l3_induction1

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6.042/18.062JMathematicsforComputerScience February8,2005SriniDevadasandEricLehman LectureNotes

InductionI1 InductionAprofessorbringstoclassabottomlessbagofassortedminiaturecandybars.Sheofferstoshareinaccordancewithtworules. First,shenumbersthestudents0,1,2,3,andsoforthforconvenientreference.Nowherearethetworules:

1. Student0getscandy.2. Foralln N,ifstudentn getscandy,thenstudentn + 1 alsogetscandy.

Youcanthinkofthesecondruleasacompactwayofwritingawholesequenceofstate-ments,oneforeachnaturalvalueofn:

Ifstudent0getscandy,thenstudent1alsogetscandy. Ifstudent1getscandy,thenstudent2alsogetscandy. Ifstudent2getscandy,thenstudent3alsogetscandy,andsoforth.Nowsupposeyouarestudent17.Bytheserules,areyouentitledtoaminiaturecandy

bar? Well,student0getscandybythefirstrule. Therefore,bythesecondrule,student1alsogetscandy,whichmeansstudent2getscandyaswell,whichmeansstudent3getcandy,andsoon.Sotheprofessorstworulesactuallyguaranteecandyforeverystudent,nomatterhowlargetheclass.Youwin!

Thisreasoninggeneralizestoaprinciplecalledinduction:

PrincipleofInduction.LetP(n) beapredicate.If P(0) istrue,and foralln N,P(n) impliesP(n + 1),thenP(n) istrueforalln N.


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2 InductionIHeresthecorrespondencebetweentheinductionprincipleandsharingcandybars.

SupposethatP(n) isthepredicate,studentn getscandy.ThentheprofessorsfirstruleassertsthatP(0) istrue,andhersecondruleisthatforalln N,P(n) impliesP(n + 1).Giventhesefacts, theinductionprinciplesaysthatP(n) istrueforalln N. Inotherwords,

everyone

gets

candy.

Theintuitivejustificationforthegeneralinductionprincipleisthesameasforevery-

onegettingacandybarundertheprofessorstworules. Mathematiciansfindthisintu-itionsocompellingthatinductionisalwayseithertakenasanaxiomorelseprovedfrommoreprimitiveaxioms,whicharethemselvesspecificallydesignedsothatinductionisprovable.Inanycase,theinductionprincipleisacoreprincipleofmathematics.

2 UsingInductionInductionisbyfarthemostimportantprooftechniqueincomputerscience. Generally,inductionisusedtoprovethatsomestatementholdsforallnaturalvaluesofavariable.Forexample,hereisaclassicformula:Theorem1. Foralln N:

n(n + 1)1 + 2 + 3 + . . . + n =

2

Theleftsideoftheequationrepresentsthesumofallthenumbersfrom1ton.Youresupposedtoguessthepatternandmentallyreplacethe. . . withtheotherterms.Wecouldeliminatetheneedforguessingbyrewritingtheleftsidewithsummationnotation:

n

i or i or ii=1 1in i{1,...,n}

Eachof theseexpressionsdenotes thesumofallvalues takenonby theexpression totherightofthesigmaasthevariablei rangesfrom1ton. ThemeaningofthesuminTheorem1isnotsoobviousinacouplespecialcases:

Ifn = 1,thenthereisonlyoneterminthesummation,andso1+2+3+ . . .+n = 1.Dontbemisledbytheappearanceof2and3andthesuggestionthat1 andn aredistinctterms!

Ifn 0,thentherearenotermsatallinthesummation,andso1+2+3+. . .+n = 0.The. . . notationisconvenient,butwatchoutforthesespecialcaseswherethenotationismisleading!

