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Efficient Fourier Transforms on Hexagonal Arrays

Permanent Link: http://ufdc.ufl.edu/UFE0017518/00001

Material Information

Title: Efficient Fourier Transforms on Hexagonal Arrays
Physical Description: 1 online resource (147 p.)
Language: english
Creator: Zheng, Xiqiang
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007

Subjects

Subjects / Keywords: array, dimension, fourier, fractal, grid, hexagonal, lattices, pyxis, tiling, transform
Mathematics -- Dissertations, Academic -- UF
Genre: Mathematics thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

Notes

Abstract: A main concern of my research is the discrete Fourier transform (DFT) on two sequences of arrays, each of which consists of a finite number of lattice points (pixels) on a hexagonal grid. There are efficient addressing schemes for these arrays that allow for zooming in and out on an image in a hexagonal grid to view fine image details or global image features. We consider the formulation and the efficient computation of the DFT on those arrays. Some related problems such as the arithmetic for the labels of those lattice points are studied as well. Each array in the first sequence consists of all lattice points of a hexagonal grid enclosed in a regular hexagon and has the same axes of symmetry as the enclosing hexagon. It is shown that the DFT on such an array is amenable to a standard Fast Fourier Transform and can be computed as a one dimensional DFT. We also provide an efficient method for evaluating the DFT of a function defined on that array based on the corresponding one dimensional standard DFT. The second sequence is called a Pyxis structure, which originated with Pyxis Innovation Inc. to create an efficient sampling scheme for the earth. Each lattice point in the $n^{th}$ array of the Pyxis structure is assigned a special label for quick data retrieval. We provide a recursive definition of the Pyxis structure, and show how such a label is assigned based on a certain unique algebraic representation of the corresponding lattice point. Also, we implement an efficient algorithm to determine the label of the vector sum of any two lattice points whose labels are given. The recursive definition and algebraic labeling scheme is used to show that, for any integer $n > 2$, the DFT on the $n^{th}$ array of the Pyxis structure is not amenable to any standard DFT. Furthermore, the fractal dimension of the limit boundary of the Pyxis structure is shown to be $\frac {\ln{4}}{\ln{3}}$.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Xiqiang Zheng.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Local: Adviser: Vince, Andrew.
Local: Co-adviser: Ritter, Gerhard.
Electronic Access: RESTRICTED TO UF STUDENTS, STAFF, FACULTY, AND ON-CAMPUS USE UNTIL 2008-12-31

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Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0017518:00001

Permanent Link: http://ufdc.ufl.edu/UFE0017518/00001

Material Information

Title: Efficient Fourier Transforms on Hexagonal Arrays
Physical Description: 1 online resource (147 p.)
Language: english
Creator: Zheng, Xiqiang
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007

Subjects

Subjects / Keywords: array, dimension, fourier, fractal, grid, hexagonal, lattices, pyxis, tiling, transform
Mathematics -- Dissertations, Academic -- UF
Genre: Mathematics thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

Notes

Abstract: A main concern of my research is the discrete Fourier transform (DFT) on two sequences of arrays, each of which consists of a finite number of lattice points (pixels) on a hexagonal grid. There are efficient addressing schemes for these arrays that allow for zooming in and out on an image in a hexagonal grid to view fine image details or global image features. We consider the formulation and the efficient computation of the DFT on those arrays. Some related problems such as the arithmetic for the labels of those lattice points are studied as well. Each array in the first sequence consists of all lattice points of a hexagonal grid enclosed in a regular hexagon and has the same axes of symmetry as the enclosing hexagon. It is shown that the DFT on such an array is amenable to a standard Fast Fourier Transform and can be computed as a one dimensional DFT. We also provide an efficient method for evaluating the DFT of a function defined on that array based on the corresponding one dimensional standard DFT. The second sequence is called a Pyxis structure, which originated with Pyxis Innovation Inc. to create an efficient sampling scheme for the earth. Each lattice point in the $n^{th}$ array of the Pyxis structure is assigned a special label for quick data retrieval. We provide a recursive definition of the Pyxis structure, and show how such a label is assigned based on a certain unique algebraic representation of the corresponding lattice point. Also, we implement an efficient algorithm to determine the label of the vector sum of any two lattice points whose labels are given. The recursive definition and algebraic labeling scheme is used to show that, for any integer $n > 2$, the DFT on the $n^{th}$ array of the Pyxis structure is not amenable to any standard DFT. Furthermore, the fractal dimension of the limit boundary of the Pyxis structure is shown to be $\frac {\ln{4}}{\ln{3}}$.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Xiqiang Zheng.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Local: Adviser: Vince, Andrew.
Local: Co-adviser: Ritter, Gerhard.
Electronic Access: RESTRICTED TO UF STUDENTS, STAFF, FACULTY, AND ON-CAMPUS USE UNTIL 2008-12-31

Record Information

Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0017518:00001


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Thoughthisresearchisanindividualwork,Icouldneverhavereachedtheheightsorexploredthedepthswithoutthehelp,support,guidanceandeortofmanypeople.Thankmyadvisors,Dr.AndrewVinceandDr.GerhardX.Ritter,fortheirinvaluablesupervisionandforthegenerousamountoftimetheyspentonthisresearch.Throughoutmydoctoralworktheyencouragedmetodevelopindependentthinkingandresearchskills.Inparticular,Dr.Vinceprovidedimportantinsight,applyingalgebraicandcombinatorialtechniquestosimplifymanyproofs.Dr.GerhardX.Ritterintroducedmetothisexcitingandchallengingresearcharea,andguidedthisresearchuntilthetimeofhissurgery.Ialsothanktheothermembersofmycommittee:Dr.DavidC.Wilson,Dr.TimOlson,andDr.JosephN.Wilsonfortheirhelpfuldiscussionsandencouragement.Dr.DavidWilsonregularlyattendedresearchmeetingsanddiscussions,andprovidedmanyimportantobservations.Dr.Olsongavemanyusefulsuggestions.IexpressmyappreciationtotheDepartmentofMathematicsforoeringmeafullteachingassistantshipandaGrinterfellowship,andtotheDepartmentofComputerandInformationScienceandEngineering(CISE)foroeringmeapartialresearchassistantship.AlsomyappreciationgoestoMrs.JaneSmithandMrs.RonnieKhuri,fortheenjoymentofteachingwiththem,andtothestaofboththeMathematicsDepartmentandtheCISEDepartmentfortheirkindhelpwheneverneeded.Furthermore,thanksgotoPyxisInnovationInc.fortheirnancialsupportofthisresearch,andtothepeopletherefortheirfriendshipandhelp.SpecialthankstoDr.CharlesHerringforservingasthementorofthisresearchprojectandforhisimportantreferences.Iammostgratefultomywife,LihuaYang,forherlove,patienceandencouragementduringtheseyearsofmygraduatestudy. 4

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page ACKNOWLEDGMENTS ................................. 4 LISTOFFIGURES .................................... 7 ABSTRACT ........................................ 10 CHAPTER 1INTRODUCTION .................................. 12 2DISCRETEFOURIERTRANSFORM(DFT) ................... 19 2.1DiscreteFourierTransformontheQuotientGroupofTwoLattices .... 19 2.2ConvolutionandCorrelation .......................... 22 2.3DFTonaLatticeandtheCorrespondingPeriodicityMatrix ........ 24 2.4RelationBetweentheDFTandtheContinuousFourierTransform ..... 25 3DFTONSOMEPREVIOUSLYSTUDIEDHEXAGONALARRAYS ...... 27 4REGULARHEXAGONALSTRUCTUREANDITSTWOSPECIALTYPES 31 4.1RegularHexagonalStructures ......................... 31 4.2TypeARegularHexagonalStructure ..................... 31 4.3TypeBRegularHexagonalStructure ..................... 36 4.4RelationBetweentheTypeAandBRHSandSomePreviouslyStudiedArrays ...................................... 44 5FASTALGORITHMSFORCOMPUTINGTHEDFTONTHETWOSPECIALTYPESOFTHERHS ................................ 46 5.1ConverttheDFTonaTileofaTilingbyTranslationsbyaSublatticetoaStandardDFT ................................ 46 5.2FastAlgorithmsfortheDFTanditsInverseontheTypeARHS ..... 48 5.3FastAlgorithmsfortheDFTanditsInverseontheTypeBRHS ..... 58 5.4ComputationalComplexityandCooley-TukeyFactorizationfortheDFTontheTypeAandTypeBRHS ....................... 59 6PYXISSTRUCTURE ................................ 62 6.1DenitionandLabelingofthePyxisStructure ................ 62 6.1.1TheDenitionofthePyxisStructure ................. 62 6.1.2TheLabelingofthePyxisStructure .................. 67 6.1.3AdditionoftheLabelsofthePyxisStructure ............. 70 6.2PyxisP(n)DoesNotTiletheUnderlyingLatticebyTranslationsbyaSublatticeforanyn>2 .................................. 91 5

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...................... 91 6.2.2PyxisP(2n)DoesNotTiletheUnderlyingLatticebyTranslationsbyaSublatticeforanyn>1 ...................... 99 7FRACTALDIMENSIONOFTHEBOUNDARYOFTHEPYXISSTRUCTURE 112 7.1TheLimitofthePyxisStructure ....................... 112 7.2FractalDimensionoftheBoundaryofthePyxisStructure ......... 114 8SUMMARYOFTHISRESEARCH ......................... 141 REFERENCES ....................................... 143 BIOGRAPHICALSKETCH ................................ 147 6

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Figure page 1-1ArraysoftypeAandtypeBRHS.a).ThethirdlevelofthetypeARHS.b).ThesecondlevelofthetypeBRHS. ........................ 13 1-2ThePyxisstructureatlevelonethroughfourwhereP(1)consistsofsevenredhexagons,P(2)consistsof13bluehexagons,P(3)consistsof55greenhexagons,andP(4)consistsof133blackhexagons.Thethreedashedvectorsshowthelabeladdition05062005=1040 .......................... 14 1-3Tessellatingsphereusinghexagonsand12pentagonsinmultiresolutions. .... 15 1-4Flattenedpolygonsusedtotessellatethesphere.(a)showsthe20hexagonsand12pentagonsinthetessellation.(b)displayseachpentagoninFigure(a)asahexagonwithoneofitssixdirectionsempty. .................... 15 1-5The(dashed)divisionlinesofthenextlevelaregeneratedfromthose(solid)divisionlinesofthepreviouslevel.(a)and(b)showthedivisionlinesnearahexagonandapentagonofthepreviouslevelrespectively. ............ 16 1-6ThegreenpolygonsobtainedfromthedivisionofthesphereatleveloneusingtheschemeofFigure1-5. ............................... 16 1-7TheredpolygonsobtainedfromthedivisionofthesphereatleveltwousingschemeofFigure1-5. ................................. 17 3-1TherstthreelevelsoftheGBT2aggregates. ................... 27 3-2ArraysstudiedinEhrhardt[ 12 ],andSunandYao[ 40 ].(a)Hexagonalsamplingofarectangularregionusedin[ 12 ]andFitzandGreen[ 14 ].(b)ThearrayusedinSunandYao[ 40 ]consistingof3n2latticepoints,wheren=9. ........ 28 3-3ArraysstudiedinAnterrieuetal.[ 2 ].(a)AnarrayshowninFigure3onPage2533ofAnterrieuetal.[ 2 ].(b)ThearrayobtainedfromthearrayinFigure(a)byomittingtheboundarypointsonthetoprowandthetwouppersides.(c)ThearrayobtainedfromthearrayinFigure(a)byomittingoneofitstwoconsecutiveboundarypoints.(d)TheperiodicextensionfromthearrayinFigure(c)toanarrayofarhombusshape. .............................. 30 4-1ArraysofthetypeARHS.(a)Thelatticepointsof
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............... 44 4-5ComparisonofapreviouslystudiedarraystructurewiththetypeBRHS.(a)Thelatticepointsof3.(b)Thelatticepointsof
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........................................ 117 7-4Thedistancefromapointxtoagivenlatticepointyandthedistancefromxtothesixneighborsofyinthelattice. ....................... 120 7-5ThecontainmentrelationamongVoronoicellsofP(2n)andP(2n+1).(a)showsthatthe(blue)VoronoicellV2n(y)iscontainedinthe(black)VoronoicellV2n1(y),whereV2n(y)hasahorizontalsideandV2n1(y)hasaverticalside.(b)showsthaty2P(2n)isthecentroidofthetrianglewithverticesq2P(2n1),r2P(2n1),andt2P(2n1).Italsoshowsthatthe(blue)Voronoicellofy2L2niscontainedintheunionofthe(black)Voronoicellsofq2L2n1,r2L2n1,andt2L2n1. ..................................... 121 7-6ThegreenVoronoicellV2n+1(y)iscontainedintheblackVoronoicellV2n1(y0). 124 7-7ForanytwohexagonsofP(2k+2)eachhavingboundaryindex2or4suchthatthedistancebetweenthemis2k+2,thereexistsahexagonofP(2k+2)thatisnexttobothofthem,whereQ,RandSarehexagonsofP(2k).(a)ThehexagonSisnexttoQandR,anditscentroidliesbelowthelineconnectingthecentroidsofQandR.(b)showsthat,foranytwobluehexagonsofP(2k+2)(eachhavingboundaryindex2or4)suchthatthedistancebetweenthemis2k+2,thereexistsahexagonofP(2k+2)thatisnexttothesetwobluehexagons. .......... 133 7-8Thelatticepointsq;r;s2P(2n2)andthelatticepointsinq+2n,r+2n,ands+2n,wheresisnexttobothqandrinthelatticeL2n2. ........ 138 9

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AmainconcernofmyresearchisthediscreteFouriertransform(DFT)ontwosequencesofarrays,eachofwhichconsistsofanitenumberoflatticepoints(pixels)onahexagonalgrid.Thereareecientaddressingschemesforthesearraysthatallowforzoominginandoutonanimageinahexagonalgridtoviewneimagedetailsorglobalimagefeatures.WeconsidertheformulationandtheecientcomputationoftheDFTonthosearrays.Somerelatedproblemssuchasthearithmeticforthelabelsofthoselatticepointsarestudiedaswell. Eacharrayintherstsequenceconsistsofalllatticepointsofahexagonalgridenclosedinaregularhexagonandhasthesameaxesofsymmetryastheenclosinghexagon.ItisshownthattheDFTonsuchanarrayisamenabletoastandardFastFourierTransformandcanbecomputedasaonedimensionalDFT.WealsoprovideanecientmethodforevaluatingtheDFTofafunctiondenedonthatarraybasedonthecorrespondingonedimensionalstandardDFT. ThesecondsequenceiscalledaPyxisstructure,whichoriginatedwithPyxisInnovationInc.tocreateanecientsamplingschemefortheearth.EachlatticepointinthentharrayofthePyxisstructureisassignedaspeciallabelforquickdataretrieval.WeprovidearecursivedenitionofthePyxisstructure,andshowhowsuchalabelisassignedbasedonacertainuniquealgebraicrepresentationofthecorrespondinglatticepoint.Also,weimplementanecientalgorithmtodeterminethelabelofthevectorsumofany 10

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ln3. 11

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Traditionalimageprocessingalgorithmsanddigitalimagetransformsareusuallycarriedoutonpixelsofsquaregrids.However,asshowninAllen[ 1 ],physicalpixelssuchasprinterdotsandelectronbeamsusuallyhavecircularshapesandthusoperatemoreeectivelyonhexagonalgrids,whereahexagonalgridisatessellationoftheplanebyregularhexagons.Thepixelsofhexagonalgridsalsoprovideforhigherpackingdensityofdiscsandgiveamoreaccurateapproximationofcircularregionsthanthatofsquaregrids.Furthermorethepixelsofhexagonalgridsareuniformlyconnectedinthesensethatthedistancefromagivenpixeltoanyadjacentpixelisthesame.Hence,hexagonalgridsareusedbyawidevarietyofresearchersinareassuchasimageprocessing(MiddletonandSivaswamy[ 30 ],Balasubramaniyametal.[ 3 ],andStrand[ 39 ]),computergraphics(Tytkowski[ 42 ]),geoscience(Carretal.[ 6 ])andecology(JurasinskiandBeierkuhnlein[ 20 ]).Forexample,theyarebeingusedinthesoilmoistureandoceansalinityspacemission(Anterrieuetal.[ 2 ],andCampsetal.[ 5 ]).F.MorganandR.Boltonhaveshownin[ 32 ]that,fortheeciencyofthedistributionofcentersofproduction,regularhexagonsaresuperiortoanyothercollectionofshapes. Thesetconsistingofallcentersofpixelsofahexagonalgridiscalledahexagonallattice.Ingeneral,ad-dimensionallatticeinRdisthesetofallintegerlinearcombinationsofdindependentvectors.Theelementsofalatticearecalledlatticepoints.TheVoronoicellofalatticepointofad-dimensionallatticeconsistsofthosepointsofRdwhichareclosesttothatpointthananyotherlatticepointofthelattice.BecauseoftheobviousonetoonecorrespondencebetweenVoronoicellsandlatticepointsofalattice,sometimeswetreatasetoflatticepointsasthecorrespondingsetofVoronoicellsintheplottingtogetabettervisualizationeectwhichcanbeseeninFigure1-1.Anarrayofalatticeisanitesubsetofthelatticeandanarraystructureisasequenceofarrays.Thentharrayofanarraystructureiscalledthenthlevelofthearraystructure.Forexample,thearray 12

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(a)(b) Figure1-1. ArraysoftypeAandtypeBRHS.a).ThethirdlevelofthetypeARHS.b).ThesecondlevelofthetypeBRHS. Aregularhexagonalstructure(RHS)isahexagonalarraystructuresuchthat,atanygivenlevel,theunderlyinghexagonallatticeisadisjointunionoftranslatedcopies,andthesetoflatticepointswhosecoordinatesarethesameasthecoordinatesofthosetranslationsalsoformsahexagonallattice.AnalgebraicdenitionofaRHSaswellastwoparticulartypesofRHSs,calledtypeAandtypeB,areprovidedinChapter4.EacharraywithinthetypeA(typeB)RHSconsistsofallhexagonallatticepointsenclosedinaregularhexagonandhasthesamecentroidandaxesofsymmetryastheenclosinghexagon.Figure1-1showsonearrayofeachtype.Pyxisstructure,denotedP,isahexagonalarraystructuretogetherwithanaturalmethodforlabelingthelatticepoints(orhexagons).LetP(n)denotethenthlevelofPforanyintegern0.TheprecisedenitionofPandP(n)aregiveninChapter6,andP(1)throughP(4)areshowninFigure1-2.ThenameandconceptofthePyxisstructureoriginatedwithPYXISInnovationInc.(Peterson[ 33 ]),aCanadabasedcompanywhosegoalisanecient 13

