Combinatorial GameTheory
ASpecialCollectioninHonorofElwynBerlekamp,John H.ConwayandRichardK.Guy
Editedby
RichardJ.Nowakowski,BruceM.Landman,FlorianLuca, MelvynB.Nathanson,JaroslavNešetřil,andAaron Robertson
MathematicsSubjectClassification2010
05A,05C55,05C65,05D,11A,11B,11D,11K,11N,11P,11Y,91A46
Editors
RichardJ.Nowakowski DalhousieUniversity Dept.ofMathematics&Statistics ChaseBuilding HalifaxNSB3H3J5 Canada r.nowakowski@dal.ca
BruceM.Landman UniversityofGeorgia DepartmentofMathematics Athens,GA30602 USA
Bruce.Landman@uga.edu
FlorianLuca UniversityofWitwatersrand SchoolofMathematics 1JanSmutsAvenue Johannesburg2000 RepublicofSouthAfrica Florian.Luca@wits.ac.za
MelvynB.Nathanson LehmanCollege(CUNY) DepartmentofMathematics 250BedfordParkBoulevardWest BronxNY10468 USA
Melvyn.Nathanson@lehman.cuny.edu
JaroslavNešetřil CharlesUniversity ComputerScienceInstitute(IUUK) Malostranskenam.25 11800Praha CzechRepublic Nesetril@iuuk.mff.cuni.cz
AaronRobertson ColgateUniversity DepartmentofMathematics 219McGregoryHall HamiltonNY13346 USA arobertson@colgate.edu
ISBN978-3-11-075534-3 e-ISBN(PDF)978-3-11-075541-1 e-ISBN(EPUB)978-3-11-075549-7
LibraryofCongressControlNumber:2022934372
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Preface
Whatis1 + 1 + 1?
JohnH.Conway,1973
Individually,eachofElwynR.Berlekamp,JohnH.Conway,andRichardK.Guyhave receivedmuch,rightlydeserved,praise.Eachmadelastingcontributionstomanyareasofmathematics.Thisvolumeisdedicatedtotheirworkincombinatorialgame theory.Itisduetotheireffortsthatcombinatorialgametheoryexistsasasubject.
BriefHistoryofhow WinningWays cametobe
Boutonfirstanalyzednim[67],littlerealizinghowcentralnimwastobe.Inthenext twodecades,otherresearcherscontributedtheanalysisofafewother,specificgames. ThechesschampionEmanuelLaskercameclosetoacompletetheoryofimpartial games.Itwasinthe1930sthatGrundy[68]andSprague[72]gaveacompleteanalysis,nowknownastheSprague–Grundytheory.Despitebeinganeleganttheoryand easytoapply,thesubjectlanguishedbecausetherewasnocleardirectioninwhich todevelopthetheory.Inthelate1940s,RichardK.Guyrediscoveredthetheoryand definedtheoctalgames.In1956,GuyandC.A.B.SmithpublishedTheG-valuesofvariousgames [42].Thisgavetheworldaninfinitenumberofimpartialgamesandledto manyinteresting,easytostate,andyetstillunsolvedconjectures.
Theanalysisofpartizangameslookedoutofreach.TheFields’medalistJohnMilnor[70]in1953publishedSumsofpositionalgames.Thisonlycoveredgamesinwhich playersgainedwhentheyplayedandwasnoteasytoapply.In1960,JohnConway metMichaelGuy,Richard’sson.Throughthisfriendship,JohnmetRichardandasked aboutpartizangames.Thisturnedouttobearecurringthemeintheirworkinthe nexttwodecades.Alsoin1960,ElwynBerlekampgotropedintoplaying3 × 3dots&-boxesgameagainstacomputer.Helost,butknowingabouttheSprague–Grundy theory,heanalyzedthegame.(Recently,Elwynclaimedthathehadneverlostagame since.)ElwynmetRichardatthe1967ChapelHillconferenceandsuggestedthatthey writeabook.Richardagreed,gotJohnandElwyntogetherin1969,andworkbegan. Theanalysisofeachnonimpartialgamewaswellthoughtoutbutadhoc.John,with histraininginsettheory,startedtoseeastructureemergingwhengamesweredecomposedintocomponents.Hegavethenamesof1and1/2totwoabstractgamesand wasdelighted(giggledlikeababywasthephraseheused)whenhediscoveredthat, asgames,1/2 + 1/2 = 1.Hewrote OnNumbersandGames [28]inaweek.Thiscaused somefrictionamongthethree,but,eventually,workrestartedonWinningWays[3,4].
R.Austin,S.Devitt,D.Duffus,andmyself,asgraduatestudentsatCalgary, scouredtheearlypage-proofs.Wesuggestednumerousjokesandpuns.Fortunately, theauthorsrejectedallofthem.
https://doi.org/10.1515/9783110755411-201
Oneotherpersondeservestobementioned,LouiseGuy,Richard’swife.Agraciousladymadeeveryvisitortotheirhousefeelwelcome.Somepeoplehaveasked whythecombinatorialgameplayers,LeftandRight,arefemaleandmale,respectively. Theoriginalreasonshavebeenforgotten,butafterWinningWaysappeared,itbecame amarkofrespecttorememberthemas Louiseand Richard.
WhyElwyn,John,andRichardareimportant
Manybooksarewritten,enjoyalittlesuccess,andthenareforgottenbyallbuta few. OnNumbersandGames butespecially WinningWays [3,4]arestillpopulartoday.Thispopularityisduetothepersonalitiesandtheirapproachtomathematics. Allweregreatambassadorsformathematics,writingexplanatoryarticlesandgiving manypubliclectures.Morethanthat,theyunderstoodthatmathematicsneedsahumantouch.Thesedays,itiseasytogetacomputertoplayagamewell,buthowdoyou getapersontoplaywell?Thiswasoneoftheiraims. WinningWays is800+ pagesof puns,humor,easy-to-remembersayings,andverses.Theseprovidegreatandmemorableinsightsintothegamesandtheirstructures,andthebookisstillarichsourceof materialforresearchers.MathscinetreportsthatWinningWaysiscitedbyover300articles,GoogleScholarreportsover3000citations.Yet,anyreaderwillbehardpressed tofindasinglemathematicalproofinthebook.Elwyn,John,andRichardwroteitto entertain,drawinareader,andgivethemanintuitivefeelingforthegames.
Afterthepublicationof WinningWays,eventhoughallwerewellknownfortheir researchoutsideofcombinatorialgametheory,theyremainedactiveinthesubject. Eachwasinterestedinmanypartsofthesubject,but,veryloosely,theirmaininterests were:
–ElwynBerlekampconsideredtheproblemofhowtodefineandquantifythenotionofthe“urgency”ofamove.Hemadegreatstrideswithhisconceptofan enrichedenvironment[11,24,25].Hewasalsofascinatedbygo[7,8,9,10,12,11] anddots-&-boxes[13,18,23].
–JohnConwayremainedinterestedinpushingthetheoryofsurrealnumbers,particularlyinfinitegames[30,37,41],gamesfromgroupsandcodes[32,39],and misèregames[35].
–RichardK.Guyretainedaninterestinsubtractionandoctalgames,writingabook forinquisitiveyoungsters[52].Hecontinuedtopresentthetheoryasitwas[54,57, 58,59,61]andalsosummarizedtheimportantproblems[56,60,62,64].