Now letsuse the inductionprinciple toproveTheorem1. Suppose thatwedefinepredicateP(n) tobe1 + 2 + 3 + . . . + n = n(n + 1)/2.Recastintermsofthispredicate,thetheoremclaimsthatP(n) istrueforalln N. Thisisgreat,becausetheinductionprincipleletsusreachpreciselythatconclusion,providedweestablishtwosimplerfacts:


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InductionI 3 P(0) istrue. Foralln N,P(n) impliesP(n + 1).Sonowourjobisreducedtoprovingthesetwostatements. Thefirstistruebecause

P(0) assertsthatasumofzerotermsisequalto0(0 + 1)/2 = 0.Thesecondstatement ismorecomplicated. But remember thebasicplan forprov-

ingthevalidityofanyimplication: assumethestatementontheleftandthenprovethestatementontheright.Inthiscase,weassumeP(n):

n(n + 1)1 + 2 + 3 + . . . + n =

2

inordertoproveP(n + 1):(n + 1)(n + 2)

1 + 2 + 3 + . . . + n + (n + 1) =2

These twoequationsarequitesimilar; infact,adding(n + 1) tobothsidesof thefirstequationandsimplifyingtherightsidegivesthesecondequation:

n(n + 1)1 + 2 + 3 + . . . + n + (n + 1) = + (n + 1)

2(n + 2)(n + 1)

=2

Thus,ifP(n) istrue,thensoisP(n + 1).Thisargumentisvalidforeverynaturalnumbern,sothisestablishesthesecondfactrequiredbytheinductionprinciple. Ineffect,wevejustprovedthatP(0) impliesP(1),P(1) impliesP(2),P(2) impliesP(3),etc. allinonefellswoop.

Withthesetwofactsinhand, theinductionprinciplesaysthatthepredicateP(n) istrueforallnaturaln.Andsothetheoremisproved!

2.1 ATemplateforInductionProofsTheproofofTheorem1wasrelativelysimple,buteventhemostcomplicatedinductionprooffollowsexactlythesametemplate.Therearefivecomponents:

1. Statethattheproofusesinduction.Thisimmediatelyconveystheoverallstructureoftheproof,whichhelpsthereaderunderstandyourargument.

2. DefineanappropriatepredicateP(n). Theeventualconclusionof the inductionargumentwillbe thatP(n) is trueforallnaturaln. Thus,youshoulddefine thepredicateP(n) sothatyourtheoremisequivalentto(orfollowsfrom)thisconclu-sion. Often thepredicatecanbe liftedstraightfrom theclaim,as in theexampleabove.ThepredicateP(n) iscalledtheinductionhypothesis.


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4 InductionI3. ProvethatP(0) istrue. Thisisusuallyeasy,asintheexampleabove. Thispartof

theproofiscalledthebasecaseorbasisstep.4. ProvethatP(n) impliesP(n + 1) foreverynaturalnumbern. Thisiscalledthe

inductivesteporinductionstep. Thebasicplanisalwaysthesame: assumethatP(n) istrueandthenusethisassumptiontoprovethatP(n + 1) istrue.Thesetwo statements shouldbe fairly similar,butbridging thegapmay require someingenuity.Whateverargumentyougivemustbevalidforeverynaturalnumbern,sincethegoalistoprovetheimplicationsP(0) P(1),P(1) P(2),P(2) P(3),etc.allatonce.

5. Invokeinduction.Giventhesefacts,theinductionprincipleallowsyoutoconcludethatP(n) istrueforallnaturaln.Thisisthelogicalcapstonetothewholeargument,butmanywritersleavethisstepimplicit.

Explicitlylabelingthebasecaseandinductivestepmaymakeyourproofsmoreclear.2.2 ACleanWriteupTheproofofTheorem1givenaboveisperfectlyvalid;however,itcontainsalotofextra-neousexplanationthatyouwontusuallyseeininductionproofs. Thewriteupbelowisclosertowhatyoumightseeinprintandshouldbepreparedtoproduceyourself.Proof. Weuseinduction.LetP(n) bethepredicate:

n(n + 1)1 + 2 + 3 + . . . + n =

2

Basecase:P(0) istrue,becausebothsidesoftheequationarezero.Inductivestep:AssumethatP(n) istrue,wheren isanynaturalnumber.ThenP(n + 1) isalsotrue,because:

n(n + 1)1 + 2 + 3 + . . . + n + (n + 1) = + (n + 1)

2(n + 1)(n + 2)

= 2

ThefirststepusestheassumptionP(n),andthesecondfollowsbysimplification.There-fore,P(n) istrueforallnaturaln byinduction.

