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THEHIDDENSUBGROUPPROBLEM By ANALESDEBHAUMIK ADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOL OFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENT OFTHEREQUIREMENTSFORTHEDEGREEOF DOCTOROFPHILOSOPHY UNIVERSITYOFFLORIDA 2010
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c r 2010AnalesDebhaumik 2
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Idedicatethistoeveryonewhohelpedmeinmyresearch. 3
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ACKNOWLEDGMENTS FirstandforemostIwouldliketoexpressmysincerestgrati tudetomyadvisorDr AlexandreTurull.Withouthisguidanceandpersistenthelp thisthesiswouldnothave beenpossible. Iwouldalsoliketothanksincerelymycommitteemembersfor theirhelpand constantencouragement. FinallyIwouldliketothankmywifeManishaBrahmacharyfor herloveandconstant support. 4
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TABLEOFCONTENTS page ACKNOWLEDGMENTS .................................. 4 ABSTRACT ......................................... 6 CHAPTER 1INTRODUCTION ................................... 8 1.1Preliminaries .................................. 8 1.2QuantumComputing .............................. 11 1.3QuantumFourierTransform .......................... 12 1.4HiddenSubgroupProblem .......................... 14 1.5Distinguishability ................................ 18 2HIDDENSUBGROUPSFORALMOSTABELIANGROUPS ........... 20 2.1Algorithmtondthehiddensubgroups .................... 20 2.2AnAlmostAbelianGroupofOrder 3 2 n ................... 22 3KEMPE-SHALEVDISTINGUISHABLITY ..................... 33 4ALGORITHMICDISTINGUISHABILITY ...................... 43 4.1Analgorithmfordistinguishability ....................... 43 4.2AbelianGroups ................................. 45 4.3FrobeniusGroups ............................... 46 5SOMECALCULATIONS ............................... 55 5.1Kempe-Shalevdistinguishability ........................ 55 5.2AlgorithmicDistinguishability ......................... 56 APPENDIX ATABLESFORCHAPTER4 ............................. 59 BTABLEFORCHAPTER5 .............................. 70 REFERENCES ....................................... 71 BIOGRAPHICALSKETCH ................................ 73 5
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AbstractofDissertationPresentedtotheGraduateSchool oftheUniversityofFloridainPartialFulllmentofthe RequirementsfortheDegreeofDoctorofPhilosophy THEHIDDENSUBGROUPPROBLEM By AnalesDebhaumik May2010 Chair:AlexandreTurullMajor:Mathematics Thetopicofmyresearchisthe HiddenSubgroupProblem .Theproblemcanbe statedasfollows: Problem. (HiddenSubgroupProblem)Let G beanitegroupand H asubgroup.Given ablack-boxfunction f : G S whichisconstanton(left)-cosets gH of H andtakes differentvaluesfordifferentcosets,determineasetofge neratorsfor H EfcientclassicalalgorithmsfortheHiddenSubgroupProb lemarenotknown. However,efcientquantumalgorithmsareknownforthispro bleminsomecases. OnesuchalgorithmimpliesShor'sfamousefcientmethodfo rbreakingtheRSA cryptosystem. Anefcientsolutionofthisprobleminallcaseswouldhavew ideimplicationsin theeldoftheoreticalcomputerscience.Forexampleitwou ldmostlikelysolvesome classicalproblemswhichareneither NP completenorarein P Asolutionwouldalso implyasolutionforGraphIsomorphismproblemwhichisalon gstandingproblemin computerscience. Inthepresentthesis,westudysomequantumalgorithmsfort he HiddenSubgroupProblem .WediscussQuantumFourierTranformanditsapplicationst othe HiddenSubgroupProblem.WediscussAlmostAbeliangroupsa ndshowthatthere isaquantumalgorithmthatsolvestheHiddenSubgroupProbl emforthem.Wealso studythedecisionversionoftheproblem.Wecomparetwodif ferentformalizationsof 6
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thisconcept.Weshowthattheseformalizationscoincidein thecaseofabelianand Frobeniusgroups.Weconcludewithafamilyofgroupswheret heformalizationsmay notcoincide. 7
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CHAPTER1 INTRODUCTION 1.1Preliminaries Aquantumcomputerisan,atpresenttheoretical,machinewh ichperforms computationsonthebasisofthelawsofquantummechanics.S uchadevicecan preparequantumstatesandperformmeasurementsonthem.As withclassical computers,numericaldataformtheinputandtheoutputfort hequantumcomputer.The processingofthisdatainaquantumcomputermayinvolveact ionsonthemwhichare notclassical.Asaresult,theoutputofaquantumcomputeri snotfullydeterminedbyits input.Givenaparticularinput,variousoutputsmayoccurw ithdifferentprobabilities.In thissense,quantumcomputersbehavedifferentlythanclas sicalones. Thoughquantumcomputingasaeldofresearchisstillinits infancysignicant amountofworkhasalreadybeendoneinthiseld.In1985Deut schgaveaquantum algorithmforaverysimpleproblemandshowedthatitworked betterthanaclassical one.Inhisproblemwearegivenablackboxwhichcomputesasi mplefunction.The boxtakestwobitsasinputandoutputstwobits.Itimplement soneofthefourfollowing functionsfrom F 22 F 22 : f 1 ( x y )=( x y ) f 2 ( x y )=( x y +1) f 3 ( x y )=( x x + y ) f 4 ( x y )=( x x + y +1) Thegoalistodeterminewhetherthefunctionimplementedby theboxisintheset f f 1 f 2 g or f f 3 f 4 g Onaclassicalcomputeritwilltake2queriestondtheanswe rto thequestion.Butonaquantumcomputeritwilltakejust1que rytondtheanswer.In 1992DeutschandJozsa( 7 )gavetherstnon-trivialquantumalgorithm.Theirproble m isthegeneralizationoftheaboveproblem.Inthisproblemw earegivenafunction f : F n2 F 2 Thegoalistondouthowmanyqueriestotheboxareneededint he 8
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worstcasetodeterminewhetherthefunctionisaconstantor balanced(0ononehalf ofthedomainstatesand1ontheotherhalf).Fortheclassica lcasetheproblemhas exponentialquerycomplexity.Butforthequantumcaseaswa sshownbyDeutschand Jozsaonlyasinglequeryisneeded.MotivatedbythisBerste inandVazirani( 3 )gavea quantumalgorithmwhichworkedsignicantlyfaster(super -polynomialspeedup)than thebestclassicalalgorithm.TheBersteinVaziraniproble mhasanon-recursiveanda recursiveversion.Inthenon-recursiveversionwearegive nafunction f : F n2 F 2 Thefunctionis f s ( x )= x s forsomeunknown s where x s denotesthedotproduct. Thegoalistond s Intheclassicalcasethequerycomplexityisatleast n whereasfor thequantumcaseonlyasinglequeryisneeded.Therecursive versionisalittlemore complexthanthenon-recursiveone.Simon( 24 )thengaveaquantumalgorithmwhich performedexponentiallyfasterthanitsclassicalcounter part.InSimon'sproblemweare givenaninteger m 1 andafunction f : F m2 R where R isaniteset.Wealso knowthatthereexistsanonzeroelement s 2 F m2 suchthatforall x y 2 F m2 f ( x )= f ( y ) ifandonlyif x = y or x = y + s Thegoalistond s AllthesepavedthewayforShor's celebratedpaperonfactoringanddiscretelog( 23 ).Thereisnoknownefcientclassical algorithmforfactoringalargenumber n ThesecurityandrobustnessofRSApublic-key cryptosystemisbasedonthisfact.Shorshowedinhispapert hatgivenaquantum computerfactoringofanumber n canbedoneinpolynomialtime. Simon'sandShor'salgorithmslaidtheframeworkforaveryi mportantproblemof quantumcomputationknownastheHiddenSubgroupProblem.T heproblemcanbe statedasfollows. Problem. (HiddenSubgroupProblem)Let G beanitegroupand H asubgroup.Given ablack-boxfunction f : G S whichisconstanton(left)-cosets gH of H andtakes differentvaluesfordifferentcosets,determineasetofge neratorsfor H TheHiddenSubgroupProblemisattheheartofShor'sandSimo n'sproblems. InShor'sproblemgiven n let a beanyintegersuchgcd( a n )=1.Let r betheorderof 9
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a 2 Z n Thegoalistond r .Inthisproblem G is Z and H is r Z Forpracticalpurposes wecanworkwith Z m where m =2 l > n Thefunction f isgivenby f ( x )= a x + n Z Simon'sproblemreducestoaspecialcaseoftheHiddenSubgr oupProblemifwetake G tobe F m2 and H = f 0, s g EfcientquantumalgorithmsforTheHiddenSubgroupProble mareknownfor abeliangroupsthankstotheeffortsofShor,Simon,Deutsch etc( 7 ).Thenon-abelian casestillposesachallenge.Somepositiveresultshavebee nobtainedforDihedral groupbyM.EttingerandP.Hoyer( 8 ),G.Kuperberg( 19 ),Regev( 21 ),DaveBacon, AndrewM.ChildsandWimvanDam( 1 ).Efcientquantumalgorithmshavebeen obtainedforsemidirectproductoftheforms Z p r oZ p byYoshifumiInuiandFrancois LeGall( 15 ),for Z N oZ p where N = p r 1 1 p r 2 2 ... p r n n andprime p doesnotdivide p j 1 by DongPyoChi,JeongSanKimandSoojoonLee( 5 ),forsomemetacyclicgroupsand allgroupsoftheform Z rp oZ p foranyprime p andaxed r byDaveBacon,AndrewM. ChildsandWimvanDam( 1 ),forWreathproductoftheform Z n2 o Z 2 byM.RottelerandT. Beth( 4 ),forsomesolvablegroupsbyK.Friedl,G.Ivanyos,F.Magni ez,M.Santha,and P.Sen( 10 )andforgroupswithsmallcommutatorsbyG.Ivanyos,F.Magn iez,andM. Santha( 17 ).Unfortunatelyauniedsolutionhasstillnotbeenfound. Anefcientsolutionofthisproblemwouldhavewideimplica tionsintheeldof theoreticalcomputerscience.Forexampleitwouldmostlik elysolvesomeclassical problemswhichareneither NP completenorarein P Asolutionwouldalsoimplya solutionforGraphIsomorphismproblemwhichisalongstand ingproblemincomputer science. InthispaperwestudydistinguishabilityoftheHiddenSubg roupProblemfor Frobeniusgroupsoftheforms Z g ( ) ( ) oZ Z g ( i ) oZ m ( i ) oZ n ( i ) Z g ( i ) o ( SL (2,5) ( Z m ( i ) o Z n ( i ) ). Forgroupsoftheform Z g ( ) ( ) oZ wegiveanecessaryandsufcientconditionfor thedistinguishabilityofthesubgroupsgiventhat g ( ) > 1 c forsome c > 0 andforall sufcientlylarge Wealsogiveanecessarysufcientconditionforthedisting uishability 10