Standingontheirshoulders
Mostofthepapersinthisvolumecanbetraceddirectlybackto WinningWays and OnNumbersandGames,ortothecontinuinginterestsofthethree.Severalthough,
illustratehowfarthesubjecthasdeveloped.Ageneralapproachofimpartialmisère gameswasonlystartedbyPlambeck[71].A.Siegel(astudentofBerlekamp),amajor figuredevelopingthistheory,pushesthisfurtherinChapter20.Thetheoryofpartizanmisèregameswasonlystartedin2007[69].Whilstplayinginthecontextofall misèregames,Chapter10analyzesaspecificgame.Chapter16containsimportantresultsforanalyzingmisèredead-endinggames.In WinningWays,dots-&-boxesand top-entailsdonotfitintothetheory,eachinaseparateway.Theyareonlypartially analyzedandthatviaadhocmethods.Chapter17findsanormalplayextensionthat coversbothtypesofgames.(Theauthorsthinkthiswouldhaveintriguedthembutare notsureiftheywouldhavefullyapproved.)
Chapters1,5–9,12,15,18,and19eitherdirectlyextendthetheoryorconsidera relatedgametoonesgiveninWinningWays.AsisevidencedbyRichardK.Guy’searly contributions,itisalsoimportanttohavenewsourcesofgames.Thesearepresented inChapters2,3,11,13,and14.
SerendipitygaveChapter4.ThispaperisthefoundationofChapters1and5.It givesasimple,effect-for-humans,testforwhengamesarenumbers.Theauthorsare surethatElwyn,John,andRichardwouldhavestarteditwitharhymingcoupletthat everyonewouldthenremember.
Elwyn,John,andRichardgavefreelyoftheirtime.Manypeoplewillrememberthe coffee-timeandeveningsattheMSRIandBIRSWorkshops.Eachwouldbeatalarge tablefullyoccupiedbyanyonewhowishedtobethere,discussingandsometimes solvingproblems.Studentswereespeciallywelcome.Allcombinatorialgamesworkshopsnowfollowthisinclusivemodel.Alargenumberofpapersoriginateatthese workshops,haveseveralcoauthors,andincludestudents.Theysharedtheirtimeoutsideofconferencesandworkshops.Manystudentswillrememberthoseoffhandmoments,withoneormoreofthem,thatoftenstretchedtohours.Iwasasecond-year undergraduatestudentwhenonmeetingJohn,heimmediatelyaskedmewhatwas 1 + 1 + 1?EvenafterIanswered“3”,hestilltookthetimetoexplaintheintricaciesof 3-playergames.(Thequestionisstillunanswered.)
Theirwit,wisdom,andwillingnesstoplayprovidedpeoplewithpleasure.They willbesorelymissed,buttheirlegacyliveson.
RichardJ.Nowakowski
BooksandPapersinCombinatorialGameTheory
authoredbyElwynR.Berlekamp,JohnH.Conway,
andRichardK.Guy,plussixotherseminalpapers
[1]ElwynR.Berlekamp.SomerecentresultsonthecombinatorialgamecalledWelter’sNim.In Proc.6thAnn.PrincetonConf.InformationScienceandSystems,pages203–204,1972.
[2]ElwynR.Berlekamp.Thehackenbushnumbersystemforcompressionofnumericaldata. Inform.andControl,26:134–140,1974.
[3]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.1.AcademicPressInc.[HarcourtBraceJovanovichPublishers],London,1982.Games ingeneral.
[4]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.2.AcademicPressInc.[HarcourtBraceJovanovichPublishers],London,1982. Gamesinparticular.
[5]ElwynR.Berlekamp.Blockbustinganddomineering. J.Combin.Theory(Ser.A),49:67–116, 1988.Anearlierversion,entitled Introductiontoblockbustinganddomineering,appearedin: TheLighterSideofMathematics,Proc.E.StrensMemorialConf.onRecr.Math.anditsHistory, Calgary,1986,SpectrumSeries(R.K.GuyandR.E.Woodrow,eds.),Math.Assoc.ofAmerica, Washington,DC,1994,pp.137–148.
[6]ElwynR.Berlekamp.Two-person,perfect-informationgames.In TheLegacyofJohnvon Neumann (HempsteadNY,1988), Proc.Sympos.PureMath.,volume50,pages275–287.Amer. Math.Soc.,Providence,RI,1990.
[7]ElwynR.Berlekamp.IntroductoryoverviewofmathematicalGoendgames.In Proceedingsof SymposiainAppliedMathematics, CombinatorialGames,volume43,pages73–100.American MathematicalSociety,1991.
[8]ElwynR.Berlekamp.Introductoryoverviewofmathematicalgoendgames.InR.K.Guy,editor, Proc.Symp.Appl.Math., CombinatorialGames,volume43,pages73–100.Amer.Math.Soc., Providence,RI,1991.
[9]ElwynR.BerlekampandDavidWolfe. MathematicalGo:ChillingGetstheLastPoint.AKPeters, Ltd.,Wellesley,Massachusetts,1994.
[10]ElwynR.BerlekampandDavidWolfe. MathematicalGoEndgames:Nightmaresforthe ProfessionalGoPlayer.IshiPressInternational,SanJose,London,Tokyo,1994.
[11]ElwynR.BerlekampandY.Kim.Whereisthe“thousand-dollarko?”.InR.J.Nowakowski, editor, GamesofNoChance,Proc.MSRIWorkshoponCombinatorialGames,July,1994, Berkeley,CA, MSRIPubl.,volume29,pages203–226.CambridgeUniversityPress,Cambridge, 1996.
[12]ElwynR.Berlekamp.Theeconomist’sviewofcombinatorialgames.InRichardJ.Nowakowski, editor, GamesofNoChance, MSRI,volume29,pages365–405.CambridgeUniv.Press, Cambridge,1996.
[13]ElwynR.Berlekamp. TheDotsandBoxesGame:SophisticatedChild’sPlay.AKPeters,Ltd., Natick,MA,2000.
[14]ElwynR.Berlekamp.Sumsof N × 2amazons.InF.T.BrussandL.M.LeCam,editors, Lecture Notes–MonographSeries,volume35,pages1–34.InstituteofMathematicalStatistics, Beechwood,Ohio,2000.PapersinhonorofThomasS.Ferguson.
[15]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.1,secondedition.AKPeters,Ltd.,2001.
[16]ElwynR.Berlekamp.The4g4g4g4g4problemsandsolutions.InRichardJ.Nowakowski, editor, MoreGamesofNoChance, MSRIPublications,volume42,pages231–241.Cambridge UniversityPress,2002.
[17]ElwynR.Berlekamp.FourgamesforGardner.InD.WolfeandT.Rodgers,editors, Puzzler’s Tribute:AFeastfortheMind,pages383–386.AKPeters,Ltd.,Natick,MA,2002.Honoring MartinGardner.
[18]ElwynR.BerlekampandK.Scott.Forcingyouropponenttostayincontrolofaloony dots-and-boxesendgame.InRichardJ.Nowakowski,editor, MoreGamesofNoChance, MSRI Publications,volume42,pages317–330.CambridgeUniversityPress,2002.
[19]ElwynR.Berlekamp.Idempotentsamongpartisangames.InRichardJ.Nowakowski,editor, MoreGamesofNoChance, MSRIPublications,volume42,pages3–23.CambridgeUniversity Press,2002.
[20]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.2,secondedition.AKPeters,Ltd.,2003.
[21]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.3,secondedition.AKPeters,Ltd.,2003.
[22]ElwynR.Berlekamp,JohnH.Conway,andRichardK.Guy. WinningWaysforYourMathematical Plays.Vol.4,secondedition.AKPeters,Ltd.,2004.
