ALSOBYRAYMONDSMULLYAN

TheoryofFormalSystemsFirstOrderLogic

Gödel’sIncompletenessTheoremsGödelianPrinciplesinMetamathematics

TheTaoIsSilentWhatIstheNameofThisBook?ThisBookNeedsNoTitle

TheChessMysteriesofSherlockHolmesTheChessMysteriesoftheArabianKnights

TheLadyortheTiger?AliceinPuzzle-Land

5000B.C.ToMockaMockingbirdForeverUndecided

THISISABORZOIBOOKPUBLISHEDBYALFREDA.KNOPF,INC.

Copyright©1992byRaymondSmullyan

AllrightsreservedunderInternationalandPan-AmericanCopyrightConventions.PublishedintheUnitedStatesbyAlfredA.Knopf,Inc.,NewYork,andsimultaneouslyinCanadabyRandomHouseofCanadaLimited,Toronto.DistributedbyRandomHouse,Inc.,NewYork.

“TheLieDetective,”“WhenIWasaBoy,”“TheAbductionofAnnabelle,”“HowKazirWonHisWife,”“APlagueofLies,”and“OntheOtherHand”wereoriginallypublishedinThe

Sciences.

LibraryofCongressCataloging-in-PublicationData

Smullyan,RaymondM.Satan,Cantor,andinfinity,andothermind-bogglingpuzzles/

RaymondSmullyan.—1sted.p.cm.

eISBN:978-0-307-81982-61.Mathematicalrecreations.2.Gödel’stheorem.I.Title.

QA95.S5199292-52893793.7′4—dc20

v3.1

Onceagain,Iwouldliketothankmyeditor,AnnClose,andtheproductioneditor,MelvinRosenthal,foralltheirexpertassistance.MythanksgoalsotoMatinaBilliasforherexcellentsecretarialhelp.

Contents

CoverOtherBooksbyThisAuthorTitlePageCopyrightAcknowledgmentsPrefaceANotetotheReader

PARTI•LOGICALSORCERY

1TheLieDetective2WhenIWasaBoy3TheAbductionofAnnabelle4HowKazirWonHisWife5APlagueofLies6OntheOtherHand7TheIslandofPartialSilence

PARTII•PUZZLESANDMETAPUZZLES

8MemoriesoftheSorcerer’sUncle9ThePlanetOg10Metapuzzles

PARTIII•SELF-REPRODUCINGROBOTS

11TheIslandofRobots

12TheQuaintSystemofProfessorQuincy13FromtheRidiculoustotheSimple

PARTIV•GÖDELIANPUZZLES

14Self-ReferenceandCross-Reference15TheSorcerer’sMiniatureGödelianLanguage

PARTV•HOWCANTHESETHINGSBE?

16SomethingtoThinkAbout!17OfTimeandChange

PARTVI•AJOURNEYINTOINFINITY

18WhatIsInfinity?19Cantor’sFundamentalDiscovery20ButSomeParadoxesArise!21Resolutions22TheContinuumProblem

PARTVII•HYPERGAME,PARADOXES,ANDASTORY

23Hypergame24Paradoxical?25Satan,Cantor,andInfinity

Preface

Fewthingshavestirredthe imaginationasmuchas Infinity! Ithasall sorts of curious propertieswhich at first seemparadoxical, butthen turn out not to be. As such, it provides ideal material for apuzzlebook.As with my earlier puzzle books, this one starts out with newpuzzlesabout truth-tellersand liars (“knights”and“knaves”),but Ihave added a remarkable character known as the Sorcerer,who isconsideredamagicianbythosearoundhim,althoughheisreallyalogicianwhouseslogicsocleverlythatitseemslikemagictothosenotintheknow.Aftermanyexhibitionsofhis“logicalsorcery,”heescortsusthroughahostofunusualadventures,includingavisittoan island where intelligent robots create other robots and endowthemwithenoughintelligencetocreateotherintelligentrobots,whoin turn create other intelligent robots, and so forth, ad infinitum.Then,aftersomespecialpuzzlesrelatedtoGödel’sfamoustheoremandsomecuriousparadoxesaboutprobability,time,andchange,theSorcerergivesusaguidedtourofInfinity,explainingthepioneeringdiscoveriesof thegreatmathematicianGeorgCantor,whowas thefirsttoputthesubjectonalogicallysoundbasis.TheSorcerer,inhistypically humorous fashion, ends upwith a delightful tale of howSatanhimselfisoutwittedbyacleverstudentofCantor’s.On the serious side, it must be wondered at that the wholefascinatingsubjectofInfinityissolittleknowntothegeneralpublic!Why isn’t it taught in high schools? It is no harder to understandthan algebra or geometry, and it is so rewarding! The last fewchaptersofthisbookprovideaninviting(andeasy)introductiontothesubject.EvenaneophytecanunderstandthenatureofInfinity,Cantor’samazingcontribution,andanaccountofwhatmaybethegreatestmathematicalproblemofalltime—whichremainsunsolvedtothisveryday!

ANOTETOTHEREADER

Theseveralpartsofthebookdonothavetobereadintheorderinwhichtheyappear.Thus,thereaderprimarilyinterestedinInfinitycanreadPartsVIandVIIquiteindependentlyoftherestofthebook.PartsIIIandIVlikewiseformaseparateunit,andPartsI,II,andVcaneachbereadindependently.Allthatareaderstartingwithlaterchaptersneedstoknowisthatthe principal characters are the Sorcerer and his two students,AnnabelleandAlexander.

PARTI

LOGICALSORCERY

1THELIEDETECTIVE

WITH A TWINGE of apprehension such as he had never felt before, ananthropologist named Abercrombie stepped onto the Island ofKnightsandKnaves.Heknewthatthisislandwaspopulatedbymostperplexing people: knights, who make only true statements, andknaves,whomakeonly falseones.“How,”Abercrombiewondered,“amIever to learnanythingabout this island if Ican’t tellwho islyingandwhoistellingthetruth?”Abercrombie knew that before he could find out anything hewould have to make one friend, someone whom he could alwaystrusttotellhimthetruth.Sowhenhecameuponthefirstgroupofnatives, three people, presumably named Arthur, Bernard, andCharles,Abercrombiethoughttohimself,“Thisismychancetofindaknight formyself.”Abercrombie firstaskedArthur, “AreBernardandCharlesbothknights?”Arthurreplied,“Yes.”Abercrombiethenasked:“IsBernardaknight?”Tohisgreatsurprise,Arthuranswered:“No.”IsCharlesaknightoraknave?

Abercrombie knew that he must first determine what type (knight orknave)Arthur andBernard are. Arthur is obviously a knave, since noknightwouldclaimthatBernardandCharlesarebothknightsyetdenythatBernard is a knight. Therefore both ofArthur’s answerswere lies.SincehedeniedthatBernardisaknight,Bernardreallyisaknight.Sincehe affirmed that both Bernard and Charles are knights, it is false thattheyarebothknights;atleastoneofthemmustbeaknave.ButBernardisnotaknave(aswehaveproved),therefore,Charlesmustbeaknave.

Abercrombiewastheninformedbytheoneofthethreeheknewtobeaknightthattheislandhadasorcerer.“Oh, good!” Abercrombie exclaimed. “We anthropologists are

particularly interested in sorcerers, witch doctors, medicine men,shamans,andthelike.WheredoIfindhim?”“YoumustasktheKing,”camethereply.Well,theanthropologistwasabletoobtainanaudiencewiththeKingandtoldhimthathewishedtomeettheSorcerer.“Oh,youcan’tdothat,”saidtheKing,“unlessyoufirstmeethisapprentice.IftheSorcerer’sApprenticeapprovesofyou,thenhewillallowyoutomeethismaster;ifhedoesn’t,thenhewon’t.”“TheSorcererhasanapprentice?”askedtheanthropologist.“Hecertainlydoes!”repliedtheKing.“Thereisafamousmusicalcompositionabouthim—IbelievethecomposerwasDukas.Anyway,ifyouwishtomeettheSorcerer’sApprentice,heisnowathishome,which is the third house on Palm Grove. At the moment he isentertainingtwoguests.If,whenyouarrive,youcandeducewhichof the threepresent is theSorcerer’sApprentice, Ibelieve thatwillimpresshimsufficientlythathewillallowyoutomeettheSorcerer.Goodluck!”A shortwalk brought the anthropologist to the house.When heentered,therewereindeedthreepeoplepresent.“WhichofyouistheSorcerer’sApprentice?”askedAbercrombie.“Iam,”repliedone.“IamtheSorcerer’sApprentice!”criedasecond.Butthethirdremainedsilent.“Canyoutellmeanything?”Abercrombieasked.“It’s funny,” answered the third onewith a sly smile. “Atmost,onlyoneofthethreeofusevertellsthetruth!”CanitbededucedwhichofthethreeistheSorcerer’sApprentice?

Hereishowtheanthropologistreasoned:Ifthethirdoneisaknave,thenhisstatementisfalse,whichmeansthatatleasttwoofthemareknights.But the first and second guests cannot both be knights, since theirstatementsconflict.Thereforethethirdguestcannotbeaknave;hemustbe a knight. This means his statement is true: he is the only knightpresent.Sincetheothertwoareknaves,theirclaimsarebothfalse;henceneither of them is really theSorcerer’sApprentice.Ergo, itmust be the

thirdone.

TheApprenticewas delightedwithAbercrombie’s reasoning andinformedhimthathecouldmeettheSorcerer.“He is now upstairs in the tower conferring with the islandAstrologer,” said the Apprentice. “You may go up and interviewthemifyoulike,butpleaseknockbeforeentering.”Theanthropologistwentupstairs, knockedon thedoor, andwasbiddentoenter.Whenhedid,hesawtwoverycuriousindividuals,onewearingagreenconicalhatandtheotherablueone.HecouldnottellfromtheirappearancewhichwastheAstrologerandwhichwastheSorcerer.Afterintroducinghimselfheasked,“IstheSorcerera knight?” The one in the blue hat answered the question (heanswered either yes or no), and the anthropologist was able todeducewhichwastheSorcerer.WhichonewastheSorcerer?

This type of puzzle is very different from the preceding two; it is ametapuzzle, for the reader isnot givenall thepiecesof thepuzzlebutinformationabouttheprocessofsolvingthepuzzle.Thereader,inotherwords, is not toldwhat answerwas given by theman in the blue hat;neverthelessheistoldthattheanthropologistcouldsolvethepuzzleaftergettingananswer;itisthisinformationthatisvital.Letusseehowthissortofpuzzleworks:Supposethemaninthebluehathadansweredyes;couldtheanthropologist thenhaveknownwhichonewastheSorcerer?Certainlynot; themanwhoansweredcouldbeaknight, in which case all that would follow is that the Sorcerer is aknight; but the twomight both be knights and either one could be theSorcerer.Oragain, themanwhoansweredcouldbeaknave, inwhichcasetheSorcererisaknaveandcouldbeeitherofthetwo(asfarastheanthropologistcouldknow).Soifyeswastheanswer,theanthropologistcouldnothavededucedwhichmanwas theSorcerer.ButwearegiventhattheanthropologistdiddeducewhichwastheSorcerer;therefore,hemusthavegottentheanswerno.Nowweknowthat thespeaker(theonewith thebluehat)answeredno.Ifthespeakerisaknight,hisanswerwastruthful;hencetheSorcerer

isreallynotaknight.Andsincethespeakerisaknight,thenheisnottheSorcerer.Ontheotherhand,ifthespeakerisaknave,hisanswerwasalie,whichmeans that the Sorcerermust be a knight; hence, again, thespeaker cannot be theSorcerer.This proves that anoanswer indicatesthatthespeakerisnottheSorcerer,regardlessofwhethertheanswerwasthetruthoralie.AndsothemaninthebluehatmustbetheAstrologerandtheoneinthegreenhatmustbetheSorcerer.Insummary,anoanswerprovesthatthemaninthegreenhatistheSorcerer, whereas a yes answer proves nothing at all. Since theanthropologistwasable todeduce theSorcerer’s identity,hemusthavegottenanoansweranddeduced that theman in thegreenhatwas theSorcerer.

While the anthropologist had deduced which one was theSorcerer,hedidnotyetknowwhethertheSorcererwasaknightora knave.With one more question he discovered that the SorcererwasaknightandtheAstrologeraknave.AndtheAstrologer,abitembarrassed, arose and left, saying: “According to the planetaryconfigurations,Ishouldbehomenow.”“Those astrologers,” said the Sorcerer with a laugh, “humbugseveryoneofthem.Now,withmeitisdifferent;mysorceryisreal.”“To tellyou the truth,” saidAbercrombie, “Iamrather skepticalabouttheexistenceofmagic.”“Oh,youdon’tunderstand,”saidtheSorcerer.“Mysorcerydoesn’tusemagic—though it seems to those aroundhere that it does.Mysorcery involves the clever use of logic. With my logic I amconstantlyfoolingthesefellows.”“Canyougivemeanexample?”askedAbercrombie.“Whycertainly.Areyouabettingman?”“Occasionally,”repliedAbercrombiewithsomecaution.“Oh,itdoesn’thavetobealargebet;we’llwagerjustonecoppercoin.Iwillaskyouaquestiontowhichyoumustansweryesorno.Even though the question has a definite correct answer, I’llwagerthatyouwillbepowerlesstogiveit.Anybodybutyoumightbeabletogivethecorrectanswer,butyoucannot.Infactitwillbelogically

impossible for you to give the correct answer, even though thequestionhasone.Doesn’tthissoundlikesorcery?”“It certainly does,” replied Abercrombie, who was enormously

intrigued. “I’ll take the bet—mainly because I’m so curious.Whatquestiondoyouhaveinmind?”The Sorcerer then asked Abercrombie a yes or no question that

definitely had one and only one correct answer. AndAbercrombiesoonrealized,tohissurpriseandamusement,thattheSorcererwasright.Itwaslogicallyimpossibleforhimtogivethecorrectanswer,eventhoughheknewwhatitwas.CanyouguesswhatquestiontheSorcererasked?

TheSorcereraskedAbercrombie,“Willyouanswernotothisquestion?”IfAbercrombiewere to answer no, hewould be denying that no is hisanswer; hence he would be wrong. If he were to answer yes, then hewouldbeaffirmingthatnoishisanswer;hencehewouldagainbewrong.ThusitislogicallyimpossibleforAbercrombietogivethecorrectanswer.

Clever,thesesorcerers!

INSPECTORCRAIGPAYSAVISIT

A fewweeks after the anthropologist set sail for home,my friendInspectorCraig,ofScotlandYard,paidavisit totheisland.Onthefirsteveningof thisparticularvisit,Craigwas invited todinewiththeChiefJustice,aknight.“Ah,yes,”saidthejudgeproudly,“yesterdayIcaughtaknaveata

trialandsentencedhimto threemonths forperjury.Oneshouldn’tlieunderoath.”“Areyouimplyingthatitisallrighttoliewhennotunderoath?”

askedCraig.“No, no,” cried the judge. “One should never lie at all—but

especiallyunderoath.”“Tellmewhathappened,”saidCraig,whotookagreatinterestin

suchmatters.

“Why,thereweretwodefendants,namedBarabandZork.IknewthatBarabpersonallydislikedZork,butthatdoesn’texcusehislyingabouthim.”“Whatliedidhetell?”askedCraig.“He claimed that a fewminutes before the trial, he heard Zork

confidetoafriend:‘Iliedyesterday.’”“Sowhat?”askedCraig.“SoIobviouslyconvictedBarabforlying.”“Howdoyouknowhewaslying?”“Oh, comenow,” said the judgewith some irritation. “I thought

you were a good logician. Obviously, Zork could never have saidthat he lied yesterday, because a knight would never falsely statethatheliedyesterdayandaknavewouldnevertruthfullyadmitthathe lied yesterday. Therefore Barab clearly lied when he said thatZorkmadethatstatement.”“Notnecessarilyso,”repliedCraig.“Youshouldbrushuponyour

ownlogicand,ofmoreimmediateconcern,youshouldhaveBarabreleased right away, since you have sentenced him without justcause.”SubsequentinvestigationrevealedthatCraigwasright.Thejudge

made a rather natural error in his reasoning, but an errornevertheless.Whatistheerror?

Subsequent investigation revealed that Barab was in fact a knight andwastellingthetruth:Zorkhadinfactmadethatstrangestatement.HowcouldZorkhaveclaimedthatheliedyesterday?Well,itturnedoutthatZorkhadlaryngitisthedaybeforeand,hence,madenostatementsatallonthatday.SoZorkwasaknavewholiedwhenheclaimedtohaveliedthedaybefore;hehadreallybeensilent.

The next day Craig was asked to preside as judge at a trialconcerning a stolen watch. The defendant was named Gary.InspectorCraigwasnotinterestedinfindingoutwhetherGarywasaknightoraknave;allhewantedtoknowwaswhetherGarydidordidnotstealthewatch.Hereisatranscriptofthetrial:

Craig:Isittruethatsometimeaftertherobbery,youclaimedthatyouwerenottheonewhostolethewatch?Gary:Yes.Craig: Did you ever claim that youwere the one who stole the

watch?Gary thenanswered (yesorno),andCraigknewwhetherornot

hewasinnocentofthetheft.DidGarystealthewatch?

This puzzle is another example of a metapuzzle. Suppose Gary hadansweredyes toCraig’s secondquestion.Then itwouldbeobvious thatGary isaknave,becauseaknightcouldneverclaimtohavemade twocontradictory claims.SinceGary isaknave (still under the suppositionthatheansweredyes),thenbothhisanswerswerelies,whichmeansthatheneverclaimedthathewasnotthethief,norhasheeverclaimedthathe was the thief, and so Craig could have no grounds for knowingwhetherGarywasinnocentorguilty.ButCraigdidknow.HenceGary’ssecondanswercouldnothavebeenyes;itmusthavebeenno.Now that we know that Gary’s second answer was no, we can

determine his innocence or guilt. Gary is either a knight or a knave.Supposeheisaknight.Thenbothhisanswersweretruthful,whichmeanshedidclaimoncethathewasnotthethief,butheneverclaimedthathewasthethief.Sinceheonceclaimedthathewasnotthethiefandheisaknight, then he is innocent.On the other hand, suppose he is a knave.Thenbothhisanswerswerelies,whichmeansthatheneverdidclaimhewasnotthethief,buthedidclaimhewasthethief.Then,beingaknaveandclaimingthathewasthethiefmeansthat inrealityhewasnot thethief,andsoagainheisinnocent.ThisprovesthatregardlessofwhetherGaryisaknightoraknave,heisinnocentofthecrime.

2WHENIWASABOY

WHEN THE ANTHROPOLOGIST Abercrombie returned from the Island ofKnightsandKnaves,hecalled in thepress to relatehisadventureswith the Sorcerer’s Apprentice and his audiencewith the Sorcererhimself.AreporternamedBillRyanwassointriguedthathedecidedtovisit the islandandinterviewtheSorcerer.Aftersettingsailonewintry day from Baltimore, he arrived on the island and trackeddownthemysteriousSorcererinhismountaincastle.“Tell me,” said Ryan, to the Sorcerer, pencil and notebook inhand, “when did you first get interested in logic?”—for theSorcerer’s “magic,” as Abercrombie had discovered, consisted ofnothingmorethanthecleveruseoflogic.“ItstartedwhenIwasaboy,”camethereply.“MyuncletoldmeaboutamythicalIslandofKnightsandKnaves.(Inowhavereasontobelievethatheknewsuchanislandactuallyexisted,butwantedtotestmebeforeraisingmyhopesabouttravelingthere.)Anyway,hefirsttoldmeaboutashipwreckedtravelerwhocameacrossthreeislandnatives namedAnthony, Bertrand, andClive.He asked one:‘Areyouaknightoraknave?’Anthonyanswered,but ina foreigntongue. The traveler then asked Bertrandwhat Anthony had said.Bertrandreplied:‘Anthonysaidthatheisaknave.’ButClivechimedin:‘Don’tbelieveBertrand;heislying.’“The traveler (who I now believe was really my uncle) wasperplexed.Andthenitsuddenlydawnedonhimwhichtype(knightorknave)Clivewas.Butmyunclewouldnottellmetheanswer. Ihad to figure it out myself. Can you guess whether Clive was aknightoraknave?”Ryanwasstumped,sotheSorcererexplained.

“I looked up at my uncle,” recalled the Sorcerer, “and said that no

inhabitant of a knight-knave island could possibly claim that hewas aknave,becauseaknightwouldneverlieandclaimtobeaknave,andaknavewouldnevertruthfullyadmittobeingaknave.IreasonedthenthatBertrand clearly lied when he said that Anthony had claimed to be aknave, and hence Clive had told the truthwhen he said that Bertrandlied.Therefore,BertrandwasaknaveandClivewasaknight.”

“NowthatIliveonthisisland,Ioftenremembermyuncle’stalesasItrytounmaskknightsandknaves.”“After all these years on the island, don’t you know who is aknightandwhoisaknave?”askedRyan.“Why,not twoweeksago,” said theSorcerer, “Iwaswalkingonthebeachand cameacross someone I didn’t know.Therewerenovisitorsonthe islandat thetime,so Iknewhehadtobeanative.ButIhadnoideawhetherhewasaknightoraknave.Hemumbledafewwordsaswepassed.Ithoughtforamoment,andthenshoutedafterhim:‘Ifyouhadn’tmadethatstatement,Icouldhavebelievedit!Beforeyouspoke,Ihadnoideawhetheryouwereaknightoraknave, nor did I have the slightest idea ofwhetherwhat you justsaidwastrueorfalse.NowIknowthatyourstatementisfalseandthatyoumustbeaknave.’”“Whatcouldhepossiblyhavesaidthatmadeyoureactlikethat?”askedRyan.“Whatdoyouthinkhesaid?”challengedtheSorcerer.“Well, hemight have said ‘Two plus two equals five,’ ” repliedRyan. “Wouldn’t that be enough to convince you that he was aknave?”“Ofcourseitwould!”exclaimedtheSorcerer.“ButIseeyoudon’tunderstandtheproblem!Hadhesaidthat,IwouldhaveknownthathewasaknaveonlybecauseIknewbeforehespokethat‘Twoplustwoequalsfive’isafalsestatement.ButItoldyouthatitwasonlyafterhemade thestatement that Icoulddeduce itwas false—falsebyvirtueoftheveryfactthathehadmadeit.Nowcanyousupplysuchastatement?”Again,thereporterwasbaffled.

“WhenI firstmet thenative,”explained theSorcerer,“Ihadno ideawhetherhewasaknightoraknave,norwhetherhewasmarried.Butthen he said, ‘I am a married knave.’ Obviously a knight could neverclaim to be a married knave (or any other kind of knave, for thatmatter),andsothenativewascertainlyaknave.Therefore,hisstatementwas false.Hewasnot reallyamarriedknave, sohemusthavebeenasingleknave.Afterhespoke,IknewtwothingsabouthimthatIhadn’tknownbefore—thathewasaknave,andthathewasnotmarried.”

“Now, just a minute!” exclaimed Ryan. “I can’t accept yoursolution as valid. Indeed, I can’t see how that incident could everhave taken place. And I am deeply disturbed that you, a knight,couldhaveevertoldmethisfalsestory.Orperhapsyouareaknaveandeverythingyouhavetoldmetodayisuntrue.”“Why do you say that the story is false?” asked the Sorcerer insurprise.“Becausenoinhabitantofthisislandcouldpossiblyclaimtobeaknave. If he claims to be a married knave, then he is certainlyclaimingtobeaknave,whichcannotbedone,asyouruncle(ifhereallyexisted)pointedout.Yourstorydoesn’tholdwater.”“Not so fast, young man,” said the Sorcerer. “You have justcommitted a rather common fallacy. Let me ask you a question:Suppose a person claims to know both French and German. Is henecessarilyclaimingthatheknowsFrench?”“Well,ofcourse,”repliedRyan.“Whatasillyquestion.”“Not so silly, if you stop and think about it. Let me put thequestion thisway: A person has just told you that he knows bothFrench and German. Then you ask him, ‘Do you know French?’Wouldhenecessarilyclaimthathedoes?”“Ofcourse,”repliedRyan.“Whywouldn’the?”“Ah,there’swhereyou’rewrong.Hemightandhemightnot—hemightevendenythatheknowsFrench.”RyanscratchedhisheadastheSorcererexplained.

“IfatruthfulpersonclaimedtoknowbothFrenchandGerman,thenhe

wouldofcoursealsoclaimtoknowFrench.Butwithaliar,it’sdifferent;ifhehappenedtoknowFrenchandnotGerman,thenhewouldmakethefalseclaimthatheknowsbothFrenchandGerman.Then if youaskedwhetherheknowsFrench,hewould lieand sayno.Similarly,anativeislandercouldclaimtobeamarriedknave,yetdenythatheisaknave.Thisisacuriousfactaboutthelogicoflyingandtruth-tellingthattakessomegettingusedto.”

Itdidn’ttakeRyanverylongtorealizethattheSorcererwasright.“Onanotheroccasion,”saidtheSorcerer,“IcameacrossanativewhomadeastatementfromwhichIcoulddeducethatthestatementmustbetrue,butIdidnotknowthetruthofthestatementbeforeitwasmade,nordidIhaveanypriorknowledgethathewasaknight.Now,ifyoureallyunderstandwhatIjusttaughtyou,Ryan,youwillknowwhatthenativecouldhavesaid.”

Ryan thought a bit and then answered boldly: “The native could havesaid, ‘I am not a married knight.’ Clearly a knave couldn’t say that(becauseaknaveisindeednotamarriedknight).Sinceheisaknighthisstatementistrueandsohemustbeanunmarriedknight.”

“Suppose,”saidtheSorcerer,“thatthenativehadinsteadsaid, ‘Iamanunmarriedknight.’Couldyoudeducewhetherheisaknightoraknaveorwhetherheismarried?”

Ryanwasreadytoansweryes,whenhecaughthimself.“No,Icouldnot.Any knave could claim to be an unmarried knight (since he is not anunmarriedknight)andanyunmarriedknightcouldalsomakethatclaim.All we can deduce from his claim is that if he is a knight then he isunmarried,whichdoesn’ttellusmuch.”

“Whew,”saidRyan.“Ialmostconfusedthetwostatements:‘Iamanunmarried knight’ and ‘I amnot amarried knight.’ They reallysayquitedifferentthings.”“Youareprogressingwell,muchbetterthanIwouldhavethoughtwhenyou firstwalked in.But I’mafraid,” said the Sorcerer, ashe

lookedupathisgreatantiqueclock,“thatImustgoofftoatrial.Itpromises to be a very interesting one. The presiding judge is aformer student of the remarkable logician Inspector Craig. Wouldyouliketocomewithme?”“Withpleasure,”repliedRyan.The two descended from the Sorcerer’s tower andwended their

waytothecourthouse.Alongtheway,theSorcerertoldRyanwhatwasalreadyknownaboutthecase.“The case concerns a stolen horse. There are four suspects—

Andrew,Bruce,Clayton,andEdward.Theauthoritiesknowforsurethatoneandonlyoneofthesefouristhethief.Thefirstthreehavealreadybeenfoundandputincustody,butEdwardcannotbefoundanywhere.Thetrialwillhavetoproceedwithouthim.”Ryan and the Sorcerer arrived at the courthouse none too soon;

the trial commenced just as they were seated. And, since almosteveryone on the island was interested in the case, the house wasfilledtocapacity.First the judge pounded his gavel and asked a highly relevant

question:“Whostolethehorse?”Hegotthefollowingreplies:Andrew:Brucestolethehorse.Bruce:Claytonstolethehorse.Clayton:ItwasEdwardwhostolethehorse.Then,quiteunexpectedly,oneof the threedefendantssaid,“The

othertwoarelying.”Thejudgethoughtforabit,thenpointedtooneofthethreeand

said, “Obviously, you didn’t steal the horse, so youmay leave thecourt.”The acquitted man was happy to comply, and so only two

defendantswereleftontrial.Thejudgethenaskedoneoftheremainingtwowhethertheother

wasaknight,and,afterreceivingananswer(yesorno),knewwhostolethehorse.Whatdidthejudgedecide?

Firstwemustdeterminewhomthejudgeimmediatelyacquitted.Supposeit was Andrew. If Andrew is a knight, then Bruce must be guilty and

Andrew innocent. IfAndrew isaknave, then it is false thatBruceandClayton both lied; at least one of them told the truth. Thismeans thateitherClayton isguilty(ifBruce told the truth),orEdward isguilty(ifClaytontoldthetruth);ineithercase,Andrewwouldbeinnocent.Andsoif it was Andrew who made the second statement, he is innocent,regardless of whether he is a knight or a knave. The judge, of course,wouldhaverealizedthisandacquittedhim.However, if either Bruce or Claytonmade the second statement, the

judgecouldnothavefoundgroundstoacquitanyone.IfBrucespoke,thejudgecouldonlytellthateitherBruceisaknightandClaytonisguilty,orBruceisaknaveandeitherBruceorEdwardisguilty.IfitwasClaytonwhospoke,thenthejudgecouldonlytellthateitherClaytonisaknightandEdwardisguilty,orClaytonisaknaveandeitherBruceorClaytonis guilty. Since the judge did make an acquittal, it must have beenAndrewwhospokeandwasacquitted.Thus the remaining defendants were Bruce and Clayton. One of the

two,byansweringthejudge’slastquestion,eitherclaimedthattheotherdefendantwasaknight,orthattheotherwasaknave.Iftheformer,thenthe twodefendantsare the same type (bothknightsorbothknaves); ifthe latter then the two are different types. Suppose the latter. ThenClaytonmaybe a knight andBruce a knave, inwhich caseEdward isguilty (because Clayton said he was), or Bruce may be a knight andClayton a knave, in which case Clayton is guilty. However, the judgecouldn’t have known which, and hence couldn’t make a conviction.Therefore,oneofthetwomusthaveclaimedthattheotherwasaknight(byansweringyestothejudge’squestion).Thejudgethenknewthattheyareofthesametype.Theycan’tbothbeknights(sincetheiraccusationsconflict), so they are both knaves and their accusations are both false.NeitherClaytonnorEdward stole thehorse.Andrew,asweknow,hasalreadybeenacquitted.SoitwasBrucewhostolethehorse.

“Speaking of horses,” said the Sorcerer to Ryan as they wereleaving the courthouse after the trial, “I must tell you of anextremelyfunnyincidentthathappenedmanyyearsagowhenIwasliving in a small town in another land. An individual named

Archibald sold a horse to an individual named Benjamin. I knewbothmen,andwasquitecurioustoknowwhichofthemhadmadetheshrewderbargain.FirstIaskedBenjaminhowmuchhehadpaidforthehorse.Henamedasurprisinglylowfigure.ThenIturnedtoArchibald and asked, ‘Whydid you sell such amagnificent animalforsolittle?’Archiereplied, ‘Hegotnobargain;thehorseislame.’Next I asked Benjamin, ‘How come you paid so much for a lameanimal?’Benjaminreplied, ‘He isn’t really lame.Yousee there isanail sticking in his foot, whichmakes him limp and appear lame.Archibald, nodoubt, thoughthewas lame, and that’swhyhe soldhimforsolittle.ButwhenItakethehorseaway,I’llpullthenailoutandhe’llbeasgoodasnew.’“Turning toArchibald, I said, ‘Aha! I seehegot thebestofyou.

Thehorse isn’t really lame.’ ButArchie replied, ‘No, no; the horsereally is lame. I simply stuck a nail in his foot tomake Benjaminthinkthat itwascausing thehorse to limp.Butwhenhe takes thenailout,he’llseethatthehorselimpsasmuchasever.’“Atthis,IsaidtoBenjamin,‘So,hereallycheatedyou.Thehorse

islame;Archibalddeliberatelyputthenailinthehorse’sfootjusttomislead you.’ Upon which Benjamin replied, ‘I considered thatpossibility,andthat’swhyIpaidhimincounterfeitmoney.’”

3THEABDUCTIONOFANNABELLE

ONE NOVEMBER NIGHT, on an uncharted atoll a few hundred leaguesdistant from the IslandofKnights andKnaves, PrincessAnnabelle,theKing’syoungerdaughter,waskidnapped.RumorhaditthatshehadbeentakenbyboattotheIslandofKnightsandKnaves,butnoonereallyknewwhetheritwastrue.Annabelle’ssuitor,astrappingyouthnamedAlexander,immediatelysetsailfortheisland,hopingtofindoutifshewasimprisonedthere.Heassumed(andrightlyso)thatifshehadbeentakentotheisland,shewouldstillbeonit.Healso rightly reasoned that the island’sWitchDoctorwouldknow ifAnnabelle was still being held captive. The only trouble was thatAlexanderdidnotknowwhetherthewitchdoctorwasaknightoraknave!Alexander arrived safely on the island, sought out the WitchDoctor,andasked,“IsPrincessAnnabelleonthisisland?”TheWitchDoctoransweredeitheryesorno.Thenthesuitorasked,“HaveyouseenPrincessAnnabelleonthisisland?”TheWitchDoctoransweredeitheryesorno,andthesuitorthenknewwhetherAnnabellewasontheisland.Wasshe?

SupposethatboththeWitchDoctor’sanswerswereyes.ThenitcouldbethattheWitchDoctorisaknightandthatAnnabelleisontheisland,orit could be that he is a knave and Annabelle is not there—Alexandercouldhavenowayofknowing.IfboththeWitchDoctor’sanswerswereno, then it couldbe that theWitchDoctor isaknaveandAnnabelle isthere,oritcouldbethatheisaknightandAnnabelleisnotthere—again,Alexanderwould beunable todecide.But sincehedoes knowwhetherAnnabelleisontheisland,hemusthavegottenoneyesanswerandonenoanswer.Let us see what would happen if the first answer was yes and the

secondanswerno.IftheWitchDoctorwereaknave,thenbothanswerswouldbelies,meaningthatAnnabellewasneverthere,butthattheWitchDoctor had seen her there, which is not possible. Therefore, theWitchDoctormustbeaknightandAnnabellemustbeontheisland(thoughtheWitchDoctorneversawherthere).On the other hand, suppose the first answerwas no and the secondansweryes.IftheWitchDoctorwereaknight,thenwewouldagainhavethe impossible situation thatAnnabellewas never on the island, yet hehad seen her there. Therefore theWitch Doctormust be a knave, andAnnabellemustbeontheisland.Ofcourse,wehavenowayofknowingwhatanswerstheWitchDoctoractuallygave(otherthanthatonewasyesandonewasno),norcanwetellwhethertheWitchDoctorisaknightoraknave(thoughAlexanderknew).Butwehaveseenthatonlytwosetsofanswersarepossible(giventhat Alexander did know whether she was there), and in either case,Annabellemustbeontheisland.

Needless to say, Alexander was overjoyed to discover that hisbelovedAnnabellewasonthe island.Thenextstepwastobargainforherrelease.Withthisinmind,heobtainedanaudiencewiththeKing,whowasknowntobeaknightandwhosenamewasZorn.“What ransom do you demand for the release of PrincessAnnabelle?”Alexanderboldlyasked.“Oh,heavens,” laughed theKing, “I neverhadher broughthereforransom!”“You mean you had worse motives?” asked Alexander, in somealarm.“Ohno,dearboy,”repliedtheKing,inareassuringtone.“PrincessAnnabelle is indeedas lovelya ladyascanbe,andshewillbetheidealbrideforyou,ifyouarecleverenoughtowinherback.”“Thenwhydidyouhaveherkidnapped?”Alexanderasked.“The reasonwill surprise you,” replied Zorn. “You have quite areputation for solving puzzles. I deliberately had the princessbroughthere to testyour skill.Youhavedonewell indeterminingthat your princess is on the island, but the difficult part is yet to

come.”“Andwhatisthat?”askedAlexander.“Ah!” said the King, “your next task is to find out whethermyGrand Vizier is a knight or a knave. If you succeed, I will haveAnnabelle released. Youmay ask the Vizier as many questions asyoulike,buttheymustallbeanswerablebyyesorno.”“Butthat’sridiculouslyeasy!”criedAlexander.“Ihavemerelytoaskonequestion—onequestionwhoseanswerIalreadyknow,suchas whether two plus two equals four. From his answer, I will ofcourseknowwhetherheisaknightoraknave.”“You shouldn’t have interrupted!” said theKing. “Of course youcanfindoutbyaskingjustonequestionwhoseansweryoualreadyknow.But Iwasabout to say that youarenot allowed toaskanyquestionwhoseansweryoualreadyknow.”Thesuitorstoodlostinthought.“Letmebemoreexplicit,”saidtheKing.“Youdon’thavetoplanthesequenceofquestionsinadvance;atanystage,thequestionyoudecide toaskmaydependon theanswersalreadygiven,butatnostageareyouallowedtoaskaquestionwhosetruthfulanswercouldbeknowntoyou.”Thesuitorthoughtaboutthissomemore.“Areyoucertainthatthispuzzlecanbesolved?”hefinallyasked.“Ineversaiditwaspossible,”repliedtheKing.“Oh,comenow!”saidAlexanderingreatagitation.“Isn’titunfairofyoutogivemeanimpossibletask?”“Ineversaiditwasimpossibleeither,”repliedtheKing.“It’suptoyoutofindoutwhetherit’spossible.Ifitispossible,thenyoumustsolve thepuzzle towinbackyourprincess. If it is impossible, andyou can prove tome that it is impossible, then again I promise toreleasePrincessAnnabelle.Eitherwayyouwinherback.Thosearemyterms.”Alexander thought over the problem for many days, thenrequestedanaudiencewithKingZorn.“I have a strategy,” said Alexander. “I need at most twoquestions!”

“Whatarethequestions?”askedtheKing.“Well,” repliedAlexander, “first Iwouldask theVizier ifhe isamarriedknight.Asofnow,Ihavenomeansofknowingwhetherheismarried,orwhetherheisaknight.Ifheshouldanswerno,thenno furtherquestionsarenecessary;hemustbeaknight,becauseaknave is certainly not a married knight, hence could not give thetruthfulanswernotothequestion.”“Butsupposeheanswersyes?”askedtheKing.“Inthatcase,Iwouldknowthatheiseitheramarriedknightoraknave—possibly married, possibly unmarried—because if he is aknight,his answerwouldbe correct,whichmeanshe is amarriedknight; if he is not a knight, then he is a knave.And so a secondquestion would be necessary. I would then ask him, ‘Are you anunmarried knight?’ If he also answers yes to that question, then ofcourse he is a knave (since a knight would never make twoincompatibleclaims). Ifheanswersno, thenhemustbeamarriedknight(sinceaknaveisnotanunmarriedknightandhencecouldn’tgivethetruthfulanswerno).Andsoanoanswerwouldindicatethatheisaknight.Therefore,Iwouldknowwhetherheisaknightoraknave.”“Haveyouactually tried this strategyon theVizier?”askedKingZorn.“Notyet,”repliedthesuitor.“Butit’swhatIplantodo.”“It’sagoodthingyoudidn’t,”saidtheKing.“Yourquestionsdon’tfulfilltheconditionsIhavestipulated!”The King was right. Why does this strategy fail to meet hisrequirements?

Thesuitor’sstrategymighthappentoworkifthesuitorwerelucky,butitisnotboundtowork—forthisreason.SupposeAlexanderhadtriedthisstrategy on the Vizier and had gotten no for the answer to his firstquestion. Then (asAlexander correctly explained) hewould know thattheVizierwasaknight.However, if theVizierhadansweredyes, thenalthough Alexander would certainly know after the second questionwhether theVizierwasaknightoraknave, the secondquestionwould

not satisfy the King’s requirement that the suitor not know the correctanswer before asking the question. The Vizier could answer yes to thefirstquestiononlyifhewereeitheramarriedknightoraknave(marriedornot)—inotherwordsonlyifhewerenotanunmarriedknight!Andsothe truthful answer (no) to the second question would already bedetermined!

“Does this mean I will never see my Annabelle again?” askedAlexander mournfully after the King had explained the strategy’sinadequacy.“Ididn’tsaythat!”saidKingZorn.“Youhaven’tactuallytakenthe

test yet; you havemerely toldmewhat youwould have done hadyoutakenit.Gobackandthinkaboutitsomemore.Thenwhenyouareready,requestaformalaudiencewithmeandtheGrandVizier,and either determine in my presence, under the terms I havespecified, whether he is a knight or a knave, or else prove tomysatisfactionthatthetaskisimpossible.”Alexander thanked the King, withdrew, and thought the matter

overforseveralmoredays.Hethencametotheconclusionthatthetask was impossible. However, he was afraid to bring his proofbeforetheKingbecauseitmightcontainsomesubtleerror.“IfonlyIhadsomewisepersonwithwhomIcoulddiscussthisbeforetakingthetest,”hethought.Fortunately forhimandAnnabelle,Alexander struckupawarm

friendshipwithaworld-famousbiologistnamedProfessorBacteriuswho was visiting the island at the time. Though a scientist bytraining, Bacterius was an enormously erudite personwith a keeninterestinlogic.ThesuitorexplainedhisdilemmatoBacterius.“I believemy proof is correct,”Alexander said, “but Iwould be

infinitelygratefultohaveyouropinion.”“Withpleasure,”repliedBacterius.“Well,” said Alexander, “here is how I see it: I am asking the

Vizierquestions.Let’sconsidermylastquestion.Thisquestionmustbesuchthatayesanswerwillindicatethatheisofonetypeandanoanswerwillindicatethatheisoftheother.”

“Rightsofar,”saidBacterius.“Furthermore, I would surely know this fact before I asked the

question,”continuedthesuitor.“Let’ssupposethatayesanswerwillreveal thathe is aknightandanoanswerwill reveal thathe is aknave.SinceIwillknowthataknightwillansweryesandaknavewillanswerno,thenIwillknowthatyesisthetruthfulanswer,andthereforeIwillknowthetruthfulanswertothequestionbeforeIaskit,whichispreciselythethingIamforbiddentodo!Therefore,thetaskcannotbepossible.”Bacteriusknithisbrowand thought for some timebeforegiving

hisanswer.“I still don’t knowwhether the task is possible,” he finally said.

“ButifIwereyou,IwouldnotgivethisprooftotheKing.Ithasasubtlethoughdefiniteweakness,andI’mnotsurewhetheritcanberemedied.”He then explained the gap in the suitor’s proof, and Alexander

realizedthatBacteriuswasright.Canyouspotthegap?

There are two fallacies in the proof. In the first place,when the suitorcomestoaskthelastquestion,whywouldhenecessarilyknowthatitwillbe the last question? The questionmight be of such a nature that if itwere answered one way, Alexander could tell whether the Vizier is aknight or a knave, but if itwere answered the otherway, then furtherquestionswouldbenecessary.However,evenifthesuitordidknowthatthis was to be the last question, there is a difficultymore serious yet!There is a way that Alexander could know that a yes answer wouldindicatethattheVizierwasofonetypeandanoanswerwouldindicatethathewasof theother.And itmight seemvery strange, buthe couldknow thisonly aftergettingananswer.There isaquestionhaving thiscuriousproperty,asyouwillsoonsee!

Alexander was, of course, crestfallen to realize that his proofdidn’tholdwater.Moreover,hewasinaquandaryastowhattodonext; he could not find an airtight argument that the task was

impossible, nor could he see any way of accomplishing it. EvenProfessorBacterius,withhis impeccable logic,wasunable to solvetheproblem!At this point, fate proved helpful. Alexander, through a

combination of craft and bribery, was able to find out whereAnnabellewas imprisonedand, in thedeadofnight, sneaked in tovisither.Now,IshouldtellyouthatAnnabelle,thoughpossessedoflittle formaleducation,waswidely read,extremely intelligent, andhad that most precious mathematical gift of all—a remarkableintuition!WhenAlexanderexplainedtheproblemtoher,Annabellesolveditinstantly.“Dearboy,”shesaid,“youhavenothingtoworryabout.Youcan

find out in only one question whether the Vizier is a knight or aknave, and if you followmy plan, youwill not know the truthfulanswertothequestionatthetimeyouaskit!”She then explained her plan, and Alexander was delighted. The

very next day he obtained a formal audience with the King. Theentirecourtwaspresentinfullregaliatoseehowthesuitorwouldfare.“I am ready,” saidAlexander. “I claim the task is possible and I

shallaccomplishitbyaskingtheVizierjustonequestion!”“Now,thisImustsee!”saidtheKingwithachuckle.(Hechuckled

because he happened to believe that the task was impossible.)“Proceed!”The suitor (in accordance with Annabelle’s instructions) took a

deckofcardsoutofhispocket,shuffledthemthoroughly,tookoutacard at random and,without looking at its face, showed it to theVizier.“Isthisaredcard?”heasked.AssoonastheVizieranswered(eitheryesorno), the suitor lookedat the faceof thecard for thefirsttimeandthenheknewwhethertheVizierhadliedortoldthetruth.“Amazing!” said the King. “I would never have thought of

anythinglikethat!Butsomehowitseemslikecheating!”“Notreally,”repliedAlexander.“WhenIaskedtheVizierwhether

the card was red, I hadn’t looked at it yet, hence I didn’t know

whetheritwasred.Therefore,myquestionwasperfectlylegitimateaccordingtoyourterms.”Well, theKinghadtoacknowledgethatAlexanderhadwon,andsoAnnabellewasreleased.“Good luck to both of you,” said theKing. “Whydon’t you stayanotherdayorsoonthisisland?Youhaven’tyetmetourSorcerer,andheisamostamazingcharacter!Heknowsaboutyoubothandwouldliketomeetyou.Whynotpayhimavisit?”Thehappycouple thought thisagood idea, anddecided tovisitthe Sorcerer that very afternoon. But that’s a story for the nextchapter.

4HOWKAZIRWONHISWIFE

ANNABELLEANDALEXANDERweretiredaftertheirlong,windingclimbtotheSorcerer’s castle. But the Sorcerer, whom they discovered to be amost delightful andhospitable chap, served themadeliciousbrewconsistingofequalpartsofCeylonteaandChinesecinnamonwine,andtheywereinstantlyrevived.“Tell me,” said Annabelle, whose interests were generally quitepractical, “how did you ever establish your reputation here as aSorcerer?”“Ah,that’sanamusingtale,”saidtheSorcerer,rubbinghishands.“Istartedmycareerherejusttwelveyearsago.ButIreallyowemybeginnings to the philosopher Nelson Goodman,who taughtme acleverlogicaltricksomefortyyearsago.”“What’sthetrick?”askedAlexander.“Do you realize,” said the Sorcerer, “that despite the fact thatevery inhabitantof this island iseitheraknightwhoonly tells thetruthor aknavewhoonly tells lies, you candiscover the truthorfalsityofanypropositionsimplybyaskinganyinhabitantonlyonequestion?Andtheoddpartisthatafterheanswers,youwon’tknowwhetherhisansweristhetruthoralie.”“Thatdoessoundclever,”saidAlexander.“Well,that’showIgotmypositionhere,”saidtheSorcererwithalaugh.“Yousee,thefirstdayIsetfootonthisisland,IdecidedtogototheKing’spalaceandapplyforthejobofsorcerer.TheproblemwasIdidn’tknowjustwherethepalacewas.AtonepointIcametoa fork in the road; Iknew thatoneof the twobranches led to thepalace,butIdidn’tknowwhichone.Therewasanativestandingatthefork;surelyhewouldknowwhichroadwascorrect,butIdidn’tknowwhetherhewasaknightoraknave.Nevertheless,using theGoodmanprinciple,Iwasabletofindthecorrectroadbyaskinghim

onlyonequestionanswerablebyyesorno.”WhatquestiondidtheSorcereraskhim?