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ofthesubgroupsofFrobeniusgroupsofthetype Z g ( i ) oZ m ( i ) oZ n ( i ) giventhat g ( i ) c m ( i ) n ( i ) forsome c > 0 andsufcientlylarge i Basedontheconditionthat g ( i ) c m ( i ) n ( i ) forsome c > 0 andsufcientlylarge i wegiveanecessaryandsufcient conditionforthedistinguishabilityofthesubgroupsof Z g ( i ) o ( SL (2,5) ( Z m ( i ) oZ n ( i ) ). In( 11 )M.Grigni,L.J.Schulman,M.Vazirani,U.Vaziranigaveane fcientalgorithmfor ”almostabeliangroups.”Theyalsogiveanoutlineofhowthe algorithmworksfor”almost abeliangroups” Z 3 oZ m where m isapowerof 2 .Insection 7 wedescribethealgorithm forthesegroupsandgiveanestimateofthenumberofstepsre quiredtoobtainthe hiddensubgroupswithprobabilitybiggerthan 1 2 1.2QuantumComputing Thetheoryofquantumcomputingwaslaunchedinthebeginnin gof1980.The famousphysicistFeynmaninanarticle( 9 )proposedthatitisnotpossibletosimulate aquantumsystemonaclassicalcomputerwithoutexponentia lslowdown.Healso suggestedthataquantumcomputercouldbeawaytoavoidsuch slowdown.In1985 Deutsch( 6 )rstgaveadescriptionofauniversalquantumcomputerwhi chwasrened laterbysomeothercomputerscientists.Thoughaquantumco mputerisstilloutofreach alotoftheoreticalprogresshasbeenmadeeversince. Tounderstandaquantumcomputerwersthavetoknowafewthi ngsabout quantummechanicsandquantumsystem.Inquantummechanics astateinan n -levelsystemisaunitvectorinan n dimensionalcomplexvectorspace H n withan orthonormalbasischosenas j x 1 i j x 2 i ,..., j x n i .Thechoiceofsuchabasisisarbitrary. Anystateinaquantumsystemcanthusbewrittenas p 1 j x 1 i + p 2 j x 2 i +...+ p n j x n i where the p i 2 C arecalledtheamplitudes( 13 )withtherequirementthat P ni =1 j p i j 2 =1 The amplitudessignifythattheprobabilitythataproperty x i isobservedofthesystemis j p i j 2 Thelinearcombinationsofthebasisstatesarecalled superpositionofstates Thebasicunitofinformationforaclassicalcomputerisabi t.Similarlyaquantum computeroperatesonquantumbitsknownas qubits Itisbasicallya 2 -levelquantum 11
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systemandcanbeidentiedwitha 2 -dimensionalHilbertspacewithbasis j 0 i j 1 i .A generalstateofaquantumsystemisthusequalto c 1 j 0 i + c 2 j 1 i wherethe c i arecomplex numbersand j c 1 j 2 + j c 2 j 2 =1. Ifwehavetwosystemswithbasisstates j x 1 i j x 2 i ,..., j x n i and j y 1 i j y 2 i ,..., j y m i then thecompoundsystemwillhavebasisstatesas( j x i i j y i i ).Thusasystemoftwoqubits isa 4 -dimensionalHilbertspacewithanorthonormalbasis fj 0 ij 0 i j 0 ij 1 i j 1 ij 0 i j 1 ij 1 ig Aquantumregisteroflength m isanorderedsystemof m qubitsassociatedwiththe 2 m dimensionalspace H 2 N H 2 .... N H 2 Anoperationon m qubitscanberepresentedbyaunitarymap U : H H where H isa 2 m -dimensionalHilbertspacerepresentingthequantumsyste m.Itcanalsobe representedbyaunitarymatrix.Asaverysimpleexamplewec anconsiderthenot operationonasinglequbit.Theoperationtakes j 0 i to j 1 i and j 1 i to j 0 i Thematrixthat denesthisoperationis 0B@ 0110 1CA Theprocessofgettinginformationoutofaquantumstateisk nownasmeasurement. Thereareseveralwaystothinkaboutmeasurement.Thesimpl estoneismeasurement inthecomputationalbasis.Supposeweareinaquantumsyste mwith n -dimensional basisvectorsand j i = P n 1 0 p j j j i isaquantumstate.Measurementinthecomputational basiswillreturnthestate j j i withprobability j p j j 2 andafterthemeasurementtheoutput statebecomes j 0 i = j j i Thusthegivenstatecollapsestotheonereturnedbythe measurementandallotherinformationaboutthestateisdes troyed. 1.3QuantumFourierTransform Onereasonwhyaquantumcomputercanworkmoreefcientlyth anaclassical one,asdemonstratedbytheperformancesoftheefcientqua ntumalgorithms,isthat quantumfouriertransformcanbeimplementedonit.Itisthe keyingredientforsome veryinterestingquantumalgorithms.Italsoaccountsfort heefciencyoftheperiod 12
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ndingalgorithmsliketheonesdevisedforvariousinstanc esoftheHiddenSubgroup Problem. Let G beanitegroupoforder n andlet R = f 1 2 ,..., g beacompletesetof inequivalentcomplexirreduciblerepresentationsof G withdegrees d i i =1,..., We choosetheserepresentationstobeunitary(( 16 ), Theorem 4.17 ).Let S time beacomplex vectorspaceofdimension j G j withanorthonormalbasis B time = fj g 1 i j g 2 i ,..., j g n ig wherethe g i aretheelementsofthegroup G Let S space beanothercomplexvector spacewithanorthonormalbasis B space = fj 1 ,1,1 i j 1 ,1,2 i ,.. j k i j i ,..., j d d ig formedbyconsideringthe ( i j ) th entryofthematrix ( g ) foreach 2 R and g 2 G Sincethereare d 2 matrixentriesinthematrix ( g ) foreach 2 R thereare P i =1 d 2 i = j G j numberofbasisvectors.Sothedimensionof S space is j G j whichisthesameas S time WearenowreadytodeneQuantumFourierTransform. Denition 1 QFTorquantumfouriertransformoveragroup G isthelinearmap Q G : S time S space suchthat Q G ( j g i )= 1 p j G j X i j p d ( g ) i j j i j i where 2 R hasdimension d ,and ( g ) i j isthe ( i j ) thentryofthematrix ( g ) for g 2 G and 1 i j d WewillnowshowthattheQFTisaunitarytransformation.Let g p g q 2 G .Then,sincethebasis S space isorthonormal,wehave h Q G ( j g p i ), Q G ( j g q i ) i = 1 j G j X i j d ( g p ) i j ( g q ) i j hj i j i j i j ii Since isunitary, ( g ) ij = ( g 1 ) ji andthepreviousexpressionbecomes 1 j G j X i j d ( g p ) i j ( g 1 q ) j i = 1 j G j X d ( g p g 1 q ) 13
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where ischaracteraffordedby Bythesecondorthogonalityrelation(( 16 ), Theorem 2.18 ) 1 j G j X d ( g p g 1 q )= 8>><>>: 1 for p = q 0 otherwise Sowehaveshownthat hj g p i j g q ii = h Q G ( j g p i ), Q G ( j g q i ) i HenceQFTisunitary. QuantumFourierTransformisanextremelypowerfultoolwhi chhasbeenused todeviseefcientquantumalgorithmsfortheHiddenSubgro upProblem.QFTrunsin polynomialtimeforabeliangroupsandthisfacthasbeenexp loitedtondanefcient generalalgorithmfortheabelianHSP.Extensiveresearchi sbeingdonetondefcient QFTinthecaseofnon-abeliangroups.Zalka( 25 )givesanefcientHSPalgorithmfor wreathproductgroupsoftheform G = Z n2 o Z 2 wherehecomputesQFTefcientlyforthe group.PeterHoyer( 14 )givesconstructionofQFTforQuaternionsandsomeMetacyc lic groups.Beth,Puschel,Rotteler,( 4 )showhowtodoQFTefcientlyonaclassofgroups -solvable2groupscontainingacyclicnormalsubgroupofin dex 2. Beals( 2 )shows howtocomputeQFTover S n intime r ( poly ( n )). ChristopherMoore,DanielRockmore, AlexanderRussell( 20 )giveefcientQFT-thatiscircuitsofpoly( log( j G j ) )size-forgroups liketheCliffordgroup,symmetricgroup,metabeliangroup sincludingmetacyclicgroups suchasthedihedralandafnegroups. 1.4HiddenSubgroupProblem InthispaperwefocusontheHiddenSubgroupProblem.Therea retwoversionsof theproblem. HiddenSubgroupProblem :Let G beanitegroupand H asubgroup.Givena black-boxfunction f : G S whichisconstanton(left)-cosets gH of H andtakes differentvaluesfordifferentcosets,determineasetofge neratorsfor H Analgorithmisaproceduretosolveacomputationalproblem .Aclassicalalgorithm isonewhichperformsanitenumberofstepsandoutputsthea nswertotheproblem. Sincethebehaviorofaquantumcomputerisnotfullydetermi nedbyitsinput,the 14
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outcomeofaquantumalgorithmisnecessarilyuncertain,bu twemaycalculatethe probabilitythatitproducesthecorrectanswertotheprobl em.Wesaythataquantum algorithmsolvestheproblemifforeveryinputitreturnsth ecorrectanswertothe problemafteranitenumberofstepswithprobability p forsome p > 1 2 Eventhoughan algorithmperformsanitenumberofstepsitmayrunforalon gtimedependingonthe inputs. Itisbelievedthatagoodwaytoestimatetherunningtimeofa nyalgorithmtosolve theHiddenSubgroupProblemsistocountthenumberofquerie smadetotheblackbox function.Thecomplexityoftheproblemsisestimatedby j G j Wesaythatanalgorithm foraclassofproblemsis efcient ifthereissomepolynomial p ( x ) 2 R [ x ] suchthatthe numberofqueriestotheblackboxfunctionrequiredbytheal gorithmisalwaysatmost p (log j G j ). AnobviousalgorithmtosolvetheHiddenSubgroupProblemwo uldbetoevaluate f ( g ) forall g 2 G andtonoticethat H = f g 2 G : f ( g )= f (1) g .Thisrequires j G j queriestothefunction f andsothisalgorithmisnotefcient.Moreefcientquantum algorithmshavebeenobtainedinvolvingQuantumFourierSa mpling. QuantumFouriersamplingisamethodwhichcanbeimplemente dbyaquantum computertoproducewithasinglequerytotheblackboxfunct ionanirreducible representation of G whichhas core G ( H ) initskernelwhere core G ( H ) isthelargest subgroupof H normalin G If affordsthecharacter thentheprobabilityof QuantumFouriersamplingyielding is P H ( )= d j H j j G j h ( ) H ,1 H i where h ( ) H ,1 H i = 1 j H j P h 2 H ( h ). QuantumFouriersamplingcanbeimplementedonaquantumcom puterasfollows. Thesetupisalmostthesameasintheprevioussection.Wehav e S time withbasis B time indexedbytheelements g i 1 i n ofthegroup G FourierTransform Q G sends S time to S space whichisanothercomplexvectorspacewithbasis B space Letusdene acomplexvectorspace S M withorthonormalbasis fj g 1 s 1 i j g 1 s 2 i ,..., j g n s n ig where 15
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g i 2 G and s i 2f 0 g S S Wealsodeneaunitarymap Q f : S time S M givenby Q f ( j g i )= j g f ( g ) i ThemethodofFourierSamplingisnowdetailedbelow: Step 1: Firstprepareastatein S M asbelow. 1 p j G j Xg 2 G j g ,0 i Step 2: Compute Q f denedon S time andgetthestate 1 p j G j Xg 2 G j g f ( g ) i whichisin S M Step 3: Measurethesecondregister.Ifthemeasuredvalueis f m wegetthestate j cH i = 1 p j H j X h 2 H j ch f m i2 S M Nowforeach f m 2 S S time isisomorphictothesubspaceof S M withbasis fj g 1 f m i j g 2 f m i ,..., j g n f m ig Hence j cH i = 1 p j H j P h 2 H j ch f m i canbeidentiedwiththestate j cH i = 1 p j H j P h 2 H j ch i Step 4: QFTisperformedoverthestate j cH i whichyields 1 p j G jj H j X i j p d X h 2 H ij ( ch ) j i j i Step 5: Measure Theprobabilitytoobserveanyparticular underQuantumFouriersamplingis P H ( )= X i j d j G jj H j j X h 2 H ij ( ch ) j 2 Foran n n matrix M welet k M k denotethematrixnormgivenby k M k 2 = trace ( M M )= X i j j M ij j 2 where,foranymatrix M M denotesthetransposeofitscomplexconjugate.Then P H ( )= d j G jj H j k X h 2 H ( ch ) k 2 16