[23]ElwynR.Berlekamp.Yellow-brownhackenbush.InMichaelH.AlbertandRichardJ. Nowakowski,editors, GamesofNoChance3, MSRI,volume56,pages413–418.Cambridge Univ.Press,2009.
[24]ElwynR.BerlekampandRichardM.Low.Entrepreneurialchess. Internat.J.GameTheory, 47(2):379–415,2018.
[25]ElwynR.Berlekamp.Temperaturesofgamesandcoupons.InUrbanLarsson,editor, Gamesof NoChance5, MathematicalSciencesResearchInstitutePublications,volume70,pages21–33. CambridgeUniversityPress,2019.
[26]JohnH.Conway.Allnumbersgreatandsmall.Res.PaperNo.149,Univ.ofCalgaryMath.Dept., 1972.
[27]JohnH.ConwayandH.S.M.Coxeter.Triangulatedpolygonsandfriezepatterns. Math.Gaz., 57:87–94;175–183,1973.
[28]JohnH.Conway. OnNumbersandGames.AcademicPress,1976.
[29]John H.Conway.Allgamesbrightandbeautiful. Amer.Math.Monthly,84(6):417–434,1977.
[30]JohnH.Conway.Loopygames. Ann.DiscreteMath.,3:55–74,1978.Advancesingraphtheory (CambridgeCombinatorialConf.,TrinityCollege,Cambridge,1977).
[31]JohnH.Conway.Agamutofgametheories. Math.Mag.,51(1):5–12,1978.
[32]JohnH.ConwayandN.J.A.Sloane.Lexicographiccodes:error-correctingcodesfromgame theory. IEEETrans.Inform.Theory,32(3):337–348,1986.
[33]JohnH.Conway.Morewaysofcombininggames.InR.K.Guy,editor, Proc.Symp.Appl.Math., CombinatorialGames,volume43,pages57–71.Amer.Math.Soc.,Providence,RI,1991.
[34]JohnH.Conway.Numbersandgames.InR.K.Guy,editor, Proc.Symp.Appl.Math., CombinatorialGames,volume43,pages23–34.Amer.Math.Soc.,Providence,RI,1991.
[35]W.L.SibertandJ.H.Conway.MathematicalKayles. Internat.J.GameTheory,20(3):237–246, 1992.
[36]JohnH.Conway.Onnumbersandgames.In SummerCourse1993:TheRealNumbers(Dutch), CWISyllabi,volume35,pages101–124.Math.CentrumCentrumWisk.Inform.,Amsterdam, 1993.
[37]JohnH.Conway.Thesurrealsandthereals.Realnumbers,generalizationsofthereals, andtheoriesofcontinua.In SyntheseLib.,volume242,pages93–103.KluwerAcad.Publ., Dordrecht,1994.
[38]JohnH.Conway.Theangelproblem.InR.J.Nowakowski,editor, GamesofNoChance,Proc. MSRIWorkshoponCombinatorialGames,July,1994,Berkeley,CA, MSRIPubl.,volume29, pages3–12.CambridgeUniversityPress,Cambridge,1996.
[39]JohnH.Conway. m13.In SurveysinCombinatorics, LondonMath.Soc.,LectureNoteSer., volume241,pages1–11.CambridgeUniv.Press,Cambridge,1997.
[40]JohnH.Conway. OnNumbersandGames,2ndedition.AKPeters,Ltd.,2001.Firstedition publishedin1976byAcademicPress.
[41]JohnH.Conway.Moreinfinitegames.InRichardJ.Nowakowski,editor, MoreGamesofNo Chance, MSRIPublications,volume42,pages31–36.CambridgeUniversityPress,2002.
[42]RichardK.GuyandCedricA.B.Smith.The G-valuesofvariousgames. Proc.Camb.Phil.Soc., 52:514–526,1956.
[43]RichardK.Guy.TwentyquestionsconcerningConway’ssylvercoinage. Amer.Math.Monthly, 83:634–637,1976.
[44]RichardK.Guy.Gamesaregraphs,indeedtheyaretrees.In Proc.2ndCarib.Conf.Combin.and Comput.,pages6–18.LetchworthPress,Barbados,1977.
[45]RichardK.Guy.Partisanandimpartialcombinatorialgames.In Combinatorics(Proc.Fifth HungarianColloq.,Keszthely,1976),Vol.I, Colloq.Math.Soc.JánosBolyai,volume18,pages 437–461.North-Holland,Amsterdam,1978.
[46]RichardK.Guy.Partizanandimpartialcombinatorialgames. Colloq.Math.Soc.JánosBolyai, 18:437–461,1978.Proc.5thHungar.Conf.Combin.Vol.I(A.HajnalandV.T.Sós,eds.), Keszthely,Hungary,1976,North-Holland.
[47]RichardK.Guy.Partizangames.In ColloquesInternationauxC.N.R.No.260—Problèmes CombinatoiresetThéoriedesGraphes,pages199–205,1979.
[48]RichardK.Guy.Anyonefortwopins?InD.A.Klarner,editor, TheMathematicalGardner,pages 2–15.WadsworthInternat.,Belmont,CA,1981.
[49]RichardK.Guy.Graphsandgames.InL.W.BeinekeandR.J.Wilson,editors, SelectedTopicsin GraphTheory,volume2,pages269–295.AcademicPress,London,1983.
[50]RichardK.Guy.Johnisbell’sgameofbeanstalkandJohnConway’sgameofbeans-don’t-talk. Math.Mag.,59:259–269,1986.
[51]RichardK.Guy. FairGame,COMAPMath.ExplorationSeries.Arlington,MA,1989.
[52]RichardK.Guy. FairGame:Howtoplayimpartialcombinatorialgames.COMAP,Inc.,60Lowell Street,Arlington,MA02174,1989.
[53]RichardK.Guy,editor. CombinatorialGames, ProceedingsofSymposiainApplied Mathematics,volume43.AmericanMathematicalSociety,Providence,RI,1991.Lecturenotes preparedfortheAmericanMathematicalSocietyShortCourseheldinColumbus,Ohio,August 6–7,1990,AMSShortCourseLectureNotes.
[54]RichardK.Guy.Impartialgames.In CombinatorialGames (Columbus,OH,1990), Proc. Sympos.Appl.Math.,volume43,pages35–55.Amer.Math.Soc.,Providence,RI,1991.
[55]RichardK.Guy.Mathematicsfromfun&funfrommathematics;aninformalautobiographical historyofcombinatorialgames.InJ.H.EwingandF.W.Gehring,editors, PaulHalmos: Celebrating50YearsofMathematics,pages287–295.SpringerVerlag,NewYork,NY,1991.
[56]RichardK.Guy.Unsolvedproblemsincombinatorialgames.In AmericanMathematicalSociety ProceedingsoftheSymposiumonAppliedMathematics,volume43,1991.Checkmyhomepage foracopy,http://www.gustavus.edu/~wolfe.
[57]RichardK.Guy.WhatisaGame?In CombinatorialGames,ProceedingsofSymposiainApplied Mathematics,volume43,1991.
[58]RichardK.Guy.Combinatorialgames.InR.L.Graham,M.Grötschel,andL.Lovász,editors, HandbookofCombinatorics,volumeII,pages2117–2162.North-Holland,Amsterdam,1995.