If youask thenativewhether the left road is the one that leads to thepalace,thequestionwillbeuseless,sinceyouhavenoideawhetherheisaknightoraknave.Therightquestiontoaskis“Areyouthetypewhowould claim that the left road leads to the palace?” After getting ananswer,youwillhaveno ideawhetherhe isa liarora truth-teller,butyouwillknowwhichroadtotake.Morespecifically, ifheanswersyes,youshouldtaketheleftroad;ifheanswersno,youshouldtaketherightroad.Theproofofthisisasfollows.Supposeheanswersyes.Ifheisaknight,thenhehastoldthetruth;heisthetypewhowouldclaimthattheleftroadleadstothepalace.Hencetheleftroadistheroadtotake.Ontheotherhand,ifheisaknave,hisanswerisalie,whichmeansheisnotthetypewhowouldclaimthattheleft road leads to thepalace;onlyoneof theopposite type—aknight—would claim that the left road leads to the palace. But since a knightwould claim that the left road leads to the palace, then again the leftroadreallydoesleadtothepalace.Regardlessofwhethertheyesansweristhetruthoralie,theleftroadleadstothepalace.Suppose now that the native answers no. If he is truthful, thenhe isreallynotthetypewhowouldclaimthattheleftroadleadstothepalace;only a knave would claim that. And since a knave would make thatclaim,theclaimisfalse,whichmeansthattheleftroaddoesnotleadtothepalace.Ontheotherhandifheislying,thenhereallywouldclaimthat the left road leads to the palace (since he says hewouldn’t), but,beingaknave,hisclaimwouldbefalse,whichmeansthattheleftroaddoesn’t lead to the palace.This proves that if heanswersno, the rightroadistheonetotake,regardlessofwhetherthenativeliedortoldthetruth.TheGoodmanprincipleisreallyaremarkablething.Usingit,onecanextractanyinformationfromonewhoeitheralways liesoralwaystellsthetruth.Ofcourse,thisstrategywon’tworkonsomeonewhosometimestellsthetruthandsometimeslies.

“Andso,”saidtheSorcerer,“Ifoundthecorrectroadandtookit.Igotlostafewmoretimesalongtheway,buteventhatwasforthebest.ThenativeIaskedwassoastoundedatwhatIhaddonethatheimmediately told many of the island’s other inhabitants, and thenews reached the King before I did. The King was absolutelydelighted with the incident and hired me on the spot! I’ve beendoinggoodbusinesseversince.“Andnow,” continued the Sorcerer, as he removedabeautifullybound volume from a shelf, “I have here an exceedingly rare andcurious old book, written in Arabic, known as the TellmenowIsitsöornot.IlearnedofitfromtheAmericanauthorEdgarAllanPoe,in his story entitled ‘The Thousand-and-Second Tale ofScheherazade.’ Poe found this odd tale of Scheherazade in theIsitsöornot,buttherearemanyotherremarkabletalesthathenevermentionedand thatare still comparativelyunknown.Forexample,haveyouheardofthe‘FiveTalesofKazir’?”Thetwoguestsshooktheirheads.“Ithoughtnot.Thesestoriesareofparticularinteresttologicians.Here,Iwilltranslatethemforyou.“OncetherewasayoungmannamedKazirwhosemainambitioninlifewastomarryaking’sdaughter.Heobtainedanaudiencewiththekingofhiscountryandfranklyconfessedhisdesire.“‘Youseemlikeapersonableyoungman,’saidtheKing,‘andIamsuremyunmarrieddaughterwilllikeyou.Butfirstyoumustpassatest.IhappentohavetwodaughtersnamedAmeliaandLeila;oneismarried and the other is not. If you can pass the test, and if myunmarrieddaughterapprovesofyou,thenyoumaymarryher.’“‘IsAmeliaorLeilathemarriedone?’askedthesuitor.“ ‘Ah, that is foryouto findout,’ replied theKing. ‘That isyourtest.’“ ‘Letme explain further,’ he continued. ‘My two daughters areidentical twins, yet temperamentally poles apart: Leila always liesandAmeliaalwaystellsthetruth.’“‘Howextraordinary!’exclaimedthesuitor.“ ‘Mostextraordinary, indeed,’repliedtheKing. ‘Theyhavebeen

likethisfromearlychildhood.Anyway,whenIstrikethegong,bothdaughterswillappear.YourtaskistodeterminewhetherAmeliaorLeila is themarried one. Of course you will not be told which isAmelia andwhich is Leila, norwill you be toldwhich of them ismarried.Youareallowedtoaskjustonequestiontojustoneofthetwo;thenyoumustdeducethenameofmymarrieddaughter.’”“Oh, I get it,” interrupted Annabelle. “Kazir used the Goodmanprinciple.Isthatit?”“No,” replied the Sorcerer. “The suitor happened to know theGoodmanprinciple—thoughnotbythatname,ofcourse.Histutor,avenerable old dervish, had taught it to him years before. And soKazir was overjoyed and thought, ‘All I need do is ask eitherdaughterwhether she is the typewhowould claim that Amelia ismarried. Ifsheanswersyes, thenAmelia ismarried; ifsheanswersno,thenLeilaismarried.It’sassimpleasthat.’“But itwasn’t as simple as that,” continued the Sorcerer. “It sohappened that the King could tell from the suitor’s triumphantexpressionthatheknewtheGoodmanprinciple.SotheKingsaid,‘Iknowwhat you’re thinking, but I’m not going to let you use thatlogicaltrickonme.Ifyourquestioncontainsmorethanthreewords,I’llhaveyouexecutedonthespot!’“‘Onlythreewords?’criedKazir.“‘Onlythreewords,’repeatedtheKing.“Thegongwasstruck,andthetwodaughtersappeared.”Whatthree-wordquestionshouldthesuitorasktodeterminethenameoftheKing’smarrieddaughter?

Thesuitorshouldask,“Areyoumarried?”Supposethedaughtertowhomheputsthequestionanswersyes.SheiseitherAmeliaorLeila,butwedon’tknowwhich.SupposesheisAmelia.Thenheransweristruthful;Ameliareallyismarried.ButsupposesheisLeila.Thenheranswerisalie;Leilaisnotmarried,soitmustbeAmeliawho is married. Regardless of whether she is Amelia or Leila, if sheanswersyes,thenAmeliamustbemarried.Suppose, on the other hand, that the daughter answers no. If she is

Amelia, then her answer is truthful, which means that Amelia is notmarried;henceLeilaismarried.Ontheotherhand,ifLeilaanswers,thenheranswerisalie,whichmeansthatshe,Leila,ismarried.Regardlessofwhethertheanswernoistrue,Leilaisthemarrieddaughter.

“Didthesuitorpassthetest?”Annabelleasked.“Alas,no,”repliedtheSorcerer.“Ifhehadn’tbeenrestrictedtoa

three-word question, he would have had no trouble at all. But,confrontedwithatotallynewsituation,hewascompletelyflustered.Hejuststoodthere,unabletoutteraword.”“Sowhathappened?”askedAlexander.“Hewasdismissedfromcourt.Butthenacuriousthinghappened.

The unmarried daughter had taken a liking to Kazir, and pleadedwithher father to summonhimback thenextday to takeanothertest.Somewhatreluctantly,theKingagreed.“The next day, when the suitor entered the throne room, he

exclaimed,‘I’vethoughtoftherightquestion.’“ ‘Too latenow,’said theKing. ‘You’llhaveto takeanother test.

WhenIstrikethegong,againmytwodaughterswillappear(veiled,ofcourse).Onewillbedressedinblueandtheotheringreen.Yourtask now isnot to find out thename ofmymarried daughter, butwhichofthetwo—theoneinblueortheoneingreen—isunmarried.Again you may ask only one question, and the question may notcontainmorethanthreewords.’“Thegongwasstruck,andthetwodaughtersappeared.”Whatthree-wordquestionshouldthesuitoraskthistime?

Thequestion“Areyoumarried?”willbeofnohelpinthissituation;thequestion to ask is “Is Amelia married?” If the daughter addressedanswersyes,thensheismarried,regardlessofwhethersheliesortellsthetruth;ifsheanswersno,thensheisnotmarried.Supposesheanswersyes.IfsheisAmelia,thenheransweristruthful;

Ameliaismarried.Ontheotherhand,ifsheisLeila,thenheranswerisalie; Amelia is not married, so Leila is married. Thus, a yes answerindicates that the daughter who answers is married. (I leave it to the

reader toverify thatanoanswer indicates that thepersonaddressed isnotmarried.)Ofcourse,thequestion“IsLeilamarried?”wouldserveequallywell;a

yes answer would then indicate that the daughter who answers is notmarried,anoanswerthatsheis.AstheSorcererexplainedtoAnnabelleandAlexander,thereisapretty

symmetrybetweenthisproblemandthe last: tofindout if thedaughteraddressedismarried,youask,“IsAmeliamarried?”;whereasifyouwishto find out whether Amelia is married, you ask, “Are youmarried?”Thesetwoquestionshavethecuriouspropertythataskingeitheronewillenableyoutodeducethecorrectanswertotheother.

“Again the suitor failed,” continued the Sorcerer, “but theunmarrieddaughterwasmorefondofhimthanever.TheKingcouldnotresistherpleadings,andsoheagreedtogiveKazirathirdtestthenextday.“ ‘Thistime,’saidtheKingtothesuitor, ‘whenIstrikethegong,

justonedaughterwillappear.Yourtasknowistofindoutherfirstname.Againyoumayaskonlyonequestion,anditmaycontainnomorethanthreewords.’”Whatquestionshouldthesuitorask?

Thisproblemissimplerthantheprecedingtwo.Allthesuitorneedaskisanythree-wordquestionwhoseanswerhealreadyknows,suchas“DoesLeila lie?” To this particular question, Amelia would obviously answeryesandLeilawouldanswerno.

“‘Nowreally,’saidtheKingtohisdaughterafterthesuitorfailedthethirdtest,‘areyousureyouwanttomarryhim?Hestrikesmeasquiteasimpleton.Thelasttestwasridiculouslyeasy,andyouknowit.’“ ‘Hewas justnervous,’ responded thedaughter. ‘Please tryhim

oncemore.’“Well,”saidtheSorcerer,“forthenexttestthesuitorwastoldthat

afterthegongwasstrucktheunmarrieddaughterwouldappear.Thesuitorwastoaskherjustonethree-wordquestionanswerablebyyes

or no. If she answered yes, the suitor could marry her. If sheansweredno,hecouldnot.”Whatquestionshouldthesuitorask?

Thequestion“AreyouAmelia?”wouldworkperfectlywell.Amelia,whoistruthful,wouldansweryes,andLeila,wholies,wouldalsoansweryes—thatis,shewouldfalselyclaimtobeAmelia.

“ ‘This istryingmypatience,’saidtheKingtohisdaughterafterthesuitorhadfailedafourthtime.‘Nomoretests!’“ ‘Just onemore,’ begged thedaughter. ‘I promise itwill be the

last.’“‘Allright,theverylastone,youunderstand?’“Thedaughterpromisednottoaskforanymoretests,andsothe

Kingagreed.“ ‘Now then,’ said theKingquite sternly to the suitor (whowas

tremblinglikealeaf),‘youhavealreadyfailedfourtests.Youseemtohavedifficultywiththesethree-wordquestions,andsoIwillrelaxthatrequirement.’“‘Whatarelief!’thoughtthesuitor.“‘WhenIsoundthegong,’saidtheKing,‘againonlyonedaughter

willappear;shemaybethemarriedoneortheunmarriedone.Youare to askheronlyonequestionanswerablebyyesorno,but thequestionmaycontainasmanywordsasyoulike.Fromheranswer,youmustdeducebothhernameandwhethersheismarried.’”Canyouthinkofaquestionthatwillwork?

The king (evidently sick by now of the whole business) has given thesuitoranimpossibletask.Ineachofthefirstfourtests,thesuitorhadtodeterminewhichof twopossibilities held. In this test, however, he is todeterminewhichoffourpossibilitiesholds.(Thefourpossibilitiesare:1.the daughter addressed is Amelia and married; 2. she is Amelia andunmarried;3.she isLeilaandmarried;4.she isLeilaandunmarried.)However, there are only two possible responses to the suitor’s question(eitheryesorno,sincethequestionisrequiredtohaveayes/noanswer).Andwithonlytwopossibleresponses,itisimpossibletodeterminewhich

offourpossibilitiesholds.WhenIsaythatthetaskisimpossible,Imerelymeanthatthereisnoyes/no question that is bound to work. There are several possiblequestions(atleastfour)thatmighthappentoworkifthesuitorislucky.Forexample, consider thequestion“Is it thecase thatyouaremarriedand that your name is Amelia?” Leila would answer yes to this(regardless ofwhether she ismarried because a compound statement—onewith twoparts joinedby“and”—is false ifbothpartsoronlyonepartof the statement is false);Amelia, if she ismarried,wouldansweryesand, if she isunmarried,wouldanswerno.Soayesanswerwouldleavethesuitorinthedark,whereasanoanswerwouldindicateforsurethat theoneaddressed isAmeliaandunmarried.So if the suitoraskedthatquestion,hewouldhaveatwenty-fivepercentchanceoffindingoutwhichoneofthefourpossibilitiesactuallyholds.Butnosinglequestioncouldensureacertaintyoffindingoutwhichoneofthefourpossibilitiesholds.

“Sotheyneverdidgetmarried?”askedAnnabelle.“The King never consented,” replied the Sorcerer. “But thedaughterwas so furious at the unfairness of the last question thatshe felt perfectly justified in marrying Kazir without her father’spermission. The two eloped and, according to the Isitsöornot, livedhappilyeverafter.“Thereisavaluablelessontobelearnedfromallthis,”continuedtheSorcerer.“Neverrelytoomuchongeneralprinciplesandroutinemechanicalmethods. Certain classes of problems can be solved bysuch methods, but they completely lose their interest once thegeneralprincipleisdiscovered.It is,ofcourse,goodtoknowthesegeneral principles—indeed, science and mathematics couldn’tadvancewithoutthem.Buttodependonprincipleswhileneglectingintuitionisashame.TheKingwasveryclevertoconstructhistestssothatthemereknowledgeofageneralprinciple—inthiscasetheGoodmanprinciple—wouldbeofnoavail.Eachofhistestsrequireda bit of ingenuity. Here the suitor failed, because of a lack ofcreativethinking.”

“I would like to know one more thing,” said Annabelle. “Is itrecordedwhichofthetwodaughtersKazirmarried?”“Oh,yes,”repliedtheSorcerer.“Fortunatelyforthesuitor itwas

Amelia, the truthfulone.This, Ibelieve,contributedto their livinghappilyeverafter.“ThestoryofLeila’smarriage,”hecontinued,“wasalsorecorded

intheIsitsöornot,andIfindthisstoryparticularlydroll.ItseemsthatLeila detested her suitor, butwhen he asked her one day, ‘Wouldyouliketomarryme?’,she,beingaperpetualliar,saidyes.Andsotheyweremarried.“Soyousee,”concludedtheSorcerer,“perpetuallyingsometimes

hasitsdangers!”Well,theyallhadaheartylaughoverthatstory.ThenAnnabelle

andAlexanderroseandthankedtheSorcererforamostentertainingandinstructiveafternoon,explainingthattheyhadtogetbackandmakepreparationsfortheirdeparturethenextday.“Have you heard about the epidemic that hit this island three

yearsago?”askedtheSorcerer.Theyshooktheirheads.“Oh,Iwouldliketotellyouallaboutit,butunfortunatelyIcan’t

today. I’mexpectingavisit fromthe islandAstrologeranyminute.Whydon’tyoustayafewmoredays?”“Myparentswillbeworriedaboutme,”explainedAnnabelle.“Notso,”repliedtheSorcerer.“I’vealreadydispatchedamessage

to your island explaining that you are here and well. Your fatherknowsthatIamaknight.”Thisreassuredthecouple,andtheyagreedtovisittheSorcererthe

nextday.“WhoisthisAstrologer?”askedAlexander,astheywereleaving.“Oh,he’sacompleteimbecile,aswellasaknaveandacharlatan.

Ihavetoputupwithhim;it’samatterofislandpolitics.ButIadviseyoutohavenothingtodowithhim.”Justthen,theAstrologerenteredtheroom,noddedtothecouple

(whomhehadneverseenbefore),andsaidtotheSorcerer,“SheistheQueenofShebaandheisKingSolomon,youknow.”

“Typical,” said the Sorcerer to the departing couple with aknowingwink.

5APLAGUEOFLIES

“IT CAME LIKE A SIROCCO WIND and struck about half of the island’sinhabitants.Fortunately, Iwasamongthosewhowerespared,” theSorcererexplainedtoAnnabelleandhersuitor,Alexander.TheyhadbothmadeaspecialtriptotheSorcerer’stowertohearhisaccountofthestrangeepidemicthathadsweptacrosstheIslandofKnightsandKnavesaboutthreeyearsbefore.“Nobodycouldreallydiagnoseit,”continuedtheSorcerer.“EvenProfessor Bacterius, with all his knowledge of immunology, wastotallybaffled.”“Wasitviralorbacterial?”askedAnnabelle.“Even that was unknown!” exclaimed the Sorcerer. “There wasabsolutelynoobservablechangeinthebodychemistry.Infact,therewere no physical symptoms at all; the effects were purelypsychological.Thewholeepidemiclastedonlyaweek,duringwhichtime pandemonium reigned over the island. Then, quite suddenly,thingsreturnedtonormal.”“You say the symptoms were purely psychological,” saidAlexander.“Justwhatwerethey?”“Well,” replied theSorcerer,“all thosewhowerestruckreversedtheirnormallyingortruth-tellingrole.Nolongerwasittruethatallknightstoldthetruthandthatallknaveslied.Instead,sickknightsliedandsickknavestoldthetruth,whilehealthyknightscontinuedtotellthetruthandhealthyknavescontinuedtolie.Soduringthisstrangeweek,therewerefourtypesontheisland—healthyknights,sickknights,healthyknaves, and sickknaves.Healthyknights andsickknavestoldthetruth;sickknightsandhealthyknaveslied.Soifyou came across a native whomade a statement you knew to befalse,ingeneralyoucouldnotknowwhetherhewasasickknightorahealthyknave.”

“Thatmusthavebeenfrightfullyconfusing,”saidAnnabelle.“At first,yes,” replied theSorcerer. “But then Iwentaround theislandinterviewingnativesIdidn’tknowtoseewhatIcouldlearn,andIlearnedsomeinterestingthingsindeed.”Alexanderthoughtforabit,andthencommented,“Icanseethatingeneralyoucouldn’t tella sickknight fromahealthyknave.Forexample, if aperson said that twoplus twoequals five, you couldcertainlynot tellwhetherhewasasickknightorahealthyknave.Butwerethereexceptions?Wasitsometimespossibletotellfromjustone false statement whether the speaker was a sick knight or ahealthyknave?”“Goodquestion,”saidtheSorcerer.“Iactuallyencounteredsuchasituation.IcameacrossanativeIdidn’tknow.Hemadeastatementfrom which I could deduce not only that it was false, but evenwhetherhewasasickknightorahealthyknave.”Canyouthinkofonesuchstatement?

One statement thatwouldwork is “I ama sick knight.”The statementcan’t be true, sincea sickknightwouldn’t truthfully claim to bea sickknight.Therefore,heislying,whichmeansheiseitherasickknightorahealthyknave (since theseare theonly typeswho lie).Buthe is not asickknight.Hence,heisahealthyknave.Ofcourse,thestatement“Iamahealthyknave”wouldworkequallywell;onlyasickknightcouldmakethatstatement.

“Onanother occasion,” continued the Sorcerer, “I cameacross anative whomade a statement fromwhich I could deduce that hemust be a knave, though I couldn’t tell whether he was sick orhealthy.”Canyousupplysuchastatement?

A simple statement that works is “I am sick.” A healthy knight wouldnever lieandclaimtohesick,andasickknightwouldnever truthfullyadmittobeingsick.Hence,anativewhowouldsaythatmustbeaknave.Hecouldbeeitherahealthyknavewho liedaboutbeing sickora sickknavewhotruthfullyclaimedtobesick.

“Soon after, I met another native who made a statement fromwhichIcoulddeducethathemustbesick,butnotwhetherhewasaknightoraknave,”saidtheSorcerer.Whatstatementwouldwork?

The statement “I am a knave” would work. (I leave the proof to thereader.)

“InextcameacrossanativewhomadeastatementfromwhichIcoulddeducethathewaseitherahealthyknight,asickknight,orahealthy knave, but it was impossible to know which of the threecasesheld.”Whatstatementwouldwork?

Thestatement“Iamahealthyknight.”Ahealthyknightcouldtruthfullysaythat;asickknightorahealthyknavecould lieandsaythat;butasick knave could never lie and say that he is a healthy knight. Thus,anyonebutasickknavecouldmakethatstatement.

“Finally,Icameacrosstwonativesunknowntome,namedAstorand Benedict. First Astor said something I could not understand,since he used the island’s native tongue, which I had not fullymastered. Then I asked Benedict what Astor had said. Benedictreplied, ‘Hesaidthatheiseitherasickknightorahealthyknave,’Astor protested, ‘I never said that!’ But Benedict said, ‘Astor is aknave.’AndfinallyAstorsaid,‘Benedictissick!’“ThisenabledmetocompletelyclassifyAstorandBenedict,”saidtheSorcerer.Whatarethey?

Duringthisstrangeweek,noinhabitantcouldpossiblyclaimtobeasickknight or a healthy knave, because if he is a sick knight or a healthyknave, he lies; hence, he would never truthfully admit to being a sickknightorahealthyknave.Andifhewerenotasickknightorahealthyknave,thenhewouldbetruthful;hence,hewouldneverfalselyclaimtobe a sick knight or a healthy knave. Therefore, Benedict lied when he

said that Astor claimed to be a sick knight or a healthy knave. Thus,Astor must be truthful, because he correctly denied having said that.Now,Benedict,wholies,saidthatAstorisaknave,soAstorisreallyaknight.SinceAstor isaknightand is truthful,Astormustbeahealthyknight.Also,Astor,whoistruthful,saidthatBenedictissick.Therefore,Benedictreallyissick.SinceBenedictliesandissick,hemustbeasickknight.SotheansweristhatAstorisahealthyknightandBenedictisasickknight.

“It so happens,” continued the Sorcerer, “that the islandAstrologerwasattackedbythedisease.Asyouknow,heisaknave,and usually a rather stupid one. But the disease had amiraculouseffect on him.Not only did he tell the truth allweek, but hewaseminently sensible during the entire period. It’s really a pity herecovered.“Anyway, during this time, we did a good deal of researchtogether.Ononeparticulardaywewerewalkingonthebeachwhenwe spiedanativewalking inourdirection. ‘Oh, I knowhim,’ saidtheAstrologer.‘Icantellyouwhetherheisaknightoraknave,butnotwhetherheissick.’“Asthenativepassedus,thenativesaid,‘Iamahealthyknight,’TheAstrologer then said tome, ‘Good. I nowknowwhether he issick.’”The Sorcerer askedAnnabelle andAlexander, “Was the native aknightoraknave,andwashesickorhealthy?”“Just a minute,” said Annabelle. “Are you sure you gave usenoughinformation?”“Icertainlyam,”repliedtheSorcerer.Whatisthesolution?

From the fact that the native claimed to be a healthy knight, all thatfollowsisthatheisnotasickknave.(Wesawinanearlierproblemthatahealthyknight,asickknight,orahealthyknavecouldallmake thatstatement,but thatasickknavecouldnot.)Now,before theAstrologerheard the native talk, he knew whether the native was a knight or a

knave,butwearenot toldwhich.Supposehepreviouslyknew that thenative was a knight. Then after the native made his statement, theAstrologerwouldhavenowayofknowingwhetherthenativewasasickknightorahealthyknight:hewouldhaveknownnomorethanheknewbefore.Butsinceweare told that theAstrologerdid finallyknow, thentheonlypossibility is thathepreviouslyknew thenative to beaknaveandfinallyknewhimtobeahealthyknave.

…

“More interesting yet,” continued the Sorcerer, “while on ourwalk, the Astrologer and I came across another native talking tohimself, about whom we each had partial information. ‘I knowwhetherheisaknightoraknave,’ItoldtheAstrologer,‘butIdon’tknowwhetherheissick.’“ ‘That’s curious,’ replied the Astrologer. ‘I happen to know

whether he is sick, but I don’t knowwhether he is a knight or aknave. Don’t tellmewhat you know, and I won’t tell youwhat Iknow.Instead,let’sgooverandhearwhathehastosay.’“Sowewent over to the native, only to hear himmumbling to

himself,‘Iamnotahealthyknave.’“TheAstrologerandIboththoughtforawhile,butIcouldnotyet

tellwhetherhewassickorhealthy,norcouldtheAstrologeryettellwhetherhewasaknightoraknave.Was thenativeaknightoraknave,andwashehealthyorsick?”“Now,wait aminute,” saidAnnabelle. “If you don’t know, then

howdoyouexpectustoknow?”“I didn’t say I don’t know,” replied the Sorcerer. “I said Ididn’t

knowatthetime.ItwasonlywhentheAstrologerlatertoldmethathestilldidn’tknowthatIknew.”Whatisthesolution?

Ahealthy knave could lie and claim he is not a healthy knave; a sickknavecouldtruthfullyclaimheisnotahealthyknave;ahealthyknightcouldtruthfullyclaimheisnotahealthyknave;butasickknightcouldnevermakethe truestatement thathe isnotahealthyknave.So,what

followsfromthenative’sstatementissimplythatheisnotasickknight.After thenativespoke, then,both theSorcererandtheAstrologerknewthathewasnotasickknight.Now,theSorcererhadpreviousknowledgeofwhetherthenativewasa

knightoraknave.Ifhehadpreviouslyknownhimtobeaknight, thenafterknowingthathewasnotasickknight,hewouldhaveknownhimtobeahealthyknight.ButwearetoldthattheSorcerercouldnottellfromthe native’s statement whether he was sick or healthy; therefore, theSorcerermusthavepreviouslyknownthenativetobeaknave.TheAstrologer,ontheotherhand,previouslyknewwhetherthenative

washealthyorsick.Hadhepreviouslyknownhimtobesick,thenafterthenativespoke,hewouldhaveknownthathewasasickknave(sinceheknewthatthenativewasnotasickknight),butifhehadpreviouslyknownhimtobehealthy,hewouldhavestillbeeninthedarkafterthenative spoke. Since he was still in the dark, then it must be that hepreviouslyknewthatthenativewashealthy.Therefore,thenativewasahealthyknave.

“Duringthisepidemic,”saidtheSorcerer,“avaluableringofminewasstolen.Threesuspects,Jacob,Karl,andLouie,wereimmediatelyarrested,andtherewasa trial thatveryday.Everybodyknewthatone of the threewas guilty, but before the trial the court did notknowwhichof the three itwas.Here isa transcriptof the trial. (Ishouldmentionthatthethreedefendantswereclosefriends,anditwas correctly assumed that the two innocent ones knew who theguiltyonewas.)”

Judge(toJacob):Whatdoyouknowaboutthetheft?Jacob:Thethiefisaknave.Judge:Ishehealthyorsick?Jacob:Heishealthy.Judge(toKarl):WhatdoyouknowaboutJacob?Karl:Jacobisaknave.Judge:Healthyorsick?

Karl:Jacobissick.

“ThejudgethoughtforawhileandthenaskedLouie,‘Areyoubyany chance the thief?’ Louie answered (yes or no), and the judgethendecidedthecase.Whostolethering?”“Justaminute,”saidAlexander.“Youhaven’ttolduswhatLouie

answered.”“You can solve the problem without being told,” replied the

Sorcerer.Whostolethering?

Karl’s two answers were either both truthful or both lies. If they wereboth truthful, then Jacob is a sick knave; if they were both lies, thenJacob isahealthyknight.SoJacob is eithera sickknaveorahealthyknight,and,ineithercase,heistruthfulinhispresentcondition.Hence,Jacob’s answers were both truthful, so the thief is actually a healthyknave and hence a liar. Since Jacob is truthful and the thief is a liar,Jacobisnotthethief.ThejudgeknewthisevenbeforesheinterrogatedLouie.Thensheasked

Louiewhetherhewasguilty,butwearenottoldwhatLouieanswered.IfLouieansweredyes,thenhemustbeinnocent,forwealreadyknowthatthe real thiefwould lieabout being guilty.On theotherhand, if Louieanswered no, then there is noway of tellingwhether he is innocent orguilty.Ifheansweredtruthfully,hewouldbeinnocent,whichisperfectlyconsistentwiththefactthatthethieflies.Butifheansweredfalsely,thenheisthethief,whichisagainconsistentwiththethief’sbeingaliar.SoifLouieansweredno, the judgecouldn’thavemadeaconviction.But thejudge did make a conviction. Thus, Louie must have answered yes,provingtothejudgethatLouiewasinnocent.SoKarlstolethering.By a curious turn of fate, Louie acquitted himself by claiming to be

guilty!

6ONTHEOTHERHAND

THE WEDDING of Princess Annabelle and her suitor, Alexander, on theIslandofKnightsandKnaveswentsmoothlyuntilthetimecameforthecoupletotaketheirvows.Thenthejusticeofthepeacedeclared:“I donot now pronounce youman andwife.” Everyone gasped inhorror.EvenAnnabelleandAlexander—whowereused to theoddcustomsoftheIslandofKnightsandKnaves,whereeveryinhabitantiseitheraknight,whoalwaystellsthetruth,oraknave,whoalwayslies—held their breath. Then, at last, among the fainting guests, aknightintheweddingpartyassuredeveryonethatthejusticewasinfact a legally certified knave and that his statement thereforeconstituted legal evidence that the couple was married. A sigh ofreliefwentup.Andthenachorusofcheers.KingZorn,whohadgrownquite fondof the couple, provided amagnificentbanquetatthepalace.Everyoneofimportanceattended,including,ofcourse,theSorcerer,andeventhedoddering,grumpyAstrologer.Afterthefeastcamethedancers,jugglers,fire-eaters,andmagicians. But the evening’s real entertainment began when theSorcerergatheredeveryoneabouthimtohearofhistravels.“I visited an odd country somemonths ago,” he began, “whereeveryinhabitantiseitherright-handedorleft-handed.”“What’ssooddaboutthat?”interruptedtheAstrologer.“Nothing,” replied the Sorcerer, “except that whatever right-handed inhabitantswritewith therighthand is true,butwhatevertheywritewiththelefthandisfalse.Left-handedinhabitantsaretheopposite: whatever they write with the left hand is true, andwhatevertheywritewiththerighthandisfalse.”“That’s not strange,” said the Astrologer. (The Astrologer,remember,isaknave.)“It certainly is strange,” said theKingquite sharply, “and Iwish

youwouldstopinterrupting!”Therewasatensesilence.“Go on with your story,” said the King. “The Astrologer won’tinterruptagain,orI’llhavehimthrownintothedungeon!”“Well,” said the Sorcerer, “this country provided marvelousopportunities for logical detective work. The problems that arosewerejustthesortthatinterestYourMajesty.”“Suchas?”askedtheKing.“Onmy very first day there, I came across a scrap of paper onwhich an inhabitant had written a single sentence. From thatsentenceIcoulddeducethattheinhabitantmustbeleft-handed.”“Whatwasthesentence?”askedtheKing.“Ah!That’s thequestion Iwould likeyouall toanswer,” repliedtheSorcerer.

Asentencethatworksis“Iwrotethiswithmylefthand.”Thesentenceiseithertrueorfalse.Ifitistrue,thenthewriterreallydidwriteitwithhisleft hand. And since only left-handed inhabitants write true sentenceswith the left hand, hemust be left-handed.But if the sentence is false,then the writer wrote it with his right hand. Since only left-handedinhabitants write false sentences with the right hand, again the writermustbeleft-handed.

“Thenextday,”saidtheSorcerer,“Icameacrossasentencejottedonaslipofpaper fromwhich Ideducedthat thewriterwasright-handed and had written the sentence with his left hand. Whatsentencedidhewrite?”“Iknow!”saidoneoftheguests.“Thesentencewassomethinglike‘Two plus two is five,’ That could have been written by a right-handedpersonusinghislefthand.”“Ofcourseitcouldhavebeen,”saidtheSorcerer,“butitalsocouldhavebeenwrittenbyaleft-handedpersonusinghisrighthand.Thesentence I came across could have been written only by a right-handedpersonusinghislefthand.”Whatwasthesentence?

Thesimplestsolutionisthesentence“Iamleft-handed,andIwrotethis

sentencewithmy righthand.”The sentencemustbe false,because if itwere true, that would mean that a left-handed person wrote a truesentencewith his right hand,which is impossible. Since the sentence isfalse, and itwas notwritten by a left-handed inhabitantwith his righthand,itmusthavebeenwrittenbyaright-handedinhabitantwithhislefthand.

“AfterIhadspentaweekinthiscountry,”boastedtheSorcerer,“Igainedquiteareputationasahandwritingexpert.OnedaytheChiefof Police consulted me on a case. Two scraps of paper had beenfound.Ononewaswritten:‘Ialwayswritewithmyrighthand,’Ontheotherwaswritten:‘Isometimeswritewithmylefthand.’“Thepoliceknewwhothewriterwas,andforsomereason,aboutwhich itwas notmybusiness to inquire, itwas very important tofindoutwhichofthetwosentenceswastrue.Wasthereanywayofdeterminingwhetherthewriteralwayswrotewithhisrighthandorwhetherhesometimeswrotewithhisleft?”

The two sentences contradict each other; therefore, one of them is trueandtheotherfalse.Thatmeansthatthewritersometimesusesonehandand sometimes the other. And so the second sentence is true: hesometimeswriteswithhislefthand.

“Onanotheroccasion,”saidtheSorcerer,“thepoliceconsultedmeabout a robbery. It was known that the thief was left-handed. Asuspect had been arrested, and the first thing to be done was todeterminewhether hewas right-handedor left-handed. Thepolicehadsearchedhishouseandfoundanotebookwithentriesthatthesuspect admittedhavingwritten. I havehere facsimiles of the firsttwopages.”TheSorcererheldupthefirstpage,whichread:“Thesentenceonpage2isfalse.”Andthenheheldupthesecond,whichread:“Thesentenceonpage1waswrittenwithmylefthand.”“The police wanted to know whether this notebook was at allrelevant to the case,” explained the Sorcerer, passing around thepages.“Wasit?”

Thereisnowayofdeterminingwhichofthesentencesistrue.Butitcanbe determined whether the suspect is right-handed or left-handed.Suppose the sentenceonpage1 is true.Then the sentenceonpage2 isfalse,whichmeansthatthesentenceonpage1waswrittenwiththerighthand.Thewritermustthereforeberight-handed,sincehewrotethetruesentence on page 1with his right hand.Now, suppose the sentence onpage1 is false.Then the sentenceonpage2mustbe true(because thesentence on page 1 falsely says that the sentence on page 2 is false),whichmeansthatthesentenceonpage1waswrittenwiththelefthand.Again,thesuspectmustberight-handed,sincehewrotethefalsesentenceon page 1 with his left hand. Thus, the suspect is right-handed andthereforeinnocentoftherobbery.

“Theeveningisgettingon,”saidtheSorcerer.“SoIwill tellyouonlyonemorestory.“OneafternoonIwassittingatacaféwithafriendofminenamedRobertSmith.Ichancedtoaskhimwhetherhewasright-handedorleft-handed. Insteadof answering,he smiled, tore a sheet fromhisnotebook,andwrote:‘Iamleft-handed,andIwrotethiswithmylefthand.’Of course I sawwhichhandheused, but Iwasnot able todeducewhether hewas right-handed or left-handed. Just then hiswife joined us. He showed her the sentence he had written andaskedwhethershecouldtellwhichhandhehadwrittenitwith.Sheknewwhetherherhusbandwasright-handedorleft-handed,andshewasagoodlogician.Still,shecouldn’ttellwhichhandhehadused.Isheright-handedorleft-handed?”“Nowjustaminute,”saidtheKing.“YousaidthatyousawwhichhandMr. Smithused.Have you forgotten to tell uswhichhand itwas?”“Ididn’t forget,”saidtheSorcerer.“Fromthe informationIhavegivenyou,itispossibletodeducewhichhandheused.”Which hand did Mr. Smith use, and is he right-handed or left-handed?

If Mr. Smith were right-handed, he couldn’t possibly have written the

sentencewithhisrighthand.Therefore,thereareonlythreepossibilities:heisright-handedandfalselywrotethesentencewithhislefthand;heisleft-handedand falselywrote the sentencewithhis righthand;orhe isleft-handedandtruthfullywrotethesentencewithhislefthand.Now,theSorcerersawwhichhandMr.Smithused.Ifitwerehisright

hand,thentheSorcererwouldhaveknownthatonlythefirstpossibilitycould be true and that Mr. Smith was right-handed. However, theSorcerer didn’t know this. Hence, Mr. Smith must have used his lefthand. (The Sorcerer then couldn’t tell whether the first or the thirdpossibilitywasthetruth.)Mrs. Smith, on the other hand, knew whetherMr. Smith was right-

handed or left-handed, but she didn’t see which hand he had used towritethesentence.IfMr.Smithwereright-handed,hiswifewouldknowthis,and shewouldhavebeenable todeduce thatonly the firstoptionwastheactualcase.Thatis,shewouldhavefiguredthathehadwrittenthe sentence with his left hand. Since she was unable to do this, herhusbandmustbeleft-handed.Eventhoughsheknewhewasleft-handed,shecouldn’ttellwhetherthesecondorthethirdpossibilitywastheactualcase. This proves thatMr. Smith is left-handed and that he wrote thesentencewithhislefthand.

“Andnowatoasttothecouple!”saidKingZorn.“Maytheiryearsbe long and happy, and may they pay many more visits to thisisland,bringingnewandchallengingpuzzleswiththem!”Anotherroundofcheersthundered.“And now for the main event of the evening—the possible

weddinggifttothebrideandbridegroom.”“Possiblegift?”mutteredtheAstrologer.“Ishall lettheSorcerermakethepresentation,”saidtheKing.“I

believeyouwillallfinditamusing.”The Sorcerer rose andwalked over to twomagnificent treasures

thatthecompanyhadbeeneyeingallevening.“The bridegroom is to rise and make a statement,” said the

Sorcerer.“HisMajestyhasgraciouslypromisedthatifthestatementistrue,thenhewillgivethecoupleoneofthesetwotreasuresand

possibly both. But if the statement is false, thenHisMajesty flatlyrefusestogivethecoupleanyweddinggiftatall!”The Sorcerer then gaveAlexander a knowingwink, as if to say:

“Come on now, old boy, use your head and come up with astatement such that the king’s only option is to give you bothtreasures!”Alexandercleverlymadesuchastatement.Whatdidhesay?

Alexandersaid:“ItisnotthecasethatHisMajestywillgiveusoneandonlyoneofthetreasures.”Suppose the statement is false. Then it is the case that theKingwill

giveoneandonlyoneofthetreasures.ButtheKingcan’tgiveoneofthetreasuresforafalsestatement.Therefore,thestatementcan’tbefalse;itmust be true. Since the statement is true, then the King, to keep hispromise, must give at least one of the treasures. Since Alexander’sstatementistrue,though,theKingcan’tgiveonlyoneofthetreasures.(Ifhedid,thiswouldfalsifythestatement!)Hence,theKingmustgivebothofthetreasures.Andsohedid.

7THEISLANDOFPARTIALSILENCE

A FEW DAYS after the wedding our happy couple, Princess AnnabelleandAlexander,setsailforhome.Butalas,theywereinfortrouble!They had only been at sea a few hours when their vessel wasattackedbypirates;thetwowerecapturedandsoldasslavestothekingofanotherisland—agrimplaceknownastheIslandofPartialSilence.“Letmeexplainthisislandtoyou,”saidthegrimKingtothetwocaptives. “As on the island of King Zorn, every inhabitant here iseither a knight or a knave. However, on this island, people don’talways answer questions asked of them! If you ask a knight aquestion,ifheanswersatall,hisanswerwillbetruthful,buthemayrefusetoanswer.Ifyouaskaquestionofaknaveandheanswersatall, his answer will be a lie, but again he may simply refuse toanswer. If you cook up a special question that a native cannotanswerwithoutviolatinghisknighthoodorknavehood,as thecasemaybe, thenhewill certainly refuse toanswer.You seenowwhythisislandhasitsname.“Two plus two equals four,” continued the King. “Now you seethatIamaknight.Unfortunatelyforyou,Iamnotaspleasantasoulas King Zorn. On the other hand, I am not quite as evil as I amcrackeduptobe,andsoIwillgiveyoutwoasportingchance.Iwillgiveyoufiveextremelydifficulttests.Ifyoupassallofthem,Iwillsetyoufree;ifyoufailsomuchasoneofthem,thegentleman’sheadwillbeforfeitandthePrincesswillbemine!”“Fiveextremelydifficult testsandwemustpassallof them,andyoucallthatasportingchance?”askedAlexander.“Thosearemyterms!”saidtheKingsternly.Herearethefivetests.

Test1•Designaquestiontowhichaknavecouldanswereitheryesorno,butwhichaknightcannotansweratall.

Test2•Designaquestiontowhichaknightcouldanswereitheryesorno,butwhichaknavecannotansweratall.

Test3 •Designaquestion towhichaknight couldanswernobutnotyes,andwhichaknavecouldnotansweratall.

Test4 •Designaquestiontowhichaknightcouldansweryesbutnotno,andwhichaknavecouldnotansweratall.

Test 5 • Design a question that neither a knight nor a knave canansweratall.

SolutionstoFirstFiveTests

Test 1. A question that works is: “Is no your answer to thisquestion?”(Thequestiondoesn’tmean:“Isnothecorrectanswertothisquestion?”Itmeans:“Isnotheansweryouwillactuallygivetothisquestion?”Perhapsamoreprecisewayofphrasingitis:“Afteryouhaveansweredthisquestion,willtheanswerhavebeenno?”)As the Sorcerer explained to Abercrombie in Chapter 1, no onewho is asked this question can answer it correctly, hence a knightcannotanswer it.Aknavecananswereitheryesorno, sincebotharewronganswers.

Test2.Thequestionis:“Isyesyouranswertothisquestion?”Ayesanswerandanoanswerarebothcorrect,henceonlyaknightcouldanswerthequestion,andhecananswereitherway.

Test3. Idoubt that any simple questionwillwork; the question (Ibelieve)mustbecompound.Aquestionthatworksis:“Isitthecasethatyouareaknaveandthatyesisyouranswertothisquestion?”If theoneaddressed is aknight, then it certainly isnot thecasethatheisaknaveandthatyesishisanswer,becauseitisnoteven

thecasethatheisaknave!Soifheisaknight,thecorrectanswertothe question is no; hencehe, being a knight,will answerno. So aknightcananswerno,buthecannotansweryes.Now, suppose the one addressed is a knave. Suppose that heanswers yes. Then it is true that he is a knaveand that yes is hisanswer,hencethecorrectanswertothequestionisyes,whichmeansthat the knave answers correctly, which a knave cannot do!Thereforeaknavecannotansweryes to thisquestion.Suppose theknave answers no. Then it is false that he is a knave and that heansweredyes (because it is false thathehas answeredyes); hencehis no answerwas correct,which is again impossible for a knave!Thereforeaknavecannotanswereitheryesornotothisquestion.

Test4.Aquestionthatworksis:“Isitthecasethateitheryouareaknightorthatyesisyouranswertothisquestion?”Note: Any statement of the form: “Either this or that” is to becountedtrueifthe“this”andthe“that”arebothtrue.Forexample,if a college entrance requirement states that an entering freshmanmust have had either a year of mathematics or a year of foreignlanguage, thecollegecertainlywon’texcludeanapplicantwhohashadboth!Andsoasusedthroughoutthisbook,either-orwillmeanatleastone(andpossiblyboth).Nowlet’sseewhythequestionworks:Supposetheonequestionedisaknight.Thenitiscertainlytruethateitherheisaknightorhisanswerisyes.Henceyesisthecorrectanswertothequestion,andsotheknightcanansweryes(buthecan’tanswerno!).Suppose,ontheotherhand,youaredealingwithaknave.Ifheanswersyes,thenitistruethateitherheisaknightoryesishisanswer(becauseyesishis answer), hence the knave answered correctly, which isimpossible. Suppose the knave answers no. Then it is neither thecasethatheisaknightnorthatheansweredyes,andsothecorrectanswertothequestionisno,whichmeansthattheknaveansweredcorrectly!Soweseethataknavecouldnotanswereitheryesorno.

Test5.Aquestionthatworksis:“Isitthecasethatyouareeithera

knight who will answer no to this question or a knave who willansweryes?”Supposeheisaknight.Ifheanswersnothenheisaknightwhohasansweredno;henceyesisthecorrectanswertothequestionandthe knight has given the wrong answer, which cannot be. If theknightanswersyes,thenheisneitheraknightwhohasanswerednonoraknavewhohasansweredyes(sinceheisnotaknaveatall),andthereforenoisthecorrectanswerto thequestion. Itcannotbethat theknighthas againgivena false answer, so aknight cannotanswerthequestionatall.Asforaknave,supposeheanswersyes.Thenheisaknavewhohas answered yes, so that means he is either a knight who hasanswered no or a knave who has answered yes (he is in fact thelatter).Thismeansthatyesisthecorrectanswertothequestion,andtherefore a knave has answered correctly, which is not possible.Supposeheisaknavewhoanswersno.Thenheisneitheraknightwhohasanswerednonoraknavewhohasansweredyes;hencethecorrect answer to the question is no, so again the knave hasanswered correctly, which he cannot do. And so a knave can’tanswerthequestioneither.Note: An alternative question that does the job is: “Are you thetypewhocouldclaimthatnoisyouranswertothisquestion?”Those readers who recall the Nelson Goodman principle,explained in Chapter 4, can quickly see why this question works.(The Goodman principle is that whenever you ask a knight or aknavewhetherheisthetypewhocouldclaimsuch-and-such,ifheanswersyes, thesuch-and-suchmustbe true; ifheanswersno, thesuch-and-suchmustbefalse.)Supposethisprincipleisappliedtotheabovequestion(wherenowthesuch-and-suchis“Noisyouranswertothisquestion”).Ifheanswersyes,thennoishisanswer,whichisabsurd. If he answers no then yes is his answer, which is absurd.Thereforehemustremainsilent.

