststack.Thisallowstomovepk+1tothethirdstack.Lastlybringthei1smallestentriesbacktothethirdstack.3Thislemmagivesusallweneedtonowprovetheearlierproposition.Proof:Sincewecanputalltheentriesinonestack,thismeansthatwecanputallnentriesofpintothestacksandgetthemallinthelaststack.Sincethestacksareindescendingorder,weneedmerelytoemptythethirdstackintotheoutputtogettheidentitypermutation.Alsonotethathavingtstackswhent>3doesnothurttheprocessaswecanalwayssaythatnoentrystaysinthestacks3;4;:::;t1.Thatis,ifanentryenters

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anyofthesestacksitkeepsmovingleftuntilitendsupinstackt.Thenweareessentiallyusingonlythreestacks.3 3{3 and 3{4 demonstrateonewaythiscanbedone. Figure3{3: PartIofsorting23451bythreestacksinacircularseries.

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Figure3{4: PartIIofsorting23451bythreestacksinacircularseries.

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4.1 ParityoftheNumberofPermutationsofLengthn 9 ]usesacollectionoflatticepathswhicharealsocountedbytheCatalannumbers.Wepresenthisproofhere.Proof:OnetypeofcombinatorialobjectthatthenthCatalannumbercountsisasetoflatticepathsDn.Theselatticepathsstartat(0;0)andendat(2n;0)byusingonlythesteps(1;1)and(1;1)andneverleavetherstquadrant.Deneaninvolution:Dn!Dnbyreectingthepathsacrossthelinex=n.ThentheonlypathsofDnthathavenotbeenpaireduparethosewhicharesymmetricwithrespecttothelinex=n.Tocompletetheproof,wewillneedtoshowthatthereareanoddnumberofthesesymmetricpathsifandonlyifn=2r1forsomenaturalnumberr.NotethateverylatticepathinDntakesexactly2nsteps.IfwisasymmetricpathofDn,thenwcanbebrokendowninsuchawaythatw=w1uuw2wherew1

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andw2arepathsthattaken1stepseach.Thenuwouldbeeitherthestep(1;1)orthestep(1;1),anduwouldbetheothertypeofstep.Weusethisdecompositiontodeneanotherinvolution.Ifw1(andthusw2)isnotapathofDn1 2,thenweknowthatw1stopsabovethex-axis.ThereforeinthiscaseitissafetotheswapstepsuandutogetanothersymmetricpathofDn.Wedeneinthefollowingway.(w)=8><>:w1uuw2ifw1=2Dn1 2welse:Nowifniseven,thenDn1 2doesnotexistandsotheinvolutioncompletesthepairing.HencejDnjisevenwhenniseven.Weuseinductiontoprovetheremainderofthetheorem.Thetheoremisclearlytruewhenn=1.Supposeitistrueforanynaturalnumberlessthann.Whenprovingthisistrueforn,wemayassumenisodd.TheonlyunpairedpathsofDnarethesymmetricpathsw=w1uuw2wherew12Dn1 2.Noticethatwiscompletelydeterminedbyw1sinceumustbethestep(1;1)sothatthepathcanremainintherstquadrantandw2simplytakestheoppositestepsofw1inthereverseorder.Thusf:Dn!Dn1 2wheref(w)=w1isabijection.HencejDnjjDn1 2j(mod2).IfjDnjisodd,thenwehavejDn1 2jisalsooddandthusn1 2=2r1forsomenaturalnumberr.Hencen=2r+11.Likewise,ifn=2r1forsomenaturalnumberr,thenn1 2=2r11andsojDn1 2jisoddandthusjDnjisoddaswell.SinceCn=jDnjthiscompletestheproof.3

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Thenumberoftimesthatthenumberof1-stacksortablepermutationsofagivenlengthisoddisinnite.Interestingly,ifweconsiderthefunctionCn=8><>:0ifCniseven1ifCnisoddwecanseethatthedensityofCniszero.Inthecaseof2-stacksortablepermutationsoflengthn,ithasbeendeter-minedcombinatoriallybyBona[ 4 ]exactlywhichvaluesofnhaveW2(n)odd.Startingwithn0=1,anumbermcanbereachedbysomecombinationofthefollowingsteps:1.ni+1=2ni1or2.ni+1=4ni+1ifandonlyifthevalueofW2(m)isodd.Infact,Bona[ 4 ]showedthatexactlyFk+1ofthevaluesnlessthan2karesuchthatW2(n)isodd.NotethatFk+1denotesthe(k+1)stFibonaccinumber.IfweconsiderthefunctionW2(n)=8><>:0ifW2(n)iseven1ifW2(n)isoddthedensityofW2(n)isagainzero.ThisissobecausetheasymptoticgrowthoftheFibonaccinumbersisroughly(1:61)n.Guiberthascomputedthenumberof3-stacksortablepermutationsoflengthnforn21;2;3;:::;9.Hefoundthattheonlytwocaseswherethisnumberisoddiswhenn=1orn=9.Theonlycaseswherethenumberisoddfor1-stacksortablepermutationsoccurwhenn=2n+1,andtheonlycaseswhenthenumberisoddfor2-stacksortablepermutationsoccurwhenn=4n+1.Thesefactssuggestthatperhapstheonlycaseswhenthenumberisoddfor3-stacksortablepermutationsoccurwhenn=8n+1.