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Contents
Preface V
AnthonyBonato,MelissaA.Huggan,andRichardJ.Nowakowski
Thegameofflippingcoins | 1
KyleBurke,MatthewFerland,MichaelFisher,ValentinGledel,andCraig Tennenhouse
Thegameofblockingpebbles 17
KyleBurke,MatthewFerland,andShang-HuaTeng
TransverseWave:animpartialcolor-propagationgameinspiredbysocialinfluence andQuantumNim 39
AldaCarvalho,MelissaA.Huggan,RichardJ.Nowakowski,andCarlosPereirados
Santos
Anoteonnumbers | 67
AldaCarvalho,MelissaA.Huggan,RichardJ.Nowakowski,andCarlosPereirados Santos
Ordinalsums,clockwisehackenbush,anddominoshave | 77
AlexanderClowandStephenFinbow
Advancesinfindingidealplayonposetgames | 99
ErikD.DemaineandYevheniiDiomidov
Strings-and-CoinsandNimstringarePSPACE-complete | 109
EricDuchêne,MarcHeinrich,RichardNowakowski,andAlineParreau
Partizansubtractiongames 121
MatthieuDufour,SilviaHeubach,andAnhVo CircularNimgamesCN(7, 4) | 139
AaronDwyer,RebeccaMilley,andMichaelWillette Misèredomineeringon2 × n boards 157
ZacharyGatesandRobertKelvey
Relatorgamesongroups | 171
L.R.Haff
PlayingBynum’sgamecautiously | 201
MelissaA.HugganandCraigTennenhouse Geneticallymodifiedgames 229
DouglasE.IannucciandUrbanLarsson Gamevaluesofarithmeticfunctions | 245
YukiIrie
Abase-pSprague–Grundy-typetheoremfor p-calmsubtractiongames:Welter’s gameandrepresentationsofgeneralizedsymmetricgroups 281
UrbanLarsson,RebeccaMilley,RichardNowakowski,GabrielRenault,andCarlos Santos Recursivecomparisontestsfordicotanddead-endinggamesundermisère play | 309
UrbanLarsson,RichardJ.Nowakowski,andCarlosP.Santos Impartialgameswithentailingmoves | 323
JamesB.Martin
ExtendedSprague–Grundytheoryforlocallyfinitegames,andapplicationsto randomgame-trees 343
RyoheiMiyaderaandYushiNakaya
Grundynumbersofimpartialthree-dimensionalchocolate-bargames | 367
AaronN.Siegel
Onthestructureofmisèreimpartialgames 389
AnthonyBonato,MelissaA.Huggan,andRichardJ.Nowakowski
Thegameofflippingcoins
Abstract:Weconsiderflippingcoins,apartizanversionoftheimpartialgameturningturtles,playedonlinesofcoins.Weshowthatthevaluesofthisgamearenumbers,andthesearefoundbyfirstapplyingareduction,thendecomposingtheposition intoaniteratedordinalsum.Thisisunusualsincemovesinthemiddleofthelinedo noteliminatetherestoftheline.Moreover,if G isdecomposedintolines H and K, then G = (H : KR).Thisisincontrasttohackenbushstrings,where G = (H : K).
1Introduction
In WinningWays,Volume3[3],Berlekamp,Conway,andGuyintroducedturning turtlesandconsideredmanyvariants.Eachgameinvolvesafiniterowofturtles, eitheronfeetorbacks,andamoveistoturnoneturtleoverontoitsback,withthe optionofflippinganumberofotherturtles,totheleft,eachtotheoppositeofitscurrentstate(feetorback).Thenumberdependsontherulesofthespecificgame.The authorsmovedtoplayingwithcoinsasplayingwithturtlesiscruel.
ThesegamescanbesolvedusingtheSprague–Grundytheoryforimpartialgames [2],butthestructureandstrategiesofsomevariantsareinteresting.Thestrategyfor moebius(flipuptofivecoins)playedwith18coins,involves Möbiustransformations; formogul(flipuptosevencoins)on24coins,itinvolvesthe miracleoctadgenerator developedbyR.CurtisinhisworkontheMathieugroup M24 andtheLeechlattice [6,7];ternups[3](flipthreeequallyspacedcoins)requiresternaryexpansions;and turningcorners[3],atwo-dimensionalversionwherethecornersofarectangleare flipped,needsnim-multiplication.
Weconsiderasimplepartizanversionofturningturtles,alsoplayedwith coins.Wegiveacompletesolutionandshowthatitinvolvesordinalsums.Thisis somewhatsurprisingsincemovesinthemiddleofthelinedonoteliminatemovesat theend.Comparethiswithhackenbushstrings[2]anddominoshave[5].
Acknowledgement: AnthonyBonatowassupportedbyanNSERCDiscoverygrant.MelissaA.HugganwassupportedbyanNSERCPostdoctoralFellowship.TheauthoralsothankstheDepartmentof MathematicsatRyersonUniversity,whichhostedtheauthorwhiletheresearchtookplace.RichardJ. NowakowskiwassupportedbyanNSERCDiscoverygrant.
AnthonyBonato, DepartmentofMathematics,RyersonUniversity,Toronto,Ontario,Canada,e-mail: abonato@ryerson.ca
MelissaA.Huggan, DepartmentofMathematicsandComputerScience,MountAllisonUniversity, Sackville,NewBrunswick,Canada,e-mail:mhuggan@mta.ca
RichardJ.Nowakowski, DepartmentofMathematicsandStatistics,DalhousieUniversity,Halifax, NovaScotia,Canada,e-mail:r.nowakowski@dal.ca
https://doi.org/10.1515/9783110755411-001
Wewilldenoteheadsby0andtailsby1.Ourpartizanversionwillbeplayedwith alineofcoins,representedbya0–1sequence d1d2 ... dn,where di ∈ {0, 1}.Withthis position,weassociatethebinarynumber ∑n i=1 di2i 1.Leftmovesbychoosingsomepair ofcoins di, dj, i < j,where di = dj = 1,andflipsthemoversothatbothcoinsare0s. Rightalsochoosesapair dk, dℓ, k < ℓ,with dk = 0and dℓ = 1,andflipsthemover.If j isthegreatestindexsuchthat dj = 1,then dk, k > j,willbedeleted.Forexample,
1011 = {0001, 001, 1 | 1101, 111}.
Thegameeventuallyendssincetheassociatedbinarynumberdecreaseswithevery move.Wecallthisgameflippingcoins.
Anotherwaytomodelflippingcoinsistoconsidertokensonastripoflocations.Leftcanremoveapairoftokens,andRightisabletomoveatokentoanopen spacetoitsleft.Weusethecoinflippingmodelforthisgametobeconsistentwiththe literature.
ThegameisbiasedtoLeft.Ifthereareanonzeroevennumberof1sinaposition, thenLeftalwayshasamove;thatis,shewillwin.Leftalsowinsanynontrivialpositionstartingwith1.However,therearepositionsthatRightwins.Thetwo-partmethod tofindtheoutcomesandvaluesoftheremainingpositionscanbeappliedtoallpositions.First,applyamodificationtotheposition(unlessitisall1s),whichreduces thenumberofconsecutive1stoatmostthree.Afterthisreduction,buildaniterated ordinalsum,bysuccessivelydeletingeverythingafterthethirdlast1,thisdeletedpositiondeterminesthevalueofthenexttermintheordinalsum.Asaconsequence, theoriginalpositionisaRightwinifthepositionremainingattheendisoftheform 0 01,andthevalueisgivenbytheordinalsum.
ThenecessarybackgroundfornumbersisinSection2.Section3containsresults aboutoutcomesandalsoincludesourmainresults.First,weshowthatthevaluesare numbersinTheorem3.2.Next,analgorithmtofindthevalueofapositionispresented, andTheorem3.3statesthatthevaluegivenbythealgorithmiscorrect.
