EFFICIENT FOURIER TRANSFORMS ON HEXAGONAL ARRAYS
By
XIQIANG ZHENG
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOLOF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
2007
1
c© 2007 Xiqiang Zheng
2
To my mom Yuehua Su, for her continuous encouragement
3
ACKNOWLEDGMENTS
Though this research is an individual work, I could never have reached the heights
or explored the depths without the help, support, guidance and effort of many people.
Thank my advisors, Dr. Andrew Vince and Dr. Gerhard X. Ritter, for their invaluable
supervision and for the generous amount of time they spent on this research. Throughout
my doctoral work they encouraged me to develop independent thinking and research skills.
In particular, Dr. Vince provided important insight, applying algebraic and combinatorial
techniques to simplify many proofs. Dr. Gerhard X. Ritter introduced me to this exciting
and challenging research area, and guided this research until the time of his surgery.
I also thank the other members of my committee: Dr. David C. Wilson, Dr. Tim
Olson, and Dr. Joseph N. Wilson for their helpful discussions and encouragement. Dr.
David Wilson regularly attended research meetings and discussions, and provided many
important observations. Dr. Olson gave many useful suggestions.
I express my appreciation to the Department of Mathematics for offering me a full
teaching assistantship and a Grinter fellowship, and to the Department of Computer
and Information Science and Engineering (CISE) for offering me a partial research
assistantship. Also my appreciation goes to Mrs. Jane Smith and Mrs. Ronnie Khuri,
for the enjoyment of teaching with them, and to the staff of both the Mathematics
Department and the CISE Department for their kind help whenever needed.
Furthermore, thanks go to Pyxis Innovation Inc. for their financial support of this
research, and to the people there for their friendship and help. Special thanks to Dr.
Charles Herring for serving as the mentor of this research project and for his important
references.
I am most grateful to my wife, Lihua Yang, for her love, patience and encouragement
during these years of my graduate study.
4
TABLE OF CONTENTS
page
ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
CHAPTER
1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 DISCRETE FOURIER TRANSFORM (DFT) . . . . . . . . . . . . . . . . . . . 19
2.1 Discrete Fourier Transform on the Quotient Group of Two Lattices . . . . 192.2 Convolution and Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 DFT on a Lattice and the Corresponding Periodicity Matrix . . . . . . . . 242.4 Relation Between the DFT and the Continuous Fourier Transform . . . . . 25
3 DFT ON SOME PREVIOUSLY STUDIED HEXAGONAL ARRAYS . . . . . . 27
4 REGULAR HEXAGONAL STRUCTURE AND ITS TWO SPECIAL TYPES . 31
4.1 Regular Hexagonal Structures . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 Type A Regular Hexagonal Structure . . . . . . . . . . . . . . . . . . . . . 314.3 Type B Regular Hexagonal Structure . . . . . . . . . . . . . . . . . . . . . 364.4 Relation Between the Type A and B RHS and Some Previously Studied
Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5 FAST ALGORITHMS FOR COMPUTING THE DFT ON THE TWO SPECIALTYPES OF THE RHS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.1 Convert the DFT on a Tile of a Tiling by Translations by a Sublattice toa Standard DFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.2 Fast Algorithms for the DFT and its Inverse on the Type A RHS . . . . . 485.3 Fast Algorithms for the DFT and its Inverse on the Type B RHS . . . . . 585.4 Computational Complexity and Cooley-Tukey Factorization for the DFT
on the Type A and Type B RHS . . . . . . . . . . . . . . . . . . . . . . . 59
6 PYXIS STRUCTURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
6.1 Definition and Labeling of the Pyxis Structure . . . . . . . . . . . . . . . . 626.1.1 The Definition of the Pyxis Structure . . . . . . . . . . . . . . . . . 626.1.2 The Labeling of the Pyxis Structure . . . . . . . . . . . . . . . . . . 676.1.3 Addition of the Labels of the Pyxis Structure . . . . . . . . . . . . . 70
6.2 Pyxis P (n) Does Not Tile the Underlying Lattice by Translations by a Sublatticefor any n > 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
5
6.2.1 Pyxis P (2n−1) Does Not Tile the Underlying Lattice by Translationsby a Sublattice for any n > 1 . . . . . . . . . . . . . . . . . . . . . . 91
