The Mathematics of Various Entertaining Subjects: Research in Recreational Math

290 Pages·2016·16.03 MB·English

by  Jennifer Beineke, Jason Rosenhouse, Raymond M. Smullyan

TheMathematicsofVariousEntertainingSubjects
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T M V
HE ATHEMATICS OF ARIOUS
E S
NTERTAINING UBJECTS
RESEARCH IN RECREATIONAL MATH
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EDITEDBY
Jennifer Beineke & Jason Rosenhouse
WITHAFOREWORDBYRAYMONDSMULLYAN
NationalMuseumofMathematics,NewYork • PrincetonUniversityPress,Princeton&Oxford
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Contents
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ForewordbyRaymondSmullyan vii
PrefaceandAcknowledgments x
PARTIVIGNETTES
1
ShouldYouBeHappy? 3
PeterWinkler
2
One-MovePuzzleswithMathematicalContent 11
AnanyLevitin
3
MinimalistApproachestoFigurativeMazeDesign 29
RobertBosch,TimChartier,andMichaelRowan
4
SomeABCsofGraphsandGames 43
JenniferBeinekeandLowellBeineke
PARTIIPROBLEMSINSPIREDBYCLASSICPUZZLES
5
SolvingtheTowerofHanoiwithRandomMoves 65
MaxA.AlekseyevandTobyBerger
6
GroupsAssociatedtoTetraflexagons 81
JulieBeierandCarolynYackel
7
ParallelWeighingsofCoins 95
TanyaKhovanova
8
AnalysisofCrosswordPuzzleDifficultyUsinga
RandomGraphProcess 105
JohnK.McSweeney
9
FromtheOutsideIn:SolvingGeneralizationsofthe
Slothouber-Graatsma-ConwayPuzzle 127
DerekSmith
PARTIIIPLAYINGCARDS
10
GalliaEstOmnisDivisainPartesQuattuor 139
NeilCalkinandColmMulcahy
vi • Contents
11
HeartlessPoker 149
DominicLanphierandLauraTaalman
12
AnIntroductiontoGilbreathNumbers 163
RobertW.Vallin
PARTIVGAMES
13
Tic-tac-toeonAffinePlanes 175
MaureenT.CarrollandStevenT.Dougherty
14 (cid:1)
ErrorDetectionandCorrectionUsingSET 199
GaryGordonandElizabethMcMahon
15
ConnectionGamesandSperner’sLemma 213
DavidMolnar
PARTVFIBONACCINUMBERS
16
TheCookieMonsterProblem 231
LeighMarieBraswellandTanyaKhovanova
17
RepresentingNumbersUsingFibonacciVariants 245
StephenK.Lucas
AbouttheEditors 261
AbouttheContributors 263
Index 269
Foreword
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RaymondSmullyan
Recreational mathematics can provide a most pleasant introduction to deep
resultsinmathematicsandlogictothegeneralpublic.Forexample,takethe
field known as propositional logic, which deals with the way simple proposi-
tionsarecombinedtoformmorecomplexpropositions,usingtheconnectives
not,and,or(where“or”shouldbetakeninthesenseof“atleastone,”)if-then,
and if and only if. I have found that a most useful way of approaching this
subjectisbythelogicoflyingandtruth-telling,whichpeoplecanrelatetoquite
easily.AndsoletusturntoaplaceIhavedubbedtheIslandofKnightsand
Knaves,inwhichknightsalwaystellthetruthandknavesalwayslie,andeach
nativeoftheislandiseitheraknightoraknave.Supposeyouvisittheisland
andcomeacrosstwonatives,AlanandBob,andAlansays,abouthimselfand
Bob,“Bothofusareknaves.”IsAlanaknightoraknave,andwhatisBob?
This problem is pretty easy: No knight could falsely claim that he and
someone else were both knaves, hence Alan is obviously a knave. Since his
statement is false, it is not true that both are knaves, and so at least one is a
knight,andsinceitisnotAlan,itmustbeBob.
Foraslightlyharderproblem,supposethatinsteadofsaying,“Bothofus
areknaves,”Alanhadsaid,“Atleastoneofusisaknave,”(or,whatamounts
tothesamething,hecouldhavesaid,“EitherI’maknaveorBobisaknave.)
Now,whatareAlanandBob?Wouldyoucaretotrysolvingthisbeforereading
further?
Hereistheanswer:IfAlanwereaknave,thenitwouldbetruethatatleast
oneofthemwasaknave,butknavesdon’tmaketruestatements.HenceAlan
cannotbeaknave;hemustbeaknight.ItfurtherfollowsthatasAlantruthfully
saidatleastoneisaknave,butsinceitisnotAlan,itmustbeBob.Hencethe
answeristheoppositetothatofthefirstproblem—nowAlanisaknightand
Bobisaknave.
Theabovetwoproblemswererespectivelyconcernedwithandandor.Now
foroneconcernedwithif-then.
SupposeAlansaid,“IfIamaknight,thensoisBob.”CanthetypesofAlan
andBobnowbeestablished?Trythisone!
Yes,bothtypescanbeestablished,andtheargumentisdelightfullysubtle!
Suppose Alan is a knight. Then, as he truthfully says, if Alan is a knight,
so is Bob, and since Alan is a knight (by supposition), Bob is also a knight.
viii • Foreword
Thisproves,notthatAlanisaknight,northatBobisaknight,butonlythat
if Alanisaknight,soisBob.ThuswenowknowthatifAlanisaknight,so
isBob,butsinceAlansaidjustthat,hespoketruthfully,henceheisaknight.
Furtheritisalsotruethatifheisaknight,soisBob(ashetruthfullysaid),and
sinceheisaknight,soisBob.Thus,theansweristhattheyarebothknights.
Finally, consider the case in which Alan says: “Bob and I are of the same
type—both knights or both knaves.” Can Alan’s type be determined? What