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Now k X h 2 H ( ch ) k 2 = trace (( X h 2 H ( ch )) X h 2 H ( ch ))= trace (( X h 2 H ( h )) ( c ) ( c ) X h 2 H ( h ))= k X h 2 H ( h ) k 2 since ( c ) isunitary.Hence P H ( )= d j H j j G j k 1 j H j X h 2 H ( h ) k 2 Now H isarepresentationof H notnecessarilyirreducible.Itcan,however,be decomposedintoirreduciblerepresentationsover H Withproperchoiceofbasiswecan make ( h ) blockdiagonalwitheachblockcorrespondingtoanirreduci blerepresentation t If m = thenumberofconjugacyclassesof H then 1 t m Henceeachdiagonal entryof P h 2 H ( h ) is P h 2 H t ( h ) forsome 1 t m By(( 22 ), Corollary 3 Chapter 2 ) X h 2 H t ( h )= 8>><>>: j H j for t =1 H 0 otherwise Henceweseethat k 1 j H j P h 2 H ( h ) k 2 = h ( ) H ,1 H i where istheirreduciblecharacter affordedby Thus P H ( )= d j H j j G j h ( ) H ,1 H i AclassofquantumalgorithmstosolvetheHiddenSubgroupPr oblemusing quantumFourierSamplinghavebeendeveloped.Theyarecall edtheWeakStandard Method. WeakStandardMethod :GiventheinputoftheHiddenSubgroupProblem,select somenumber n = Q (log j G j ). ApplyquantumFouriersampling n timesandobtain representations 1 2 ... n Outputthesubgroup T ni =1 ker ( i ). Ithasbeenshownin( 12 )that n =4log j G j isenoughtoretrieve core G ( H ) with probabilitybiggerthan 1 2 Ifthehiddensubgroup H G then core G ( H )= H andtheWeak StandardMethodoutputsthehiddensubgroup H withhighprobability. 17
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Thereisanothermethodwhichisbelievedtobemorepowerful thanWeakStandard Method.Itisknownasthestrongstandardmethod.Inthe StrongStandardMethod both anditsentries i j aresampled.Itisbelievedthatincertaincasessuchsampli nggives moreinformationabout H thantheWeakStandardMethod. 1.5Distinguishability ThereisadecisionversionoftheHiddenSubgroupproblem.I tisstatedasfollows. DecisionVersionoftheHiddenSubgroupProblem :Let G beanitegroupand H asubgroup.Givenablack-boxfunction f : G S whichisconstanton(left)-cosets gH of H andtakesdifferentvaluesfordifferentcosets,determine whether H = f e g or not. Certainsubgroups H of G mayproduceprobabilitiesofrepresentationsunder QuantumFourierSamplingthatareveryclosetothosethatar isefromthetrivial subgroup.Inthiscasewewouldsaythat H isindistinguishablefromthetrivialsubgroup. In2005KempeandShalev( 18 ),proposedthefollowingdenitionofindistinguishable subgroup. Denition 2 Let G beanitegroup.Let Irr( G ) bethesetofirreducible(complex) representationsof G .Wedenoteby thecharacterassociatedtoa 2 Irr( G ) andlet d beit'sdegree.For H < G wedene D H = 1 j G j X 2 Irr( G ) d j X h 6 = e h 2 H ( h ) j Wesaythatasubgroup H isKempe-Shalevdistinguishableif D H log ( j G j ) c forsome c > 0, c beingindependentof G Usingthisdenition,KempeandShalevclassiedthedistin guishablesubgroupsof S n Inchapter 3 weprovethatallthenon-trivialsubgroupsofanabeliangro upare Kempe-Shalevdistinguishable.WealsoclassifytheKempeShalevdistinguishable subgroupsofsomeFrobeniusgroups. 18
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WhiletheKempeShalevdenitioncapturesthedifcultyofd istinguishingthe hiddensubgroupfromthetrivialgroupinsomecases,itisno talwaysobvioushow totranslateitsanswerstopracticalalgorithms.Inchapte r 4 wedene”Algorithmic distinguishability”whichdenesdistinguishabilityofa subgroup H ofanynitegroup G fromthetrivialoneonthebasisofwhenanaturalalgorithmt hatusestheweak standardmethodsucceedsinpolynomialtime.Weprovethatt henon-trivialsubgroups ofanabeliangrouparealgorithmicdistinguishable.Weals oshowthattheKernelsof theFrobeniusgroups,whichareKempe-Shalevdistinguisha ble,arealsoalgorithmic distinguishableandthecomplements,whicharenotKempe-S halevdistinguishable,are alsonotalgorithmicdistinguishable. EventhoughincaseofFrobeniusgroupsthetwoconceptsseem tocoincide,we showinchapter 5 thattheymightnotbethesame.Thoughwedon'thaveageneral proof,computationsshowthatthetwoconceptsmaybediffer entfor G = S 3 S 3 ... S 3 ( ncopies ). 19
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CHAPTER2 HIDDENSUBGROUPSFORALMOSTABELIANGROUPS In( 11 )thenotionofanalmostabeliangroupwasintroduced. Denition 3 (AlmostAbelianGroup):Let G beanitegroup.Let N G ( H ) denotethe normalizerofasubgroup H in G Considerthenormalsubgroup K ( G )= T H N ( H ) of G Wecall G almostabelianif [ G : K ( G )] 2 exp ( r (lg 1 2 ( n ))) where n =lg( j G j ) In( 11 )theauthorsgiveanalgorithmtondthehiddensubgroupsof analmost abeliangroup.Inthischapterwegiveaverysmallvariation oftheiralgorithm.We provethatthisalgorithmsucceedswithasmallnumberofite rationswithprobability greaterthan 1 = 2 oneveryalmostabeliangroup.Weanalyseindetailtheoutco mes ofrunningthisalgorithminthecaseofthegroup G = Z 3 oZ m where m =2 n and anynon-trivialhiddensubgroup.Wendtheprobabilitytha ttheprocesswillyieldany particularsubgroupafter i iterations. 2.1Algorithmtondthehiddensubgroups Thealgorithmtodeterminethehiddensubgroupsforthealmo stabeliangroupsis asfollows: RepeatWeakStandardMethod n = s (log j G j ) times,where s ( x ) 2 R [ x ] is anon-zeropolynomial,forallthesubgroupsof G containing K ( G ). Considerthe intersectionsofthekernelsoftherepresentationsobserv edineachcaseafter n repetitions.Thealgorithmreturnsthelargestsuchinters ection. Thealgorithmcanbeformallywrittenasfollows:StepI:RepeatweakStandardMethod n = s (log j G j ) times,where s ( x ) 2 R [ x ] isa nonzeropolynomialforallsubgroupsof G containing K ( G ). StepII:Taketheintersectionsofthekernelsofthereprese ntationsobservedin eachcaseafter n repeats. StepIII:Returnthelargestsuchintersection. 20
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NoticethatthisalgorithmextendstheWeakStandardMethod tondanyhidden subgroupandnotjustthenormalones.Wenowestimatethenum berofiterations neededsothattheprobabilityofretrievingthehiddensubg roupismorethan 1 = 2. Lemma2.1. Let G beanitegroupoforder a and H thehiddensubgroup.Supposefor eachsubgroup M of G theprobabilityofgetting core M ( H ) be 1 2 e lg( j K ( G ) j ) k Thenthe productoftheseprobabilitiesisgreaterthan 1 = 2 if k < lg j K ( G ) j ln(4)+(lg( a )) 2 Proof. (1 2 e lg( j K ( G ) j ) k ) 2 lg 2 a > 1 2 lg 2 a 2 e lg j K ( G ) j k > 1 e lg 2 a 2 e lg j K ( G ) j k > 1 = 2 implies e (lg( a )) 2 e lg( j K ( G ) j ) k < 1 = 4. Hence ln(4) < lg( j K ( G ) j ) k (lg( a )) 2 Hence k < lg j K ( G ) j ln(4)+(lg( a )) 2 Lemma2.2. Suppose G isanitegroupand H isthehiddensubgroup.Thenwecan retrieve core G ( H ) after m =2 l lg( j G j ) stepswithprobability 1 e lg ( j G j ) 4 l ( l 1) 2 Proof. TheproofisthesameastheproofofTheorem5in( ? )with k =2 l lg( j G j ) and =( l 1)lg( j G j ). Followingtheproofweconcludethattheprobabilityofobta ining core G ( H ) is 1 e lg ( j G j ) 4 l ( l 1) 2 Theorem2.3. Suppose G isanitegroupand a =[ G : K ( G )]. IfwerepeatQuantum FourierSampling m =2 l lg( j G j ) timeswhere l =(8 d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e +8) forallsubgroups of G containing K ( G ) wecanretrieve H inthe”almostabelian”algorithmwithprobability greaterthan 1 = 2. Proof. Werstnotethat 4 l ( l 1) 2 = 32 d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e +32 (8 d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e +7) 2 < 32 d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e +32 49( d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e +1) 2 < 32 49( d ln(4)+lg 2 ( a ) lg( j K ( G ) j ) e ) < 32 49 lg j K ( G ) j ln(4)+(lg( a )) 2 Ifthealgorithmisrunfor m stepsforallsubgroupsof G containing K ( G ) thenthe probabilityofretrieving H G foreachofthesesubgroupswillbe 1 e lg ( j G j ) 4 l ( l 1) 2 which,bythe abovecomputation,isbiggerthan 1 e lg ( j G j ) l 0 where l 0 = 32 49 lg j K ( G ) j ln(4)+(lg( a )) 2 Sincethenumber 21
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ofsubgroupsofagroupoforder a cannotexceed 2 lg 2 ( a ) ,byLemma4.1itisclearthatwe canretrieve H attheendofthe”almostabelian”algorithmwithprobabilit ybiggerthan 1 = 2. 2.2AnAlmostAbelianGroupofOrder 3 2 n Wenowanalyzethequantumalgorithmforndingthehiddensu bgroupsofthe almostabeliangroup G = Z 3 oZ m where m =2 n .Thesegroupsarementionedas examplesofalmostabeliangroupsin( 11 ).Foreachnon-trivialsubgroup,wecalculate theprobabilitiesthatthealgorithmyieldsanyparticular subgroupafter i steps.These probabilitiesaregiveninAppendix A Thesubgroupsofthegroup G areasfollows.Wedenoteby Z 0 theuniquenormal subgroupofthesylow2-subgroupsofindex2.Wedenoteby T thesylow3-subgroupof G by SY i where i =1,2,3 thethreesylow2-subgroups.Apartfromthesethesubgroups of TZ 0 and Z 0 arealsosubgroupsof G If N ( H ) denotethenormalizerofasubgroup H of G wehavethat T H N ( H )= Z 0 G has m linearcharactersand m = 2 non-linearcharactersofdegree2inducedfrom thecharactersof TZ 0 where T isthesylow3-subgroupof G Inthissectionweexplicitly calculate,givenanyhiddensubgroup,theprobabilitiesof gettingthesubgroupafter i stepsastheintersectionofthepossiblekernelsoftheirre duciblerepresentations measuredcorrespondingtoallthesubgroupsof G containing Z 0 Theorem2.4. Let G = Z 3 oZ m beanalmostabeliangroup.Let H beanon-trivial subgroupof G Runthealmostabelianalgorithm i times.Thentheprobabilityofthe processyieldinganyparticularsubgroupof G isgivenbytheTablesinAppendix A ( A i where i =1,...15 ). Proof. 1)When H = SY 1 and G = TZ 0 Theprobabilityofgettinganirreducible representation isgivenby d j G j P h 2 H T G ( h ). Hencetheprobabilityisequalto 1 3 m = 2 m = 2=1 = 3 when Z 0 isinthekernelof otherwisetheprobabilityisequalto 22