Alexander was allowed to consult Annabelle on these problems,and together theyworked out all five of them correctly. The next

daytheybroughttheiranswerstotheKing.“Hm!” he said, after he had checked the correctness of their

solutions. “I’m not sure I should let you go yet! Tomorrow I haveelevenmoretestsforyou.”Well,itsohappenedthattheKingdiedthatnight.Somehistorians

haveclaimedthathediedofaheartattack;others,thathediedofabadconscienceforplayingfoul.Athirdgroupmaintainsthathediedof a heart attack caused by a bad conscience. This is not an easymattertosettle!Atanyrate,thenewkingwhoascendedthethronethenextdaywasafair-mindedindividualwhoinsistedonhonoringthe old king’s promise, and so the couple were released. Theyhappilysetsailonceagainforhome,arrivedwithoutfurthermishap,andwerejoyfullywelcomedbythepeopleoftheirisland.Now, there are several historianswho havewondered justwhat

theeleventestswerethatthedeceasedKinghadinmind!Ofcoursethe matter can never be settled (unless there is an afterlife—as alogicianImustthinkofallpossibilities!),butIdohaveatheorythatIwouldliketotellyou.In reading the chronicles where I found this story, I thought it

ratheroddthattheKinggavejustfiveteststothecoupleinsteadofsixteen;thereareelevenotherpossibilitiesinasimilarveinthatheleft out, and my theory is that these are the very ones he wasplanningtospringthenextday.Yousee,therearefourpossibilitiesforwhataknightcoulddowiththequestion(hecouldansweryesbutnotno,orhecouldanswernobutnotyes,orhecouldanswereither yes or no, or he could answer neither yes or no), andwitheach of these possibilities there are four possibilities for what aknave can do. Thus there are sixteen possibilities altogether. Theinteresting thing is that each of these sixteen possibilities can berealizedbyaquestion!Fiveof themhavealreadybeenconsidered,and Ihavechecked that for eachof theeleven requirements listedbelow (I have numbered them 6 through 16), there is indeed aquestionsatisfyingthatrequirement.

6. A knave could answer yes but not no, and a knight could not

answeratall. 7. A knave could answer no but not yes, and a knight could notansweratall.

8.Aknavecouldansweryesbutnotno,andaknightcouldanswereitherway.

9.Aknavecouldanswernobutnotyes,andaknightcouldanswereitherway.

10.Aknightcouldansweryesbutnotno,andaknavecouldanswereitherway.

11.Aknightcouldanswernobutnotyes,andaknavecouldanswereitherway.

12.Bothknightsandknavescananswereitherway.13.Bothtypescananswerno;neithertypecanansweryes.14.Bothtypescanansweryes;neithertypecananswerno.15.Aknightcananswernobutnotyes;aknavecanansweryesbutnotno.

16.Aknightcanansweryesbutnotno;aknavecananswernobutnotyes.

Thereadermighthavefuntryingtheseasexercises.Igiveanswersat the end of this chapter (but I do not give proofs that theseanswers work; by now the reader should be able to supply theseproofs).

DISCUSSION

Twelveofthequestionsinvolved(includingthefirstfive,whichthereader already knows) have the curious property that their correctanswer cannot be determined until the actual answer is alreadygiven. (I can’t see how questions without this property couldpossiblywork!) They also have the property of self-reference—eachquestionisaquestionabouttheveryquestionitself!

Another example of a self-referential question is: “Does thisquestioncontainexactlysevenwords?”Theanswerisobviouslyyes.Curiouslyenough,yes isnot the answer to thequestion “Does thisquestioncontainsevenwords?”Anotherexampleofaself-referentialquestionis:“Isthisquestionasillyquestion?”(Myguessisthatthecorrectanswerisyes.)ThereisastorytoldthattheGreekphilosopherEpimenidesonce

made a pilgrimage to meet the Buddha. Epimenides asked him:“What is thebestquestion thatcanbeasked,andwhat is thebestanswerthatcanbegiven?”TheBuddhareplied:“Thebestquestionthatcanbeaskedisthequestionyouhavejustasked,andthebestanswerthatcanbegivenistheanswerIamgiving.”Self-referenceplaysaprominentroleinmodernlogicaltheoryand

computer science. It is also addictively fascinating! It is central intheproofofGödel’s famous incompleteness theorem.(TheSorcererwillhavemuchtosayaboutthislaterinthebook.)

AnswerstoExercises6through16

6.Isitthecasethatyouareaknightandthatnoisyouranswertothisquestion?

7.Isitthecasethateitheryouareaknaveorthatnoisyouranswertothisquestion?

8.Isitthecasethatyouareaknightandthatyouwillansweryes?

9.Isitthecasethateitheryouareaknaveoryouwillansweryes?

10.Isitthecasethateitheryouareaknightoryouwillanswerno?

11.Isitthecasethatyouareaknaveandthatyouwillanswerno?

12.Willyouansweryes?

13.Areyouaknave?

14.Areyouaknight?

15.Doestwoplustwoequalfive?

16.Doestwoplustwoequalfour?

PARTII

PUZZLESANDMETAPUZZLES

8MEMORIESOFTHESORCERER’SUNCLE

AFEWMONTHSafterarrivinghome,ourcouplefelttheurgetovisitKingZorn’s islandagain—particularlytoseetheSorcerer,ofwhomtheyhadbecomequite fond.Theyplanned to stayaweekor two, littlerealizingthatthingswouldbecomesointerestingthattheywouldbeunabletotearthemselvesawayforseveralmonths!Once arrived on the island, they lost no time in calling on theSorcerer, who was delighted to see them. He was in a happilynostalgicmood,andspenttheafternoonreminiscingabouthisuncle.“You really should have met him,” he said to Annabelle andAlexander. “Hewasoneof themost interestingpeople Ihaveeverknown.Itwashe,infact,whogotmeinterestedinlogic.”“Howdidhedothat?”askedAlexander.“Quite naturally,” replied the Sorcerer. “He gaveme all sorts offascinatingpuzzlesthroughoutmyboyhoodandadolescence.Irecallwhen Iwas extremelyyoung—about six—myunclehad fourdogs,withwhomIusedtoplayagreatdeal.Onedayhegavemeapuzzleabout them,and Idon’tknowwhether thestorywas trueormadeup—butthisisthepuzzle:

•1•

“‘Ioncesetoutabowlofbiscuitsformydogs.First,theoldestonecamebyandatehalfofthemandonemore.Then,theseconddogcame by and ate half of what he found and onemore. Then, thethird one came by and ate half ofwhat she found and onemore.Then, the fourth and littlest one came by and ate half ofwhat hefoundandonemore,andthebiscuitswerethenallgone.Howmanybiscuitswereinthebowltobeginwith?’“Thatistheproblemmyunclegaveme.”

What is theanswer? (Fromthispointon, solutionsaregenerallygivenattheendofeachchapter.)

•2•

“I remember,” the Sorcerer continued, “once my uncle asked me:‘Whichismore,sixdozendozenorahalfadozendozen?’”“That’sobvious,”saidAnnabelle.“Ofcourseitis,”saidAlexander.“True,” said the Sorcerer, “but many people nevertheless get itwrong.”Whatisthecorrectanswer?

•3•

“Welivednearafarm,”saidtheSorcerer.“Thefarmersoldmostofhis produce to wholesalers, and some he sold retail at a littlevegetablestand.Myuncletoldmethatthefarmersold90percentofhis produce wholesale and 10 percent retail, but he got twice asmuchperitemretailaswhenhesoldtheitemwholesale.Myunclethen asked me if I could figure out what percentage, or whatfraction,ofhisgrossincomecamefromthestand.”Whatistheanswer?

•4•

“Anothersimplearithmeticalone:SupposeyouandIhavethesamenumberofcoppercoins.HowmanymustIgiveyousothatyouhavetenmorethanI?”

•5•

“Amanoncebrought toa jewelersixchainsof five linkseach.Hewanted tomake themall into one large, closed circular chain andasked the jeweler how much it would cost. The jeweler replied:‘EverylinkIcutopenandthenclosecostsadollar.Sinceyouwanta

circular chain and you have six chains, then it will cost you sixdollars.’“‘No,’saidtheman,‘thejobcanbedoneforless,’“Themanwasright,”saidtheSorcerer.“Why?”

•6•

“Myuncleoncefooledmebytellingmethatacertainbeggarhadabrother and the brother died. Butwhile the brotherwas alive, heneverhadabrother.Howisthistobeexplained?”

•7•

“My uncle also told me about a very absentminded professor heknewwhohadthreedaughters.Heonceaskedtheprofessorhowoldhisdaughterswere.Theprofessorreplied:‘I’mnottoosure.Iknowthatoneofthethreeistheyoungest.’“ ‘That’snot too surprising,’myuncle replied. ‘Whichone is theyoungest?’“‘Ireallydon’tknowforsure;it’seitherAliceorMabel.’“‘Well,whichoneistheoldest?’“ ‘I’mnot sure about that either. I recall that eitherAlice is theoldestorLillianistheyoungest,butIcan’trememberwhich.’”“Which one,” asked the Sorcerer, “is the oldest and which theyoungest?”

•8•

“Another family problemmy uncle toldme: A certain boy had asmanybrothersassisters.HissisterGracehadtwiceasmanybrothersassisters.Howmanybrothersandsisterswerethereinthefamily?”

•9•

“Here’satrickyone,”saidtheSorcerer.“If5catscancatch5micein5minutes, howmany cats are required to catch 100mice in 100

minutes?”

•10•

“Here is a simple problem my uncle told me that many peopleneverthelessgetwrong.Acertainmillertookastollone-tenthoftheflour thatheground foracustomer.Howmuchdidhegrind foracustomerwhohadjustonebushelafterthetollhadbeentaken?”

•11•

“MyuncleoncetoldmeofanancientpuzzlepropoundedinA.D.310by someone namedMetrodorus. It concerns oneDemochares,whohas livedone-fourthofhis life as aboy,one-fifthas ayouth,one-thirdasamaninhisprime,andthirteenyearsinhisadvancingage.Howoldishe?”

•12•

“MyuncleoncegavemeaproblemIspentmanyhoursover.HadIusedmyhead,Iwouldhaverealizedthetruthofthematterinaboutaminute.Theproblemwasthis:“We are given an 8× 8 board divided into 64 squares, and 2diagonallyoppositecornersareremoved.

“Wearegivenapileofdominoes,andeachdominocancover2squares. The problem is to pave the surface. The whole of each

dominomustbeused—that is,eachdominomust lieon2squares.Howcanthisbedone?“The principle involved in the solution provides a beautiful

exampleofwhatconstitutesamathematicalproof.”

•13•

“My uncle told a delightful incident about the American puzzle-masterSamLloyd.Lloydwasalsoamagician,andhedidonetrickwithhistwelve-year-oldsonaboardshipthatfooledevenmagicians.Theboywasblindfoldedandhadhisback turned to theaudience.One of the spectators, not a confederate, took a deck of cards,shuffled it, and showed the faces one by one to the father. Eachtime,theboycorrectlynamedthecard.Thiswasdonebeforeradiowas invented, so radio signalsarenot the solution.As I said, evenmagicianswerefooled.OfallthetricksIknow,thisonestrikesmeasthecleverest.Howwasitdone?”

•14•

“Do you know of the logician Raymond Smullyan?” asked theSorcerer.“Neverheardofhim,”repliedAnnabelle.“Whoishe?”“I’veneverheardofhimeither,”saidAlexander.“No matter,” replied the Sorcerer. “He invented many puzzles

whichmyunclewasfondof.Oneofthemisaboutamanwhohadtwoten-gallonjars.Sixgallonsofwinewasinone,andsixgallonsofwater in theother.Hepouredthreegallonsofwine into thewaterjarand stirred; thenpoured threegallonsof themixtureback intothewinejarandstirred;thenthreegallonsofthemixturenowinthewine jar into the water jar, and so on, back and forth, until theconcentration of wine in the two jars was the same. How manypouringswerenecessary?”

•15•

“My uncle also toldme of a nice variant Smullyan devised of theclassical colored-hats problem. Do you know the colored-hatsproblem?”“I think I once heard it, or something like it, but I don’t really

rememberit,”saidAnnabelle.“Idon’tbelieveI’veeverheardit,”saidAlexander.“Well, there are many versions,” said the Sorcerer. “One of the

simplerones is this:Threemen—A,B,andC—areblindfoldedandtold thateithera redoragreenhatwillbeplacedon theheadofeach.Theblindfoldsarethenremoved,soeachofthethreecanseethecolorsofthehatsoftheothertwo,butneithercanseethecolorof his own.Actually, all three are given green hats. The three arethenaskedtoraiseahandif theyseeat leastonegreenhat,soallthree raise their hand. Then they are asked to lower their hand ifthey knowwhat they have. After some pause the cleverest of thethreelowershishand.Howdoesheknowthecolorofhishat?”

•16•

“Now,hereisSmullyan’svariant.Thistimethethreepeople—A,B,andC—areofequalintelligence;infactallthreeareperfectlogiciansinthattheycaninstantlydeduceallconsequencesofanygivensetofpremises. It is also common knowledge among the three that allthreeareperfectlogicians.Therearefourredstampsandfourgreenstampsavailable.Whiletheireyesareclosed,twostampsarepastedontheforeheadsofeachandtheremainingtwostampsareputintoadrawer.Afterthethreeopentheireyesandeachseestheforeheadsoftheothertwo,Aisaskedwhetherheknowswhathehas,andhesaysno.ThenBisaskedifheknowswhathehas,andhesaysno.ThenCisasked,andhesaysno.ThenAisaskedasecondtime,andhe says no. Then B is asked a second time, and he says yes.HowdoesBknowwhathehas?”

•17•

“Oneofmyuncle’sfavoritesamongSmullyan’sproblemswasthis:In

1918,onthedaythatthearmisticeofWorldWarIwassigned,threemarriedcouplescelebratedbyhavingdinnertogether.Eachhusbandisthebrotherofoneofthewivesandeachwifeisthesisterofoneofthe husbands; that is, there are three brother-sister pairs in thegroup.Wearegiventhefollowingfivefacts:

1.Helenisexactlytwenty-sixweeksolderthanherhusband,whowasborninAugust.2.Mr.White’ssister,whoismarriedtoHelen’sbrother’sbrother-

in-law,marriedhimonherbirthday,whichwasinJanuary.3.MargueriteWhiteisnotastallasWilliamBlack.4.Arthur’ssisteristallerthanBeatrice.5.Johnisfiftyyearsold.

“WhatisthefirstnameofMrs.Brown?”

Solutions

1.Thisproblemismosteasilysolvedbyworkingitbackwards.Howmanybiscuits did the last dog find in order that by eating half ofthemandonemore,allweregone?Theonlypossibilityistwo.Thedogbeforethatmusthavefoundsix;thedogbeforethatmusthavefoundfourteen,andthefirstdogmusthavefoundthirty.

2. Some people guess that they are the same. This, of course, iswrong;sixdozendozenis6×144,whereashalfadozendozenisthesameassixdozen(notsixdozendozen),whichis72.Actually,sixdozendozenisthesameashalfadozendozendozen.

3. Let’s say each item sells for a dollarwholesale and two dollarsretail. Then, out of ten items sold, nine are sold for one dollar,yieldingninedollars, andone is sold for twodollars. So ten itemsbring in eleven dollars, two of which come from the retail stand.And so two-elevenths of his gross income comes from the retailstand.

4.Theansweris5,not10!

5.Whatthejewelerhadinmindwastoopenoneendlinkofeachofthesixchainsandthen join themtogether to forma largecircularchain.Thiswouldindeedcostsixdollars.Buteachofthesixchainshadonlyfivelinksapiece,andsoonecouldcutopenthefivelinksofanyone chainanduse those five loose links to join the remainingfivechainsintoacircle.Thus,thejobcouldbedoneforfivedollars.

6.Thebeggarwasawoman.

7. Since either Alice is the oldest or Lillian is the youngest, it isimpossiblethatAliceistheyoungest,forifshewere,itcouldneitherbetruethatAliceistheoldestnorLilliantheyoungest.Thus,Aliceisnottheyoungest.SinceeitherAliceorMabelistheyoungest,thenitis Mabel who is the youngest. Therefore, Lillian cannot be theyoungest,but either she is theyoungestorAlice is theoldest, andsincesheisnottheyoungest,thenAliceistheoldest.Andso,Mabelistheyoungest,Lillianisthemiddleone,andAliceistheoldest.

8.Therewerefourbrothersandthreesisters.

9.Theanswerisfive.

10.Thecommonwrongansweris11⁄10;thecorrectansweris11⁄9.Letuscheck:1⁄10of11⁄9is1⁄10×10⁄9=1⁄9.Sothemanbrought11⁄9bushelstothemiller,gavehim1⁄10ofit,whichis1⁄9ofabushel,andkepttheremaining9⁄10of10⁄9forhimself,whichisonebushel.

11. Democharesmust have been sixty. This can be found by trialanderror,orbysolvingthealgebraicequationx⁄4+x⁄5+x⁄3+13=x,wherexishisage.

12. Imagine the squares colored alternately red and black like acheckerboard.The twocorner squares thathavebeen removedare

ofthesamecolor—say,red.Thus,therearenowmoreblacksquaresthanredsquares—32blacksquaresandonly30redsquares.Now,adominocoversoneblackandoneredsquare,sothetotalnumberofblacksquarescoveredmustbethesameasthetotalnumberofredsquares covered. Since there are more black squares than redsquares,asolutionisimpossible!Thisisoneofthesimplestandneatestproofsoftheimpossibility

of doing something that I know. Of course, if the corner squareswererestoredandanytwosquaresofthesamecolorwerecutout,asolutionwouldstillbeimpossible.Ithasbeenproved(Ihear)thatifanytwosquaresofdifferentcolorsareremoved,regardlessofwheretheyare,asolutionispossible.

13.Thechildneversaidaword;SamLloydwasaventriloquist!I have been a magician for many years and know all sorts of

ingeniousmagicalsecrets.ButthisstrikesmeasthecleverestthinginmagicthatIknow!ThefunnythingisthatanelderlygentlemansaidtoSamLloyd:“Youshouldn’tstraintheboy’smindthatmuch;it’snotgoodforhim!”

14. It cannot be done in any finite number of pourings! To beginwith,theconcentrationofwineinthewinejarisobviouslygreaterthantheconcentration in thewater jar.After the firstpouring, thewaterjarisweaker(inwine)thanthewinejar.Inthenextpouring,theweakerisbeingpouredintothestronger,andsothewaterjarisstillweakerthanthewinejar.Thensomeofthestrongerispouredinto theweaker,andso thewater jar is stillweaker than thewinejar.Atanystage,thewaterjarisweakerinwinethanthewinejar,and so at the next stage, regardless of which jar is poured intowhich,thewaterjarisstillweakerthanthewinejar.Of course, this analysis is purely mathematical, assuming that

wine is a totally homogeneous substance, rather than ultimatelycomposedof discreteparticles.As far as the real physicalworld isconcerned,Idon’tknowhowmanypouringswouldbenecessaryforthetwoconcentrationstobeobservablyindistinguishable.

15.Let’ssayitwasAwhowasthemostintelligent.Hereasonedthisway:“Supposemyhatwerered.ThenBwouldknowthatifhetoowerered,Ccouldn’thaveraisedhishand,indicatingthathesawatleastonegreen.AndsoifIwerered,Bwouldknowthatheisgreen.ButBdoesn’tknowthatheisgreen.Hence,Ican’tbered;Imustbegreen.”

16.Thelogicinthisproblemisfarmoreintricate.Tobeginwith,itisobviousthatCdidnotseefourstampsofthesamecolor,sincehewould thenhave knownwhathehad.Nor couldChave seen tworedsononeofAorBandtwogreensontheother, forsupposehesawtworedsonBandtwogreensonA.Then,Cwouldhaveknownthathecouldn’thave tworeds, sinceAwouldhave thenseen fourredsandknownwhathehad.NorcouldChavetwogreens,sinceBwouldhavethenseenfourgreensandknownthathehadtworeds.ThisprovesthatitisnotpossiblethatbothofA’sstampsarethe

samecolorand thatbothofB’s stampsarealso the samecolor; atleastoneofthetwohastohaveredandgreenstamps.AandBbothrealizedthisafterCsaidthathedidn’tknowwhathehad,andeachrealizedthattheotherrealizedthis.Therefore,onthesecondround,B realized that if his colors were the same, then A would haveknownonthesecondroundthathewasred/green,whichhedidn’t.Andso,Brealizedthathemusthimselfbered/green.

17.Step 1. SinceMr.White’s sister ismarried toHelen’s brother’sbrother-in-law, then she is not married to Helen’s brother. Also,MargueriteisMrs.White.Usingthesetwofacts,wewillseethatMr.White’ssistermustbeHelen.Now, Helen’s brother is married either to Marguerite or to

Beatrice.IfitisMarguerite,thenHelen’sbrotherisMr.White(sinceMr. White is married to Marguerite), in which case Helen is Mr.White’s sister. On the other hand, suppose it is Beatrice to whomHelen’s brother ismarried. Then Beatrice is notMr.White’s sister(becauseMr.White’s sister isnotmarried toHelen’s brother), andsinceMarguerite is notMr.White’s sister (she is his wife) then it

must again be Helen who is Mr. White’s sister. This proves thatHelenisMr.White’ssister.Step2.HeleniseitherMrs.BrownorMrs.Black(sinceMarguerite

isMrs.White).WenowshowthatifHelenisMrs.Brown,thenshemustbemarriedtoJohn,whichwewilllatershowcannotbe.Well, supposeHelen isMrs. Brown. Then Beatrice isMrs. Black

(sinceMarguerite isMrs.White).HenceBeatrice isnotMr.Black’ssister,norissheMr.White’ssister(sinceHelenis).HenceBeatriceisMr.Brown’ssister.Now,Arthur’ssisterisnotBeatrice(sheistallerthanBeatrice),butMr.Brown’ssisterisBeatrice;henceArthurisnotMr.Brown.HeisalsonotMr.Black(Williamis);henceArthurisMr.White.ThenJohn,whoisnotMr.WhitenorMr.Black(Williamis),must be Mr. Brown. Then, by our assumption that Helen is Mrs.Brown,HelenmustbemarriedtoJohn.ThisprovesthatifHelenisMrs.Brown,sheismarriedtoJohn.Step 3. But Helen can’t be married to John for the following

interestingreason:WehavealreadyprovedthatHelenisMr.White’ssister(Step1).

AndwehavebeengiventhatHelen(asMr.White’ssister)wasborninJanuary.ThusHelenwasborninJanuary,herhusbandwasbornin August, and Helen is exactly twenty-six weeks older than herhusband. Using a calendar, we can see that the only way this ispossibleisthatHelenwasbornonJanuary31andherhusbandwasbornonAugust1,andthereisnoFebruary29inbetween!Therefore,Helen and her husband were not born in a leap year. But John,beingfiftyin1918,wasborninaleapyear.ThereforeJohncannotbeHelen’shusband.But we saw (Step 2) that if HelenwereMrs. Brown, then John

wouldbeHelen’shusband.Sinceheisnot,thenHelencan’tbeMrs.Brown.Also,MargueriteisnotMrs.Brown(sheisMrs.White),andsoitisBeatricewhoisMrs.Brown.

9THEPLANETOG

“DIDYOURUNCLEinventanypuzzlesofhisown?”askedAlexander.“Heavens,yes!Averygreatmany,”repliedtheSorcerer.“Didhepublishanyofthem?”askedAlexander.“You remind me,” said the Sorcerer, “of the story of the twoAmerican professors standing in Florence before a statue of Jesus.‘Now,therewasagreatteacher!’saidone.‘Yes,’saidtheother,‘butheneverpublished!’“This obsession with publishing! No, my uncle never publishedanyofhisproblems;unfortunately,heneverevenwrotethemdown.He would improvise them on the spot—mainly, I think, to amuseme. Then he would forget them completely. Fortunately, I haverememberedmanyofthem.”“Willyoutellussome?”askedAnnabelle.“Surely,”repliedtheSorcerer.“Oneparticulargroupoccurstomenow.TheyareallaboutaplanetmyunclecalledOg.”“Istherereallysuchaplanet?”askedAlexander.“I doubt it,” replied the Sorcerer. “I’m pretty sure hemade thename up. Anyway, I think he was influenced by science fiction,particularlybyThePrincessofMars,abookbyEdgarRiceBurroughs.AsinBurroughs’sMartianfantasy,myuncle’splanetOgisinhabitedbytworaces—thegreenpeopleandtheredpeople.Also,thepeoplefromthenorthernhemisphere—thatis,thosewhowerebornthere—areverydifferent fromthesoutherners, thoseborn in thesouthernhemisphere.”“Howaretheydifferent?”askedAlexander.“Well,”repliedtheSorcerer,“thecuriousthingaboutthisplanetisthat the green northerners always tell the truth and the rednorthernersalwayslie,whereasthegreensouthernerslieandtheredsouthernerstellthetruth.”

“Oh,dear!”saidAnnabelle,“Icanseethatthesepuzzlesaregoingtobeverycomplicated!”“Somearecomplicatedandsomearesimple,”repliedtheSorcerer.“Let’sstartwithasimpleone.”HerearesomeofthepuzzlestheSorcererrecalled.

1•OneDarkNight

Suppose youmeet an inhabitant on the street one dark night andcannot seewhetherhe is redor green.Also, youdon’t know fromwhichhemispherehehails.What singleyes-noquestion couldyouaskhimthatwoulddeterminehiscolor?

2•AnotherDarkNight

OnceatravelerfromourplanetEarthvisitedtheplanetOgandmetanativeonthestreetoneverydarknight.“Areyoured?”heasked.Thenativedidn’tanswer.“Areyouasoutherner?”thetravelerasked.Againthenativedidn’tanswer.“Areyounotgoingtosayanything?”askedthetraveler.Beforewalkingawaythenativereplied:“IfIanswerednotobothofyourfirsttwoquestions,Iwouldbelyingatleastonce.”Is it possible from this to determine the native’s color andwhathemisphereheisfrom?

3•AnotherDarkNight

Onanotherdarknight,atravelerfromEarthmetanativeandaskedhim:“Areyoured?”Thenativesaidthathewas.Thenthestrangeraskedhim fromwhichhemispherehe came.Thenative replied, “Irefusetotellyou,”andwalkedaway.Fromwhichhemispherewashe?

4•InBroadDaylight

Onebrightday,atravelerfromEarthmetanOgnativewhosaid,“Iam a green northerner.” The traveler, though good at logic, couldnot deduce whether the native was a northerner or a southerner(though,ofcourse,hesawwhatcolorhewas).Whatcolorwasthenative?

5•APointofLogic

There is adifferencebetweenanativeofOg saying “I amagreennortherner,”andhismakingtwoseparatestatements—onethatheisgreen, and the other that he is a northerner. In the one case, itcannotbededucedwhatthenativereallyis,butintheothercase,itcanbe.Inwhichcasecanitbe,andinthatcase,whatisthenative?

6•AnotherQuestionofLogic

Supposeanativesays:“IfIamgreen,thenI’masoutherner.”Canitbededucedwhatcolorheis?Canitbededucedfromwhichhemispherehehails?

7•ADuo

ThisproblemconcernstwoinhabitantsofOg,arednorthernerandasoutherner (possibly red and possibly green). One of the two isnamedArkandtheotherBark.Theymadethefollowingstatements:

Ark:BarkandIarethesamecolor.Bark:ArkandIaredifferentcolors.

Which one is right?Which one is the northerner?What color isthesoutherner?

8•AnotherDuo

TwonativesnamedOrkandBorkmadethefollowingstatements:

Ork:Borkisfromthenorth.

Bork:Orkisfromthesouth.Ork:Borkisred.Bork:Orkisgreen.WhatcolorisOrkandwhereishefrom?AndBork?

9•TwoBrothers

AnytwosiblingsontheplanetOgarenecessarilythesamecolor,butnotnecessarilyfromthesamehemisphere.(Themothermighthavegivenbirthtooneandthen,sometimelater,crossedtheequatorandgivenbirthtotheother.)Buttheydefinitelyarethesamecolor.Twobrothers,ArgandBarg,oncemadethefollowingstatementsaboutthemselves:

Arg:Wewerebornindifferenthemispheres.Barg:Thatistrue.Weretheylyingortellingthetruth?

10•Brothers?

Inthismorecomplicatedsituation,twonatives,OrgandBorg,madethefollowingstatements:

Org:Borgisfromthenorth.Borg:Infactbothofusarefromthenorth.Org:Thatisnottrue!Borg:OrgandIarebrothers.

Are the two really brothers?What color is each?Where is Orgfrom?WhataboutBorg?

11•ATrial

Anativewasbeingtriedforrobbery.Itwasimportantforthecourttodecidewhetherhewasanorthernerorasoutherner,sinceitwas

known that the theftwas committed by a northerner. The defenseattorneydidnotwanthis own color to be known, sohe appearedmaskedandglovedincourt.Toeveryone’ssurprise,heclaimedthathe and the defendant were both northerners. (One would haveexpected him to have made a statement proving the defendant’sinnocence!) To everyone’s even greater surprise, when theprosecutor cross-examined the defense attorney, the latter claimedthathehimselfwasnotanortherner.Howdoyouexplainthis?Canthecolorofthedefenseattorneybe

established?Canitbedeterminedwhetherthedefendantwasguiltyorinnocent?

12•WhatAreThey?

Twoinhabitants,AandB,wereofdifferentcolorsandfromdifferenthemispheres.Theymadethefollowingstatements:

A:Bisanortherner.B:Aisred.

WhatcolorsareAandBandwherearetheyfrom?

13•WhatColorIsSnarl?

A southerner named Snarl once claimed that he overheard aconversationbetweentwobrothers,AandB,inwhichAsaidthatBwasanorthernerandBsaidthatAwasasoutherner.WhatcolorisSnarl?

14•AnotherDuo

Two natives, A and B, of different colors made the followingstatements:A:Bisanortherner.B:Bothofusarenortherners.WhatareAandB?

15•IsThereaKing?

TheanthropologistAbercrombieoncevisitedtheplanetOgandwascurious to find out if the planet had a king.Hehad the followingconversationwithanative:Abercrombie:Iwastoldthatyouonceclaimedthatthisplanethasnoking?Isthattrue?Native:No,Ineverclaimedthat!Abercrombie: Well, have you ever claimed that this planet does

haveaking?Native:Yes,Ihaveclaimedthat.Can it be determined whether the native was truthful or lying?

Canitbedeterminedwhetherornottheplanethasaking?

16•IsThereaQueen?

Abercrombiewas also interested in findingoutwhether theplanethas a queen. He found the answer when he came across twobrothers,AandB,whomadethefollowingstatements:A:Iamanortherner,andthisplanethasnoqueen.B:Iamasoutherner,andthisplanethasnoqueen.Isthereaqueenornot?

17•WhatColorIstheKing?

ThispuzzleisoneoftheSorcerer’sfavorites.TheKingofOgalwayswearsamaskandgloves,andnoneofhis

subjects ever learns his color. His brother, the Duke of Snork, issimilarlydisguised.One day, Abercrombie came to the court and the King and his

brotherdecidedtotesttheirvisitor’sintelligence.First,Abercrombiehad to swear that if he should find out the color of the King, hewouldnevertellanyoftheinhabitants.Thenhewasusheredintoaroom in which the King and his brother were both seated—bothmasked and gloved, of course—and one of them said: “If I am agreennortherner,thenIamtheKing.”Theotheronethensaid:“IfI

am either a green northerner or a red southerner, then I am theKing.”WhatcoloristheKing?

Solutions

1.Aquestion thatworks is:“Areyouanortherner?” Ifheanswersyes,heisgreen;ifheanswersno,heisred.Ileavetheprooftothereader.Of course, the traveler could equallywell have found out if the

nativewasanorthernerorasouthernerbyasking:“Areyougreen?”Thesituationhastheprettysymmetrythattofindoutifthenativeisgreen, you ask him if he is northern, and to find out if he isnorthern,youaskhimifheisgreen.

2. The native said in effect that he is either red or southern (andpossiblyboth)—inotherwords,thatheisnotagreennortherner.Agreennorthernerwouldnot lie and say that.Also, neither a greensouthernernora rednortherner couldmake the truthful statementthatheisnotagreennortherner.Andsothenativemusthavebeenaredsoutherner.

3.Whenthenativesaid“Irefusetotellyou,”hespoketruly,sincehereallydidrefusetotellthetraveler.Andsothenativeistruthful.Hence,hisfirstanswerwastruthful;sohereallyisred.Sinceheistruthfulandred,hemustbefromthesouth.

4.Ifthenativehadbeenred,thetravelerwouldhaveknownthathewasanortherner(sincearedsouthernerwouldneverclaimtobeagreen northerner), but since he didn’t know, then the nativemusthavebeengreen(possiblyatruthfulgreennorthernerorpossiblyalyinggreensoutherner).

5. In the case of a native who claims to be a green northerner,nothingcanbededucedotherthanthatheisnotaredsoutherner.

Butitisaverydifferentstoryifanativefirstclaimstobegreenandthenmakesaseparateclaimthatheisanortherner.Forsupposehemakesthesetwoseparateclaims.Then,theyareeitherbothtrueorbothfalse.Iftheyarebothfalse,thenhemustbered(sincethefirstclaimisfalse)andhemustbeasoutherner(sincethesecondclaimisfalse). It is impossible for a red southerner to make false claims;therefore, both claims must be true. So the native is a greennortherner.

6.Tosaythatifheisgreenthenheisasoutherner,istantamounttosayingthatheisnotagreennortherner.Andsothenative,ineffect,claimedthathewasnotagreennortherner.Onlya redsouthernercouldmakesuchaclaim.(ThisisreallythesameasProblem2;onlythewordingisdifferent.)

7. Since the twodisagree, one is telling the truth and theother islying. The red northerner must be lying; hence the southerner istellingthetruth,andthereforehemustbearedsoutherner.Sothetworeallyare thesamecolor;whichmeansArktold thetruthandBark lied. So Ark is the red southerner and Bark is the rednortherner.

8.IfOrk’sstatementsaretrue,thenBorkisarednortherner,andifOrk’sstatementsarefalse,Borkisagreensoutherner.IneithercaseBorkisaliar.Thus,Bork’sstatementsarebothfalse,andOrkisarednortherner.HenceOrklied,sohisstatementsarebothfalseandBorkisagreensoutherner.

9. If itwerereally true that theywere fromdifferenthemispheres,thenwewouldhavetheimpossiblesituationofanorthernerandasoutherner of the same color agreeing with each other, which isimpossible.Sowhattheysaidwasnottrue.

10.SinceOrgcontradictedBorg,oneislyingandtheotheristellingthe truth. Suppose Borg is telling the truth. Then both arenortherners (by Borg’s first statement) and both are brothers (by

Borg’s second statement); hence both are the same color and wehave the impossible situationof twonorthernersof the samecolordisagreeing.Therefore, Borgwasnot telling the truth; he is a liar,anditisOrgwhoistruthful.ThenBorgisanorthernerasOrgtruthfullysaid,buttheyarenot

bothnortherners(sinceBorgliedandclaimedtheywere);henceOrgisa southerner.ThusOrg isa truthful southernerandmustbe redandBorg is a lyingnortherner andalso red.Org andBorg arenotbrothers(sinceBorgsaidtheywere),eventhoughtheyarethesamecolor.

11. Obviously, the defense attorney’s claims cannot both be true,henceheisaliar.Sincehedeniedbeinganortherner,hereallyisanortherner, andmust be a red northerner. Since his claim that heandhisclientarebothnorthernerswasfalseandheisanortherner,then the defendantmust be a southerner.And so the defendant isnotguilty.Thedefenseattorney,despitethestrangenessofhisbehavior,was

notsostupid;hegothisclientacquitted.

12.SinceAandBaredifferentincolorandarealsofromdifferenthemispheres,theymusteitherbothbetellingthetruthorbothlying.Supposetheirstatementswerebothfalse.Then,BwouldbesouthernandAwouldbegreen.SoBwouldbered(sinceAisgreen)andalsoAwouldbenorthern(sinceBissouthern).Hencewewouldhaveared southern B (as well as a green northern A) making falsestatements,which is not possible. Therefore, both statementsmustbe true.HenceB isanorthernerandA is red,and soB isagreennorthernerandAisaredsoutherner.

13. If Snarl’s account were true, we would have the followingcontradiction:SinceAandBarebrothers, theyarethesamecolor.Suppose they are red. IfA is a northerner, he is a red northerner;hence his statement is false and B is really a southerner, a redsoutherner who is therefore truthful and couldn’t have made thefalsestatement thatA isasoutherner.Ontheotherhand, ifA isa

southerner,heisaredsoutherner;hencehisstatementistrueandBreallyisanortherner,arednortherner,thereforeuntruthful.YethewouldtruthfullyhavesaidthatAisasoutherner.Thisisimpossible;hence the brothers cannot be red.A symmetrical argument,whichweleavetothereader,revealsthattheycannotbegreeneither.Anotherwayoflookingatit(whichIbelieveismoreinstructive)

istofirstrealizethatinanylandinwhicheveryinhabitantiseithera constant liar or a constant truth-teller, it is impossible to haveinhabitantXsayingthatinhabitantYislyingandYsayingthatXistruthful (for then Y is assenting to X’s claim that Y lies, whichneitheratruth-tellernoraliarcando).Now,inthepresentproblem,sinceA andB are the same color, ifA calls B a northerner andBcallsA a southerner, then one of the two is effectively calling theothera liarandtheother iseffectivelycallingthe first truthful. (Ifthey are red, A is effectively calling B a liar and B is effectivelycallingA truthful. If theyaregreen,wehave thereverse.)But thiscannotbe.Atanyrate,Snarl’saccountwasfalse,andsinceheisasoutherner,

hemustbegreen.

14. Suppose B told the truth. Then the two are both northerners;henceA’sstatementthatBisanortherneristrue.Wethenhavetheimpossibility of two northerners of different colors both telling thetruth. Thus B did not tell the truth, so at least one of them is asoutherner. Suppose B is a northerner. Then A must be thesoutherner. Also, A told the truth that B is a northerner, hence AmustbearedsouthernerandBwouldbegreen,andwewouldhavetheimpossibilityofagreensouthernertellingthetruth.Therefore,Bisnotanorthernerbutasoutherner.SinceBisasouthernerandlied,hemustbeagreensoutherner.Also,sinceBisasoutherner,Alied,andAisred(sinceBisgreen).Thus,Aisarednortherner.So A is a red northerner and B is a green southerner, and both

lied.

15. It cannot be determined whether the native was truthful or

lying,butitcanbedeterminedwhetherornotthereisaking.Supposethenativeistruthful.Thenhereallydidonceclaimthat

the planet had a king and his claimmust have been true, so theplanetdoeshaveaking.Ontheotherhand,ifthenativeisaliar,thenitisnottruethathe

neverclaimedtherewasnoking;henceheactuallydidonceclaimtherewasnoking,andbeingaliar,thisclaimwasfalse.Hencetheremustbeaking.Thus,regardlessofwhetherornotthenativeistruthful,theplanet

doeshaveaking.

16.Thetwobrothersarethesamecolor.Supposetheyarered.ThenA’sstatementcan’tbetrue,sinceifitwere,thenAwouldhavetobeanortherner (as he claimed) andwewouldhave a rednorthernertellingthetruth,whichcannotbe.Therefore,A’sstatementisfalse.Thus A is red and makes false statements; so A is a northerner.Therefore, if therewere no queen on the planet, then itwould betrue that A is a northerner and there is no queen; hence A’sstatementwouldbetrueafterall,whichitisn’t!Thisprovesthatifthebrothersarered,thentheplanetmusthaveaqueen.Byasymmetricalargument,usingB’sstatementinsteadofA’s, if

both brothers are green, then the planet has a queen. And so theplanethasaqueen.

17. Let A be the first one who spoke and B the second. Is A’sstatementnecessarilytrue?Thatis,ifAisagreennortherner,musthe then necessarily be the King? Well, suppose he is a greennortherner.Then,heistruthful,soitreallyistrue,ashesaid,thatifhe isagreennortherner,he is theKing.Thisprovesthat ifhe isagreennortherner, thenheis theKing.Sincehesaid just that,he istruthful. So nowwe know that A is truthful. From this it doesn’tfollowthatAmustbetheKing;allweknowisthatifheisagreennortherner,thenheistheKing.Butwedon’tknowwhetherornotheisagreennortherner.(Weknowthatheistruthful,buthecouldbearedsoutherner.)

Now,withB’sstatementitisdifferent.Again,B’sstatementmustbe true, because if he were either a green northerner or a redsoutherner,hisstatementwouldhavetobetrue.Henceitwouldbetrue that ifhe is eitheragreennorthernerora red southerner,hewouldhavetobetheKing.Andso,ifheiseitheragreennortherneroraredsoutherner,thenheistheKing.Sincehesaidjustthis,heistruthful; hence he must be either a green northerner or a redsoutherner. But we have already seen that if he is either a greennortherneroraredsoutherner,thenhemustalsobetheKing.AndthisprovesthatheistheKing.Since B is the King, then A is not; hence A cannot be a green

northerner (because if hewere, then hewould be the King, aswehaveshown).SinceAisnotagreennorthernerbutistruthful,thenAmust be a red southerner. Since A is red, so is his brother, theKing.ThustheKingisred(andlikehisbrother,aredsoutherner).

10METAPUZZLES

•1•

“Here isanotherpuzzleabout theplanetOg. It isofaveryspecialtype,”saidtheSorcerer.“Itmightaptlybecalledametapuzzle.“Anativeonthisplanetclaimstobeamarriednortherner.IfItoldyouwhatcolorheis,wouldyouhaveenoughinformationtodeducewhetherornotheismarried?”Annabelle and Alexander went to work on this, and sometimelaterAlexandersaid:“Wedon’tknow;thereisnowayoftelling.”“Youareright,”saidtheSorcerer,“andnowIwilltellyouthatifItoldyouthenative’scolor,youwouldthenhaveenoughinformationtobeabletodeterminethenative’smaritalstatus.”“Great!”saidAnnabelle.“NowIknowwhetherornotthenativeismarried.”Isthenativemarriedornot?

2•AnotherMetapuzzle

“HereisanothermetapuzzleabouttheplanetOg,”saidtheSorcerer.“AlogicianfromourplanetvisitedOgandmetanativeonedarknightandaskedhimwhetherhewasagreennortherner.Thenativeanswered(yesorno),butthelogiciancouldn’ttellfromhisanswerwhathewas.“Asecondlogicianmetthenativeonanotherdarknightandaskedhimwhetherhewasagreensoutherner.Thenativeanswered(yesorno),butthelogiciancouldn’tfigureoutwhathewas.“Onanotherdarknight,athirdlogicianmetthenativeandaskedhimwhetherhewasaredsoutherner.Thenativeanswered(yesorno),butthelogiciancouldn’tfigureoutwhathewas.“Whatwashe?”askedtheSorcerer.

3•AMetametapuzzle

“Ilovedthelasttwopuzzles,”saidAnnabelle,“butwhydoyoucallthemmetapuzzles?Justwhatdoesthewordmean?”“The term was invented by my uncle,” replied the Sorcerer.“Metapuzzles are, so to speak,puzzles aboutpuzzles.One solves ametapuzzleon thebasisofknowing thatacertainotherpuzzle,orpuzzles,canorcannotbesolved.Suchpuzzlescanbequiteintricate!“Some puzzles go one level deeper and can be solved only byknowingwhether certainmetapuzzles can be solved! Such puzzlesmyunclecalledmetametapuzzles.Myunclewasamasteratinventingthese.”“Can you give us an example of a metametapuzzle?” askedAlexander.The Sorcerer thought amoment. “All right,” he said, “my uncleoncegavemeametapuzzleaboutalogicianwhovisitedtheplanetOg and met a native one dark night and was curious to knowwhether or not the native was truthful. He then asked the nativewhich of the four types hewas (green northerner, red northerner,greensoutherner,redsoutherner)andthenativethennamedoneofthefourtypesandclaimedtobethat.”“Oh,” said Annabelle, “was the logician then able to determinewhetherornotthenativewastruthful?”“Goodquestion,”repliedtheSorcerer,“andIaskedmyunclejustthat.”“Didheansweryou?”askedAlexander.“Yes.”“Whatanswerdidhegive?”askedAlexander.“I’mnottellingyou.”“Wellthen,”saidAnnabelle,“afteryouruncletoldyouwhetherornot the logician could solve his problem, were you then able todeterminewhetherornotthenativewastruthful?”“I’mnottellingyouthateither,”repliedtheSorcerer.“IfItoldyouthat,thenyouwouldbeabletodeterminewhetherornotthenativewastruthful.”

“Thenwhywon’tyoutellus?”askedAlexander.“Don’tyouwantustosolvetheproblem?”“It’s no longer necessary,” said the Sorcerer with a smile. “Youhave enough information to determine whether or not the nativewastruthful.”At this point, the reader has been given enough information toanswerthefollowingquestions:

1.Wasthenativetruthful?2.WastheSorcererabletosolvethemetapuzzlegivenhimbyhisuncle?3.CouldtheSorcererdeterminewhichofthefourtypesthenativewas?4. Did the logician determine whether or not the native wastruthful?5.Couldthelogiciandeterminewhichofthefourtypesthenativewas?

Ibelievethereaderwillfindthispuzzlequiteinstructive!

4•AKnight-KnaveMetapuzzle

“Thatwasquiteingenious,”saidAnnabelle.“Haveyoueverinventedanymetapuzzlesormetametapuzzlesyourself?”“Oh yes,” said the Sorcerer. “I learned the art frommy uncle. Ihave even invented metametametapuzzles—and gone even abovethese. I have constructed puzzles of all degrees of complexityrangingfromverysimpletoverydifficult.”“AretheyalsoabouttheplanetOg?”askedAnnabelle.“Ohno.Theyaremainlyaboutordinaryknight-knaveislandslikethisone—islandsinwhichallknightstellthetruthandallknaveslieandeveryinhabitantiseitheraknightoraknave.“Let’s see now,” the Sorcerer continued, “here’s one you mightlike: A logician visited a knight-knave island and met twoinhabitants,AandB,andaskedA: ‘Is it true thatBonce said thatyouareaknave?’Aanswered(yesorno).Thenthe logicianasked

one of the twowhether the otherwas a knave. Hewas answered(yesorno).Itisnotgivenwhetherornotthelogiciancouldthentellwhattheywere.“AsecondlogiciancameacrossthesamepaironanotheroccasionandaskedAwhetherBhadonce claimed thatA andBwerebothknaves.Aanswered(yesorno).Thenthelogicianaskedoneofthemwhethertheotherwasaknave,andwasanswered(yesorno).Itisnot givenwhether or not the second logician could then solve theproblem.“What isgiven is thatoneof thetwo logicianswasable tosolvethe problem and the otherwasn’t, butwe are not toldwhich onesolvedit.AndnowyourproblemistodeterminewhatAandBare,andwhichlogiciansolvedtheproblem.”

5•ACourtCase

“Wow! That was a hard one,” said Annabelle, after the two hadsolvedit,“butquiteintriguing.Doyouhaveanymore?”“Thenextisametametapuzzle,”saidtheSorcerer.“Itconcernsacourttrial.Adifficultfeatureofthistrialwasthatitwasnotknownwhether the prosecutor was a knight or a knave, nor what thedefense attorney was. Anyway, the two made the followingstatements:

Prosecutor: The defendant is guilty and has committed othercrimesinthepast.DefenseAttorney:Thedefendantisinnocent,andtheprosecutorisaknave.

“Thejudgethenaskedaquestionwhoseanswerhealreadyknew,andoneofthetwoattorneysansweredit,thusrevealingtothejudgewhether he was a knight or a knave, and the judge then knewwhetherthedefendantwasinnocentorguilty.“Ahighlyintelligentreporterwaspresentatthetrialandlatertoldalogicianfriendofhisall Ihavetoldyousofarandnomore.Thelogicianworkedontheproblemforawhileandrealizedhedidnot

have enough information to solve it. ‘Tell me,’ he said to thereporter, ‘if the other attorney had answered the question instead,would the judge then have known whether the defendant wasguilty?’ The reporter thought for a while and said: ‘I really don’tknow.Infact,thereisnowayofknowing.’“ ‘Oh, good!’ said the logician. ‘Now I know whether the

defendantwasinnocentorguilty.’“Was the defendant guilty or innocent? Also, which attorney

answeredthejudge’squestion,andwasheaknightoraknave?”