6.2.2 Pyxis P (2n) Does Not Tile the Underlying Lattice by Translationsby a Sublattice for any n > 1 . . . . . . . . . . . . . . . . . . . . . . 99
7 FRACTAL DIMENSION OF THE BOUNDARY OF THE PYXIS STRUCTURE 112
7.1 The Limit of the Pyxis Structure . . . . . . . . . . . . . . . . . . . . . . . 1127.2 Fractal Dimension of the Boundary of the Pyxis Structure . . . . . . . . . 114
8 SUMMARY OF THIS RESEARCH . . . . . . . . . . . . . . . . . . . . . . . . . 141
REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
BIOGRAPHICAL SKETCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
6
LIST OF FIGURES
Figure page
1-1 Arrays of type A and type B RHS. a). The third level of the type A RHS. b).The second level of the type B RHS. . . . . . . . . . . . . . . . . . . . . . . . . 13
1-2 The Pyxis structure at level one through four where P (1) consists of seven redhexagons, P (2) consists of 13 blue hexagons, P (3) consists of 55 green hexagons,and P (4) consists of 133 black hexagons. The three dashed vectors show thelabel addition 0506⊕ 2005 = 1040 . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1-3 Tessellating sphere using hexagons and 12 pentagons in multiresolutions. . . . . 15
1-4 Flattened polygons used to tessellate the sphere. (a) shows the 20 hexagons and12 pentagons in the tessellation. (b) displays each pentagon in Figure (a) as ahexagon with one of its six directions empty. . . . . . . . . . . . . . . . . . . . . 15
1-5 The (dashed) division lines of the next level are generated from those (solid)division lines of the previous level. (a) and (b) show the division lines near ahexagon and a pentagon of the previous level respectively. . . . . . . . . . . . . 16
1-6 The green polygons obtained from the division of the sphere at level one usingthe scheme of Figure 1-5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1-7 The red polygons obtained from the division of the sphere at level two usingscheme of Figure 1-5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3-1 The first three levels of the GBT 2 aggregates. . . . . . . . . . . . . . . . . . . . 27
3-2 Arrays studied in Ehrhardt [12], and Sun and Yao [40]. (a) Hexagonal samplingof a rectangular region used in [12] and Fitz and Green [14]. (b) The array usedin Sun and Yao [40] consisting of 3n2 lattice points, where n = 9. . . . . . . . . 28
3-3 Arrays studied in Anterrieu et al. [2]. (a) An array shown in Figure 3 on Page 2533of Anterrieu et al. [2]. (b) The array obtained from the array in Figure (a) byomitting the boundary points on the top row and the two upper sides. (c) Thearray obtained from the array in Figure (a) by omitting one of its two consecutiveboundary points. (d) The periodic extension from the array in Figure (c) to anarray of a rhombus shape. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4-1 Arrays of the type A RHS. (a) The lattice points of <A2 . (b) The lattice points
of <A3 . The coordinates inside each hexagon are for the basis
vA
1 ,vA2
. . . . . . 32
4-2 Arrays of the type B RHS. (a) The lattice points of <B2 . (b) The lattice points
of <B3 . The coordinates inside each hexagon are for the basis
vB
1 ,vB2
. . . . . . 38
4-3 The lattice points of <B2n (red) and LB (blue) where n = 6. . . . . . . . . . . . . 42
7
4-4 Comparison of a previously studied array structure with the type A RHS. (a)The lattice points of Γ4. (b) The lattice points of <A
3 . . . . . . . . . . . . . . . . 44
4-5 Comparison of a previously studied array structure with the type B RHS. (a)The lattice points of Υ3. (b) The lattice points of <B
3 . . . . . . . . . . . . . . . 45
5-1 The lattice points enclosed by the blue dashed polygon form the set of coset
representatives U of (LA3 )∗/(LA)∗. . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6-1 The generators of the lattices L1 and L2, and the lattice points contained inβ1 and β2. In this figure, the two red vectors are v1,1 and v1,2, the two dashedgreen vectors are v2,1 and v2,2, β1 consists of the six black * points, and β2 consistsof the six blue o points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6-2 The first 4 levels of the Pyxis structure where P (1) consists of 7 blue hexagons,P (2) consists of 13 red hexagons, P (3) consists of 55 green hexagons, and P (4)consists of 133 black hexagons. In this figure, the black vector is the sum of thetwo blue vectors. It follows that the coordinates of the lattice point labeled 0205is the sum of the coordinates of the two lattice points labeled 0030 and 0001,respectively. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