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0.Thepossiblekernelsare TZ 0 and Z 0 Soaftertherstiterationtheprobabilitiesof gettingkernelssuchthat j ker ( ) j =3 a 2 N are p ( a N )= 8>>>><>>>>: 1 = 3 N = n 1, a =1 2 = 3 N = n 1, a =0 0 otherwise After i iterationstheprobabilitiesaregivenby p ( a N )= 8>>>><>>>>: (1 = 3) i N = n 1, a =1 1 (1 = 3) i N = n 1, a =0 0 otherwise When G = Z 0 Theprobabilityofgettinganon-trivialirreduciblerepre sentationis0. Thetrivialrepresentationismeasuredwithprobability1. Theonlypossiblekernelis Z 0 After i iterationstheprobabilityofgetting Z 0 is1. When G = SY 1 Theprobabilityofgettinganynon-trivialirreduciblerep resentation is0.Theprobabilityofmeasuringthetrivialrepresentati onis1.Thepossiblekernelis SY 1 After i iterationstheprobabilityofgetting SY 1 is1. When G = SY 2 Theprobabilityofmeasuringanirreduciblerepresentatio n is givenby1/2.Soafter i stepstheprobabilitiesofgettingtheintersectionofthek ernelsto beoforder 2 N aregivenby p ( N )= 8>>>><>>>>: (1 = 2) i N = n 1 (1 = 2) i N = n 1 0 otherwise When G = SY 3 Theprobabilitiesthatwegetarethesameasabove. 23
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When G = G If islinearthen d j G j P h 2 H ( h )=1 = 3 if =1 G and0otherwise. Suppose isnon-linear.Then = G where 2 Irr ( TZ 0 ). Now d j G j X h 2 H ( h )=2 j H j j G j (( ) H ,1 H )=2 = 3(( ) H ,1 H )=2 = 3(( G ) H ,1 H ) =2 = 3(( H T TZ 0 ) H ,1 H )=2 = 3( H T TZ 0 ,1 Z 0 ) =2 = 3( Z 0 ,1 Z 0 )=0 if Z 0 6 =1 Z 0 and2/3if Z 0 =1 Z 0 Thepossiblekernelsare G and Z 0 Aftertherstiterationtheprobabilitiesofgetting j ker ( ) j =3 a 2 N regivenby p ( a N )= 8>>>><>>>>: (1 = 3) N = n a =1 (2 = 3) N = n 1, a =0 0 otherwise After i iterationstheprobabilitieswillbe p ( a N )= 8>>>><>>>>: (1 = 3) i N = n a =1 1 (1 = 3) i N = n 1, a =0 0 otherwise H = TZ 0 1)When G = Z 0 Thentheprobabilityofmeasuringthetrivialrepresentati onis1. Thepossiblekernelis Z 0 .Theprobabilitythattheintersectionofthekernelswillb e Z 0 after i iterationsis1. 2)When G = TZ 0 Theprobabilityofmeasuringthetrivialrepresentationis 1.The onlypossiblekernelis TZ 0 .Theprobabilitythatafter i iterationstheintersectionofthe kernelsis TZ 0 is1. 24
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3)When G = SY 1 Theprobabilityofmeasuringanon-trivialrepresentation is givenby 1 j SY 1 j X h 2 Z 0 ( h ). Sotheprobabilityis 1 = 2 if Z 0 2 ker ( ) and0otherwise.Sothepossiblekernelsare Z 0 and SY 1 .Soaftertherstiterationtheprobabilitiesthatthe j ker ( ) j =2 N are p ( a N )= 8>>>><>>>>: (1 = 2) N = n (1 = 2) N = n 1 0 otherwise After i iterationstheprobabilitiesare p ( a N )= 8>>>><>>>>: (1 = 2) i N = n (1 = 2) i (2 i 1) N = n 1 0 otherwise 4)When G = G Thentheprobabilityofmeasuringanirreduciblerepresent ation is 1/2.Henceifthekernelsaregivenby j ker ( ) j =3.2 N thentheprobabilitiesofgetting theintersectiontobeoneofthemafter i iterationsare p ( a N )= 8>>>><>>>>: (1 = 2) i N = n (1 = 2) i (2 i 1) N = n 1 0 otherwise When H = Z 0 1) G = TZ 0 Thentheprobabilityofmeasuringanirreduciblerepresent ation of TZ 0 is1/3when Z 0 2 ker ( ) and0otherwise.Thepossiblekernelsare TZ 0 and Z 0 25
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After i iterationstheprobabilitiesofgetting j ker ( ) j =3 a 2 N are p ( a N )= 8>>>><>>>>: (1 = 3) i N = n 1, a =1 1 (1 = 3) i N = n 1, a =0 0 otherwise 2) G = Z 0 Theprobabilityofmeasuringthetrivialrepresentationis 1and0 otherwise.Soafter i iterationstheprobabilitiesthat j ker ( ) j =2 N are p ( a N )= 8><>: 1 N = n 1 0 otherwise 3) G = SY 1 Theprobabilityofmeasuringanirreduciblerepresentatio n of SY 1 is 1/2if Z 0 isinkernelof and0otherwise.Thepossiblekernelsof are SY 1 and Z 0 theprobabilitiesafter i iterationsofgetting j ker ( ) j =2 N are p ( a N )= 8>>>><>>>>: (1 = 2) i N = n (1 = 2) i (2 i 1) N = n 1 0 otherwise If G = g 1 SY 1 ( g 1 ) 1 theprobabilitiesaregoingtobethesame. 4) G = G Theprobabilityofmeasuringanirreduciblerepresentatio n is1/6if islinearandcontains Z 0 initskerneland2/3ifitisnon-linearandcontains Z 0 inits kerneland0otherwise.Thepossiblekernelsare Z 0 TZ 0 and G Sotheprobabilities that j ker ( ) j =3 a 2 N are p ( a N )= 8>>>>>>><>>>>>>>: (1 = 6) N = n a =1 (1 = 6) N = n 1, a =1 2 = 3 N = n 1, a =0 0 otherwise 26
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Soafter i iterationstheprobabilitiesare p ( a N )= 8>>>>>>><>>>>>>>: (1 = 6) i N = n a =1 (1 = 6) i (2 i 1) N = n 1, a =1 (1 1 3 i ) N = n 1, a =0 0 otherwise When H < TZ 0 and j H j =3.2 N 1)When G = TZ 0 Theprobabilityofmeasuringanirreduciblerepresentatio n is 1 j TZ 0 j X h 2 H ( h )=2 k +1 = 2 n =2 k +1 n if H 2 ker ( ) andequalto0otherwise.Sotheprobabilitiesthat j ker ( ) j =3.2 N are p ( N )= 8>>>><>>>>: 0 N < k 2 k +1 n N = n 1 1 2 N k +1 k N < n 1 After i iterationstheprobabilitythattheintersectionoftheker nelsis H isgivenby 1 [ p ( N 6 = k )] i =1 (1 = 2) i Theprobabilitythattheintersectionis TZ 0 isgivenby [ p ( N = n 1)] i Theprobability thattheintersectionisasubgroupof TZ 0 oforder 3.2 m where k < m < n 1 isgivenby p ( N m ) p ( N > m )=[ 2 2 m k +1 ] i [ 1 2 m k +1 ] i =[ 1 2 m k +1 ] i (2 i 1) Hencewehaveafter i iterations p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (2 k +1 n ) i N = n 1 ( 1 2 m k +1 ) i (2 i 1) k < m < n 1 (1 [1 = 2] i ) N = k 27
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2)When G = Z 0 Theprobabilityofmeasuringanirreduciblerepresentatio n is 1 j Z 0 j X h 2 H T Z 0 ( h )=2 k +1 = 2 n =1 = 2 n k 1 if H T Z 0 2 ker ( ) and0otherwise.Sotheprobabilitiesthat j ker ( ) j =2 N are p ( N )= 8>>>><>>>>: 0 N < k 2 k +1 n N = n 1 1 2 N k +1 k N < n 1 After i iterationstheprobabilitythattheintersectionoftheker nelsis Z 0 is [1 = 2 n k 1 ] i Theprobabilitythattheintersectionisasubgroupof Z 0 oforder 2 m where k < m < n 1 is [1 = 2 m k +1 ] i (2 i 1) .Henceafter i iterationswehave p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (2 k +1 n ) i N = n 1 ( 1 2 m k +1 ) i (2 i 1) k < m < n 1 (1 [1 = 2] i ) N = k 3)When G = SY 1 Theprobabilityofmeasuringanirreduciblerepresentatio n is givenby 1 j SY 1 j X h 2 H T SY 1 ( h )=2 k = 2 n =1 = 2 n k if H T SY 1 2 ker ( ) and0otherwise.Sotheprobabilitiesthat j ker ( ) j =2 N are p ( N )= 8>>>><>>>>: 0 N < k 1 = 2 n k N = n 1 2 N k +1 k N < n 28
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Theprobabilitiesafter i iterationsare p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (1 = 2 n k ) i N = n ( 1 2 m k +1 ) i (2 i 1) k < m < n (1 [1 = 2] i ) N = k 4)When G = G Thentheprobabilityofmeasuringanirreduciblerepresent ation of G is 2 k = 2 n =1 = 2 n k if H 2 ker ( ) and0otherwise.Theprobabilitiesthat j ker ( ) j =3.2 N aregivenby p ( N )= 8>>>><>>>>: 0 N < k 1 = 2 n k N = n 1 2 N k +1 k N < n After i stepstheprobabilitiesare p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (1 = 2 n k ) i N = n ( 1 2 m k +1 ) i (2 i 1) k < m < n (1 [1 = 2] i ) N = k When H < Z 0 and j H j =2 k 1)When G = TZ 0 Theprobabilityofmeasuringanirreduciblerepresentatio n is givenby d j TZ 0 j X h 2 H ( h )= 1 3 2 k n +1 if H 2 ker ( ) and0otherwise.Theprobabilitiesthat j ker ( ) j =3 a 2 N aregivenby 29
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p ( a N )= 8>>>>>>>>>><>>>>>>>>>>: 0 N < k 1 3.2 n k 1 N = n 1, a =1 2 k N 1 3 k N < n 1, a =1 2 k N 3 k N < n 1, a =0 1 3.2 n k 2 N = n 1, a =0 Thisprobabilitydistributionisaproductoftwoindepende ntprobabilitydistributions p 0,3 ( i ) and p 0,2 ( i ) where p 0,3 ( i )= 8><>: 2 = 3 i =0 1 = 3 i =1 and p 0,2 ( i )= 8>>>><>>>>: 0 i < k 1 = 2 n k 1 i = n 1 2 k i 1 i < n 1 After i iterationstheprobabilitythattheintersectionoftheker nelsis TZ 0 is 1 3 i 1 (2 n k 1 ) i Theprobabilitythattheintersectionis Z 0 is (1 1 3 i )( 1 (2 n k 1 ) i Theprobability thattheintersectionis H is (1 1 3 i )(1 1 2 i ). Theprobabilitythattheintersectionisa subgroupof TZ 0 oforder 3.2 m isgivenby 1 3 i (2 i 1) 1 (2 m k +1 ) i andtheprobabilitythatthe intersectionofkernelsisasubgroupof Z 0 oforder 2 m isgivenby (1 1 3 i )(2 i 1) 1 (2 m k +1 ) i Hencewehave p ( a N )= 8>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>: 0 N < k 1 3 i 1 (2 n k 1 ) i N = n 1, a =1 (1 1 3 i ) 1 (2 n k 1 ) i N = n 1, a =0 (1 1 3 i )(1 1 2 i ) N = k a =0 1 3 i (1 1 2 i ) N = k a =1 1 3 i (2 i 1)( 1 2 m k +1 ) i N = m a =1, k < m < n 1 (1 1 3 i )(2 i 1)( 1 2 m k +1 ) i N = m a =0 30
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2)When G = SY 1 Theprobabilityofmeasuringanyirreduciblerepresentati on is 1 = 2 n k if H isinthekernelof and0otherwise.Sotheprobabilitiesthat j ker ( ) j =2 N aregivenby p ( N )= 8>>>><>>>>: 0 N < k (1 = 2 n k ) N = n ( 1 2 N k +1 ) k N < n After i iterationstheprobabilitiesthattheintersectionofthek ernelshaveorder 2 N are givenby p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (1 = 2 n k ) i N = n (1 1 2 i ) N = k 1 (2 m k +1 ) i (2 i 1) k < m < n 3)When G = Z 0 Theprobabilityofmeasuringanyirreduciblerepresentati on of Z 0 isgivenby 1 = 2 n k 1 if H iscontainedinthekernelof and0otherwise.The probabilitiesthat j ker ( ) j =2 N aregivenby p ( N )= 8>>>><>>>>: 0 N < k (1 = 2 n k 1 ) N = n 1 1 2 N k +1 k N < n 1 After i stepstheprobabilitiesoftheintersectionofthekernelsa re p ( N )= 8>>>>>>><>>>>>>>: 0 N < k (1 = 2 n k 1 ) i N = n 1 (1 1 2 i ) N = k 1 (2 m k ) i (1 1 2 i ) k < m < n 1 4)When G = G Theprobabilityofmeasuringanirreduciblerepresentatio n is givenby d j G j X h 2 H ( h ). 31