TheactualanalysisisinSection4.Itstartsbyidentifyingthebestmovesforboth playersinTheorem4.2.ThisleadsdirectlytothecoreresultLemma4.5,whichshows thatthevalueofapositionisanordinalsum.Theordinalsumdecompositionof G isfoundasfollows.Let GL bethepositionaftertheLeftmovethatremovestherightmost1s.LetH bethestringG \GL;thatis,thesubstringeliminatedbyLeft’smove.Let HR betheresultofRight’sbestmovein H.Nowwehavethat G = GL : HR.Incontrast, theordinalsumsforhackenbushstringsanddominoshave[5]involvethevalueof H not HR .
TheproofofTheorem3.3isgiveninSection4.1.Thefinalsectionincludesabrief discussionofopenproblems.
Finally,weposeaquestionforthereader,whichweanswerattheendofSection4.1:Whowins0101011111 + 1101100111 + 0110110110111andhow?
2Numbers
Allthevaluesinthispaperarenumbers,andthissectioncontainsallthenecessary backgroundtomakethepaperself-contained.Forfurtherdetails,consult[1,8].Positionsarewrittenintermsoftheiroptions;thatis, G = {Gℒ | Gℛ}.
Definition2.1 ([1,2,8]).Let G beanumberwhoseoptionsarenumbers,andlet GL , GR betheLeftandRightoptionsofthecanonicalformof G.
1.Ifthereisaninteger k, GL < k < GR,orifeither GL or GR doesnotexist,then G is theinteger,say n,closesttozerothatsatisfies GL < n < GR .
2.Ifboth GL and GR existandthepreviouscasedoesnotapply,then G = p 2q ,where q istheleastpositiveintegersuchthatthereisanoddinteger p satisfying GL < p 2q < GR .
Thepropertiesofnumbersrequiredforthispaperarecontainedinthenextthree theorems.
Theorem2.2 ([1,2,8]). LetGbeanumberwhoseoptionsarenumbers,andletGL and GR betheLeftandRightoptionsofthecanonicalformofG.IfG′ andG′′ areanyLeft andRightoptions,respectively,then
Theorem2.2showsthatifweknowthatthestringofinequalitiesholds,thenwe needtoonlyconsidertheuniquebestmoveforbothplayersinanumber.
Weincludethefollowingexamplestofurtherillustratetheseideas:
(a)0 = {|} = { 9 |} = { 1 2 | 7 4};
(b) 2 = {| 1} = { 5 2 | 31 16};
(c)1 = {0 |} = {0 | 100};
(d) 1 2 = {0 | 1} = {3 8 | 17 32}.
Forgames G and H,toshowthat G ⩾ H,weneedtoshowthat G H ⩾ 0,meaning thatweneedtoshowthat G H isaLeftwinmovingsecond.Formoreinformation, seeSections5.1,5.8,and6.3of[1].
Let G and H begames.The ordinalsum of G,the base,and H,the exponent,is
Intuitively,playingin G eliminates H,butplayingin H doesnotaffect G.Foreaseof reading,ifanordinalsumisaterminanexpression,thenweencloseitinbrackets. Notethat x : 0 = x = 0 : x sinceneitherplayerhasamovein0.Wedemonstrate howtocalculatethevaluesofotherpositionswiththefollowingexamples:
(a)1 : 1 = {1 |} = 2;
G′ ⩽ GL < G < GR ⩽ G′′
(b)1 :
=
Notethatinallcases,whenbaseandexponentarenumbers,theplayersprefertoplay intheexponent.Intheremainderofthispaper,alltheexponentswillbepositive. OneofthemostimportantresultsaboutordinalsumswasfirstreportedinWinning Ways.
Theorem2.3 (ColonPrinciple[2]). IfK ⩾ K′,thenG : K ⩾ G : K′ . TheColonPrinciplehelpsproveinequalitiesthatwillbeusefulinthispaper.
Theorem2.4. LetGandHbenumbersallofwhoseoptionsarealsonumbers,andlet H ⩾ 0.
1. IfH = 0,thenG : H = G.IfH > 0,then (G : H) > G. 2. GL < (G : HL) < (G : H) < (G : HR) < GR .
Proof. Foritem(1),theresultfollowsimmediatelybyTheorem2.3. Foritem(2),if H ⩾ 0andalltheoptionsof G and H arenumbers,then GL < G = (G : 0) ⩽ (G : HL) < (G : H) < (G : HR).Thesecond,third,andfourthinequalitieshold since H isanumberandthus0 ⩽ HL < H < HR andbyapplyingtheColonPrinciple. Tocompletetheproof,weneedtoshowthat (G : HR) < GR.Todoso,wecheckthat GR (G : HR) > 0;inwords,wecheckthatLeftcanalwayswin.Leftmovingfirstcan moveinthesecondsummandto GR GR = 0andwin.Rightmovingfirsthasseveral options:
1.Movingto GR GL > 0,since G anditsoptionsarenumbers.HenceLeftwins.
2.Movingto GR (G : HRL) > 0byinduction.
3.Movingto GRR G : HR,butLeftcanrespondto GRR GR > 0since G andits optionsarenumbers.
Inallcases,Leftwinsmovingsecond.Theresultfollows.
Toprovethatallthepositionsarenumbers,weuseresultsfrom[4].Asetofpositionsfromarulesetiscalleda hereditarilyclosedsetofpositionsofaruleset ifitis closedundertakingoptions.Thisgamesatisfiesrulesetpropertiesintroducedin[4]. Inparticular,thepropertiesarecalledtheF1property andtheF2property,whichboth highlightthenotionof First-move-disadvantage innumbersandaredefinedformally asfollows.
Definition2.5 ([4]).Let S beahereditarilyclosedruleset.Givenaposition G ∈ S,the pair(GL , GR) ∈ Gℒ ×Gℛ satisfiestheF1propertyifthereisGRL ∈ GRℒ suchthatGRL ⩾ GL orthereis GLR ∈ GLℛ suchthat GLR ⩽ GR .
Definition2.6 ([4]).Let S beahereditarilyclosedruleset.Givenaposition G ∈ S,the pair (GL , GR) ∈ Gℒ × Gℛ satisfiesthe F2property ifthereare GLR ∈ GLℛ and GRL ∈ GRℒ suchthat GRL ⩾ GLR .
Asprovenin[4],ifgivenanyposition G ∈ S,allpairs (GL , GR) ∈ Gℒ × Gℛ satisfy oneoftheseproperties,thenthevaluesofallpositionsarenumbers.Furthermore, satisfyingtheF2propertyimpliessatisfyingtheF1property,anditwasshownthatall positions G ∈ S arenumbersifandonlyifforany G ∈ S,allpairs (GL , GR) ∈ Gℒ × Gℛ satisfytheF1property.Combiningtheseresultsgivesthefollowingtheorem.
Theorem2.7 ([4]). LetSbeahereditarilyclosedruleset.AllpositionsG ∈ Sarenumbers ifandonlyifforanypositionG ∈ S,allpairs (GL , GR) ∈ Gℒ × Gℛ satisfyeithertheF1or theF2property.
3Mainresults
Beforeconsideringthevaluesandassociatedstrategies,weconsidertheoutcomes, thatis,wepartiallyanswerthequestion“Whowinsthegame?”Thefullanswerrequiresananalogousanalysistofindingthevalues.
Theorem3.1. LetG = d1d2 dn.Ifd1d2 dn containsanevennumberof1s,orifd1 = 1 andthereareleasttwo 1s,thenLeftwinsG.