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ThePyxisstructureatlevelonethroughfourwhereP(1)consistsofsevenredhexagons,P(2)consistsof13bluehexagons,P(3)consistsof55greenhexagons,andP(4)consistsof133blackhexagons.Thethreedashedvectorsshowthelabeladdition05062005=1040 samplingschemeforthesurfaceoftheearth(Figure1-3).Thefollowingshowsthestructureofthisecientsamplingscheme.Consideraspherewhichistessellatedby20regularhexagonsand12regularpentagons.Figure1-4showsthespherewiththose32polygonsattenedontotheplane.Foreachsideofthosepolygons,makealinesegment(asthedashedbluelinesinFigure1-5)withlengthbeingequaltothelengthofthegivensidedividedbyp 14

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Tessellatingsphereusinghexagonsand12pentagonsinmultiresolutions. (a)(b) Figure1-4. Flattenedpolygonsusedtotessellatethesphere.(a)showsthe20hexagonsand12pentagonsinthetessellation.(b)displayseachpentagoninFigure(a)asahexagonwithoneofitssixdirectionsempty. appliedrecursively,thenthesphereisdividedintosmallerandsmallerpolygonalregionsbutthenumberofpentagonalregionsisalways12.Thedivisionofthespherebyapplyingsuchrecursionntimesiscalledthedivisionofthesphereatleveln.Figure1-6and1-7 15

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Figure1-5. The(dashed)divisionlinesofthenextlevelaregeneratedfromthose(solid)divisionlinesofthepreviouslevel.(a)and(b)showthedivisionlinesnearahexagonandapentagonofthepreviouslevelrespectively. Figure1-6. ThegreenpolygonsobtainedfromthedivisionofthesphereatleveloneusingtheschemeofFigure1-5. showtheattenedversionsofsuchdivisionsatleveloneandleveltwo,respectively.Foranyintegern>1,thesetofallcellsofthesphereatthenthlevelisadisjointunionof20copiesofP(n1)and12copiesofP(n)byomittingoneofitssixdirections.Forexample, 16

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TheredpolygonsobtainedfromthedivisionofthesphereatleveltwousingschemeofFigure1-5. Figure1-7showsthatthesetofallcellsofthesphereatleveltwoisadisjointunionof20copiesofP(1)whichareinblueand12copiesofP(2)whichareinredbyomittingoneofthesixdirections. Certainpropertiesofthosetwohexagonalarraystructures,inparticularthediscreteFouriertransform(DFT),arestudied.TheDFTonarraysofhexagonallatticesisanimportanttopicinhexagonalimageprocessingasexplainedinMiddletonandSivaswamy[ 30 ].WeprovideanecientmethodtocomputetheDFTonanyarrayoftypeAortypeBRHSandshowthatthecomputationalcomplexityisO(NlogN),whereNisthenumberofthelatticepointsofthearray.WealsoprovidearecursivedenitionofthePyxisstructureanduseittoshowthattheDFTasdenedinChapter2cannotbeappliedtoanyarrayofthePyxisstructurewhenthelevelislargerthantwo. AsshowninFigure1-2,thecellsofP(1)arelabeled0,1,2,...,6,andarrangedinacertainorder.ThecellsofP(2)arelabeledijwherei;j=0;1;2;:::;6andeitheriorj

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ln3. 18

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1. (Associativity)Foranyx;y;z2G,(x+y)+z=x+(y+z). 2. (Existenceofazero)Thereexistsanelement02Gsuchthat,foreachx2G,x+0=x=0+x. 3. (Existenceofaninverse)Foreachx2G,thereexistsanelementofG,denotedx,suchthatx+(x)=0=(x)+x. 4. (Commutativity)Foranyx;y2G,x+y=y+x. Inthisresearch,allgroupswillbeabelianandwewritex+(y)asxyforanytwoelementsxandyofagroupG.LetG0beanonemptysubsetofagroupG.IfG0alsoformsagroupundertheoperation+ofG,thenG0iscalledasubgroupofG.NowassumethatG0beasubgroupofagroupG.Foranyp;g2G,ifpg2G0,thenwesaythatpiscongruenttogmoduloG0,denotedbypgmodG0.Foranyp2G,letp=fu2G:upmodG0g.Obviouslyp=p+G0wherep+G0denotesthesetfp+y2G:y2G0g.ThesetpiscalledacosetofG0inG.Foranypairp;g2G,itiseasytoshowthateitherp=gorpTg=;.Denep+g= 19

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7 ].ThenotationL0LisusedtodenotethatL0isasublatticeofLwhichhasthesamedimensionasL.Theinnerproductoftwovectorsr;s2Cdisdenedashr;si=Pdj=1rjsjwhererjandsjarethejthcomponentofrandsrespectively,andrjdenotesthecomplexconjugateofrj.Ifr;s2Rd,thenhr;si=rTs=sTr.ThedualofalatticeLisL=s2Rd:hr;si2Zforallr2L. Inthisresearch,thesuperscriptmeansthedualwhenitisappliedtoalattice,andmeansthecomplexconjugatewhenitisappliedtoacomplexnumberorvector.ThecardinalityofasetSisdenotedjSj.Furthermore,fortwogroupsAandB,thesymbol=meansthatAandBareisomorphic.Thenextlemma,whoseproofappearsinZapata[ 47 ],andConwayandSloane[ 9 ],providessomeusefulpropertiesofduallattices. 1.

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3. IfLL0,thenL=L0=L0=LandhencejL=L0j=jL0=Lj. TheinverseFouriertransformisthefunctionF1:CbG!CGdenedby InthedenitionofthediscreteFouriertransform,thedomainsGandbGarecalledthespatialandfrequencydomainoftheFouriertransformFrespectively. Proof. ToprovethattheDFTisinvertible,i.e.,F1F(a)=aforanya2CG,weneedthefollowinglemma:

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2.1.2 ,h(r)iswelldened.Ifr=0,thenXs2bGe2ihr;si=Xs2bGe2ih0;si=jbGj: Proof. ByLemma 2.1.3 ,wehave ItfollowsthatF1(ba)(t)=a(t)foranyt2G.HenceF1(ba)=a. 22

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48 ]. Proof. 2.2.1 .Sinceg(s)=a(s)foranys2G,wehavebg(t)=Ps2Gg(s)e2ihs;ti=Ps2Gg(s)e2ihs;ti 23

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AmatrixViscalledasamplingmatrixofalatticeLifitscolumnsareasetofgeneratorsforL.Obviouslyinsuchacase,L=Vn:n2Zd=VZd.Itfollowseasilyfromthefollowinglemmathatthedualofahexagonallatticeisalsohexagonal. Proof. AnonsingularmatrixMwithintegerentriesiscalledaperiodicitymatrix.AsetofcosetrepresentativesofthequotientgroupZd=MZdisalsocalledasetofcosetrepresentativesassociatedwiththeperiodicitymatrixM. Proof. LetL0L,G=L=L0,bG=L0=L,andF:CG!CbGbetheDFT.IfVisasamplingmatrixofLandbVasamplingmatrixofL0,thenViscalledasamplingmatrixinthe

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2{1 forcomputingtheDFTonL=L0becomesthefollowingequationforcomputingtheDFTonP. Nowforanyr2Pands2Q,let[r]bethecoordinatesofrwithrespecttothebasisthatarethegeneratorsofL,andlet[s]bethecoordinatesofswithrespecttothebasisthatarethegeneratorsofL0.Thenr=V[r]ands=bV[s].ByLemma 2.3.2 ,thereisaperiodicitymatrixMsuchthatthesamplingmatrixofL0isVMandthesamplingmatrixofL0isbV=[(VM)T]1.LetIP=f[p]:p2PgZdandIQ=f[q]:q2QgZd.Thenhr;si=sTr=(bV[s])TV[r]=[s]TbVTV[r]=[s]T(VM)1V[r]=[s]TM1V1V[r]=[s]TM1[r].HenceEquation 2{5 becomesthefollowingequation. Byreplacing[r]and[s]withmandkrespectively,Equation 2{6 becomesthefollowingsimplerequation. 8 ]alsoholdsforthe2-dimensionalcase.Foranycomplex-valued,Lebesgueintegrablefunctionf(x)withx2R2,byPinsky[ 34 ]itsFouriertransformisthecomplex-valuedfunctionbf(y)dened 25

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LetLbea2-dimensionallattice,L0L,andPasetofcosetrepresentativesofthequotientgroupL=L0.LetePbetheunionoftheVoronoicellsofthelatticepointsofPandAtheareaofaVoronoicellofthelatticeL.Intheniteanalogue,wereplaceR2inEquation 2{8 byeP.ThecorrespondingdiscretizedversionofEquation 2{8 isthefollowingequation. NowletQbeasetofcosetrepresentativesofthequotientgroupL0=LandeQbetheunionoftheVoronoicellsofthelatticepointsofQ.SincePisasetofcosetrepresentativesofL=L0,ePtilesR2bytranslationsbythesublatticeL0.SimilarlyeQtilesR2bytranslationsbythesublatticeLofL0.WeclaimthatbfPisperiodicinthesensethatbfP(y)=bfP(y+s0)foranys02L.Sinces02Landp2L,wehavehp;s0i2Z.Thenhp;y+s0i=hp;yi+hp;s0ifollowsthate2ihp;y+s0i=e2ihp;yi.Hence,byEquation 2{9 ,bfP(y+s0)=bfP(y).ThereforebfP(y)isperiodic.BythedenitionofeQ,QisthesetofthesampledpointsofeQinthelatticeL0.InEquation 2{9 ,theexpressionPp2Pf(p)e2ihp;yiistheDFTasdenedinEquation 2{5 or 2{7 26

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The2-dimensionalgeneralizedbalancedternary,denotedGBT2,hasbeenusedinZapataandRitter[ 48 ]throughGibsonandLucas[ 17 ]fortheprocessingofhexagonallysampledimages.ThenthleveloftheGBT2,denotedGBT2(n),consistsof7nhexagons.Foranyn2N,theGBT2(n+1)istheunionof7translatedcopiesoftheGBT2(n)asshowninFigure3-1.TheGBT2isperhapsthemoststudiedexamplebecauseoftheelegantmethodofindexingitscells(latticepoints)describedinGibsonandLucas[ 17 ],Kitto[ 21 ],andMiddletonandSivaswamy[ 30 ].AperiodicitymatrixfortheGBT2(n) Figure3-1. TherstthreelevelsoftheGBT2aggregates. isBnwhereB=0B@21131CA.TheDFTontheGBT2(n)anditsfastalgorithmshavebeensuccessfullydevelopedinZapataandRitter[ 48 ],andMiddletonandSivaswamy[ 29 ].SomedecienciesoftheGBT2havebeenpointedoutbythepeopleworkinginthePyxisInnovationInc.(theshareddocumentofPyxisInnovationInc.,2003).Forexample,the 27

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TheDFTonarraysofrectangularshapeofahexagonallattice(showninFigure3-2(a))wasutilizedinEhrhardt[ 12 ],andFitzandGreen[ 14 ].Foranysucharray,thesamplingmatrixofthelatticeLis0B@11 20p 21CAandtheperiodicitymatrixisoftheformM=0B@n1n2 12 ]. (a)(b) Figure3-2. ArraysstudiedinEhrhardt[ 12 ],andSunandYao[ 40 ].(a)Hexagonalsamplingofarectangularregionusedin[ 12 ]andFitzandGreen[ 14 ].(b)ThearrayusedinSunandYao[ 40 ]consistingof3n2latticepoints,wheren=9. 28

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40 ],afastalgorithmwasdevelopedforcomputingtheDFTonanarraywhichconsistsof3n2latticepointsofahexagonallattice.Figure3-2(b)showsthisarrayforn=9wherethetoprowconsistsofnlatticepointsbutthebottomrowconsistsofn+1latticepoints.AperiodicitymatrixforthisarrayisM=0B@2nnn2n1CA. InAnterrieuetal.[ 2 ],thehexagonalarrayshowninFigure3-3(a)wasstudied.BecausethehexagonalarrayinFigure3-3(a)doesnottiletheunderlyinghexagonallatticebytranslationsbyanysublattice,theDFTonsuchanarraycannotbeformulatedusingthemethodinChapter2.HencetheauthorsinAnterrieuetal.[ 2 ]removehalfoftheboundarypointstoapplya2-dimensionalstandardDFT.AfterhalfoftheboundarypointsofthearrayinFigure3-3(a)areomittedinacertainway,wegetthearrayshowninFigure3-3(b)orFigure3-3(c),whichconsistsofn2latticepoints.ThearraysinFigure3-3(b)orFigure3-3(c)dotiletheunderlyinghexagonallatticebytranslationsbyasublattice.AperiodicitymatrixforthearrayineitherFigure3-3(b)orFigure3-3(c)isM=0B@nn0n1CA.SinceMisalsoaperiodicitymatrixoftherhombusshapedarrayshowninFigure3-3(d),theDFToneachofthearraysinFigure3-3(b)andFigure3-3(c)isthesameastheDFTonthatrhombusshapedarrayandhencecanbecomputedusingthefastalgorithmsfora2-dimensionalstandardDFT. ForanyarrayinFigure3-2(b)orFigure3-3(b),thenumberoflatticepointsonitstoprowisdierentfromthatonthebottomrow.Hencetheshapeofthosearraysisnotaregularhexagon.Inthenextchapter,wewillconsiderhexagonalarrayswhose6sideshavethesamenumberoflatticepointsandhavethesamecentroidandsymmetricaxesasaregularhexagon. 29

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(c)(d) Figure3-3. ArraysstudiedinAnterrieuetal.[ 2 ].(a)AnarrayshowninFigure3onPage2533ofAnterrieuetal.[ 2 ].(b)ThearrayobtainedfromthearrayinFigure(a)byomittingtheboundarypointsonthetoprowandthetwouppersides.(c)ThearrayobtainedfromthearrayinFigure(a)byomittingoneofitstwoconsecutiveboundarypoints.(d)TheperiodicextensionfromthearrayinFigure(c)toanarrayofarhombusshape. 30

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Ifwetaker=2nands=n(r=nands=0)inthematrixMabove,thenwegetaperiodicitymatrixforthearrayinSunandYao[ 40 ](Anterrieuetal.[ 2 ]).HencethearraysinSunandYao[ 40 ]orAnterrieuetal.[ 2 ]constituteaRHS.Inthefollowing,westudyarraysinthetypeAandtypeBRHSwhicharecloselyrelatedtothoseinSunandYao[ 40 ],andAnterrieuetal.[ 2 ]. 2p 21CA,andLA=n1(vA1)+n2(vA2):n1;n22Z: 31

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Figure4-1. ArraysofthetypeARHS.(a)Thelatticepointsof
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Proof. Foranyn2N,letgVA1;n=(2n+1)vA1+(n+1)vA2,gVA2;n=n(vA1)+(2n+1)vA2,andfLAn=nn1(gVA1;n)+n2(gVA2;n):n1;n22Zo. Proof. 4.2.3 ,themaximaldistanceofthelatticepointsof
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Proof. 4.2.5 theorderofthequotientgroupLA=fLAnisthedeterminantofMAn.ItfollowsthattheorderofthequotientgroupLA=fLAnis3n2+3n+1sinceitisthedeterminantofMAn.ByLemma 4.2.1 ,theorderof
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Proof. 4.2.6 ,wehavefVAn=(VA)(MAn).Itfollowsthat[(fVAn)T]1=[(VA)T]1[(MAn)T]1.Hence[(VA)T]1=[(fVAn)T]1(MAn)T.SinceVAisasamplingmatrixofthelatticeLA,[(VA)T]1isasamplingmatrixoftheduallattice(LA).Forasimilarreason,[(fVAn)T]1isasamplingmatrixoftheduallattice(fLAn).Since(MAn)T=0B@2n+1n+1n2n+11CA,byaproofsimilartotheproofofTheorem 4.2.6 itcanbeshownthatc<>:0B@jk1CA2Z2:j(vA1)+k(vA2)2=>;: 35

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4.2.6 and 4.2.7 21 21CA,vB2=0B@p 21 21CAandLB=n1(vB1)+n2(vB2):n1;n22Z: 21 21CAandvB2=0B@p 21 21CA.Theset0,wedene
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Inthefollowing,wearegoingtoshowthat
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(b) Figure4-2. ArraysofthetypeBRHS.(a)Thelatticepointsof
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Since~s=s1+s2=(3q1+r1)+(3q2+r2)=3(q1+q2)+(r1+r2)=3(q1+q2)+3+~r=3(q1+q2+1)+~r=3~q+~r,wehave~r=mod(~s;3)and~q=~s~r 4{2 andInequality 4{3 ,a1+a22r10andhencer21.Let~q=q1q21and~r=3+r1r2.Then2nr2+q12q22nr2+q1q22n1+q1q2=2n+~q,and2n+r1+2q1q22n+r1+2q1q2(q22r2)=2n+~x+2~q.Obviouslys1s2=3(q1q2)+(r1r2)=3(q1q21)+(3+r1r2)=3~q+~r.ThenbyInequality 4{4 itfollowsthat2n+~r+2~q~x2n+~q.Hence,bythedenitionof
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Proof. 4.3.2 wehavethata1+a22
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LetVBbethematrixwhosetwocolumnsarevB1andvB2.Foranyintegern>0,letMBn=0B@3n+13n13n+11CA,fVBn=(VB)(MBn),andfLBnbethelatticegeneratedbythetwocolumnvectorsoffVBn.ThefollowinglemmawillbeusedinTheorem 4.3.7 Proof. SincefLBnisahexagonallatticegeneratedbyfb1;b2gandtheanglebetweenb1andb2is

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4.3.5 ,themaximaldistanceofthelatticepointsof
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4.3.7 thattheDFTon0,letd<>:0B@jk1CA2Z2:j(vB1)+k(vB2)2=>;: 4.3.7 and 4.3.8

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40 ](alsoFigure3-2(b)).Ifwedeletethepointsmarkedinred,thenthenumberoflatticepointsreducesto37=34234+1.Thereducedarraycorrespondstothearray
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a)(b) Figure4-5. ComparisonofapreviouslystudiedarraystructurewiththetypeBRHS.(a)Thelatticepointsof3.(b)Thelatticepointsof
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2{1 forcomputingtheDFTonatileofalatticeisreducedtoEquation 2{7 .TheuseoftheSmithnormalformtoconvertEquation 2{7 toastandardDFTappearedpreviouslyinProblem20,Chapter2,ofDudgeonandMersereau[ 10 ]andinGuessoumandMersereau[ 19 ].Inthisresearch,byaunimodularmatrixwemeanasquarematrixwithintegerentriesanddeterminant+1or1.ApplyingtheSmithnormalformtotheperiodicitymatrixMinEquation 2{7 ,wecanobtainunimodularmatricesEandFsuchthatM=EDFforsomediagonalperiodicitymatrixD=0B@d100d21CA,wheredj0forj=1;2,andd1dividesd2.Letgk;l=E0B@kl1CA,andhk;l=FT0B@kl1CA.Also,letG=fgk;l:0k
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2{7 wehave Since(hs;t)TM1gk;l=(FT"s;t)T(EDF)1E"k;l="s;tTD1"k;l=sk d1+tl d2,wehave d1+tl d2)forallhs;t2H:(5{3) Thus,bfcanbecomputedusingalgorithmsofthestandardDFT. 5.1.1 ,andM=0B@rssrs1CAbeaperiodicitymatrixsuchthatrandsarerelativelyprime.ThenthereexistunimodularmatricesEandFsuchthatM=EDFwhereD=0B@100N1CAandN=jdet(M)j.HencetheDFTbfoffcanbeconvertedtoa1-dimensionalstandardDFT. Proof. 47