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When islinearthentheprobabilityis 2 k = 3.2 n = 1 3 2 k n if ker ( ) contains H andit is 1 32 n k 2 if isnonlinearanditskernelcontains H Sowehavethefollowingforthe probabilitiesofgetting j ker( ) j =3 a 2 N p ( a N )= 8>>>>>>>>>>>>>><>>>>>>>>>>>>>>: 0 N < k 1 3 2 k N 1 k N n 1, a =1 1 3 2 k n N = n a =1 0 N = n a =0 1 3 2 k n +2 N = n 1, a =0 1 3 2 k N k N < n 1, a =0 After i iterationstheprobabilitythatintersectionofkernelsha veorder 3 a 2 N is ( p ( a N )) i ( p ( a +1, N )) i ( p ( a N +1)) i +( p ( a +1, N +1)) i Hencewehave p ( a N )= 8>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>: 0 N < k ( 2 k n +1 3 ) i ( 2 k n 3 ) i N = n 1, a =1 (2 k m ) i ( 2 k m 3 ) i (2 k m 1 ) i +( 2 k m 1 3 ) i k m < n 1, a =0 ( 2 k m 3 ) i ( 2 k m 1 3 ) i k m < n 1, a =1 (2 k n +1 ) i ( 2 k n +1 3 ) i N = n 1, a =0 0 N = n a =0 1 3 i ( 1 2 n k ) i N = n a =1 32
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CHAPTER3 KEMPE-SHALEVDISTINGUISHABLITY Fortheremainderofthiswork,weconsiderthedecisionvers ionoftheHidden Subgroupproblem.Aswesaw,itisstatedasfollows. DecisionVersionoftheHiddenSubgroupProblem :Let G beanitegroupand H asubgroup.Givenablack-boxfunction f : G S whichisconstanton(left)-cosets gH of H andtakesdifferentvaluesfordifferentcosets,determine whether H = f e g or not. Therearetwowaystoformalizetheexistenceofsolutionsto thisproblem.Inthis chapter,wediscussthedenitiongivenbyKempeandShalev( 18 ).Adifferentway todenethiswillbediscussedinthenextchapter.Kempeand Shalevproposethatif thesubgroup H of G producesprobabilitiesofrepresentationsunderQuantumF ourier Samplingthatareveryclosetothosethatarisefromthetriv ialsubgroup,thenthe subgroupshouldbecalledindistinguishablefromthetrivi alsubgroup.Moreprecisely, thedenitionproposedbyKempeandShalev( 18 )in2005isasfollows. Denition 4 Let G beanitegroup.Let Irr( G ) bethesetofirreducible(complex) representationsof G .Wedenoteby thecharacterassociatedtoa 2 Irr( G ) andlet d beit'sdegree.For H < G wedene D H = 1 j G j X 2 Irr( G ) d j X h 6 = e h 2 H ( h ) j Wesaythatasubgroup H isdistinguishableif D H log ( j G j ) c forsome c > 0, c being independentof G Webeginthissectionbyprovingthatallsubgroupsofanabel iangroupare Kempe-Shalevdistinguishable. Theorem3.1. Subgroupsofanabeliangroup G areKempe-Shalevdistinguishable. 33
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Proof. Let H beahiddensubgroupoftheabeliangroup G Then D H = 1 j G j X d j X h 6 = e ( h ) j = 1 j G j X j X h 6 = e ( h ) j = 1 j G j [ X ker ( ) H j X h 6 = e ( h ) j + X ker ( ) + H j X h 6 = e ( h ) j ] = 1 j G j [ X ker ( ) H jj H j < ( ) H ,1 H > (1) j + X ker ( ) + H jj H j < ( ) H ,1 H > (1) j ] = 1 j G j [ X ker ( ) H jj H j (1) (1) j + X ker ( ) + H j (1) j ] = 1 j G j [ X ker ( ) H (1)[ j H j 1]+ X ker ( ) + H (1)] = 1 j G j [ X ker ( ) H [ j H j 1]+ X ker ( ) + H 1] = 1 j G j [ j G j = j H j [ j H j 1]+( j G jj G j = j H j )]=2 1 j G j ( j G jj G j = j H j )=2(1 1 j H j ) > 1 > (log( j G j ) c foreach c > 0 as j G j!1 Inthischapterwestudythedistinguishabilityofsubgroup sofFrobeniusgroups usingtheolddenitionofdistinguishability.Herewecons iderFrobeniusgroupswith abeliankernel.Wegivenecessaryandsufcientconditions forthedistinguishabilityof thesubgroupsoftheFrobeniusgroupsundercertainconditi ons. Denition 5 Let G beanitegroup.Let H G and H isanontrivialpropersubgroupof G .Assumethat H \ H g =1 whenever g 2 G H .Then H isaFrobeniuscomplementin G .AgroupwhichcontainsaFrobeniuscomplementiscalledaFr obeniusgroup. 34
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Theorem3.2 (Frobenius) Let G beaFrobeniusgroupwithcomplement H .Thenthere exists K G with HK = G and H \ K =1 Proof. See( 16 ). Thissubgroup K iscalledtheFrobeniuskernelof G .ItisclearthataFrobenius groupwithkernel K andcomplement H canbewrittenas K o H .Italsofollowsthatany non-trivialelementof H xesnoelementof K i.e H actssemiregularlyon K Theorem3.3. Let G = Z ( ) g ( ) oZ beaFrobeniusgroupwhere g ( ) isafunctionof and ( ) iscoprimeto ,andforsome c > 0 andforallsufcientlylarge g ( ) > 1 c .Let H beasubgroupoforder ( ) f ( ) or ( ) f ( ) where f ( ) isafunctionof such that 0 f ( ) g ( ) then H isKempe-Shalevdistinguishableifandonlyif f ( ) > 0 for all >> 0. Proof. Suppose f ( ) > 0 for >> 0, andtake sufcientlylarge.Then f ( ) > 0 and H hasorderdivisibleby whichiscoprimeto Furthermoreforeach 2 Irr ( G ) j X h 6 = e h 2 H ( h ) j = jj H j [( ) H ,1 H ] (1) j 1, since alpha divides j H j andiscoprimeto (1). Hence X = non-linear j X h 6 = e h 2 H ( h ) j ( ) g ( ) 1 Hence D H ( j H j 1)+ ( g ( ) 1 ) ( ) g ( ) > ( ( ) g ( ) 1)+1 ( ) g ( ) = 1 > 1 ( log j G j ) c forsome c > 0. Hence H isdistinguishable.If f ( ) =0then H hasordereither f ( ) =1 or ( ) f ( ) = .Inthiscase D H 2( 1) ( ) g ( ) 35
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andforeach c > 0 andsufcientlylarge ,wehave 2( 1) ( ) g ( ) < ( log ( j G j )) c Therefore,if f ( )=0 innitelyoften, H isindistinguishable.Hencetheproof. Corollary3.4. TheFrobeniuscomplementof G isindistinguishable. Proof. TheFrobeniuscomplementof G hasorder .Hence f ( )=0 .Fromtheprevious theoremthecomplementisindistinguishable. Corollary3.5. TheFrobeniuskernelof G isKempe-Shalevdistinguishable. Proof. TheorderoftheFrobeniuskernelis ( ) g ( p ) .Hencethecorollaryfollowsfrom theprevioustheorem. Proposition3.6. Forthefrobeniusgroup Z 2 p 1 oZ p thesmallest D S for S anon-trivial subgroupisobtainedwhenthesubgroupSistheSylow p –subgroupofthegroup. Proof. Fromthecharactertableitisevidentthatwhen S istheSylow p –subgroupthen P d j P h 6 = e h 2 S ( ) j =2 ( p 1) thecontributionfromnon-linearcharactersbeingzero. So D S = 2( p 1) 2 p 1 p .NowifHisasubgroupwith j H j =2 k then D H = p (2 k 1)+.... 2 p 1 p where p (2 k 1) isthecontributionfromlinearcharacters.This D H isobviouslybiggerthan D S when k > 1.If k =1then j H j =2and D H = ( p 1)+ p x 2 p 1 p where x = P = non-linear j P h 6 = e h 2 H ( ) j > 0. Thissumisequalto P = non-linear jj H j [( ) H ,1 H ] (1) j whichisbiggerthan1 because j H j [( ) H ,1 H ] (1) isneverequaltozeroandbiggerthan1.Hencethis isalsobiggerthan D S .Anyothersubgroupof G willbeoftheform H w Z 2 k oZ p This subgroup H has 2 k 1 elementsoforder2andso D H = p (2 k 1)+.... 2 p 1 p .Since k > 1thisis greaterthan D S Lemma3.7. Let > 1 beaconstant,andlet m n g benon-negativerealvalued functionsof N .Assumethat m ( i ), n ( i ) !1 as i !1 .Alsoassumethatthereexists 36
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some c 0 > 0 suchthat g ( i ) c 0 > m ( i ) n ( i ) forallsufcientlylarge i .Thenforeach c > 0 g ( i ) m ( i ) n ( i ) > g ( i ) c forallsufcientlylarge i Proof. Let c > 0 .Forallsufcientlylarge i ,wehave g ( i ) > g ( i ) c + c 0 .Henceforall sufcientlylarge i wehave g ( i ) > g ( i ) c + c 0 = g ( i ) c g ( i ) c 0 > g ( i ) c m ( i ) n ( i ) Hence g ( i ) m ( i ) n ( i ) > g ( i ) c Theorem3.8. Let G = Z g ( i ) oZ m ( i ) oZ n ( i ) beaFrobeniusgroupwhere ( m ( i ), n ( i ))=1 and iscoprimeto m ( i ) and n ( i ). Assumethat m ( i ), n ( i ) !1 as i !1 Alsoassume thatforsome c > 0 g ( i ) c m ( i ) n ( i ) andforsufcientlylarge i .Let H beasubgroup oforder f ( i ) or f ( i ) m ( i ) or f ( i ) n ( i ) or f ( i ) m ( i ) n ( i ) .Then H isdistinguishableiff f ( i ) > 0 forallsufcientlylarge i > 0 Proof. Let f ( i ) > 0 andtake i sufcientlylarge.Then H hasorderdivisibleby whichis coprimeto m ( i ) and n ( i ) .Thereforeforeach 2 Irr ( G ) j X h 6 = e h 2 H ( h ) j = jj H j [( ) H ,1 H ] (1) j 1. Hence X = non-linearofdegreem(i)n(i) j X h 6 = e h 2 H ( h ) j g ( n ) 1 mn Hence D H = P = linear j P h 6 = e h 2 H ( h ) j + P = non-linear d j P h 6 = e h 2 H ( h ) j g ( i ) m ( i ) n ( i ) Hence D H 1+ P = non-linear d j P h 6 = e h 2 H ( h ) j g ( i ) m ( i ) n ( i ) 1+ m ( i ) n ( i ) P = non-linearofdegreem(i)n(i) 1 g ( i ) m ( i ) n ( i ) 37
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1+ m ( i ) n ( i ). g ( i ) 1 m ( i ) n ( i ) g ( i ) m ( i ) n ( i ) = 1 m ( i ) n ( i ) whichisevidentlybiggerthan log ( j G j ) c fromhypotheses.If f ( i )=0 then j H j = m ( i ) or n ( i ) or m ( i ) n ( i ). If j H j = m ( i ) then D H n ( i )( m ( i ) 1)+ n ( i )( m ( i ) 1) n ( i )( m ( i ) 1) g ( i ) m ( i ) n ( i ) = n ( i )( m ( i ) 1) g ( i ) m ( i ) n ( i ) + n ( i ) 2 ( m ( i ) 1) 2 g ( i ) m ( i ) n ( i ) 1 g ( i ) + m ( i ) n ( i ) g ( i ) ( log j G j ) c foreach c > 0 andsufcientlylarge i fromLemma 3.7 .If j H j = n ( i ) then D H 2( n ( i ) 1)+ n ( i )( m ( i ) 1) n ( i )( n ( i ) 1) g ( i ) m ( i ) n ( i ) = 2 g ( i ) m ( i ) 2 g ( i ) m ( i ) n ( i ) + n ( i ) 2 g ( i ) ( log j G j ) c foreach c > 0 andsufcientlylarge n ( i ). If j H j = m ( i ) n ( i ) then D H ( m ( i ) n ( i ) 1)+( n ( i ) 1) m ( i ) n ( i )+ n ( i ) m ( i ) 1 n ( i ) n ( i ) g ( i ) m ( i ) n ( i ) n ( i ) g ( i ) + 1 g ( i ) 1 g ( i ) m ( i ) ( log j G j ) c foreach c > 0 andsufcientlylarge i fromLemma 3.7 .Hencetheproof. Theorem3.9. Let G = Z g ( i ) o ( SL (2,5) ( Z m ( i ) oZ n ( i ) )) beaFrobeniusgroupwhere i 2 N ( m ( i ), n ( i ))=1 and > 1 isaconstantandcoprimeto m ( i ) and n ( i ). Assume that m ( i ), n ( i ) !1 as i !1 Alsoassumethatforsome c > 0 g ( i ) c m ( i ) n ( i ) forallsufcientlylarge i .Let H beasubgroupoforder f ( i ) or f ( i ) m ( i ) or f ( i ) n ( i ) or f ( i ) m ( i ) n ( i ) or f ( i ) 120 m ( i ) or f ( i ) 120 n ( i ) .Then H isKempe-Shalevdistinguishableiff f ( i ) > 0 forallsufcientlylarge i 38