6-3 Voronoi cells of the lattice points of P(2t+1) which are generated from the latticepoints of P(2t −1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
6-4 The boundary hexagons of Q0 and Q. (a) The black hexagon Q0 has boundaryindex 1. (b) The boundary hexagon Q should be one of the three red hexagons. 97
6-5 The 13 lattice points in the set (q + β2n)⋃
(q + β2n−1) and their Voronoi cells. . 100
6-6 Generation of the boundary hexagons of P (2n). . . . . . . . . . . . . . . . . . . 107
6-7 Tiling of P (2). (a) Uj is a hexagon and next to Bj but not in P (2n) for j =1, 2. (b) Showing where is W3 located. . . . . . . . . . . . . . . . . . . . . . . . 110
7-1 The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 4, and the boundary index of R and T is 1. (b) The boundary hexagonsof P(2n+2) which have boundary index 2 or 4. Their Voronoi cells overlap theinterior of the boundary hexagon Q of P (2n), where A4, B4, and C4 have boundaryindex 4, and D2 and E2 have boundary index 2. . . . . . . . . . . . . . . . . . . 115
7-2 The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 2, and the boundary index of R and T is 1. (b) The boundary hexagonsof P(2n+2) which have boundary index 2 or 4, and whose Voronoi cells overlapthe interior of the boundary hexagon Q of P (2n), where A4 has boundary index4, and B2 and C2 have boundary index 2. . . . . . . . . . . . . . . . . . . . . . 116
8
7-3 The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 1, and the boundary index of R and T is 2 or 4. (b) The boundary hexagonsof P(2n+2) which have boundary index 2 or 4, and whose Voronoi cells overlapthe interior of the boundary hexagon Q of P (2n), where A and B have boundaryindex 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
7-4 The distance from a point x to a given lattice point y and the distance from xto the six neighbors of y in the lattice. . . . . . . . . . . . . . . . . . . . . . . . 120
7-5 The containment relation among Voronoi cells of P (2n) and P(2n+1). (a) showsthat the (blue) Voronoi cell V2n(y) is contained in the (black) Voronoi cell V2n−1(y),where V2n(y) has a horizontal side and V2n−1(y) has a vertical side. (b) showsthat y∈ P (2n) is the centroid of the triangle with vertices q∈ P(2n −1), r∈P(2n −1), and t∈ P(2n −1). It also shows that the (blue) Voronoi cell of y∈L2n is contained in the union of the (black) Voronoi cells of q∈ L2n−1, r∈ L2n−1,and t∈ L2n−1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7-6 The green Voronoi cell V2n+1(y) is contained in the black Voronoi cell V2n−1(y0). 124
7-7 For any two hexagons of P(2k+2) each having boundary index 2 or 4 such thatthe distance between them is ρ2k+2, there exists a hexagon of P(2k+2) that isnext to both of them, where Q, R and S are hexagons of P (2k). (a) The hexagonS is next to Q and R, and its centroid lies below the line connecting the centroidsof Q and R. (b) shows that, for any two blue hexagons of P(2k+2) (each havingboundary index 2 or 4) such that the distance between them is ρ2k+2, there existsa hexagon of P(2k+2) that is next to these two blue hexagons. . . . . . . . . . . 133
7-8 The lattice points q, r, s∈ P (2n−2) and the lattice points in q + β2n, r+β2n,and s + β2n, where s is next to both q and r in the lattice L2n−2. . . . . . . . . 138
9
Abstract of Dissertation Presented to the Graduate Schoolof the University of Florida in Partial Fulfillment of theRequirements for the Degree of Doctor of Philosophy
EFFICIENT FOURIER TRANSFORMS ON HEXAGONAL ARRAYS
By
Xiqiang Zheng
December 2007
Chair: Andrew VinceCochair: Gerhard X. RitterMajor: Mathematics
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 nth 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
10
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 nth 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 ln 4ln 3
.
11
CHAPTER 1INTRODUCTION
Traditional image processing algorithms and digital image transforms are usually
carried out on pixels of square grids. However, as shown in Allen [1], physical pixels
such as printer dots and electron beams usually have circular shapes and thus operate
more effectively on hexagonal grids, where a hexagonal grid is a tessellation of the plane
by regular hexagons. The pixels of hexagonal grids also provide for higher packing
density of discs and give a more accurate approximation of circular regions than that
of square grids. Furthermore the pixels of hexagonal grids are uniformly connected
in the sense that the distance from a given pixel to any adjacent pixel is the same.