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4.2 ParityofthePermutationsofLengthnwhichare SortablebyTwoStacksinSeriesTheparityofpermutationsoflengthnthataresortablebytwostacksinseriesisnotdiculttondfromtheaforementionedformulainTheorem 1.28 forthenumberofsuchpermutations,Sn(1342)=7n23n2 2(1)n1+3nXi=22i+1(2i4)! 1 ].Whatwewillshowisthatthereisanoddnumberofplaneforestsof(0;1)treesonnverticesexactlywhenniscongruenttozerooronemodulofour.Thenumberof(0;1)treesonnverticesisgivenby32n2(2n2)! (n1)!(n+1)!aswasprovedbyCori,Jacquard,andSchaeer[ 7 ].

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Thisproofisonlysemi-combinatorialduetothefactthatweusethisformulatoseethatthenumberof(0;1)treesonnverticesisevenexactlywhenn>3.Wedeneftobeaninvolutionon(0;1)treesthattakesone(0;1)treeonkverticestoanother(0;1)treeonkvertices.Furtherwemayassumethatfhasnoxedpointswhenk>3.Inthecaseswhenk=1andk=2,thereisjustone(0;1)treeforeach,sofwillxthesetwotrees.Finallyinthecasewhenk=3,wehavethree(0;1)treesasshowninFigure 4{1 Figure4{1: Thethree(0;1)treesonthreevertices. Inthiscasewewillhaveftaketree(a)totree(b)andxthetreesoftype(c).Nowweconsiderplaneforestsof(0;1)treeswithatotalofkvertices.Wepairmostoftheseforestswithothersuchforestsbythefollowingsteps.First,pairforestsbyapplyingftoeachtree.Theremainingunpairedforestshavesomecombinationofonlysinglevertextrees,treeswithtwovertices,orthethreevertextree(c).Theforeststhatremainareexactlythosewhichcontainatmostonetypeoftreewiththreevertices,atmostonetypeoftreewithtwovertices,andatmostonetypeoftreewithonevertex.Whatwearereallylookingatareorderedpartitionsofnintopartsnolargerthanthree.Letanbethenumberofsuchpartitionsofn.Itiscleartoseethata0=1,a1=1,a2=2,anda3=4.Byusingarecursiononthenumberofthesepartitions,wewillprovethefollowingclaim.Claim:Thenumberoforderedpartitionsofnintopartsthatarenolargerthanthreeisoddexactlywhenniscongruenttozerooronemodulofour.

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Incontrasttothepreviouscases,ifwedeneSn(1342)=8><>:0ifSn(1342)iseven1ifSn(1342)isoddthedensityofSn(1342)notzero.Infact,sinceSn(1342)isoddexactlywhenn0or1(mod4),thedensityofSn(1342)isexactlyonehalf.Thisalsomeansthatthenumberofpermutationsoflengthnthatavoid1342isoddforexactlyhalfthevaluesofn.Thisisrstinstancewherethenumberofpermutationsagivenlengththatavoidapatternisknowntohaveanequalnumberofoddandevencases.

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5.1 IntroductionTheideaofgraphreconstructionbeganwithaquestionofUlam[ 21 ]inhisproblembook. 51

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Thisconjecturecanbestatedalternativelytosaythatanyunlabeledgraphwithmorethanthreeverticescanbereconstructed(uptoisomorphism)fromthesetofminorsthatresultfromtheremovalofasinglevertexandallitsincidentedges.Onevariationonthisproblemistryingreconstructagraphbylookingattheminorsthatresultwhenasingleedgeisremovedinsteadofavertex.Anotheristoxanumberkandseeifgraphscanbereconstructedifoneconsiderstheremovalofeachsetofkverticesandtheirincidentedgestoformtheminors.SinceUlammadehisconjecture,manypartialresultshavebeenfoundingraphtheorythatareconsistentwiththis.However,thecompleteproblemremainsunsolved.Wenowconsiderthisproblemappliedtopermutationsinsteadofgraphs.