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5.1.1 theDFTbfoffcanbeconvertedtoa1-dimensionalstandardDFT. 2{5 ,itsDFTbfsatisesthefollowingequation: AsinSection 2.3 ,fcanbeextendedperiodicallytothewholelatticeLA.ItfollowsfromEquation 2{5 that,foranysetofcosetrepresentativesPofLA=fLAn,theset
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5.1.1 thatP1isasetofcosetrepresentativesassociatedwithMAn,P2isasetofcosetrepresentativesassociatedwith(MAn)T,andbf(0B@2kk1CA)=P3n2+3nm=0f(0B@0m1CA)e2ilm BytheSmithnormalformof(MAn)T,wehave(MAn)T=E(DAn)F,whereE=0B@103n+211CAandF=0B@2n+1n+1211CAareunimodular.ItnowfollowsfromProposition 5.1.1 thatthesetP1isasetofcosetrepresentativesassociatedwith(MAn)TandthesetP2isasetofcosetrepresentativesassociatedwithMAn. ThenextCorollaryfollowsfromProposition 5.2.1 andtheuseofchangeofbases. 49

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5.2.1 togetthevaluesofbfonC(A;n),weneedanecientcorrespondencebetweenC(A;n)andP1inthespatialdomain,andanecientcorrespondencebetweenC(A;n)andP2inthefrequencydomain. (a)ThecorrespondencebetweenC(A;n)andP1inthespatialdomain Leta=0B@uv1CA2C(A;n).BythedenitionofC(A;n),wehavejujn,jvjnandjuvjn.Weconsiderthefollowingthreecasesforu. 50

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(b)ThecorrespondencebetweenP2andC(A;n)inthefrequencydomain 5.2.1 andCorollary 4.2.8 ,eachofthesetsP2andC(A;n)isasetofcosetrepresentativesassociatedwiththematrix(MAn)T.HencethereexistsabijectionbetweenP2andC(A;n)suchthatg(g)mod(MAn)Tforanyg2P2.SuchabijectioniscalledthecorrespondencebetweenP2andC(A;n)inthefrequencydomain.Inthefollowing,weshowanecientmethodtondthebijection.Tomakeexpressionssimpler,weassumethatnisodd,andletn=2l+1forsomeintegerl>0.Whenniseven,thecorrespondencecanbesimilarlyshown.Themethodconsistsofthefollowingtwosteps.InStep1,wendthecorrespondencebetweenP2andasetUwhichwillbedenedinStep1.InStep2,wendthecorrespondencebetweenUandC(A;n). 51

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2+1+3n+1 2(2n+1)=3n2+3n+1latticepointseachofwhichwillbeshowntobecongruenttoanelementofC(A;n)withrespecttothemodulorelationassociatedwiththematrix(MAn)T.Whenn=3,thelatticepointsenclosedbythebluedottedpolygoninFigure5-1areexactlythelatticepointsofU.Inthefollowing,weshowanecientmethodtondabijectionbetweenP2andUsuchthatg(g)mod(MAn)Tforanyg2P2. Sinceg2P2,thereexistsm2Zsatisfying0m3n2+3nsuchthatg=0B@2mm1CA.Tond(g),weconsiderthefollowingtwocases. When1t2l+2,leth=tl2.Thent=h+l+2.Sincem=l+(4l+3)(k1)+t,itfollowsthatm=2l+2+(4l+3)(k1)+h.HencegAk;h=0B@2mm1CA0B@2h+1h+k1CA=0B@4l+4+(8l+6)(k1)+2h2l+2+(4l+3)(k1)+h1CA

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When2l+3t4l+3,leth=t3l3.Thent=h+3l+3.Sincem=l+(4l+3)(k1)+t,itfollowsthatm=(4l+3)(k1)+h+4l+3=k(4l+3)+h.HencegBk;h=0B@2mm1CA0B@2hh+k1CA=0B@2k(4l+3)+2hk(4l+3)+h1CA0B@2hh+k1CA=0B@2k(4l+3)k(4l+2)1CA=0B@4l+32l+22l+14l+31CA0B@2k01CA=0B@2n+1n+1n2n+11CA0B@2k01CA.ThusgBk;hmod(MAn)T.SinceAk;h2U,itfollowsthat(g)=Bk;h. Wehaveshownthat,withrespecttothemodulorelationassociatedwiththematrix(MAn)T,theelementsofP2correspondsequentiallytotheelementsofD0,A1,B1,A2,B2,...,A3l+2,andB3l+2. BythedenitionsofD0andC(A;n),wehaveD0C(A;n).Itfollowsthat(u)=uforanyu2D0. Foranyu2Ak,bythedenitionofAk,thereish2Zsatisfyingl1hlsuchthatu=Ak;hwhereAk;h=0B@2h+1h+k1CA.Weconsiderthefollowingthreecasesforktond(u). 53

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ThelatticepointsenclosedbythebluedashedpolygonformthesetofcosetrepresentativesUof(fLA3)=(LA). 54

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Weclaimthateu2C(A;n).Sincel+2k2l+1and(l+1)hk2l3,wehavea=2h+n+12(k2l3)+n+12(2l+12l3)+n+1=n3n.Hencejaj0>n.Hencejabjn.Thereforeeu2C(A;n). Becauseueu=0B@nn+11CA=0B@2n+1n+1n2n+11CA0B@111CA,wehaveueumod(MAn)T.Itfollowsthat(u)=eu. 55

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Weclaimthateu2C(A;n).Sincel+2k2l+1and2lk+2hl,wehavea=2hn2ln<2ln.Thusjaj2l>2l1=n.Thusjabjn.Thereforeeu2C(A;n). Becauseueu=0B@n+12n+11CA=0B@2n+1n+1n2n+11CA0B@011CA,wehaveueumod(MAn)T.Itfollowsthat(u)=eu. Weclaimthateu2C(A;n).Since2l+2k3l+2and(l+1)h1,wehavea=2h+n+12+n+1n.Hencejaj
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Becauseueu=0B@nn+11CA=0B@2n+1n+1n2n+11CA0B@111CA,wehaveueumod(MAn)T.Itfollowsthat(u)=eu. Weclaimthateu2C(A;n).Since2l+2k3l+2and0hl,wehavea=2hn2ln<2ln.Hencejabjn.Thereforeeu2C(A;n). Becauseueu=0B@n+12n+11CA=0B@2n+1n+1n2n+11CA0B@011CA,wehaveueumod(MAn)T.Itfollowsthat(u)=eu. Wehaveshownthat(u)iseitheru,oru0B@nn+11CA,oru0B@n+12n+11CA.Foranyk2Zsatisfying1k3l+2andh2Zsatisfyinglhl,bythedenition 57

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2{5 ,itsDFTbfsatisesthefollowingequation: Asintheprevioussection,theset
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Proof. 5.1.1 thatS1isasetofcosetrepresentativesassociatedwithMBn,S2isasetofcosetrepresentativesassociatedwith(MBn)T,andbf(0B@ll1CA)=P9n2+3nm=0f(0B@0m1CA)e2ilm ThecorrespondencebetweenC(B;n)andS1inthespatialdomaincanbeestablishedinalikewisefashion. 4.2.1 ,N=3n2+3n+1.FromtheresultsofSection 5.2 ,thecomputationalcomplexityoftheDFTon
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2{5 ,itsDFTbfsatisesthefollowingequation: Wegeneratedafunctionfdenedon5122=262144latticepoints.Althoughthenumber267307isprime,ona2.8GHzPC,theMatlabcomputationaltimeoftheDFToffislessthanfoursecondsusingthemethodprovidedinthischapter.AccordingtoMoler[ 31 ],Version6ofMatlabusestheFastFourierTransformintheWest(FFTW)whosewebpageiswww.tw.org.AsshowninFrigoandJohnson[ 15 ],FFTWcomputestheDFTofarbitraryinputsizeandutilizesRader'salgorithm(Rader[ 35 ])forprimesizes.TheRader'salgorithmcomputestheDFTofsizeNbywayoftwoDFTsandoneinverseDFTofsizeN1.FortheDFTon2,theproductn(n+1)hasatleastthreenontrivialintegerfactorsandhenceN1hasatleastfournontrivialintegerfactors.Therefore,whenNisprime,theRader'salgorithmcanbeappliedecientlyforcomputingtheDFTon
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Since3n2+3n+1=ab,wehaveTAn=0B@103n1ab1CA=GHwhereG=0B@103n1a1CAandH=0B@100b1CA.ItiseasytocheckthatP=8><>:0B@0j1CA:0j=>;isasetofcosetrepresentativesassociatedwithGandthatQ=8><>:0B@0k1CA:0k=>;isasetofcosetrepresentativesassociatedwithH.LetE=fGq+p:p2P;q2Qg.ThenEisasetofcosetrepresentativesassociatedwithTAn.Bydirectcomputation,wehave ItiseasytoverifythatallconditionsofTheorem 5.4.1 aresatised. 61

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6.1.1TheDenitionofthePyxisStructure 33 ]andmentionedinChapter 1 .Togivearecursivedenitionbasedonvectoradditions,weneedthefollowingnotation.FirstrecallfromChapter 4 thatvA1=0B@101CA;vA2=0B@1 2p 21CA;vB1=0B@p 21 21CA;vB2=0B@p 21 21CA;LA=n1(vA1)+n2(vA2):n1;n22ZandLB=n1(vB1)+n2(vB2):n1;n22Z. Let1beaxedpositivenumber.Foranynsatisfying11letP(n)=P(n1)[(P(n2)+n):

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ThegeneratorsofthelatticesL1andL2,andthelatticepointscontainedin1and2.Inthisgure,thetworedvectorsarev1;1andv1;2,thetwodashedgreenvectorsarev2;1andv2;2,1consistsofthesixblack*points,and2consistsofthesixblueopoints. Foranyintegern0,thesetP(n)iscalledthePyxisstructureatleveln.WewillshowinTheorem 6.1.4 thatP(n1)andP(n2)+naredisjoint.ThenextlemmashowsthecontainmentrelationshipamongP(n)andLnforanyn2N. Proof. 2p 21CA,vB1=0B@p 21 21CAandvB2=0B@p 21 21CA,wehavevA1=1 63

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Bythedenitionofn,wehavenLnforanyn2N.AlsobythedenitionofP(1),wehaveP(1)=1.ItfollowsthatP(1)L1.ObviouslyP(0)=0L0.Usinginduction,assumethatP(k)Lkforanyksatisfying0k
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Therst4levelsofthePyxisstructurewhereP(1)consistsof7bluehexagons,P(2)consistsof13redhexagons,P(3)consistsof55greenhexagons,andP(4)consistsof133blackhexagons.Inthisgure,theblackvectoristhesumofthetwobluevectors.Itfollowsthatthecoordinatesofthelatticepointlabeled0205isthesumofthecoordinatesofthetwolatticepointslabeled0030and0001,respectively. LetD=f(1;1);(0;1);(1;0);(1;1);(0;1);(1;0)g.Itfollowseasilyfromthedenitionofnthatn=fn1vn;1+n2vn;2:(n1;n2)2Dg: BycombiningEquations 6{1 and 6{2 ,wehave(n1~n1)vn;1+(n2~n2)vn;2=l1vn2;1+l2vn2;2:

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Proof. Sincenisassumedtobeeven,vn+1;k=n+1vAkandvn;k=nvBkfork=1;2.ThenEquation 6{3 becomes Sincen=p 1. 2. 3. 5[3n+2(2)n+2].

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6.1.3 2.SinceP(n)LnLn+1andP(n+1)Ln+1,ifP(n+1)T(P(n)+n+2)6=;,wehaveLn+1Tn+26=;whichcontradictsLemma 6.1.3 3.Theconclusionisobviouslytrueforn=0;1.Whenn>1,bythedenitionthePyxisstructure,wehaveP(n)=P(n1)S(P(n2)+n).SinceP(n1)T(P(n2)+n)=;byItem2,wehavejP(n)j=jP(n1)j+jP(n2)+nj.ByLemma 6.1.2 wehavejP(n2)+nj=jP(n2)nj=6jP(n2)j.HencejP(n)j=jP(n1)j+6jP(n2)j.WealsohaveinitialconditionsjP(0)j=1andjP(1)j=7.ThuswecangetjP(n)j=1 5[3n+2(2)n+2]. Letn;0=0foranyn2N.Sincen=n[f0g,wehaven=fn;j:j=0;1;2;:::;6g.Todenethelabeling,weprovethat,foranya2P(n),thereexistsuniquelydeterminedi2f0;1;2;:::;6gforeachisatisfying1insuchthata=Pni=1i;i,andeitheri=0ori+1=0foreachisatisfying1i
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6.1.4 ,wehaveP(t1)P(t).Hencea2P(t).Ift6=0,thent1=0bytheconditionofthislemma.Hence TheassumptionofthisinductionfollowsthatPt2i=1i;i2P(t2).Sincet6=0,wehavet;t2t.HencebyEquation 6{5 ,wehavea2t+P(t2)P(t). Proof. Toshowtheuniqueness,supposethatahasanothersuchexpressiona=Pni=1i;~i,where~i2f0;1;2;:::;6gforeachi=1;2;:::;nandeither~i=0or~i+1=0foranyisatisfying1i
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6{6 becomes ByLemma 6.1.5 ,wehavePn1i=1i;i2P(n1).Bytheassumptionofthisinduction,itfollowsfromEquation 6{7 thati=~iforanyisatisfying1in1.Thustheexpressionofaisuniqueinthiscase.Ifn6=0but~n=0,thenitfollowsfromEquation 6{5 that ByLemma 6.1.1 ,wehaveiLn1foreachi=1;2;:::;n1.ByEquation 6{8 ,itfollowsthatn;n2Ln1.Sincen6=0,wehaven;n2n.Hencen;n2nTLn1=;byLemma 6.1.3 ,whichisacontradiction.Ifn=0but~n6=0,thenacontradictioncanbesimilarlyshown.Ifn6=0and~n6=0,thenn1=0and~n1=0bytherequirementofsuchexpressions.ByLemma 6.1.5 ,wehavePn2i=1i;i2P(n2)andPn2i=1i;~i2P(n2).Sincen;n2nandn;~n2n,byLemma 6.1.2 ,itfollowsfromEquation 6{6 thatPn2i=1i;i=Pn2i=1i;~iandn;n=n;~n.SincePn2i=1i;i=Pn2i=1i;~i,bytheassumptionofthisinduction,wehavei=~iforeachisatisfying1in2.Furthermore,sincen;n=n;~n,wehaven=~n.Thereforetheuniquenessisprovedforallcases. Foranyn2Nanda2P(n),theuniqueexpressiona=Pni=1i;iinTheorem 6.1.6 iscalledthestandardexpressionofainP(n).Thestring12:::nofintegersinthatstandardexpressioniscalledthelabelofainP(n).IfVadenotestheVoronoicellofa2Ln,thenthestring12:::nisalsocalledthelabelofVa.ThelabelofeachlatticepointofP(n)alsoservesasthelabeloftheVoronoicellofthatlatticepoint.Itiseasytoverifythat,inFigure6-2oftheprevioussubsection,thecenterhexagonofP(1)islabeled0,andothersixhexagonsofP(1)arelabeled1,2,3,4,5,6goingcounter-clockwise.Forany 69

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1. 6.1.5 andTheorem 6.1.6 ,wehaven=f12:::n:i2f0;1;2;:::;6gforeachi,andii+1=0foreachi
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001 104010002000006060103205003005603105 002 010205020003000001206204306004006104 003 002020306030004000205301305401005001 004 000003030401040005002306402406502006 005 006000004040502050001003401503501603 006 060001000005050603104002004502604602 PartialadditiontableforP(4). 0001 010400200002000000060010010501030205000300050603 0002 002002050030000300000001010402060204030600040006 0003 000200300306004000040000000102050301030504010005 0004 000000030040040100500005000600020306040204060502 0005 000600000004005005020060060300010003040105030501 0006 001000010000000500600603060201040002000405020604 0010 010501040001000606030602104001000020000000600600 0020 010302060205000200010104010020500200003000000010 0030 020502040301030600030002002002003060030000400000 0040 000303060305040204010004000000300300401004000050 0050 000500040401040605030502006000000040040050200500 0060 060300060005050205010604060000100000005005006030 BytheplotofthelatticepointsofP(3)(P(4)respectively)inFigure6-2,itiseasytoverifyTables6-1(6-2respectively)whichgivespartialadditiontablesforP(3)(P(4)respectively)andwillbeusedinouralgorithm.ThefollowinglemmagivessomepropertiesoflabelsofthePyxisstructurewhichwillbeusedveryoftenfortheadditionoflabels. 1. If12:::n2nandk
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Foranya1a2:::an2n,b1b2:::bn2nandc1c2:::cn2n,wehavea1a2:::anb1b2:::bn=c1c2:::cninnifandonlyif0a1a2:::an0b1b2:::bn=0c1c2:::cninn+1ifandonlyifa1a2:::an0b1b2:::bn0=c1c2:::cn0inn+1. Proof. 6.1.7 ,wehaveeitheri=0ori+1=0foranyisatisfying1i
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Letan+1=bn+1=cn+1=0.ThenPni=1i;ai+Pni=1i;bi=Pni=1i;ciifandonlyifPn+1i=1i;ai+Pn+1i=1i;bi=Pn+1i=1i;ci.Itfollowsthata1a2:::anb1b2:::bn=c1c2:::cnifandonlyifa1a2:::an+1b1b2:::bn+1=c1c2:::cn+1ifandonlyifa1a2:::an0b1b2:::bn0=c1c2:::cn0. LetLbeahexagonallattice.Foranya;b2L,iftheirVoronoicellsshareexactlyoneside,thenwesaythataandbarenexttoeachotherinthelatticeL,andeachofthemiscalledaneighboroftheanotherinthelatticeL.Thefollowingthreelemmaswillbeappliedinsubsequentlemmas. Proof. Proof. 1. Ifa6=b,theneithera+b=0ora+b2n1Sn.