6•AVeryAbstractHyper-Metapuzzle

“I recently thought of a very abstract metapuzzle,” laughed theSorcerer.“Actually,thesolutionisnotatallcomplicated;infact,itisverysimple.Still,itistrickyandoneisapttooverlookit.“Byaperfectlogician,Imeanonewho,givenanyproblem,ifheis

givenenoughinformationtosolvetheproblem,willthenknowthatthereisenoughinformationandwillsolvetheproblem,butifheisnot given enough information, then he will know that it is notenough.Now,acertainperfectlogicianwasgivenacertainproblem—call it Problem P. I am not telling you what Problem P is orwhetherhewasgivenenoughinformationtosolveit.AllIwilltellyouisthatifIdidtellyouwhatProblemPwasandwhetherornotthe logician had enough information to solve it, then you wouldhave enough information to solve Problem P. By you, I of coursemean anybody. That is, the information of what Problem P is,togetherwithwhetherthelogicianhadenoughinformationtosolveit—thatinformationisenoughtosolveProblemP.“And now your problem is: Did the logician solve Problem P or

didn’the?”

“Thetimehascome,”saidtheSorcerer,“foracompletechangeofpace.IsuggestthatwevisittheIslandofRobots.Ibelievethatitwillinterestyouenormously!Tomorrowissupposedtobeanunusuallyclearandbeautifulday,andIthinkweshouldleavebeforesunrisesowecanhaveanentiredayontheisland.Whydon’tyoustaythe

nightsothatwecanleavereallyearly?”This sounded like an attractive idea, and the couple happily

agreed.Whathappenedon the IslandofRobotswillbe told in thenextchapter.

Solutions

1.Thepointisthatifthenativeisred,hecannotbemarried,butifheisgreen,hemightormightnotbemarried.Hereiswhy.Supposeheisred.Ifheistruthful,heisamarriednortherner(as

he claimed); hence he is a red northerner who is truthful, whichcannotbe.Therefore (assuminghe is red)he isnot truthful;henceheisarednortherner,butnotamarriednortherner(sincehefalselyclaimedhewas).Thisprovesthatifheisred,hemustbeunmarried.On the other hand, if he is green, then he could be a green

married northerner who speaks truly, or a married or unmarriedgreensouthernerwhospeaksfalsely.Thus,theknowledgethatheisgreendoesnotsolvethequestionofwhetherheismarried,buttheknowledge that he is red does. Now, the Sorcerer said that theknowledge of the native’s color would solve the question of hismarital status, and so it must be that the native is red andunmarried.

2.Letuslabelthequestionsaskedofthenative.Q1.Areyouagreennortherner?Q2.Areyouagreensoutherner?Q3.Areyouaredsoutherner?Asthereadercaneasilycheck,herearetheanswersthateachof

thefourtypeswouldgiveifaskedthequestion.

In each case, three would have answered one way and one theother.Hadthefirstanswerbeenno,thelogicianwouldhaveknownthatthenativewasaredsoutherner,butsincehedidn’tknow,thenative didn’t answer no and hencewas not a red southerner.Alsothenativewasn’tarednortherner(sincethesecondlogiciandidn’tknow) and he wasn’t a green northerner (since the third logiciandidn’tknow).Therefore,thenativewasagreensoutherner.

3.Weleaveittothereadertoverifythefollowingfoursimplefacts.(1) Ifanativeclaimstobeagreennortherner, thenhecanbea

green northerner, a red northerner, or a green southerner (but hecan’tbearedsoutherner).(2) If a native claims to be a red northerner, he can be only a

greensoutherner.(3) If he claims to be a red southerner, then he can be a red

southerner,agreensoutherner,orarednortherner(buthecan’tbeagreennortherner).(4)Ifheclaimstobeagreensoutherner,thenhecanbeonlyared

northerner.Therefore, if the native had claimed either to be a green

northerneroraredsoutherner,thelogicianwouldhavehadnowayofknowingwhichofthethreetypeshemightbe,hencethelogiciancouldn’thaveknownwhetherthenativewastruthful.On the other hand, if the native claimed either to be a red

northerner or a green southerner, the logician would have knownthat the native was lying—indeed, he would have even knownwhether the native was a red northerner or a green southerner(whicheverheclaimedtobe,hemustreallybetheother).Andsoin

short,ifthelogicianwasabletodeterminethenative’sveracity,thenative must have lied, but if the logician couldn’t determine hisveracity,thenthenativemighthaveliedorhemighthavetoldthetruth—thereisnowayofknowing.Now, the Sorcerer’s uncle told him whether or not the logiciansucceededinhisquest(todeterminewhetherornotthenativewastruthful). If the uncle answered affirmatively, then the Sorcererwould know that the native must have lied, but if the uncleanswered negatively (if he had told him that the logician didn’tknow),thentheSorcererwouldhavenowayofknowingwhetherornot the native was truthful (all he would have known is that thenative either claimed to be a green northerner and was thusanything but a red southerner) or claimed to be a red southerner(andwasthusanythingbutagreennortherner).Atthispoint,weknowthatiftheSorcerersolvedthemetapuzzlethat his uncle gavehim, then thenativemust be a liar, but if theSorcererdidn’tsolvehismetapuzzle,thenthereisnowayoftellingwhether thenativewas a liar. But the Sorcerer further toldus (orratherhis two students) that a knowledgeofwhether the Sorcerersolvedhismetapuzzlewouldbeenoughtodeterminetheveracityofthe native. Well, the knowledge that the Sorcerer didn’t solve hismetapuzzlewouldnotbeenough,whereasaknowledgethathedidsolve it would be. And so the Sorcerer must have solved hismetapuzzleandthenativemusthavelied.In summary, the native lied. The Sorcerer knew that the nativelied, but couldn’t determine which of two types of liar he was,whereas the logician not only knew that the native lied, but alsowhetherhewasarednortherneroragreensoutherner(thoughwehavenowayofknowingwhich).

4. Step 1. Let us see what we can learn from the first logician’sencounter. A answered either yes or no to the logician’s firstquestion.CaseA1.Heansweredyes.ThusAaffirmedthatBhadoncecalledhimaknave. IfA is aknight, thenBdidonceclaim thatAwasa

knave and B must be a knave. Thus A and B could not both beknights.Ontheotherhand,ifAisaknave,thenBcouldbeeither,and there is noway of tellingwhich (since B thenmade no suchclaim, nothing can be deduced about B). And so all the logicianwouldknowatthisstagewasthatAandBwerenotbothknights.CaseA2.Aansweredno.ThusAdeniedthatBoncecalledhima

knave.Ofcourse,AcouldbeaknightandBcouldbeeitheraknightorknave,butifAisaknave,thenBmustbeaknight(becauseifA’sdenialisfalse,thenBreallydidoncecallAaknave;henceBmustbeaknight).Andso,inthiscase,allthelogicianwouldknowatthisstageisthatAandBwerenotbothknaves.Step2.Now, theanswer to the logician’s secondquestionwould

revealwhetherAandBwere the same type,because ifonenativeaffirmsthattheotheroneisaknave,theymustbedifferenttypes(aknightwouldnotclaimanotherknight tobeaknave,norwouldaknaveclaimanotherknavetobeaknave),andifonenativeaffirmsanother tobeaknight, the twomustbeof thesametype.Andso,after the second question was answered, the logician would thenknowwhetherthetwowerealike(ofthesametype).Ifhefoundoutthattheywerealike,thenhewouldhavesolvedtheproblemofwhateachwas,regardlessofwhetherCaseA1orCaseA2held,becauseinCaseA1, hewould have known that theywere alike but not bothknights,hencebothknaves,andinCaseA2,hewouldhaveknowntheywerebothknights.So,regardlessofwhatA’sfirstanswerwas,ifthesecondanswerrevealedthetwotobealike,thenthelogiciancould solve the problem.On the other hand, if the second answerrevealedthatAandBweredifferenttypes,theninneitherCaseA1norCaseA2couldthelogicianhaveknownwhichofAorBwastheknightandwhichtheknave.Wehavethusproved:

(1) If the twoare the same type, then the first logicianhas solvedtheproblem.

(2) If the two are different types, then the first logician has notsolvedtheproblem.

Step3.Wenowconsiderthesecondlogician’sencounter.Thistime,A, inanswering the firstquestion,eitheraffirmedordenied thatBclaimedthattheywerebothknaves.CaseB1.SupposeAaffirmed thatBclaimed that theywereboth

knaves.IfAwasaknight,thenBmusthavebeenaknave.IfAwasaknave,Bcouldbeeither.SoallthatthesecondlogicianwouldknowinthiscaseisthatAandBwerenotbothknights.Case B2. Suppose A denied that B claimed that they were both

knaves.ThenAcouldn’tbeaknave,forifhewere,thenBreallydidonceclaimthatbothwereknaves,whichisnotpossible(becauseifBwere a knight, hewouldn’t falselymake such a claim, and if Bwere a knave, he would never claim the true fact that both areknaves).Andso,Amustbeaknight;BneverdidmakesuchaclaimandBcouldbeeither.Thus,ifCaseB2istheonethatoccurred,allthesecondlogicianwouldknow(andhewouldknowthis)isthatAisaknight.Step 4. Now, suppose after the logician’s second question was

answered, he knew whether or not A and B were the same type.Suppose he found that they were the same. Then, he must havesolvedtheproblem(inCaseB1hewouldknowthatbothareknaves,andinCaseB2hewouldknowthatbothareknights).Suppose,ontheotherhand,hefoundoutthattheyweredifferent.TheninCaseB1, he couldn’t solve the problem (he would know that not bothwerenotknightsandalso that theyweredifferent,buthecouldn’tknowwhichwaswhich),whereasinCaseB2,hewouldknowthatAwasaknightandBaknave(sincehewouldknowthatAisaknightandthetwoaredifferent).ThisprovesthatifAandBaredifferent,thentheonlywayhecouldsolvetheproblemisforAtobeaknightandB a knave (because ifA andB are different,CaseB1 couldn’tholdorthe logicianwouldn’thavesolvedtheproblem,aswehaveseen),henceCaseB2mustholdandthereforethelogicianknewthatAisaknightandBaknave.Wehavethusprovedthefollowing:

(3)IfAandBarethesametype,thesecondlogicianhassolvedtheproblem.

(4) If A and B are different types, then the only way the secondlogiciancouldsolve theproblem is thatA isaknightandB isaknave.

Step5.Nowwehaveallthenecessarypieces!SupposeAandBarethesametype.Thenby(3),thesecondlogiciansolvedtheproblemandby(1)thefirstlogiciansolvedtheproblem,andthisiscontrarytothegivenconditionthatoneandonlyoneofthelogicianssolvedtheproblem.ThereforeAandBaredifferenttypes.Since A and B are different types, then the first logician didn’t

solvetheproblem,by(2);hencethesecondlogiciandid.Henceby(4),AisaknightandBisaknave.5. We must consider all four possible cases—which of the twoattorneysansweredthequestionandwhethereachrevealedhimselftothejudgeasaknightoraknave.Case 1. The prosecutor revealed himself to be a knight. Then, of

course,thejudgeknewthatthedefendantwasguilty(andalsothatthedefenseattorneywasaknave).Case2.Theprosecutorrevealedthathewasaknave.Thenthejudge

couldn’tknowwhatthedefendantwas(hecouldbeinnocent,orhecould be guilty but, contrary to the prosecutor’s claim, he mightneverhavecommittedanycrimesinthepast).Case 3. The defense attorney revealed that he was a knight. Then,

obviously, the judgewould know that the defendantwas innocent(andalsothattheprosecutorwasaknave).Case4.Thedefenseattorneyrevealedhimselftobeaknave.Thenthe

judge would know that the defendant must be guilty as follows:Sincethedefenseattorney’sstatementisfalse,thenitcannotbethatthedefendantisinnocentandthattheprosecutorisaknave;henceeitheritisfalsethatthedefendantisinnocentoritisfalsethattheprosecutorisaknave.Iftheformer,thenofcoursethedefendantisguilty.Ifthelatter,thentheprosecutorisaknight,inwhichcasethedefendant is again guilty (since the prosecutor said he was). Thisproves that if thedefenseattorney revealedhimself tobeaknave,thejudgewouldknowthatthedefendantmustbeguilty(thoughhe

wouldn’tknowwhethertheprosecutorwasaknightoraknave).

Now let’s see why the reporter didn’t know the answer to hisfriend’squestion.Thereporterwaspresentatthetrial,soheknewwhichattorney

answeredthejudge’squestionandwhetherherevealedhimselfasaknight or a knave. Suppose it was the prosecutor who answered.Then, had the defense attorney answered instead, the judgewouldhavesolvedthecase(aswesawinouranalysisofCases3and4),and so the reporter would have answered yes to his friend’squestion.Sincehedidn’tansweryes,butinsteadsaidthathedidn’tknow, then it was not the prosecutor who answered the judge’squestion.Itwasthedefenseattorneywhoansweredandrevealedhistype.Suppose the defense attorneywas a knight. Then the prosecutor

must be a knave, and therefore had he, instead of the defenseattorney,answeredthe judge, the judgewouldn’thavebeenabletosolvethecase(aswesawinouranalysisofCase2)andthereporterwould know this and would have answered no to his friend’squestion.Thisleavesonlythepossibilitythatthedefenseattorneyrevealed

himselftobeaknave.Thenthereportercouldnothavetoldwhethertheprosecutorwasaknightoraknave,andhencedidnotknowtheanswer to his friend’s question. (All he knew was that if theprosecutorhadbeentheonetohaveansweredthejudge,andifhehadrevealedhimselftobeaknight,thejudgecouldhavesolvedthecase,andifhehadrevealedhimselftobeaknave,thejudgecouldn’thave. But the reporter had no way of knowing whether theprosecutorwouldhaverevealedhimselftobeaknightoraknave.)

6.Obviously, the logicianknewwhatProblemPwas,andsincehewasperfect,heknewwhetherornothehadenoughinformationtosolvetheproblem.Thisinformationisenoughtosolvetheproblem;hence the logician had enough information to solve the problemand,beingperfect,hesolvedtheproblem.

PARTIII

SELF-REPRODUCINGROBOTS

11THEISLANDOFROBOTS

THE SORCERER and his friends Annabelle and Alexander retired earlyandroseintheweehoursofthemorningandsetsailfortheIslandofRobots.Theyarrived justaboutas the sunwas risingand spentthe timebefore lunchwalkingaroundobserving the strange thingsthatwent on. This islandwas the noisiest place that Annabelle orAlexanderhadeverbeentointheirlives.Theclingingandclangingand binging and banging and din and clatter were almostunbearable. And the things that went on! The whole island wasbustling with metallic robots moving about, some apparentlyaimlessly, others creating new robots from various parts lyingaround, andothersdismantling robotswhosepartswouldoftenbeusedforcreatingotherrobots.Eachrobotborealabelconsistingofastringofcapitalletters.AnnabelleandAlexanderatfirstthoughtthelettersweremerely for identification, but later found theywere aprogram determiningwhat the robot should do—whether it wouldwalkaroundaimlesslyorwhetheritwouldcreateotherrobots,andifso,whatprogramthenewrobotswouldhave,orwhetheritwouldbeadestructiverobot,andifso,whatrobotsitwoulddestroy.One thing struck the couple as quite peculiar: They sawa robotpickupabunchofpiecesandconstructarobotthatlookedidenticalto its creator; indeed it had the same program. Having the sameprogram, this duplicate constructed a duplicate of itself, which inturn constructed a duplicate of itself, which in turn constructed aduplicate of itself (as a matter of fact, from parts of its great-grandparent, which had been dismantled by then). Indeed, thisprocesscouldgoonforever,unlesssomethingstoppedit.Next, the couple saw another curious thing: a robot—call it x—constructedaroboty,quitedifferentfromx,whichthenconstructedaduplicateofx,whichthenconstructedaduplicateofy,andthis2-

cyclecouldgoonforever!Then they saw a 3-cycle: x constructed y, which constructed z,whichconstructed(aduplicateof)x,whichconstructed(aduplicateof)y,…Nexttheysawsomethingveryupsetting.Arobotxconstructedarobot y and the first thing y did when it was completed was todestroy its creator. This struck the couple as the height ofingratitude.Then they saw a robot x destroy a robot y. Later on, y gotreassembled and met x and immediately destroyed it. Then x gotreassembled sometime later anddestroyedy, and this could goonforever!Next they saw two very different-looking robots, x and y, walkbriskly towardeachotherand immediatelybegindismantlingeachother,andsoonallthatwasleftwasarubbleofparts.Thentheysawarobotxconstructaroboty,whichconstructedarobotz,whichthendismantledx.Thusxwasdestroyedbyitsowngrandchild!Next, they saw something quite sad: a suicidal robot dismantleditself,andallthatwasleftwasapileofparts.(Atacertainstageofdismantling,therobotpressedabuttonwhichcausedwhatwasleftto fall completely apart.) Then another robot came by andreassembled thedismantled robot, butwhen the second robot left,the firstone, stillhaving the sameprogram,destroyed itself again.Thereseemedtobenohopeforthispoorrobot.Ifitwereeveragainreassembled, still with the same program, then it would have todestroy itself again; no robotwith this program could possibly bestable.“I’mcompletelybewilderedbyallthis,”saidAnnabelle.“Ihavenoideawhatintheworldisgoingon!”“NorI,”saidAlexander.“Oh, youwill findout this afternoon,” said the Sorcerer. “Thereare several robot stations on this island, each with its ownprogramming system.After lunch,wewill visit the laboratories ofsomeoftheengineers incharge.Theywillexplaintheirsystemsto

you.”

I.THESYSTEMOFCHARLESROBERTS

After an excellent lunch, cooked and served by skilled robots, theSorcerer took our two friends to the Northeast Station, whosedirector was a pleasant individual named Charles Roberts. Afterintroducing his students to Roberts, the Sorcerer took his leave,explainingthathehadsomethingstoattendtoontheisland.“Nowforthedetailsofmysystem,”saidRobertswithasmile.“Asyou have observed, each robot is labeled with a string of letters.These letters constituteaprogramdetermining justwhat the robotwilldo.“Let me explain to you my terminology and notation,” hecontinued.“Byanexpressionoracombination, Imeananystringofcapitalletters—forexample,MLBPisanexpression;soisLLAZLBA;soisasingleletterlikeGstandingalone.Iusesmallletterslikex,y,z,a,b,ctostandforparticularstringsofcapitalletters,andbyxyImean the combination x followed by the combination y. Forexample, if x is the expressionMBPandy is the expression SLPG,then xy is the expressionMBPSLPG.Also,Ax is thenAMBP; xA isMBPA;GxHisGMBPH;CxLyisCMBPLSLPG.Doyougettheidea?”Hisvisitorshadnotroublegraspingthis.“Now,”explainedRoberts, “myprogramming system isbasedontheideaofcertainexpressionsbeingnamesofothers.AndIhavetworulesconcerningthenamingofexpressions.Myfirstruleis:

RuleQ.Foranyexpressionx,theexpressionQxnamesx.“Thus, for example, QBAF names BAF; QQH names QH; QDCDnamesDCD.“WemightrefertoQxastheprincipalnameofx,but,asyouwillsee, an expression x may have other names as well. My secondnamingruleisthefollowing:

RuleR.Ifxnamesy,thenRxnamesyy.

“Thus,forexample,RQBnamesBB,sinceQBnamesB.Oragain,RQBR names BRBR, since QBR names BR. In general, for anyexpression x, the expressionRQx names xx, sinceQx names x.Donot make the common mistake of believing that Rx names xx; ingeneralitdoesnot.ItisRQxthatnamesxx.“IcallRuleRtherepetitionrule,becauseforanyx,theexpressionxxiscalledtherepeatofx—thus,ABCABCistherepeatofABC.AndsoRuleRtellsusthatifxnamesy,thenRxnamestherepeatofy.“Toseeifyouhavegraspedtheserules,whatdoesRRQBHname?”Afteramoment’sthought,Alexandersaid:“ItnamesBHBHBHBH.”“Why?”askedRoberts.“Because RQBH names BHBH, hence RRQBH must name therepeatofBHBH,whichisBHBHBHBH.”“Good!”saidRoberts.“You said before,” said Alexander, “that an expression can havemorethanonename.Howcanthathappen?”“Oh,” said Roberts, “for example, xx has RQx as one name andQxxasanother.Bothnamexx.Oragain,Qxxxxnamesxxxx,andsodoesRQxx,andsodoesRRQx.ThusQxxxx,RQxx,RRQxare threedifferentnamesforthesameexpression.”“Oh,ofcourse!”saidAlexander.“How is all this related to some robots creating others?” askedAnnabelle.“Oh, my creation rule is a very simple one,” said Roberts.“Remember that whenwe say that x creates y wemean that anyrobotwithprogramxwillcreatearobotwithprogramy.Hereismycreationrule.

RuleC.Ifxnamesy,thenCxcreatesy.“Thus,foranyy,robotCQy(thatis,anyrobotwithprogramCQy)createsroboty(arobotwithprogramy).“Italso follows fromour three rules thatCRQxcreatesxx (sinceRQxnamesxx)andCRRQxcreatesxxxx(sinceRRQxnamesxxxx).Andnow,forsomeinterestingapplications.”

•1•

“You have seen self-producing robots on this island, robots thatcreateduplicatesofthemselves?”“Oh,yes,”saidAnnabelle.“Well,canyounowgivemeaprogramforone.Thatis,canyou

findanxsuchthatxcreatesx?”

•2•

“Very good,” said Roberts, after the couple showed him theirsolution.“Andnow,”hecontinued,“beforeweturntoanymoreproblems

about creation, there are some basic principles about naming thatyou should grasp. For one thing, can you find some x that namesitself?”

•3•

“Excellent,”saidRoberts.“Canyoufindanotherxthatnamesitself,oristhereonlyone?”

•4•

“You said before,” said Alexander, “that in general, Rx does notname xx. Can it ever happen, by some sort of coincidencemaybe,thatRxdoes name xx? That is, is there any x such thatRx namesxx?”“What a curious question,” saidRoberts. “I never thought about

thatbefore.Let’sseenow.”At thispoint,Roberts tookoutpencilandpaperandmadesome

calculations.“Yes,”hesaidafterawhile,“theredoeshappentobesuchanexpressionx.Canyoufindit?”

5•SomeMoreCuriosities

“Ifyoulikecuriosities,”saidRoberts,“itmightamuseyoutoknow

thatforeachofthefollowingconditionsthereissomexthatsatisfiesthatcondition.(a)Rxnamesx.(b)RQxnamesQRx.(c)RxnamesQx.(d)RRxnamesQQx.(e)RQxnamesRRx.“But these are only curiosities,” added Roberts, “and have no

applicationsthat Iknowof torobotprogramming.Youmighthavefunworkingthemoutsometimeatyour leisure, ifyoulikepuzzlesfortheirownsake,regardlessofpracticalapplications.”

Some Fixed-Point Principles “We saw some robots destroyingothers,”saidAlexander.“Doesyourprogramprovideforthese?”“No,”repliedRoberts,“thedestructionprogramsareinthehands

ofmybrother,DanielChaunceyRoberts,whoisalsoaroboteeronthisisland.Hisprogramsareveryinteresting,andyoushouldmakeapointofvisitinghimtoday.“Andnow,Iwilltellyouofabasicprincipleunderlyingmanyof

myprogramsaswellasthoseofmybrother.”

6•TheFixed-PointPrinciple

“Givenan expression a,we call x a fixedpoint of a if x names ax.(Rememberthata,likexandy,representsanygivencombinationofcapitalletters.)The fixed-point principle is that every expression a has a fixed

point. Moreover, there is a simple recipe whereby, given anyexpression a, one can find a fixed point of a. Can you prove thefixed-pointprincipleandgivemetherecipe?Forexample,whatxissuchthatxnamesABCx?”

•7•

“The fact that there isaself-creatingrobot isanapplicationof thefixed-pointprinciple.Canyouseewhy?No?Letmeputthematter

moreprecisely.SupposethatweconsideranothernamingsysteminwhichRuleQandRuleRare replacedbyothernaming rules,andwealsohavetherulethatifxnamesy,thenCxcreatesy.Ifalsothisother naming system had the fixed-point property—if for everyexpression a, therewere some x that names ax—then therewouldhavetobeaself-creatingrobot.Canyouseewhy?”

•8•

“As another application of the fixed-point principle,” said Roberts,“theremustbe somex thatnamesxx.Canyou findone?Also, foranyexpressiona, theremustbe somex thatnamesaxax.Canyouseewhy?”

•9•

“Asafurtherapplication,findanxthatcreatesxx.”

•10•

“Nowshowthatforanyexpressiona,thereissomexthatcreatesax.Givearecipeforfindingone,giventheexpressiona.Thisrecipewillhavefurtherapplications,andIwillcallittheCRecipe.”

11•TheDoubleFixed-PointPrinciple

“As another applicationof the fixed-point principle, it follows thatforanyexpressionsaandb,thereareexpressionsxandysuchthatxnamesayandynamesbx. I refer to thisas thedouble fixed-pointprinciple.Thereare, in fact, twodifferentwaysof findingxandy,given a and b. I shall refer to these recipes as double fixed-pointrecipes.Whatarethey?”

•12•

“Isawtwodistinct-lookingrobots,”saidAnnabelle,“suchthateachcouldcreatetheother.Doesyourprogramallowforthis?”

“Ofcourse,”saidRoberts,“Youshouldnowbeabletofindsuchanxandy.Whatarethey?”

•13•

“Itisalsopossible,”saidRoberts,“tofindexpressionsxandysuchthateachnamestheother,yettheyaredistinct.Canyoufindthem?”

•14•

“Alsoitispossibletofindxandysuchthatxcreatesyandynamesx.Therearetwosolutions.Canyoufindboth?”

•15•

“There is also a triple fixed-point principle,” said Roberts. “Givenanythreeexpressions,a,b,andc,thereareexpressionsx,y,andzsuchthatxnamesay,ynamesbzandznamesex.Therearethreedifferentrecipesforfindingthem.Whatarethey?”

•16•

“Nowfindexpressionsx,y,andzsuchthatxcreatesy,ynamesz,andzcreatesx.I’llbesatisfiedwithjustonesolution.”“And now,” said Roberts after our couple had solved the lastproblem, “I think you are ready to visit my brother, DanielChauncey.”He then gave them directions to Daniel Chauncey’s laboratory,and after thanking him, they went off to visit the other ProfessorRoberts.

II.THESYSTEMOFDANIELCHAUNCEYROBERTS

“As my brother has told you,” said Daniel Chauncey, “I haveprograms for robots that destroy others. I also have programs forrobotsthatcreateothers.

“I say thatxdestroysy ifany robot labeledxdestroysany robotlabeledy.Now,mycreationrulesareexactly thesameas thoseofmybrother.As for thedestruction rules, Ineedonlyone,which isthefollowing:

RuleD.Ifxcreatesy,thenDxdestroysy.“Forexample,ifxisanyexpression,thenDCQxdestroysx,since

CQxcreatesx.AlsoDCRQxwilldestroyxx,sinceCRQxcreatesxx.”

•17•

“What happens,” askedAnnabelle, “if x destroys y,where x and yarethesameexpression?Inotherwords,supposexdestroysx.Doesthismeanthatrobotxwilldestroyitself?”“Exactly!” said Daniel Chauncey. “Such robots we call suicidal

robots.Andnow,canyoufindmeanxsuchthatrobotxissuicidal?Myprogramprovidesone.”“That isobvious,” saidAlexander.Wealreadyknowhowto find

anxsuchthatxcreatesx.ThenDxisself-destroying.”“Notso!”saidDanielChauncey.“Ifxisself-creating,thenDxwill

destroyx,but thatdoesn’tmean thatDxwilldestroyDx,which iswhatwewant.”“Oh,ofcourse!”saidAlexander.Whatisthecorrectsolution?

•18•

“Another useful principle,” said Daniel Chauncey, “is that for anyexpression a there is some x that destroys ax. Can you giveme arecipeforone?IcallthistheDRecipe.”

•19•

“Itisalsotruethatforanyexpressionathereissomexthatdestroystherepeatofax—thatis,xdestroysaxax.Canyouseewhy?”

•20•

“We also saw another sad situation,” said Annabelle. “We saw arobotxcreatearobotywhothenungratefullydestroyeditscreator,robotx.Doesyoursystemprovideforsuchaprogram?”“Yes indeed,” replied Daniel Chauncey. “You should be able to

findanxandsomeysuchthatxcreatesyandydestroysx.Infact,therearetwosolutions.”Whatarethey?

•21•

“We also saw two distinct robots destroy each other,” saidAlexander.“Doesyoursystemprovideforthis?”“Certainly,”repliedDaniel.“There issomexandsomeydistinct

fromxsuchthatxandydestroyeachother.”Whatisthesolution?

•22•

“Somethingelsewesaw,”saidAnnabelle.“Wesawarobotxcreatearobotywhocreatedarobotzwhothendestroyedx.”“Thatcanbedoneinmysystem,”saidDaniel.“Trytofindsuchan

x,y,andz.”(Actually,thereismorethanonesolution.)

FriendsandEnemies.“SomethingIobserved,”saidAlexander,“isthatseveraloftherobotshadlabelsbeginningwithFandseveralwithE.Dothesetwolettershaveanyspecialsignificance?”“Oh,yes indeed!” repliedDanielChauncey. “You see,our robots

haveformedvariousfriendshipsandenmities.Ifxcreatesy,thenFxisafriendofyandExisanenemyofy.Also,thebestfriendofyisFCQyandtheworstenemyofyisECQy.“Thisleadstosomeinterestingresults,”saidDaniel.“I’llgiveyou

someexamples.”

•23•

“A robot who is a friend of himself is called narcissistic,” Danielcontinued.“Thereisanxthatisnarcissistic.Canyoufindit?“Also,thereisaprogramforarobotxwhoisanenemyofhimself.

Such a robot is called neurotic. If you can find a program for anarcissistic robot, then of course you can find one for a neuroticrobot.”

•24•

“It isobviously impossible foranx tobehisownbest friend(noxcanequalFCQx),butthereisanxwhocreateshisownbestfriend.Findone.”

•25•

“Nowfindanxthatdestroyshisworstenemy.”

“Of course,” continued Daniel, “there must also be some x thatdestroyshis best friend—just replaceE by F in the solution to thelastproblem.”“Whywouldanyrobotwanttodosuchahorriblethingasdestroy

hisbestfriend?”askedAnnabelle,quiteshocked.“Unfortunately,someofourrobotsareinsane,”repliedDaniel.“Then why do you have programs for insane robots?” persisted

Annabelle.“Wecan’tavoidit,”repliedDaniel.“Sofar,wehavebeenunable

to find an interesting program system that doesn’t have some badsideeffects.Thesituationisanalogoustothehumangeneticcode.Itunfortunatelyallowsforpathologiestoarise.”“Tellmethis,”saidAnnabelle.“Isitpossibleforarobotxtobea

friendofarobotywhoisanenemyofx?”“Ohyes,”saidDaniel.“SuchrobotsarecalledChristlike.Afterall,

Jesus said thatwe should love our enemies. In fact, there is an xsuch that he is a friend of hisworst enemy. Such an x is called amessiah.“Also, a robotwho is an enemyof someonewho ishis friend is

calledevil,andifheisanenemyofsomeonewhoishisbestfriend,thenheiscalledsatanic.”

•26•

“Thereare,infact,twoChristlikerobots,”continuedDaniel,“oneofwhomisevenamessiah;andtherearetwoevilrobots,oneofwhomisevensatanic.Canyoufindprogramsforthem?”

•27•

“There is also an xwhich creates some ywhich destroys the bestfriendofx.Canyoufindit?”

•28•

“There is also an x who creates the best friend of some y thatdestroystheworstenemyofx.Canyoufindone?”

•29•

“And then there is somexwho is thebest friendof someonewhodestroysx’sworstenemy.Findsuchanx.”

“This is only the beginning of the possibilities,” said DanielChauncey. “The combinations are endless, and that iswhatmakesthisislandsointeresting.”“I’mreallyintriguedbytheseprograms,”saidAnnabelle,“butone

thing puzzles me. I can understand your interest in the software,since it presents problems that are of combinatorial interest. Butwhatisthepointinactuallyconstructingtheserobots?Whyareyounotsatisfiedwithjusttheprograms?Whydoyoutakethetroubletocarrythemout?”“There are several reasons,” replied Daniel. “For one thing, it’s

nice to have empirical confirmation that our programs actuallywork.Butmoreimportant,it’sofgreatinteresttoseehowall theserobotswith their various programs interactwith each other. Since

we have several hundred robots on this island, it is virtuallyimpossibletopredictwhatwillhappeninthefuture.Canthiscolonyof robots survive indefinitely, orwill themembers oneday all getdestroyed and nothing remain but dismantled parts? All sorts ofinterestingsociologicalquestionsarise,andweareabouttoengagearobotsociologisttostudythesociologyoftherobotcommunity.It’sverypossible that thismay throw lighton the sociologyofhumancommunities.”“Well, it all sounds very interesting,” saidAnnabelle, “and I am

verymuchintriguedbyyoursystem.”“Mysystemisamodernizedversionofanoldersystemdevisedby

ProfessorQuincy,”repliedDaniel.“Hewasthefirstroboteeronthisisland.Heisnowretired,butisstillactivelydoingresearch,andhelikestoreceivevisitorswhoareinterestedinseeinghissystem.Whydon’tyoupayhimavisit?”AtthispointtheSorcererenteredthelaboratory.“ThoughtI’dfind

you here,” he said, “and I certainly agree it’s a good idea to visitProfessorQuincy. I’mnot familiarwithhis system, so I’ll takeyouthere.”ThethreethankedProfessorRobertsforhistimeandwentoff to

visitProfessorQuincy.

Solutions

1.Foranyexpressionx,theexpressionCRQxcreatestherepeatofx(sinceRQxnamestherepeatofx),andsowetakeCRQforxandwehaveCRQCRQcreatetherepeatofCRQ,whichisCRQCRQ.AndsoCRQCRQcreatesCRQCRQ.ThusCRQCRQisoursolution.

2. RQRQ names the repeat of RQ, which is RQRQ. Thus RQRQnamesitself.

3.Theonly expressions thatnameanythingmustbeof oneof theformsQy,RQy,RRQy,RRRQy,andsoforth.Thatis,any“name”iseitherof the formQy, for somey,orQyprecededbyoneormore

R’s.Wewant a “namer” x that names itself. Could it be of the form

Qy?CouldQy nameQy?Certainly not;Qy names y and y cannotequalQy(itisshorter).WhataboutanexpressionoftheformRQy?Well,RQynamesyy,

andwewantRQytonameRQy,andsothiscanhappenifandonlyifyy=RQy.Isthispossible?Yes,takey=RQ.ThusRQRQnamesRQRQ, which is the solution we got before. Moreover, the only ysuch that yy=RQy is RQ. Thus theonly x of the formRQy thatnamesitselfisRQRQ.(DoIhearanyR-Q-ment?)WhataboutanxoftheformRRQy?Couldthatwork?No,because

RRQy names yyyy, which is necessarily different from RRQy(becauseiftheywerethesame,wewouldhavetohavey=Randy=Q,whichisimpossible).RRRQycouldn’tworkeither,becauseitnamesyyyyyyyy,whichis

necessarilylongerthanRRRQy.With four ormoreR’s at the beginning, the disparity in lengths

wouldbegreaterstill,andsotheonlyexpressionthatnamesitselfisRQRQ.

4.Tryx=RRQRQ.5.(a)QQ(b)QR(c)QQQ(d)QQ(e)RR

6.RQaRQnamestherepeatofaRQ,whichisaRQaRQ.Andsoifwetakex=RQaRQ,thenxnamesax.Again recall that I am using the small letter a to represent any

combinationofcapitalletters.Thus,forexample,takingABCfora,anexpressionxthatnamesABCxisRQABCRQ.Perhaps it is easiest to thinkof it thisway:Howeverone fills in

theblanks,thefollowingistrue:anxthatcreates—xisRQ—RQ.

7.Inanynamingsystem,ifthereissomeythatnamesCy,thenCycreates Cy, and Cy is then a self-creating expression (Robot CycreatesRobotCy).Inthisparticularnamingsystem,anexpressionythat names Cy is RQCRQ (by the fixed-point recipe), and thusCRQCRQisself-creating(whichisthesolutionweoriginallyhad).

8.IfwecangetsomeythatnamesRy,thenRywillnametherepeatof Ry, andwe can then take x to be Ry.Well, by the fixed-pointrecipe,y=RQRRQ(thisisaspecialcaseoftherecipeinwhichaisthe letterR). Sowe take x=RRQRRQ, and the reader can checkthatitnamesRRQRRQRRQRRQ.Also,givenanyexpressiona,tofindanxthatnamestherepeatof

ax,wefirstlookforsomeythatnamesaRy—suchanexpressionyisRQaRRQ(by the fixed-point recipe, takingaR for a)—and thenRynamestherepeatofaRy,andsowetakex=Ry.Thusoursolutionisx=RRQaRRQ.

9.Bythesolutionofthelastproblem,takingCfora,thereissomeythat names CyCy—namely, y = RRQCRRQ. Then Cy must createCyCy,andsowetakex=CRRQCRRQ.

10.Whatweneed is somey thatnamesaCy, and thenCy createsaCy,sowethentakex=Cy.Usingthefixed-pointrecipe,wetakey=RQaCRQ.AndsooursolutionisCRQaCRQ.ThisisourCRecipe.

11.Wewantxtonameayandytonamebx.Therearetwowayswecangoabout it.Onewayis togetanxthatnamesaQbxandthentakeytobeQbx.(Then,obviously,xnamesayandy,whichisQbx,namesbx.)Weusethefixed-pointrecipeforgettingsuchanx,andwegetthesolution

x=RQaQbRQy=QbRQaQbRQ

AnotherwayisfirsttogetsomeythatnamesbQayandtakex=Qay.Thisgivesthefollowingsolution:

x=QaRQbQaRQy=RQbQaRQ

12.AllweneedisanxthatcreatesCQx.ThenCQxinturncreatesx.WeusetheCRecipe,takingCQfora,andwegetx=CRQCQCRQ,whichcreatestherepeatofCQCRQ,whichisCQCRQCQCRQ,whichis CQx. Thus two distinct expressions that create each other areCRQCQCRQandCQCRQCQCRQ.

13.RQQRQnamesQRQQRQ,whichinturnnamesRQQRQ.

14.OnesolutioncanbeobtainedbytakinganxthatcreatesQxandtakingy=Qx.Thisgivesthefollowingsolution:

x=CRQQCRQy=QCRQQCRQ

Another solution is obtained by taking some y that names CQyandtakingxtoequalCQy.Thisgivesthefollowingsolution:

y=RQCQRQx=CQRQCQRQ

15.OnesolutionistotakeanxthatnamesaQbQcx,andthentakez=Qcxandy=Qbz.Thisgivesthefollowingsolution:

x=RQaQbQcRQz=QcRQaQbQcRQy=QbQcRQaQbQcRQ

AnothersolutionisobtainedbytakingythatnamesbQcQayandthentakingx=Qayandz=Qcx.Weleavethistothereader.Another solution is to take somez thatnames cQaQbzand then

takey=Qbzandx=Qay.Weleavethistothereader.

16. One way to go about it is this: Hold x in abeyance for the

momentandwhateverwefinallydecideforx,wewilltakeztobeCQx(whichcreatesx).Thenyistonamez,sowewilltakey=Qz.Thusy=QCQx.AndsowewantxsuchthatitcreatesQCQx.Usingthefixed-pointrecipe,wegetthefollowingsolution:

x=CRQQCQRQy=QCQCRQQCQRQz=CQCRQQCQRQ

There are two other solutions that we leave to the interestedreader.

17.DCRQDCRQ.

18.Takex=DCRQaDCRQ.

19.Takey=DCRRQaDCRRQ.

20.OnesolutionisobtainedbytakinganxthatcreatesDCQxandtakingy thatequalsDCQx.Weuse theCRecipe(Problem10)andgetthefollowingxandy:

x=CRQDCQCRQy=DCQCRQDCQCRQ

AnothersolutionisobtainedbytakingaythatdestroysCQyandtakinganxthatequalsCQy.WeusetheDRecipe(Problem18)andgetthefollowingsolution:

y=DCRQCQDCRQx=CQDCRQCQDCRQ

21.WewantanxthatdestroysDCQx.WeusetheDRecipeandgetx=DCRQDCQDCRQ.ThisxdestroysDCQDCRQDCQDCRQ,whichinturndestroysx.

22.Onesolution isobtainedbygettinganx thatcreatesCQDCQx,which we take for y, and then taking DCQx for z. Using the CRecipe,wegetthefollowingsolution:

x=CRQCQDCQCRQy=CQDCQCRQCQDCQCRQz=DCQCRQCQDCQCRQ

23. FCRQFCRQ is a friend of the repeat of FCRQ which isFCRQFCRQ.ThusFCRQFCRQisnarcissistic.Obviously,then,ECRQECRQisneurotic(anenemyofhimself).

24.This isobvious:WewantxtocreateFCQx,sobytheCRecipewetakex=CRQFCQCRQ.

25.Takex=DCRQECQDCRQ.

26.AmessiahisanxwhoisafriendofECQx(theworstenemyofx).SoxmustbeoftheformFzforsomezthatcreatesECQFz(whichisECQx).Weuse theCRecipeand takez=CRQECQFCRQ.Thenwe takex=FCRQECQFCRQandx isourmessiah.Now,zcreatesECQFz,whichisECQFCRQECQFCRQ,whichwewillcally.Althoughxisafriendofy,itisnotthebestfriendofy,andsoyisevil(beinganenemy—infact,theworstenemy—ofx),butnotsatanic.Of course, by interchanging E and F, we get a satanic x and a

Christlikeywhoisnotamessiah.Insummary:FCRQECQFCRQisamessiah.ECQFCRQECQFCRQisevilbutnotsatanic.ECRQFCQECRQissatanic.FCQECRQFCQECRQisChristlikebutnotamessiah.

27. We will take an x that creates DCQFCQx. Such an x isCRQDCQFCQCRQ.

28. We will take an x that creates FCQDCQECQx. Let y =DCQECQx.ThenxcreatesFCQy,thebestfriendofy,andydestroys

ECQx, the worst enemy of x. And so we get the solution x =CRQFCQDCQECQCRQ.

29.Wewouldlikexandytobesuchthatx=FCQy(thebestfriendofy)andydestroysECQx.ThuswewantytodestroyECQFCQy.Wethus take y = DCRQECQFCQDCRQ, and so our desired x isFCQDCRQECQFCQDCRQ.

12THEQUAINTSYSTEMOFPROFESSORQUINCY

I.QUINCY’SSYSTEM

“AHYES,”saidQuincy,afterhisthreeguestswerecomfortablyseated,“mysystemmaybeold-fashioned,butitreallyworks!Yes,itreallyworks,youknow,”herepeatedwithachuckle.“My system is based on the idea of quotation—you know, onenamesanexpressionbyenclosingitinquotationmarks,theopeningquoteontheleftandtheclosingquoteontheright.Onlyinsteadofusing opening and closing quotes, I use the symbol Q1 for theopeningquoteandQ2fortheclosingquote.Andsobythequotationofanexpressionx, ImeantheexpressionQ1xQ2—forexample, thequotationofHFUGisQ1HFUGQ2.Myfirstruleisthefollowing:

RuleQ1Q2.Q1xQ2namesx.

“Thus the quotation of x names x. Compare this with the firstnaming ruleofRoberts’s system—Qxnamesx.His system isbasedonthemoremodernideaofone-sidedquotation,whichtodayisusedin many computer programming languages such as LISP. Suchsystemsuseonlyanopeningquote,whichforRobertsisthesymbolQ.“Next, by the norm of an expression, I mean the expressionfollowedbyitsownquotation.Thus,thenormofxisxQ1xQ2.Thisnotion of norm I have borrowed from the logician RaymondSmullyan.Theoperationof taking thenormofanexpressionplaysthe same fundamental role inmy system as repetition does in theRoberts system. And so, instead of Roberts’s Rule R, my basicnamingruleis:

RuleN.Ifxnamesy,thenNxnamesthenormofy.

“Thus,ifxnamesy,thenNxnamesyQ1yQ2.Inparticular,NQ1yQ2namesyQ1yQ2.Ihaveothernamingrulestoo,butfirstIwouldliketoshowyousomethingsthatcanbedonewithjustthesetwo.”Professor Quincy then gave his three guests the followingproblems:

•1•

Findanxthatnamesitself.

•2•

Thefirstfixed-pointprincipleofRoberts’ssystemalsoholdsforthepresent system—that is, for any expression a, there is some x thatnamesax.Givearecipeforfindingsuchanx,givena.

•3•

There is an expression that names its ownnorm, and another thatnamesthenormofitsnorm.Doyouseehowtofindthem?

•4•

“MycreationruleisthesameasintheRobertssystem,”saidQuincy.“Ifxnamesy(inmysystem)thenCxcreatesy.AlsomydestructionruleisthesameasRuleDoftheRobertssystem—ifxcreatesy,thenDxdestroysy.“Nowfindanxthatcreatesitselfandanxthatdestroysitself.”

•5•

“Also,givenanyexpressiona,thereisanxthatcreatesaxandanxthatdestroysax.Canyouseewhy?”

•6•

“AswiththeDanielRobertssystem,ifxcreatesy,thenFxisafriendofyandExisanenemyofy.Butinmysystem,thebestfriendofyisFCQ1yQ2,andtheworstenemyofyisECQ1yQ2.“Nowfindanxthatisafriendofxandanxthatisanenemyofx.”

“In theRoberts system,” saidAnnabelle, “onecan findanx thatcreatesafriendofx.Canthisbedoneinyoursystem?”“Probablynot,with the rules Ihavegivenyou so far.Therearetwomorenamingrulestocome;thefirstis:

RuleM.IfxnamesythenMxnamesQ1yQ2(thequotationofy).“With this rule, one can do several things that apparently onecouldnotdobefore.Iwillgiveyousomeexamples.”

•7•

Findanxthatnamesitsownquotation.

•8•

Findanxthatnamesthenormofitsownquotation,andanotherxthatnamesthequotationofitsnorm.

•9•

Show that for any expression a there is some x that names thequotationofax.

•10•

Show that for anyexpressiona thereareexpressionsxandy suchthatxnamesyandynamesax.

“Isitalsotrue,”askedAnnabelle,“thatforanyexpressionsaandb,thereareexpressionsxandysuchthatxnamesayandynamesbx,asintheRobertssystem?”

“I doubt it,” saidQuincy, “and that’swhy I needmy final rule,whichIamabouttogiveyou.“Foranyexpressionxof twoormore letters,byx# I shallmeantheresultoferasingthefirstletterofx—forexample,ifxisBFGH,thenx#isFGH.Ifxconsistsofonlyoneletter,thenwewilltakex#to be x itself (this is for ‘waste’ cases).My final rule, then, is thefollowingerasurerule:

RuleK.Ifxnamesy,thenKxnamesy#.

“For example, KQ1BFGHQ2 names FGH (sinceQ1BFGHQ2 namesBFGH).“With this ruleadded,myprogrammingsystem iscomplete,andexceptforafewcuriosities,wecandojustabouteverythingthatcanbedoneintheRobertssystem.I’llgiveyousomeexamples.”Quincy thenpresented theproblems that follow.The firstone isreallythebasisfortheothers.