Hence, hexagonal grids are used by a wide variety of researchers in areas such as image
processing (Middleton and Sivaswamy [30], Balasubramaniyam et al. [3], and Strand [39]),
computer graphics (Tytkowski [42]), geoscience (Carr et al. [6]) and ecology (Jurasinski
and Beierkuhnlein [20]). For example, they are being used in the soil moisture and
ocean salinity space mission (Anterrieu et al. [2], and Camps et al. [5]). F. Morgan and
R. Bolton have shown in [32] that, for the efficiency of the distribution of centers of
production, regular hexagons are superior to any other collection of shapes.
The set consisting of all centers of pixels of a hexagonal grid is called a hexagonal
lattice. In general, a d-dimensional lattice in Rd is the set of all integer linear combinations
of d independent vectors. The elements of a lattice are called lattice points. The Voronoi
cell of a lattice point of a d-dimensional lattice consists of those points of Rd which are
closest to that point than any other lattice point of the lattice. Because of the obvious
one to one correspondence between Voronoi cells and lattice points of a lattice, sometimes
we treat a set of lattice points as the corresponding set of Voronoi cells in the plotting to
get a better visualization effect which can be seen in Figure 1-1. An array of a lattice is
a finite subset of the lattice and an array structure is a sequence of arrays. The nth array
of an array structure is called the nth level of the array structure. For example, the array
12
structure whose nth level is a square shaped array of n× n lattice points of a square lattice
is used for image zooming on square grids. An array (structure) of the hexagonal lattice is
called a hexagonal array (structure). This research concerns the following two hexagonal
array structures which are used for hexagonal image processing.
(a) (b)
Figure 1-1. Arrays of type A and type B RHS. a). The third level of the type A RHS. b).The second level of the type B RHS.
A regular hexagonal structure (RHS) is a hexagonal array structure such that, at
any given level, the underlying hexagonal lattice is a disjoint union of translated copies,
and the set of lattice points whose coordinates are the same as the coordinates of those
translations also forms a hexagonal lattice. An algebraic definition of a RHS as well as
two particular types of RHSs, called type A and type B, are provided in Chapter 4. Each
array within the type A (type B) RHS consists of all hexagonal lattice points enclosed
in a regular hexagon and has the same centroid and axes of symmetry as the enclosing
hexagon. Figure 1-1 shows one array of each type. Pyxis structure, denoted P , is a
hexagonal array structure together with a natural method for labeling the lattice points
(or hexagons). Let P (n) denote the nth level of P for any integer n ≥ 0. The precise
definition of P and P (n) are given in Chapter 6, and P (1) through P (4) are shown
in Figure 1-2. The name and concept of the Pyxis structure originated with PYXIS
Innovation Inc. (Peterson [33]), a Canada based company whose goal is an efficient
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Figure 1-2. The Pyxis structure at level one through four where P (1) consists of seven redhexagons, P (2) consists of 13 blue hexagons, P (3) consists of 55 greenhexagons, and P (4) consists of 133 black hexagons. The three dashed vectorsshow the label addition 0506⊕ 2005 = 1040
sampling scheme for the surface of the earth (Figure 1-3). The following shows the
structure of this efficient sampling scheme. Consider a sphere which is tessellated by 20
regular hexagons and 12 regular pentagons. Figure 1-4 shows the sphere with those 32
polygons flattened onto the plane. For each side of those polygons, make a line segment
(as the dashed blue lines in Figure 1-5) with length being equal to the length of the given
side divided by√
3 such that this line segment and the given side are perpendicular to
and bisect each other. After any two neighboring ends are connected using lines as the
dashed red lines in Figure 1-5, we get a set of polygons at the next level which consist
of 80 hexagons and 12 pentagons dividing the sphere into smaller polygonal regions.
Figure 1-6 displays the flattened versions of those 80 hexagons and 12 pentagons (taken as
hexagons with one direction empty) on a plane. If the division rule shown in Figure 1-5 is
14
Figure 1-3. Tessellating sphere using hexagons and 12 pentagons in multiresolutions.
−60 −40 −20 0 20 40 60 80
−60
−40
−20
0
20
40
60
−60 −40 −20 0 20 40 60 80
−60
−40
−20
0
20
40
60
(a) (b)
Figure 1-4. Flattened polygons used to tessellate the sphere. (a) shows the 20 hexagonsand 12 pentagons in the tessellation. (b) displays each pentagon in Figure (a)as a hexagon with one of its six directions empty.
applied recursively, then the sphere is divided into smaller and smaller polygonal regions
but the number of pentagonal regions is always 12. The division of the sphere by applying
such recursion n times is called the division of the sphere at level n. Figure 1-6 and 1-7
15
−20 −15 −10 −5 0 5 10 15 20
−20
−15
−10
−5
0
5
10
15
20
−15 −10 −5 0 5 10 15
−15
−10
−5
0
5
10
15
(a) (b)
Figure 1-5. The (dashed) division lines of the next level are generated from those (solid)division lines of the previous level. (a) and (b) show the division lines near ahexagon and a pentagon of the previous level respectively.