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Ifa6=b,thenaisnexttoeitherborbinthelatticeLn. Proof. 6.1.9 ,a+b2n1.IfbisnexttoainthelatticeLn,thenbyLemma 6.1.10 ,wehaveb(a)2n.Hencea+b2n. 2.Sincea;b2nandncontainsexactlysixlatticepointswhichareallneighborsof0inthelatticeLn,ifa6=b,thenamustbenexttoeitherborbinthelatticeLn. ThefollowingtwolemmaswillbeusedinTheorem 6.1.14 ,whichisoneofthemainresultsinthissubsection. 3a+1 3b+1 3c2P(3). Proof. 3a+1 3b+1 3candletthelabelsofa;bandcbea1,b1,andc1,respectively.ByapplyingItem3ofLemma 6.1.8 ,itfollowsthatthelabelsof1 3a;1 3band1 3care00a1,00b1,and00c1,respectively. Ifeithera=0,orb=0,orc=0,withoutlossofgenerality,weassumethatc=0.Itfollowsthatu=1 3a+1 3bandthelabelofuis00a100b1.ByapplyingTable6-1,wehave00a100b1=d10d3forsomelabeld10d323.Hencethelabelofuisd10d323andthusu2P(3). Otherwise,ifa=b=c,thenu=a2P(1)P(3). Otherwise,ifeithera=b,ora=c,orb=c,withoutlossofgenerality,weassumea=b.Itfollowsthatu=1 3c2P(3). Otherwise,ifeithera=b,ora=c,orb=c,withoutlossofgenerality,weassumethata=b.Sincec6=aanda;c21,itfollowsthataisnexttoeithercorcinL1.IfaisnexttocinL1,then1 3aisnextto1 3cinL3.Bylemma 6.1.9 ,itfollowsthat1 3a1 3c23.Henceu=a+1 3(ca)2P(1)+3P(3).IfaisnexttocinL1,then 74

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3aisnextto1 3cinL3.Itfollowsthat1 3a+1 3c23,and1 3aisnextto1 3a+1 3cinL3.Hence,byLemma 6.1.9 ,u=1 3a+1 3(c+a)22P(2)P(4). Otherwise,a;bandcaredistinctandeachofthosethreeelementsisnexttoatleastoneoftheothertwoelementsinL1.Withoutlossofgenerality,weassumethataisnexttobandbisnexttocinL1.Itfollowsthata+c=b.Henceu=1 3(a+b+c)=2 3b=b1 3b2P(1)+3P(3).Thereforethislemmaisvalidforallcases. 3a+1 3b+1 3c2P(4). Proof. 3a+1 3b+1 3candletthelabelsofa;bandcbea1a2,b1b2,andc1c2,respectively.ByapplyingItem3ofLemma 6.1.8 ,itfollowsthatthelabelsof1 3a;1 3band1 3care00a1a2,00b1b2,and00c1c2,respectively.Sincea;b;c2P(2)andP(2)=1S2Sf0g,weverifythislemmabyconsideringthefollowingcases. 6.1.12 ,wehave00a100b100c123.ByItem4ofLemma 6.1.8 ,itfollowsthat00a1000b1000c1024,i.e.,00a1a200b1b200c1c224.Henceu2P(4). 3a+1 3bandthelabelofuis00a1a200b1b2.Ifa16=0andb16=0,thenbyItem1ofCorollary 6.1.7 wehavea2=b2=0.ByapplyingTable6-1,thereexistsd10d323suchthat00a100b1=d10d3in3.ByItem4ofLemma 6.1.8 ,itfollowsthat00a1000b10=d10d30in4.Ifeithera1=0orb1=0,thenbyTable6-2,wehave00a1a200b1b2in4.Itfollowsthatu2P(4). Ifb26=c2,thenb6=c.ByItem1ofLemma 6.1.11 ,itfollowsthatb+c2P(2).Henceeitherb+c22orb+c21orb+c=0.ByItem3ofLemma 6.1.8 ,itfollowsthat 75

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3(b+c)24or1 3(b+c)23or1 3(b+c)=0.Becausea21,wehave1 3a23.Sinceu=1 3a+1 3(b+c),itfollowsthateitheru23+4oru23+3oru23P(4).ByTable6-2,wehave3+4P(4).ByTable6-1,wehave3+3P(3)P(4).Henceu2P(4). Ifb2=c2,withoutlossofgenerality,weassumethatb2=c2=1.Then,byTable6-2,thelabelof1 3(b+c)is00010001=0104in4.Sincea21,thelabelofain2isa20.Hencethelabelofuis00a200104in4.ByTable6-2,thereexists0d10d324suchthat00a200004=0d10d3in4whered12f0;3;4;5g.Sinced12f0;3;4;5g,byTable6-1,wehave00d1001=00t1in3forsomet12f0;1;2;6g.ByItem4ofLemma 6.1.8 ,itfollowsthat0d1000100=0t100in4.Hencethelabelofuis00a200104=0d10d30100=0t100000d3=0t10d3in4.Thusu2P(4). 6.1.8 ,wehave1 3a;1 3b23and1 3c24. Ifa6=b,then,byItem1ofLemma 6.1.11 ,wehaveeither1 3a+1 3b22,or1 3a+1 3b23,or1 3a+1 3b=0.Henceeitheru22+4P(2)+4P(4),oru23+4P(4)byTable6-2,oru24P(4).Thusu2P(4). Ifa=b,withoutlossofgenerality,weassumethata1=b1=1.Thenthelabelof1 3a+1 3binP(4)is00100010=1040in4.Letthelabelofcbe0c2.Itfollowsthatthelabelofuis1040000c2=10000040000c2in4.ByTable6-2,wehave0040000c2=0d10d2in4whered12f0;3;4g.Hencethelabelofubecomes10000d10d2=10000d100000d2in4.Sinced12f0;3;4g,byTable6-2,wehave100d12f10;01;06g2.ByapplyingItem4ofLemma 6.1.8 twice,itfollowsthat10000d1002f1000;0100;0600g4.Hencethelabelofubelongstothesetf1000000d2;0100000d2;0600000d2g=f100d2;010d2;060d2g4.Thusu2P(4). 76

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3a+1 3b2P(n+2). Proof. 6.1.8 that1 3a;1 3b2P(n+2),andthelabelsof1 3aand1 3bare00a1a2:::an2n+2and00b1b2:::bn2n+2,respectively.LetA=00a1a2:::anandB=00b1b2:::bn.Inthefollowing,weuseinductiontoshowthatAB2n+2. Ifn=2,thenthisstatementfollowsfromLemma 6.1.13 .Assumethisstatementistrueforallintegerslessthann.Wethenhave00a3a4:::an00b3b4:::bn2n.Letc1c2:::cn=00a3a4:::an00b3b4:::bninn.Itfollowsthat,inn+2,wehave 00c1c2:::cn=0000a3a4:::an0000b3b4:::bn:(6{9) ByItem1ofLemma 6.1.8 ,inn+2,wehave ByapplyingEquation 6{9 ,Equation 6{10 becomes Tonishtheproof,weconsiderthefollowingcases. 6.1.13 ,wehave00a1a200b1b200c1c2in4.Letd1d2d3d4=00a1a200b1b200c1c2.ByapplyingItem4ofLemma 6.1.8 forn2times,itfollowsthatd1d2d3d4000:::0=00a1a2000:::000b1b2000:::000c1c2000:::0inn+2.Hence 77

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6{11 becomesAB=d1d2d3d4000:::00000c3c4c5:::cninn+2.Sincec3=0,byItem2ofLemma 6.1.8 ,itfollowsthatAB=d1d2d3d4c3c4c5:::cninn+2. 6.1.7 ,wehavec2=c4=0inthiscase.InEquation 6{11 ,weclaimthatwecanassumea2=b2=0aswell. Supposea26=0.ByItem1ofCorollary 6.1.7 ,itfollowsthata1=0.Hence00a1a2000000c3c4=000a2000000c30in6.ByTable6-1,in3,wehave0a2000c3=x0yin3forsomex;y2f0;1;2;:::;6gwithy6=0.ByapplyingItem4ofLemma 6.1.8 forn1times,itfollowsthat000a2000:::00000c300:::0=00x0y00:::0inn+2,i.e.,00a1a2000:::00000c3c40:::0=00x0y00:::0inn+2.HenceEquation 6{11 becomes whichiswhatwewant.Similarlywecanassumeb2=0.HenceEquation 6{11 becomes ByLemma 6.1.12 wehave00a100b100c1in3.Letxyz=00a100b100c123.ByapplyingItem4ofLemma 6.1.8 forn1times,itfollowsthatxyz0000:::0=00a10000:::000b10000:::000c10000:::0inn+2.HenceEquation 6{13 becomesAB=xyz0000:::00000c3c4c5:::cn=xyz0c3c4c5:::cninn+2byItem2ofLemma 6.1.8 .ThusAB2n+2. Thefollowingpropositiontellswhetherornota+b2P(n)basedonthelabelof1 3a+1 3b2P(n+2). 3a+1 3b.Ifs1s2:::snsn+1sn+22n+2isthelabelofs,thena+b2P(n)ifandonlyifs1=s2=0. Proof. 6.1.6 thereexistsuniquelydeterminedci2f0;1;2;:::;6gforeachi=1;2;:::;nsuchthata+b=Pni=1i;ciandeitherci=0or 78

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3a+1 3b=Pni=11 3i;ci=Pni=1i+2;ci.Sinces=1 3a+1 3bands1s2:::snsn+1sn+22n+2isthelabelofs,itfollowsthats1=s2=0. Converselyifs1=s2=0,sinces1s2:::snsn+1sn+22n+2isthelabelofs,wehaves=Pn+2i=1i;si=Pn+2i=3i;si=Pn+2i=31 3i2;si=1 3Pn+2i=3i2;si.Sinces=1 3a+1 3b,itfollowsthata+b=Pn+2i=3i2;si=Pnj=1j;sj+2.Becauses1s2:::snsn+1sn+22n+2,eithersi=0orsi+1=0foranyisatisfying1i0,weusethefollowingalgorithmtoaddanytwolabelsa1a2:::an2nandb1b2:::bn2n.Fromthediscussionofthepreviousparagraph,weneedtocompute00a1a2:::an00b1b2:::bntogetthesumc1c2:::cncn+1cn+22n+2.RecallthatanylabelofthePyxisstructurehasthepropertythat,foranytwoconsecutivedigits,atleastoneofthemiszero.Iftheadditionproceedsfromtherightdigitsanandbntotheleftdigitsa1andb1,theideaofremainder,carryandsumofaddingtwodigitsissimilartothatoftheusualadditionexceptthat,fortheadditionofaiandbi,weuseTable6-1wheniisoddanduseTable6-2wheniiseven.Furthermorethecarryofaddingaiandbiconsistsoftwodigitsandthesumoftwolabelsobtainedintheusualwaymaynolongerhavethepropertythat,foranytwoconsecutivedigits,atleastoneofthemiszero.Forexample,ifweaddtwolabels10and05in2usingtheusualdigitby 79

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Input:Anintegern>0,andtwolabelsa1a2:::an2nandb1b2:::bn2n. 1. IftherstdigitoftempRemis0,letremd(meansremainder)betempRem. (a) Ifk1=0,applySubroutine1toaddtheelementsinthesetcarrySet2toobtainacarryc1c2andaremainderr10.LettempCarry=c1c2andtempRem=r10. (b) Ifk1=1,let0c2denotetheelementinthesetcarrySet1. 80

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Ifk2=0,lettempCarry=00andtempRem=0c2. ii. Elseifk2=1,letd10denotetheelementinthesetcarrySet2.ApplyTable6-2tocompute000c200d10=0e10e2.LettempCarry=0e1andtempRem=0e2. iii. Else,letd10ande10denotethetwoelementsinthesetcarrySet2.ByTable6-2,000c200d10=0f10f2forsome0f10f224withf26=0.Similarly,wehave000f200e10=0g10g2forsome0g10g224.ByTheorem 6.1.14 ,wehave0f10g1=h1h222.LettempCarry=h1h2andtempRem=0g2. (c) Ifk1=2,let0c1and0d1denotethetwoelementsinthesetcarrySet1. i. Ifk2=0,byTable6-2,000c1000d1=0f1f2f3forsome0f1f2f324.LettempCarry=0f1andtempRem=f2f3. ii. Else,lete10denotetheelementinthesetcarrySet2. A. Ifc1=d1,byTable6-2,wehave000c1000d1=0f10f3forsome0f10f324.ByTable6-2again,wehave000f300e10=0g10g3forsome0g10g324.ByTheorem 6.1.14 ,wehave0f10g1=h1h2forsomeh1h222.LettempCarry=h1h2andtempRem=0g3. B. Else,sincec16=d1,byLemma 6.1.11 wehave0c10d1=f1f222.ByTable6-2,wehave00f1f200e10=g1g2g3g4forsomeg1g2g3g424.LettempCarry=g1g2andtempRem=g3g4. (d) Ifk1=3,addthethreeelementsinthesetcarrySet1usingthefollowingSubroutine2toobtain0h2h3h424.LettempCarry=0h2andtempRem=h3h4. 2. Else(nowtherstdigitofremdisnonzero), (a) ifk1=0,letremd=tempRemandaddtheelementsofthesetcarrySet2usingSubroutine1togetacarryc1c2andaremainderr1r2.LettempCarry=c1c2andtempRem=r1r2. 81

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Elselete10denotetempRemandlet0f2denotetherstelementofthesetcarrySet1.Compute00e10f20usingtable6-1togetg10g323.LetcarrySet2=carrySet2Sfg10g. i. Ifk1=1,letremd=g30andaddtheelementsofthesetcarrySet2usingSubroutine1togetacarryc1c2andaremainderr10.LettempCarry=c1c2andtempRem=r10. ii. Else,let0y2denotethesecondelementofthesetcarrySet1andcompute00g30y20usingTable6-1togeth10h323.LetcarrySet2=carrySet2Sfh10g. A. Ifk1=2,letremd=h30andaddtheelementsofthesetcarrySet2usingSubroutine1togetacarryc1c2andaremainderr10.LettempCarry=c1c2andtempRem=r10. B. Else,let0z2denotethethirdelementofthesetcarrySet1andcompute00h30z20usingTable6-1togetl10l323.LetcarrySet2=carrySet2Sfl10g,remd=l30,andaddtheelementsofthesetcarrySet2usingSubroutine1togetacarryc1c2andaremainderr10.LettempCarry=c1c2andtempRem=r10. Letkbethesizeoftheset.Sincecontainsatmostthreeelements,weconsiderthefollowingfourcases. 82

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1. Ifa1=b1=c1,thenreturnC=a10andR=00. 2. Elseifa1=b1,thencompute00b100c1usingTable6-1toget0r1r2andcompute0r1r200a1usingTable6-1togetd1d2d323,andletC=d1d2andR=d30,andreturnCandR. 3. Elsecompute00a100b1usingTable6-1toget0r1r2andcompute0r1r200c1usingTable6-1togetd1d2d323,andletC=d1d2andR=d30,andreturnCandR. Letkdenotethesizeoftheset.Sincecontainsatmostthreeelements,weconsiderthefollowingfourcases. 1. Ifa2=b2=c2,thenreturnC=0a2andR=00. 2. Elseifa2=b2,thencompute0b20c2usingTable6-2togetacarry00andaremainderr1r2,andcomputer1r20a2usingTable6-2togetacarryC=0x222andaremainderR=y1y222,andreturnCandR. 83

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Elsecompute0a20b2togetacarry00andaremainderr1r2,andcomputer1r20c2usingTable6-2togetacarryC=0x222andaremainderR=y1y222,andreturnCandR. AteachiterationinStep5ofAlgorithmaddLabels,thesetcarrySetcontainsatmostthreeelements.ItfollowsthatthecomputationalcomplexityofaddinganytwolabelsinnusingAlgorithmaddLabelsisofordern.Inthefollowing,wegivetwoexamplesshowingtheapplicationofAlgorithmaddLabels.InExample1,foreachiterationinStep5ofthealgorithm,therstdigitoftempRemis0.InExample2,foreachiterationinStep5ofthealgorithm,therstdigitoftempRemisnonzero.HencetheadditioninExample2isactuallymorecomplicatedthanthatinExample1. 1. Wheni=2,letcarrySetconsistofelementsofthesetfaPairf2g;bPairf2g;tempCarryg 84

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2. Wheni=3,letcarrySetconsistofelementsofthesetfaPairf3g;bPairf3g;tempCarryg Noticethat,inExample1,therstdigitoftempRemis0foreachiterationinStep5ofthealgorithm.Inthenext,weshowanotherexamplewheretherstdigitoftempRemisnonzeroforeachiterationinStep5ofthealgorithm. 85

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1. Wheni=2,letcarrySetconsistofelementsofthesetfaPairf2g;bPairf2g;tempCarryg 2. Wheni=3,letcarrySetconsistofelementsofthesetfaPairf3g;bPairf3g;tempCarryg 86

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Proof. 6.1.8 ,wejustneedtoconsiderthecasewhenniseven.Alsowhenn2,thistheoremisobviouslytrue.Henceweassumethatnisanevenintegerwhichisbiggerthan2.Letn=2MforsomeintegerM>1,A=00a1a2:::a2M2n+2,andB=00b1b2:::b2M2n+2.Fori=1toM,letAi=00a2(Mi)+1a2(Mi)+2:::a2M1a2M22i+2,andBi=00b2(Mi)+1b2(Mi)+2:::b2M1b2M22i+2.ByTheorem 6.1.14 ,wehaveAiBi22i+2.LetSum1,tempCarry1andtempRem1betheSum,tempCarryandtempRemoutputfromStep4ofAlgorithmaddLabels.Fori=1toM,letTi=AiBi,andletremdi,Sumi,tempCarryi,andtempRemibetheremd,Sum,tempCarry,andtempRemoutputfromtheithiterationofStep5inAlgorithmaddLabels. Inthefollowing,weuseinductiontoprovethatTiistheconcatenationoftempCarryi,tempRemi,andSumiforanyisatisfying1iM.Wheni=1,A1=00a2M1a2MandB1=00b2M1b2M.InAlgorithmaddLabels,obviouslyaPairf1g=a2M1a2MandbPairf1g=b2M1b2M.FromStep4ofAlgorithmaddLabels,Sum1=;andhenceA1B1istheconcatenationoftempCarry1,tempRem1,andSum1.ItfollowsthatT1istheconcatenationoftempCarry1,tempRem1,andSum1.NowassumethatTi1istheconcatenationoftempCarryi1,tempRemi1,andSumi1forsomeisatisfying 87

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and wehave Byourinductionassumption,Ti1istheconcatenationoftempCarryi1,tempRemi1,andSumi1.HenceEquation 6{16 becomes InStep5ofAlgorithmaddLabels,let Nowweconsiderthefollowingtwocases. 6.1.13 ,wehave 00a2(Mi)+1a2(Mi)+200b2(Mi)+1b2(Mi)+200tempCarryi1=x1x2x3x4(6{19) 88

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6{19 becomes 00aPairfig00bPairfig00tempCarryi1=x1x2x3x4:(6{20) ItfollowsthatthecarryandremainderoftheadditionofthethreeelementsinthesetcarrySetarex1x2andx3x4,respectively.HencetempCarryi=x1x2andtempRemi=x3x4.ByEquations 6{17 and 6{19 ,wehave ItfollowsthatTiistheconcatenationoftempCarryi,tempRemi,andSumi. Ifk1=0,similartoCase1,wecanshowthatTiistheconcatenationoftempCarryi,tempRemi,andSumi. Ifk16=0,wecomputethesumS=00a2(Mi)+1a2(Mi)+20000b2(Mi)+1b2(Mi)+20000tempCarryi1tempRemi1