11•TheQuincyRecipes

Givenanyexpressiona,onecanfindanxthatnamesaxQ2,alsoanxthatnamesaxQ2Q2,alsoanx thatnamesaxQ2Q2Q2, and the sameforanynumberofQ2’s.

•12•

Now one can establish the double fixed-point principle: For any aandbtherearexandysuchthatxnamesayandynamesbx.Andtherearetworecipesforfindingxandy.Whatarethey?Andwhataboutthetriplefixed-pointprinciple?

13•MoreQuincyRecipes

Givena,itisusefultohaverecipesonhandtofindanxthat(1)createsaxQ2(2)createsaxQ2Q2

(3)createsaxQ2Q2Q2These will be used in subsequent problems; we shall call them

QuincyCRecipes.Itisalsousefultohaveonhandarecipeforfindinganxthat(4)destroysaxQ2This, however, can be got from (1) by replacingCwithDC;we

callthisaQuincyDRecipe.Now,findtheserecipes.

•14•

Findanxandysuch thatxcreatesyandydestroysx. (Therearetwosolutions.)

•15•

Findanxthatcreatesitsbestfriendandanxthatdestroysitsworstenemy.

•16•

Find an x that is a messiah (a friend of its worst enemy) and asatanicx(anenemyofitsbestfriend).

•17•

Findanxwhoisthebestfriendofsomeonewhodestroysx’sworstenemy.

•18•

Findanxwhocreates thebest friendof someonewhodestroysx’sworstenemy.

“This should give you an idea of how my system works,” saidQuincy.“Thepossibilitiesarereallyendless.”

II.CANYOUFINDTHEKEY?

“Ifyoulikecuriositiesandcomplexproblems,”saidQuincy,“ImusttellyouofacurioussystemdevisedbyProfessorCudworth,whoisanotherroboteeronthisisland.“Hewascurious toknowwhatwouldhappenwithmysystem if

thenormalization rule—RuleN,which is reallyat theheartofmysystem—wasreplacedbytherepetitionrule—RuleRofRoberts.Hetried this and got nowhere. Yes, he could solve a few trivialproblems,but theywereuseless forconstructingprograms. Indeed,hecouldn’tevengetaprogramforaself-duplicatingrobot—thatis,hecouldn’tevengetanxsuchthatxcreatesx.Then,afriendofhis,Inspector Craig of Scotland Yard, who takes a great interest incombinatorial puzzles suggested that Cudworth add the reversalrule, and when he did so, the system worked fine! It could doeverything that could be done in my system, and also everythingthatcouldbedoneinRoberts’ssystem,andmore!”“Whatisthereversalrule?”askedAlexander.“Oh, by the reverse of an expression is meant the expression

writtenbackwards.Forexample,thereverseofABCDisDCBA.Well,Cudworth took the letter V and added the rule that if x names y,thenVxnamesthereverseofy.Then,asIsaid,thesystemworkedperfectly.”QuincythenreviewedtherulesofCudworth’ssystemforhisthree

guests:

RuleQ1Q2.Q1xQ2namesx.

RuleR.Ifxnamesy,thenRxnamesyy.

RuleM.Ifxnamesy,thenMxnamesQ1yQ2.

RuleK.Ifxnamesy,thenKxnamesy#.

RuleV.Ifxnamesy,thenVxnamesthereverseofy.

“The creation, destruction, friendship and enmity rules are thesameasinmysystem,”continuedQuincy.“Now,thefirstproblemistofindanxthatnamesitself.Thisisnotsimple!TheshortestoneIknow has eighteen letters. There is also an x that names its ownnorm.TheshortestonethatIknowhasthirty-fourletters.“Thereisalsoanxthatnamesitsownreverse,onethatcreatesits

ownreverse,onethatdestroysthereverseofitsrepeat,onethatisafriendofanenemyofthereverseofthenormofitsownquotation,and I could go on endlessly mentioning other combinations. Thesystem also obeys the fixed-point principle, the double fixed-pointprinciple,andsoforth.“Youmighthavefunworkingtheseproblemsoutatyourleisure.I

believethatonceyousolvethefirstproblem,youwillhavethekeytoalltheothers.”

•19•

ProfessorQuincywasright insayingthat theessential ideabehindtheconstructionofanxthatnamesitselfisreallythekeytosolvingalltheothers.Whatisthekey?WhenourthreefriendsleftProfessorQuincy,theyfeltthatitwas

too late to set sail for home that evening, so they decided to stayovernightataninn.“Tomorrow,” said the Sorcerer, “before we leave this island, I

would like us to visit the station of Simon Simpson, who is theyoungestandmostprogressiveroboteeronthisisland.I’veheardheisabitonthevainside,butalsothathisprogramsystemisaverysimpleandelegantone,anddefinitelyworthlearning.”The three stayed up about half the night trying to decipher

Cudworth’s strange system, but they finally managed to solve it.Exhausted,theywenttosleepanddidnotrisetilllatenextmorning.After breakfast they went off to visit Simon Simpson. Whathappenedwillberelatedinthenextchapter.

Solutions

1.NQ1NQ2namesthenormofN,whichisNQ1NQ2.AndsoNQ1NQ2namesitself.

2. Take x = NQ1aNQ2. It names the norm of aN, which isaNQ1aNQ2,whichisax.

3.IfynamesNy,thenNywillnamethenormofNy.Thenwetakex=Ny,andxwillthennameitsownnorm.AccordingtothesolutiontoProblem2 (which is the fixed-point recipe forQuincy’s system)wecantakey=NQ1NNQ2 (that is,wetakeNfora),andsoanxthatnamesitsownnormisNNQ1NNQ2.AnxthatnamesthenormofitsnormisNNNQ1NNNQ2.

4.CNQ1CNQ2createsitself(sinceNQ1CNQ2namesCNQ1CNQ2).DCNQ1DCNQ2destroysitself.

5.Anxthatcreatesaxis foundbyfirst findingsomeythatnamesaCy,andthentakingxtoequalCy.Wetakey=NQ1aCNQ2,andsooursolutionisx=CNQ1aCNQ2.AnxthatdestroysaxisDCNQ1aDCNQ2.

6.WeneedsomeythatcreatesFy,andthenFywillbeafriendofFy. We take x= Fy, and x is then a friend of itself. By the lastproblem (using F for a), we take y = CNQ1FCNQ2, and soFCNQ1FCNQ2isafriendofitself.AnenemyofitselfisECNQ1ECNQ2.

Discussion. Before giving further solutions, let us pause and notethat so far these solutions are not verydifferent from those of theRobertssystem—justreplaceNbyRandbothQ1andQ2byQandthe above solutions become solutions in the Roberts system. Forexample,compareNQ1NQ2,whichnames itself inQuincy’ssystem,toRQRQ,whichnamesitselfintheRobertssystem.Oragain,inthe

lastproblem,ECNQ1ECNQ2isanenemyofitselfinQuincy’ssystem,whereasECRQECRQisanenemyofitselfintheRobertssystem.Butintheproblemsthatfollow,thesituationchangesquitenoticeably.

7.MNQ1MNQ2works, becauseNQ1MNQ2 names the normofMN,which isMNQ1MNQ2. HenceMNQ1MNQ2 names the quotation ofMNQ1MNQ2.

8.AnxthatnamesthenormofitsownquotationisNMNQ1NMNQ2.AnxthatnamesthequotationofitsnormisMNNQ1MNNQ2.

9.Takex=MNQ1aMNQ2.

10. This follows from the last problem: Since there is some x thatnamesthequotationofax,wecantakeytobethequotationofax,andsoythennamesax.Hencethesolutionisx=MNQ1aMNQ2;y=Q1aMNQ1aMNQ2Q2.Wenote thatwithQuincy’s final rule (RuleK) there is a second

solutionthatwewilldiscusslater.

11.AnxthatnamesaxQ2isKMNQ1aKMNQ2,becauseNQ1aKMNQ2namesaKMNQ1aKMNQ2,which is ax, henceMNQ1aKMNQ2 namesQ1axQ2,henceKMNQ1aKMNQ2namesaxQ2.

Remark.This gives a second solution to the lastproblem:Wecantake y such that it names aQ1yQ2 (by taking aQ1 instead of a)—namely, y = KMNQ1aQ1KMNQ2. Then, we take x to be thequotationofy—i.e.,x=Q1KMNQ1aQ1KMNQ2Q2.

AnxthatnamesaxQ2Q2isKKMMNQ1aKKMMNQ2.AndanxthatnamesaxQ2Q2Q2isKKKMMMNQ1aKKKMMMNQ2,andsoforth.

12.OnewayistogetanxthatnamesaQ1bxQ2andthentakey=

Q1bxQ2.Weuse theQuincyrecipe(takingaQ1b inplaceofa)andget x = KMNQ1aQ1bKMNQ2. We then take y =Q1bKMNQ1aQ1bKMNQ2Q2. (Thereadercanverifydirectlythatthisxandyworks.)AnothersolutionisobtainedbyfirsttakingythatnamesbQ1ayQ2

and then taking x = Q1ayQ2. We then get the solution x =Q1aKMNQ1bQ1aKMNQ2Q2andy=KMNQ1bQ1aKMNQ2.As for3-cycles,wecantakexsuchthat itnamesaQ1bQ1cxQ2Q2

(namely, x = KKMMNQ1aQ1bQ1cKKMMNQ2) and then take z =Q1CxQ2 andy=Q1bzQ2.We leave it to the reader towriteout zandyexplicitly.Anothersolutioncanbeobtainedby takingysuch that itnames

bQ1cQ1ayQ2Q2,thentakingx=Q1ayQ2andz=Q1cyQ2.AnothersolutioncanbeobtainedbytakingzthatnamescQ1aQ1bzQ2Q2,andtakingy=Q1bzQ2andx=Q1ayQ2.Thesesolutionscanbewrittenoutexplicitly,ifsodesired.

13. To find an x that creates axQ2, we need some y that namesaCyQ2 and then take x = Cy. We use the first Quincy recipe ofProblem10andgety=KMNQ1aCKMNQ2.AndsoourdesiredxisCKMNQ1aCKMNQ2.AnxthatcreatesaxQ2Q2isCKKMMNQ1aCKKMMNQ2.AnxthatcreatesaxQ2Q2Q2isCKKKMMMNQ1aCKKKMMMNQ2.AnxthatdestroysaxQ2isDCKMNQ1aDCKMNQ2.

14. Oneway is to take an x that createsDCQ1xQ2,which in turndestroysx.UsingtheQuincyCRecipe,wegetthesolution

x=CKMNQ1DCQ1CKMNQ2y=DCQ1CKMNQ1DCQ1CKMNQ2Q2

AnotherwayistotakesomeythatdestroysCQ1yQ2andtakex=

CQ1yQ2.Wethenget

x=CQ1DCKMNQ1CQ1DCKMNQ2Q2y=DCKMNQ1CQ1DCKMNQ2

15.Thisisnowsimple:BytheQuincyCRecipe,anxthatcreatesitsbestfriendFCQ1xQ2isCKMNQ1FCQ1CKMNQ2.AnxthatdestroysitsworstenemyisDCKMNQ1EDCQ1DCKMNQ2.

16.Toget amessiah,weneed somez that createsECQ1FzQ2 (theworst enemyof Fz) and Fz is then amessiah.Using theQuincyCRecipe,wetakez=CKMNQ1ECQ1FCKMNQ2.Andso,ourmessiahisFCKMNQ1ECQ1FCKMNQ2.AsatanicxisECKMNQ1FCQ1ECKMNQ2.

17.Letybethis“someone”whodestroysx’sworstenemy.Thenxisthe best friend of y, so x = FCQ1yQ2. The worst enemy of x isECQ1xQ2,whichintermsofyisECQ1FCQ1yQ2Q2.AndsoweneedytocreateECQ1FCQ1yQ2Q2.WeuseaQuincyCRecipeandgety=CKKMMNQ1ECQ1FCQ1CKKMMNQ2. And so our x isFCQ1CKKMMNQ1ECQ1FCQ1CKKMMNQ2Q2.

18.AgainweletybethesomeonewhodestroysECQ1xQ2(theworstenemyofx).Wewill takey tobeDCQ1ECQ1xQ2Q2.Then thebestfriend of y is FCQ1DCQ1ECQ1xQ2Q2Q2. And sowe need an x thatcreatesFCQ1DCQ1ECQ1xQ2Q2Q2.WeuseaQuincyCRecipeandwegetx=CKKKMMMNQ1FCQ1DCQ1ECQ1CKKKMMMNQ2.

19.ThekeytoalltheproblemsthatariseinCudworth’ssystemistofindanexpressions thatdoes the same jobas the letterNdoes inQuincy’s system—that is, we want s to be such that for anyexpressionsxandy,ifxnamesy,thensxwillnamethenormofy.Suchanexpressionsmightaptlybecalledanormalizer.

Iwillusethenotation tomeanthereverseofy.So,howcanweget from an expression y to its norm yQ1yQ2 by a sequence ofoperationsallofwhichcanbeprogrammedinthissystem?Onesuchsequenceisthefollowing:(1)Takethequotationofy,thusgettingQ1yQ2.(2)Reversewhatyounowhave,gettingQ2 Q1.(3)ErasetheQ2,getting Q1.(4)Reversewhatyouhave,gettingQ1y.(5)Repeatwhatyounowhave,gettingQ1yQ1y.(6)ErasetheleftmostQ1,gettingyQ1y.(7)Nowtakeitsquotation,gettingQ1yQ1yQ2.(8)ErasetheleftmostQ1,gettingyQ1yQ2.

Andsosupposexnamesy.Then(1)MxnamesQ1yQ2.(2)VMxnamesQ2 Q1.(3)KVMxnames Q1.(4)VKVMxnamesQ1y.(5)RVKVMxnamesQ1yQ1y.(6)KRVKVMxnamesyQ1y.(7)MKRVKVMxnamesQ1yQ1yQ2.(8)KMKRVKVMxnamesyQ1yQ2.And so, KMKRVKVM is a normalizer. All the problems of this

chaptercanbesolvedinCudworth’ssystembytakingKMKRVKVMinplaceofthesingleletterN.For example, an expression that names itself is

KMKRVKVMQ1KMKRVKVMQ2. (Is there a shorter one?) AnexpressionthatcreatesitselfisCKMKRVKVMQ1CKMKRVKVMQ2.Iwill leave to the reader, as an exercise, the construction of an

expression that names its own reverse. Also, what about anexpression that creates its own reverse? Andwhat about one thatcreatestherepeatofitsreverse?

13FROMTHERIDICULOUSTOTHESIMPLE

“THOSEOLD-FASHIONEDSYSTEMSreallygetme,”saidSimonSimpson,shakinghis head. “Do you know, for example, the weird system ofCudworth?”“Yes,” said the Sorcerer, “Professor Quincy showed it to usyesterday.”“Now,ifthatisn’tthecraziestsystemI’veeverseeninmylife!Togotoall that trouble just togetanormalizer!SometimesIwonderaboutthesanityofsomeofmycolleagues!”“Oh, I don’t know,” said the Sorcerer. “I found it quite aninterestingchallengetofindthekeytothesystem.”“But the system is so unnecessarily complicated!” said Simpson.“Cudworthseemstodelightinmakingthingsasdifficultaspossible.Myphilosophyistheveryopposite—Iliketomakethingsassimpleaspossible.“Actually,Quincy’ssystemisnotallthatbad,”Simpsonwenton,“andconsideringthatitwasdevisedbeforeone-sidedquotationwasdiscovered,itisunderstandable.Still,thelastrule—theerasurerule—seemsartificialandnothingmorethananadhocdevicetoremedyanone-too-good situation. But evenhere,Quincy couldhave donebetter:hecouldhavereplacedthetworules—RuleMandRuleK—by a single rule: if x names y, then Lx names yQ2. That one rulewouldhaveyieldedthedoubleandtriplefixed-pointprinciples,andalltheproblemsthatheprobablygaveyouwouldbesolvable.“Ofcourse,theRobertssystemismuchcleanerandmorenatural.But even this system is more complex than it need be for theproblemsthatreallymatterfromapracticalpointofview.Iamnotinterestedinsuchacademicquestionsaswhetheranxcanbefoundthatcreates itsownrepeat;ofwhatsignificance is that torobotry?Myapproachispurelypragmatic.I’minterestedonlyinquestionsof

sociologicalimportance:Whichrobotscreatewhich?Whichdestroywhich?Whichare friends,best friends, enemies,worst enemies,ofwhich?And formatters like these,myprogramsystem is themostefficientofall.”“Do you use one-sided or two-sided quotation?” asked theSorcerer.“Neither;my system is quotationless. I don’t botherwith certainexpressionsnamingothers.”“That’s interesting,” said the Sorcerer. “I have also beenexperimenting of latewith quotationless systems—not for robotry,butforcertaingeneralproblemsthatarisewithself-reference.Iamreallycurioustohearaboutyoursystem.”“My rules are direct, short, sweet, and to the point. I use thesymbolsC,D,F,E,Ċ,Ḋ,Ḟ,Ėandmyrulesarethese:

RuleC.Cxcreatesx.

RuleĊ.Ċxcreatesxx.

RuleD.Dxdestroysx.

RuleḊ.Ḋxdestroysxx.

RuleF.Fxisthebestfriendofx.

RuleḞ.Ḟxisafriendofxx.

RuleE.Existheworstenemyofx.

RuleĖ.Ėxisanenemyofxx.“What couldbemoredirect than these?The solutions to all theproblems I will now give you are shorter than any you have yetseen.Forexample, it isobvious thataself-reproducingrobot isĊĊanda self-destroyingone isḊḊ.Sohereare someproblems, and Iam sure you will solve them quite easily. And incidentally, thesolutionsplacethesolutionsinRoberts’ssysteminaclearerlight,in

amannerIwill laterexplain.Butfornow,let’sconcentrateonjusttheproblems.”ProfessorSimpson’sproblemsfollow.

•1•

Findadistinctxandysuchthateachcreatestheother.

•2•

Findanxandysuchthatxcreatesyandydestroysx.Therearetwosolutions.

•3•

Showthatforanyexpressionathereissomexthatcreatesax,andsomexthatdestroysax.

•4•

Givenanyexpressionsaandb,show:

(a) There are an x and y such that x creates ay and y creates bx.(Therearetwosolutions.)

(b) There are an x and y such that x destroys ay and ydestroysbx.(Twosolutions.)(c) There are an x and y such that x creates ay and ydestroysbx.

•5•

Findanxthatisafriendofitself.

•6•

Findanxthatcreatesitsbestfriend.

•7•

Findanxthatcreatesafriendthatisnotitsbestfriend.

•8•

Findanxthatisafriendofitsworstenemy.

•9•

Findanxthatisthebestfriendofoneofitsenemies.

•10•

Find an x and y such that x creates the best friend of y and ydestroystheworstenemyofx.

•11•

Findanxthatisthebestfriendofonethatdestroysitsworstenemy.

•12•

Findanxthatcreatessomeythat isa friendofsomezthat is theworst enemy of somew that destroys the best friend of theworstenemyofx.

“And so you see,” said Simpson proudly, “that all sorts ofcomplicated sociological situations can be programmed very easilyinmysystem.”“Your system is indeed neat and economical,” said the Sorcerer,“and I like it very much. It has many similarities to a system ofmine,andbyastrangecoincidenceyouusethedotoveraletterinmuchthewayIdo.“One thingyoumaynot realize is thatall your solutions canbeeasily transformed into solutions in the Roberts system simply byreplacingCbyCQ,ĊbyCRQ,DbyDCQ,ḊbyDCRQ,FbyFCQ,Ḟ

by FCRQ, E by ECQ, and Ė by ERCQ. For example, your x thatcreatesitselfisĊĊ.IfwereplaceĊbyCRQ,wegetCRQCRQ,whichisthexoftheRobertssystemthatcreatesitself.Thesameholdsforallyoursolutions.”“I certainly do realize that,” said Simpson, “and that is what I

meantbeforewhenIsaidthatmysolutionsputtheRobertssolutionsinaclearerlight.”

The reader caneasily check that ineachof the twelveproblemsabove,thesolutionsintheSimpsonsystemcanbetransformedintosolutions in the Roberts system by just the method stated by theSorcerer.

Solutions

1.ĊCĊandCĊCĊ

2.Solution1:x=ĊDĊ,y=DĊDĊ

Solution2:x=CḊCḊ,y=ḊCḊ

3.AnxthatcreatesaxisĊaĊ.AnxthatdestroysaxisḊaḊ.

4.(a)Solution1:x=ĊaCbĊ,y=CbĊaCbĊ

Solution2:x=CaĊbCaĊ,y=ĊbCaĊ(b)Same,usingḊinplaceofĊandDinplaceofC(c)Solution1:x=ĊaDbĊ,y=DbĊaDbĊSolution2:x=CaḊbCaḊ,y=ḊbCaḊ

5.ḞḞ

6.ĊFĊ

7.CḞCḞ

8.ḞEḞ

9.FĖFĖ(RobotĖFĖissatanic)

10.OneSolution:x=ĊFDEĊ,y=DEĊFDEĊ

AnotherSolution:x=CFḊECFḊ,y=ḊECFḊ

11.x=FḊEFḊ

12.Thereareseveralsolutions.Oneofthemisx=ĊFEDFEĊ(whichcreates FEDFEx). Another is x = CḞEDFECḞ. Another isCFEḊFECFEḊ.Ineachsolution,they,z,andwcaneasilybefound.

PARTIV

GÖDELIANPUZZLES

14SELF-REFERENCEANDCROSS-REFERENCE

AFEWDAYSAFTERtheirreturnfromtheIslandofRobots,ourcouplepaidanothervisittotheSorcerer.“Iwasreallyintriguedbythoseprograms,”saidAnnabelle.“Howdidthoserobotengineerseverthinkthemup?”“They are all related to problems in designation,” replied theSorcerer,“particularlytoquotationaldesignation.”“Justwhatisthat?”“I’ll have to explain it to you from the beginning,” said theSorcerer.“Doyouknowthedifferencebetweentheuseandmentionofwords?”NeitherAnnabellenorAlexanderhadeverheardofthis.“I’ll illustrate with an example,” said the Sorcerer, and he thenwrotedownthefollowingsentence.

(1)ICEISFROZENWATER.

“Isthatsentencetrueorfalse?”“Obviouslytrue,”theybothreplied.“Allright.Now,whataboutthefollowingsentence?”

(2)ICEHASTHREELETTERS.

“Alsotrue,”saidAnnabelle.“Ofcourseit’strue,”saidAlexander.“Icedoeshavethreeletters.”“Noitdoesn’t!”saidtheSorcerer.“Frozenwaterdoesn’thaveanylettersatall!It’stheword‘ICE’thathasthreeletters.Sosentence(2)asitstandsissimplywrong!Thecorrectversionisthefollowing.

(2)“ICE”HASTHREELETTERS.

“Sentence (1) talks about ice the substance; Sentence (2) talksabout ice the substance, but what it says is wrong. Sentence (2)’talksabouttheword‘ICE’andwhatitsaysistrue.Onetalksaboutthewordbyenclosingitinquotes—atleastthat’stheolderwaythatwewillworkwith forawhile.Anyway, inSentence (1), theword‘ICE’isused,sinceittalksaboutthesubstance,notabouttheword.ButinSentence(2),theword‘ICE’ismentioned,ortalkedabout,butnotused,sincethesentencetalksabouttheword,notabouticethesubstance.”“Thatseemsclearenough,”saidAnnabelle.“WhatisusedinSentence(2),”continuedtheSorcerer,“isnottheword‘ICE’butthenameorquotationofthatword.Itisusedtotalkabout theword. I know you grasp the general idea, yet beginnersoften lapse into confusing use withmention. Here, let me try thefollowing.”TheSorcerer thenwrotedown the followingandaskedthecouplewhetheritwastrueorfalse.

(3)“““ICE”””HASTHREEPAIRSOFQUOTES.

“That’sobviouslytrue,”saidAnnabelle.“Ofcourseitis!”saidAlexander.“I’msorry,butyouarebothwrong,”saidtheSorcerer.“Yes,whatIwrotedowndoesindeedhavethreepairsofquotes,butwhatittalksabout has only two pairs of quotes. The correct rendition is thefollowing:

(3)“““ICE”””HASTWOPAIRSOFQUOTES.

“It’sabitconfusing,”saidAnnabelle.“Allright,maybethiswillhelp.Isn’tthefollowingtrue?”

(4)ICEHASNOPAIRSOFQUOTES.

“Yes,”saidAlexander,“thesubstanceicehasnopairsofquotes.”“Thenwhataboutthefollowing?”askedtheSorcerer.

(5)“ICE”HASNOPAIRSOFQUOTES.

“That’sfalse,”saidoneofthetwo.“Iseeapairofquotes.”“That’swhatyou see,” said theSorcerer, “butwhat the sentencetalksabout is theword ‘ICE’withoutanyquotesaround it.Andsosentence(5)istrue!Whatisusedhasonepairofquotes,butwhatistalkedabout,ormentioned,hasnone.”“IthinkIbegintosee,”saidAnnabelle.“Good,thenconsiderthefollowingsentence.”

(6)““ICE””HASONEPAIROFQUOTES.

“Iseenowthatthatistrue,”saidAlexander.“Good.Andwhataboutthefollowing?”

(7)“““ICE”””HASTWOPAIRSOFQUOTES.

“Yes,itistrue,”saidAnnabelle.“Good,”saidtheSorcerer.“AndnowIwouldliketoillustratethedifference between ‘use’ and ‘mention’ in an amusing way. Is thefollowingsentencetrueorfalse?”

(8)ITTAKESLONGERTOREADTHEBIBLETHANTOREAD“THEBIBLE.”

“Itcertainlyistrue!”laughedAnnabelle.“Ittakesmuchlonger!”“That’s a case,” said the Sorcerer, “in which ‘the Bible’ is bothusedandmentionedinthesamesentence.“Now,letustrythisone.”

(9)THISSENTENCEISLONGERTHAN“THISSENTENCE.”

“Alsotrue,”theybothsaid.“Good.Nowtellmeinwhatlanguageisthefollowingsentence.IsitFrenchorEnglish?”

(10)“LEDIABLE”ISTHENAMEOFLEDIABLE.

“Iwouldsayboth,”saidAnnabelle.“ItcontainsbothEnglishandFrenchwords.”“That is right,” said the Sorcerer. “Now what about thefollowing?”

(11)“LEDIABLE”ISTHEFRENCHNAMEOFTHEDEVIL.

“Also both,” said Alexander. “It also contains both French andEnglishwords.”“Wrongthistime,”saidtheSorcerer.“TheFrenchwordsinitareboth inquotes, hence talked aboutbutnotused.Thepoint is thatanyEnglish-readingperson,evenifhedidnotknowasinglewordofFrench, could understand the sentence perfectly (and also learn atiny bit of French in the bargain). On the other hand, the sameperson could not understand Sentence (10), since he would notknowwhatitisthat‘lediable’isthenameof.”

SELF-REFERENTIALSENTENCES

“And now,” said the Sorcerer, “I would like to get to themoreinteresting problems of constructing sentences that refer tothemselves.“Supposewewishtoconstructasentencethatascribestoitselfacertainproperty.Tobeexact,let’stakethepropertyofbeingreadbysomeonenamedJohn.HowdoweconstructasentencethatassertsthatJohnisreadingthatverysentence?Ofcourse,oneobviousway

todosoistoconstructthefollowingsentence.

(12)JOHNISREADINGTHISSENTENCE.

“Obviously,sentence(12)istrueonlyifJohnisreadingSentence(12).However,thatsentencecontainsthe indexicalword ‘this,’andwhatwewant is toaccomplish thesameself-referencewithout theuseofanindexical.”“Whatdoyoumeanbyanindexical?”askedAnnabelle.“Oh, an indexical is a term whose designation depends on its

context.Forexample, ‘JamesSmith’ isnotan indexical, for inanycontext itdenotes thepersonJamesSmith,whereas theword ‘I’ istypicallyanindexicalbecausewhenusedbyonepersonitdenotesadifferentpersonthanwhenusedbyanother.WhenJamesSmithsays‘I,’ he means James Smith, whereas when Paul Jones says ‘I,’ hemeans Paul Jones. Another indexical is the word ‘you,’ whosedesignationdependsonthepersontowhomitisaddressed.Anotherindexical is theword ‘now,’whichdesignatesdifferentmomentsoftimewhenutteredatdifferentpointsintime.“The logicianRaymond Smullyan has half-humorously given the

name chameleonic terms to indexicals, which he wrote about in apapercalled‘ChameleonicLanguages.’Ashesaysinthispaper:‘Likeachameleonwhosecolordependsonitssurroundings,thesewordschange their denotation from context to context,’ A friend of hiswith a good sense of humor, who had heard of Smullyan’s paperbefore it was published, wrote to him: ‘I have heard of yourchameleonic languages. I do know what they are, except that Iassumetheyarenotwhattheyappeartobe.’“At any rate, I think you now understand what chameleonic or

indexical terms are. The phrase ‘this sentence’ is obviously anindexical;itsdenotationdependsonthesentenceinwhichitappears—providing, of course, that it is used and not mentioned in thesentence.”“Idon’tthinkIfollowthat,”saidAnnabelle.“Well, consider the followingsentence,and tellmewhether it is

trueorfalse.”

(13)THISSENTENCEHASFIVEWORDS.

“Thatsentenceistrue,”saidAlexander.“Itdoeshavefivewords.”“Right.”saidtheSorcerer.“Nowwhataboutthefollowing?”

(14)THISSENTENCEHASEXACTLYTWOWORDS.

“That’sobviouslyfalse,”saidAnnabelle.“Right.Nowwhataboutthefollowing?”

(15)“THISSENTENCE”HASEXACTLYTWOWORDS.

“Oh, I begin to see what you mean,” said Annabelle. “The lastthing you wrote is indeed true. Sentence (15) doesn’t say thatSentence(15)hasexactlytwowords,whichitobviouslyhasn’t,butthat the phrase ‘this sentence’ has exactly two words, which isindeedtrue.Ontheotherhand,Sentence(14)doesn’tsaythat‘thissentence’hasexactlytwowords,butthatthewholeofSentence(14)hasexactlytwowords,whichisclearlyfalse—ithassixwords.”“Very good!” said the Sorcerer. “I see you are becoming able to

distinguish between use and mention. You now realize that thephrase‘thissentence’whensurroundedbyquotesisnotindexical;italwaysdenotesthetwowordsthatappearwithinthequotes.“And now Iwant to explain how self-reference can be obtained

withoutusingindexicals.”“Whywouldonewantto?”askedAnnabelle.“Whatiswrongwith

indexicals?Theyseemveryuseful!”“Of course they are,” said the Sorcerer. “They work fine for

languageslikeordinaryEnglishthathaveindexicals,butforformalmathematical systemsof the type studiedbyKurtGödel—which iswhatI’mleadingupto—therearenoindexicals;henceGödelhadtoachieveself-referencewithouttheuseofindexicals.”

“Howdidhedothat?”askedAlexander.“That’swhatI’mcomingto.Youknowthatinalgebraoneusesthe

letters x andy as standing forunknownnumbers, and that on theisland we just visited, the robot engineers used these letters asstanding for arbitrary expressions of their programming languages.Well,IwillnowusetheletterxasstandingforunknownexpressionsofordinaryEnglish.“And now, using an essential idea of Gödel’s, I will define the

diagonalization of an expression to be the result of replacing thesymbol x by the quotation of the entire expression. For example,considerthefollowingexpression:

(1)JOHNISREADINGx.

“Expression(1)isnotasentence;it isneithertruenorfalseasitstands, since we don’t know what the symbol x stands for. If wereplace x by the name of some expression, then (1) becomes anactual sentence, trueor falseas thecasemaybe.Wecould indeedreplace x by the quotation of Expression (1), thus getting itsdiagonalizationwhichis:

(2)JOHNISREADING“JOHNISREADINGx.”

“Now, (2) is an actual sentence; it asserts that John is readingExpression (1), and it is true if and only if John is reading (1).However, (2) is not self-referential; it does not assert that John isreading(2),butthatJohnisreading(1).Toachieveaself-referentialsentence, insteadof startingwithExpression (1),we startwith thefollowing.

(3)JOHNISREADINGTHEDIAGONALIZATIONOFx.

“Now let’s see what the diagonalization of Expression (3) lookslike.

(4)JOHNISREADINGTHEDIAGONALIZATIONOF“JOHNISREADINGTHEDIAGONALIZATIONOFx.”

“At first, (4) may not appear to make much sense, but a littlethought shows that it makes a good deal of sense and,moreover,reveals something very interesting! Sentence (4) says that John isreadingthediagonalizationof(3),but thediagonalizationof (3) is(4) itself.AndsoSentence(4)assertsthatJohnisreadingtheverySentence (4)!Thus (4) is a self-referential sentence.Credit for thisbasicideagoestoGödel.“Ibelieveitiseasiertoseethissymbolically.LetususetheletterJ as an abbreviation of “John is reading,” and let us use D toabbreviate“thediagonalizationof.”Thus(1)inabbreviatednotationisthis:

(5)Jx

“Itsdiagonalizationis:

(6)J“Jx”

“What (6) says is that John is reading the two-letter expression‘Jx’—thus(6)saysthatJohnisreading(5).Itisnotself-referential;(6)doesn’tsaythatJohnisreading(6).Nowconsiderthefollowing.

(7)JDx

“Itsdiagonalizationis:

(8)JD“JDx”

“What(8)saysisthatJohnisreadingthediagonalizationof(7),butthediagonalizationof(7)is(8)itself.Thus(8)isself-referential

—Sentence(8)assertsthatJohnisreadingSentence(8)!“Self-reference comes up in a crucial way in Gödel’s famous

incompleteness theorem,which Iwill later tellyouabout.And theidea of diagonalization comes very close to the technique used byGödeltoachieveself-reference.However,therearesimplermethods—subsequently discovered by the logicians Alfred Tarski, WillardQuine,andRaymondSmullyan—oneofwhichIwillnowshowyou.“In a paper called Languages in Which Self-Reference Is Possible,

Smullyan defines the norm of an expression to be the expressionfollowedbyitsownquotation.Letusconsideranexample.Webeginwiththefollowingexpression:

(9)JOHNISREADING

“Itsnormis:

(10)JOHNISREADING“JOHNISREADING”

“Sentence (10) is not self-referential; it doesn’t say that John isreading(10),butthatJohnisreading(9).Butnow,let’sconsiderthefollowinginsteadof(9).

(11)JOHNISREADINGTHENORMOF

“Itsnormis:

(12)JOHNISREADINGTHENORMOF“JOHNISREADINGTHENORMOF”

“Sentence(12)isself-referential;itassertsthatJohnisreadingthenormof(11),butthenormof(11)is(12)itself.“Let’s look at the symbolic version. As before, we will use J to

abbreviate ‘John is reading,’ andwewill nowuseN to abbreviate

‘thenormof.’So(11)isabbreviatedthus:

(13)JN

“And(12)isabbreviatedthus:

(14)JN“JN”

“Now, (14) is the norm of (13). Also (14) asserts that John isreading the norm of (13). Thus (14) asserts that John is reading(14),so(14)isaself-referentialsentence.“Comparetheself-referentialsentenceJN‘JN,’whichisbasedon

normalization, with the sentence JD ‘JDx’ which is based ondiagonalizationandwhichcameearlier.“Using diagonalization, can we construct an expression that

denotes itself? Yes; such an expression is D‘Dx’which denotes thediagonalizationofDxand thediagonalizationofDx isD‘Dx’.ThusD‘Dx’denotesitself.“But using normalization, the solution is even simpler: The

expressionN‘N’denotesthenormoftheletterN,whichisN‘N.’ThusN‘N’ denotes itself. Of course this is the same as NQ1NQ2 ofProfessor Quincy’s system, only using Q1 in place of the openingquote and Q2 in place of the closing quote. That’s what Quincymeant when he said that his system was based on two-sidedquotation.”“Oh, I’m really glad tounderstand thatnow,” saidAnnabelle. “I

wasactuallypuzzledwhenhesaidthat,butIdidn’twishtointerrupthim.Healsomentionedsystemsbasedonone-sidedquotation.Whatarethey?”“One uses opening quotes but not closing quotes,” said the

Sorcerer.“Ithascertainadvantagesandcertaindisadvantagesovertwo-sided quotation. One takes a symbol—not the usual one foropening quotes—say the symbol°, and given an expression x, oneuses °x instead of x as a name for x. As I said, there are certain

advantages to this, which I will shortly explain. Certain machinelanguagessuchasLISPuseone-sidedquotation.“Following Professor McCulloch’s terminology (from Smullyan’s

book The Lady or the Tiger?), we will define the associate of anexpressiontobetheexpression,followedbyanasterisk(thesymbol“°”), followed again by the expression. And now, let us use thesymbolAtoabbreviate‘theassociateof.’Then,insteadofJN‘JN,’asa sentence that asserts that John is reading it, we can use thefollowing:

(15)JA°JA

“Sentence(15)assertsthatJohnisreadingtheassociateofJA,butthe associate of JA is the very sentence JA°JA. And so we haveanothermethodofachievingself-reference.“Also, an expression that denotes itself is A°A. It denotes the

associateofA,whichisA°A.“Anadvantageofone-sidedquotationisthatitaffordsarelatively

easy method of achieving cross-reference—the construction of twosentenceseachofwhichtalksabouttheother.Suppose,forexample,we consider two individuals—John and Paul—and we wish toconstructtwosentencesxandysuchthatxsaysthatJohnisreadingy, and y says that Paul is reading x. How can this be done? Let’scontinue to let J abbreviate ‘John is reading,’ and let’s use P toabbreviate‘Paulisreading,’Now,howcanthesentencesxandybeconstructed?”

•1•

Howcanthey?Therearetwosolutions.

•2•

Suppose nowwe consider also a third person—call himWilliam—andusetheletterWtoabbreviate“Williamisreading.”

Construct sentences x, y, and z such that x says that John isreadingy,y says thatPaul is readingz,andz says thatWilliam isreadingx.

“This notion of associate,” said the Sorcerer, “was the basis ofseveral of McCulloch’s systems reported in The Lady or the Tiger?Now, it sohappens that theoperationof repeating canbeused forachieving self-reference and cross-reference just as well as theoperationoftakingtheassociate.Idon’tbelievethatSmullyanwasaware of this when he wrote The Lady or the Tiger?, but hedemonstrated this in a later paper called Quotation and Self-Reference.Nowlet’stakealookattheideasinvolved.“Bytherepeatofanexpressionwemeantheexpressionfollowed

byitself.LetususetheletterRtoabbreviate ‘therepeatof,’UsingthesymbolsJ,R,and°,howdoweconstructasentencethatassertsthat John is reading it? A common wrong guess is JR°JR. This iswrong because JR°JR says that John is reading the repeat of JR,whichisJRJR.ThusJR°JRsaysthatJohnisreadingJRJR;itdoesn’tsaythatJohnisreadingJR°JR.“The correct solution is JR°JR°. This sentence says that John is

readingtherepeatofJR°,whichistheverysentenceJR°JR°.“Letusnowcomparethefourwayswenowhaveofconstructinga

sentence that asserts that John is reading it. We can use eitherdiagonalization, normalization, association, or repetition, and werespectivelyhavethefollowingfoursentences.(i)JD“JDx”(ii)JN“JN”(iii)JA°JA(iv)JR°JR°“Of course the repetition method is the whole basis of Charles

Roberts’sprogrammingsystem[Chapter11]. Idon’tknowwhetherRobertsthoughtofthishimselforgottheideafromSmullyan.“Also, repetition can be used as well as association to achieve

cross-reference.”

•3•

UsingthesymbolsJ,P,R,°,constructsentencesxandysuchthatxassertsthatJohnisreadingyandyassertsthatPaulisreadingx.

SELF-REFERENCEUSINGGÖDELNUMBERING

“I have heard the phrase Gödel numbering,” said Annabelle, “andhave been told that Gödel used it to achieve self-reference. Couldyouexplainthistous?”“Certainly. You see, in the mathematical systems studied by

Gödel, the sentences talk about things like numbers and sets, notaboutsentences.Thesystemshavenothing likequotationmarksorother devices for talking directly about expressions. And so Gödelcleverly got around this by assigning to each sentence a number,calledtheGödelnumberofthesentence.Then,roughlyspeaking,theGödelnumberofthesentenceplayedtheroleofthequotationofthesentence.“Asacrudeillustration,supposeIassignnumberstoallsentences

in theEnglish languageandcan findanumbern that is theGödelnumberofthefollowingsentence.

JohnisreadingthesentencewhoseGödelnumberisn.

“But n is the Gödel number of the very sentence, and so thissentenceissayinginaroundaboutwaythatJohnisreadingit.“Now,howcanthisbedone?Iwillshowyoutwoways—thefirst

uses Gödel’s diagonalization method. Let’s go back to using J for‘John is reading,’ and D for ‘the diagonalization of,’ but a newmeaningwillnowbeusedfortheword‘diagonalization.’“Let’susethefivesymbolsJ,D,x,1,0.Andlet’sassigntothese

five symbols the respective Gödel numbers 10, 100,1000, 10000,100000.Foreasierreference,letmerewritethesefivesymbolswiththeirGödelnumbersunderneaththem.

“Then, to any compound expression made up of these fivesymbols, if we replace each symbol by its Gödel number, theresulting numberwill be the Gödel number of the expression. Forexample,theexpressionxJ1DhasGödelnumber10001010000100;theexpressionDJhasGödelnumber10010.“We now redefine the diagonalization of an expression to be the

result of replacing the symbol x not by the quotation of theexpression(thereisn’tanyinthissystem),butbytheGödelnumberoftheexpression(written,ofcourse,inordinarybase10notation).For example, the diagonalization of Jx is J101000. ThediagonalizationofDxJisD100100010J.“Andnow,foranyGödelnumbern,weinterpretJntomeannot

that John is reading the number n, but that John is reading theexpressionwhoseGödelnumberisn.Forexample,J10000100assertsthatJohnisreadingtheexpression‘1D.’Or,J10assertsthatJohnisreading the letter J.Or again, J10010 asserts that John is readingDJ.“And now we interpret JDn to mean that John is reading the

diagonalization of the expression whose Gödel number is n. Forexample, JD10100010 asserts that John is reading thediagonalizationoftheexpressionwhoseGödelnumberis10100010.Well, theexpressionwhoseGödelnumber is10100010 is JxJ, andthe diagonalization of JxJ is J10100010J. And so, JD10100010asserts that John is reading the (meaningless) expressionJ10100010J.“Andnow it shouldbeobvioushowtoconstructa sentence that

assertsthatJohnisreadingthatverysentence.”

•4•

Findthesentence.

“Now, the principle of Gödel numbering can be used with

normalizationinsteadofdiagonalization,aswasdonebySmullyan.Hedidsomethinglikethefollowing.“Letusnowuse the four symbolsJN10—weno longerneedx.

Let’s assign them the respective Gödel numbers 10, 100, 1000,10000. Again, the Gödel number of a compound expression isobtainedbyreplacingeachofthefoursymbolsbyitsGödelnumber.Andnow,weredefinethenormofanexpressiontobetheexpressionfollowed by its Gödel number. For example, the norm of J1JN isJ1JN10100010100. We now interpret JNn to mean that John isreadingthenormoftheexpressionwhoseGödelnumberisn.(IfnisnottheGödelnumberofanyexpression,wetakeJNntobefalse.)”

•5•

NowwhatsimplesentencesaysthatJohnisreadingit?

THESORCERER’SSPECIALSYSTEM

“Recently,”saidtheSorcererwitharatherproudsmile,“Ithoughtofanotherschemeofself-referencewhichusesneitherquotationmarksnorGödelnumbering,andthatworksveryneatlyforcross-referenceaswellasself-reference.ItisremarkablysimilartoSimonSimpson’smethodofprogramming.“Foranyexpressionx, IshallnowwriteJxtomeanthatJohnis

readingtheveryexpressionx.Iwon’tputquotesaroundx,norastarbeforethex,norusetheGödelnumberofx.IshallboldlywriteJxtomeanthatJohnisreadingx.AndnowIshallwrite xtomeanthatJohnisreadingtherepeatofx.Thus xmeans thatJohn is readingxx.(Also,Jxxmeansthesamething.)“Self-reference is now completely trivial; the sentence asserts

thatJohnisreadingtherepeatof ,whichis .Thus saysthatJohnis reading thevery sentence .That’saboutas simpleamethodofgettingself-referenceascanbeimagined.“Cross-reference,thoughnotquitethatsimple,isstillrelativelyso

comparedtotheothermethodswehavediscussed.IwillwritePxto

meanthatPaulisreadingx,andṖxtomeanthatPaulisreadingxx.Andlikewise, forWilliam,IwilluseWand similarly.Then,canyou see how to construct sentences x and y such that x says thatJohnisreadingy,andysaysthatPaulisreadingx?Andcanyouseehowtoconstructsentencesx,y,andzsuchthatxsaysthatJohnisreadingy,y says thatPaul is readingz,andz says thatWilliam isreadingx?”

•6•

Whatarethesolutions?

Solutions

15THESORCERER’SMINIATUREGÖDELIANLANGUAGE

“TODAY,”saidtheSorcerer,“IwanttoshowyouaminiatureversionofGödel’sfamousincompletenesstheorem.Itwillserveasabridgefromwhatwe did last time towhatwe’ll get into a bit later. ThesystemIwillnowpresent isamodernizedandstreamlinedversionof a Smullyan ‘language,’ I shall use the quotationless method IshowedyoulasttimeinplaceofSmullyan’sone-sidedquotation.“Inthesystem,varioussentencescanbeproved.ThesystemusesthefoursymbolsP,Ṗ,Q, .ThesymbolPmeansprovabilityinthesystem—thus,foranyexpressionXinthelanguageofthesystem,PXasserts that X is provable in the system and accordingly will becalledtrueifandonlyifXisprovableinthesystem.ThesymbolQstandsfornonprovabilityinthesystemandforanyexpressionX,QXassertsthatXisnotprovableinthesystemandQXiscalledtruejustinthecasethatXisnotprovableinthesystem.Next,ṖXmeansthatXXisprovable inthesystem,andisaccordinglytrue ifandonly ifthis is the case. Lastly, X means that XX is not provable in thesystem, and is called true if and only if XX is not provable in thesystem. By a sentence is meant any expression of one of the fourforms PX, ṖX, QX, X, where X is any combination of the foursymbols.Ihenceforthusethewordprovabletomeanprovableinthesystem.Letusreviewthebasicfacts.(1)PXassertsthatXisprovable.(2)QXassertsthatXisnotprovable.(3)ṖXassertsthatXXisprovable.(4) XassertsthatXXisnotprovable.“Weseethatthesystemisself-referentialinthatitprovesvarioussentencesthatassertwhatthesystemcanandcannotprove.Wearegiven that the system is wholly accurate in that every sentenceprovable in the system is true—in other words the following four

conditionshold(whereXisanyexpression).C1:IfPXisprovablesoisX.C2:IfQXisprovablethenXisnotprovable.C3:IfṖXisprovablesoisXX.C4:If XisprovablethenXXisnotprovable.“Now,justbecauseeverysentenceprovableinthesystemistrue,itdoesn’tnecessarily followthatevery truesentence isprovable inthesystem.Asamatteroffact,therehappenstobeasentencethatistruebutnotprovableinthesystem.Canyoufindit?”