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Proof. Let f ( i ) > 0 andtake i sufcientlylarge.Then H hasorderdivisibleby whichis coprimeto m ( i ) and n ( i ). Thereforeforeach 2 Irr ( G ) j X h 6 = e h 2 H ( h ) j = jj H j [( ) H ,1 H ] (1) j 1. Hence X = non-linearofdegree120m(i)n(i) j X h 6 = e h 2 H ( h ) j g ( i ) 1 120 m ( i ) n ( i ) Hence D H = P = linear j P h 6 = e h 2 H ( h ) j + P = non-linear d j P h 6 = e h 2 H ( h ) j g ( i ) .120 m ( i ) n ( i ) Hence D H 1+ P = non-linear d j P h 6 = e h 2 H ( h ) j g ( i ) .120 m ( i ) n ( i ) 1+ m ( i ) n ( i ) P = non-linearofdegree120m(i)n(i) 1 g ( i ) .120 m ( i ) n ( i ) 1+ m ( i ) n ( i ). g ( i ) 1 120 m ( i ) n ( i ) g ( i ) .120 m ( i ) n ( i ) = 1 120 m ( i ) n ( i ) whichisevidentlybiggerthan log ( j G j ) c forsufcientlylarge i .Nowlet f ( i ) > 0 .If theorderof H is 120 m ( i ) thensincethereare n ( i ) linearcharacters P = linear P h 6 = e h 2 H j ( h ) j n ( i )(120 m ( i ) 1) .Also jj H j [( ) H ,1 H ] (1) j (1)(120 m ( i ) 1) ( D H ) g ( i ) 120 m ( i ) n ( i ) n ( i )(120 m ( i ) 1)+2 n ( i ) 2( m ( i ) 1) n ( i ) 2 n ( i )(120 m ( i ) 1) +3 n ( i )2( m ( i ) 1)3 n ( i )(120 m ( i ) 1) +4 n ( i )2( m ( i ) 1)4 n ( i )(120 m ( i ) 1) +5 n ( i )2( m ( i ) 1)5 n ( i )(120 m ( i ) 1) 39
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+6 n ( i )2( m ( i ) 1)6 n ( i )(120 m ( i ) 1). Hence D H < 1 g ( n ) + (8+18+16+25+36) m ( i ) n ( i ) g ( i ) < ( log j G j ) c foreach c > 0 andsufcientlylarge i .Iforderof H is 120 n ( i ) then jj H j [( ) H ,1 H ] (1) j (1)(120 n ( i ) 1). Sincethereare n ( i ) linearcharactersthecontributionfromthemislessthanor equalto n ( i )(120 n ( i ) 1) .Then ( D H )120 g ( i ) m ( i ) n ( i ) n ( i )(120 n ( i ) 1) +2 n ( i )2( m ( i ) 1)2 n ( i )(120 n ( i ) 1) +3 n ( i )2( m ( i ) 1)3 n ( i )(120 n ( i ) 1) +4 n ( i )2( m ( i ) 1)4 n ( i )(120 n ( i ) 1) +5 n ( i )2( m ( i ) 1)5 n ( i )(120 n ( i ) 1) +6 n ( i )2( m ( i ) 1)6 n ( i )(120 n ( i ) 1). Hence ( D H )120 g ( i ) m ( i ) n ( i ) < 120( n ( i ) 3 +8 n ( i ) 3 m ( i )+18 n ( i ) 3 m ( i ) +32 n ( i ) 3 m ( i )+25 n ( i ) 3 m ( i )+36 n ( i ) 3 m ( i )) < 120 2 n ( i ) 3 m ( i ). Hence D H < 120 n ( i ) 3 m ( i ) g ( i ) m ( i ) n ( i ) < ( log j G j ) c foreach c > 0 40
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andsufcientlylarge i fromLemma 3.7 .Iforderof H is m ( i ) then jj H j [( ) H ,1 H ] (1) j (1)( m ( i ) 1). Thecontributionfromlinearchararctersislessthanorequ alto n ( i )( m ( i ) 1) .Thus ( D H )120 g ( i ) m ( i ) n ( i ) n ( i )( m ( i ) 1) +2 n ( i )2( m ( i ) 1)2 n ( i )( m ( i ) 1) +3 n ( i )2( m ( i ) 1)3 n ( i )( m ( i ) 1) +4 n ( i )2( m ( i ) 1)4 n ( i )( m ( i ) 1) +5 n ( i )( m ( i ) 1)5 n ( i )( m ( i ) 1) +6 n ( i )( m ( i ) 1)6 n ( i )( m ( i ) 1) ( D H )120 g ( i ) m ( i ) n ( i ) < n ( i ) m ( i )+8 m ( i ) 2 n ( i ) 2 +18 m ( i ) 2 n ( i ) 2 +32 m ( i ) 2 n ( i ) 2 +25 m ( i ) 2 n ( i ) 2 +36 m ( i ) 2 n ( i ) 2 Hence D H < 1 120 g ( i ) + 119 m ( i ) n ( i ) 120 g ( i ) < ( log j G j ) c foreach c > 0 andsufcientlylarge i fromLemma 3.7 .Iforderof H is n ( i ) thenthecontributionfromlinearcharactersislessthanor equalto 2( n ( i ) 1) ( D H )120 g ( i ) m ( i ) n ( i ) 2( n ( i ) 1) +2 n ( i )2( m ( i ) 1)2 n ( i )( n ( i ) 1) +3 n ( i )2( m ( i ) 1)3 n ( i )( n ( i ) 1) +4 n ( i )2( m ( i ) 1)4 n ( i )( n ( i ) 1) +5 n ( i )2( m ( i ) 1)5 n ( i )( n ( i ) 1) +6 n ( i )2( m ( i ) 1)6 n ( i )( n ( i ) 1). 41
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Hence ( D H )120 g ( i ) m ( i ) n ( i ) < ( n ( i ) 3 +8 n ( i ) 3 m ( i )+18 n ( i ) 3 m ( i ) +32 n ( i ) 3 m ( i )+25 n ( i ) 3 m ( i )+36 n ( i ) 3 m ( i )) < 120 n ( i ) 3 m ( i ). Hence D H < n ( i ) 3 m ( i ) g ( i ) m ( i ) n ( i ) < ( log j G j ) c foreach c > 0 andsufcientlylarge i fromLemma 3.7 Iftheorderof H is m ( i ) n ( i ) then jj H j [( ) H ,1 H ] (1) j (1)( m ( i ) n ( i ) 1). Sothecontributionfromlinearcharactersislessthanoreq ualto n ( i )( m ( i ) n ( i ) 1) ( D H )120 g ( i ) m ( i ) n ( i ) n ( i )( m ( i ) n ( i ) 1) +2 n ( i )2( m ( i ) n ( i ) 1)2 n ( i ) +3 n ( i )2( m ( i ) n ( i ) 1)3 n ( i ) +4 n ( i )2( m ( i ) n ( i ) 1)4 n ( i ) +5 n ( i )( m ( i ) n ( i ) 1)5 n ( i )+6 n ( i )( m ( i ) n ( i ) 1)6 n ( i ) Hence D H < n ( i ) 2 g ( i ) < ( log j G j ) c foreach c > 0 and sufcientlylarge i fromLemma 3.7 .Henceinallthesecases H isindistinguishable. Hencetheproof. 42
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CHAPTER4 ALGORITHMICDISTINGUISHABILITY InChapter 3 wehavediscussedindetailKempe-Shalevdistinguishabili tyas denedbyKempeandShalevin( 18 ).WhiletheKempeShalevdenitioncapturesthe difcultyofdistinguishingthehiddensubgroupfromthetr ivialgroupinsomecases, itisnotalwaysobvioushowtotranslateitsanswerstopract icalalgorithms.Inthis chapterwedene”Algorithmicdistinguishability”.Algor ithmicdistinguishabilitydenes distinguishabilityofasubgroup H ofanynitegroup G fromthetrivialoneonthe basisofwhenanaturalalgorithmthatusestheweakstandard methodsucceedsin polynomialtime.Westudythisnewdenitionasappliedtoan umberofexamples.We alsoshowthatthenewconceptcoincidesfortheFrobeniusgr oupswiththeKempe Shalevdenitiondiscussedinthepreviouschapter. 4.1Analgorithmfordistinguishability Anaturalalgorithmtotellwhetherornotthehiddensubgrou pistrivialisas follows.ApplyQuantumFourierSampling m times.Considertheresultingsequenceof representations 1 ,.., m Askwhetheritismorelikelytoobtainthisparticularseque nce ifthehiddensubgroupis H orifitistrivial.Return 1 ifitis H andreturn 0 ifitistrivial. Thisalgorithmisgivenbyafunction 4 : Irr ( G ) m !f 0,1 g whichcanbeobtainedasfollows: Let G beanitegroup,andlet H beanon-trivialsubgroupof G .Wedenote P H ,1 :Irr( G ) [0,1] thefunctiondenedby P H ,1 ( )= (1) j H j j G j j H ,1 h j 43
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forall 2 Irr( G ) .Foreachpositiveinteger m ,wedenoteby P H m :Irr( G ) m [0,1] thefunctiondenedby P H m ( 1 ,..., m )= m Y i =1 P H ,1 ( i ). If m isunderstoodfromthecontext,wedenote P H m simply P H .Let 4 :Irr( G ) m !f 0,1 g beafunctiondenedby 4 ( 1 ,.., m )= 8><>: 0 P 1, m ( 1 ,.., m ) P H m ( 1 ,.., m ) 1 P 1, m ( 1 ,.., m ) < P H m ( 1 ,.., m ) Thustheformalalgorithmisasfollows:Algorithm : StepI:RepeatQuantumFourierSampling m times. StepII:Getasequenceof i where 1 i m StepIII:Applythefunction 4 tothetuple ( 1 ,.., m ). StepIV:Ifthefunctionreturns 1 then H isnon-trivialandifitreturns 0 then otherwise. Denition 6 Let G beanitegroup,let H beanon-trivialsubgroupof G ,andlet A and b beconstants.Wesaythat H is distinguishablewithconstants A and b > 0 ifthereexists some m suchthatthefollowinghold. 1. m A log 2 ( j G j ) b 2. X ( 1 ,..., m ) 24 1 (0) P 1 ( 1 ,..., m ) > 1 = 2, 44
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3. Wealsohavethefollowing: X ( 1 ,..., m ) 24 1 (1) P H ( 1 ,..., m ) > 1 = 2. Hence,wearesayingthatthereisanalgorithm f whichgivenasequenceofresults returnsitsguessofeither trivial or nontrivial ,andthisfunctioniscorrectwithmorethan probability 1 = 2 bothwhenthehiddensubgroupis H andwhenthehiddensubgroupis 1 .Thisdenitionseemstocapturetheessenceofwhatitmeans forasubgrouptobe distinguishable .MostauthorsonQuantumComputingthinkintermsofsequenc esof groups,andforthisreasontheconstants A and b neednotbeexplicitlymentioned. Denition 7 Let ( G i ) 1i =1 beasequenceofnitegroupsandlet ( H i ) 1i =1 beasequenceof non-trivialsubgroupswhere H i G i Wesaythat ( H i ) 1i =1 isdistinguishableifthereexists someconstants A and b ,suchthat H i isdistinguishablewithconstants A and b inthe previoussenseforall i 4.2AbelianGroups Usingthisdenitionwewillrstshowthatthenon-trivials ubgroupsofanabelian group G areallalgorithmicallydistinguishable. Theorem4.1. Allnon-trivialsubgroups H ofanabeliangroup G arealgorithmically distinguishable. Proof. Let G beanabeliangroupand H anynon-trivialsubgroupof G Since G is abelianallirreduciblecharactersof G arelinear. H isnormalin G andiscontainedinthe kernelsof j G = H j linearcharacters.Now P H ( )= (1) j H j < H ,1 H > j G j = 8><>: 0 H ker ( ) 1 [ G : H ] otherwise So P H m ( 1 ,.., m )= 8><>: 0 H ker ( i ) forsomei ( 1 [ G : H ] ) m otherwise 45
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Also P 1, m ( 1 ,.., m )= m Y i =1 1 j G j =( 1 j G j ) m Weset 4 ( 1 ,.., m )= 8><>: 0 P 1, m ( 1 ,.., m ) > P H m ( 1 ,.., m ) 1 P 1, m ( 1 ,.., m ) < P H m ( 1 ,.., m ) Weobservethat X ( 1 ,.., m ) 24 1 (1) P 1, m ( 1 ,.., m )=( 1 j G j ) m [( j G = H j ) m ]=( 1 j H j ) m < 1 = 2 forall m > 0. Hence X ( 1 ,.., m ) 24 1 (0) P 1, m ( 1 ,.., m ) > 1 = 2 forall m > 0. Also X ( 1 ,.., m ) 24 1 (1) P H m ( 1 ,.., m )=1 forall m > 0. So H isdistinguishable. 4.3FrobeniusGroups Inthissectionwestudythealgorithmicditinguishability ofthekernelsand complementsoftheFrobeniusgroupsdiscussedinChapter3. Wehavethefollowing lemma. Lemma4.2. If 0 < x < 1 and k > 1 then (1 x ) k > 1 kx Proof. Let f ( x )=(1 x ) k (1 kx ). Differentiatingbothsideswithrespectto x weget g ( x )= k (1 x ) k 1 + k = k [1 (1 x ) k 1 ] > 0. 46
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Hence f ( x ) isincreasingandalso f (0)=0. Hence (1 x ) k > 1 kx Theorem4.3. Let G = Z g ( ) ( ) o Z beaFrobeniusgroupwhere ( ), g ( ) arefunctions of and ( ) iscoprimeto and ( ), g ( ) !1 as !1 ThentheKernelof G is algorithmicallydistinguishableandthecomplementisnot Proof. Let H bethekernelof G Then H hasorder ( ) g ( ) Then P H ( )= (1) j H j j G j = (1) ( ) g ( ) ( ) g ( ) = (1) if H ker ( ) andisequalto 0 otherwise.Since H isinthekernelofonlylinear characterswehave P H ( )= 1 when H ker ( ) andisequalto 0 otherwise.So P H m ( 1 ,.., m )= 8><>: ( 1 ) m i (1)=1 andH ker ( i ) 0 otherwise Also P 1, m ( 1 ,.., m )= m Y i =1 i (1) 2 ( ) g ( ) Soweseethat P 1, m ( 1 ,.., m ) > P H m ( 1 ,..., m ) if P H m ( 1 ,.., m )=0. and P 1, m ( 1 ,.., m ) < P H m ( 1 ,..., m ) otherwise.Weset 4 ( 1 ,.., m )= 8><>: 1 P H m ( 1 ,.., m ) > P 1, m ( 1 ,.., m ) 0 P H m ( 1 ,.., m ) < P 1, m ( 1 ,.., m ) Now X ( 1 ,.., m ) 24 1 (1) P 1, m ( 1 ,.., m )=( 1 ( ) g ( ) ) m m < 1 = 2. 47