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6{17 ,SconsistsoftherstsixdigitsofTi.Bytheassumptionofthisinduction,Ti122iistheconcatenationoftempCarryi1,tempRemi1,andSumi1.HencetheconcatenationoftempRemi1andSumi1isalegitimatePyxislabel.SinceTi=AiBi22i+2andSumi122i4,byItem2ofLemma 6.1.8 ,itfollowsthatS26.IntheithiterationofStep5ofAlgorithmaddLabels,weapplyTable6-1toconverttheexpression00a2(Mi)+1a2(Mi)+20000b2(Mi)+1b2(Mi)+20000tempCarryi1tempRemi1totheexpression00x00000y00000z0000000r0forsomex;y;z;r2f0;1;2;3;4;5;6g.LetcarrySet2consistofnonzeroelementsofthesetfx0;y0;z0gandaddtheelementsincarrySet2.BySubroutine2,thesumoftheelementsincarrySet2equalsc1c2c30forsomec1c2c3024.ItfollowsthatS=c1c2c30r0.HencetempCarryi=c1c2,tempRemi=c30,remdi=r0,andSumiistheconcatenationofremdi=r0andSumi1.NowEquation 6{17 becomes HenceTiistheconcatenationoftempCarryi,tempRemi,andSumi. Byinduction,TiistheconcatenationoftempCarryi,tempRemi,andSumiforanyisatisfying1iM.ItfollowsthatTMistheconcatenationoftempCarryM, 90

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RecallfromChapter2that,ifanarraytilesitsunderlyinglatticebytranslationsbyasublattice,thenwecancomputetheDFTonit.ForthenthlevelofthePyxisstructureP(n),toconsideritsDFT,weneedtoknowwhetherornotittilestheunderlyinglatticeLnbytranslationsbyasublattice.Inthissection,weshowsomepropertiesofthePyxisstructureandapplythemtoprovethatP(n)doesnottiletheunderlyinglatticebytranslationsbyasublatticeforanyn>2.HencetheDFTonthePyxisstructurecannotbedenedusingthemethodinChapter2.Forany2
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Proof. Proof. 6.2.2 .Supposeitistrueforalln
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6.2.6 andsomeotherresults. Proof. 6.1.8 ,wehavea1a2:::an=a1a2:::an20000:::0an.Itfollowsthata=a+an.Supposea2Ln1.Thenan=aa2Ln1\n=;,whichisacontradiction.Thusa62Ln1. Proof. 6.1.10 ,wehaveba;ca;bc2n.Itfollowsthatu;v;uv2n,anduandvarenexttoeachotherinthelatticeLn.Thusn=fu;v;(uv)g.Leta2P(n)andan2nbelatticepointsofLnsuchthattheirlabelsarea1a2:::an200and00:::0an,respectively.ByItem1ofLemma 6.1.8 ,wehavea1a2:::an=a1a2:::an20000:::0an.Itfollowsthata=a+an.Nowitfollowsfroman2nthateitheran=uoran=voran=(uv). Ifan=u,thenanandvarenexttoeachotherinthelatticeLn.ByLemma 6.1.9 ,itfollowsthatan+v2n1Ln1.Hencean+ca2Ln1.Sincethelabelofaisa1a2:::an200,wehavea2Pn2Ln1.Itfollowsthatana=a2Ln1.Hencec2Ln1whichcontradictsLemma 6.2.5 Ifan=v,similartothepreviouscase,thereisacontradiction. 93

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6.2.5 Ifan=uv,thenanisnexttouinthelatticeLn.ByLemma 6.1.9 ,itfollowsthatan+u2n1Ln1.Hencean+ba2Ln1.Sinceana=a2Ln1,itfollowsthatb2Ln1whichalsocontradictsLemma 6.2.5 .Fortheremainingtwocases,contradictionscanbesimilarlyshown.Thereforeeitheran=0orbn=0orcn=0. Proof. 6.2.6 ,wecanassumethata2P(n1).Sincetisthecentroidofthetrianglewithverticesa;b;c,wehavet=1 3(a+b+c)=a+1 3u+1 3v.Leteu=1 3uandev=1 3v.Sinceu;v2nandtheyarenexttoeachotherinthelatticeLn,wehaveeu;ev2n+2andarenexttoeachotherinthelatticeLn+2.ByLemma 6.1.9 ,itfollowsthateu+ev2n+1.Thust=a+(eu+ev)2P(n1)+n+1P(n+1). 3(a+b+c)anda62P(2n1),thent62P(2n+1). Proof. 3a;1 3b;1 3c2L2n+1.Itfollowsthatt2L2n+1.Becausea;b;c2L2n1andtheyaremutuallynexttoeachotherinthelatticeL2n1,byLemma 6.1.10 ,thereexistu;v22n1suchthatb=a+u,c=a+v,anduandvarenexttoeachotherinthelatticeL2n1.Sincet=1 3(a+b+c),similartotheproofofLemma 6.2.7 ,wehavet=a+wwithw=1 3(u+v)22n.Supposethatt2P(2n+1).LetthelabeloftbeT=t1t2:::t2n+122n+1.Thenweconsiderthefollowingtwocasesaboutt2n+1. 94

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6.2.5 6.1.3 ,whichisacontradiction.Thust2n6=0.Itfollowsthatt2n1=0.Lett1;2n22P(2n2)bethelatticepointwhoselabelist1t2:::t2n222n2andt2n2P(2n)bethelatticepointwhoselabelis00:::0t2n22n.Itfollowsthatt=t1;2n2+t2n.Sincet=a+w,wehavea=tw=t1;2n2+t2nw.Ift2n=w,thena=tw=t1;2n22P(2n2)P(2n1)whichcontradictstheconditionofthislemmathata62P(2n1).Hencet2n6=w.Sincebotht2n22nandw22n,itfollowsfromTable6-2thatt2nw22nS2n1S(2n2+2n). Ift2nw22n,thent2nw=at1;2n222nTL2n1=;,whichisacontradiction.Ift2nw22n1,thena=t1;2n2+t2nw2P(2n2)+2n1.ByLemma 6.2.3 ,P(2n2)+2n1P(2n1).Hencea2P(2n1)whichagaincontradictstheconditiona62P(2n1).Ift2nw22n2+2n,thentherearex22n2andy22nsuchthatt2nw=x+y.Hencea=t1;2n2+t2nw=t1;2n2+x+y.Itfollowsthaty=at1;2n2x22nTL2n1=;,whichisacontradictionaswell.Thereforet62P(2n+1). Let;6=TL,andletq2TbeaboundarylatticepointofT,andQtheVoronoicellofq.IfSisasideofQbutSisnotasideofanyotherVoronoicellofT,thenSiscalledaB-sideofQinT.TheboundarylatticepointqiscalledaC-boundarylatticepointofTiftheunionofallB-sidesoftheVoronoicellQisaconnectedsetofR2.

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Proof. 6.2.4 ,qisnotaboundarylatticepointofP(2t+1)whichleadstoacontradiction.Henceq2P(2t1)+2t+1.Letq=q0+rforsomeq02P(2t1)andr22t+1. Supposeq0isnotaboundarylatticepointofP(2t1).Thenthesixneighborsofq0inthelatticeL2t1arealsolatticepointsofP(2t1).Letfqi2P(2t1):i=1;2;:::;6gbethesetconsistingofthesixneighborsofq0inthelatticeL2t1suchthatqiisnexttoqi+1foreachi=1;2;3;4;5.AlsotheVoronoicellofqiisdenotedQiforeachi=1;2;:::;6.Sinceq0,q1,andq2arelatticepointsofP(2t1)andmutuallynexttoeachother,byLemma 6.2.7 ,thecentroidofthetrianglewhoseverticesareq0,q1,andq2isinP(2t+1).ItfollowsthatthehexagonAinFigure6-3isaVoronoicellofP(2t+1).SimilarlythehexagonsB,C,D,E,andFareVoronoicellsofP(2t+1).EachgreenhexagonexceptA,B,C,D,E,andFistheVoronoicellofalatticepointinthesetSfqi+2t+1:i=1;2;:::;6g.Sinceqi+2t+1P(2t1)+2t+1P(2t+1),eachgreenhexagoninFigure6-3isaVoronoicellofP(2t+1)whichcontradictsthatqisaboundarylatticepointofP(2t+1).Thusq0isaboundarylatticepointofP(2t1). Bytheassumptionofthisinduction,q0isaC-boundarylatticepointofP(2t1)andtheboundaryindexofq0(inP(2t1))iseither1or3.Inthefollowing,weassumethattheboundaryindexofq0is1becausetheproofcanbesimilarlydoneiftheboundaryindexofq0is3.ItfollowsthattheVoronoicellQ0ofq0issurroundedbyvehexagonsofP(2t1)asshowninFigure6-4(a).Sinceq0anditsveneighborsinthelatticeL2t1arelatticepointsofP(2t1)andsinceP(2t1)+2t+1P(2t+1),byLemma 6.2.7 ,all 96

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VoronoicellsofthelatticepointsofP(2t+1)whicharegeneratedfromthelatticepointsofP(2t1). (a)(b) Figure6-4. TheboundaryhexagonsofQ0andQ.(a)TheblackhexagonQ0hasboundaryindex1.(b)TheboundaryhexagonQshouldbeoneofthethreeredhexagons. greenandredhexagonsinFigure6-4(b)areVoronoicellsofP(2t+1).Becauseq=q0+rforsomer22t+1andbecauseqisaboundarylatticepointofP(2t+1),theVoronoicellQofqshouldbeoneofthethreeredhexagonsinFigure6-4(b).ByLemma 6.2.8 ,thehexagonWinFigure6-4(b)isnotaVoronoicellofP(2t+1).Hence,inFigure6-4(b),thehexagonChasboundaryindex1.Forasimilarreason,wecanshowthatthehexagonAhasboundaryindex1,andthehexagonBhasboundaryindex3andtheunionofits 97

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LetLbealattice,;6=TL,andletpandqbetwoC-boundarylatticepointsofT.IftheunionoftheB-sidesoftheVoronoicellsofpandqisaconnectedsetofR2,thenwesaythatpandqareB-connectedinT. Proof. 6.2.9 ,thereexistsoneboundarylatticepointq0ofP(2n3)whoseboundaryindexis1.LetQ0betheVoronoicellofq0.LetWbethehexagonshowninFigure6-4(b).AsshownintheproofofLemma 6.2.9 ,WisnotaVoronoicellofP(2n1),andallgreenandredhexagonsinFigure6-4(b)areVoronoicellsofP(2n1).LetwbealatticepointofL2n1suchthatthehexagonWisaVoronoicellofw.Thenw62P(2n1)andwhasatleastfourneighborsbelongingtothesetP(2n1). Themainresultinthissubsectionisthefollowingtheorem. Proof. 6.2.10 ,foranyintegern2,thereexistsw2L2n1nP(2n1)suchthatwhasatleastfourneighborsbelongingtothesetP(2n1).SupposethatP(2n1)tilesL2n1bytranslationsbyasublatticeLs.Thenthereisr2Lssuchthatw2r+P(2n1).Sincew62P(2n1),bythedenitionoftiling,wehaveP(2n1)T(r+P(2n1))=;.Hencetheboundaryindexofwinr+P(2n1)isatleast4becausewhasatleastfourneighborsbelongingtoP(2n1).ByLemma 6.2.1 ,r+P(2n1)hasthesamegeometryasP(2n1).HencebyLemma 6.2.9 ,theboundaryindexofwinr+P(2n1)iseither1or3.Howeverwehaveshownthattheboundaryindexofwinr+P(2n1)isatleast4,whichisacontradiction.ThereforeP(2n1) 98

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Inthissubsection,werstshowsomepropertiesofP(2n)andapplythemtoprovethatP(2n)doesnottiletheunderlyinglatticeL2nbytranslationsbyasublatticeforanyn2. Proof. 6.1.4 ,P(2n1)P(2n).ItfollowsthatP(2n1)+2n+1P(2n)+2n+1.HenceP(2n)S(P(2n1)+2n+1)P(2n)S(P(2n)+2n+1),i.e.,P(2n+1)P(2n)S(P(2n)+2n+1).HenceP(2n+1)P(2n)S(P(2n)+2n+1)S(P(2n)+2n+2).Obviously(P(2n)+2n+2)P(2n)S(P(2n)+2n+1)S(P(2n)+2n+2).ThusP(2n+1)S(P(2n)+2n+2)P(2n)S(P(2n)+2n+1)S(P(2n)+2n+2),i.e.,P(2n+2)P(2n)S(P(2n)+2n+1)S(P(2n)+2n+2).Ontheotherhand,itfollowsfromLemma 6.2.3 that(P(2n)+2n+1)P(2n+1)P(2n+2).FurthermoreP(2n+2)=P(2n+1)S(P(2n)+2n+2)impliesthatP(2n)+2n+2P(2n+2).HenceP(2n)S(P(2n)+2n+1)S(P(2n)+2n+2)P(2n+2).ThereforeP(2n+2)=P(2n)S(P(2n)+2n+1)S(P(2n)+2n+2). Foranyq2P(2n),byLemma 6.2.12 ,wehave(q+2n+1)S(q+2n+2)P(2n+2).The13latticepointsintheset(q+2n+1)S(q+2n+2)andtheirVoronoicellsareshowninFigure6-5,wherethesixlatticepointsinthesetq+2n+2surroundqandthesixlatticepointsinthesetq+2n+1aretheverticesofthe(red)Voronoicellofq2P(2n).

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The13latticepointsintheset(q+2n)S(q+2n1)andtheirVoronoicells. 6.2.3 ,wehavep+b2P(2n+1).Itfollowsthat12:::2n000:::0x22n+1. Proof. 6.2.12 ,P(2n)=P(2n2)S(P(2n2)+2n1)S(P(2n2)+2n).Itfollowsthatq+2nP(2n2)+2nP(2n).HenceallneighborsofqinP(2n)arestillinP(2n),whichcontradictsthatqisaboundarylatticepointofP(2n).Thuseitherq2n16=0orq2n6=0. LeteqbethelatticepointofP(2n2)labeledq1q2:::q2n2.ToshoweqisaboundarylatticepointofP(2n2),itsucestoshowthatthereexistsb22n2suchthateq+b62P(2n2),i.e.,q1q2:::q2n3q2n200:::0j622n2forsomej2f1;2;3;4;5;6g,where00:::0j22n2isthelabelofb.SinceqisaboundarylatticepointofP(2n)withlabelq1q2:::q2n2q2n1q2n22n,thereexistsk2f1;2;3;4;5;6gsuchthatq1q2:::q2n2q2n1q2n

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Sincewehaveshowninthepreviousparagraphthateitherq2n16=0orq2n6=0,wesplittheproofintotwocases. 6.1.7 ,wehaveq2n=0.Sinceq2n16=0andq2n=0,byTable6-2,wehaved3=0.Supposed2=0.Then,byEquation 6{23 ,wehaveq1q2:::q2n2q2n1q2n00:::00k=q1q2:::q2n20000:::00d4=q1q2:::q2n20d422n,whichisacontradiction.Thusd26=0.Nowsupposeq1q2:::q2n3q2n200:::0d222n2.Lety1y2:::y2n3y2n2=q1q2:::q2n3q2n200:::0d2: 6{23 becomes whichcontradictsthatqisaboundarylatticepointofP(2n)withlabelq1q2:::q2n2q2n1q2n.Thusq1q2:::q2n3q2n200:::0d2622n2.Thereforethelatticepointlabeledq1q2:::q2n2isaboundarylatticepointofP(2n2). 101

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6.1.7 ,wehaveq2n1=0.Supposed2=0.ThenEquation 6{23 becomes Ifd36=0,then,byItem1ofCorollary 6.1.7 ,wehaved4=0.Sinced36=0andd4=0,byLemma 6.2.13 ,wehaveq1q2:::q2n3q2n20000:::00d3d422n.ByEquation 6{25 ,itfollowsthatq1q2:::q2n3q2n2q2n1q2n00:::000k22n,whichcontradictsthatqisaboundarylatticepointofP(2n)withlabelq1q2:::q2n2q2n1q2n.Ifd3=0,thenq1q2:::q2n3q2n20000:::00d3d4=q1q2:::q2n3q2n20d422n,whichalsocontradictsthatqisaboundarylatticepointofP(2n)withlabelq1q2:::q2n2q2n1q2n.Thusd26=0,whichimpliesthatd3=0.Sinceq1q2:::q2n2q2n1q2n00:::00k622n,byEquation 6{23 ,wehaveq1q2:::q2n3q2n20000:::0d2d3d4622n,i.e.,q1q2:::q2n3q2n20000:::0d20d4622n.ByItem2ofLemma 6.1.8 ,itfollowsthatq1q2:::q2n3q2n200:::0d2622n2.Thereforethelatticepointlabeledq1q2:::q2n2isaboundarylatticepointofP(2n2). ThefollowinglemmashowssomeinvariantpropertiesofP(n)withrespecttotherotationofR2byanangleofk Proof. 3k1;jand2k+2;j=1 3k2;j,wehaveM2k+1;j=1 3kM1;j22k+1andM2k+2;j=1 3kM2;j22k+2.ObviouslyMi;0=M0=0=i;0for 102

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Sinceeitheryi=0oryi+1=0foreachisatisfying1i2n1,byLemma 6.1.5 ,itfollowsthaty2P(2n). Furthermore,ifxisaboundarylatticepointofP(2n)withboundaryindexbforsomeb2f1;2;3;4;5;6g,thenthereexistsBf1;2;3;4;5;6gsuchthatjBj=band,foreachj2f1;2;3;4;5;6g,x+2n;j62P(2n)ifandonlyifj2B.Bytheresultofthepreviousparagraph,wehavex+2n;j2P(2n)ifandonlyifMx+M2n;j2P(2n).HenceyisaboundarylatticepointofP(2n)withboundaryindexb.Conversely,ifyisaboundarylatticepointofP(2n)withboundaryindexbforsomeb2f1;2;3;4;5;6g,similarlywecanshowthatxisalsoaboundarylatticepointofP(2n)withboundaryindexbsincex=M1y=0B@cos()sin()sin()cos()1CAy. ThefollowinglemmashowsthateachboundarylatticepointofP(2n)isaC-boundarylatticepoint,anditsboundaryindexis1,2,or4.