−60 −40 −20 0 20 40 60 80
−60
−40
−20
0
20
40
60
Figure 1-6. The green polygons obtained from the division of the sphere at level one usingthe scheme of Figure 1-5.
show the flattened versions of such divisions at level one and level two, respectively. For
any integer n > 1, the set of all cells of the sphere at the nth level is a disjoint union of 20
copies of P (n− 1) and 12 copies of P (n) by omitting one of its six directions. For example,
16
−60 −40 −20 0 20 40 60 80
−60
−40
−20
0
20
40
60
Figure 1-7. The red polygons obtained from the division of the sphere at level two usingscheme of Figure 1-5.
Figure 1-7 shows that the set of all cells of the sphere at level two is a disjoint union of 20
copies of P (1) which are in blue and 12 copies of P (2) which are in red by omitting one of
the six directions.
Certain properties of those two hexagonal array structures, in particular the discrete
Fourier transform (DFT), are studied. The DFT on arrays of hexagonal lattices is
an important topic in hexagonal image processing as explained in Middleton and
Sivaswamy [30]. We provide an efficient method to compute the DFT on any array of
type A or type B RHS and show that the computational complexity is O(N log N), where
N is the number of the lattice points of the array. We also provide a recursive definition
of the Pyxis structure and use it to show that the DFT as defined in Chapter 2 cannot be
applied to any array of the Pyxis structure when the level is larger than two.
As shown in Figure 1-2, the cells of P (1) are labeled 0,1,2,...,6, and arranged in a
certain order. The cells of P (2) are labeled ij where i, j = 0, 1, 2, ..., 6 and either i or j
17
is 0. Furthermore, the cell of P (2) labeled i0 has the same centroid as the one of P (1)
labeled i for any i = 0, 1, 2, ..., 6, and the cells labeled 0i for i = 1, 2, ..., 6 surround the
one labeled 00. The labeling system of P (n) for any integer n ≥ 0 appears in Chapter
6. Those labels are important for quick data retrieval. The vector addition of any two
Pyxis labels as defined in Chapter 6 (Figure 1-2) is useful in proving some properties of
the Pyxis structure in Chapter 6, and a research topic proposed by the Pyxis Innovation
Inc. We provide an efficient algorithm to perform such addition. Finally we compute the
fractal dimension of the boundary of the limit of the Pyxis structure P (n) as n →∞ since
it measures how convoluted that boundary is. Our computation shows that the dimension
is ln 4ln 3
.
18
CHAPTER 2DISCRETE FOURIER TRANSFORM (DFT)
2.1 Discrete Fourier Transform on the Quotient Group of Two Lattices
Throughout this research, N, Z, R, and C denote the set of positive integers, integers,
real numbers, and complex numbers, respectively. For any d ∈ N, let Zd, Rd, and Cd
denote the set of d-dimensional column vectors whose components are integers, real
numbers, and complex numbers, respectively. An abelian group is a set G together with a
closed binary operation + defined on G with the following properties:
1. (Associativity) For any x,y, z ∈ G, (x + y) + z = x + (y + z).
2. (Existence of a zero) There exists an element 0 ∈ G such that, for each x ∈ G,
x + 0 = x = 0 + x.
3. (Existence of an inverse) For each x ∈ G, there exists an element of G, denoted −x,
such that x + (−x) = 0 = (−x) + x.
4. (Commutativity) For any x,y ∈ G, x + y = y + x.
In this research, all groups will be abelian and we write x + (−y) as x− y for any
two elements x and y of a group G. Let G0 be a nonempty subset of a group G. If G0
also forms a group under the operation + of G, then G0 is called a subgroup of G. Now
assume that G0 be a subgroup of a group G. For any p,g ∈ G, if p − g ∈ G0, then we
say that p is congruent to g modulo G0, denoted by p ≡ g mod G0. For any p ∈ G, let
p = u ∈ G : u ≡ p mod G0. Obviously p = p + G0 where p + G0 denotes the set
p + y ∈ G : y ∈ G0. The set p is called a coset of G0 in G. For any pair p,g ∈ G,
it is easy to show that either p = g or p⋂
g = ∅. Define p + g = p + g and let
G/G0 = p : p ∈ G. It is easy to check that the set G/G0 together with the binary
operation + defined on G/G0 is an abelian group. This group is called the quotient group
of the group G by the subgroup G0. By choosing one representative from each coset of G0
in G, we get a set of coset representatives of the quotient group G/G0.