Proof. ARightmovedoesnotdecreasethenumberof1sintheposition.Thus,ifin G, Lefthasamove,thenshestillhasamoveafteranyRightmovein G.Consequently, regardlessofd1,ifthereareanevennumberof1sinG,thenitwillbeLeftwhoreduces thegametoall0s.Similarly,ifd1 = 1andthereareanoddnumberof1s,thenLeftwill eventuallyreduce G toapositionwithasingle1,thatis,to d1 = 1and di = 0for i > 1. Inthiscase,Righthasnomoveandloses.
Theremainingcase,d1 = 0andanoddnumberof1s,ismoreinvolved.Theanalysisofthiscaseisthesubjectoftheremainderofthepaper.Wefirstprovethefollowing: Theorem3.2. All flippingcoins positionsarenumbers.
Proof. LetGbeaflippingcoinsposition.Ifonlyoneplayerhasamove,thenthegame isaninteger.Otherwise,letLbetheLeftmovetochange (di, dj) from (1, 1) to (0, 0).Let RbetheRightmovetochange (dk, dℓ) from (0, 1) to (1, 0).Nootherdigitsarechanged. Ifallfourindicesaredistinct,thenboth L and R canbeplayedineitherorder.Inthis case, GLR = GRL.ThustheF2propertyholds.Ifthereareonlythreedistinctindices, thentwoofthebitsareones.IfLeftmovesfirst,then di = dj = dk = 0.IfRightmoves first,thentherearestilltwoonesremainingafterhismove.AfterLeftmoves,wehave di = dj = dk = 0,andhence GL = GRL.TheF1propertyholds.
Therearenomorecasessincetheremustbeatleastthreedistinctindices.Since everypositionsatisfieseithertheF1ortheF2property,byTheorem2.7itfollowsthat everypositionisanumber.
Givenaposition G,thefollowingalgorithmreturnsavalue.
Algorithm Let G beaflippingcoinsposition.Let G0 = G.
1.Set i = 0.
2.Reductions:Let α and β bebinarystrings,andeithercanbeempty.
(a)If G0 = α013+jβ, j ⩾ 1,thenset G0 = α101jβ.
(b)If G0 = α013β with β containinganevennumberof1s,thenset G0 = α10β.
(c)Repeatuntilneithercaseapplies;thengotoStep3.
3.If Gi is0r1, r ⩾ 0,or1a0pi10qi1, a ⩾ 0,and pi + qi ⩾ 0,thengotoStep5.
Otherwise, Gi = α01a0pi10qi1, pi + qi ⩾ 1, a > 0,andsome α.Set Qi = 0pi10q
GotoStep4.
4.Set i = i + 1.GotoStep3.
5.IfGi = 0r1,thensetvi =−r.IfGi = 1a0pi10qi1,thenset
6.For j from i 1downto0,set vj = vj+1 : 1 2
+qj 1 .
.GotoStep6.
7.Returnthenumber v0.
Thealgorithmimplicitlyreturnstwodifferentresults:
1.ForStep3,thesubstrings Q0, Q1,..., Qi 1, Gi partitionthereducedversionof G; 2.Thevalue v0.
First,weillustratethealgorithmwiththefollowingexample.Considertheposition G = 10011110110110111011110011.Wehighlightateachstepwhichreductionisbeing appliedtotheunderlineddigits;2(a)isdenotedby †,whereas2(b)isdenotedby ‡. Thealgorithmgivesthat
10011110110110111011110011 = 10011110110110111011110011(†) = 100111101101101111010011(†) = 1001111011011101010011(‡) = 10011110111001010011(‡) = 100111110001010011(†) = 1010110001010011.
Step3partitionsthelastexpressioninto101(011)(000101)(0011) sothattheordinal sumisgivenby v0 = ((1 2 : 1 2) : 1 64) : 1 8 = 10257 16348.
Nowlet H = 01001110110111011101.Thereductionsgivethat 01001110110111011101 = 01001110110111011101 = 010011101110011101 = 0100111100011101 = 01010100011101.
Thelastexpressionpartitionsinto01(0101)(00011)(101) sothat
0 = (( 1 : 1 4) : 1 32) : 1 =− 893 1024
Thenexttheoremisthemainresultofthepaper.
Theorem3.3 (Valuetheorem). LetGbea flippingcoins position.Ifv0 isthevalue obtainedbythealgorithmappliedtoG,thenG = v0.
Inthenextsection,wederiveseveralresultsthatwillbeusedtoproveTheorem3.3.TheproofofTheorem3.3willappearinSection4.1.
4Bestmovesandreductions
Theproofsinthissectionuseinductionontheoptions.Analternatebutequivalent approachistoregardthetechniquesasinductionontheassociatedbinarynumberof thepositions.Theproofsrequiredetailedexaminationofthepositions,andwewill usenotationsuitabletothecasebeingconsidered.Often,atypicalpositionwillbe writtenasacombinationofgenericstringsandthesubstringunderconsideration. Forexample,111011000110101mightbeparsedas (11101)(100011)(0101) andwritten as α100011β ormorecompactlyas α10312β.
WerequireseveralresultsbeforebeingabletoproveTheorem3.3.Webeginby provingasimplifyingreduction,followedbythebestmovesforeachplayer,andthen theremainingreductionsusedinthealgorithm.
AsanimmediateconsequenceofTheorems3.2and2.2,wehavethefollowing:
Corollary4.1. Letα,β,andγbearbitrarybinarystrings.Wethenhavethatα1β0γ > α0β1γ.Moreover,foranintegerr ⩾ 0,wehavethatβ10r1 > β.
Proof. RecallthatbyTheorem3.2allflippingcoinspositionsarenumbers.ThusTheorem2.2applies.
ARightoptionof α0β1γ is α1β0γ,andsowehavethat α1β0γ > α0β1γ.Similarly,a Leftoptionof β10r1is β,andsowehavethat β10r1 > β.
Next,weprovethebestmovesforeachplayer.Rightwantstoplaythezerofurthest totherightandthe1adjacenttoit.Leftwantstoplaythetwoonesfurthesttotheright.
Theorem4.2. LetGbea flippingcoins position,whereinG,randn,r n,arethe greatestindicessuchthatdr = dn = 1.Letsbethegreatestindexsuchthatds = 0.Left’s bestmoveistoplay (dr, dn),andRight’sbestmoveistoplay (ds, ds+1).
Proof. Weprovethistheorembyinductionontheoptions.Notethatweusetheequivalentbinaryrepresentationofthegameposition.Iftherearethreeorfewerbits,then, byexhaustiveanalysis,thetheoremistrue.
Let G be d1d2 ... dn.WebeginbyprovingLeft’sbestmoves.Let r and n bethetwo largestindices,wheredr = dn = 1,andthusdk = 0forr < k < n.Letiandj,i < j,betwo indiceswithdi = dj = 1.WeusethenotationG(di, dj, dr, dn) tohighlightthesalientbits. TheclaimedbestLeftmoveisfromG(1, 1, 1, 1) toG(1, 1, 0, 0).Thismustbecomparedto anyotherLeftmove,representedbymovingfrom G(1, 1, 1, 1) to G(0, 0, 1, 1).Thatis,we needtoshowthat G(1, 1, 0, 0) G(0, 0, 1, 1) ⩾ 0.
Forthemovestobedifferent,atleastthreeofi,j,r,naredistinct.Wefirstassume thatthefourindicesaredistinct.Inthiscase,wehavethat i < j < r < n.Byapplying Corollary4.1twicewehavethat
G
Wemayassumethen,withoutlossofgenerality,that j = r or j = n.If j = n,then i < r,sincetherearetwodistinctmoves.Nowconsider G(di, dr, dn) = G(1, 1, 1).By Corollary4.1wehavethatif j = r,then G(1, 0, 0) > G(0, 0, 1),andif j = n,then G(1, 0, 0) > G(0, 1, 0).