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6.1.8 twice,itfollowsthat0j0k=x1x222.Hencewehave Whenx1=0,Equation 6{27 becomes Whenx2=0,Equation 6{27 becomes ByLemma 6.2.13 ,12:::2n2000:::0x122n1.Itfollowsthat12:::2n20000:::0x1022n.HencebyEquation 6{29 ,wehave12:::2n20j00:::00k22n.ThusatleastveneighborsofqinthelatticeL2narestillinP(2n).Thereforetheboundaryindexofqis1. Ifthelabelofqisj0forsome=12:::2n222n2andj2f1;2;3;4;5;6g,thenthelabelsofthesixneighborsofqinthelatticeL2nconstitutethesetf12:::2n2j000:::00k:k=1;2;:::;6g: 6.2.15 ,theboundaryindexwouldbethesameifj2f2;3;4;5;6g.Ifk=3,byTable6-2,wehave00j0000k= 104

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Itfollowsthatn32P(2n).Similarly00j00004=0006impliesthat12:::2n2j000:::00422n.Thusn42P(2n).Nowconsidern1andn2.ByTable6-2,00j00001=00100001=0105and00j00002=00100002=0104.Itfollowsthatn162P(2n)ifandonlyif12:::2n2j000:::001622nifandonlyif12:::2n32n20000:::0105622nifandonlyif12:::2n32n200:::01622n2ifandonlyif12:::2n32n20000:::0104622nifandonlyifn262P(2n).Similarlywecanshowthatn562P(2n)ifandonlyifn662P(2n).Thustheboundaryindexofqiseither2or4.Obviouslyn5isnextton6,n6isnextton1,andn1isnextton2.ItfollowsthatqisaC-boundarylatticepointofP(2n). Proof. 6.2.14 ,wehaveq2n1q2n=0iorq2n1q2n=i0forsomei2f1;2;3;4;5;6g. Let=q1q2:::q2n222n2andletqbealatticepointofP(2n2)suchthatthelabelofqis.Thenthelabelofqisq2n1q2n.SinceqisaboundarylatticepointofP(2n),byLemma 6.2.14 ,qisaboundarylatticepointofP(2n2).Furthermore,by 105

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6.2.12 ,wehaveP(2n)=P(2n2)S(P(2n2)+2n1)S(P(2n2)+2n).Itfollowsthatq+2nP(2n),andq+2n1P(2n).Foranyi2f1;2;3;4;5;6g,byLemma 6.2.13 ,wehave Leti=q1q2:::q2n2000:::0i.WhenqistakenasanelementofP(2n1),thelabelofqisq1q2:::q2n20.ByEquation 6{31 ,itfollowsthatq+2n1;i2P(2n1)andthelabelofq+2n1;iisi22n1.Hence,ifq+2n1;iistakenasanelementofP(2n),itslabelisi022n.WhenqistakenasanelementofP(2n),thelabelofqisq1q2:::q2n200.Itfollowsthatthelabelofq+2n;iisq1q2:::q2n20000:::00i.ByItem2ofLemma 6.1.8 ,wehaveq1q2:::q2n20000:::00i=q1q2:::q2n20i22n.Let=q1q2:::q2n2.Thenthelabelofq+2n;iis0i.ByLemma 6.2.16 ,theboundaryindexofqis1,2,or4,andqisaC-boundarylatticepointofP(2n2).Withoutlossofgenerality,weassumethattheboundaryindexofqis4sincetheproofsfortheothertwocasescanbesimilarlydone.ThentheVoronoicellofqinthelatticeL2n2andtheVoronoicellsoflatticepointsofL2nintheset(q+2n)S(q+2n1)areshowninFigure6-6,wheretheVoronoicellsoflatticepointsinthesetq+2narelabeledifori=0;1;2;:::;6,andtheVoronoicellsoflatticepointsinthesetq+2n1arelabeled0ifori=1;2;:::;6.ByLemma 6.2.16 ,theboundaryindexofqis1,2,or4.Henceweconsiderthefollowingcases. Iftheboundaryindexofqis1,then,byLemma 6.2.16 ,wehaveq2n1q2n=0iforsomei2f1;2;:::;6g.Hencethelabelofqis0i.AsshowninFigure6-6,thereexisttwoboundarylatticepointsc1andc2ofP(2n)suchthat,foreachj=1;2,cjisB-connectedtoqinP(2n)andtheboundaryindexofcjiseither2or4. Similarlyiftheboundaryindexofqis2or4,thenbyLemma 6.2.16 ,wehaveq2n1q2n=i0forsomei2f1;2;:::;6g.Sinceq2n1=i6=0,byItem1ofCorollary 6.1.7 ,wehaveq2n2=0.ByItem2ofLemma 6.1.8 ,itfollowsthatq1q2:::q2n2q2n1q2n=q1q2:::q2n20000:::0i0=i0whereiisdenedinthesecondparagraphofthisproof.Hencethelabelofqisi0wherei2f1;2;3;4;5;6g.AsshowninFigure6-6,there 106

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GenerationoftheboundaryhexagonsofP(2n). existtwoboundarylatticepointsc1andc2ofP(2n)suchthat,foreachj=1;2,cjisB-connectedtoqinP(2n)andtheboundaryindexofcjis1.Byinduction,thislemmaistrueforanyn2N. Proof. 6.1.8 twice,itfollowsthat1001622.Hence12;1622andtheclaimistruefori=1.AssumetheclaimistrueforanyiN0forsomeN02N.Inthefollowing,weshowthattheclaimisalsotrueifi=N0+1.Obviously N0+12N0+2;1=1010:::10100000:::00001=1010:::10000000:::00100000:::0001=N0000000:::00100000:::0001:(6{32) 107

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6.1.8 for2N02times,wehave0000:::00100000:::0001=0000:::010522N0+2.HenceEquation 6{32 becomes N0+12N0+2;1=N0000000:::0105:(6{33) Bytheassumption,wehaveN02N0;1622N0,i.e.,N00000:::01622N0.ByEquation 6{33 andItem2ofLemma 6.1.8 ,itfollowsthatN0+12N0+2;1622N0+2.Thustheclaimistrueforanyi2N. NowweproveLemma 6.2.18 .Ifn=2,bytheplotofVoronoicellsofP(4)asshowninFigure6-2,thislemmaisobviouslytrue.Ifn>2,letQ=1010:::103022nandletqbealatticepointofP(2n)whoselabelisQ.BythedenitionofthelatticeL2nandthedenitionof2n,thesixneighborsofqinthelatticeL2nareexactlytheelementsofthesetq+2n.Bythedenitionoflabels,wehaveq+2n;k2P(2n)ifandonlyifQ2n;k22n.ByItem1ofLemma 6.1.8 ,wehaveQ=1010:::1000000000:::001030=n200000000:::001030.Hence Inthefollowing,weshowthatQ2n;k622nfork=1;2,andQ2n;k22nfork=3;4;5;6. ByTable6-2,wehave00300001=0205.Hence000030000001=000205.Since001030=001000000030,itfollowsthat 001030000001=001000000030000001=001000000205=001000000200000005:(6{35) ByTable6-2,wehave00100002=0104.Itfollowsthat001000000200=010400.ThenEquation 6{35 becomes001030000001=010400000005=010405.ByEquation 6{34 108

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Bytheclaimintherstparagraph,wehaven22n4;1622(n2).ByItem2ofLemma 6.1.8 andEquation 6{36 ,itfollowsthatQ2n;1622n.SimilarlywecanshowthatQ2n;2622n. ByTable6-2,wehave00300003=0301.Since10300003=100000300003,itfollowsthat10300003=10000301.Because1003=0122,itfollowsthat10300003=010124.ByEquation 6{34 ,itfollowsthatQ2n;3=n200000000:::0010300000:::000003=n2030122n.SimilarlywecanshowthatQ2n;k22nfork=4;5;6. WehaveshownthatQ2n;k622nfork=1;2,andQ2n;k22nfork=3;4;5;6.Bythedenitionofthesumoftwolabels,itfollowsthatq+2n;k62P(2n)fork=1;2,andq+2n;k2P(2n)fork=3;4;5;6.ThusqisaboundarylatticepointofP(2n)andtheboundaryindexofP(2n)is2.Furthermore,sinceq+2n;1isnexttoq+2n;2inthelatticeL2n,qisaC-boundarylatticepointofP(2n). Inthefollowing,weusethepreviousresultstoprovethemainresultofthissubsection. Proof. 6.2.18 ,thereexistsaC-boundarylatticepoint,denoteda,ofP(2n)withboundaryindex2.ByLemma 6.2.17 ,thereexisttwoboundarylatticepointsbjofP(2n)forj=1;2suchthataisB-connectedtobjinP(2n)andbjhasboundaryindex1foreachj=1;2.Letb3=aandletBjbetheVoronoicellofbjforj=1;2;3.Becauseb3hasboundaryindex2andisB-connectedtob1andb2inP(2n),wecanassumethat 109

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(a)(b) Figure6-7. TilingofP(2).(a)UjisahexagonandnexttoBjbutnotinP(2n)forj=1;2.(b)ShowingwhereisW3located. SupposethatP(2n)tilesL2nbytranslationsbyasublatticeLs.Then[fu+P(2n):u2Lsg=L2n

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6.2.17 ,u1isB-connectedtotwoboundaryboundarylatticepointsinr1+P(2n)eachhavingboundaryindex1.Itfollowsthattheboundaryindexofu2inr1+P(2n)shouldbe1,whichcontradictsthattheboundaryindexofu2inr1+P(2n)isatleast2. 6.2.16 ,theboundaryindexofu2islessthanorequalto4,whichleadstoacontradiction.ThusP(2n)doesnottileitsunderlyinglatticeL2nbytranslationsbyasublattice. 111

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4 ].Foranyx2R2andB2H(R2),letd(x;B)=minfd(x;y):y2Bg.Thenumberd(x;B)iscalledthedistancefromthepointxtothesetB.ForanyA;B2H(R2),dened(A;B)=maxfd(x;B):x2Ag.Thenumberd(A;B)iscalledthedistancefromthesetA2H(R2)tothesetB2H(R2).Alsoleth(A;B)=maxfd(A;B);d(B;A)g.Thenumberh(A;B)iscalledtheHausdordistancebetweenAandB. ByTheorem1onpage37ofBarnsley[ 4 ],(H(R2);h)isacompletemetricspace.RecallfromChapter 6 thatvA1=0B@101CA;vA2=0B@1 2p 21CA;vB1=0B@p 21 21CA;vB2=0B@p 21 21CA;LA=n1(vA1)+n2(vA2):n1;n22Z,LB=n1(vB1)+n2(vB2):n1;n22Z,1isaxedpositivenumber,andn=(1 6 thatP(n)LnandP(n)P(n+1).NowletsndenotethelengthofonesideofaVoronoicellofthelatticeLnandletePnbetheunionofallVoronoicellsofthelatticepointsofP(n).SinceLnisahexagonallatticewithgeneratorsvn;1andvn;2eachhavinglengthn,wehavesn=n 1.

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3. Proof. 2.Foranyx2ePn,thereexistsy2P(n)suchthatxliesintheVoronoicellofy.SinceLnisahexagonallatticeandsnisthelengthofonesideofaVoronoicellofthelatticeLn,wehaved(x;y)sn.Itfollowsthatd(x;P(n))sn.Henced(ePn;P(n))sn.BecauseP(n)ePn,wehaved(P(n);ePn)=0.Thush(ePn;P(n))sn. 3.ByItem2,wehaveh(ePn;P(n))snforanyn2N.Bythetriangleinequality,wehave Sincesn=1 113

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Proof. 7.1.1 ,wehaveh(P(n);P(n+1))n+1foranyn2N.Sincen=(1 Similarly,byItem2ofLemma 7.1.1 ,thesequencefePn:n2NgisaCauchysequenceinthespace(H(R2);h).ThereforethesequencefePn:n2NgalsoconvergesinH(R2). LetPbethelimitofthesequencefP(n):n2Ng.Forany>0,thereexistsn12Nsuchthath(P(n);P)<1 2foranyn>n1.ByItem3ofLemma 7.1.1 ,wehaveh(ePn;P(n))1 2foranyn>n2.Letn3=maxfn1;n2g.Then,foranyn>n3,wehaveh(ePn;P)0,letN(A;)denotethesmallestnumberofclosedballsofradius>0neededtocoverA.IfD=lim!0fln(N(A;)) ln(1=)gexists,thenDiscalledthefractaldimensionofA. ln(1=n)gexists,thenAhasfractaldimensionD. 22n+1=31 3)n.Foranyn2N

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Proof. 6.2.17 ,eachvertexinthesetV1isB-connectedtoexactlytwoverticesinthesetV2.ItfollowsthatjE(G)j=2jV1j.Also,byLemma 6.2.17 ,eachvertexinthesetV2isB-connectedtoexactlytwoverticesinthesetV1.ItfollowsthatjE(G)j=2jV2j.ThusjV1j=jV2jandhencebn;1=bn;2+bn;4. (a)(b) Figure7-1. TheboundaryhexagonsofP(2n)andP(2n+2).(a)TheblackhexagonsQ,R,andTofP(2n)suchthatQisB-connectedtoRandT,theboundaryindexofQis4,andtheboundaryindexofRandTis1.(b)TheboundaryhexagonsofP(2n+2)whichhaveboundaryindex2or4.TheirVoronoicellsoverlaptheinterioroftheboundaryhexagonQofP(2n),whereA4,B4,andC4haveboundaryindex4,andD2andE2haveboundaryindex2.

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7.2.2 ,wehavethefollowingEquation. Henceitsucestocomputebn;2andbn;4.LetQbeaboundaryhexagonofP(2n).ByLemma 6.2.16 ,theboundaryindexofQis1,2,or4.IftheboundaryindexofQis4,thenitfollowsfromLemma 6.2.17 that,asshowninFigure7-1(a),thereareexactlytwoboundaryhexagonsRandTofP(2n)suchthateachofRandTisB-connectedtoQandhasboundaryindex1.ByLemma 6.2.12 ,itfollowsthat,asshowninFigure7-1(b),thereexistthreeboundaryhexagonsA4,B4,andC4ofP(2n+2)withboundaryindex4andthereexisttwoboundaryhexagonsD2andE2ofP(2n+2)withboundaryindex2suchthateachofthemhasinterioroverlappingtheinteriorofQ.Similarlyiftheboundary (a)(b) Figure7-2. TheboundaryhexagonsofP(2n)andP(2n+2).(a)TheblackhexagonsQ,R,andTofP(2n)suchthatQisB-connectedtoRandT,theboundaryindexofQis2,andtheboundaryindexofRandTis1.(b)TheboundaryhexagonsofP(2n+2)whichhaveboundaryindex2or4,andwhoseVoronoicellsoverlaptheinterioroftheboundaryhexagonQofP(2n),whereA4hasboundaryindex4,andB2andC2haveboundaryindex2. indexofQis2,thenasshowninFigure7-2(a),thereareexactlytwoboundaryhexagonsRandTofP(2n)suchthateachofRandTisB-connectedtoQandhasboundaryindex1.ByLemma 6.2.12 ,itfollowsthatthereexistsoneboundaryhexagonA4withboundaryindex4andthereexisttwoboundaryhexagonsB2andC2ofP(2n+2)withboundary 116

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6.2.17 that,asshowninFigure7-3(a),thereareexactlytwoboundaryhexagonsRandTofP(2n) (a)(b) Figure7-3. TheboundaryhexagonsofP(2n)andP(2n+2).(a)TheblackhexagonsQ,R,andTofP(2n)suchthatQisB-connectedtoRandT,theboundaryindexofQis1,andtheboundaryindexofRandTis2or4.(b)TheboundaryhexagonsofP(2n+2)whichhaveboundaryindex2or4,andwhoseVoronoicellsoverlaptheinterioroftheboundaryhexagonQofP(2n),whereAandBhaveboundaryindex2. suchthateachofRandTisB-connectedtoQandhasboundaryindex2or4.ByLemma 6.2.12 ,itfollowsthatthereexisttwoboundaryhexagonsAandBofP(2n+2)withboundaryindex2suchthateachofthemhasinterioroverlappingtheinteriorofQasshowninFigure7-3(b).ButboundaryhexagonsAandBarealreadycountedfromhexagonsRandTwhichhaveboundaryindex2or4.Thereforewehavethefollowingsystemofequations. LetXj=0B@bj;4bj;21CAforanyj2N,andletM=0B@31221CA.BySystem 7{3 ,Xn+1=MXnforanyn2N.Thereforewehave 117

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3)0B@11211CA=(1 3)0B@1+2(4n)4n12(4n)24n+21CA.ThenbyEquation 7{4 wehave 3)0B@1+2(4n)4n12(4n)24n+21CA0B@601CA=0B@4n+1+24n+141CA:(7{5) Itfollowsthatb2n+2;4=4n+1+2andb2n+2;2=4n+14.Henceb2n+2;1=b2n+2;2+b2n+2;4=2(4n+1)2.Thereforeb2n;4=4n+2,b2n;2=4n4,andb2n;1=2(4n)2. ThefollowinglemmawillbeusedintheproofofLemma 7.2.7 Proof. 6.2.12 ,wehaveP(2n+2)=P(2n)[(P(2n)+2n+1)[(P(2n)+2n+2) foranyn2N.Itfollowsthat,foranyq2P(2n),wehaveq+2n+1Sq+2n+2P(2n+2).AsshowninFigure6-5ofChapter 6 ,theVoronoicellofq2P(2n)iscontainedinthe 118

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ToproveLemma 7.2.7 ,weneedthefollowinglemma. Proof. 7.1.2 ,PisthelimitofthesequenceneP2n:n2Nointhespace(H(R2);h).ByTheorem1onPage37throughpage38fromBarnsley[ 4 ],wehaveP=nx2R2:thereisaCauchysequencenxn2eP2nothatconvergestoxo.Foranyx2eP2k,letxn=xforanynk,andletxn=02eP2nforanynd(x;p)andd(x;p)d(x;Ny).Henced(x;y)>d(x;Ny). 119

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Thedistancefromapointxtoagivenlatticepointyandthedistancefromxtothesixneighborsofyinthelattice. Ifp62Ny,thenpliesonexactlyoneside,saythesideconnectingthetwoverticesaandb,ofthehexagonHN.Bysymmetry,withoutlossofgenerality,weassumethatd(x;a)d(x;b)asshowninFigure7-4.Considerthetrianglewithverticesa,x,andy.Thethreeanglesofthattrianglewithverticesa,x,andyaredenotedA,X,andY,respectively.Sinced(x;a)d(x;b),wehaveY 22n+1. Proof. 7.1.1 ,wehaveh(P(i);P(i+2))i+1foranyi2N.SincePisthelimitofthesequencefP(2i):i2N)g,wehaveh(P(2n);P)P1i=nh(P(2i);P(2i+2)).Thush(P(2n);P)P1i=n2i+1.Ifweletj=in,thenitfollowsthath(P(2n);P)P1j=02j+2n+1=P1j=01 3j2n+1=3 22n+1.Henced(P;P(2n))3 22n+1.SincePBP,itfollowsthatd(PB;P(2n))3 22n+1.Foranyx2PB,wehaved(x;P(2n))d(PB;P(2n))3 22n+1.BecauseP(2n)isaniteset,thereexistsy2P(2n)suchthatd(x;P(2n))=d(x;y).WeclaimthatyisaboundarylatticepointofP(2n).LetNy

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7.2.5 ,itfollowsthatxliesoutsidetheVoronoicellofy.HencebyLemma 7.2.6 ,d(x;y)>d(x;Ny)d(x;P(2n)),whichcontradictsthatd(x;P(2n))=d(x;y).ThusyisaboundarypointofeP2nandd(x;y)=d(x;P(2n))3 22n+1. TogetthefractaldimensionofPB,westillneedthefollowinglemmas.Foranyn2Nandx2Ln,letVn(x)denotetheVoronoicellofx. Proof. (a)(b) Figure7-5. ThecontainmentrelationamongVoronoicellsofP(2n)andP(2n+1).(a)showsthatthe(blue)VoronoicellV2n(y)iscontainedinthe(black)VoronoicellV2n1(y),whereV2n(y)hasahorizontalsideandV2n1(y)hasaverticalside.(b)showsthaty2P(2n)isthecentroidofthetrianglewithverticesq2P(2n1),r2P(2n1),andt2P(2n1).Italsoshowsthatthe(blue)Voronoicellofy2L2niscontainedintheunionofthe(black)Voronoicellsofq2L2n1,r2L2n1,andt2L2n1. 121