19
If v1,v2, ...,vd are linearly independent vectors in Rd, then the set defined by L =∑d
j=1 njvj : nj ∈ Z, 1 ≤ j ≤ d
is a d-dimensional lattice and v1,v2, ...,vd is called
a set of generators of L. If d = 2, v1 =
1
0
and v2 =
0
1
, then L is the
standard square lattice commonly used in image processing. A hexagonal lattice has a set
of generators v1,v2 such that ‖v1‖ = ‖v2‖ and the angle between v1 and v2 is π3, where
the notation ‖x‖ denotes the norm of the x for any x ∈ R2. Let L0 be a nonempty subset
of a lattice L. If L0 itself is a lattice, then L0 is called a sublattice of L. Obviously a lattice
is an abelian group and a sublattice is a subgroup. For any ∅ 6= T ⊆ L and x ∈ L, to avoid
confusion in some expressions, we let Tx = x + T . If T is a set of coset representatives of
the quotient group L/L0, then T tiles the lattice L by translations by the sublattice L0 in
the sense that⋃ Tp : p ∈ L0 = L and Tp = Tg whenever Tp
⋂Tg 6= ∅ for p,g ∈ L0.
Hence T is called a tile of L. Each tiling (tile) involved in this research is a tiling (tile)
by translations by a sublattice. Also we just consider those sublattices of L which have
the same dimension as L. In this case, the quotient group L/L0 is finite by Lemma 1 of
Section I.2.2 of Cassels [7]. The notation L0 ≺ L is used to denote that L0 is a sublattice
of L which has the same dimension as L. The inner product of two vectors r, s ∈ Cd is
defined as 〈r, s〉 =∑d
j=1 r∗jsj where rj and sj are the jth component of r and s respectively,
and rj∗ denotes the complex conjugate of rj. If r, s ∈ Rd, then 〈r, s〉 = rT s = sT r. The
dual of a lattice L is L∗ =s ∈ Rd : 〈r, s〉 ∈ Z for all r ∈ L
.
In this research, the superscript ∗ means the dual when it is applied to a lattice, and
means the complex conjugate when it is applied to a complex number or vector. The
cardinality of a set S is denoted |S|. Furthermore, for two groups A and B, the symbol ∼=means that A and B are isomorphic. The next lemma, whose proof appears in Zapata [47],
and Conway and Sloane [9], provides some useful properties of dual lattices.
Lemma 2.1.1. Let L0 and L be two lattices. Then we have the following results.
1. L0 ≺ L if and only if L∗ ≺ L∗0.
20
2. (L∗)∗ = L for any lattice L.
3. If L ≺ L0, then L/L0∼= L∗0/L
∗ and hence |L/L0| = |L∗0/L∗|.For any finite abelian group G, let CG denote the set of all complex-valued functions
defined on G. Let L0 ≺ L, G = L/L0, and G = L∗0/L∗. The discrete Fourier transform
(DFT) on the quotient group G is a function F : CG → CG defined by
a(s) = F(a)(s) :=∑r∈G
a(r) · e−2πi〈r,s〉. (2–1)
The inverse Fourier transform is the function F−1 : CG → CG defined by
F−1(a)(r) =1
|G|∑
s∈G
a(s) · e2πi〈r,s〉. (2–2)
In the definition of the discrete Fourier transform, the domains G and G are called
the spatial and frequency domain of the Fourier transform F respectively.
Proposition 2.1.2. The discrete Fourier transform is well defined.
Proof. If r1, r2 ∈ G, s1, s2 ∈ G such that r1 = r2 and s1 = s2, then r1 − r2 = 0 ∈ G
and s1 − s2 = 0 ∈ G. It follows that r1 − r2 ∈ L0 and s1 − s2 ∈ L∗. Obviously
〈r1, s1〉 = 〈r2, s2〉 + 〈r2, s1 − s2〉 + 〈r1 − r2, s1〉. Since s1 − s2 ∈ L∗ and r2 ∈ L, we have
〈r2, s1 − s2〉 ∈ Z. Also r1 − r2 ∈ L0 and s1 ∈ L∗0 imply that 〈r1 − r2, s1〉 ∈ Z. Hence
e−2πi〈r1,s1〉 = e−2πi〈r2,s2〉. Therefore the DFT is well defined.