WenowproveRight’sbestmove.Therearemorecasestoconsider.Let s bethe largestindexsuchthat ds = 0andtherefore ds+1 = 1.Let i, j, i < j beindiceswith di = 0and dj = 1.Theclaimedbestmoveis ds, ds+1,andthismustbecomparedtothe arbitraryRightmove di, dj.Forthemovestobedifferent,theremustbeatleastthree distinctindices.
Theoriginalpositioniseither
G(di, dj, ds, ds+1) = G(0, 1, 0, 1), i < s,
or G(ds, ds+1, dj) = G(0, 1, 1), i = s, j > s + 1
WeneedtoshoweitherD = G(1, 0, 0, 1) G(0, 1, 1, 0) ⩾ 0orD = G(1, 1, 0) G(1, 0, 1) ⩾ 0, respectively.SupposeRightplaysinthefirstsummandof D.Notethat,byinduction, thebestmovesofLeftandRightareknown.
1.First,suppose j < s.ByinductionRight’sbestmoveinthefirstsummandof D is to D′ = G(1, 0, 1, 0) G(0, 1, 1, 0).Since i < j,itfollowsthat G(1, 0, 1, 0) isaRight optionof G(0, 1, 1, 0),andthus D′ ispositivebyCorollary4.1.
2.If j = s + 1,thenthereareonlythreedistinctindices.Theoriginalgameis G(di, ds, ds+1) = G(0, 0, 1) and D = G(1, 0, 0) G(0, 1, 0).Since G(1, 0, 0) isaRight optionof G(0, 1, 0),itfollowsthat D ispositivebyCorollary4.1.
3.Suppose j > s + 1.
If i < s,thentheoriginalgameisoftheform
and
TwoapplicationsofCorollary4.1(appliedtothehighlightedterms)give
If i = s,then
and
OneapplicationofCorollary4.1(relevanttermsagainhighlighted)gives
Thus D ⩾ 0.
Next,weconsiderRightmovinginthesecondsummandofD=G(1, 0, 0, 1) G(0, 1, 1, 0). Notethatbythechoicesofthesubscripts, dℓ = 1if n ⩾ ℓ ⩾ s + 1.
1.If n > s + 2,thenRight’sbestmoveinthesecondsummandistochange dn 1, dn from (1, 1) to (0, 0).Leftcopiesthismoveinthefirstsummand,andtheresulting differencegameisnonnegativebyinduction.
2.Suppose n = s + 2.
i.If j < s + 1,then G(di, dj, ds, ds+1, ds+2) = G(0, 1, 0, 1, 1) and D = G(1, 0, 0, 1, 1) G(0, 1, 1, 0, 1).Right’sbestmoveisto G(1, 0, 0, 1, 1) G(0, 1, 0, 0, 0).Leftmoves toG(1, 0, 0, 0, 0) G(0, 1, 0, 0, 0).ThisispositivebyCorollary4.1,andLeftwins. Forthenexttwosubcases,exactlytwo1swilloccupytwoofthefourindexed positions.SinceRightismovinginthesecondsummand,heischangingtwo 1stotwo0s.ThusLeft’sbestresponseforeachcaseistomoveinthefirst summand,bringingthegameto G(0, 0, 0, 0) G(0, 0, 0, 0) = 0,andshewins. Forthesecases,weonlylisttheoriginalposition.Thestrategyforbothcases isasjustdescribed.
ii.If j = s + 1,then G(di, ds, ds+1, ds+2) = G(0, 0, 1, 1) and D = G(1, 0, 0, 1)
G(0, 1, 0, 1).
iii.If j = s + 2,then G(di, ds, ds+1, ds+2) = G(0, 0, 1, 1) and D = G(1, 0, 1, 0)
G(0, 1, 0, 1).
3.Nowsuppose n = s + 1.
i.If j < s + 1,thenlet ℓ < s + 1bethelargestindexsuchthat dℓ = 1.
Ifj < ℓ,thenwehaveG(di, dj, dℓ, ds, ds+1)=G(0, 1, 1, 0, 1) andD=G(1, 0, 1, 0, 1)
G(0, 1, 1, 1, 0).Right’sbestmoveisto G(1, 0, 1, 0, 1) G(0, 1, 0, 0, 0).Leftmoves toG(1, 0, 0, 0, 0) G(0, 1, 0, 0, 0),whichispositivesinceG(1, 0, 0, 0, 0) isaRight optionof G(0, 1, 0, 0, 0).
If j = ℓ,then G(di, dj, ds, ds+1) = G(0, 1, 0, 1) and D = G(1, 0, 0, 1) G(0, 1, 1, 0). Right’sbestmoveisto G(1, 0, 0, 1) G(0, 0, 0, 0).Leftmovesto G(0, 0, 0, 0)
G(0, 0, 0, 0) = 0,andLeftwins.
ii.If j = s + 1,then G(di, ds, ds+1) = G(0,
).Thisis positivebyCorollary4.1.
Inallcases,Leftwins D movingsecond,provingtheresult.
SupposeinapositionthatthebitsofthebestRightmovearedifferentfromthose ofthebestLeftmove.Thenextlemmaessentiallysaysthatthepositionsbeforeandafteronemovebyeachplayerareequal.Itisphrasedinawaythatisusefulforreducing thelengthoftheposition.RecallthatanontrivialpositionlookslikeG = α
a0p10q1β, where a, p,and q arenonnegativeintegers,and α and β arearbitrarybinarystrings. Forthealgorithm,itsufficestoprovetheresultforβbeingempty.However,itisuseful, certainlyforahuman,toreducethelengthofthepositionasmuchaspossible.
Lemma4.3. Letαbeanarbitrarybinarystring.Ifa ⩾ 0,thenwehavethatα01111a = α101a .
Proof. Let H = α01111a α101a.Weneedtoshowthat H = 0.Tosimplifytheproof, insomecasesthesecondplayerwillplaysuboptimalmoves.Wehaveseveralcasesto consider.
1.If a ⩾ 2,thenplayingthesamemoveintheothersummandisagoodresponse. Aftertwosuchmoves,wehaveeither
or
2.If a = 1,then H = α01111 α101.Thecasesare:
i.Leftplaysinthefirstsummandto α011 α101;thenRightmovesto α101
α101 = 0.
ii.Rightplaysinthesecondsummandtoα01111 α;thenLeftmovestoα011 α. Since (α011)L = α,wehave α011 > α.
iii.Rightplaysinthefirstsummandtoα10111 α101;thenLeftrespondstoα101
α101 = 0.
iv.Leftplaysinthesecondsummandtoα01111 α11;thenRightmovestoα10111 α11 = α11 α11 = 0byinduction.
3.If a = 0,then H = α0111 α1.Thereareseveralcasestoconsider.
i.IfLeftorRightplaysinthefirstsummand,thentheresponseisinthefirst summandgiving α1 α1 = 0.
ii.IfLeftplaysinthesecondsummand,thensincethereisaLeftmove,wehave α = β01b , b ⩾ 0.If b > 0,thenwehavethat β01b0111 β01b1,andLeft movesto β01b013 β101b.HereRightrespondsto β101b 1013 β101b,which byinductionisequalto β101b 11 β101b = 0.If b = 0,thenwehavethat β01b0111 β01b1 = β00111 β01,andwewanttoshowthatRightcanwinmovingsecond.Leftplaysto β00111 β10,andRightcanrespondto β01110 β1, which,byinduction,isequalto β1 β1 = 0.
iii.Rightplaysinthesecondsummand.ThenforaRightmovetoexist,α = β10a , a ⩾ 0.Thus H = β10a0111 β10a1,andRightmovesto β10a0111 β.Left respondsbymovingto β00a011 β.Wethenhavethat (β00a011)L = β,and thus β00a011 > β.Hencewefindthat β00a011 β > 0.