•1•

Findatruesentencethatisnotprovableinthesystem.

RefutableSentences“Foreachsentence,wedefineitsconjugateasfollows.TheconjugateofPXisQX,andtheconjugateofQXisPX.TheconjugateofṖXis Xandtheconjugateof XisṖX.ThusthesentencesPXandQXareconjugatesofeachother,andthesentencesṖXand Xareconjugatesofeachother.Givenanyconjugatepair,itisobviousthatoneofthepairistrueandtheotherfalse.“A sentence is called refutable (in the system) if its conjugate isprovable (in thesystem).Thus,PX is refutable ifandonly ifQX isprovable,andPXisprovableifandonlyifQXisrefutable.LikewisewithṖXand X.”

•2•

Findasentencethatassertsthatitisrefutable.

•3•

Findasentencethatassertsthatitisnotrefutable.

•4•

Whatsentenceassertsthatitisprovable?

Undecidable Sentences. “A sentence is called undecidable (in thesystem)ifitisneitherprovablenorrefutableinthesystem,”saidtheSorcerer. “Now, as you saw in the solution of Problem 1, thesentence istruebutnotprovableinthesystem.Sinceit is true,thenitsconjugateṖ isfalse,hencealsonotprovableinthesystem.Thusthesentence isundecidableinthesystem.“My argument has appealed to the notion of truth, but evenwithoutappealtothisnotiononecanobtaintheundecidabilityof as a direct consequence of conditions C1 through C4 as follows:Suppose wereprovable.ThenbyC4,taking forX,therepeatofisnotprovable,whichmeansthat isnotprovable.Soif isprovable,thenitisnotprovable,whichisacontradiction.Therefore, is not provable. If its conjugate Ṗ were provable, then by C3

(taking forX) wouldbeprovable,whichwejustsawisnotthecase. And so Ṗ is not provable either. Thus the sentence isundecidableinthesystem.”“Tellmethis,”saidAnnabelle.“Is theonlysentencethatistruebutunprovable,orarethereothers?”“Thesentence ,”repliedtheSorcerer,“istheonlysentencethatI know of having the property that for every system satisfyingconditionsC1throughC4,itistrueforthatsystemandunprovableinthatsystem.But,asyouwill see later, foranysystemsatisfyingC1throughC4,thereareothersentencesthataretruebutunprovableinthat system.The sentence is, as I said, theonly sentence that IknowthatsimultaneouslyworksforallsystemssatisfyingC1throughC4.”TheSorcererthengavethefollowingproblems:

5•SomeFixed-PointProperties

ShowthatforanyexpressionEthereisasentenceXthatassertsthatEXisprovable(XistrueifandonlyifEXisprovable)andthereissomeXthatassertsthatEXisnotprovable.

6•SomeAnti-Fixed-PointProperties

ForanysentenceX,let betheconjugateofX.Show that for any expression E there is some sentence X thatassertsthatE isprovableandasentenceXthatassertsthatE isnotprovable.

Next,theSorcererpresentedsomeproblemsincross-reference.

•7•

FindsentencesXandYsuchthatXassertsthatYisprovableandYassertsthatXisnotprovable.(Therearetwosolutions.)ThenshowthatatleastoneofthesentencesX,Ymustbetruebutnotprovable(thoughthereisnowaytotellwhich).

•8•

Now find sentencesXandY such thatXasserts thatY is refutableandYassertsthatXisnotrefutable.(Therearetwosolutions.)Thenshow that at least one of the twomust be false but not refutable(thoughthereisnowaytotellwhich).

•9•

FindsentencesXandYsuchthatXassertsthatYisprovableandYassertsthatXisrefutable.(Therearetwosolutions.)Thenshowthatoneof them is trueandnotprovable,or theother is falsebutnotrefutable.WhichofX,Yiswhich?

•10•

FindsentencesXandYsuchthatXassertsthatYisnotprovableandYassertsthatXisnotrefutable.Doesitfollowthatoneofthemmustbeundecidable?

•11•

FindsentencesX,Y,andZsuchthatXassertsthatYisrefutable,YassertsthatZisnotrefutableandZassertsthatXisprovable.Isoneofthethreenecessarilyundecidable?

•12•

“I have said before,” said the Sorcerer, “that for any systemsatisfyingconditionsC1throughC4,therearesentencesotherthan that are true but unprovable in the system. You are now in apositiontoprovethis.Doyouseehow?”

Regularity.“Ishallcallasystemregular,”saidtheSorcerer,“iftheconversesofconditionsC1andC3hold—thatis,ifXisprovable,soisPXandifXXisprovable,so isṖX.This togetherwithC1andC3tell us that PX is provable if and only if X is provable, and ṖX isprovableifandonlyifXXisprovable.ImightremarkthatregularityistheanalogueofaconditionthatdoesholdforthetypeofsystemsstudiedbyGödel,butI’llsaymoreaboutthatanothertime.Regularsystemshavesomeinterestingproperties,asyouwillsoonsee.“LetmedefineapositivesentenceasoneoftheformPXorṖXand

anegativesentenceasoneoftheformQXor X.Positivesentencesassertthatcertainsentencesareprovable;negativesentencesassertthatcertainsentencesarenotprovable.Letusnownotethat if thesystemisregular,thenalltruepositivesentencesareprovable,andconversely, if all true positive sentences are provable, then thesystemisregular.”

•13•

Why is it that the system is regular if andonly if all truepositivesentencesareprovable?

“Andso,”continuedtheSorcerer,“weseethatinaregularsystem,only negative sentences can be true but unprovable. Any sentencethat asserts that something is provable, if true, must itself beprovable.”

•14•

If a system is regular, does it necessarily follow that every falsenegativesentenceisrefutable?“Regular systems have some interesting features,” said the

Sorcerer,“asyouwillnowsee.”

•15•

“Forone thing, ina regular system, theambiguitiesofProblems7through10disappear—that is, ifwe assume the system is regular,then inProblem7wecantellwhether it isXorYthat is truebutunprovable.Whichisit?AndinProblem8,isitXorYthatisfalsebut not refutable? And for Problem 9, is it X that is true butprovable,orisitYthatisfalsebutnotrefutable?AndforProblem10, is it X that is undecidable? All this, of course, with theassumptionofregularity.”

“Letusnote,”saidtheSorcerer,“thatforanysystemsatisfyingourgiven conditions C1 through C4, whether the system is regular ornot, if E is any string of P’s, then if EX is provable, so is X. Thisfollows by repeated applications of C1. For example, if PPPX isprovable,soisPPX(byC1);hencesoisPX(againbyC1);hencesoisX (again by C1). You can readily see that the same holds if EcontainsfourormoreP’s—orifEcontainstwoP’sorjustoneP.Andso ifE isany stringofP’s, ifEX isprovable, so isX.Fora regularsystem, theconversealsoholds—that is, ifX isprovable, so isEX,whereE isanystringofP’s.For ifX isprovableand thesystemisregular,thenPXisprovable(byregularity);hencesoisPPX,andsoforth.Andsoforaregularsystem,ifEisanystringofP’s,thenEXisprovableifandonlyifXisprovable.“Another thing about regular systems is this: For any system

satisfyingC1 throughC4, ṖX is true if and only if PXX is true, foreach is true if and only if XX is provable. However, withoutregularity, there is no reason to believe that ṖX is provable if and

onlyifPXXisprovable.Ifeitheroneisprovable,thentheotheroneis true, but that doesn’t mean that the other one is provable.However,ifthesystemisregular,thenṖXisprovableifandonlyifPXXisprovable.”

•16•

Whyisitthatinaregularsystem,ṖXisprovableifandonlyifPXXisprovable?

“Now comes a particularly interesting thing about regularsystems,” said the Sorcerer. “We have already seen that in anysystemsatisfyingconditionsC1throughC4,thereareinfinitelymanysentencesthataretrueforthesystembutnotprovableinthesystem.ButthisdoesnotmeanthatthereareinfinitelymanysentencessuchthateachoneistrueforallsystemssatisfyingC1throughC4andatthe same timeunprovable forall suchsystems.However, thereareinfinitely many sentences X such that for every regular systemsatisfying C1 through C4, each X is true for the system butunprovableinthesystem.”

•17•

Canyouprovethis?

“What I have shown you today,” said the Sorcerer, “hasapplications to the field known asmetamathematics—the theory ofmathematicalsystems.MyminiaturesystemprovidesoneapproachtoGödel’sfamousincompletenesstheorem:“Let us consider a mathematical system (M) in which there are

well-definedrulesspecifyingcertainsentencesastrueandothersasprovablein(M),andsupposethatwewishtoknowwhether(M)iscomplete in thesense thatall truesentencesof (M)areprovable in(M).Nowitcanbeshownthatif(M)isanyoneofawidevarietyofsystems investigated by Kurt Gödel, it is possible to translate mysysteminto(M)inthesensethatcorrespondingtoeachsentenceX

ofmysystem,thereisasentenceX°ofthesystem(M)suchthatXistrueinmysystemifandonlyifthecorrespondingsentenceX°of(M)isatruesentenceof(M),andalso,XisprovableinmysystemifandonlyifX°isprovablein(M).Doyourealizetheramificationsofthis?It means that for every such system (M), there must be a truesentenceof(M)thatisnotprovablein(M)—itstruthcanbeknownonlybygoingoutsideofthesystem.Thus,nosystem(M)intowhichmysystemistranslatablecanpossiblybecomplete.Doyouseewhythisisso?”

•18•

Whyisthisso?

Allthisismostremarkable!”saidAnnabelle.“Itcertainlyis!”agreedAlexander.“Whatwillyoutellusaboutnexttime?”askedAnnabelle.“On your next visit,” replied the Sorcerer with a mischievoussmile,“Ihaveaveryspecialparadoxpreparedforyou.”“I’m looking forward to it,” said Annabelle. “I’ve always beenintriguedbyparadoxes.”

Solutions

1.Thesentenceis .Itassertsthattherepeatof isnotprovable,but the repeat of is .Hence is true if and only if it is notprovable in the system. This means that it is either true and notprovableorfalseandprovable.Thelatteralternativecontradictsthegivenconditionthatonlytruesentencesareprovableinthesystem.Therefore,theformeralternativeholds—thesentenceistruebutnotprovable in the system. (This sentence is a transcriptionofGödel’sfamoussentencethatassertsitsownnonprovability.)

2.Ṗ assertsthattherepeatof —whichis —isprovable.Butis the conjugate of Ṗ . And so Ṗ asserts that its conjugate is

provable,or,whatisthesamething,thatitisitselfrefutable.

3.Thesentenceis Ṗ.ItassertsthatthesentenceṖṖ—whichistheconjugateof Ṗ—isprovable.

4.ṖṖassertsthatitisprovable.

5.AsentenceXthatassertsthatEXisprovableisṖEṖ,whichassertsthattherepeatofEṖisprovable.ButtherepeatofEṖisEṖEṖ,whichisEX.AsentenceXthatassertsthatEXisnotprovableis E .

6.AsentenceXthatassertsthatE isprovableisṖE .AsentenceXthatassertsthatE isnotprovableis EṖ.

7.OnesolutioncanbeobtainedbytakingXsuchthatitassertsthatQXisprovableandthentakingY=QX(whichassertsthatXisnotprovable).Thisgivesthesolution

X=ṖQṖ,Y=QṖQṖ

Another solution can be obtained by taking some Y that assertsthatPYisnotprovable—namely,Y= P andthentakingX=PY.Wethushavethealternativesolution

X=P P ,Y= P

IneithersolutionXassertsthatYisprovableandYassertsthatXisnotprovable.Therefore,XistrueifandonlyifYisprovableandYistrueifandonlyifXisnotprovable.Now,ifXandYareany twosentences bearing these two relationships with each other, one ofthem must be true but unprovable by the following argument:SupposeX is provable. ThenX is true; henceY is provable, hencetrue;henceXisnotprovable,whichisacontradiction.Therefore,Xcannotbeprovable.ItthenfollowsthatYmustbetrue.Andso,XisdefinitelynotprovableandYisdefinitelytrue.Now,eitherXistrueor it isn’t. IfX is true, thenX is truebutnot provable. IfX is not

true, then Y is not provable (because X is true if and only if Y isprovable);henceYisthentruebutnotprovable.Insummary,XisnotprovableandYistrue.IfXistrue,thenitis

Xthatistruebutnotprovable;ifXisfalse,thenitisYthatistruebutnotprovable.

(IobtainedthesesolutionsbytakingforXtheconjugateoftheYof the last problem, and for Y, the conjugate of the X of the lastproblem.)Now, X asserts that Y is refutable; hence X asserts that is

provable;hence assertsthat isnotprovable.Also,YassertsthatX isnot refutable;henceYasserts that is not provable; henceassertsthat isprovable.Thenbythelastproblem,taking forXand forY,weseethatatleastoneof , istruebutnotprovable;hence one of X, Y is false but not refutable. (Of course,we couldhaveprovedthisfromscratch—andifthereaderhasanydoubts,heorshemighttryit—butwhyduplicateworkalreadydone?)

9.OnewayistotakeanXthatassertsthatP isprovableandthentake Y = P . Another is to take some Y that asserts that QY isprovableandthentakeX=PY.Wethusgetthefollowingsolutions:

Solution1:X=ṖP Y=P P

Solution2:X=PṖQṖ Y=ṖQṖ

SupposeX is true. ThenY is provable (as X asserts); hence Y istrue; hence X is refutable (as Y asserts), which is not possible.Therefore,Xcannotbetrue;itmustbefalse.ThenYisnotprovable(as X asserts), and so we now know that X is false and Y is notprovable.IfXisnotrefutable,thenXisfalsebutnotrefutable.Ontheotherhand,ifXisrefutable,thenwhatYassertsistrue;henceYis then true but not provable. And so, either X is false but not

refutable,orYistruebutnotprovable.

10.We can simply take the conjugates of the X and Y of the lastproblemandinterchangethem,thusgettingthesolutions:

Solution1:X=Q P Y= P

Solution2:X= QṖ Y=QṖQṖ

Byapplyingouranalysisofthelastproblemto and ,insteadofXandYrespectively,weseethateither isfalsebutnotrefutable,or is truebutnotprovable.Thismeans thateitherX is falsebutnotrefutableorYistruebutnotprovable.

11.Weshallgivebutonesolution,whichwegetbytakinganXthatasserts thatPQXisprovable, thentakingYtobeQQXandZ tobePX. Thus X asserts that Y is refutable, Y asserts that QX is notprovable, or equivalently that PX is not refutable, but PX is Z. Ofcourse, Z asserts that X is provable. We thus have the followingsolution:

X=ṖPQṖY=QQṖPQṖZ=PṖPQṖ

Now,supposeZistrue.ThenXisprovable,hencetrue;henceYisrefutable,hencefalse;henceZisrefutable,hencefalse;andwehaveacontradiction.Therefore,Zcannotbetrue;Zisfalse.Alsothen,Xisnotprovable.IfXistrue,thenXistruebutnotprovable.SupposeXisfalse.ThenYisnotrefutable.IfYisfalse,thenYisfalsebutnotrefutable. If Y is true, then Z is not refutable. And so, one of thefollowingthreethingsmusthold:(1)Xistruebutnotprovable;(2)Yisfalsebutnotrefutable;(3)Zisfalsebutnotrefutable.

12.ThisisimmediatefromProblem7.WeknowthatinanysystemsatisfyingconditionsC1throughC4,atleastoneofthetwosentencesṖQṖ,QṖQṖistruebutnotprovableinthesystem.Likewise,withthetwosentencesP P , P .And,ofcourse,ouroldstandby istrue

butnotprovableinthesystem.Andso,thesystemcontainsatleastthree sentences that are true but not provable in the system.(Actually, thereare infinitelymany—forexample,oneof the threesentences PP PP , P PP , PP must be true but not provable.Also,oneofthefoursentencesPPP PPP ,PP PPP ,P PPP , PPPistruebutunprovable,andsoforth.)

13.Supposethesystemisregular.ConsiderapositivesentenceX.ItiseitheroftheformPYorṖY.IfPYistrue,thenYisprovable;hencebyregularityPYisprovable.IfṖYistrue,thenYYisprovable;hencebyregularity,ṖYisprovable.Thisshowsthatregularityimpliesthatalltruepositivesentencesareprovable.Conversely,supposethatall truepositivesentencesareprovable.

Well, suppose Y is provable. Then PY is true, hence provable (byhypothesis).Also, ifYY isprovable,ṖY is true,henceprovable (byhypothesis),andsothesystemisregular.

14.Ofcourse itdoes!Suppose thesystemis regular.NowletXbeany falsenegativesentence.Then itsconjugate isa truepositivesentence,henceprovable.Hence,Xisrefutable.

15.Supposethesystemisregular.Then,aswehaveshown,alltruepositivesentencesareprovable inthesystemandall falsenegativesentencesarerefutable.InProblem7,Xisapositivesentence;henceitisnotpossiblethat

X is true and unprovable; hence it is definitely Y that is true andunprovable. This goes for either of the two solutions for X andY.Thus,thesentencesQṖQṖand P arebothtrueandunprovableineveryregularsystem.InProblem8,itcannotbeYthatisfalsebutnotrefutable,because

Yisanegativesentence,soitmustbeX.InProblem9,itcannotbethatYistrueandnotprovable,sinceY

isapositivesentence,soitmustbethatXisfalsebutnotrefutable.InProblem10,itcannotbeXthatisfalseandnotrefutable,since

Xisanegativesentence,soitisYthatistruebutnotprovable.

16.Supposethesystemisregular.ThenṖXisprovableifandonlyifṖX is true,which in turn is the case if andonly ifXX isprovable,whichinturnisthecaseifandonlyifPXXistrue,whichinturnisthecaseifandonlyifPXXisprovable.More simply,weknow thatṖX is true if andonly ifPXX is true

(the truth of either is equivalent to the provability ofXX), but ṖXandPXXarebothpositivesentences,andforaregularsystem,truthandprovabilitycoincideforpositivesentences.

17.IfEisanystringofP’s,thesentence E mustbebothtrueandunprovableforallregularsystems.Tobeginwith,evenforasystemnotnecessarilyregular,E E cannotbeprovable,forifitwere, Ewould be provable (by repeated applications of C1); hence E Ewouldnotbeprovable(byC3),andwewouldhaveacontradiction.Therefore,E E isnotprovable.Hencealso E mustbe true.Butnow,ifweaddtheassumptionthatthesystemisregular,thenif Ewereprovable,thenE E wouldalsobeprovable,whichisnotthecase.Therefore, E isnotprovableinanyregularsystem—yetitistrueforallregularsystems(satisfyingC1throughC4,ofcourse).

18.SincethereisatruesentenceXoftheSorcerer’ssystemthatisnotprovableinhissystem(forexample,thesentence ,aswesawinthesolutionofProblem1),thenitstranslationXinto(M)mustbeatruesentenceof(M)thatisnotprovablein(M).

PARTV

HOWCANTHESETHINGSBE?

16SOMETHINGTOTHINKABOUT!

ON THEIR NEXT VISIT, Annabelle and Alexander were introduced to themostbafflingparadoxtheyhadeverheardintheirlives!“BeforeIgiveyouthisparadox,”saidtheSorcerer,“Iwouldfirstlike to offer you a little problem in probability that causes muchcontroversy.Manypeoplegivethewronganswerandinsisttheyareright, andnoargument canconvince them theyarewrong.Wouldthisinterestyou?”“Ofcourse,”theybothreplied.“Well,theproblemisthis:Youhavethreeclosedboxes—callthemA, B, and C. One of them contains a prize and the other two areempty. You pick one of the three boxes at random—say, Box A.Beforeyouopen it to seewhetheryouhavewon theprize, Iopenone of the other two boxes—say, Box B—and show you that it isempty.ThenyouaregiventheoptionofkeepingthecontentsofboxA, theboxyouareholding,orof trading it for thecontentsof thethird box—Box C. The question is: Is there any probabilisticadvantageinyourtradingBoxAforBoxCornot?”Aftersomethought,Annabellereplied:“No,itmakesnodifferencewhether I trade itornot.When I firstpickBoxA, thechancesareoneoutofthreethatmyboxhastheprize.ButwhenIseethatBoxBisempty,thechancesarethenoneoutoftwothatmyboxhastheprize.Thatis,thechancesarenoweventhattheprizeisinBoxAorin Box C, and so there is neither any advantage nor anydisadvantageinmytrading.”“Ithoroughlyagree,”saidAlexander.TheSorcerersmiled.“Yes,that’showmostpeopleseeit,”hesaid,“buttheyarewrong.Itdefinitelyistoyouradvantagetotrade.Youwillbeincreasingyourchancesofwinningfromoneoutofthreetotwooutofthree.”

“Ican’tseethatatall!”saidAnnabelle.“Howcanthatbe?Don’tyouagreethatinitiallytheprizeiswithequalprobabilityineachofthethreeboxes?”“Ofcourse,”repliedtheSorcerer.“Thenonceyouknowit’snotinBoxB,itiswithequalprobabilityinBoxAorBoxC.Isn’tthatobvious?”“No, it’s not obvious,” replied the Sorcerer. “It’s not even true.WhatyouforgetisthatitwasIwhoopenedthebox.IdeliberatelyopenedaboxthatIknewwasempty.”“Sowhat?”saidAlexander.“SupposeyouhadopenedBoxBbeforewechoseBoxAandshowedusthatitwasempty.YoumeantotellusthatthechancesarethengreaterthattheprizeisinBoxCthaninBoxA?”“No,” replied theSorcerer. “In that case, the chances are clearlyequal.”“You are confusing me more every minute!” cried Alexander.“Look,letmeputitthisway.SupposeIchooseonebox—sayBoxA—andAnnabellechoosesanother—sayBoxC.ThenyouopenBoxBand showus that it is empty.According to your logic, it is tomyadvantagetotradeboxeswithAnnabelle,butalsoitisequallytoheradvantagetotradeboxeswithme,andthatisclearlyabsurd!”“Itcertainlywouldbe,ifitfollowedfromwhatIsaidbefore,butitdoesn’t.Thisisacompletelydifferentsituation!IfyouchooseBoxAandyourwifechoosesBoxCandIthenopenBoxBandshowyouthatitisempty,thenofcoursethechancesareequalthattheprizeisinBoxAorinBoxC.”“Idon’tseethedifference,”saidAnnabelle.“Thedifferenceisthatinthesecondcase—whenyoueachchooseaboxinadvance,youleavemenochoiceastowhatremainingboxtoopen—thereisonlyone.One-thirdofthetimeIwillbeunable toopenBoxBand show it is empty,becauseone-thirdof the time itwon’tbe.But in the first case,whenyoubothchooseBoxAand IhavethechoiceofopeningforyouBoxBorBoxCandshowingitisempty, I can always be sure of showing you an empty box.RegardlessofwhetheryourBoxAhastheprizeornot,atleastone

of the two remaining boxes is empty, and since I knowwhere theprize is, I simply open one of them that I know to be empty.Myshowingyouthisemptyboxdoesn’tgiveyoutheslightestadditionalinformation,sinceIcanalwaysmanagetoshowyouanemptybox.“Now,itwouldbedifferent,”continuedtheSorcerer,“ifImyselfdidn’tknowwheretheprizewas.IfIsimplyopenedBoxBorBoxCatrandomandyousawthattheboxwasempty,thentheprobabilityofBoxAhavingtheprizewouldindeedjumpfromone-thirdtoone-half, and in that case it would not be to your advantage ordisadvantagetotradeyourboxforBoxC.Or,toputitanotherway,supposeyoupickBoxA.ThenacoinistossedtodeterminewhetherBoxBorBoxC is tobeopened—sayheadsmeansweopenBandtailsmeansweopenC.Thecoinistossedandheadscomesup.ThusBox B is to be opened. But now, mark this well! Before Box B isopened,thereisarealpossibilitythattheprizewillbeinit—infactthechancesareone-thirdthatitwillbe.Thentheboxisopenedandseen to be empty. In that case, you have really gained moreinformation, and the chances are now equal that your Box Acontainstheprize.Butifinsteadoftossingacoin,Ihavethechoiceof opening the box, and I know where the prize is, then it is acertainty that thebox I choose toopenwill be empty. If theprizehappened tobe inBoxB, Iwouldn’thaveopened it; IwouldhaveopenedtheemptyBoxCinstead.Andsomyshowingyouanemptyboxgivesyounohelpfulinformationwhatsoever.“Thecorrectwaytolookatitisthis:YouchooseBoxA,andthereis a one-third probability that you are holding the prize. I thendeliberatelyopenaboxIknowtobeemptyandshowyouthatitisempty.Youhavegainednonewinformationwhatsoeverconcerningtheprobabilitythatyourboxholdstheprize;itisstillone-third,butyouhavegainedinformationaboutBoxC;theprobabilitythatithastheprizeisnowtwo-thirds;before,itwasonlyone-third.Andsoyoudefinitelyshouldtrade.”“IthinkIbegintoseewhatyouaresaying,”saidAnnabelle,“buttruthfully, Iamnotyetconvinced. I’llhave to thinkabout it somemore.”

“Well,perhapsthiswillhelpyou,”saidtheSorcerer.“Toillustratemy point more dramatically, “suppose instead of three boxes wehave a hundred boxes and just one of them contains a prize. Youpickoneoftheboxesatrandom.Thechancesthatyourboxcontainstheprizeareoneoutofahundred,isn’tit?”“Certainly,”theybothagree.“Allright,”saidtheSorcerer,“thatleavesninety-nineboxes,andIknowwheretheprizeis.Ithendeliberatelypickninety-eightemptyboxes,openthem,andshowyouthattheyareempty.Doyoureallybelieve that the chances of your box holding the prize have risenfromoneoutofahundredtoone-half?”“That’sagoodwayofputtingit,”saidAnnabelle,“butIstillwanttothinkaboutitsomemore.”

Discussion. Of course the Sorcerer is right, but it is amazing thenumberofpeoplewhowillneverbeconvinced!Iamsurethatsomeof youwho are reading thiswill never be convinced.Most of youwill, but a few of youwon’t. Those of youwho don’t believe theSorcerer’s argument, Iwould love toplaya fewdozengameswithyouusingahundredboxesandgivingyouoddsoftentoone.Youwouldsoonfindthatyouarelosingyourshirt!

THEENVELOPEPARADOX

“NowI’dliketotellyouofaverybafflingparadoxthathasmadetheroundsthesepastfewyears,”saidtheSorcerer.“There are two sealed envelopes on the table. You are told thatoneof themcontains twiceasmuchmoneyas theother.Youpickoneofthemandopenittoseehowmuchmoneyisinside.Let’ssayyoufind$100init.Thenyouaregiventheoptionofkeepingitortradingitfortheotherenvelope.Now,theotherenvelopehaseithertwice as much or half as much with equal probability. Thus thechancesareequalthattheotherenvelopehas$200or$50,andthechancesofyourgainingorlosingareequal.Andsotheoddsareinyourfavorifyoutrade.”

“Thatsoundsperfectlylogical,”saidAnnabelle.Alexanderagreed.“Butnowcomesthecuriousthing,”saidtheSorcerer.“Beforeyou

open the envelope, you know thatwhatever the amount you find,yourreasoningwillbethesame,andsotherationalthingtodoistotrade your envelope for the other one immediately, withoutbothering toopen it.For, letnbe theamount in theenvelopeyouareholding.Then theother envelope contains either n⁄2 dollars or2ndollarswithequalprobability,andsoyouhaveanequalchanceof winning n dollars or losing n⁄2 dollars, and so it is to youradvantage to trade your envelope for the other one. But had youoriginally chosen the other one, then, by the same reasoning, youshould trade it for theoneyouarenotholding,andclearly this isabsurd!Thatistheparadox.”Theythoughtaboutthisforsometime.“I certainly can see the absurdity of the Situation,” Annabelle

finallysaid,“butIcan’tfigureoutthefallacyinthereasoning.Whatisit?”“To tell you the truth, I have not yet heard a completely

satisfactory answer to your question,” said the Sorcerer. “I havegiven this problem to several experts on probability theory; somewereaspuzzledasI,andothersgavemeanexplanationintermsoftherebeingnosuchthingasaprobabilitymeasureontheinfinitesetofpositiveintegers.ButIsuspectthatprobabilityisnottheheartofthe matter, and I have thought of a new version of the paradoxwhichdoesn’tinvolveprobabilityatall.”“Oh?”saidAnnabelle.“Yes,myversionisthis:Youpickuponeofthetwoenvelopesand

youdecide thatyouaregoing to trade it for theother.Eitheryouwillgainoryouwillloseonthetrade.Iwillnowprovetoyoutwocontradictorypropositions:“Proposition1. The amount that youwill gain, if youdo gain, is

greaterthantheamountyouwilllose,ifyoudolose.“Proposition2.Theamountsarethesame.

“Obviously the two propositions cannot both be true, yet I willprovetoyouthateachoneistrue.“TheproofofthefirstpropositionisessentiallyoneIhavealready

givenyou.Letnbe thenumberofdollars in theenvelopeyouarenowholding.Thentheotherenvelopehaseither2norn⁄2dollars.”“Withequalprobability,”saidAlexander.“Probabilityisnowirrelevant,”saidtheSorcerer.“Iwanttoleave

outprobabilityaltogether.Theimportantthingnowisthattheotherenvelopehaseither2ndollarsorn⁄2dollars;wedon’tknowwhich.”“Allright,”saidAlexander.“Thenifyougainonthetrade,youwillgainndollars,butifyou

loseonthetrade,youwilllosen⁄2dollars.Sincenisgreaterthann⁄2,then the amount you gain, if you do gain—which is n—is greaterthan the amount you will lose, if you do lose—which is n⁄2. ThisprovesProposition1.“Now for the proof of Proposition 2. Let d be the difference

between the amounts in the two envelopes, or what is the samething, let d be the lesser of the two amounts. If you gain on thetrade,youwillgainddollars,andifyouloseonthetrade,youwilllose d dollars. And so the amounts are the same after all. Forexample, suppose the lesser envelope has $50; then the greaterenvelope has $100. If you gain on the trade, that means you areholdingthelesserandsoyouwillthengain$50;butifyouloseonthetrade,thatmeansyouareholdingthe$100envelopeandsoyouwilllose$50.Thus$50istheamountyoustandtogainanditisalsothe amount you stand to lose. The same argument goes for any dthat is the lesserof thetwoamounts.Thenumberd is theamountyou stand to gain and is also the amount you stand to lose. ThisprovesProposition2;theamountsareequalafterall.“Well,”saidtheSorcerertoathoroughlybaffledcouple,“whichof

thetwopropositionsisthecorrectone?Theyobviouslycan’tbothberight!”

Epilogue.ItmightamusethereadertoknowthatAnnabelleandher

husband never did agree on thematter. Annabelle was absolutelysurethatProposition1wascorrect,andAlexanderwasequallysurethatitwasProposition2.“Buthowcanyousaythat?”askedAnnabelle.“Supposeyouopen

your envelope and find $100. Then you know that the otherenvelopecontainseither$50or$200.Andsoyougain$100ifyoudo gain, and lose $50, if you do lose. So isn’t it obvious that theamountyoustand togain isgreater than theamountyoustand tolose? Isn’t it obvious that $100 is more than $50? How can youpossiblyhaveanydoubtsonthematter?”“Youare lookingat it thewrongway,” insistedAlexander. “The

amountsinthetwoenvelopesarendollarsand2ndollarsforsomenwhosevaluewedon’tknow.Ifyouwinonthetrade,youwillbemovingupfromnto2ndollars,hencegainingndollars,andifyoulose on the trade, youwill bemoving down from2n to n dollars,hencelosingndollars.Clearlytheamountsarethesame.”Thetwoneverdidagreeonthis!Whichdoyoubelievetobethe

trueone,Proposition1orProposition2?

17OFTIMEANDCHANGE

THENEXTTIMEourcoupleclimbedtheSorcerer’stower,theyfoundtheirhostinaparticularlyphilosophicalmood.HehadspentthemorningreadingearlyGreekandChinesephilosophy.“Time is really a strange thing,” he said. “Some mystics haveclaimedthattimeisunreal,andattimesIaminclinedtoagreewiththem!”“Isee,”saidAnnabelle.“Attimesyouagreewiththemandatothertimesyoudon’t,isthatit?”“That’sabout it,” laughed theSorcerer. “And speakingof time, Imust read you a delightful passage I just found in a work by theancientChinesephilosopherChuangtse.”

There was a beginning. There was a time before thatbeginning.Andtherewasatimebeforethetimewhichwasbefore that beginning. There was being. There was non-being. There was a time before that non-being. And therewasa timebefore the time thatwasbefore thatnon-being.Suddenly there is beingand there is non-being, but I don’tknowwhichofbeingandnon-beingisreallybeingorreallynon-being. I have just said something, but I don’t know ifwhatIhavesaidreallysayssomethingorsaysnothing.*

Histwolistenershadagoodlaugh—especiallyatthelastline.“I’malsoremindedofSmullyan’srecipeforimmortality,”saidtheSorcerer.“Areyoufamiliarwithit?”“No,”saidAlexander.“ThisImusthear!”“It’sreallyverysimple:Tobeimmortal,allyouhavetodoarethefollowingtwothings:(1)Alwaystellthetruth;nevermakeanyfalsestatements in the future. (2) Just say: ‘I will repeat this sentence

tomorrow!’Ifyoudothesetwothings,Iguaranteethatyouwillliveforever!”Of course, the Sorcererwas right: If today you truthfully say: “Iwill repeat this sentence tomorrow,” then you will repeat thesentence tomorrow.Assuming you remain truthful tomorrow, thenyouwill repeat the sentence thenextday, then thenextday, thenthenextday,…“Theoretically,it’saperfectplan,”saidAnnabelle,“butit’shardlythemostpracticalplanintheworld!”“ItremindsmeoftheWhiteKnight’splanforgettingoveragate,”remarkedAlexander.“Whatplanisthat?”askedtheSorcerer(whoevidentlyhadneverreadAlice’sAdventuresinWonderlandandThroughtheLookingGlass).“As the White Knight explained it,” said Alexander, “the onlydifficulty iswith the feet: thehead ishighenoughalready.Soyoufirst put your head on the top of the gate—then the head’s highenough; then you stand on your head—then the feet are highenough,yousee;thenyou’reover,yousee.”“Very good,” laughed the Sorcerer. “And concerningAnnabelle’sobjection to Smullyan’s plan as not being practical, Smullyanpresented his plan in a very amusing story of aman in search ofimmortalitywhovisitedagreatsageintheEastwhowassaidtobeanexpertonthissubject.Thesageexplainedtheplanofhowtobeimmortal, but themanobjected, as didAnnabelle, on the groundsthat the plan was not practical. He said to the sage: ‘How can ItruthfullysaythatIwillrepeatthissentencetomorrowwhenIdon’tknowwhether I’ll evenbealive tomorrow?’ ‘Oh,’ replied the sage,‘youwantedapracticalsolution!No,I’mnotverygoodatpractice;Idealonlyintheory.’”AnnabelleandAlexanderhadagoodlaughoverthatone.“Speakingofimmortality,”saidtheSorcerer,“IrecallthatwhenIwasa lad,myuncle—whotookakeeninterest inall topics logicaland philosophical—gaveme a remarkable proof that it is logicallyimpossibleforanyonetodie.Wouldyouliketohearit?”“Oh,yes!”theycriedsimultaneously.

“Well,myuncle put it thisway: If a person dies,whendoes hedie?Doeshediewhileheislivingorwhileheisdead?Hecan’tdiewhileheisdead,becauseonceheisdead,hecannolongerdie—heisalreadydead.Ontheotherhand,hecan’tdiewhileheisstillalivebecausehewould thenbedeadandalive simultaneously,which isnotpossible.Therefore,hecan’tdieatall.”“I would say,” said Annabelle, after some thought, “that at theinstant he dies, he is neither alive nor dead. The instant is thetransitioninstantbetweenlifeanddeath.”“Thatseemsreasonable,”saidAlexander.“No,” said the Sorcerer, “you can’t get out of it that easily! By‘dead’ I mean no longer alive. In fact, to avoid any semanticconfusion, I’ll phrase the argument differently: at the instant of aperson’sdeath,ishealiveornotaliveatthatinstant?”Theythoughtsomemoreaboutthis.“Is this related to Zeno’s proofs of the impossibility ofmotion?”askedAnnabelle.“There is some connection,” replied the Sorcerer. “In fact, myuncle’s argument can be generalized to give a new proof of theimpossibilityofmotion.”“WhatareZeno’sarguments?”askedAlexander.“Ihaveheardofthem,butIhaveneverheardthem.”“Ihavereadthem,”Annabellesaid,“butIcouldneverquitefigureout where the fallacies lie. There must be fallacies, since thingsobviouslydomove.Justwhatarethefallacies?”“Zeno had three proofs,” replied the Sorcerer. “There is also afourthproof,butthisproofwaspoorlyrecordedandtheredoesnotseemtobeuniformagreementastojustwhatthatproofis.SoIwillconfinemyselftojustthefirstthree.“Thefirstargumentisthis:SupposeanobjectmovesfromapointA to a point B. Before it can get toB, itmust get to the pointA1midwaybetweenA andB; call this the first step.After completingthe first step, the bodymust get fromA1 to the point A2midwaybetweenA1 andB; call this the second step. Then itmust take thethirdstep—gettingfromA2tothepointA3midwaybetweenA2and

B,andsoforth.

“AfternofinitenumberofstepswillthebodyhavereachedB;thatis,foreachpositiveintegern,afterithasperformedthefirstnstepsand isatapointAn, it stillhas to takeanother stepandgo to thepoint An+1 midway between An and B. So the body must takeinfinitelymany steps to get to B and since this is impossible in afinitelengthoftime,theobjectcannotmovefromAtoB.“Lookingatitanotherway,beforetheobjectcangetfromAtoB,itmust first get to themidpointB1; but before it can get toB1, itmustfirstgettothepointB2midwaybetweenAandB1;butbeforeitcandothat,itmustgettothepointB3midwaybetweenAandB2,andsoforth.Therefore,thebodycan’tevengetstarted!

“Zeno’s second argument is about Achilles trying to overtake atortoise.Tobeginwith,letussaythatAchillesis100yardsbehindthetortoiseandthatheisrunningtentimesasfastasthetortoise.Hisfirststepistoruntothespotwherethetortoisenowis—thatis,he runs100yards.Whenhe reaches that spot, the tortoisewillnolongerbethere;itwillhavemoved10yardsforward.ThenAchillestakesthesecondstep—heruns10yardstoreachthespotwherethetortoisewasattheendofthefirststep,butwhenhegetsthere,thetortoisewillhaverun(orratherwalked)1yardfurther.ThenwhenAchillesrunsthisyardforward,thetortoisewillstillbe1⁄10ofayardaway, and so forth. In otherwords,wheneverAchilles reaches theplacewhere the tortoise had been, the tortoise is no longer there.HenceAchillescanneverovertakethetortoise.“Zeno’s thirdproofstrikesmeas themostsophisticated; it is theproof about the flying arrow. Suppose an arrow is flyingcontinuously forward during a certain time interval. Take anyinstant in that time interval. It is impossible that the arrow is

movingduringthatinstant,becauseaninstanthasdurationzero,andthe arrow cannot be at two different places at the same time.Therefore,ateveryinstantthearrowismotionless,hencethearrowis motionless throughout the entire interval, which means that itcan’tmoveduringtheinterval!”“Yes, I remember thosearguments,now thatyouhave reminded

me of them,” saidAnnabelle, “and I am asmuch puzzled as ever.Surely,somethingmustbewrongwith them, since thingsobviouslydo move, but what is it that is wrong? Is there no logicalexplanation?Oristheresomethingwrongwithlogicitself?”“No, there is certainly nothingwrongwith logic itself,” laughed

theSorcerer,“andtherecertainlyisaperfectlylogicalexplanationasto just what the fallacies are in Zeno’s arguments. I am reallysurprisedthatpeoplearefooledbythefirsttwoarguments,thoughit ismoreunderstandablewiththethird,whichismoredifficulttoseethrough.”“Whatarethefallacies?”askedAlexander.“Well, let’s start with the first one,” said the Sorcerer. “In any

argumentthatleadstoafalseconclusion,theremustbeafalsestepsomewhere,hence theremustbea first false step.So,where is thefirstfalsestepinZeno’sfirstargument?”Thetwoofthemthoughtaboutthisforawhile.“I can’t see any false step,” said Alexander; “every step seems

correcttome.”“Tomealso,”saidAnnabelle.“Really now!” said the Sorcerer. “Even the conclusion seems

correcttoyou?”“Of coursenot!” saidAnnabelle. “Theconclusion is clearly false.

Obviouslythingsdomove.”“Well then, if the conclusion is false and every step before the

conclusion is correct, then obviously the first false step is theconclusionitself.”“Ineverthoughtofthat!”saidAnnabelle.“Most people don’t, and that’s the surprising thing!” said the

Sorcerer. Of course, the conclusion is the first wrong statement!

EveryobjectmustgothroughtheinfinitenumberofstepsthatZenodescribed,but it is completely specious toconclude: ‘Therefore theobject can’tmove.’Where is the warrant for that inference? Zenotacitlyassumed thatonecannotperform infinitelymany steps inafinitelengthoftime,andthatassumptionistotallyunwarranted.”“That assumption seems perfectly reasonable to me!” said

Annabelle. “How can the sum of infinitely many quantities befinite?”“Thathappensallthetimeinmathematics,”repliedtheSorcerer.

“Forexample,theinfiniteseries1+1⁄2+1⁄4+1⁄8+1⁄16+1⁄32+…+1⁄2n+…alladdsupto2.Suchaninfiniteseriesiscalledconvergent;itconvergesto2.Ontheotherhand,theinfiniteseries1+ 1⁄2+ 1⁄3+ 1⁄4+ 1⁄5 +…+ 1/n+… is what is called adivergent series; ifyou takeenough termsof it,you’llgetbeyondamillion; take enough more terms, you’ll get beyond a trillion;whatever number you pick, no matter how large, that serieseventuallygetsbeyondthatnumber.Suchadivergentseriesisalsosaidtoapproachinfinity.However,theseriesrelevanttoZeno’sfirstproblemis1⁄2+1⁄4+1⁄8+…,whichconvergesto1,andsoZenowastotallyunjustifiedinconcludingthattheobjectcouldnevergetfromAtoB.AsimilaranalysisappliestoZeno’ssecondargument—in fact the series is 1⁄10+1⁄100+ 1⁄10004…+ 1⁄10n+…,whichconvergesevenmorerapidly.“Thethirdargumentoftheflyingarrowismoresubtle,andIam

afraid you don’t know enough mathematics to understand theresolution. Those who have studied calculus know that at anyinstant during the interval, the arrow is inmotion at that instant,andthatZenowaswronginsayingthatifthearrowwereinmotionataninstant,thenitwouldbeinmorethanoneplaceattheinstant.Ofcourse, thearrow isonlyatoneplaceatanygiven instant,butthatdoesnotimplythatthearrowisatrestatthatinstant.”“I don’t understand,” said Annabelle. “If an arrow in motion is

onlyatoneplaceatagiveninstant,howisitanydifferentfroman

arrowatrest,whichisalsoatonlyoneplaceataninstant?Howcanonetellthedifferencebetweenthetwocases?”“You ask a great and profound question,” replied the Sorcerer,

“and thepurposeof thedifferential calculus—orat leastone of itsmainpurposes—istoanswer just thissortofquestion.Forthefirsttime in the history of mankind, the inventors of the calculus—NewtonandLeibniz—gaveaperfectlyprecisedefinitionofwhat itmeansforabodytobeinmotionataninstantandwhatitmeanstosay that the velocity of the body is such and such at an instant. Icannot give you these definitions without first teaching you somefundamentalnotionsof thecalculus.Fornow,suffice it tosay thatthestatementthatabodyhasacertainvelocityataninstantisreallyastatementnotaboutjusttheinstantbutaboutsuccessivelysmallerand smaller time intervals centeredaround the instant. I hopeoneday I can give you amore complete and technical account of thematter.”“You said before,” said Alexander, “that your uncle’s argument

canbegeneralizedtoprovideyetanotherproofoftheimpossibilityofmotion.Howisthat?”“Myuncle’sargumentcanbegeneralizedtoshowthatnothingcan

ever change—which surely would have pleased the philosopherParmenides!Suppose somethingchanges frombeing inone state—callitstateA—tobeingoutofstateA.Atwhatinstantcanitmakethe change? It can’t make the change when it is out of state A,because it is already out of stateA.And it can’tmake the changewhileitisstillinstateA,becauseitwouldthenbeinstateAandoutofstateAatthesametime.Anddon’ttellmethatattheinstantofchange,itisneitherinstateAnoroutofstateA,becausebyoutofstate A I mean not in state A, and it is logically impossible forsomething to be neither in state A nor not in state A at the sametime.”“How does that give another proof of the impossibility of

motion?”askedAlexander.“Well,obviouslyifanobjectmovesawayfromagivenposition,it

changes its state from being in that position to not being in that

position.Andsoifchangingfromonestatetoanotherisimpossible,thenmotionisimpossible.…“Yes, time and change are curious things! What really is time,anyway?AsAugustinesaid: ‘Whenaskedwhattimeis,Iknownot;whenaskednot,Iknow!’“Andspeakingof time, Iunderstandyouwillbothbe leavingusforawhile?”“Yes,”repliedAnnabelle,“tomorrowwesail forhome.Mysister,PrincessGertrude,isgettingmarried.”“Howlongwillyoubegone?”askedtheSorcerer.“Probably a couple of months,” replied Annabelle. “There areseveral things Alexander and Imust attend to back home. Butwedefinitely plan to come back. In fact, we are thinking of movinghere.Wearebothabsolutelyintriguedbythiswholesubjectoflogic.No one at home seems to knowmuch about it. You have no ideahow grateful we are for all that you have taught us. But I ambeginning to see that we have so much to learn!When we comeback,couldyoutellussomethingaboutinfinity?That’sonesubjectthathasalwaysintriguedandpuzzledme!”“Ah, yes!” said the Sorcerer. “Infinity. That is a subject indeed!Yes, when you come back, I promise to give you a guided tourthroughinfinity.Docomebacksoon,andbonvoyage!”

*Wing-TsitChan,ASourceBookinChinesePhilosophy(Princeton,N.J.:PrincetonUniversityPress,1963),pp.185,186.

PARTVI

AJOURNEYINTOINFINITY

18WHATISINFINITY?