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Ifi2f1;3;5g,thenleteq=2n;i+2n;2,er=2n;i+2n;4,andet=2n;i+2n;6.Assumei=1becausetheprooffori=3ori=5iscompletelysimilartothatfori=1.ByTable6-2,00010002=002024.ByapplyingItem4ofLemma 6.1.8 for2n4times,itfollowsthat 00:::000100:::0002=00:::002022n:(7{6) Bythedenitionof2n1;2,thelabelof2n1;2is00:::00222n1.ByItem3ofLemma 6.1.8 ,thelabelof2n1;2inP(2n)is00:::002022n.ByEquation 7{6 ,itfollowsthat2n;1+2n;2=2n1;2,i.e.,eq=2n1;2.Similarly,byTable6-2,wehave00010004=000024and00010006=001024.Thenwecanshowthat2n;1+2n;4=0and2n;1+2n;6=2n1;1,i.e.,er=0andet=2n1;1.Henceeq;er;et22n1;1andtheyaremutuallynexttoeachotherinthelatticeL2n1.Sinceeq+er+et=32n;i+2n;2+2n;4+2n;6andsince2n;2+2n;4+2n;6=0,wehave2n;i=1 3(eq+er+et).Letq=y0+eq,r=y0+er,andt=y0+et.Then1 3(q+r+t)=y0+1 3(eq+er+et)=y0+2n;i=y.Itfollowsthatyisthecentroidofthetrianglewithverticesq,r,andt.Sincey02P(2n2)andeq;er;et22n1;1,byLemma 6.2.3 ,wehaveq;r;t2P(2n1).Becauseeq,er,andetaremutuallynexttoeachotherinthelatticeL2n1,thethreelatticepointsq,r,andtarealsomutuallynexttoeachotherinthelatticeL2n1.Henceyisthecentroidofthetrianglewithverticesq2L2n1,r2L2n1,andt2L2n1whicharemutuallynexttoeachotherinthelatticeL2n1.AsshowninFigure7-5(b),itfollowsthatV2n(y)iscontainedintheunionofV2n1(q),V2n1(r),andV2n1(t),wheretheVoronoicellsofthelatticeL2n1haveaverticalsideandtheVoronoicellsofthelatticeL2nhaveahorizontalside.HenceV2n(y)eP2n1.Sincex2V2n(y),itfollowsthatx2eP2n1. 122

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Proof. 7.2.9 ,thereexistthreelatticepointsq2L2n1,r2L2n1,andt2L2n1whicharemutuallynexttoeachotherinthelatticeL2n1suchthatyisthecentroidofthetrianglewithverticesq,r,andt.ItfollowsthatV2n(y)eP2n1asshowninFigure7-5(b).SinceV2n(y)andV2n+1(y)havethesamecentroidandsincethelengthofonesideoftheVoronoicellV2n(y)isp 32n1,asshowninFigure7-6,wehaveV2n+1(y)V2n1(y0).Hencex2V2n1(y0)eP2n1.ThereforeeP2n+1eP2n1. Proof. 4 ],wehaveP=nx2R2:thereisaCauchysequencenxn2eP2n:nkothatconvergestoxo:

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ThegreenVoronoicellV2n+1(y)iscontainedintheblackVoronoicellV2n1(y0). ByLemma 7.2.8 ,wehaveeP2neP2n1foranynk.ByLemma 7.2.9 ,foranynk,wehaveeP2n1eP2k1.HenceeP2neP2k1foranynk.Foranynk,sincexn2eP2n,itfollowsthatxn2eP2k1.BecauseeP2k1isaclosedsetofR2andx=limn!1xn,itfollowsthatx2eP2k1.ThereforePeP2k1. Foranyn2Nandi2f1;2;3;4;5;6g,recallfromChapter 6 thatn;i=0000:::0i2nisthelabelofn;iwheren;i2nP(n). Proof. 6.2.15 ,weonlyneedtoconsiderthecasei=1becauseanyoftheothercasescanbeprovenbyrotatingthevectorsqandeqbyanangleofk 6.1.8 twice,itfollowsthat1003=0122,i.e.,i003=0122.ByapplyingItem4ofLemma 6.1.8 for2n2times,i.e.,byattaching2n2zerostotheleftsideoneachtermofthepreviousequation,wehave00:::0i000:::003=00:::00122n:

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6.1.8 ,wehaveq1q2:::q2n2i0=q1q2:::q2n20000:::0i0: Similarlyitfollowsfrom00100004=000624thatq1q2:::q2n2102n;4=q1q2:::q2n20622n.Sincetheboundaryindexofqis4andsinceq1q2:::q2n21000:::000i22nfori=3;4,wehaveq1q2:::q2n21000:::000i622nforeachi2f1;2;5;6g. Similartothepreviouscomputation,wecanshowthatq1q2:::q2n210102n+2;3=q1q2:::q2n2100122n+2 ByTable6-2,wehave00100001=010524.ByItem4ofLemma 6.1.8 for2n2times,itfollowsthat00:::00i000:::0001=00:::010522n+2.Hence Sinceq1q2:::q2n21000:::001=q1q2:::q2n2102n;1622n,byItem4andItem2ofLemma 6.1.8 ,wehaveq1q2:::q2n2100000:::00105622n+2.ByEquation 7{8 ,itfollowsthatq1q2:::q2n2i0i02n+2;1622n+2. 125

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Nowconsiderq1q2:::q2n2i0i02n+2;5.ByTable6-2,wehave00100005=060324.ByapplyingItem4ofLemma 6.1.8 for2n2times,itfollowsthat00:::001000:::0005=00:::060322n+2.Hence Sinceq1q2:::q2n21000:::006=q1q2:::q2n2102n;6622n,byItem4andItem2ofLemma 6.1.8 ,wehaveq1q2:::q2n2100000:::00603622n+2.ByEquation 7{9 ,itfollowsthatq1q2:::q2n2i0i02n+2;5622n+2.Similarlywecanshowthatq1q2:::q2n2i0i02n+2;6622n+2. Wehaveshownthatq1q2:::q2n2i0i02n+2;j22n+2forj=3;4,andq1q2:::q2n2i0i02n+2;j622n+2foranyj2f1;2;5;6g.Sinceq1q2:::q2n2i0i0isthelabelofeq,itfollowsthattheboundaryindexofeqis4. Proof. 6.1.8 ,wehaveq1q2:::q2n1q2n=q1q2:::q2n1000:::0q2n.Itfollowsthatq=r+2n;i.Hence2n;i=qr.Becauseq2L2n1andr2P(2n1)L2n1,itfollowsthatqr2L2n1,i.e.,2n;i2L2n1.Since2n;i22n,itfollowsthat2n;i2L2n1T2n=;byLemma 6.1.3 ,whichisacontradiction.Thusq2n=0.Itfollowsthatq2P(2n1). 126

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3(q+uj+uj+1)foreachj2f1;2;3;4;5;6g.Obviouslytj=q+1 3(2n1;j+2n1;j+1)=q+1 32n2;j=q+2n;j: 6.2.7 ,wehavetj2P(2n),i.e.,q+2n;j2P(2n).Thusq+2nP(2n).ItfollowsthatqisnotaboundarylatticepointofP(2n),whichcontradictsthegivenconditionofthisLemma.ThusqisalsoaboundarylatticepointofP(2n1). Proof. 6.2.14 ,eitherq2n36=0orq2n26=0.Becausetheboundaryindexofeqis1,byLemma 6.2.16 ,q2n3=0andq2n26=0.Becauseq1q2:::q2n22nandq2n26=0,byItem1ofCorollary 6.1.7 ,wehaveq2n1=0.SinceqisaboundarylatticepointofP(2n),byLemma 6.2.14 ,itfollowsthatq2n6=0.Hence,byLemma 6.2.16 ,theboundaryindexofqis1. Proof. 127

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6.2.14 ,eitherq2k16=0orq2k6=0,andeqisaboundarylatticepointofP(2k2).Ifq2k6=0,thenbyLemma 6.2.16 ,theboundaryindexofqis1whichcontradictsthegivenconditionthattheboundaryindexofqiseither2or4.Thusq2k16=0andq2k=0.ByLemma 7.2.13 ,iftheboundaryindexofeqis1,thentheboundaryindexofqisalso1,whichagaincontradictsthegivenconditionthattheboundaryindexofqiseither2or4.Thustheboundaryindexofeqiseither2or4.Bytheassumptionofthisinduction,wehaveq2i=0andq2i16=0foranyisatisfying1ik1.Thusthislemmaistruewhenn=k.Byinduction,itistrueforanyn2N. Recallthat,foranyn2N,B2n;4denotesthesetconsistingofallboundarylatticepointsofP(2n)eachhavingboundaryindex4.Foranyq2B2n;4,letq1q2:::q2n22nbethelabelofq.Sinceqhasboundaryindex4,byLemma 7.2.14 ,wehaveq2n=0andq2n1=aforsomea2f1;2;3;4;5;6g.Letn(q)=q+1 22n1;a: Proof. 22n1;awherea=q2n1.Sinceq2B2n;4,theboundaryindexofqis4.ByLemma 7.2.14 ,wehaveq2i=0andq2i16=0foreachisatisfying1in.Becauseq2n=0,wehaveq2P(2n1).SinceP(2n1)+2n+1P(2n+1),itfollowsthatq+2n+1;a2P(2n1)+2n+1P(2n+1).Byinduction,itiseasytoshowthatq+Pki=02n+2i+1;a2P(2n+2k+1)foreachk0.SinceP(2n+2k+1)P(2n+2k+2)eP2n+2k+2,wehaveq+Pki=02n+2i+1;a2eP2n+2k+2foreachk2N.ByLemma 7.2.5 ,wehaveeP2n+2k+2P.Itfollowsthatq+Pki=02n+2i+1;a2Pforeachk2N. 128

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3i2n+1;aforeachi0,thesequencenPki=02n+2i+1;a:k2NoisaCauchysequenceinthespace(R2;d).Hencethelimitofthissequenceexists.Asusual,thelimitofthesequencenPki=02n+2i+1;a:k2NoisdenotedP1i=02n+2i+1;a.Obviouslywehave 3i2n+1;a=3 22n+1;a=1 22n1;a:(7{10) Itfollowsthatq+P1i=02n+2i+1;a=q+1 22n1;a.Henceq+1 22n1;aisalimitpointofthesequencenq+Pki=02n+2i+1;a:k2No.Sinceq+Pki=02n+2i+1;a2Pforeachk2NasshowninthepreviousparagraphandsincePisaclosedsubsetofR2,itfollowsthatq+1 22n1;a2P,i.e.,n(q)2P. Weclaimthatn(q)2PB.Supposen(q)62PB.Thenthereexistsanopen-neighborhoodN(n(q))=fy2R2:d(n(q);y)0suchthatN(n(q))P.Because>0andlimn!1n=0,thereexistsc2Nsuchthat2c+1< 7.2.11 thatq+Pki=02n+2i+1;aisaboundarylatticepointofP(2n+2k+2).Tomakethefollowingexpressionssimpler,letsk=q+Pki=02n+2i+1;a.Becausesk=q+Pki=02n+2i+1;a2P(2n+2k+1)asshowninthepreviousparagraphandbecauseskisaboundarylatticepointofP(2n+2k+2),itfollowsfromLemma 7.2.12 thatskisalsoaboundarylatticepointofP(2n+2k+1).Hencethereexistsl2f1;2;3;4;5;6gsuchthatsk+2n+2k+1;l62P(2n+2k+1).Sincesk+2n+2k+1;l2L2n+2k+1,itfollowsthatsk+2n+2k+1;l62eP2n+2k+1.ByCorollary 7.2.10 ,wehavePeP2n+2k+1.Itfollowsthatsk+2n+2k+1;l62P.Sincen(q)=q+1 22n1;a,wehaven(q)sc=(q+1 22n1;a)(q+cXi=02n+2i+1;a)=1 22n1;acXi=01 3i2n+1;a: 22n1;aPci=01 3i2n+1;ak=k1 22n1;a3 2(11 3c+1)2n+1;ak=k1 22n1;a1 2(11 3c+1)2n1;ak=k1 2(3c+1)2n1;ak=1 2(3c+1)2n1=1 22n+2c+1. 129

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22n+2c+1,itfollowsthatd(sc+2n+2c+1;l;n(q))2n+2c+1+1 22n+2c+1<22n+2c+1=2 3n2c+122c+1<.Hencesc+2n+2c+1;l2N(n(q))P.Thiscontradictsthatsk+2n+2k+1;l62Pforanyk2N.Thereforen(q)2PB. ThenextthreelemmaswillbeappliedintheproofofLemma 7.2.20 Proof. 7.2.14 ,wehavex2n=0.Itfollowsthatx2P(2n1)L2n1.Similarlywehavey2L2n1. Ifd(x;y)=2n,thenxisnexttoyinthelatticeL2n.Itfollowsthatx=y+zforsomez22n.Sincez=xyandx;y2L2n1,wehavez2L2n1T2n.However,byLemma 6.1.3 ,wehaveL2n1T2n=;.Itfollowsthatz2;whichisacontradiction.Thusd(x;y)6=2n. Ifd(x;y)=22n,thenkxyk=22n.SinceL2nisahexagonallattice,itfollowsthatxy=22n;iforsomei2f1;2;3;4;5;6g.Hencex=y+22n;i.ByTable6-2,wehave000i000i=0j0k24andk6=0.ByapplyingItem4ofLemma 6.1.8 for2n4times,itfollowsthat00:::000i00:::000i=00:::0j0k22n.Hence2n;i+2n;i=2n2;j+2n;k,i.e.,22n;i=2n2;j+2n;k.Sincex=y+22n;i,itfollowsthatx=y+2n2;j+2n;k.Hence2n;k=xy2n2;j.Becausex;y2L2n1and2n2;j2L2n2L2n1,itfollowsthat2n;k=xy2n2;j2L2n1.Since2n;k22n,itfollowsthat2n;k2L2n1T2n=;,whichisacontradiction.Thusd(x;y)6=22n. 130

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Proof. Ifm<0orn<0,wecanrotatexby2 7{11 ,wealsoassumethatmn. Ifn=0orn=m,thenkxk=rm.Since0
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7.2.14 ,wehavex2i=y2i=0,x2i16=0,andy2i16=0foranyisatisfying1ik+1.LetqandrbelatticepointsofP(2k)whoselabelsarex1x2:::x2k1x2k22kandy1y2:::y2k1y2k22k,respectively.Alsoleta=x2k+1andb=y2k+1.ByItem1ofLemma 6.1.8 ,wehavex1x2::::x2k1x2kx2k+1x2k+2=x1x2:::x2k1x2k0000:::00x2k+1x2k+2.Sincex2k+1x2k+2=a0,itfollowsthatx=q+2k+1;a.Similarlywehavey=r+2k+1;b.Nowweconsiderthefollowingtwocases. 6.2.12 ,wehaveq+2k+1Sq+2k+22P(2k+2).InFigure6-5ofChapter 6 ,theVoronoicellsoflatticepointsinq+2k+1andq+2k+2areinblueandingreen,respectively.Sinced(x;y)=p Henced(x;y)d(q;r)22k+1.Supposed(q;r)22k.Thend(x;y)22k22k+1=62k+22p 7.2.17 ,wehaveeitherd(q;r)=2kord(q;r)=p 6.2.14 ,qandrareboundarylatticepointsofP(2k).Since 132

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6.2.16 ,theboundaryindexofqiseither2or4.Soistheboundaryindexofr.Hence,byLemma 7.2.16 ,d(q;r)6=2k.Thusd(q;r)=p (a)(b) Figure7-7. ForanytwohexagonsofP(2k+2)eachhavingboundaryindex2or4suchthatthedistancebetweenthemis2k+2,thereexistsahexagonofP(2k+2)thatisnexttobothofthem,whereQ,RandSarehexagonsofP(2k).(a)ThehexagonSisnexttoQandR,anditscentroidliesbelowthelineconnectingthecentroidsofQandR.(b)showsthat,foranytwobluehexagonsofP(2k+2)(eachhavingboundaryindex2or4)suchthatthedistancebetweenthemis2k+2,thereexistsahexagonofP(2k+2)thatisnexttothesetwobluehexagons. Sinced(q;r)=p 6.2.15 ,wecanrotatethevectorsq,r,andsbyanangleofj 6.2.12 ,itfollowsthatallblue,greenandredhexagonsareVoronoicellsofP(2k+2).Sincex=q+2k+1;aandy=r+2k+1;b,theVoronoicellofx2P(2k+2)isoneofthebluehexagonsoverlappingQandtheVoronoicellofy2P(2k+2)isoneofthebluehexagonsoverlappingR.Becaused(x;y)=p 133

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TosimplifytheproofofLemma 7.2.20 ,weneedthefollowinglemma. Proof. 6.2.15 ,wehaveR(x);R(y)2B2n;4. Letx1x2:::x2n1x2n22n,y1y2:::y2n1y2n22n,~x1~x2:::~x2n1~x2n22nand~y1~y2:::~y2n1~y2n22nbethelabelsofx,y,R(x)andR(y),respectively.Alsoleta=x2n1,b=y2n1,c=~x2n1,andd=~y2n1.Bythedenitionofn,wehaven(x)=x+1 22n1;a,n(y)=y+1 22n1;b,n(R(x))=R(x)+1 22n1;c,n(R(y))=R(y)+1 22n1;d. Sincex1x2:::x2n1x2nisthelabelofx,wehavex=P2ni=1i;xi.Hence Foranyisatisfying1i<2n,byItem1ofCorollary 6.1.7 ,wehaveeitherxi=0orxi+1=0.Henceeitheri;xi=0ori+1;xi+1=0.SinceRistherotationofR2abouttheorigin,itfollowsthateitherR(i;xi)=0orR(i+1;xi+1)=0.Foreachisatisfying1i2n,ifi;xi6=0,theni;xi2i.BecauseRistherotationofR2byanangleofk 7{13 isthestandardexpressionofR(x)inP(2n).Since~x1~x2:::~x2n1~x2n22nisthelabelofR(x),theEquation isalsothestandardexpressionofR(x)inP(2n).ByTheorem 6.1.6 fortheuniquestandardexpression,itfollowsfromEquation 7{13 andEquation 7{14 thatR(i;xi)=i;~xiforeachisatisfying1i2n.HenceR(2n1;a)=2n1;c.Sincen(R(x))= 134