Inductionwashelpfulforprovingthecorrectnessofthissummationformula,butnothelpful fordiscovering the formula in thefirstplace. Therearesome tricks forfindingsuchformulas,whichwellshowyouinafewweeks.


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6 InductionIInductivestep:AssumethatP(n) istrue,wheren isanynaturalnumber.Then:

23 (n3 n) 3 (n3 n) + 3(n + n)| |3 n3 + 3n2 + 3n + 1 n 1|3 (n + 1)3 (n + 1)|

Thefirst implication relieson the fact that3(n2 + n) isdivisibleby3. The remainingimplications involveonly rewriting theexpressionon the right. The last statement isP(n + 1),soweveprovedthatP(n) impliesP(n + 1) foralln N.

Bytheprincipleofinduction,P(n) istrueforalln N,whichprovestheclaim.Thisproofwouldlookquitemysterioustoanyonenotprivytothescratchworkwedid

beforehand.Inparticular,onemightaskhowwehadtheforesighttointroducethemagicterm3(n2 +n).Ofcourse,thiswasnotforesightatall;wejustworkedbackwardinitially!

4 AFaultyInductionProofSometimeswewanttoprovethatastatementistruefor,say,allintegersn 1 ratherthanallintegersn 0. Inthiscircumstance,wecanuseaslightvariationoninduction:proveP(1) inthebasecaseandthenprovethatP(n) impliesP(n + 1) foralln 1 intheinductivestep. Thisisaperfectlyvalidvariantofinductionandisnottheproblemwiththeproofbelow.False

Theorem

3.

All

horses

are

the

same

color.

Proof. Theproofisbyinduction.LetP(n) bethepropositionthatineverysetofn horses,allarethesamecolor.Basecase:P(1) istrue,becauseallhorsesinasetof1mustbethesamecolor.Inductivestep:AssumethatP(n) istrue,wheren isapositiveinteger;thatis,assumethatineverysetofn horses,allarethesamecolor.Nowconsiderasetofn + 1 horses:

h1, h2, . . . , hn, hn+1

Byourassumption,thefirstn horsesarethesamecolor:h1, h2, . . . , hn, hn+1

samecolorAlsobyourassumption,thelastn horsesarethesamecolor:

h1, h2, . . . , hn, hn+1

samecolor


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7InductionITherefore,horsesh1, h2, . . . , hn+1 mustallbethesamecolor,andsoP(n+1) istrue.Thus,P(n) impliesP(n + 1).

Bytheprincipleofinduction,P(n) istrueforalln 1. Thetheoremisaspecialcasewheren isequaltothenumberofhorsesintheworld.

Weveprovedsomethingfalse!Ismathbroken?Shouldweallbecomepoets?Theerrorinthisargumentisinthesentencethatbegins,Therefore,horsesh1,h2,. . .,

hn,hn+1 mustallbethesamecolor. The. . .notationcreatestheimpressionthatthesetsh1, h2, . . . , hn andh2, . . . , hn, hn+1 overlap. However,thisisnottruewhenn = 1. Inthatcase,thefirstsetisjusth1 andthesecondish2,andthesedonotoverlapatall!

Thismistakeknocksacriticallinkoutofourinductionargument.WeprovedP(1) andweprovedP(2) P(3),P(3) P(4),etc. ButwefailedtoproveP(1) P(2),andsoeverythingfallsapart:wecannotconcludethatP(3),P(4),etc.aretrue.And,ofcourse,thesepropositionsareallfalse;therearehorsesofadifferentcolor.

5 CourtyardTilingInductionservedpurelyasaprooftechniqueintheprecedingexamples. Butinductionsometimescanserveasamoregeneralreasoningtool.

MITrecentlyconstructedanewcomputersciencebuilding.Astheprojectwentfurtherand furtheroverbudget, thereweresomeradical fundraising ideas. Oneplanwas toinstallabigcourtyardwithdimensions2n 2n:

2n

2n

Oneofthecentralsquareswouldbeoccupiedbyastatueofawealthypotentialdonor.Lets

call

him

Bill.