To prove that the DFT is invertible, i.e., F−1 F(a) = a for any a ∈ CG, we need the
following lemma:
Lemma 2.1.3. Let L0 ≺ L, G = L/L0 and G = L∗0/L∗. If r ∈ G and h(r) =
∑s∈G e2πi〈r,s〉,
then
h(r) =
|G|, if r = 0,
0, otherwise.
21
Proof. Similar to Proposition 2.1.2, h(r) is well defined. If r = 0, then
∑
s∈G
e2πi〈r,s〉 =∑
s∈G
e2πi〈0,s〉 = |G|.
If r 6= 0, then r 6∈ L0 = (L∗0)∗. It follows that there exists s0 ∈ L∗0 such that 〈r, s0〉 6∈ Z
by the definition of dual lattices. Hence e2πi〈r,s0〉 6= 1. Let c = e2πi〈r,s0〉. Since h(r) =∑
s∈G e2πi〈r,s〉 =∑
s∈G e2πi〈r,s0+s〉 = e2πi〈r,s0〉 ∑s∈G e2πi〈r,s〉 = c(h(r)), we have h(r)(1−c) = 0.
Since c 6= 1, h(r) = 0.
Theorem 2.1.4. (Inversion Theorem) If L0 ≺ L, G = L/L0, and G = L∗0/L∗, then
F−1 F(a) = a for any a ∈ CG.
Proof. For any t ∈ G where t ∈ L, we have
F−1(a)(t) =1
|G|∑
s∈G
a(s) · e2πi〈t,s〉
=1
|G|∑
s∈G
∑r∈G
a(r) · e−2πi〈r,s〉· e2πi〈t,s〉
=1
|G|∑r∈G
a(r)
∑
s∈G
e2πi〈t−r,s〉
(2–3)
By Lemma 2.1.3, we have
∑r∈G
a(r)
∑
s∈G
e2πi〈t−r,s〉
= a(t) · |G|. (2–4)
It follows that F−1(a)(t) = a(t) for any t ∈ G. Hence F−1(a) = a.
2.2 Convolution and Correlation
In this section, we consider two Fourier transform relationships that constitute a basic
link between the spatial and frequency domains. These relationships, called convolution
and correlation, are of fundamental importance in image processing techniques based on
the Fourier transform.
22
Let G be a finite abelian group and let a, b ∈ CG. The convolution c ∈ CG of a and b,
written c = a ∗ b, is defined by
c(t) =∑r∈G
a(r)b(t− r) for all t ∈ G .
The importance of the concept of convolution lies in the fact that convolution in the
spatial domain corresponds to multiplication in the frequency domain and vice versa. The
proof of the following Convolution Theorem appears in Zapata and Ritter [48].
Theorem 2.2.1. (Convolution Theorem) Let L0 ≺ L and G = L/L0. If a, b ∈ CG, then
a ∗ b = a · b.A principle application of correlation in image processing is in the area of template
or prototype matching, where the problem is to find the closest match between a given
unknown image and a set of images of known origin. One approach to this problem is
to compute the correlation between the unknown image and each of the known images.
Since the resultant correlations are 2-dimensional functions, this involves searching for the
largest amplitude of each function. As in the case of convolution, the computation of the
correlation is often more efficiently carried out in the frequency domain using the Fourier
transform. Let G be a finite abelian group and a, b ∈ CG. The correlation c ∈ CG of a and
b, written c = a b, is defined as c(t) =∑
r∈G a∗(r)b(t + r) for all t ∈ G. If a = b, then the
correlation is called an autocorrelation.
Theorem 2.2.2. (Correlation Theorem) Let L0 ≺ L and G = L/L0. If a, b ∈ CG and
c = a b, then c = a∗ · b.
Proof. For all t ∈ G, c(t) =∑
r∈G a∗(r)b(t + r) =∑
s∈G a∗(−s)b(t − s). Define
g ∈ CG by g(s) = a∗(−s) for any s ∈ G. Then c(t) =∑
s∈G g(s)b(t − s) = (g ∗ b)(t).