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22n1;c,itfollowsthatn(R(x))=R(x)+1 2R(2n1;a)=R(x+1 22n1;a)=R(n(x)): 2R(2n1;b)=R(y+1 22n1;b)=R(n(y)): Proof. 7.2.14 ,wehavex2i16=0,y2i16=0,andx2i=y2i=0foreachi=1;2;:::;n.Bythedenitionofn,wehaven(x)=x+1 22n1;awitha=x2n1,andn(y)=y+1 22n1;bwithb=y2n1.Ifn=1,letq=r=0.Ifn>1,letqandrbethelatticepointsofP(2n2)whicharelabeledx1x2:::x2n2andy1y2:::y2n2,respectively.Sincex2n1=aandy2n1=b,wehavex=q+2n1;aandy=r+2n1;b.Hencewehave n(x)=x+1 22n1;a=q+2n1;a+1 22n1;a=q+3 22n1;a(7{15) and n(y)=y+1 22n1;b=r+2n1;b+1 22n1;b=r+3 22n1;b:(7{16) 135

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22n1;a;r+3 22n1;b)=d(3 22n1;a;3 22n1;b)=3 2d(2n1;a;2n1;b)3 22n1=9 22n+1>32n+1. 7{15 and 7{16 ,wehave 22n1;a;r+3 22n1;b)(7{17) BythetriangleinequalityandEquation 7{17 ,wehave 22n1;a)+d(q+3 22n1;a;r+3 22n1;b)+d(r+3 22n1;b;r)=k(q+3 22n1;a)qk+d(n(x);n(y))+kr(r+3 22n1;b)k=k3 22n1;ak+d(n(x);n(y))+k3 22n1;bk=3 22n1+d(n(x);n(y))+3 22n1=d(n(x);n(y))+32n1:(7{18) ItfollowsfromEquation 7{18 thatd(n(x);n(y))d(q;r)32n1.Sinced(q;r)p 7.2.17 ,itfollowsthateitherkqrk=2n2orkqrk=p SincexandyareboundarylatticepointsofP(2n),byLemma 6.2.14 ,qandrareboundarylatticepointsofP(2n2).Becausethelabelofqisx10x30:::x2n3022n2andx2n36=0,byLemma 6.2.16 ,theboundaryindexofqiseither2or4.Forasimilarreason,theboundaryindexofriseither2or4aswell.Hence,byLemma 7.2.16 ,wehaved(q;r)6=2n2andd(q;r)6=22n2.Therefored(q;r)=p

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7.2.18 ,itfollowsthatthereexistss2P(2n2)suchthatsisnexttobothqandrinthelatticeL2n2.Becauseq;r2P(2n2)L2n2,wehaverq2L2n2.Sincekrqk=d(q;r)=p 7.2.19 ,foranyrotationRofR2byk 7{15 and 7{16 ,wehave 22n1;a;r+3 22n1;b)=krq+3 2(2n1;b2n1;a)k:(7{19) BythesymmetryofqandrinEquation 7{19 ,withoutlossofgenerality,wealsoassumethatqliestotheleftsideofrasshowninFigure7-8.Thenthetwovectorsrqand2n1;6havethesamedirection.Becausekrqk=d(q;r)=p Sinceq;r;s2P(2n2),byLemma 6.2.12 ,itfollowsthatallblue,greenandredhexagonsinFigure7-8areVoronoicellsofP(2n).Becausex2B2n;4,theboundaryindexofxis4.ReferringtoFigure7-8,sincex=q+2n1;a,itfollowsthata2f1;2;3;4g.Similarly,becausey2B2n;4,wehaveb2f1;2;5;6g.BycombiningEquations 7{19 and 7{20 ,we 137

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2(2n1;b2n1;a)k=3 2k22n1;6+2n1;b2n1;ak:(7{21) Figure7-8. Thelatticepointsq;r;s2P(2n2)andthelatticepointsinq+2n,r+2n,ands+2n,wheresisnexttobothqandrinthelatticeL2n2. Ifa=b,thenEquations 7{21 becomesd(n(x);n(y))=3 2k22n1;6k=3k2n1;6k=32n1=92n+1>32n+1: 7{21 becomes 2k22n1;622n1;ak=3k2n1;62n1;ak:(7{22) Since2n1;6;2n1;a2L2n1,wehave2n1;62n1;a2L2n1.Becausea6=6,2n1;62n1;a6=0.Hence06=2n1;62n1;a2L2n1.Itfollowsthatk2n1;62n1;ak2n1.ThenbyEquation 7{22 wehaved(n(x);n(y))32n1=92n+1>32n+1. Ifa6=band2n1;b6=2n1;a,thenthetwovectors2n1;aand2n1;barenotparallel.Sincek2n1;ak=k2n1;bk=2n1,itfollowsthatk2n1;a2n1;bk<22n1.Howeverk22n1;6k=22n1.Hence22n1;66=2n1;a2n1;b,i.e.,22n1;6+2n1;b

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7{21 ,wehaved(n(x);n(y))3 22n1=9 22n+1>32n+1. ThenextcorollaryfollowsdirectlyfromLemma 7.2.20 ln3,wherePBistheboundaryofthelimitofthePyxisstructure. Proof. 22n+1.Since2n+1=1 3n1,wehaven=CrnwhereC=31 3.ByLemma 7.2.7 ,PBcanbecoveredbythoseclosedballswithradiusnandcenteredatboundarylatticepointsofP(2n).ByLemma 7.2.3 ,itfollowsthatN(PB;n)4n+14,whereN(PB;n)isthesmallestnumberofclosedballsofradiusnneededtocoverPB. RecallthatB2n;4isthesetconsistingofboundarylatticepointsofP(2n)eachhavingboundaryindex4.LeteB2n;4=fn(q):q2B2n;4g.ByCorollary 7.2.21 ,themapnisonetoone.ItfollowsthatjeB2n;4j=jB2n;4j.ByLemma 7.2.3 ,wehavejB2n;4j=4n+2.HencejeB2n;4j=4n+2.ByLemma 7.2.20 ,foranyx;y2B2n;4withx6=y,wehaved(n(x);n(y))>32n+1=2n.SincejeB2n;4j=4n+2,itfollowsthatN(eB2n;4;n)4n+2,whereN(eB2n;4;n)isthesmallestnumberofclosedballsofradiusnneededtocovereB2n;4.ByLemma 7.2.15 ,wehaveeB2n;4PB.ItfollowsthatN(eB2n;4;n)N(PB;n).HenceN(PB;n)4n+2. Wehaveshownthat4n+2N(PB;n)4n+14foranyn2N.Itfollowsthat ln(4n+2) ln(1=n)ln(N(PB;n)) ln(1=n)ln(4n+14) ln(1=n):(7{23) 139

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ln(1=n)=limn!1ln(4n+2) ln(3n=C)=limn!1nln4+ln(1+2(4n)) ln3: ln(1=n)=ln4 ln3.Hence,byInequality 7{23 ,wehavelimn!1ln(N(PB;n)) ln(1=n)=ln4 ln3: 7.2.1 ,itfollowsthatthefractaldimensionofPBisln4 ln3. 140

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WehaveprovidedasystematicmathematicaltheoryforcomputingtheDFTonalatticeandstudiedtheDFTonsomeimportanthexagonalarraystructuressuchastheregularhexagonalstructureandthePyxisstructure.ItisshownthateacharrayofthetypeA(typeB)RHSisasetofcosetrepresentativesofthequotientgroupoftwohexagonallattices.HencetheDFTonsuchanarrayisamenabletothestandardDFT.WedevelopedanecientmethodforcomputingtheDFTonthesearraysandhaveshowntherelationbetweenthesearraysandsomepreviouslystudiedhexagonalarrays. ThePyxisstructureconsistsofasequenceofhexagonalarrayswithaschemeforlabelingthelatticepointsofthearrays.ItoriginatedwithPYXISInnovationInc.andisappliedinadigitalearthreferencemodel(Zhengetal.[ 50 ]).DevelopingalgebraandDFTforthePyxisstructure,computingthefractaldimensionofthePyxisstructure,anddevelopingalgorithmsfortheadditionofPyxislabelsareresearchtopicsproposedbythePyxisInnovationInc.WehaveprovidedarecursivedenitionofthePyxisstructureandanalgebraicmethodtolabelthelatticepointsofthePyxisstructure.BasedontherecursivedenitionandlabelingofthePyxisstructure,wehavedevelopedanalgorithmwithcomputationalcomplexityoforderntoaddanytwolabelsoflengthn.AlsowehaveshownthatP(n),thenthlevelofthePyxisstructure,doesnottiletheunderlyinghexagonallatticebytranslationsbyasublatticeforanyn>2.HencetheDFTonP(n)cannotbeevaluatedusingthemethodinChapter2.Furthermore,wehaveshownthatthelimitofthePyxisstructureinthespaceofnonemptycompactsubsetsofthetwodimensionalEuclideanspacewithHausdormetricexists,andthefractaldimensionofitsboundaryisln4 ln3. ThemethodsforcomputingtheDFTonhexagonalarraysdevelopedinthisresearchcanbeappliedtomanygeneralperiodicallysampledimages.AsshowninWikipedia[ 45 ],theDFTiswidelyemployedinsignalprocessingandrelatedeldstoanalyzethe 141

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142

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[1] J.D.Allen/SignalSolutionsUnltd,\Filterbanksforimagesonhexagonalgrid,"2003.http://james.fabpedigree.com/hexim1.htm(accessedAug.20,2007). [2] E.Anterrieu,P.Waldteufel,andA.Lannes,\Apodizationfunctionsfor2-Dhexagonallysampledsyntheticapertureimagingradiometers,"IEEETransactionsonGeoscienceandRemoteSensing,Vol.40,No.12,p.2531,2002. [3] B.Balasubramaniyam,E.A.Edirisinghe,andH.E.Bez,\Fractalimagecodinginhexagonalgridimages,"Proceedingsofthe24thIASTEDInternationalConferenceonInternetandMultimediaSystemsandApplications,p.67,2006. [4] M.Barnsley,FractalsEverywhere,AcademicPress,SanDiego,1988. [5] A.Camps,J.Bara,I.Corbella,F.Torres,\Theprocessingofhexagonallysampledsignalswithstandardrectangulartechniques:Applicationto2-Dlargeaperturesynthesisinterferometricradiometers,"IEEETransactionsonGeoscienceandRemoteSensing,Vol.35,No.1,p.183,1997. [6] D.B.Carr,A.R.Olsen,andD.White,\Hexagonmosaicmapsforthedisplayofunivariateandbivariategeographicaldata,"CartographyandGeographicInformationSystems,Vol.19,p.228,1992. [7] T.W.S.Cassels,AnIntroductiontotheGeometryofNumbers,Springer-Verlag,Berlin,1959. [8] M.Cartwright,FourierMethodsforMathematicians,ScientistsandEngineers,EllisHorwoodLimited,1990. [9] J.H.ConwayandN.J.A.Sloane,SpherePackings,Lattices,andGroups,Springer-Verlag,NewYork,NY,1988. [10] D.E.DudgeonandR.M.Mersereau,MultidimensionalDigitalSignalProcessing,PrenticeHall:EnglewoodClis,NJ,1984. [11] C.F.Dunkl,\Orthogonalpolynomialsonthehexagon,"SiamJ.Appl.Math.,Vol.47,No.2,p.343,1987. [12] J.C.Ehrhardt,\HexagonalfastFouriertransformwithrectangularoutput,"IEEETransactionsonSignalProcessing,Vol.41,p.1469,1993. [13] K.Falconer,FractalGeometry,JohnWiley,Chichester,1990. [14] A.P.FitzandR.J.Green,\FingerprintclassicationusingahexagonalfastFouriertransform,"PatternRecognition,Vol.29,p.1587,1996. [15] M.FrigoandS.G.Johnson,\FFTW,"Jun.2006,MassachusettsInstituteofTechnology.www.tw.org/tw3.pdf(accessedAug.20,2007). 143

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[16] L.GibsonandD.Lucas,\Vectorizationofrasterimagesusinghierarchicalmethods",ComputerGraphicsandImageProcessing,Vol.20,p.82,1982. [17] L.GinsonandD.Lucas,\Spatialdataprocessingusinggeneralizedbalancedternary,"ProceedingsoftheIEEEComputerSocietyConferenceonPatternRecognitionandImageProcessing,IEEEComputerSociety,p.566,1982. [18] A.M.Grigoryan,\Ecientalgorithmsforcomputingthe2-DhexagonalFouriertransforms,"IEEETransactionsonSignalProcessing,Vol.50,p.1438,2002. [19] A.GuessoumandR.M.Mersereau,\FastalgorithmsforthemultidimensionaldiscreteFouriertransform,"IEEETransactionsonAcoustics,Speech,andSignalProcessing,ASSP-34,p.937,1986. [20] G.JurasinskiandC.Beierkuhnlein,\Spatialpatternsofbiodiversity-assessingvegetationusinghexagonalgrids,"BiologyandEnvironment:ProceedingsoftheRoyalIrishAcademy,Vol.106B,No.3,p.401,2006. [21] W.Z.Kitto,A.Vinceand,D.C.Wilson,\Anisomorphismtheorembetweenthep-adicintegersandaringassociatedwithatilingofN-spacebypermutohedra,"DiscreteAppliedMath.,Vol.52,p.39,1994. [22] W.Z.Kittoand,D.C.Wilson,\Anisomorphismtheorembetweenthe7-adicintegersandaringassociatedwithahexagonallattice,"AppliedAlgebrainEngineering,Communication,andComputing,Vol.2,p.105,1991. [23] C.V.Loan,ComputationalFrameworksfortheFastFourierTransform,Philadelphia,PA:SIAM,1992. [24] D.Lucas,\AmultiplicationinN-space,"ProceedingsoftheAmericanMathematicalSociety,Vol.74,p.1,1979. [25] B.B.Mandelbrot,TheFractalGeometryofNature,W.H.FreemanandCo,1983. [26] R.M.Mersereau,\Theprocessingofhexagonallysampledtwo-dimensionalsignals,"ProceedingsoftheIEEE,Vol.67,p.930,1979. [27] R.M.MersereauandT.C.Speake,\AunitedtreatmentofCooley-TukeyalgorithmsfortheevaluationofthemultidimensionalDFT,"IEEETransactionsonAcoustics,Speech,andSignalProcessing,ASSP-29,p.1011,1983. [28] R.M.MersereauandT.C.Speake,\Theprocessingofperiodicallysampledmultidimensionalsignals,"IEEETransactionsonAcoustics,Speech,andSignalProcessing,ASSP-31,p.188,1983. [29] L.MiddletonandJ.Sivaswamy,\TheFFTinahexagonal-imageprocessingframework,"ProceedingsofImageandVisionComputing,NewZealand,p.231,2001.

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[30] L.MiddletonandJ.Sivaswamy,HexagonalImageProcessing,SpringerVerlag,London,2005. [31] C.MolerandS.Eddins/TheMathWorksInc.,\FasterFouriertransformsMATLAB6incorporatesFFTW,"1994-2007.www.mathworks.com/company/newsletters/news notes/clevescorner/winter01 cleve.html(accessedAug.20,2007). [32] F.MorganandR.Bolton,\Hexagonaleconomicregionssolvethelocationproblem,"Amer.Math.Monthly,Vol.109,p.165,2002. [33] P.Peterson,Close-packed,UniformlyAdjacent,Multiresolutional,OverlappingSpatialDataOrdering,InternationalPatentApplication,thePyxisInnovationInc.,Kingston,Ontario,Canada,2003. [34] M.A.Pinsky,IntroductiontoFourierAnalysisandWavelets,Brooks/Cole,2002. [35] C.M.Rader,\DiscreteFouriertransformswhenthenumberofdatasamplesisprime,"proceedingsoftheIEEE,p.1107,1968. [36] R.W.Ramirez,TheFFTFundamentalsandConcepts,EnglewoodClis,NJ:Prentice-Hall,1985. [37] K.Sahr,D.White,andA.J.Kimerling,\Geodesicdiscreteglobalgridsystems,"CartographyandGeographicInformationScience,Vol.30,No.2,p.121,2003. [38] R.C.StauntonandN.Storey,\Acomparisonbetweensquareandhexagonalsamplingmethodsforpipelineimageprocessing,"ProcSPIE,Vol.1194,p.142,1989. [39] R.Strand,\Usingthehexagonalgridforthree-dimensionalimages:DirectFouriermethodreconstructionandweighteddistancetransform,"18thInternationalConferenceonPatternRecognition(ICPR'06),p.1169,2006. [40] J.Sun,andJ.Yao,\AfastalgorithmofdiscretegeneralizedFouriertransformsonhexagondomains,"ChineseJournalofNumericalMathematicsandApplications,Vol.26,No.4,p.70,2004. [41] R.Tolimieri,M.An,andC.Lu,MathematicsofMultidimensionalFourierTrans-formAlgorithms,Springer-Verlag,NewYork,1997. [42] K.T.Tytkowski,\Hexagonalrasterforcomputergraphics,"Proceedingsofthe24thIASTEDInternationalConferenceonInternetandMultimediaSystemsandApplications,p.67,2006. [43] A.Vince,\Replicatingtessellations,"SiamJ.DiscreteMath.,Vol.6,p.501,1993. [44] A.VinceandX.Zheng,\ComputingthediscreteFouriertransformonalattice,"JournalofMathematicalImagingandVision,toappearinVol.28,No.2,2007.

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[45] Wikipedia,\TheDiscreteFourierTransform,"Aug.2006.http://en.wikipedia.org/wiki/Discrete Fourier transform(accessedAug.20,2007). [46] X.G.Xia,andB.W.Suter,\ConstructionofMalvarwaveletsonhexagons,"Appl.Comput.HamonicAnal.,Vol.2,p.65,1996. [47] J.L.Zapata,\ThegeneralizedmatrixprodictandfastFouriertransformforpermutohedralaggregates,"PhDDissertationofUniversityofFlorida,May2000. [48] J.L.ZapataandG.X.Ritter,\FastFouriertransformforhexagonalaggregates,"JournalofMathematicalImagingandVision,Vol.12,p.183,2000. [49] R.Zbikowski,\Flylikeay:Thecommonhouseyexecutesexquisitelypreciseandcomplexaerobaticswithlesscomputationalmightthananelectrictoaster,"IEEESpectrum,Vol.42,p.46,2005. [50] X.Zheng,A.Vince,G.Ritter,andD.Wilson,\ArithmeticandEcientFourierTransformforaUniformandMultiresolutionalEarthModel,"toappearintheProceedingoftheFifthInternationalSymposiumonDigitalEarth,June2007. [51] H.ZhuandG.X.Ritter,\Thep-productanditsapplicationsinsignalprocessing,"SIAMJournalofMatrixAnalysisandApplications,Vol.16,p.579,1995.

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XiqiangZhengwasborninJiangxiProvinceofChina.HereceivedhisMasterofScienceinmathematicsfromJiangxiUniversityinJanuaryof1991.From1991to1999,hetaughtatNanchangInstituteofAeronauticalTechnology,China.BeforehecametoUniversityofFloridaforhisPhDdegree,hestudiedatBowlingGreenStateUniversityforoneyear.HewillreceivehisPhDdegreeonappliedmathematicsinDecemberof2007andhisresearchinterestsincludeappliedmathematics,imageprocessing,andpatternrecognition. 147