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Notethatalthoughgraphreconstructionhasbeenseriouslystudied,itcouldbethattheresultsforpermutationreconstructionarecompletelydierent.Onereasontothinkthatthismaybethecasecanbeseenbylookingatposetsorderedbyminorsofbothgraphsandpermutations. 14 ]thatPGdoesnothaveaninniteantichain.Sincethereisnoinniteantichain,thegraphsofPGaresaidtohaveawell-quasiordering.However,PPdoesnothaveawell-quasiorderingasthereareexamplesofinniteantichainsoftheposetPP.OnesuchexampleisthesetofminimalpermutationswhicharenotsortablebytwostacksinseriesgiveninChapter1.AnotherexamplecanbefoundinChapter16ofAWalkThroughCombinatoricsbyBona[ 6 ].Thisdierenceinpossibilitiesintheminorscouldmakesolvingthecorre-spondingreconstructionproblemswithpermutationsmorepromisingthanit

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hasbeenwithgraphtheory.Infact,inthenextsections,wepresentproofsthatN1=5,N2=6,aswellassomepartialresultsforlargervaluesofk. 5.2 TheCasewhenk=1 5.6 isduetoBruceSagan[ 15 ].Proof:Letpbeapermutationoflengthn5.Wewillrstshowthatwecandeterminewhichpositioninpthe1isinfromM1(p).If1isintheithpositioninp,thenconsiderthepositionof1inthe(n1)-minorsofp.Wehave (5.1) Also,onetimethepositionof1inan(n1)-minorisdeterminedby2whentheoriginal1ofpistheentrydeleted.Thuswehavethreepossibilities. 1. The1isinpositioniwherepdoesnotcontain12or21asafactor.Thishappensifandonlyifthereisapositionj6=i;i1suchthatoneofthe(n1)-minorsofphas1inpositionj. 2. The1isinpositioniwherepcontainsthefactor12.Thiscasehappensifandonlyifthepositionsiandi1aretheonlyplaceswhereanyofthe(n1)-minorshavea1andatleastn2ofthem,whichismorethanhalfofthe(n1)-minorssincen5,alsocontainthefactor12.(Then2timesarewhenneithertheoriginal1or2waseliminated.)

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3. The1isinpositioniwherepcontainsthefactor21.Thiscasehappensifandonlyifthepositionsiandi1aretheonlyplaceswhereanyofthe(n1)-minorshavea1andatleastn2ofthemalsocontainthefactor21.NowwecandeterminewhichpositioniisfromthesepossibilitiesandEquation 5.1 andthusweknowwherethe1isinp.Tocompletetheproof,weneedtondwhichminorp0wasproducedbyeliminatingtheoriginal1ofp.Thenwecansimplyinsertthe1intotheithpositionofp0andincreasealltheotherentriesbyonetogetthepermutationp.Wewillconsiderthethreecaseslistedabove. 1. Inthecasewhenpdidnotcontain12or21asafactor,p0isthe(n1)-minorwhere1isnotinpositioniori1. 2. Inthecasewhenpcontainsthefactor12,wecanseethatp0isthepermuta-tionwithshortestfactoroftheform123:::m.Thisistruesinceeliminatinganentryfromthefactorwilldecreasethelengthofitbyone.Notethatiftheoriginalpcontains123:::(m+1),theneliminatinganyoftheentries1;2;:::;m+1willresultinsuchaminor,butsincetheentriesarealladjacent,theseminorswillallbethesame,soitwillnotmatterwhichoneoftheseminorswepick. 3. Finallywhenpcontainsthefactor21,wecanseethatp0isthepermutationwithshortestfactoroftheformm(m1)(m2):::21bythesamereasoningasincase2.Hencethepropositionisproved.3