aboutBob’s?
WewillseethattheansweristhatAlan’stypecannotbedetermined,but
Bob’stypecanbe.Tobeginwith,letusnotethatnonativeofthisislandcan
claimtobeaknave,becausenoknightwouldfalselydoso,andnoknavewould
truthfullydoso.Therefore,ifBobwereaknave,thenAlancouldneverhave
claimedtobethesametypeasBob,asthiswouldbetantamounttoclaimingto
beaknave!ThusBobmustbeaknight.AstoAlan,hecouldbeaknight,who
truthfullyclaimedtobelikeBob,oraknave,whofalselyclaimstobelikeBob,
andthereisnowayoftellingwhich.
Next,Iwishtoconsiderarecreationalproblemthatisrelatedtoafunda-
mentaldiscoveryinmathematicallogic,whichIwilllatertellyouabout.
A certain logician, whom we will call Larry, is completely accurate in his
proofs—everythinghecanproveisreallytrue.HevisitstheIslandofKnights
andKnavesandcomesacrossanativenamedJal,whosaystohim,“Youcan
neverprovethatI’maknight.”IsJalaknightoraknave?Yes,thiscanreally
bedetermined!Wanttotry?
Here is the answer, which may be a bit startling: Jal must be a knight,
but Larry can never prove that he is! Here is why: If Jal was a knave, then
contrarytowhathefalselysaid,Larrycould provethatJalisaknight,which
goescontrarytothegivenconditionthatLarryneverprovesfalsepropositions.
Hence,Jalcannotbeaknave;hemustbeaknight.Hence,ashetruthfullysaid,
LarrycanneverprovethatJalisaknight.ThusJalisreallyaknight,butLarry
canneverprovethatheis.
This problem was suggested to me by the discovery of the great logician
Kurt Gödel, that for the most powerful mathematical systems yet known,
there must always be sentences which, though true, cannot be proved in the
system.Foreachofthesesystems,aswellasvarietiesofrelatedsystems,Gödel
showed how to construct a sentence—call it “G”—which asserted its own
nonprovability in the system. That is, G is either true but not provable in
the system, or G is false but provable in the system. The latter alternative is
ruledoutbythereasonableassumptionthatthemathematicalsystemsunder
considerationarecorrect,inthesensethatonlytruesentencesareprovablein
them.Thusthefirstalternativeholds:G istruebutnotprovableinthesystem.
HowdidGödelconstructasentencethatasserteditsownnonprovability?
Thefollowingproblemshouldillustratetheessentialidea.
Consider a computing machine that prints out expressions composed of
thefollowingfivesymbols:(,),P,N,andD.Anexpressioniscalledprintable
Foreword • ix
if the machine can print it. We will call an expression X regular if it is
compounded from just the three symbols, P,N, and D (thus it has no
parentheses). For any regular expression X, by the diagonalization of X is
meant the expression X(X)—for example, the diagonalization of PNPD is
PNPD(PNPD).Byasentenceismeantanyexpressionofoneofthefollowing
forms,where Xstandsforanyregularexpressionwhatever:
1. P(X)
2. NP(X)
3. PD(X)
4. NPD(X)
Thesesentencesareinterpretedasfollows(thesymbols P, N, D,respectively,
standforprintable,not,anddiagonalizationof):
1. P(X)meansXisprintable,andisaccordinglycalledtrueif(andonly
if) Xisprintable.
2. NP(X)iscalledtrueifandonlyif Xisnotprintable.
3. PD(X) is called true if and only if the diagonalization of X is
printable.(Itisread,“PrintablethediagonalizationofX,”or,inbetter
English,“Thediagonalizationof Xisprintable.”)
4. NPD(X) is called True if and only if the diagonalization of X not
printable.
We are given that the machine is totally accurate in the sense that all
printablesentencesaretrue.Forexample,if P(X)isprintable,then X isalso
printable, as P(X) truthfully says. But suppose that X is printable. Does it
follow that P(X) is printable? Not necessarily! If X is printable, then P(X)
is true, but we are not given that all true sentences are printable; only that
printablesentencesaretrue.Asamatteroffact,thereisatruesentencethatis
notprintable,andtheproblemistoexhibitone.Caretotry?
Well,hereisone:weknowthatforanyregularexpression X,thesentence
NPD(X) is true if and only if the diagonalizationof X is notprintable. We
taketheexpressionNPDforX,andsoNPD(NPD)istrueifandonlyifthe
diagonalizationof NPD isnotprintable,butthediagonalizationof NPD is
theverysentence NPD(NPD)!Thus NPD(NPD)istrueifandonlyifitis
notprintable,whichmeansitiseithertrueandnotprintable,orelsenottrue
but printable. The latter alternative is ruled out by the given condition that
onlytruesentencesareprintable,henceitmustbetruethat NPD(NPD)is
true,butthemachinecannotprintit.
Thissentence NPD(NPD)obviouslycorrespondstoGödel’ssentenceG,
whichassertsitsownnonprovability.
Thisisjustoneexamplewhererecreationalmathematicsleadsyounaturally
into the deepest topics in mathematics. In this volume, you will enjoy many
othermathematicaltopicsintroducedbyrecreationalpuzzles!