Inallcasesthesecondplayerwins H therebyprovingtheresult.
Therearereductionsthatcanbeappliedtothemiddleoftheposition,butextra conditionsareneeded.
Lemma4.4. Letαandβbearbitrarybinarystringswhereeither (a) βstartswitha 1,or (b) βstartswith 0 andhasanevennumberof 1s.Wethenhavethat
0111β = α10β
Proof. Let H = α0111β α10β.Weneedtoshowthat H = 0.Wehaveseveralcasesto consider.
1.Ifβ isemptyorβ = 1a,thenH = 0byLemma4.3.Thereforewemayassumethatβ hasatleastone1andone0.
2.If β = 1γ1(β mustendina1),theninbothsummandsthebestmovesarepairsof bitsinβand β.Ifeachplayercopiestheopponent’smoveintheothersummand, thenthisleadsto
andthelatterexpressionisequalto0byinduction.
3.If β 1γ1,then β = 0γ1,and γ1hasatleasttwo1s.Thebestmovesarein β and β andarethebestresponsestoeachother.Wethenderivethat
Inallcases, H = 0,andthisconcludestheproof.
InLemma4.4theconditionsarenecessary.Anexampleis
3/8 = 011101 1001 = 1/4
Here β startswitha0andhasanoddnumberof1s.
Thesereductionlemmasareimportantinevaluatingaposition.Thereducedpositionswillendin011or01.Byconsideringtheexactendofthestring,specifically,if thereareatleasttwo0s(inonespecialcase,three0s),thenwecanfindanordinal sumdecomposition.Thedecompositionisdeterminedbywherethethirdrightmost1 issituated.
Thenextresultisthestartoftheordinalsumdecompositionofaposition.The exponentisthevalueoftheRightoptionofthesubstringbeingremoved.
Lemma4.5. Letαbeanarbitrarybinarystring.Ifa ⩾ 1 andpandqarenonnegative integerssuchthatp + q ⩾ 1,then
Proof. Weprovethat
NotethatinTheorem2.4wehavethatplayinginthebaseofα01a : 1 22p+q 1 isworsethan playingintheexponent.Wehavetwocasestoconsider.
1.Leftplaysfirstinthefirstsummand,andRightrespondsinthesecondsummand, orRightplaysfirstinthesecondsummand,andLeftrespondsinthefirstsummand.Ineithercase,Righthasamoveintheexponent(movesto0)since2p+q 1 ⩾ 0.Ineitherorderthefinalpositionisgivenby
2.Rightplaysfirstinthefirstsummand,andLeftrespondsinthesecondsummand, orLeftplaysfirstinthesecondsummand,andRightrespondsinthefirstsummand.Ineithercase,weconsider
Wehavetwosubcases.
i.Assumethat2p + q 1 0.Afterthetwomoves,wehavetheposition
where2r +
+ q 1.Byinductionwehavethat
Thus α01a0r10s1 (α01a : 1 22p+q 2 ) = 0.
ii.Assumethat2p + q 1 = 0,thatis, q = 1and p = 0.Theoriginalpositionis α01a101 (α01a : 1)
Afterthetwomoves,wehavethepositionα01a11 α101a 1 (notethatLefthas nomoveintheexponent).ByLemma4.3, α01a11 = α101a 1.Hencewehave that α01a11 α101a 1 = 0,andtheresultfollows.
ThevaluesofthepositionsnotcoveredbyLemma4.5aregivennext.
Lemma4.6. Leta,p,andqbenonnegativeintegers.Wethenhavethat
Proof. Let G = 0p1.Lefthasnomoves,andRighthas p.Notethatin1a,Lefthas ⌊a 2 ⌋ moves,andRighthasnone.
Nowlet G = 1a0p10q1.Weproceedbyinductionon p + q.Inallcases,Left’smove isto1a,thatis,to ⌊a 2 ⌋.Ifp = 0andq = 0,thenG = 1a11,whichhasthevalue ⌊a 2 ⌋+ 1 20 = ⌊a 2 ⌋+1.Assumethatp+q = k,k > 0.Ifq > 0,thenG = {⌊a 2 ⌋| 1a0p10q 11}.Byinduction wehavethat
If q = 0,then G = {⌊a 2 ⌋|
11011}.Byinductionwehavethat
4.1Proofofthevaluetheorem
WenowhavealltoolstoproveTheorem3.3.
ProofofTheorem 3.3. Let G beaflippingcoinsposition.Step2reducesthebinary string.ThereductionsinStep2(a)arethoseofLemma4.3andLemma4.4(a).ThereductionsinStep2(b)arethoseofLemma4.4(b).Inallcases,theselemmasshowthat eachnewreducedpositionisequalto G.
InStep3,weclaim Gi β13 forany β.Thisistruefor i = 0byLemma4.3.If i > 0, thenateachiterationofStep3,thelasttwo1sareremovedfromGi 1.Nowtheoriginal reducedpositionwouldbeG0 = β13γ,whereγhasanevennumberof1s.Lemma4.4(b) wouldapplyeliminatingthethreeconsecutive1s.NoweitherGi isoneof0r1,r ⩾ 0,or 1a0pi10qi1, a ⩾ 0, pi + qi ⩾ 0,or Gi = α01a0pi10qi1, pi + qi ⩾ 1, a > 0.Inthelattercase theindexisincremented,andthealgorithmgoesbacktoStep3.
Step5applieswhenStep3nolongerapplies,i.e., Gi isoneof0r1, r ⩾ 0,or 1a0pi10qi1, a ⩾ 0, pi + qi ⩾ 0.Now vi isthevalueof Gi asgiveninLemma4.6.
Lemma4.5showsthatforeach j < i, Gj = Gj+1 : 1 22pj+qj ,theevaluationinStep6. Thusthevalueof G is v0,andthetheoremfollows.
Thequestion“Whowins0101011111+1101100111+0110110110111andhow?”from Section1cannowbeanswered.
First,wehavethat 0101011111 = 01011011 = (01011 : 1 2) = ((01 : 1 2) : 1 2) = (( 1 : 1 2) : 1 2) =− 11 16,
Thuswehavethat
Left’sonlywinningmoveisto
Herbestmoveinthesecondsummandgivesasumof
,andinthe third,itgives 11 16 + 3 4 1 8 =− 1 16.Leftlosesbothtimes.
5Futuredirections
Naturalvariantsofflippingcoinsinvolveincreasingthenumberofcoinsthatcan beflippedfromtwotothreeormore.Abriefcomputersearchsuggeststhattheonly versionwherethevaluesarenumbersisthegameinwhichLeftflipsasubsequence ofall1sandRightflipsasubsequenceof0sendedbya1.Weconjecturethatasimilarordinalsumstructurewillariseinthesevariants.Othervariantshavevaluesthat include switches,tinies,minies, andotherthree-stopgames.However,somevariants, whenthereducedcanonicalvaluesareconsidered,onlyseemtoconsistofnumbers andswitches.Amorethoroughinvestigationshouldshedlightontheirstructures.
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