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TheCasewhenk=2

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Ifin2n6k2n3(allbutatmostthree)ofthese(n2)-minors,1istotheleftof2,then3mustbetotherightofboth1and2.Sincen8,allofthesecasesaredistinct.Wecannowdeterminewhere1,2,and3arerelativetooneanother.Supposewecandeterminewhere1;2;3;:::;iarerelativetooneanother.Thenwewillshowhowtodeterminewherei+1isrelativeto1;2;3;:::;i.Weuseanotherinductionargumenthere.Firstwewillshowhowtondwherei+1isrelativeto1,2,and3.Considerwhere1isrelativetoi1inthe(n2)-minorsofp.Thesecanallbefoundbywhere1;2;3;:::;iarerelativetoeachotherexceptforthefollowingcases:i12timesthisisdeterminedbywhere1isrelativetoi+1,i2timesthisrelativeorderisdeterminedbywhere2isrelativetoi+1,andonetimethisisdeterminedbywhere3isrelativetoi+1.Sinceweknowwhere1,2,and3arerelativetooneanother,thisdetermineswherei+1isrelativeto1,2,and3.Nowsupposeweknowwherei+1isrelativeto1;2;:::;k1.Tondwherei+1isrelativetok,considerwherek2isrelativetoi1inthe(n2)-minorsofp.Exceptinthecasewherethesetwoentriescomefromkandi+1,weknowwherek2andi1shouldberelativetooneanotherinthe(n2)-minorsofp.Simplylookattheremainingminorstodeterminewherekisrelativetoi+1inp.Sincewecandeterminewherealltheentriesarerelativetooneanother,wecanndpfromits(n2)-minors.3

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Ifweletp=13524andq=14253wecancheckthatM2(p)=1233;1324;213;231;312=M2(q).3 TheCasewhenk=3InthissectionwewillusesimilartechniquesasshowninthelastsectiontoprovethatN3=7;10;11;12;or13.

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Ifin0k3n8ofthese(n3)-minors,1istotheleftof2,then3mustbetotheleftofboth1and2.Ifinn32kn32+3n8ofthese(n3)-minors,1istotheleftof2,then3mustbebetween1and2.Ifin2n32k2n32+3n8ofthese(n3)-minors,1istotheleftof2,then3mustbetotherightofboth1and2.Sincen13,allofthesecasesaredistinct.Wecannowdeterminewhere1,2,and3arerelativetooneanother.Onceagain,fromthetotalmultisetof(n3)-minorsofp,setasiden33+3n32oftheseminorsthathave1,2,and3inthesameorderaspdoes.Wedothissincetherearen33+3n32minorswhereatmostoneof1,2,and3iseliminated.Inthe(n3)-minorsthatresultfromatleasttwoof1,2,and3beingeliminated,the4actsasthe2exceptwhenthreeof1,2,3,and4areeliminated.Ifin0k4ofthese(n3)-minors,1istotheleftof2,then4mustbetotheleftof1,2,and3.Ifinn4knofthese(n3)-minors,1istotheleftof2,then4mustbetotheleftofexactlytwoof1,2,and3.Ifin2(n4)k2n4ofthese(n3)-minors,1istotheleftof2,then4mustbetotheleftofexactlyoneof1,2,and3.Ifin3(n4)k3n8ofthese(n3)-minors,1istotheleftof2,then4mustbetotherightof1,2,and3.Sincen13,allofthesecasesaredistinct.Wecandeterminewhere1,2,3,and4arerelativetooneanother.Supposewecandeterminewhere1;2;3;:::;iarerelativetooneanother.Thenwewillshowhowtodeterminewherei+1isrelativeto1;2;3;:::;i.

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Weuseanotherinductionargumenthere.Firstwewillshowhowtondwherei+1isrelativeto1,2,3,and4.Considerwhere1isrelativetoi2inthe(n2)-minorsofp.Thesecanallbefoundbywhere1;2;3;:::;iarerelativetoeachotherexceptforthefollowingcases:i13timesthisrelativeorderisdeterminedbywhere1isrelativetoi+1,i22timesthisisdeterminedbywhere2isrelativetoi+1,i31timesthisisdeterminedbywhere3isrelativetoi+1,andonetimethisisdeterminedbywhere4isrelativetoi+1.Sinceweknowwhere1,2,3,and4arerelativetooneanother,thisdetermineswherei+1isrelativeto1,2,3,and4.Nowsupposeweknowwherei+1isrelativeto1;2;:::;k1.Togetwherei+1isrelativetok,considerwherek3isrelativetoi2inthe(n2)-minorsofp.Exceptinthecasewherethesetwoentriescomefromkandi+1,weknowwherek3andi2shouldberelativetooneanotherinthe(n3)-minorsofpbecauseoftheinductionhypotheses.Nowsimplylookattheremainingminorstodeterminewherekisrelativetoi+1inp.Sincewecandeterminewherealltheentriesarerelativetooneanother,wecanndpfromits(n3)-minors.3

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5.5 FurtherResultsandDirectionsFirstitisgoodtonotethatifwehavetwopermutationswhoseminormulti-setsarethesamewhenremovingk1entriesatatime,thentheminormultisetsthatresultfromremovingkentriesatatimewillalsobethesameasisstatedinthefollowinglemma.