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Hence X ( 1 ,.., m ) 24 1 (0) P 1, m ( 1 ,.., m ) > 1 = 2 forall m > 0 and !1 Also X ( 1 ,.., m ) 24 1 (1) P H m ( 1 ,.., m )=1. Hence H isalgorithmicallydistinguishable. Let H nowbetheFrobeniuscomplement.Then j H j = and P H ( )= (1) j H j j G j < H ,1 H > = 8>>>><>>>>: j H j j G j =1 G (1) 2 j G j (1)= 0 otherwise = 8>>>><>>>>: 1 g ( ) =1 G g ( ) (1)= 0 otherwise = 8><>: (1) g ( ) =1 G (1)= 0 otherwise So P H k ( 1 ,.., k )= 8><>: Q ki =1 i (1) g ( ) i =1 G i (1)= 0 otherwise Also P 1, k ( 1 ,.., k )= k Y i =1 i (1) 2 g ( ) Set 4 ( 1 ,.., k )= 8>>>><>>>>: 0 P 1, k ( 1 ,.., k )= P H k ( 1 ,.., k ) 0 P 1, k ( 1 ,.., k ) > P H k ( 1 ,.., k ) 1 P 1, k ( 1 ,.., k ) < P H k ( 1 ,.., k ) 48
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Wenoticethat X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k )= X ( 1 ,.., k ) 24 1 (1) k Y i =1 i (1) g ( ) = 1 ( g ( ) ) k [(1+( g ( ) 1)) k ( g ( ) 1) k ]=[1 (1 1 g ( ) ) k ]. Nowsince k < c lg( j G j ) b where c > 0 and b > 0, forlarge wehave k ( ) g ( ) < 1 = 2. BythepreviousLemmaweget (1 1 g ( ) ) k > 1 k 1 1 ( ) g ( ) > 1 = 2. Thisinturnimpliesthat X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k ) < 1 = 2. Hence H isnotalgorithmicallydistinguishable. Corollary4.4. Let G = Z p o Z 2 beaFrobeniusgroup.Thekernelof G isalgorithmically distinguishableandthecomplementisnot. Corollary4.5. Let G = Z p 1 2 o Z p beaFrobeniusgroup.Thekernelof G isalgorithmicallydistinguishableandthecomplementisnot. Theorem4.6. Let G = Z g ( i ) o Z m ( i ) o Z n ( i ) beaFrobeniusgroupwhere ( m ( i ), n ( i ))=1 and iscoprimeto m ( i ) and n ( i ). Wealsoassume m ( i ), n ( i ), g ( i ) !1 as i !1 Thenthekernelof G isalgorithmicallydistinguishablebutthecomplementisn ot. Proof. If G = Z g ( i ) o Z m ( i ) o Z n ( i ) and j H j = g ( i ) Now P H ( )= (1) j H j j G j < H ,1 H > 49
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Also P 1 ( )= (1) 2 j G j So P H ( )= 8><>: (1) 2 m ( i ) n ( i ) H Ker ( ) 0 otherwise Hence P H m ( 1 ,.., k )= 8><>: Q ki =1 i (1) 2 m ( i ) n ( i ) H Ker ( i ) 0 otherwise Also P 1, k ( 1 ,.., k )= k Y i =1 i (1) 2 g ( i ) m ( i ) n ( i ) Set 4 ( 1 ,.., k )= 8><>: 0 P 1, k ( 1 ,.., k ) > P H m ( 1 ,.., k ) 1 P 1, k ( 1 ,.., k ) < P H m ( 1 ,.., k ) So X ( 1 ,.., k ) 24 1 (0) P 1, k ( 1 ,.., k )= 1 ( g ( i ) m ( i ) n ( i )) k X ( 1 ,.., k ) 24 1 (0) k Y i =1 i (1) 2 1 ( g ( i ) m ( i ) n ( i )) k X ( 1 ,.., k ) 24 1 (0) m ( i ) 2 k n ( i ) 2 k = m ( i ) k n ( i ) k g ( i ) k ( g ( i ) 1) k m ( i ) k n ( i ) k =(1 1 g ( i ) ) k > 1 = 2 as i !1 Also X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k ) > 1 = 2. Hence H isdistinguishable. If j H j = m ( i ) n ( i ) then P H ( )= (1) j H j j G j < H ,1 H > = 8>>>><>>>>: j H j j G j =1 G (1) 2 j G j (1)= m ( i ) n ( i ) 0 otherwise 50
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= 8>>>><>>>>: 1 g ( i ) =1 G m ( i ) n ( i ) g ( i ) (1)= m ( i ) n ( i ) 0 otherwise = 8><>: (1) g ( i ) =1 G (1)= m ( i ) n ( i ) 0 otherwise So P H k ( 1 ,.., k )= 8><>: Q ki =1 (1) g ( i ) =1 G (1)= m ( i ) n ( i ) 0 otherwise Also P 1, k ( 1 ,.., k )= k Y i =1 (1) 2 g ( i ) m ( i ) n ( i ) Set 4 ( 1 ,.., k )= 8>>>><>>>>: 0 P 1, k ( 1 ,.., k )= P H k ( 1 ,.., k ) 0 P 1, k ( 1 ,.., k ) > P H k ( 1 ,.., k ) 1 P 1, k ( 1 ,.., k ) < P H k ( 1 ,.., k ) Wenoticethat X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k )= X ( 1 ,.., k ) 24 1 (1) k Y i =1 i (1) g ( i ) = 1 ( g ( i ) ) k [(1+( g ( i ) 1)) k ( g ( i ) 1) k ]=[1 (1 1 g ( i ) ) k ]. Theaboveexpressionislessthan 1 = 2 for i !1 andsmall k So H isnotalgorithmically distinguishable. Theorem4.7. Let G = Z g ( i ) o ( SL (2,5) Z m ( i ) o Z n ( i ) ) beaFrobeniusgroupwhere i 2 N ( m ( i ), n ( i ))=1 and > 1 isaconstantandcoprimeto m ( i ), n ( i ). Wealsoassumethat m ( i ), n ( i ), g ( i ) !1 as i !1 ThenthekernelofthisFrobeniusgroupisalgorithmically distinguishablebutthecomplementisnot. 51
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Proof. Let H betheFrobeniuskernel.So j H j = g ( i ) Now P H ( )= (1) j H j j G j < H ,1 H > Also P 1 ( )= (1) 2 j G j So P H ( )= 8><>: (1) 2 120 m ( i ) n ( i ) H Ker ( ) 0 otherwise So P H m ( 1 ,.., k )= 8><>: Q ki =1 i (1) 2 120 m ( i ) n ( i ) H Ker ( i ) 0 otherwise Also P 1, k ( 1 ,.., k )= k Y i =1 i (1) 2 120 g ( i ) m ( i ) n ( i ) Set 4 ( 1 ,.., k )= 8><>: 0 P 1, k ( 1 ,.., k ) > P H m ( 1 ,.., k ) 1 P 1, k ( 1 ,.., k ) < P H m ( 1 ,.., k ) So X ( 1 ,.., k ) 24 1 (0) P 1, k ( 1 ,.., k )= 1 (120 g ( i ) m ( i ) n ( i )) k X ( 1 ,.., k ) 24 1 (0) k Y i =1 i (1) 2 1 (120 g ( i ) m ( i ) n ( i )) k X ( 1 ,.., k ) 24 1 (0) m ( i ) 2 k n ( i ) 2 k = m ( i ) k n ( i ) k 120 k g ( i ) k ( g ( i ) 1) k 120 k m ( i ) k n ( i ) k =( 1 120 1 120 g ( i ) ) k > 1 = 2 as i !1 Also X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k ) > 1 = 2. Hence H isalgorithmicallydistinguishable. 52
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If j H j =120 m ( i ) n ( i ) then P H ( )= (1) j H j j G j < H ,1 H > = 8>>>><>>>>: j H j j G j =1 G (1) 2 j G j (1)=120 m ( i ) n ( i ) 0 otherwise = 8>>>><>>>>: 1 g ( i ) =1 G 120 m ( i ) n ( i ) g ( i ) (1)=120 m ( i ) n ( i ) 0 otherwise = 8><>: (1) g ( i ) =1 G (1)=120 m ( i ) n ( i ) 0 otherwise So P H k ( 1 ,.., k )= 8><>: Q ki =1 (1) g ( i ) =1 G (1)=120 m ( i ) n ( i ) 0 otherwise Also P 1, k ( 1 ,.., k )= k Y i =1 (1) 2 120 g ( i ) m ( i ) n ( i ) Set 4 ( 1 ,.., k )= 8>>>><>>>>: 0 P 1, k ( 1 ,.., k )= P H k ( 1 ,.., k ) 0 P 1, k ( 1 ,.., k ) > P H k ( 1 ,.., k ) 1 P 1, k ( 1 ,.., k ) < P H k ( 1 ,.., k ) Wenoticethat X ( 1 ,.., k ) 24 1 (1) P H k ( 1 ,.., k )= X ( 1 ,.., k ) 24 1 (1) k Y i =1 i (1) g ( i ) = 1 ( g ( i ) ) k [(1+( g ( i ) 1)) k ( g ( i ) 1) k ]=[1 (1 1 g ( i ) ) k ]. Thisexpressionislessthan 1 = 2 for i !1 andsmall k Hence H isnotalgorithmically distinguishable. 53
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Wenotethatinallthecaseswehavelookedatsofar,Kempe-Sh alevdistinguishability coincideswithAlgorithmicdistinguishability. 54
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CHAPTER5 SOMECALCULATIONS EventhoughKempe-ShalevdistinguishabilityandAlgorith micdistinguishability coincideinthecasesdescribedinChapter 3 andChapter 4 thesetwoconceptsappear tobedifferent.Thefollowingisanexamplewheretheyseemn ottocoincide. Herewepicksomepositiveinteger n andset G = S 3 ... S 3 where S 3 isthe symmetricgrouponthreelettersandthereareexactly n copiesofitintheproduct.We x t tobeanelementoforder 3 in S 3 andwelet H bethesubgroupof G generatedby ( t t ,..., t ). Then H hasorder 3. Now Irr ( S 3 )= f 1 S 3 sgn g where 1 S 3 istheprincipalcharacter, sgn isthesign character,and istheuniquenon-linearcharacter.Both 1 S 3 and sgn arelinear,and (1)=2. Furthermore,wenoticethatthevaluesofthesecharacterso n t areasfollows. Bothlinearcharactershavevalue 1 on t and ( t )= 1. Now Irr ( G ) canbethoughtof asthecartesianproductof n copiesof Irr ( S 3 ). Hence j Irr ( G ) j =3 n andthedegreesof theseirreduciblecharactersareoftheform 2 i forsome i with 0 i n Anycharacterof degree 2 i canbeobtaineduniquelybychoosing i locationsoutof n ( wherethecharacter willbetakentobe ) andthenchoosingoneorotherofthelinearcharactersforal lthe otherlocations.Hence,thereareexactly n i 2 n i charactersin Irr ( G ) ofdegree 2 i 5.1Kempe-Shalevdistinguishability UsingthedenitionofKempe-Shalevdistinguishabilitywe seethat D H = 1 6 n X d j X h 6 = e ( h ) j = 1 6 n [ X d = linear j X h 6 = e ( h ) j + X = non linear d j X h 6 = e ( h ) j ]= 1 6 n [2 n 2+2( j 1 1 j n 1 2 n 1 )+4( j 1+1 j n 2 2 n 2 )+..+2 n nn ]= 1 6 n [2 n +1 +2 n +1 [ n1 +...+ nn ]]= 1 6 n [2 n +1 (1+(2 n 1))]= 2 n +1 2 n 6 n =( 2 3 ) n 2 < ( n log(6)) c forall c > 0 as n !1 55
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Hence H isKempe-Shalevindistinguishable. 5.2AlgorithmicDistinguishability Let 2 Irr ( G ) have (1)=2 i Then (( t t ,.., t ))=( 1) i Itfollowsthat P H ,1 ( )= 2 i 3 6 n 1 3 (2 i +2( 1) i ). Furthermore,wehave P 1,1 ( )= (1) 6 n < 1 ,1 1 > = (1) 2 6 n Fixsomepositiveinteger m Theabovecomputationyieldsthat,forevery ( 1 ,.., m ) 2 Irr ( G ) m wehave P H m ( 1 ,.., m )= m Y i =0 2 i 6 n (2 i +2( 1) i ), and P 1, m ( 1 ,.., m )= m Y i =0 2 2 i 6 n for i =1,.., m Itthenfollowsthat P H m ( 1 ,.., m )= P 1, m ( 1 ,.., m ) m Y i =0 (1+ ( 1) i 2 i 1 ). Inviewofthesecalculations,itisnaturaltodeneafuncti on 4 : Irr ( G ) m !f 0,1 g byfor ( 1 ,.., m ) 2 Irr ( G ), weset 4 ( 1 ,.., m )= 8>>>><>>>>: 0 P 1, m ( 1 ,.., m )= P H m ( 1 ,.., m ) 0 P 1, m ( 1 ,.., k ) > P H k ( 1 ,.., m ) 1 P 1, m ( 1 ,.., k ) < P H k ( 1 ,.., m ) Usingthisfunction f theprobabilityofitgivingthecorrectanswerwhenthehidd en subgroupis 1 is X ( 1 ,.., m ) 24 1 (0) P 1 ( 1 ,.., m ) 56
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andtheprobabilityofitgivingthecorrectanswerwhentheh iddensubgroupis H is X ( 1 ,.., m ) 24 1 (1) P H ( 1 ,.., m ). Thesenumberscanbecomputedforsmallvaluesof n and m using GAP Hereisa GAP programthatcalculatestheseprobabilities. C_1:=function(n,m)localprob,tuples,reltuples,pp,a,s;prob:=0;reltuples:=[];tuples:=Tuples([0..n],m);foraintuplesdopp:=Product(a,x->(1+(-1)^x/(2^(x-1))));ifpp<=1thenAppend(reltuples,[a]);fi;od;forainreltuplesdos:=Product(a,x->Binomial(n,x));s:=s*Product(a,x->2^(n-x));s:=s*Product(a,x->2^(2*x)/(6^n));prob:=prob+s;od;returnprob;end;C_2:=function(n,m)localprob,tuples,reltuples,pp,a,s;prob:=0;reltuples:=[];tuples:=Tuples([0..n],m);foraintuplesdopp:=Product(a,x->(1+(-1)^x/(2^(x-1)))); 57