OUR COUPLE WERE AWAY for more than two months. When they finallyreturnedtotheisland,theylostnotimeinvisitingtheSorcerer.“Welcome back!” he said. “And so you want to know aboutinfinity?”“Youhaveagoodmemory,”saidAnnabelle.“Allrightnow,”saidtheSorcerer.“Thefirstthingweshoulddoisto carefully define our terms. Just what is meant by the word‘infinite’?”“Tome,itmeansendless,”saidAlexander.“Iwouldsaythesame,”saidAnnabelle.“That’s not quite satisfactory,” said the Sorcerer. “A circle isendless, in the sense of having no beginning or end, and yet youwouldn’tsayit’sinfinite—thatis,ithasonlyfinitelength,thoughitdoeshaveinfinitelymanypoints.“I want to talk about infinity in the precise sense used bymathematicians. Of course, there are other uses of the word; forexample,theologiansoftenrefertoGodas infinite, thoughsomeofthemarehonestenoughtoadmitthatwhenappliedtoGodthewordhasadifferentmeaningthanwhenappliedtoanythingelse.Now,Ido not wish to disparage the theological or any othernonmathematicaluseof theword,but Iwant tomake itclear thatwhatI’mproposingtodiscuss is infinity inthepurelymathematicalsenseoftheterm.Andforthisweneedaprecisedefinition.“Theword ‘infinite’ isobviouslyanadjective,and the first thingwemust agreeupon is the sort of things towhich the adjective isapplicable.What sort of things can be classified as either finite orinfinite?Well, in themathematical use of the term, the answer isthat it is sets, or collections of things, that can be called finite orinfinite.We say that a set of objects has finitelymany or infinitely

manymembers,andnowwemustmakethesenotionsprecise.“Thecluehere lies in thenotionofaone-to-onecorrespondencebetween one set and another. For example, a flock of seven sheepandagroveof seven treesare related toeachother inaway thatneitherisrelatedtoapileoffivestones,forthesetofsevensheepcanbepairedwiththesetofseventrees,forexample,bytetheringeach sheep to a tree, so that each sheep and each tree belongs toexactlyoneof thepairs.Or, inmathematical terms, a set of sevensheepcanbeput intoa1 to1correspondencewitha setof seventrees.“Another example: Supposeyou look into a theater and see thatevery seat is taken and that no one is standing—and also, that noone is sittingonanyone’s lap; there isoneandonlyoneperson toeachseat.Then,withouthavingtocount thenumberofpeople,orthe number of seats, you know that the numbers are the same,becausethesetofpeopleisina1to1correspondencewiththesetofseats:eachpersoncorrespondstotheseatheisoccupying.“Now, I know that you are familiar with the set of naturalnumbers,thoughyoumightnotbefamiliarwiththembythatname.Thenaturalnumbersaresimplythenumbers0,1,2,3,4,…Thatis,by a natural number is meant either zero or any positive wholenumber.”“Istheresuchathingasanunnaturalnumber?”askedAnnabelle.“No, I haveneverheardof that,” said the Sorcerer, “and Imustsay the idea strikesme as funny! Anyway, from now on Iwill beusing the word number to mean natural number, unless I saysomethingtothecontrary.“Now, given a natural number n, justwhat does itmean to saythatacertainsethasexactlynmembers?Forexample,whatdoesitmeantosaythatthereareexactlyfivefingersonmyrighthand?ItmeansthatIcanputthesetoffingersonmyrighthandintoa1to1correspondencewiththesetofpositivewholenumbersfrom1to5—say,bylettingmythumbcorrespondto1,thenextfingerto2,themiddle finger to 3, the next finger to 4, and the little finger to 5.And, ingeneral,givenanypositivewholenumbern,wesaythata

set has (exactly) n members if the set can be put into a 1 to 1correspondencewith the setofpositive integers from1 ton.A setwithnelementsisalsocalledann—elementset.Andtheprocessofputtingann—elementsetintoa1to1correspondencewiththesetofpositive integershas apopularname—thepopularnameof thisprocessiscounting.Yes,that’sexactlywhatcountingis.“And so, I have told you what it means for a set to have nelements,wherenisapositivewholenumber.Whataboutwhenniszero;whatdoes itmean fora set tohave0elements? Itobviouslymeansthatthesethasnoelementsatall.”“Therearesuchsets?”askedAlexander.“There is only one such set,” replied the Sorcerer. “This set istechnically called the empty set, and is extremely useful tomathematicians.Withoutit,exceptionswouldconstantlyhavetobemade and things would get very cumbersome. We want, forexample,tobeabletospeakaboutthesetofpeopleinatheateratagiven instant. It may happen that there are no people there at agiveninstant,inwhichcasewesaythatthesetofpeopletheninthetheaterisempty—justaswespeakofanemptytheater.Thisisnotto be confusedwith no theater at all! The theater still exists as atheater;therejusthappentobenopeopleinit.Likewise,theemptysetexistsasaset,butithasnomembers.“Irecallacharmingincident:Manyyearsago,Itoldalovelyladymusician about the empty set. She seemed surprised and said:‘Mathematicians really use this notion?’ I replied: ‘They certainlydo!’Sheasked: ‘Where?’Ireplied: ‘It’susedallovertheplace.’Shethoughtforamomentandsaid:‘Oh,yes.Iguessit’sliketherestsinmusic.’Ithoughtthatwasquiteagoodanalogy!“Smullyan relatesanamusing incident.Whenhewasagraduatestudent atPrincetonUniversity, oneof the famousmathematicianstheresaidduringalecturethathehatedtheemptyset.Atthenextlecture,hemadeuseoftheemptyset.Smullyanraisedhishandandsaid, ‘I thought you said that you disliked the empty set.’ Theprofessorreplied:‘IsaidIdislikedtheemptyset.IneversaidthatIdon’tusetheemptyset!’”

“You haven’t yet told us,” said Annabelle, “what you mean by‘finite’and‘infinite.’Aren’tyougoingto?”“Iwasjustaboutto,”repliedtheSorcerer.“AllthethingsI’vesaidsofararejustleadinguptothedefinition.Wesaythatasetisfiniteif there exists a natural number n such that the set has exactly nelements—whichwerecallmeansthatthesetcanbeputintoa1to1correspondencewiththepositive integers from1ton. If there isno such natural number n, then the set is called infinite. It’s assimpleasthat.Thusa0-elementsetisfinite;a1-elementsetisfinite,a2-elementssetisfinite,…andann-elementsetisfinite,wherenisanynaturalnumber.Butifforeverynaturalnumbern,itisfalsethatthesethasexactlynelements,thenthesetisinfinite.Thus,ifasetisinfinite, then for any natural number n, if we remove n elementsfrom the set, there will be elements left over—in fact, infinitelymanyelementswillbeleftover.“Doyouseewhythisistrue?Let’sfirstconsiderasimpleproblem.Suppose I remove a single element from an infinite set. Is whatremainsnecessarilyinfinite?”“Itwouldcertainlyseemso!”saidAnnabelle.“Itcertainlydoes!”saidAlexander.“Well,youareright,butcanyouproveit?”Thetwothoughtaboutit,buthaddifficultyprovingit.Itseemedtooobvious toneedproof.But it isnotdifficult toprove this fromtheverydefinitionsof“finite”and“infinite.”Thesedefinitionsmustbeused.

•1•

Howisthisproved?

Ittookabitofprodding,butthetwofinallycameupwithaproofthatsatisfiedtheSorcerer.

Hubert’s Hotel. “Infinite sets,” said the Sorcerer, “have somecuriouspropertiesthatattimeshavebeenlabeledparadoxical.Theyare not really paradoxical, but they are a bit startling when first

encountered.ThisiswellillustratedbythefamousstoryofHubert’sHotel.“Supposewehaveanordinaryhotelwithonlyafinitenumberof

rooms—say a hundred. Suppose all the rooms are taken and eachroomhasoneoccupant.Anewpersonarrivesandwantsaroomforthenight,butneitherhenoranyofthehundredguestsiswillingtoshare a room. Then, it is impossible to accommodate the newarrival;onecannotput101peopleintoa1to1correspondencewith100rooms.Butwithinfinitehotels(ifyoucanimaginesuchathing)thesituationisdifferent.Hubert’sHotelhasinfinitelymanyrooms—oneforeachpositiveinteger.Theroomsarenumberedconsecutively—Room 1, Room 2, Room 3,… Room n,… and so forth. We canimaginetheroomsofthehotelarrangedlinearly—theystartatsomedefinitepointandgoinfinitelyfartotheright.Thereisafirstroom,butthereisnolastroom!It’simportanttorememberthatthereisnolast room—just as there is no last natural number.Nowagain,weassumethatalltheroomsareoccupied—eachroomhasoneguest.Anewpersonarrivesonthesceneandwantsaroom.Theinterestingthingisthatitisnowpossibletoaccommodatehim.Neitherhenoranyoftheotherguestsiswillingtosharearoom,buttheguestsareall cooperative in that they are willing to change their rooms, ifrequestedto.”

•2•

Howcanthisbedone?

“Nowforanotherproblem,”saidtheSorcerer,afterthesolutiontothislastproblemhadbeendiscussed.“Weconsiderthesamehotelasbefore.Butnow,insteadofjustonenewpersonarriving,aninfinitenumberofnewguestsarrive—oneforeachpositiveintegern.Letuscall theoldguestsP1,P2,…Pn,…, and the new arrivalsQ1,Q2,…,Qn,… The newQ-people allwant accommodations. The surprisingthingisthatitcanbedone!

•3•

How?

“Nowforastillmoreinterestingproblem,”saidtheSorcerer.“Thistimewehave infinitelymanyhotels—one foreachpositive integern.ThehotelsarenumberedHotel1,Hotel2,…Hoteln,…andeachhasinfinitelymanyrooms—oneforeachpositiveinteger.Thehotelsarearrangedinarectangularsquarearray—thus:

“The whole chain of hotels is under one management. All therooms of all the hotels are occupied. One day the managementdecidestoshutdownallthehotelsbutone,inordertosaveenergy.

Butthismeansshiftingalltheinhabitantsofallthehotelsintojustoneofthehotels—again,onlyonepersontoaroom.”

•4•

Isthispossible?

“You see what these problems reveal,” said the Sorcerer. “Theyshowthataninfinitesetcanhavethestrangepropertyofbeingabletobeput intoa1to1correspondencewithaproperpartof itself.Letmemakethismoreprecise.“AsetAiscalledasubsetofasetBifeveryelementofAisalsoan

elementofB.Forexample,ifAisthesetofnumbersfrom1to100andBisthesetofnumbersfrom1to200,thenAisasubsetofB.OrifEisthesetofallevenintegersandNisthesetofallintegers,thenEisasubsetofN.AsubsetAofBiscalledapropersubsetofBifAisa subset of B but does not contain all the elements of B. In otherwords,AisapropersubsetofBifAisasubsetofB,butBisnotasubset ofA.Now, let P be the set {1, 2, 3,…n,…}of all positiveintegersandletP–betheset{2,3,…n,…}ofallpositiveintegersother than 1.Wehave seen in the first problemofHubert’sHotelthatthatPcanbeputintoa1to1correspondencewithP–,yetP–is a proper subset of P! Yes, an infinite set can have the strangepropertyofbeingabletobeputintoa1to1correspondencewithaproper subset of itself! Thiswas known long ago. In 1638, Galileopointedoutthatthesquaresofthepositiveintegerscanbeputina1to1correspondencewiththepositiveintegersthemselves,thus

“This seemed to contradict the ancient axiom that the whole isgreaterthananyofitsparts.”“Well,doesn’tit?”askedAlexander.

“Not really,” replied the Sorcerer. “Suppose that A is a propersubset of B. Then, in one sense of theword ‘greater,’ B is greaterthan A—namely, in the sense that B contains all elements that AcontainsandsomeelementsthatBdoesn’tcontain.ButthatdoesnotmeanthatBisnumericallygreaterthanA.”“I’mnotsureIknowwhatyoumeanbynumericallygreater,”saidAnnabelle.“Goodpoint!”saidtheSorcerer.“Firstofall,whatdoyouthinkitmeansforonesetAtobeofthesamesizeasanothersetB?”“IguessitmeansthatAcanbeputintoa1to1correspondencewithB,”saidAnnabelle.“Right! Andwhatwould you guess itmeans to say that A is ofsmallersizethanB,orthat,numerically,AhasfewerelementsthanB?”“I guess it would mean that A can be put into a 1 to 1correspondencewithapropersubsetofB.”“Nice try,” said theSorcerer, “but itwon’twork.Thatdefinitionwouldbefineforfinitesets,butnotforinfinitesets.ThetroubleisthatitmightbepossibleforAtobeputintoa1to1correspondencewithapropersubsetofB,anditmightalsobepossibleforBtobeputintoa1to1correspondencewithapropersubsetofA.Inwhichcase would you want to say that each is of smaller size than theother?Forexample,letObethesetofoddpositiveintegersandEbethesetofevenpositiveintegers.Obviously,Ocanbeputintoa1to1correspondencewithE.

“ButalsoOcanbeputintoa1to1correspondencewithapropersubsetofEthus

“Atthesametime,Ecanbeputintoa1to1correspondencewithapropersubsetofOthus

“Now,youcertainlywouldn’twanttosaythatEandOareofthesamesize,yetEisofsmallersizethanOandalsoOisofsmallersizethanE!No,thatdefinitionwon’tserve.”“Thenwhatisthecorrectdefinitionof‘smallerthan’whenapplied

toinfinitesets?”askedAnnabelle.“Thecorrectdefinitionisthis:WesaythatAissmallerinsizethan

B,orthatBisgreater insize thanA if the following twoconditionsaremet:(1)Acanbeputintoa1to1correspondencewithapropersubsetofB;(2)Acannotbeputintoa1to1correspondencewithallofB.“It is crucial that both conditions be met,” emphasized the

Sorcerer,“inorderthatitcanbecorrectlysaidthatAissmallerthanB.TosaythatAissmallerthanBmeansfirstofallthatAcanbeputintoa1to1correspondencewithasubsetofBandalsothatevery1to 1 correspondence between A and a subset of Bmust leave outsomeelementsofB.“Andnowcomesafundamentalquestion,”saidtheSorcerer.“Isit

thecasethatanytwoinfinitesetsarenecessarilyofthesamesize,ordoinfinitesetscomeindifferentsizes?Thatisthefirstquestiontobeansweredinbuildingupatheoryofinfinity,andfortunatelythequestionwasansweredbyGeorgCantortowardstheendofthelastcentury.Theanswercreatedastormandstartedawholenewbranchofmathematicswhoseramificationsarefantastic!“IwillgiveyouCantor’sansweratournext session.Meanwhile,

doyouhaveaguessastowhattheansweris?Areallinfinitesetsofthesamesize,ordotheycomeindifferentsizes?”

Remark.Iraisethisproblemtomystudentsinmybeginninglogicclasses, and among those students who don’t know the answer,abouthalfguessthatallinfinitesetsareofthesamesizeandabouthalfguessotherwise.Thoseofyouwhodon’talreadyknowtheanswer,wouldyoucare

tohazardaguessbeforeturningtothenextchapter?

Solutions

1.Letusfirstshowthatwhenyouaddanelementtoafiniteset,youget a finite set.Well, suppose a set A is finite. By definition, thismeansthatforsomenaturalnumbern,thesetAhasnelements.IfweaddanewelementtoA,theresultingsetthenobviouslyhasn+1elements,hencebydefinitionisfinite.From this, it immediately follows that ifwe remove an element

fromaninfinitesetB,theresultingsetmustbeinfinite,forifitwerefinite,wecouldputbacktheelementandtheresultingset—whichisthe original set B that we started with—would be finite, which itisn’t.

2. All that the management need do is to request that eachinhabitant move one room to the right—in other words, theoccupantofRoom1goestoRoom2,theoccupantofRoom2goestoRoom3,…theoccupantofRoomnmovestoRoomn+1.Sincethehotelhasnolastroom(unlikethemorenormalfinitehotel),noonewillbeoutinthecold.(Inafinitehotel,thepersoninthelastroomwouldhavenoroomtohisright).Afteralltheguestshavebeenkindenoughtomove,Room1isnowemptyandthenewlyarrivedguestcantakeit.Mathematically,whatwehavedoneistoputthesetofallpositive

integers into a 1 to 1 correspondence with the set of all positiveintegersfrom2on.Ofcourse,themanagerofthehotelcouldhave

done a similar thing if a hundredmillion new guests had arrivedinsteadofjustone.Themanagerwouldthenhaveaskedeachguesttomoveahundredmillionroomstotheright(thepersoninRoom1would move to Room 100,000,001; the person in Room 2 wouldmovetoRoom100,000,002,andsoforth).Foranynaturalnumbern,thehotelcouldaccommodatennewguests—justhaveeachguestmovenroomstotheright, leavingthefirstnroomsemptyforthenewguests.

3. Now, if infinitelymany new guests Q1,Q2,… Qn,… arrive, thesolution is a little different. One false solution that has beensuggestedisthatthemanagerfirstaskseveryoneoftheoldgueststomoveoneroomtotheright.Then,onenewguestisputintotheemptyRoom1.Thenthemanageragainaskseveryonetomoveoneroomtotheright,andsoRoom1isagainvacantandasecondnewguest isput inRoom1.Then thisoperation is repeated,againandagain,infinitelymanytimes,andsosoonerorlatereverynewguestgetsintothehotel.But what a horribly restless solution! No one keeps a room

permanentlyandinnofinitetimewouldalltheguestsbesettled—infinitelymanyshiftingswouldberequired.No,thewholethingcanbe done cleanly with only one shifting. Can you see what thisshiftingis?Theshiftingisthateacholdguestdoubleshisroomnumber—that

is,thepersoninRoom1goestoRoom2,theoneinRoom2goestoRoom4,theoneinRoom3goestoRoom6,…theoneinRoomngoestoRoom2n.Alltheseare,ofcourse,donesimultaneously,andaftertheshifting,alltheeven-numberedroomsareoccupiedandalltheinfinitelymanyodd-numberedroomsarenowfree.Andso,thefirstnewguestQ1goestothefirstodd-numberedroom—Room1;Q2goestoRoom3;Q3goestoRoom5,andsoforth(QnmovestoRoom2n–1).

4.We first “number” all the occupants of all the rooms of all thehotelsaccordingtothefollowingplan:

And so, each of the guests is “tagged” with a positive integer.Thenallofthemvacatetheirroomsandwaitoutsideforabit.Thenthemanagementshutsdownallthehotelsbutone,andeachguestisaskedtooccupytheroomwhosenumberisthesameasthenumberwithwhichheistagged—thepersontaggedngoestoRoomn.

19CANTOR’SFUNDAMENTALDISCOVERY

“WELL,”SAIDTHESorcereratthenextsession,“haveyouthoughtaboutthematter?Have you any guess as towhether there ismore thanoneinfinityoronlyone?”Oneof thetwo(I forgetwhich)guessedoneway,andtheother,theother.“The curious thing,” said the Sorcerer, “is that Cantor originallyconjecturedthatanytwoinfinitesetsmustbeofthesamesize,andas I understand it, he spent twelve years trying to prove hisconjecture.Then,inthethirteenthyear,hefoundacounterexample—whichIliketocalla‘Cantor-example,’Yes,thereismorethanoneinfinity—thereareinfinitelymany,infact.AndthisbasicdiscoveryisduetoCantor.“Now,hereiswhatCantordid.Asetiscalleddenumerablyinfinite,or, more briefly, denumerable, if it can be put into a 1 to 1correspondencewiththesetofpositiveintegers.AndsothequestionconsideredbyCantoris:Iseveryinfinitesetdenumerable?AsIsaid,hefirstguessedthateveryinfinitesetisdenumerable,andonlylaterdiscovered the real truth. What he did, during his initialinvestigation,was toconsidersets thatappeared tobe too large tobe denumerable, but then by some clever device he was able toenumeratethemafterall.”“Whatdoyoumeanbyenumeratingaset?”askedAnnabelle.“Toenumerateasetmeanstoputit intoa1to1correspondencewith the set of positive integers. In fact, theword ‘enumerable’ isused synonymouslywith ‘denumerable.’ Anyway, as I said, Cantorconsideredonesetafteranotherwhichatfirstsightappearedtobenon-denumerable—whichmeansinfinitebutnotdenumerable—andfoundacleverwaytoenumerateit.“Toillustratehismethod, letus imaginethat IamtheDeviland

youaremyvictimsinHell.That’snottoohardtoimagine,isit?”AnnabelleandAlexanderhadagoodlaughatthatthought.“Now, I tell you that I will give you a test. I tell you: ‘I amthinkingofapositivewholenumber,whichIhavewrittendownonthis folded piece of paper. Each day, you have one and only oneguessastowhatthenumberis.Ifandwhenyoucorrectlyguessit,youcangofree,butnotbeforethen,’Now,istheresomestrategybywhichyoucanbesureofgettingoutsoonerorlater?”“Ofcourse,”saidAlexander.“OnthefirstdayIaskifit’s1,onthesecondday, if it’s2,andso forth.Sooneror later I’mboundtohityournumber.”“Right,”saidtheSorcerer.“Now,mysecondtestisatinybitmoredifficult.Thistime,Iwritedowneitherapositivewholenumberoranegativewholenumber—Iwriteeitheroneofthenumbers1,2,3,4,…oroneofthenumbers–1,–2,–3,–4,…andagaineachdayyouhaveoneandonlyoneguessastowhatthenumberis.Now,doyouhaveastrategythatwillsurelygetyououtsoonerorlater?”“Of course,” saidAnnabelle. “On the first day I askwhether thenumber is+1,on thenextday, Iask if it is–1, then Ikeepgoing+2,–2,+3,–3,+4,–4,…andsoon.SoonerorlaterImusthityournumber.”“Right,”saidtheSorcerer,“andyouseewhatthismeans.Onthesurface, itwould seem that the set of positive andnegativewholenumberstakentogethershouldbelargerthan—infacttwicethesizeof—thepositivewholenumbersalone.Yet,youhavejustseenhowtoputthesetofpositiveandnegativewholenumbersintoa1to1correspondencewith thesetofpositivewholenumbers,andso thetwosetsarereallyof thesamesizeafterall.Thesetofallpositiveand negative whole numbers taken together is denumerable. TheproblemyouhavejustsolvedissubstantiallythesameasthesecondproblemIgaveyouconcerningHubert’sHotel.Asyourecall, therewere denumerablymany people in denumerablemany rooms, andthenaseconddenumerablesetofpeoplecamealongandwewantedtoputthosetwosetstogetherandaccommodateallthemembers.“The next test I givemy victim is definitelymore difficult. This

time,Iwritedown twonumbersonapieceofpaper,ormaybethesamenumberwritten twice.Forexample, Imightwrite3,57,or Imightwrite17,206,orImightwrite23,23.Eachday,youhaveoneand only one guess as towhat the two numbers are. You are notallowed to guess one of themone day and the other another day;youmust guess themboth on the same day.Now, do you believethere is a strategy by which you can certainly get out sooner orlater?”“I doubt it,” said Annabelle. “There are infinitely manypossibilitiesforthefirstnumberthatyouwritedown,andwitheachof those possibilities there are infinitely many possibilities for thesecond number, and so it would seem that an infinity times itselfshouldbebiggerthantheinfinityyoustartwith.”“Soitmightseem,”saidtheSorcerer,“andsoitdidseemtomanyatCantor’stime,butappearanceissometimesdeceptive.Yes,thereisastrategyforcertainlygettingout.Thesetofpossibilitiesthatyouare dealingwith is really denumerable after all. Can you find thestrategy?”“Amazing!” saidAnnabelle, andAlexanderagreed.The two thenput their heads together and came upwith a simple strategy thatdefinitelyworks.

•1•

Whatstrategywillwork?

“SupposeImaketheproblemslightlyharderbynowrequiringnotonly that you guess what the two numbers are, but also in whatorder theyarewritten—say,one iswritten to the leftof theother.Canyounowgetoutforsure?”“Ofcourse,” saidAnnabelle. “Thisproblem iseasy,nowthat thelastproblemwassolved.”

•2•

Whatisthesolution?

“Thenletmeaskyouthis,”saidtheSorcerer.“Whataboutthesetofallpositivefractions?Isthatsetdenumerableornot?Youarenowinagoodposition toanswer thatquestion.Byapositivefraction issimplymeantaquotientof twopositive integers—numbers like 3⁄7or21⁄13.”

•3•

Isthesetofpositivefractionsdenumerable?

“The answer was quite a shock to many mathematicians ofCantor’stime,”saidtheSorcerer.“AndnowIhaveamoredifficultproblemforyou.Thistime,Iwritedownsomefinitesetofpositiveintegers. I’mnottellingyouhowmanynumbersareintheset,norwhatthehighestnumberofthesetis.Eachday,youhaveoneandonlyoneguessastowhatthesetis.Ifandwhenyoucorrectlyguesstheset,yougofree.Now,doyouthinkthereisastrategyforgoingfree?”Thetwofeltthatitwasquiteunlikely.“There issuchastrategy,”saidtheSorcerer.“Thesetofall finitesetsofpositiveintegersisdenumerable.”

•4•

Howcanthesetofallfinitesetsofpositiveintegersbeenumerated?Whatstrategywouldyouusetogetfree?

“Whataboutthesetofallsetsofpositiveintegers—allinfinitesetsaswellasallfiniteones?”askedAnnabelle.“Isthatsetdenumerableornot?Oristheanswerunknown?”“Ah!” said the Sorcerer. “That set is non-denumerable—that isCantor’sgreatdiscovery!”“No one has yet found a way of enumerating that set?” askedAlexander.“No one has, and no one ever will, because it is logically

impossibletoenumeratethatset.”“Howisthatknown?”askedAnnabelle.“Well, let’s first look at it this way: Imagine a book with

denumerablymanypages—page1,page2,page3,…pagen,…Oneachpageiswrittendownadescriptionofasetofpositiveintegers.You own the book. If every set of positive integers is listed in thebook,youwinagrandprize.But I tell you thatyoucan’twin theprizebecause Icandescribea setofpositive integers thatcouldn’tpossiblybedescribedonanypageofthebook.”

•5•

Describeasetofpositiveintegersthatisdefinitelynotdescribedonanypageofthebook.

“Andsoyousee,”saidtheSorcerer,afterexplainingthesolutionof the last problem, “that the set of all sets of positive integers isnon-denumerable.Itislargerthanthesetofpositiveintegers.”“Youhaven’tshownthat,”saidAnnabelle.“Youhaveshownthat

thesetofallsetsofpositiveintegers—isthereanameforthisset?”“Yes,”saidtheSorcerer.“ForanysetA,thesetofallsubsetsofA

iscalledthepowersetofAandisdenotedP(A).WecanuseNforthesetofpositiveintegers,andsothesetofallsubsetsofN—whichisthesetofallsetsofpositiveintegers—isthusthepowersetofNandisdenotedP(N).“Allright,”saidAnnabelle.“Anyway,youhaveindeedshownthat

P(N) is non-denumerable—that P(N) cannot be put into a 1 to 1correspondence with N, and of course P(N) is infinite, but toconclude thatP(N) is therefore larger thanN isunwarranted, sinceyouhaven’tshownthatNcanbeput intoa1to1correspondencewith some subset of P(N). Don’t you have to do that in order tocompleteyourargument?”“We have already put N into a 1 to 1 correspondence with a

subsetofP(N),”repliedtheSorcerer.

•6•

Whenwasthatdone?

After the Sorcerer remindedAnnabelle of Problem4, one of thetwo (I forget whether it was Annabelle or Alexander) raised aquestion: “We have seen that the set of all finite sets of positiveintegersisdenumerable;henceitcanbeenumeratedinsomeinfinitesequenceS1,S2,…Sn,…Whycan’tweapplyCantor’sargumentandget a set Sdifferent fromall the sets S1,S2,… Sn,… ?Doesn’t thiscreateaparadox?”

•7•

Doesitcreateaparadox?

“Ihaveaquestion,”saidAlexanderafterthelastmatterhadbeensettled.“Weknowthatthesetofallfinitesetsofpositiveintegersisdenumerable. What about the set of all infinite sets of positiveintegers?Isthatsetdenumerableornot?”

•8•

WhatistheanswertoAlexander’squestion?

“Another question.We have seen that P(N) is larger than N. IsthereasetlargerthanP(N)?”askedAnnabelle.“Whycertainly,”repliedtheSorcerer.“ThefactthatP(N)islarger

thanNisonlyaspecialcaseofCantor’stheorem,whichis:

TheoremC(Cantor’s theorem).Forany setA, the setP(A)of allsubsetsofAislargerthanA.

“The proof of Cantor’s theorem,” said the Sorcerer, “is notsubstantiallydifferentfromtheproofIhavegivenyouforthespecialcasewhenA is the setNofpositive integers.The ideabehind theproof has been nicely illustrated by Smullyan in the followingproblem:Inacertainuniverse,everysetofinhabitantsformsaclub.

Theregistraroftheuniversewouldliketonameeachclubafteraninhabitantinsuchawaythatnotwoclubsarenamedafterthesameinhabitantandeachinhabitanthasaclubnamedafterhim.Itisnotnecessary that an inhabitantbeamemberof the clubnamedafterhim.Now,forauniversewithonlyfinitelymanyinhabitants,thisisclearlyimpossible,fortherearethenmoreclubsthaninhabitants(ifnisthenumberofinhabitants,then2nisthenumberofclubs).Butthisparticularuniversehas infinitelymany inhabitants, and so theregistrar seesno reasonwhy this shouldnotbepossible.However,every scheme he has ever tried has failed—there are always clubsleftover.Whyistheregistrar’sschemeimpossibletocarryout?”

•9•

Explain why the registrar’s scheme is impossible and how this isrelatedtoCantor’stheorem.

“As a consequence of Cantor’s theorem,” said the Sorcerer afterAnnabelle andAlexander understood the proof, “theremust be aninfinitenumberofdifferent-size infinities, forwecan startwithN,thesetofpositiveintegers,thenwehaveitspowersetP(N)—thesetof all subsets of N—which is larger than N, but then again byCantor’s theorem the power set of this new set—in other wordsP(P(N))—is larger than P(N), and then the set P(P(P(N))) is largerstill,andsoforth.Thus,foranyset,thereisalargerset,andsothesizesofsetsareendless.”

Solutions

1. For each n, there are only finitelymany possibilities for a pairwhosehighestnumberisn—thereareexactlynsuchpossibilitiesinfact. Thus there is only one possibility for the pair whose highestnumberis1—namely(1,1).Therearetwopossibilitiesforthepairwhosehighest number is 2—namely (1, 2) and (2, 2).Then, therearethethreepossibilitiesforthepairwhosehighestnumberis3—

(1,3),(2,3),and(3,3),andsoforth.Andsoweenumeratethemintheorder(1,1),(1,2),(2,2),(1,3),(2,3),(3,3),(1,4),(2,4),(3,4),(4,4),…(1,n),(2,n),…(n–1,n),(n,n),…

2.Inthiscase,itmighttakeusroughlytwiceaslongtogetout,butwestillgetoutsoonerorlaterbyenumeratingtheorderedpairsintheorder(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,3),(3,2),(3,1),…(1,n),(2,n)…(n–1,n),(n,n),(n,n–1),…(n,2),(n,1),(1,n+1),…

3.This is really thesameproblemas the last,except thatwehaveoneintegerabovetheother(asthenumerator)insteadoftotheleftof the other. Thus,we can enumerate the positive fractions in theorder 1⁄1, 1⁄2, 2⁄2, 1⁄3, 2⁄3, 3⁄3, 3⁄2, 3⁄1, 1⁄4, 2⁄4, 3⁄4, 4⁄4, 4⁄3, 4⁄2, 4⁄1,… Of course, we could get out a bit sooner by not namingduplications,suchas2⁄2(whichisreally1⁄1),and3⁄3,and2⁄4(whichduplicates1⁄2),andsoforth.

4.Asetwhosemembersareelementsa1,a2,…aniswritten{a1,a2,… an}. Now, there is only one set whose highest number is 1,namely{1}.Therearetwosetswhosehighestnumberis2,namely{1,2} and {2}. There are four sets whose highest number is 3,namely {3}, {1, 3}, {2, 3}, {1, 2, 3}. In general, for any positiveintegern,thereare2n–1setswhosehighestnumberisn.Thereasonisthis:Foranyn,thereare2nsubsetsofthesetofpositiveintegersfrom1ton(andthisincludestheemptyset!).Andsoanysetwhosehighestnumber isnconsistsofn togetherwith somesubsetof theintegersfrom1ton–1,andthereare2n–1suchsubsets.Anyhow, the important thing is that for each n, there are onlyfinitely many sets of positive integers whose highestmember is n.And so, I first name the empty set. Then, I name the set whosehighest number is 1. Then I go through the sets whose highestnumber is 2 (the order doesn’t reallymatter), then the setswhosehighestnumberis3,andsoforth.

5.Givenanynumbern,eithernbelongstothesetlistedonpagenoritdoesn’t.WeletSbethesetofallnumbersnsuchthatndoesnotbelongtothesetlistedonpagen.ForeachnweletSnbethesetlistedonpagen.OurdefinitionofSissuchthatforeveryn,thesetSmust bedifferent fromSn, because n belongs to S if and only if ndoesn’tbelongtoSn.ThismeansthateithernisinSbutnotinSn,orn isnot inSbut inSn. Ineithercase,Smustbedifferent fromSn,becauseoneofthosetwosetscontainsnandtheotherdoesn’t.To give a more concrete idea of the construction of the set S,

supposethesetslistedonthefirsttenpagesareasfollows:Page1—SetofallevennumbersPage2—Setofall(positivewhole)numbersPage3—TheemptysetPage4—Setofallnumbersgreaterthan100Page5—Setofallnumberslessthan58Page6—SetofallprimenumbersPage7—SetofallnumbersthatarenotprimePage8—Setofallnumbersdivisibleby4Page9—Setofallnumbersdivisibleby7Page10—Setofallnumbersdivisibleby5I have just listed the first ten sets at random.Now,whatwill S

look likeas faras the first tennumbersareconcerned?Well,whatabout1;shouldScontain1?Is1amemberofthesetlistedonpage1;thatis,is1anevennumber?No,itisnot,so1doesn’tbelongtoS1;soweput1inS.Whatabout2?Well,2certainlyisinS2(2isapositivewholenumber),andsowedon’tallow2inS.Thenumber3iscertainlynotinS3(nonumberisintheemptyset),andsowetake3asamemberofS.Now,4isnotinS4(4isnotgreaterthan100),so4isinS.Weletthereadercheckthenextsixcases:5isinS,since5isnotinS5;6isnotinS6(6isn’tprime),so6isinS;7isnotinS7,so7isinS,8isinS8,so8isnotinS;9isn’tinS9,so9getsputinS;10isinS10(10isdivisibleby5),so10isleftoutofS.ThusinlistingtheentriesofS,letusputthenumberninthenthplaceifnisinS,

andletusputablankinthenthplacetosignifythatnisdefinitelyleftoutofS.Then,thefirsttenplacesofourlistlooklikethis:1,—,3,4,5,6,7,—,9,—.…WenowseethatoursetSisdifferentfromS1 since it contains 1 and S1 doesn’t. Also, S is different from S2becauseitdoesn’tcontain2andS2does.Andsoyousee,foreachn,either S contains n and Sn doesn’t, or S doesn’t contain n and Sndoes,soScannotequalSn.Thus,thesetSisdifferentfromeveryoneofthesetslistedinthebook.Ofcourse,wedon’treallyneedthebookforthisargumenttohold

true. The whole point is that given any enumeration S1, S2,… Sn,…ofsetsofpositiveintegers,thereexistsasetSofpositiveintegersthatisdifferentfromeveryoneoftheSn’s—namely,thesetofallnsuchthatndoesn’tbelongtoSn.Thus,theinfinitesequenceS1,S2,…Sn,… fails to catchevery setofpositive integers, since the set Shasbeenleftout.Andso,thesetofallsetsofpositiveintegersisnotdenumerable.

6. InProblem4weshowedthat thesetofall finite subsetsofN isdenumerable, and thusN canbeput into a1 to1 correspondencewiththesetofallfinitesubsetsofN.Obviously,thesetFofallfinitesubsets ofN is a subset of the set ofall subsets ofN—thus F is asubsetofP(N)andNcanbeputintoa1to1correspondencewithF.

7.Ofcoursenot!ThesetSisindeeddifferentfromeachofthefinitesetsSn,butallthatmeansisthatthesetSisnotfinite.

8. We know that the set of all finite sets of positive integers isdenumerable;henceitcanbeenumeratedinsomeinfinitesequenceF1,F2,…Fn.…Thatis,toeachnwecancorrespondsomefinitesetFn of positive integers, and the correspondence is such that everyfinitesetofpositiveintegersissomeFnorother.Now,supposethatthesetofallinfinitesetsofpositiveintegersweredenumerable.Thenit could be enumerated in some infinite sequence I1, I2,… In,

…whereforeachn,Inistheinfinitesetcorrespondingtotheintegern. But then, we could enumerate all sets of integers—finite andinfinite—in theorderF1, I1,F2, I2,…Fn, In,…which is contrary tothe fact that the set of all sets of positive integers is non-denumerable.

9.Supposetheregistrar’sschemecouldbecarriedout.Thenwegetacontradictionasfollows:Callaninhabitantsociableifhebelongstotheclubnamedafterhimandunsociable ifhedoesn’t.Sinceinthisuniverseevery setof inhabitantsconstitutesaclub, then the setofallunsociableinhabitantsconstitutesaclub.Thisclubisnamedaftersomeone—say John. Thus, every member of John’s Club isunsociableandeveryunsociable inhabitantbelongs toJohn’sClub.IsJohnsociableornot?Eitherway,wegetacontradiction:SupposeJohn is sociable.Thismeans thatJohnbelongs toJohn’sClub,butonlyunsociablepeoplebelongtoJohn’sClub,sothisisout.Ontheotherhand,supposethatJohnisunsociable.Sinceeveryunsociableinhabitant belongs to John’s Club, then John, being unsociable,would have to belong to John’s Club,whichmakes John sociable.And so whether John is sociable or not, we get a contradiction.Thus,theregistrar’sschemecannotbecarriedout.The relation of this problem to Cantor’s theorem should be

obvious—it isonlya specialcaseofCantor’s theorem. Insteadofauniverseofpeople,consideranarbitrarysetA.Supposewehavea1to1correspondencebetweenAandthesetP(A)ofallsubsetsofA.Wegetthefollowingcontradiction:EachelementxofAcorrespondstooneandonlyonesubsetofA,whichwemightcallx’sset.Now,letSbethesetofallelementsxofAsuchthatxdoesnotbelongtox’s set. (In application to the problem above, S is the set ofunsociableinhabitants.)SomeelementbofAcorrespondstothissetS,andsob’ssetisthesetofallx’shavingthepropertythatxdoesn’tbelongtox’sset.Ifbbelongstob’sset,thenbisoneoftheelementshaving the property of not belonging to b’s set, which is acontradiction.Ifbdoesn’tbelongtob’sset,thenbhasthepropertyofnotbelonging tob’s set;buteveryelementhaving thisproperty

belongstob’sset,andsobmustthenbelongtob’sset,andweagainhave a contradiction. This proves that there exists no 1 to 1correspondencebetweenAanditspowersetP(A).Ofcourse,Acanbeput intoa1 to1correspondencewithsome

subsetofP(A)asfollows:Foranyelementx,by{x}ismeantthesetwhose only element is x. (Such a set {x} is called a unit set or asingleton.Ithasonlyoneelement,regardlessofhowmanyelementsx itself may have.) Well then, we can let each element x of Acorrespondtothesingleton{x}.Thiscorrespondenceisobviously1to1,andofcourse{x}isasubsetofA(sinceeveryelementof{x},ofwhichthereisonlyone—xitself—isanelementofA).AndsoAisthusina1to1correspondencewithasetconsistingofsomeoftheelementsofP(A).SinceAcanbeputintoa1to1correspondencewithasubsetof

P(A)andAcannotbeput intoa1 to1correspondencewithallofP(A)(aswehaveshown),thenbydefinition,P(A)islargerthanA.ThisprovesCantor’stheorem.

20BUTSOMEPARADOXESARISE!

“SOMETHINGhasbeenbotheringme,”saidAlexander,onthenextvisit.“CantorhasprovedthatforanysetAthereisasetlargerthanA—namelyP(A).Isn’tthatright?”“Yes,”saidtheSorcerer.“Well,”saidAlexander,“supposewetakeforAthesetofallsets.ThenbyCantor’stheoremthereisasetlargerthanA,buthowcantherebeasetlargerthanthesetofallsets?ThepowersetofA—theset P(A)—is a subset ofA, sinceA containsall sets, so how can asubsetofAbelargerthanAitself?Ireallydon’tunderstand!”“Ah,youhaverediscoveredafamousparadoxthatCantorhimselffound in 1897,” said the Sorcerer. “Later, Bertrand Russell gave asimplified version of Cantor’s paradox known as Russell’s paradoxwhichgoeslikethis:Givenanysetx,eitherxisamemberofitselforitisn’t.Forexample,asetofchairsisnotitselfachair,sonosetofchairsisamemberofitself.Ontheotherhand,takesomethinglikethe set of all things conceivable to the human mind. This set isapparently something conceivable to the human mind, hence isapparently a member of itself. Sets that are not members ofthemselvesarecalledordinarysets,whereassetsthataremembersofthemselvesarecalledextraordinarysets.Whetherextraordinarysetsreallyexistisopentoquestion,butordinarysetscertainlydoexist.Justaboutallthesetsweencounterareordinary.Now,letBbetheset of all ordinary sets. Thus, everyordinary set is amemberofBandeverymemberofBisordinary—noextraordinarysetsareinB.IsBamemberofitselforisn’tit?Eitherway,wegetacontradiction.SupposeB is amemberof itself.Then sinceonlyordinary sets aremembersofB, thenBmustbeoneof theordinarysets;butontheotherhand,sinceB isamemberof itself,Bmustbeextraordinary,whichisacontradiction.Thus,itiscontradictorytoassumethatBis

extraordinary. Now suppose that B is ordinary. Since all ordinarysets belong to B, then B, being ordinary, must belong to B, thusmakingBextraordinary(sinceBthenbelongstoitself),andsoitisalso contradictory to assume that B is ordinary. This is Russell’sfamousparadox.ItisasimplificationofCantor’sparadoxinthatthenotionofsize isnot involved.IwilldiscussthepossibleresolutionsofCantor’sandRussell’sparadoxeslater.“In1919,Russellgaveapopularizationoftheparadoxintermsofthe barber of a certain village who shaves all and only thoseinhabitants of the village who don’t shave themselves. Thus, thebarber won’t shave any inhabitant who shaves himself, but anyinhabitantwhodoesnotshavehimselfisshavedbythebarber.Doesthebarbershavehimselfordoesn’the?Ifhedoes,thenheisshavingsomeone (namelyhimself)who shaveshimself,henceviolating therulethathenevershavesanyonewhoshaveshimself. Ifhedoesn’tshavehimself, thenhe isoneof the inhabitantswhodoesn’t shavehimself, but hemust shave every such inhabitant, and so hemustthenshavehimself,andweagainhaveacontradiction.Sodoesthebarbershavehimselfordoesn’the?Howdoyousolvethisparadox?”“Perhapsthebarberwasawoman,”suggestedAnnabelle.“Thatwon’thelp,”saidtheSorcerer.“Ididn’tsaythatthebarbershavedall themen in thevillagewhodidn’t shave themselves,butthat thebarber shavedall the inhabitants of the villagewhodidn’tshavethemselves.”“Thenwhatisthesolution?”askedAnnabelle.“We’lldiscuss it later,” replied theSorcerer. “First I’d like to tellyousomevariantsofthisparadox.“There is the paradox of Mannoury, about a certain country inwhich every municipality must have a mayor and no twomunicipalitiesmayhavethesamemayor.Amayormayormaynotbe a resident of the municipality of which he is mayor. A law ispassed setting aside a special municipality called Arcadia that isexclusively for nonresident mayors, and by law every nonresidentmayor is compelled to reside there. Arcadia, like every othermunicipality, must have a mayor. Should the mayor of Arcadia

resideinArcadiaornot?“Then there is the ancient dilemma of the crocodile: A certaincrocodilehasstolenachild.Thecrocodilepromisesthechild’sfathertoreturnthechildifandonlyifthefathercorrectlyguesseswhetherthecrocodilewillreturnthechildornot.Whatshouldthecrocodiledoifthefatherguessesthatthecrocodilewillnotreturnthechild?“I am reminded of a paradoxical situation that arose with thelogician Smullyan when he was once cleverly outwitted by astudent.Inhisintroductorylogicclasses,Smullyanlikedtoillustratean essential idea behindGödel’s proof as follows:Hewould put apennyandaquarteronthetableandsaytoastudent: ‘Youare tomakeastatement.Ifthestatementistrue,thenIpromisetogiveyouoneof thetwocoins,notsayingwhichone.But if thestatement isfalse, then you get neither coin.’ The problem was to find astatementthatwouldforceSmullyantogivethequarter(assuming,ofcourse,thatSmullyankepthisword).”

•1•

Forthosewhodon’tknowthispuzzle,whatstatementwouldwork?(Thesolutionisgivenbelow.)

•2•

AfterAnnabelleandAlexanderhad solvedproblem1, theSorcerersaid “Now, a clever student made a statement which made itimpossible forSmullyantokeephisword.Whatstatementcould ithavebeen?”

SolutionstoProblems1and2

1.A statement thatworks is: “Youwillnotgiveme thepenny.” Ifthestatementwere false—if itwere false that Iwon’tgiveyouthepenny—thatwouldmean that Ido give you the penny, but then Iwouldgiveyouacoinforafalsestatement,whichIsaidIwouldn’t

do.Therefore,thestatementcan’tbefalse;itmustbetrue.Sinceitistrue, thatmeansthat I reallywon’tgiveyouthepenny;yet Imustgiveyouoneofthetwocoinsformakingatruestatement.Hence,Ihavenochoiceotherthantogiveyouthequarter.The readermaywonder how this is related to Gödel’s theorem.Well,Smullyanthoughtofthequarterasrepresentingtruthandthepennyasrepresentingprovability.Thenthestatement“Youwillnotgivemethepenny,”or“Iwillnotbegiventhepenny”correspondstoGödel’ssentence“Iamnotprovable.”

2. A statement that makes it impossible for Smullyan to keep hiswordis:“Youwillgivemeneithercoin.”WhateverIdo,Iwillhavetobreakmypromise.(Incidentally,thisstoryisapocryphal;itneverreallyhappenedtome!Idon’tknowhowtheSorcererevergotthisidea,butIadmitthatthestoryisagoodone.)

“I know another amusing incident about Smullyan,” said theSorcerer.“You seem to have an eerie connection with this characterSmullyan,”saidAnnabelle.“Youhavereferredtohimseveraltimes.Doyouknowhimpersonally?”“No,thetwoofushaveneveractuallymet.Inasense,weliveindifferent realities. I have reason to believe that he believes that Idon’treallyexist—thatIamonlyafictitiouscharacter.Ibelievethathebelievesthatyoutwodon’texisteither.Reallynow,howsillycanoneget!”“Perhaps it’s Smullyan who doesn’t really exist,” suggestedAlexander.“Ihave thoughtof thatpossibility,” said theSorcerer, “inwhichcase thestories Ihaveheardabouthimareonly legends.Anyhow,whetheritisalegendorthetruth,asIheardthematter,Smullyanwasoncegivinga graduate course in axiomatic set theory.Duringthe middle of one of his lectures, a girl student of his came in,apologized for being late, and asked Smullyan whether she couldhavea copyof thenotes. Smullyan replied: ‘Youcanhavea copy,

providedthatyouaregood!’Shethenasked,‘Justwhatdoesitmeantobegood?’Smullyanreplied,‘Itmeansnotknowingwhatitmeanstobegood!’Thisgotagenerallaugh.”*“That’s very curious,” said Annabelle, “because according to

Smullyan’s definition of ‘good,’ once a person has heard thedefinition,thatpersoncanneverbegood,forheorshethenknowswhat‘good’means—itmeansnotknowingwhat‘good’means.”“I’mnotsureabout that,”said theSorcerer.“Howcanoneknow

that ‘good’ means not knowing what ‘good’ means? That seemscontradictorytome.“Anyway,let’sgetbacktotheotherparadoxes.I’dlikeyoutotry

toanswerthefollowingquestions.”1.Doesthebarbershavehimselfordoesn’the?2.DoesthemayorofArcadiaresideinArcadiaornot?3.Whatshouldthecrocodiledowhentoldthatitwillnotreturn

thechild?4.HowdoyougetoutofRussell’sparadoxandCantor’sparadox?