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Theresultsforsmallkshowninthischapterssuggestthatperhapsthefollowingmaybetrue.

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[1] M.D.Atkinson,M.M.Murphy,N.Ruskuc,Sortingwithtwoorderedstacksinseries.TheoreticalComputerScience,289(2002),205-223. [2] M.Bona,CombinatoricsofPermutations.C.R.C.Press,BocaRaton,FL,2004. [3] M.Bona,ExactEnumerationof1342-avoidingpermutations;Acloselinkwithlabeledtreesandplanarmaps.JournalofCombinatorialTheory,SeriesA80(1997),257-272. [4] M.Bona,Parityandstacksortability.PresentedattheThirdInternationalConferenceonPatternAvoidingPermutations(Gainesville,2004). [5] M.Bona,Asurveyofstacksortingdisciplines.ElectronicJournalofCombi-natorics,9(2002-2003),no.2,Article1. [6] M.Bona,AWalkThroughCombinatorics:AnIntroductiontoEnumerationandGraphTheory.WorldScientic,RiverEdge,NJ,2002. [7] R.Cori,B.Jacquard,G.Schaeer,Descriptiontreesforsomefamiliesofplanarmaps.Proceedingsofthe9thConferenceonFormalPowerSeriesandAlgebraicCombinatorics,(Vienna,1997),196-208. [8] S.Dulucq,S.Gire,J.West,Permutationswithforbiddensubsequencesandnonseparableplanarmaps.Proceedingsofthe5thConferenceonFormalPowerSeriesandAlgebraicCombinatorics,(Florence,1993),DiscreteMath.,153(1996),85-103. [9] O.Egecioglu,TheparityoftheCatalannumbersvialatticepaths.FibonacciQuart.21(1983),no.1,65-66. [10] I.P.Goulden,J.West,Raneypathsandacombinatorialrelationshipbe-tweenrootednonseparableplanarmapsandtwo-stack-sortablepermutations.JournalofCombinatorialTheory,SeriesA75(1996),220-242. [11] D.E.Knuth,FundamentalAlgorithms,TheArtofComputerProgramming,vol.1,2nded.Addison-Wesley,Reading,MA1973. [12] G.Kreweras,Suruneclassedesproblemesliesautreillisdespartitionsd'entiers.CahiersduB.U.R.O.,6(1965),5-105. 64

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[13] A.Marcus,G.Tardos,ExcludedpermutationmatricesandtheStanley-Wilfconjecture.J.Combin.TheorySeriesA107(2004),no.1,153-160. [14] N.Robertson,P.D.Seymour,GraphMinors.XX.Wagner'sConjecture.JournalofCombinatorialTheory,SeriesB92(2004),no.2,325-357. [15] BruceSagan,personalcommunication. [16] R.P.Stanley,Log-concaveandunimodalsequencesinalgebra,combina-torics,andgeometry,in\GraphTheoryandItsApplications:EastandWest."Ann.NYAcad.Sci.,576(1989),500-535. [17] Z.Stankova,J.West,AnewclassofWilf-equivalentpermutations.J.AlgebraicCombinatorics,15(2002),no.3,271-290. [18] R.Smith,Comparingalgorithmsforsortingwithtstacksinseries.AnnalsofCombinatorics,8(2004),113-121. [19] R.E.Tarjan,Sortingusingnetworksofqueuesandstacks.JournaloftheACM,19(1972),341-346 [20] J.W.Tutte,Acensusofplanarmaps.CanadianJournalofMathematics,33(1963),249-271. [21] S.M.Ulam,ACollectionofMathematicalProblems.Wiley,NewYork,NY,1960. [22] J.West,PermutationswithforbiddensubsequencesandStacksortablepermutations.PhDthesis,MassachusettsInstituteofTechnology,1990. [23] J.West,Sortingtwicethroughastack.TheoreticalComputerScience,117(1993),303-313. [24] D.Zeilberger,AproofofJulianWest'sconjecturethatthenumberoftwo-stack-sortablepermutationsoflengthnis2(3n)!/((n+1)!(2n+1)!).DiscreteMath.,102(1992),85-93.

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IwasborninRochester,NY.IlivedthereuntilImovedtoIndianaandthenFloridatopursuegraduatestudies.Iearnedmybachelor'sdegreefromNazarethCollegein1997andmymaster'sdegreefromPurdueUniversityin2000.IcametotheUniversityofFloridain2001andbeganstudyingunderthedirectionofMiklosBonain2002. 66