(In

the

special

case

n = 0,

the

whole

courtyard

consists

of

asingle

centralsquare;otherwise, therearefourcentralsquares.) Acomplicationwasthat thebuildingsunconventionalarchitect,FrankGehry,insistedthatonlyspecialL-shapedtilesbeused:


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8 InductionIAcourtyardmeetingtheseconstraintsexsists,atleastforn = 2:

B

Forlargervaluesofn,isthereawaytotilea2n 2n courtyardwithL-shapedtilesandastatueinthecenter?Letstrytoprovethatthisisso.Theorem4.Foralln 0 thereexistsatilingofa2n 2n courtyardwithBillinacentralsquare.Proof.

(doomed

attempt)

The

proof

is

by

induction.

Let

P(n) be

the

proposition

that

there

existsatilingofa2n 2n courtyardwithBillinthecenter.Basecase:P(0) istruebecauseBillfillsthewholecourtyard.Inductivestep:Assumethatthereisatilingofa2n 2n courtyardwithBillinthecenterforsomen 0.Wemustprovethatthereisawaytotilea2n+1 2n+1 courtyardwithBillinthecenter...

Nowwereintrouble!TheabilitytotileasmallercourtyardwithBillinthecenterdoesnothelptilealargercourtyardwithBillinthecenter.WecannotbridgethegapbetweenP(n) andP(n + 1).Theusualrecipeforfindinganinductiveproofwillnotwork!

Whenthishappens,yourfirstfallbackshouldbetolookforastrongerinductionhy-pothesis; that is,onewhich impliesyourprevioushypothesis. Forexample,wecouldmakeP(n) thepropositionthatforeverylocationofBillina2n 2n courtyard,thereexistsatilingoftheremainder.

Thisadvicemaysoundbizzare:Ifyoucantprovesomething,trytoprovesomethingmoregrand!Butforinductionarguments,thismakessense.Intheinductivestep,whereyouhavetoproveP(n) P(n + 1),youreinbettershapebecauseyoucanassumeP(n),whichisnowamoregeneral,moreusefulstatement. Letsseehowthisplaysoutinthecaseofcourtyardtiling.Proof.

(successful

attempt)

The

proof

is

by

induction.

Let

P(n) be

the

proposition

that

for

everylocationofBillina2n 2n courtyard,thereexistsatilingoftheremainder.Basecase:P(0) istruebecauseBillfillsthewholecourtyard.Inductivestep:AsumethatP(n) istrueforsomen 0;thatis,foreverylocationofBillina2n 2n courtyard,thereexistsatilingoftheremainder.Dividethe2n+12n+1 courtyardintofourquadrants,each2n 2n. OnequadrantcontainsBill(Binthediagrambelow).PlaceatemporaryBill(Xinthediagram)ineachofthethreecentralsquareslyingoutsidethisquadrant:


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9InductionI

XX X

B

2n

2n

2n 2n

Nowwecantileeachofthefourquadrantsbytheinductionassumption.ReplacingthethreetemporaryBillswithasingleL-shapedtilecompletesthejob.ThisprovesthatP(n)impliesP(n + 1) foralln 0.Thetheoremfollowsasaspecialcase.

Thisproofhastwoniceproperties. First,notonlydoestheargumentguaranteethatatilingexists,butalsoitgivesanalgorithmforfindingsuchatiling. Second,wehaveastronger result: ifBillwantedastatueon theedgeof thecourtyard, away from thepigeons,wecouldaccommodatehim!

Strengtheningtheinductionhypothesisisoftenagoodmovewhenaninductionproofwontgo through. Butkeep inmind that thestrongerassertionmustactuallybe true;otherwise,thereisntmuchhopeofconstructingavalidproof!Sometimesfindingjusttherightinductionhypothesisrequirestrial,error,andinsight.Forexample,mathematiciansspentalmosttwentyyearstryingtoproveordisprovetheconjecturethatEveryplanargraphis5-choosable1.Then,in1994,CarstenThomassengaveaninductionproofsimpleenough toexplainonanapkin. Thekey turnedout tobefindinganextremelycleverinductionhypothesis;withthatinhand,completingtheargumentiseasy!

15-choosability isaslightgeneralizationof5-colorability. Althougheveryplanargraph is4-colorableand therefore5-colorable,noteveryplanargraph is4-choosable. If thisallsounds likenonsense,dontpanic.Welldiscussgraphs,planarity,andcoloringintwoweeks.

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