•3•

Beforereadingfurther,howwouldyouanswerthosefourquestions?

The Sorcerer’s Explanations. “I can’t even get the first one,” saidAnnabelle.“Ican’tseehowthebarbercouldeithershavehimselfornotshavehimselfwithoutcontradiction,yethemustdooneortheother. I don’t knowwhat to think! Is there somethingwrongwithlogic?”“Ofcoursenot!”laughedtheSorcerer.“Thesolutiontothebarber

paradoxissoobviousthatit’samazingthatanyonecanbefooledbyit! And yet some very intelligent people have been taken in. Thisreveals an interesting psychological characteristic that’sunfortunatelyonlytoocommon.”“Don’tkeepusinsuspense,”saidAlexander.“Whatisthesolution

tothebarberparadox?”“I’llgiveyouahint,”saidtheSorcerer.“SupposeItoldyouthata

certainman ismore than 6 feet tall and also less than 6 feet tall.

Howwouldyouexplainthat?”“Iwouldsaythatit’simpossible,”repliedAlexander.“Well, doesn’t that give you an idea how to solve the barber

paradox?”“Don’t tell me,” said Annabelle, “that the solution simply is to

denythatthereissuchabarber?”“Ofcourse!”saidtheSorcerer.“Whatelse?HereIhavegivenyou

contradictory informationabout a certainbarber andaskedyou toexplainthecontradiction.Theonlyexplanationis thatwhat Ihavetoldyouisnottrue!”“Ineverthoughtofthat!”saidAnnabelle.“NorI,”saidAlexander.“Exactly!” said the Sorcerer. “And that’s the unfortunate

psychological characteristic to which I referred—the tendency tobelievewhatoneistold.”“Are all the other paradoxes solved the same way?” asked

Annabelle.“More or less,” replied the Sorcerer. “Let’s look at them one by

one.As for themayorofArcadia,hecannotpossiblyobey the lawbecause the law is inconsistent. If the mayor decides to reside inArcadia,heisbreakingthelaw,sinceonlynonresidentmayorsmayliveinArcadia.IfhelivesoutsideArcadia,heisagainbreakingthelaw, because his living outside Arcadia makes him a nonresidentmayor,andallnonresidentmayorsare required to live inArcadia.Thus it is logically impossible for themayor toobey the law.Thisdoesn’t constitute a paradox; it merely means that the law isinconsistent.Astothepuzzleofthecrocodile,theanswerissimplythatthecreatureisunabletodowhatitsaiditwoulddo.“Now, the Russell and Cantor paradoxes are more serious and

disturbing because they show that there is something basicallywrongwithourwayofthinking.WhatIhaveinmindisthis:Doesn’tit seemobvious that given anyproperty, there exists the set of allthingshavingthatproperty?”“Itwouldcertainlyseemso!”saidAnnabelle.“Thatseemsobviousenough,”saidAlexander.

“That’s the whole trouble!” said the Sorcerer. “This principle—called the unlimited abstraction principle—the principle that everypropertydeterminesthesetofallthingshavingthatproperty—thatprinciple indeed seems self-evident, yet it leads to a logicalcontradiction!”“Howso?”askedAnnabelle.“It leads to the Russell paradox as well as the Cantor paradox.

Suppose that itwere really true that for anyproperty, there existsthesetofallthingshavingthatproperty.Well,takethepropertyofbeinganordinaryset.ThentheremustexistthesetBofallordinarysets, and we get the Russell paradox: The set B can be neither amember of itself nor not a member of itself without presenting acontradiction. Thus, the unlimited abstraction principle leads toRussell’s paradox. It also leads to Cantor’s paradox, since we canconsiderthepropertyofbeingaset,andsothereisthenthesetofallsets,andthissetisontheonehandaslargeasanysetcanbe,buton the other hand, for any set there is a larger set (by Cantor’stheorem),andsotheremustbeaset largerthanthesetofallsets,whichisabsurd.Thus,thefallacyofCantor’sparadoxisthatthereissuchathingasthesetofallsets,andthefallacyofRussell’sparadoxisthatthereissuchathingasthesetofallordinarysets.Thesetwosets simply cannot exist. Yet, the unlimited abstraction principle,whichseemssoobvious,leadstotheseparadoxes,andhencecannotbetrue.ThefactthatitseemstrueiswhatImeantwhenIsaidthattherewassomethingbasicallywrongwithournaïvewayofthinkingaboutsets.“The discovery of these paradoxes was at first very disturbing,

since it seemedto threaten thatmathematicsmightbe inconsistentwithlogic.Areconstructionofthefoundationsofmathematicswasnecessary,whichIwilltellyouaboutnexttime.”

*Thisstoryreallyistrue,buthowtheSorcerer,whoI insist isa fictitiouscharacter,everfounditoutisbeyondme!Howdoesanonexistentpersonmanagetofindoutthings?—R.S.

21RESOLUTIONS

“YOU WERE GOING to tell us how the rise of the paradoxes led to thereconstruction of the foundations ofmathematics,” said Alexanderonthenextvisit.“Yes,” said the Sorcerer. “At the time, the most comprehensivesystemofmathematical foundationsthenknownwasthesystemofGottlobFrege. Itspurposewas toderiveallofmathematics fromafewbasicprinciplesof logicandsets. Inadditiontocertainaxiomsof logic,hetook justoneaxiomforsets—theunlimitedabstractionprinciplethateverypropertydeterminesaset—thesetofallthingshavingthatproperty.Fromjustthisoneaxiomofsettheory,Fregecouldderiveallthesetsthatmathematicsneeded.Tobeginwith,wecan takesomeproperty thatdoesn’thold foranything, suchas theproperty of something not being equal to itself, and thenwehavethesetofall thingshavingthatproperty,andthis is theemptyset(sincenothinghasthatproperty)andisdenotedØ.Next,givenanyentities x and y, we can form the set of all things having thepropertyof eitherbeing identical to x or identical to y.This set—denoted {x,y}—has x and y as elements and no others. This alsoholdsifxandyhappentobethesameentity,inwhichcasetheset{x,y}issimplythesingleton{x}—thesetwhoseonlyelementisx.“Andso,wenowhavetheemptysetØ,and,havingthis,wethenhavetheset{Ø}whoseonlymemberistheemptysetØ.Thisisnottobeconfusedwiththeemptysetitself,sincetheemptysethasnomembers,whereas the set {Ø} has amember—namely, Ø.Havingtheset{Ø},wecanthenformtheset{{Ø}}whoseonlymemberis{Ø},thentheset{{{Ø}}},thentheset{{{{Ø}}}},andsoforth,thusgetting infinitely many sets! These sets can serve as the naturalnumbers,whichwaslaterrealizedbyZermelo,whotook0tobetheemptyset,then1tobe{Ø},then2tobe{1}(whichis{{Ø}}),3to

be {2}, and so forth. It is also possible to derive from Frege’sunlimited abstractionprinciple the set of all natural numbers. Thissetisusuallydenotedbythesymbolω.“Next,givenanysetA,wecantalkaboutthepropertyofbeingasubsetofA,andsobyFrege’sprincipletherethenexiststhesetofallthingshavingthatproperty—inotherwords, thereexists thesetofallsubsetsofA,andthisisthepowersetP(A)ofA,whichplayssuchafundamentalroleinCantor’swork.“Next,wecan talkabout thepropertyofbeinganelementof atleastoneelementofA,andsobyFrege’sprinciple,thereisthesetofallelementsofallelementsofA,andthissetiscalledtheunionofAanddenotedUA.(Forexample,ifAisasetofclubs,thenUAisthesetofallthepeopleintheclubs.)“Yes,fromFrege’sunlimitedabstractionprinciple,oneneatlygetsallthesetsoneneedstodoclassicalmathematics.Thereisonlyonetroublewith Frege’s system; it is inconsistent! From the unlimitedabstractionprincipleonecangetthesetofallordinarysets,whichgives Russell’s paradox, and also the set of all sets, which givesCantor’s paradox. Sadly enough, just as Frege’s monumental workwasabouttobepublished,FregegotaletterfromRussellexplainingthathissystemwasinconsistent,andgavetheproof,usingRussell’sparadox.FregeacknowledgedthecorrectnessofRussell’sproofandwasextremelyupset, feeling thathis life’sworkhadcollapsed.Hisupsetwasreallyneedless,sincetheinconsistencyofhissystemcouldbe corrected, and hiswork contained an enormous number of thebasicideasusedbyRussellandotherslateron.Indeed,Russellhadthe highest respect for Frege. In his book The Principles ofMathematics, written in 1902, Russell says the following aboutFrege.”

The work of Frege, which appears to be far less knownthanitdeserves,containsmanyofthedoctrinessetforthinPartsIandIIofthepresentwork,andwhereitdiffersfromthe viewswhich I have advocated, the differences demanddiscussion.Frege’sworkabounds insubtledistinctions,and

avoidsalltheusualfallacieswhichbesetwritersonLogic.…In what follows, I shall try briefly to expound Frege’stheories on the most important points, and to explain mygrounds for differing where I do differ. But the points ofdisagreementarevery fewand slight compared to thoseofagreement.

“You said,” said Annabelle, “that the inconsistency of Frege’ssystemcouldbecorrected.How?”“That’s where the work of reconstruction came in,” said theSorcerer, “and this was done along two main lines. The first wasdonebyWhitehead andRussell in theirmonumental three-volumeworkPrincipiaMathematica.ThesecondlinewasadoptedbyZermeloinwhatisknownasZermelosettheory,whichwaslateramplifiedbyFraenkel toproducewhat isknownasZermelo-Fraenkelset theory—abbreviatedZ.F.—and isoneof themajormathematical systems inuse today.The systemofWhiteheadandRussell, though free frominconsistency with reasonable certainty, was a relativelycomplicatedaffairandisnotingeneralusenow,soIwouldrathertellyouofthelinesfollowedbyZermelo.“Zermelo’sbasicideawastoreplaceFrege’sunlimitedabstractionprinciple, which leads to an inconsistency, by what is called thelimited abstraction principle, or separation principle, which is this:Given any property and given any set A, there exists the set of allelementsofthesetAthathavetheproperty.Thus,wecannotspeakof the set ofall x’s having the property, as Frege did, butwe canspeak of the set of all x’s inA that have the property. The reasonwhythislimitedabstractionprincipleofZermeloissometimescalledtheseparationprincipleisthatgivenanysetA,apropertyseparatesthoseelementsofAthathavethepropertyfromthoseelementsofAthatdonot.Now,thisseparationprinciplehasneverbeenknowntolead to any contradiction and appears unlikely ever to do so. It isindeed a principle in commonuse by the everydaymathematicianwho speaks, for example, of the set of all numbers having a givenproperty,orifheisdoinggeometry,hemightspeakofthesetofall

pointsonaplanehavingagivenproperty.Hedoesn’tspeakofthesetofallthingshavingagivenproperty;the‘things’comefromsomesetAwhoseexistencehasalreadybeenestablished.“If we use Zermelo’s separation principle instead of Frege’sunlimited abstraction principle, Russell’s paradox disappears: Wecannolongerformthesetofallordinarysets,butgivenasetAinadvance,wecanformthesetBofallordinaryelementsofthesetA.(Werecallthatanordinarysetisasetthatisnotamemberofitself.)This leads to no paradox, but merely to the conclusion that B,thoughasubsetofA,cannotbeamemberofA.”“What’s the difference between being a subset and being amember?”askedAlexander.“TosaythatasetXisamemberofasetYmeansthatYisabunchofthingsandthatXisoneofthem.TosaythatXisasubsetofYistosaynotthatXitselfisamemberofY,butthatallmembersofXarealso members of Y. For example, suppose X is the set of malehumansonthisplanetandYisthesetofallhumansonthisplanet.Surely,thesetXofallhumanmalesisnotamemberofY;ititselfiscertainlynotahuman,butallitsmembersaremembersofY—everyhumanmaleisalsoahuman.Oragain,thesetofchairsinthishouseis a subset of the set of pieces of furniture of this house, but it ishardlyamemberof this set—it itself isnotapieceof furniture.Oragain,thesetEofallevenpositiveintegersisasubsetofthesetNofallpositiveintegers,butEiscertainlynotamemberofN;Eitselfisnotasinglepositiveinteger.”“Iunderstand,”saidAlexander.“Allrightthen,canyouseewhythesetBofallordinarymembersofA,thoughobviouslyasubsetofA,cannotbeamemberofA?”

•1•

Whyisthisso?

“ThustheRussellparadoxcannotbereconstructedinZermelosettheory.NorcantheCantorparadox,sincethereisnowayofprovinginZermeloset theorythat thereexists thesetofall sets. In fact, it

can be proved from the separation principle that there is no suchthingasthesetofallsets.Doyouseehow?”

•2•

Howcanthisbeproved?

•3•

Here is ananalogousproblem.Supposeyouare told thata certainbarber shavesall the inhabitantsof the townofPodunkwhodon’tshavethemselvesandthathenevershavesanyinhabitantofPodunkwho does shave himself. Does that necessarily lead to acontradiction?

“Now then,” said the Sorcerer, “as a price for having given upFrege’s unlimited abstraction principle, Zermelo had to take theexistenceof thesetsØ,{x,y},P(A),UAasseparateaxioms,andhealsohad to takeasanaxiomtheexistenceof thesetofallnaturalnumbers—thisistheso-calledaxiomofinfinity.AndsotheaxiomsofZermelo’s systemare: (1) the separation principle; (2) existence oftheemptysetØ;(3)foranysetsxandy,thereexiststheset{x,y}whosemembersare justxandy; (4) forany setA, thereexists itspowersetP(A);(5)foranysetA,thereexistsitsunionUA;(6)theaxiomofinfinity.“That is Zermelo’s entire system. Much later (in the twenties)

Abraham Fraenkel added what turned out to be an extremelypowerful axiom called the axiom of replacement, which is roughlythat, given any set A, one can form a new set by replacing eachelement of A by any element whatsoever, with the understandingthattwoormoreelementscanbereplacedbythesameelement,butnoelementmaybereplacedbymorethanoneelement(andthusthesizeofthesetAwouldn’tbeincreased).“ThatisthefamoussystemZ.F.—Zermelo-Fraenkelsettheory.Itis

in widespread use today, and it is amazing that the whole ofclassicalmathematics—number theory, algebra, calculus, topology,

etc.—canbederivedfromjustthesefewaxiomsofsettheory!Ihopeonedaytogiveyouanideaofhowthiscanbedone.”

Solutions

1.LetBbethesetofallordinaryelementsofA.ThusBconsistsofallandonlythosex’sinAsuchthatxisordinary.Thus,foreveryxwhichhappenstobeinA,xisinBifandonlyifxisordinary.(Thisisverydifferent fromsaying that foranyxwhatever, x is inB if andonlyifxisordinary.ThislatterwouldsaythatBisthesetofallx’swhateversuchthatxisordinary,whereasBisonlythesetofallx’sin A that are ordinary.) Now, if B were an element of A, then Bwouldbeoneofthex’ssuchthatxisinBifandonlyifxisordinary—inotherwords,BwouldbeinBifandonlyifBwereordinary;butthisisabsurd,sincetosaythatBisordinaryistosaythatBisnotinB! Thus, the assumption that B is an element of A leads to acontradiction (similar to that of the Russell paradox), but thecontradictionisavoidedbyBbeingoutside(notamemberof)A.

2.ThisfollowsfromProblem1:GivenanysetA,it failstocontainthesetBofallordinaryelementsofA; thussomeset isoutsideA.HencenotallsetsaremembersofA,andsonosetAisthesetofallsets.

3.No,thisleadstonocontradiction,sinceyouwerenottoldthatthebarber himself was an inhabitant of Podunk! The conclusion issimplythatthebarberisnotaresidentofPodunk,forifhewere,hecould neither shave himself nor not shave himself withoutcontradicting the given conditions. Living outside of Podunk,however,hecaneither shavehimselfornot shavehimselfwithoutcontradiction,sincenothingwastoldyouaboutthepeopleoutsideofPodunk whom he shaved or didn’t shave! (I hope you see theanalogytoProblem1!)

22THECONTINUUMPROBLEM

“ONE THING IWOULD like toknow,” saidAnnabelle. “Weknowthat theset P(N) of all sets of positive integers is larger than the set N ofpositive integers. Is there a setA that is larger thanNbut smallerthan P(N)? In other words, is there some set of size intermediatebetweenthesizesofNandP(N)orisP(N)thesetofnextlargersizethanN?”“Ah!”saidtheSorcerer,“wouldn’tyouliketoknowindeed?Andwouldn’tIliketoknow,andwouldn’tthewholemathematicalworldliketoknow!ThiswasthebasicquestionraisedbyCantorandtheanswerisnotknowntothisday!Cantorconjecturedthatthereisnosetwhose size is intermediatebetween thatofNand thatofP(N),and this conjecture is known as the continuum hypothesis. But it isonly a hypothesis or conjecture; it has not been either proved ordisproved to this day. Cantor conjectured the more generalhypothesisthatfornoinfinitesetAisthereasetofsizeintermediatebetweenthatofAandthatofP(A).Thisconjectureisknownasthegeneralizedcontinuumhypothesis.Butagain,itisonlyaconjecture;noonehasyetprovedordisprovedit.Ipersonallyregardthisunsolvedproblem as the grand unsolved problem—the most interestingunsolvedprobleminallofmathematics.ManymathematiciansandlogiciansfeelasIdoaboutthis.”“Whytheword‘continuum’?”askedAlexander.“It so happens that the set P(N) can be put into a 1 to 1correspondencewiththesetofpointsonaninfinitestraightline,andastraightlineissometimescalledacontinuum.Thus,P(N)issaidtobe of the same size as the continuum. And so, the question iswhether there is a set larger than N but smaller than thecontinuum.”“Whataretheprospectsofsolvingthecontinuumproblem?”asked

Annabelle.“That’sverydifficulttosay,”repliedtheSorcerer.“In1939,GödelprovedthatifwetakethesystemofZermelo-Fraenkel—whichisoneof the most powerful systems of mathematics yet known—thecontinuumhypothesiscannotbedisprovedinit.Andin1963,PaulCohen proved that the continuum hypothesis can never be provedfromtheseaxioms.Andsothecontinuumhypothesis is independentofthepresent-dayaxiomsofsettheory.”“Doesthismeanthatthecontinuumhypothesisisneithertruenorfalse, butdependsmerelyonwhat axiom systemyou take?” askedAnnabelle.“That’s a highly controversial question,” replied the Sorcerer.“There are those called formalists who regard the continuumhypothesisasneithertruenorfalse,butentirelydependentonwhataxiom system you take, since we can add either the continuumhypothesis or its negation to the axioms of set theory and have aconsistent system in either case—assuming, of course, that theaxioms of set theory are themselves consistent, of which there islittledoubt.Andso,theformalistdoesn’tbelievethatthecontinuumhypothesis is in itselfeithertrueor false,butmerelydependentonwhataxiomsystemonetakes.Attheotherextremearetheso-calledmathematicalrealists, orPlatonists—ofwhich I amdefinitely one—whobelievethatofcoursethecontinuumhypothesisiseithertrueorfalse,butwedon’tknowwhich.Webelievethatwedon’tyetknowenoughaboutsetstoanswerthequestion,butthisdoesn’tmeanthatthequestionhasnoanswer!“The formalist position really strikes me as most strange!Physicists and engineers certainly don’t think like that. Suppose acorpsofengineershasabridgebuiltand thenextday thearmy isgoing tomarch over it. The engineers want to knowwhether thebridge will support the weight or whether it will collapse. Itcertainlywon’tdo themonebitofgood tobe told: ‘Well, in someaxiom systems it can be proved that the bridge will hold, and inotheraxiomsystemsitcanbeprovedthatthebridgewillfall.’Theengineerswanttoknowwhetherthebridgewillreallyholdornot!

Andtome(andotherPlatonists)thesituationisthesamewiththecontinuumhypothesis:isthereasetintermediateinsizebetweenNandP(N)orisn’tthere?Iftheformalistisgoingtotellmethatinoneaxiomsystemthereisandinanotherthereisn’t,Iwouldreplythatthat does me no good unless I know which of those two axiomsystems is the correct one! But to the formalist, the very notion ofcorrectness,other thanmereconsistency, iseithermeaninglessor isitselfdependentonwhichaxiomsystemistaken.Thusthedeadlockbetween the formalist and the Platonist is pretty hopeless! I don’tthinkthateithersidecanbudgetheotheroneiota!”“Ididn’trealize,”saidAnnabelle,“thattherewasthatamountofcontroversyinafieldlikemathematics!Ithoughtthatfieldwascutanddriedandtherewasnoroomfordifferencesofopinion.”“Thedifferencesofopinionarenotsomuchinmathematicsasinthe foundations of mathematics. And the subject of mathematicalfoundations comes quite close to philosophy, in which, of course,thereisgreatdifferenceofopinion,”repliedtheSorcerer.“WhataboutGödelandPaulCohenthemselves?”askedAlexander.“AretheyformalistsorPlatonists?”“I’mnotsureaboutCohen,”repliedtheSorcerer.“Infact,I’mnotsure thatCohenhasmadeuphismindabout thematter, though Isuspecthe is somewhatclose to formalism—butpleasedon’tquotemeon this, because I reallydon’t know.Now,withGödel,hewasemphaticallyaPlatonist!Heexplicitlystatedthatwhatwasneededwastofindnewaxiomsofsettheorywhichareasself-evidentlytrueasthepresentonesandwhichwouldbestrongenoughtosettlethecontinuumhypothesisonewayor theother.Andhepredictedthatonedaysuchaxiomswouldbefoundandwhentheywere,Cantor’scontinuumhypothesiswouldbeseentobefalse!Yes,Gödelprovedthat although the continuum hypothesis—even the generalizedcontinuumhypothesis—couldneverbedisproved fromthepresent-dayaxiomsofsettheory,itisneverthelessfalse.“Well,”concludedtheSorcerer,“sofarGödel’shopehasnotbeenrealized.Nonewself-evidentlytrueaxiomshaveyetbeendiscoveredthat can settle the question. Will they ever be discovered? Who

knows?Ifandwhentheyare,itwillcertainlybeagloriousday!”

PARTVII

HYPERGAME,PARADOXES,ANDASTORY

23HYPERGAME

“DOYOUKNOWtheparadoxhypergame?”askedtheSorcereroneday.NeitherAnnabellenorAlexanderhadheardofit.“It is a lovely paradox created in the eighties by themathematicianWilliamZwicker.Besidesbeingadelightfulparadoxinitsownright,itleadstoatotallynewproofofCantor’stheorem.”“Thatsoundsinteresting!”saidAnnabelle.“Well, first for the paradox,” said the Sorcerer. “We will bediscussinggamesplayedbyjusttwopeople.Callagamenormalifithastoterminateinafinitenumberofmoves.Anobviousexampleistic-tac-toe;itmustendinatmostninemoves.Chessisalsoanormalgame; the fifty-move rule ensures that the game cannot go onforever. Checkers is also a normal game. Every card game that Iknowisnormal.Possiblychess,ifplayedonaninfiniteboard,mightbenonnormal.“Now, here is hypergame: The first move in hypergame is todeclarewhatnormalgameshouldbeplayed.Suppose, forexample,thatoneofyouisplayingagainstmeandIhavethefirstmove.ThenI must declare what normal game should be played. I might say:‘Let’splaychess,’inwhichcaseyoumakethefirstmoveinchessandwe keep playing until the chess game is terminated.Or, instead, Imight say: ‘Let’s play checkers.’ Then youmake the first move incheckers, and we continue playing until the checker game isterminated.OrImightsay:‘Let’splaytic-tac-toe’—IcanchooseanynormalgameI like.ButIamnotallowedtochooseagamethat isnotnormal;Imustchooseanormalgame.“Nowtheproblemisthis:Ishypergamenormalornot?”Thetwothoughtaboutthisforawhileandcametotheconclusionthatthegamemustbenormal.“Why?”askedtheSorcerer.

“Because,”theyexplained,“whatevernormalgameischosen,thatgamemusteventuallyterminate,sinceitisnormal.Thisterminatesthe hypergame that is played. And so regardless of what normalgameischosen, theprocesshasgot toterminate.Thus,hypergamemustbenormal.”“So far, so good,” said the Sorcerer, “but then a problemarises.Nowthatitisestablishedthathypergameisnormal,andsinceImayselectanynormalgameonmyfirstmove,thenIcansay:‘Let’splayhypergame.’Thenyoucansay:‘OK,let’splayhypergame.’AndthenI can say: ‘OK, let’s play hypergame,’ and this process can go onindefinitely. Thus, hypergame does not have to terminate, whichmeans that hypergame is not normal after all! And yet, you haveprovedthatitisnormal!Thisisaparadox.”NeitherAnnabellenorAlexandercouldsolveit.“Thewholepoint,”saidtheSorcerer,“isthatthegeneralnotionofagameisnotwelldefined.GivenasetSofwell-definedgames,onecan indeed define a hypergame of that set S, but this hypergamecannotitselfbeoneofthegamesofS.“Now,someone—IthinkHegel—oncedefinedaparadoxasatruthstandingon itshead.Veryoftenwhat firstcomesoutasaparadoxgets modified and leads to an important truth. And so it is withZwicker’s paradox of hypergame. A modification of the argumentestablishes an interesting theorem that yields a completely newproofofCantor’stheorem.“Recall brieflyCantor’s proof.Weare given a setA and to eachelementxofAisassociatedasubsetofAdenotedSx.TheideaistoconstructasubsetCofAthatisdifferentfromSxforeveryx.CantortookCtobethesetofallelementsxofAsuchthatxdoesn’tbelongtoSx.Now,whatZwickerdidwastofindanentirelydifferentsetZthat is distinct from every one of the sets Sx. It, like Cantor’sargument, shows that it is impossible to put A into a 1 to 1correspondencewiththesetofallsubsetsofA,butthenewsetZgotbyZwickerisentirelydifferentfromthesetCgotbyCantor.HereiswhatZwickerdid.

“Oncethecorrespondence(thatassignstoeachxinAthesubsetSx) is given,wedefine apath to be any finite or infinite sequencex,y,z,…ofelementsofAsuchthatforeachtermwofthesequence,eitherwisthelastterm,orthenexttermisanelementofSw.Thus,apathisgeneratedasfollows:StartwithanarbitraryelementxofA.IfSxisempty,that’stheendofthepath.Ifnot,picksomeelementyof Sx. We then have the two-term sequence (x,y). If Sy is empty,that’stheendofthepath.Ifnot,picksomeelementzofSy,andwethenhavethethree-termsequence(x,y,z). IfSz isempty,that’stheend of the path, but if Sz is nonempty, pick some element w andmake it the fourth term of the path, and keep on going in thismanner until you either come to some Sv that is empty, inwhichcasethepathends,orelsekeepgoingwithoutstop,thusgeneratingan infinitepath.[Forexample, ify isanelementofSxandx isanelementofSy,then(x,y,x,y,x,y,…)wouldbeaninfinitepath.Orifxhappens to be in Sx, then (x,x,x,x,…) would be an infinite path.]Now,givenanyx,eithertheredoesortheredoesn’texistaninfinitepaththatstartswithx.Nowdefinextobenormalifthereexistsnoinfinitepathstartingwithx.Thus,ifxisnormal,theneverypossiblepathstartingwithxmustterminate.NowletZbethesetofallthenormalelements.Wethenhave

TheoremZ—(Zwicker’stheorem).ThesetZisdifferentfromSxforeveryx.

“Theproof,”said theSorcerer“isanobviousmodificationof theargumentestablishingtheparadoxinhypergame.”

•1•

GivetheproofofZwicker’stheorem.

“Notice,”saidtheSorcerer,“thatZwicker’ssetZbearsnorelationto Cantor’s set C. The set of normal elements bears no significant

relationtothesetofx’sthatdon’tbelongtoSx.“Cantor’s proof relies essentially on the notion of negation. C isthesetofallx’ssuchthatxdoesnotbelongtoSx.Zwicker’sproofisnotbasedonnegation;itisbasedonthenotionoffinitenessinstead.”“It seems tome,” saidAnnabelle, “that thenotionofnegation issecretlyhiddeninZwicker’sproof.Hedefinesxtobenormalifthereexistsno infinite path startingwith x. Isn’t that an implicit use ofnegation?”“That’s a clever observation,” said the Sorcerer, “but that use ofnegation isnot reallyessential.Wecouldhavesimplydefinedx tobenormalifallpathsstartingwithxarefinite.”

Solution

1.WearetoshowthatthesetZofnormalelementscannotbeSxforanyx.Equivalently,wearetoshowthatfornoxisitthecasethatSxisthesetofallnormalelements.SosupposethatxissuchthatSxisthe set of all normal elements. We then get a contradiction asfollows:Wewillfirstshowthatxmustbenormal.Well,consideranypathstartingwithx.IfSxhappenstobeempty,thepathstopsrighttherewith x (since any second term ymust be amember of Sx). SowesupposeSxtobenonempty.Thenthesecondtermyofthepathmustbe chosen from Sx, hence must be normal (since only normalelementsareinSx).Sinceyisnormal,theneverypathstartingwithy must terminate; hence every path starting with (x,y…) mustterminate,andsoxmustbenormal.SincexisnormalandSx isthesetofallnormalelements, thenxmustbeamemberofSx.Hence,thereistheinfinitepath(x,x,x,…)just like the infinite game (“Let’s play hypergame,” “Let’s playhypergame,” “Let’s play hypergame,” …), and we thus get acontradiction.ThusthesetZofallnormalelementsmustbedifferentfromevery

Sx.

24PARADOXICAL?

“I RECENTLY THOUGHT of a paradox,” said Annabelle. “It concerns yourexplanationofCantor’sproofintermsofthebookoflistedsets.Youremember? You described a book with denumerably many pagesnumbered page 1, page 2,… and so on, and on each page waswrittenthedescriptionofasetofpositiveintegers.Theproblemwastodescribeasetthatwasn’tdescribedonanypageofthebook.Doyouremember?”“Yes,ofcourseIremember,”repliedtheSorcerer.“Verywell, and ifyou remember, the solutionyougavewas thedescription ‘The set of all n such that n is not a member of the setdescribedonpagen.’”“That’sright,”saidtheSorcerer.“Now, my paradox is this: Suppose that that very descriptionoccursonsomepageofthebook—saypage13.Is13thenamemberof that set or not? Just think, we are considering the set S of allnumbersnthatdon’tbelongtothesetdescribedonpagen.Thus,foranynumbern,nbelongstoSifandonlyifndoesn’tbelongtothesetdescribedonpagen.Inparticular,13belongstoSifandonlyif13doesn’tbelong to the setdescribedonpage13,butS is the setdescribedonpage13,sowehavetheabsurditythat13belongstoSjustincaseitdoesn’tbelongtoS!Howcanthatbe?”“That’s neat!” said the Sorcerer,with a broad smile. “I like thatideaverymuch!”“Butwhatisthesolution?”pleadedAnnabelle.

•1•

Beforereadingfurther,doesthereaderhaveanyideaofhowtogetoutofthisone?

SolutiontoProblem1.“Theexplanationisthis,”saidtheSorcerer.“Considerthefollowingexpression:

(1)Thesetofallnsuchthatndoesn’tbelongtothesetdescribedonpagen.

“Now,if(1)appearsononeofthepages—saypage13—thenitisnot a genuine description of any set; it is what is called apseudodescription.”“Why?”askedAnnabelle.“Because if it were, then it would lead to a contradiction—theverycontradictionyousoaptlydescribed.”“I’m not sure that explanation is quite satisfactory,” saidAnnabelle.“Lookatitthisway,”saidtheSorcerer.“Foradescriptionofasetofnumberstobegenuine,itmustprovideadefiniterule—adefinitecriterion forwhichnumbers are in the set andwhichnumbers arenot. If theexpression(1)appearsonpage13of thebook, then foreverynotherthan13,ittellsyouwhethernisinthesetornot.Butit does not tell you whether 13 is or is not in the set. Now, thefollowingisagenuinedescription,evenifitdoesappearonpage13.

(2)Thesetofallnotherthan13suchthatndoesn’tbelongtothesetdescribedonpagen.“According to this, the number 13 is not a member of the setdescribedonpage13.Alsothefollowing,ifappearingonpage13,isagenuinedescription.

(3)Thesetofallnotherthan13suchthatndoesn’tbelongtothesetdescribedonpagen,togetherwith13.

“Thisisgenuine,becauseittellsyouwhattodowith13—13istobeincludedintheset.Andso(2)and(3)aregenuinedescriptions,even if one of them appears on page 13, but (1), if it appears onpage13,isnotagenuinedescription.Thecuriousthingisthat(1)isgenuineprovidingitdoesn’tappearonanypageofthebook!Ifwritten

outside the book, it is genuine—providing, of course, that all thedescriptionsinthebookaregenuine.”At this point, the Sorcerer stared out into space for a couple ofminutes,lostinthought.Whenhecameto,hesaid:“Youknow,youhave just givenme an idea for amuchmorebafflingparadox!Allright,wenowconsideranotherbookwithdenumerablymanypagesand on each page is written either a genuine description or apseudodescriptionofasetofpositiveintegers.Unlikethelastbook,wenowallowpseudodescriptions to appear on someof thepages.Now,thefollowingdescriptionissurelygenuine:

The set of all n such that the description on page n isgenuineandndoesn’tbelongtothesetdescribedonpagen.

“Thisdescriptionmustbegenuinebecauseitprovidesadefiniterulefor which numbers belong to the set S so described and whichnumbersdonot.Consideranynumbern.Thedescriptiononpagenis either genuine or it isn’t. If it isn’t, then n is automaticallyexcluded from S. If it is, then the description on page n reallydescribesaset;henceitisreallydeterminedwhethernisinthatsetanditisaccordinglydeterminedwhethernisinSornot.Therefore,theabovedescriptionisindeedgenuine.“Now,whathappensifthatgenuinedescriptioniswrittenonpage13?Is13theninthesetsodescribedorisn’tit?Eitherway,yougetacontradiction,asAnnabellehasshownus,andwecan’tgetoutofitthistimebysayingthatthedescriptionisnotgenuine,forIhavejustprovedtoyouthatitis!Well?”“Youarecertainlymakinglifedifficult!”saidAlexander.“Irealizethat,”saidtheSorcerer,withamischievoussmile.

•2•

Well,howdoesonegetoneselfoutofthisone?(Solutionisgivenattheendofthechapter.)

RelationtotheBerryParadox.“Actually,”saidtheSorcerer,after

he gave the solution, “my paradox is closely related to the Berryparadox—itisasortofCantorianversionofit.”“WhatistheBerryparadox?”askedAnnabelle.“It is this: as numbers get larger and larger, it takes more andmorewordstodescribethem.”“Thatseemsreasonable,”saidAlexander.“In fact for any positive integer n, there must be numbers thatcannotbedescribedinlessthannwords.”“Ibelievethat,”saidAlexander.“Thenforanyn,theremustbethesmallestnumbernotdescribableinlessthannwords.Right?”“Ofcourse,”saidAnnabelle.“Allright,nowlookatthefollowingdescription.

THESMALLESTNUMBERNOTDESCRIBABLEINLESSTHANELEVENWORDS.

“Thatdescribesadefinitenumber,doesn’tit?”askedtheSorcerer.Thecoupleagreed.“Willyoupleasecountthenumberofwordsinthatdescription,”askedtheSorcerer.Theycountedthenumberandrealizedtotheirhorrorthatitwas10.“Andso, thesmallestnumbernotdescribable in less thanelevenwords is describable in tenwords.Now,will youplease explain tomehowthatcanbe?”saidtheSorcerer.

•3•

Oh,no!Howcouldthatbe?

“All these paradoxes,” said the Sorcerer, “remind me of a verydelightful story of how the Devil was once outwitted by a cleverstudent ofGeorgCantor.Do you know the story of Satan, Cantor,andInfinity?”NeitherAnnabellenorAlexanderknewthestory.“ThenIwilltellittoyouonyournextvisit.”

Solutions

1.Thissolutionisgivenimmediatelyaftertheproblem.

2.Theexplanationisthattheverynotionofagenuinedescriptionisnotwelldefined.Onecandefineagenuinedescriptiononlywithinaprecisely formulated language,andEnglish isnot sucha language.The situation is similar to that of truth.Ashas been shownby thelogicianAlfredTarski,forlanguagesofsufficientstrength,truthforsentencesofthelanguageisnotdefinablewithinthelanguage.Forexample,truthforEnglishsentencesisnotdefinableinEnglish,forifitwere,youwouldgettheparadoxicalsentence“Thissentenceisnottrue.”

3. The solution is really the sameas thatof the last problem.Thenotionofadescriptionisnotwelldefined.

25SATAN,CANTOR,ANDINFINITY

HereisthestoryastoldbytheSorcerer.“Ihavebeenhavinga lotof funwithsomeofourvictims,” saidSatantoBeelzebub,asherubbedhishandsinglee.“IneachcaseItellthevictimthatIamthinkingofoneandonlyoneobjectoutofan infinite set of objects. Each day the victim is allowed one andonly one guess as to what the object could be. If and when heguessesit,hegoesfree.Thisisthegeneralformatofthesetests.Insomecasesthevictimhasbeencleverenoughtodeviseastrategytowinhisfreedom,inothercasesnot.Well,tomorrowIamexpectinganewvictimandIwillarrangematterssohecannevergofree!”“Whatwillyoudo?”askedBeelzebub.“Ihavewrittendownthenameofasetofpositivewholenumbers.Eachdaythevictimwillbeallowedtonameoneandonlyoneset,and if he ever namesmy set, he can go free. But hewillnever gofree!”saidSatan,shriekingwithdelight.“Why?”askedBeelzebub.“Well,justlookatwhatIhavewritten!”

The set of all numbers n such that n does not belong to the setnamedonthenthday.

“Idon’tunderstand!”saidBeelzebub.“I thought you wouldn’t, blockhead!” said Satan. “He can’tpossiblynamemysetonanyday,becauseforeachpositiveintegern,thesethenamesonthenthdayisdifferentfrommyset,sinceoneofthesetwosetscontainsthenumbernandtheotherdoesn’t!It’sas

simpleasthat!”“Thatsoundslikefun!”saidBeelzebub.Well, it sohappened that thenextvictimwasaprize studentofGeorg Cantor! He not only knew his mathematics of infinityperfectlybutwasalsoanexpertinsemanticsandlaw.Infact,hehadplannedtogointolawbeforehefellunderCantor’smagneticswayanddecidedinsteadtogointologicandmathematics.“BeforeIsignanycontract,”saidthestudenttoSatan,“IwanttobesurethatI’mabsolutelyclearastotheterms.”“I’vealreadytoldyou,”saidSatan,“thatIhavewrittendownthename of some set of positive integers and it is right here in thisenvelopewithmyroyalseal.Eachdayyouareallowedtonameoneandonlyoneset.Ifandwhenyounamethesetwrittenhere,yougofree.It’sassimpleasthat!”“I already understood that,” replied the student, “but there areseveralquestionsthatneedtobeanswered.First,supposethatonagiven day I name the same set that you have written, but mydescriptionofthesetisdifferentfromyours.Anysetcanbedescribedinmanydifferentways.Forexample,supposeyouhavewritten,‘Thesetwhose onlymember is the number 2,’ but on some day I say,‘Thesetofallevenprimenumbers.’Now,thetwosetsarereallythesame,since2istheonlyevenprimenumber;yetthedescriptionsaredifferent.Whathappensthen?”“Oh,inthatcaseyouwin,”repliedSatan.“Idonotdemandthatourdescriptions be the same but only that they describe the sameset.”“But that raises a serious problem!” said the student. “It is notalwaysasimplemattertodeterminewhethertwodescriptionsnamethesameset.SupposeonacertaindayInameasetandyoureply,‘No,that’snotthesetIhaveinmind,’butIhavereasontobelievethat it really is the same set and that you have only described itdifferently.Whathappensthen?”“Inthatcase,”saidSatan,“youareallowedtochallengeme.Now,a challenge is a very serious matter and you should think verycarefullybeforeyoumakeone.Itmightwinyouinstantfreedom,or

itmightdoomyouhereforever!”“Justwhatdoyoumeanbya‘challenge’?”askedthestudent.“You challenge me to open the envelope and show you what Ihavewritten.Ifyoucanprovethatthetwodescriptions—yoursandmine—arereallyofthesameset,youwinthechallengeandgofree.ButifIcanprovethatthedescriptionsnamedifferentsets,thenyouhavelostthechallengeandyourrighttonameanymoresetsinthefuture is canceled. There is then no way you can ever escape.Rememberwellthatafterachallenge,youarenotallowedtonameanymoresets.”“That’sclearenough,”saidthestudent,“butnowcomesasecondpoint.HowdoIknowthatyouhavereallywrittenthenameofasetinthisenvelope?”“Youdoubtmyword?”askedSatan.“Oh,notatall;Idon’tdoubtforamomentthatyouhavewrittensomething in this envelope which you believe to be a genuinedescription of a set, but it has happened in the history ofmathematics that what at first sight appeared to be a genuinedescription has turned out not to describe any set at all—in otherwords,thattherereallyisnosetansweringsuchadescription.Such‘descriptions’arewhatlogicianscallpseudodescriptions.Theyappearto describe a set but really don’t. Now suppose that at a certainstage, I have reason to suspect thatwhat you havewritten in theenvelopeisnotagenuinedescriptionbutonlyapseudodescription.Whathappensthen?”“If on any day you suspect that I have written only apseudodescription,” replied Satan, “then again you may challengeme.IwillopentheenvelopeandshowyouwhatIhavewritten. Ifyou can prove that it is only a pseudodescription, you win thechallengeandgofree.But if Icanprovethat it really isagenuinedescription, then you lose the challenge and again your right toname any more sets in the future is canceled. I must earnestlyremindyouthatafterachallenge,youmaynotnameanymoresets.”“That point is now clear,” said the student. “One last thing:Areyouwillingtohaveitwritteninthecontractthatifatanytimeyou

violateanyoftheconditions,thenIgofree?”“Yes,” replied Satan, “provided that you are willing to have itwritten that if at any time you violate any of the conditions, thenyoustayhereforever.”“Agreed!”saidthestudent.ThecontractwasthendrawnupbyBeelzebubanddulyexecutedbybothparties.“Good!”saidSatan.“Whenwouldyouliketobegin?”“Today’sasgoodadayasany,”saidthestudent.“Letthisbethefirstdayofthetest.”“Verywell,then,nameasetofpositiveintegers!”“ThesetofallnsuchthatndoesnotbelongtothesetInameonthenthday,”saidthestudent.“AndnowIchallengeyoutoopentheenvelope.”“Goodgrief!”criedSatan.“Ineverthoughtofthat!”“Opentheenvelope!”demandedthestudent.Satanhadtoopentheenvelope,andofcoursehehadwrittenthesamething.“SoIgofree!”saidthestudent.“Notsofast,youngman!”saidSatan.“Youhavenotreallynameda set; youhave done justwhat you accusedme of possibly doing;youhavegivenonlyapseudodescription,notagenuinedescription!”“Why?”askedthestudent.“Because the assumption that you have named a set leads to alogicalcontradiction:Supposeyouhavenamedaset.Thenthissetisthesetyouhavenamedonthefirstday.Now,thenumber1belongsto this set if andonly if itdoesn’tbelong to the setnamedon thefirstday,butsincethissetisthesetnamedonthefirstday,then1belongs to this set if and only if it doesn’t. This is a clearcontradiction,andtheonlywayoutofthecontradictionisthatyouhavenotreallynamedaset.”“I’mgladyourealizethis,”saidthestudent,“becausebythesametoken,youhavefailedtonameaset.”“Now, just a minute!” said Satan. “The genuineness of mydescriptionispredicatedontheassumptionthatyounameoneand

onlyoneseteachday,asyou’re supposed to.So far,youhavenotyetnamedasettoday,soInowcommandyoutonameaset.”“Oh,Ihavenointentionofnaminganysetstoday.”“What?”criedSatan,unabletobelievehisears.“Itdoesn’tsayanywhereinthecontractthatImustnameaseton

eachday;itsaysthatoneachdayIamallowedtonameaset.Well,itsohappensthattodayIdon’tchoosetonameanyset.”“Oh,really!”shriekedSatan.“Yourefusetonameasettoday,eh?

WellI’llforceyoutonameasettoday,andtomorrowI’llagainforceyou tonamea set, and thedayafter and thedayafter, and soonthroughoutalleternity.Youhavenoideahowterriblemymethodsare!”“Oh,youcan’tdothat,”saidthestudent.“I’vealreadychallenged

you,anditsaysquiteexplicitlyinthecontractthatafterachallenge,I’mnotallowedtonameanymoresets.”

Epilogue.Ofcourse,Satanhadtosetthestudentfree.Thestudentimmediatelyascendedtoparadiseandembracedhisbelovedmaster,Georg Cantor. The two had a delighted chuckle over the entireaffair.“You realize,” said Cantor, “that you didn’t have to be that

elaborate; you didn’t have to start the procedure by giving apseudodescription. You could have started by simply saying: “Ichallengeyou!”Afterthechallenge,youwouldn’thavebeenallowedto name any sets, which would automatically have made Satan’s‘description’amerepseudodescription.”“Oh, I realized that,” said the student. “I just thought I’dhavea

littlefunwithhim.”

“Yourealize,”saidtheSorcerertoAnnabelleandAlexanderwhenhehadendedhis story, “thatSatanusedCantor’s famousdiagonaldevicetoprovethatthepowersetofasetnhashighercardinalitythann.Thestudent rightfullyguessed thatSatanwould try topullsuch aCantorian trick. Several peoplehave askedmewhether theexpression ‘The set of all n such that n does not belong to the set

namedondayn’isagenuinedescriptionornot.Theansweristhatitisagenuinedescriptionifandonlyifoneachdaythereisoneandonlyonesetnamedonthatday.Ifthestudentfailedtonameaseton so much as one day, that would be enough to nullify themeaningfulnessofSatan’sdescription.Orifthestudentnamedmorethan one set on a given day, that would also invalidate Satan’sdescription.Butifthestudentnamedoneandonlyonesetoneachday, then Satan’s description would be perfectly well defined. Acurious thing, though, about this description is that after no finitenumber of days can it be known that Satan wrote a genuinedescription, unless it could somehow be known that the studentwouldnameoneandonlyoneseteachday.“Satanreallymadeaverypoorcontract!Ifhehadrequiredinstead

of justallowed the student tonameoneandonlyone seteachdayandifhehadjustdeletedthisbusinessaboutthestudentnotbeingallowedtonameanymoresetsafterachallenge,hewouldobviouslyhave won. Had he done that, then it would indeed have beenlogicallyimpossibleforthestudentevertonameSatan’sset.Butasthe contract stands, a mere challenge on the part of the studentdisallowshimfromnaminganymoresets,whichinturnnullifiesthegenuinenessofSatan’s‘description.’“Themoral of the story,” said the Sorcerer, “is that even fallen

angelsmightbenefitfromagoodcourseinmathematicallogic.”