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ifpp>1thenAppend(reltuples,[a]);fi;od;forainreltuplesdos:=Product(a,x->Binomial(n,x));s:=s*Product(a,x->2^(n-x));s:=s*Product(a,x->2^(x)*(2^x+2*(-1)^x)/(6^n));prob:=prob+s;od;returnprob;end; Thecomputationappearstoshowthatthesubgroup H isdistinguishableaccording toourdenitioninthecaseswetried.Forsomecomputations seechartsinappendixB. 58
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APPENDIXA TABLESFORCHAPTER4 HerearethetablesforChapter 4. TableA-1. H = SY 1 TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 3 i 0 0 0 0 0 Z 0 11 3 i 1 0 11 2 i 11 2 i 11 3 i SY 1 0 0 1 0 0 0 SY 2 0 0 0 1 2 i 0 0 SY 3 0 0 0 0 1 2 i 0 H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 0 0 0 0 0 1 3 i TableA-2. H = SY 1 Intermediate TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 1 3 i 1 1 1 1 11 3 i Z 0 0 0 0 0 0 0 SY 1 11 3 i 1 0 11 2 i 11 2 i 11 3 i SY 2 11 3 i 1 0 1 1 2 i 1 1 2 i 1 1 3 i SY 3 11 3 i 1 0 1 1 2 i 1 1 2 i 1 1 3 i H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 1 1 1 1 1 1 1 3 i 59
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TableA-3. H = SY 1 Final H = SY 1 TZ 0 1 3 i (1 1 3 i ) Z 0 0 SY 1 (1 1 3 i ) 2 (1 1 2 i + 1 3.2 2 i ) SY 2 1 2 (1 1 3 i ) 2 (1 1 2 i ) 1 2 i + 1 3 (1 1 3 i ) 2 1 2 2 i SY 3 1 2 (1 1 3 i ) 2 (1 1 2 i ) 1 2 i + 1 3 (1 1 3 i ) 2 1 2 2 i H t ( k ) 0 H ( k ) 0 G 1 3 i TableA-4. H = TZ 0 TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 0 0 0 0 11 2 i Z 0 0 1 11 2 i 11 2 i 11 2 i 0 SY 1 0 0 1 2 i 0 0 0 SY 2 0 0 0 1 2 i 0 0 SY 3 0 0 0 0 1 2 i 0 H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 0 0 0 0 0 1 2 i 60
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TableA-5. H = TZ 0 Intermediate TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 0 1 1 1 1 0 Z 0 0 0 0 0 0 0 SY 1 0 1 1 1 2 i 1 1 2 i 1 1 2 i 0 SY 2 0 1 1 1 2 i 1 1 2 i 1 1 2 i 0 SY 3 0 1 1 1 2 i 1 1 2 i 1 1 2 i 0 H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 1 1 1 1 1 1 1 2 i TableA-6. H = TZ 0 Final H = TZ 0 TZ 0 (1 1 2 i ) Z 0 0 SY 1 0 SY 2 0 SY 3 0 H t ( k ) 0 H ( k ) 0 G 1 2 i 61
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TableA-7. H = Z 0 TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 3 i 0 0 0 0 1 3 i 1 6 i Z 0 11 3 i 1 1 1 2 i 11 2 i 11 2 i 11 3 i SY 1 0 0 1 2 i 0 0 0 SY 2 0 0 0 1 2 i 0 0 SY 3 0 0 0 0 1 2 i 0 H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 0 0 0 0 0 1 6 i TableA-8. H = Z 0 Intermediate TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 1 3 i 1 1 1 1 1 1 3 i Z 0 0 0 0 0 0 0 SY 1 11 3 i 1 1 1 2 i 1 1 2 i 1 1 2 i 1 1 3 i SY 2 11 3 i 1 1 1 2 i 1 1 2 i 1 1 2 i 1 1 3 i SY 3 11 3 i 1 1 1 2 i 1 1 2 i 1 1 2 i 1 1 3 i H t ( k ) 0 0 0 0 0 0 H ( k ) 0 0 0 0 0 0 G 1 1 1 1 1 1 1 6 i 62
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TableA-9. FinalProbabilities H = Z 0 TZ 0 1 3 i (2 1 2 i 1 3 i ) Z 0 (1 1 3 i )(1 1 2 i ) 3 (1 1 3 i ) SY 1 (1 1 3 i ) 2 (1 1 2 i ) 2 1 2 i +(1 1 3 i ) 2 (1 1 2 i ) 1 2 2 i + 1 3 (1 1 3 i ) 2 1 2 3 i SY 2 (1 1 3 i ) 2 (1 1 2 i ) 2 1 2 i +(1 1 3 i ) 2 (1 1 2 i ) 1 2 2 i + 1 3 (1 1 3 i ) 2 1 2 3 i SY 3 (1 1 3 i ) 2 (1 1 2 i ) 2 1 2 i +(1 1 3 i ) 2 (1 1 2 i ) 1 2 2 i + 1 3 (1 1 3 i ) 2 1 2 3 i H t ( k ) 0 H ( k ) 0 G 1 6 i 63
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TableA-10. H = H t ( k ) TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 ( 1 2 n k 1 ) i 0 0 0 0 ( 1 2 n k ) i (2 i 1) Z 0 0 ( 1 2 n k 1 ) i ( 1 2 n k ) i (2 i 1) ( 1 2 n k ) i (2 i 1) ( 1 2 n k ) i (2 i 1) 0 SY 1 0 0 ( 1 2 n k ) i 0 0 0 SY 2 0 0 0 ( 1 2 n k ) i 0 0 SY 3 0 0 0 0 ( 1 2 n k ) i 0 H t ( k ) 11 2 i 0 0 0 0 11 2 i H t ( m ) ( 1 2 m k +1 ) i (2 i 1) 0 0 0 0 ( 1 2 m k +1 ) i (2 i 1) H ( m ) 0 ( 1 2 m k +1 ) i (2 i 1) ( 1 2 m k +1 ) i (2 i 1) ( 1 2 m k +1 ) i (2 i 1) ( 1 2 m k +1 ) i (2 i 1) 0 H ( k ) 0 11 2 i 11 2 i 11 2 i 11 2 i 0 G 0 0 0 0 0 ( 1 2 n k ) i 64
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TableA-11. H = H t ( k ) Intermediate TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 ( 1 2 n k 1 ) i 1 1 1 1 1 ( 1 2 n k 1 ) i Z 0 1 ( 1 2 n k 2 ) i 1 ( 1 2 n k 1 ) i 1 ( 1 2 n k 1 ) i 1 ( 1 2 n k 1 ) i 1 ( 1 2 n k 1 ) i 1 ( 1 2 n k 2 ) i SY 1 1 ( 1 2 n k 1 ) i 1 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k 1 ) i SY 2 1 ( 1 2 n k 1 ) i 1 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k 1 ) i SY 3 1 ( 1 2 n k 1 ) i 1 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k ) i 1 ( 1 2 n k 1 ) i H t ( k ) 0 11 2 2 i 11 2 2 i 11 2 2 i 11 2 2 i 0 H t ( m ) 1 ( 1 2 m k ) i 1( 1 2 m k +2 ) i 1( 1 2 m k +2 ) i 1( 1 2 m k +2 ) i 1( 1 2 m k +2 ) i 1( 1 2 m k ) i H ( m ) 1( 1 2 m k 1 ) i 1 ( 1 2 m k ) i 1 ( 1 2 m k ) i 1( 1 2 m k ) i 1( 1 2 m k ) i 1( 1 2 m k 1 ) i H ( k ) 0 0 0 0 0 0 G 1 1 1 1 1 1( 1 2 n k ) i 65
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TableA-12. Finalprobabilities H = H t ( k ) Z 0 (1 ( 1 2 n k 2 ) i ) 2 (1 ( 1 2 n k ) i ) 3 (1 ( 1 2 n k 2 ) i ) 2 (1 ( 1 2 n k 1 ) i ) 4 TZ 0 (1 ( 1 2 n k ) i ) (1 ( 1 2 n k 1 ) i ) 2 SY 1 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) i (1 ( 1 2 n k ) i ) 2 +(1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 2 i (1 ( 1 2 n k ) i )+ 1 3 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 3 i SY 2 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) i (1 ( 1 2 n k ) i ) 2 +(1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 2 i (1 ( 1 2 n k ) i )+ 1 3 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 3 i SY 3 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) i (1 ( 1 2 n k ) i ) 2 +(1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 2 i (1 ( 1 2 n k ) i )+ 1 3 (1 ( 1 2 n k 1 ) i ) 2 ( 1 2 n k ) 3 i H t ( k ) (1 1 2 i ) 2 (1 1 2 2 i ) 4 H t ( m ) (1 ( 1 2 m k +1 ) i ) 2 (1 ( 1 2 m + k +2 ) i ) 4 (1 ( 1 2 m k ) i ) 2 (1 ( 1 2 m + k +2 ) i ) 4 H ( k ) 0 H ( m ) (1 ( 1 2 m k 1 ) i ) 2 (1 ( 1 2 m k +1 ) i ) 5 (1 ( 1 2 m k 1 ) i ) 2 (1 ( 1 2 m k ) i ) 4 G ( 1 2 n k ) i 66
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TableA-13. H = H ( k ) TZ 0 Z 0 SY 1 SY 2 SY 3 G TZ 0 1 3 i ( 1 2 n k 1 ) i 0 0 0 0 ( 2 k n +1 3 ) i ( 2 k n 3 ) i Z 0 (1 1 3 i )( 1 2 n k 1 ) i (1 = 2 n k 1 ) i ( 1 2 n k ) i (2 i 1) ( 1 2 n k ) i (2 i 1) ( 1 2 n k ) i (2 i 1) (2 k n +1 ) i ( 2 k n +1 3 ) i SY 1 0 0 (1 = 2 n k ) i 0 0 0 SY 2 0 0 0 ( 1 2 n k ) i 0 0 SY 3 0 0 0 0 ( 1 2 n k ) i 0 H t ( k ) 1 3 i (1 1 2 i ) 0 0 0 0 1 3 i (1 ( 1 2 ) i H t ( m ) 1 3 i (2 i 1)( 1 2 m k +1 ) i 0 0 0 0 ( 2 k m 3 ) i ( 2 k m 1 3 ) i H ( m ) (1 1 3 i )(2 i 1)( 1 2 m k +1 ) i 1 (2 m k ) i (1 1 2 i ) 1 (2 m k +1 ) i (2 i 1) ( 1 2 m k +1 ) i (2 i 1) ( 1 2 m k +1 ) i (2 i 1) (2 k m 1 ) i (1 1 3 i )(2 i 1) H ( k ) (1 1 3 i )(1 1 2 i ) (1 1 2 i ) (1 1 2 i ) (1 1 2 i ) 11 2 i (1 1 3 i )(1 1 2 i ) G 0 0 0 0 0 1 3 i ( 1 2 n k ) i 67
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TableA-14. H = H ( k ) Intermediate TZ0 Z0 SY1 SY2 SY3 G TZ0 1-1 3 i(1 2 n k 1)i 1 1 1 1 1 1 3 i(1 2 n k 1)i Z0 1 1 3 i(1 2 n k 2)i 1 (1 = 2n k 1)i 1 (1 2 n k 1)i 1 (1 2 n k 1)i 1 (1 2 n k 1)i 1 (2 k n +2 3)i SY1 1 1 3 i(1 2 n k 1)i 1 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 2 n k 2)i 1 3 i SY2 1 1 3 i(1 2 n k 1)i 1 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 2 n k 2)i 1 3 i SY3 1 1 3 i(1 2 n k 1)i 1 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 = 2n k)i 1 (1 2 n k 2)i 1 3 i Ht( k ) (1 1 3 i)(1 1 2 2 i) 1 1 2 2 i 1 1 2 2 i 1 1 2 2 i 1 1 2 2 i (1 1 2 2 i)(1 1 3 i) Ht( m ) 1 3 i(1 2 m k)i(1 1 4 i+3 i 4 i)+1 1 (1 2 m k +2)i) 1 (1 2 m k +2)i) 1 (1 2 m k +2)i) 1 (1 2 m k +2)i) 1 3 i(1 2 m k)i(1 1 4 i+3 i 4 i)+1 H ( m ) 1 3 i(1 (1 2 m k 1)i)+(1 1 3 i)(1 (1 2 m k)i) 1 (1 2 m k)i 1 (1 2 m k)i 1 (1 2 m k)i 1 (1 2 m k)i (1 2 m k)i(1+2 i 3 i1 3 i)+1 H ( k ) 0 0 0 0 0 0 G 1 1 1 1 1 1-1 3 i(1 2 n k)i 68
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TableA-15. Final H = H ( k ) Z0 (1 (2 k n +2 3)i+(2k n +1)i (2 k n +1 3)i)2 (1 (2 k n +2 3)i)2(1 (1 2 k n +1)i)4 TZ0 (1 (2 k n 3)i) (1 (2 k n +1 3)i)2 SY1 (1 (2 k n +1 3)i)2(1 2 n k)i(1+(1 2 n k)2 i (1 2 n k)i) SY2 (1 (2 k n +1 3)i)2(1 2 n k)i(1+(1 2 n k)2 i (1 2 n k)i) SY3 (1 (2 k n +1 3)i)2(1 2 n k)i(1+(1 2 n k)2 i (1 2 n k)i) Ht( k ) (1+1 2 i(1 6 i1 2 i1 3 i))2(1 1 2 2 i)4 (1 1 3 i)2(1 1 2 2 i)6 Ht( m ) (1 (1 2 m k +2)i)4((1 (2 k m 1 3)i+(2 k m 2 3)i (2k m 2)i)2 (1 (2 k m 3)i(1 1 4 i+3 i 4 i))2) H ( k ) (1 1 3 i)2(1 1 2 i)6 H ( m ) (1 (1 2 m k 1)i (2 k m +1 3)i+(2 k m 1 3)i)2(1 (1 2 m k +1)i)4 (1 (1 2 m k)i)4(1 (1 2 m k)i(1+2 i 3 i1 3 i))2 G (1 2 n k)i 69
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APPENDIXB TABLEFORCHAPTER5 TableB-1.Tablefor P P 1, m ( 1 ,.., m ) C 1 n m sum 4 3 > 1 = 2 5 4 > 1 = 2 6 5 > 1 = 2 7 6 > 1 = 2 TableB-2.Tablefor P P H m ( 1 ,.., m ) C 2 n m sum 4 3 > 1 = 2 5 4 > 1 = 2 6 5 > 1 = 2 7 6 > 1 = 2 70
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BIOGRAPHICALSKETCH AnalesDebhaumikwasbornintheyear1972inCalcutta,India .Hegraduated withbachelor'sinmathematicsfromCalcuttaUniversity.H ealsograduatedwitha master'sdegreeinappliedmathematicsfromCalcuttaUnive rsityandamaster'sdegree incomputerapplicationsfromBangaloreUniversity,India .HecametoUnitedStatesin 2003asaPhDstudentinmathematicsDepartmentofUniversit yofFlorida.Hereceived M.SinmathematicsfromUniversityofFloridaintheyear200 5.HegraduatedwithPhD inMay2010.Hisresearchinterestisnitegrouptheory. 73
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