Hence c = g · b by Theorem 2.2.1. Since g(s) = a∗(−s) for any s ∈ G, we have
g(t) =∑
s∈G g(s) · e−2πi〈s,t〉 =∑
s∈G g(−s) · e2πi〈s,t〉
=∑
s∈G a∗(s) · e2πi〈s,t〉 = (a(t))∗. It follows that g = a∗. Therefore c = a∗ · b.
23
2.3 DFT on a Lattice and the Corresponding Periodicity Matrix
In this section, we discuss the DFT on a tile of a lattice. To make the DFT easier
to be computed, we write the DFT in terms of an integer matrix and vectors which have
integer components.
A matrix V is called a sampling matrix of a lattice L if its columns are a set of
generators for L. Obviously in such a case, L =V n : n ∈ Zd
= VZd. It follows easily
from the following lemma that the dual of a hexagonal lattice is also hexagonal.
Lemma 2.3.1. Let L be a lattice. Then V is a sampling matrix of L if and only if (V T )−1
is a sampling matrix of the dual lattice L∗.
Proof. Let V be a sampling matrix of L. Then L =V n : n ∈ Zd
. Hence s ∈ L∗ if and
only if 〈V n, s〉 = (V n)T s = nT (V T s) ∈ Z for all n ∈ Z, if and only if V T s ∈ Zd, if and
only if s ∈ (V T )−1Zd. It follows that (V T )−1 is a sampling matrix of the dual lattice L∗.
Similarly, if (V T )−1 is a sampling matrix of L∗, then V is a sampling matrix of L.
A non singular matrix M with integer entries is called a periodicity matrix. A
set of coset representatives of the quotient group Zd/MZd is also called a set of coset
representatives associated with the periodicity matrix M.
Lemma 2.3.2. Let L0 ≺ L and V be a sampling matrix of L. Then there is a periodicity
matrix M such that V M is a sampling matrix of L0. Furthermore V = [(V M)T ]−1 is a
sampling matrix of L∗0.
Proof. Let B be a sampling matrix of L0 whose columns are b1,b2, ...,bd, and let vr be
the rth column of V for r = 1, 2, ..., d. Since br ∈ L for each r = 1, 2, ..., d, there are
integers ljr such that br =∑d
j=1 ljrvj. Let M be the matrix whose entries are ljr. Then
we have B = V M. It follows that V M is a sampling matrix of L0. Hence V is a sampling
matrix of L∗0 since V = (BT )−1.
Let L0 ≺ L, G = L/L0, G = L∗0/L∗, and F : CG → CG be the DFT. If V is a sampling
matrix of L and V a sampling matrix of L∗0, then V is called a sampling matrix in the
24
spatial domain and V is called a sampling matrix in the frequency domain of the DFT F .
For any a ∈ CG, let f ∈ CL be defined by f(u) = a(u + L0) for any u ∈ L. Then f is
called the periodic extension of a. Let P and Q be any set of coset representatives of the
quotient group L/L0 and L∗0/L∗ respectively. Then Equation 2–1 for computing the DFT
on L/L0 becomes the following equation for computing the DFT on P .
f(s) =∑r∈P
f(r) · e−2πi〈r,s〉, for all s ∈ Q. (2–5)
Now for any r ∈ P and s ∈ Q, let [r] be the coordinates of r with respect to the basis
that are the generators of L, and let [s] be the coordinates of s with respect to the basis
that are the generators of L0∗. Then r = V [r] and s = V [s]. By Lemma 2.3.2, there is a
periodicity matrix M such that the sampling matrix of L0 is V M and the sampling matrix
of L∗0 is V = [(V M)T ]−1. Let IP = [p] : p ∈ P ⊂ Zd and IQ = [q] : q ∈ Q ⊂ Zd.
Then 〈r, s〉 = sT r = (V [s])T V [r] = [s]T V T V [r] = [s]T (V M)−1V [r] = [s]TM−1V −1V [r] =
[s]TM−1[r]. Hence Equation 2–5 becomes the following equation.
f(V [s]) =∑r∈P
f(V [r]) · e−2πi[s]T M−1[r], for all s ∈ Q. (2–6)
By replacing [r] and [s] with m and k respectively, Equation 2–6 becomes the following
simpler equation.
f(V k) =∑m∈IP
f(V m) · e−2πikT M−1m, for all k ∈ IQ. (2–7)
2.4 Relation Between the DFT and the Continuous Fourier Transform
In this section, we show that the relation between the continuous Fourier Transform
and the DFT for the 1-dimensional case as appears in Cartwright [8] also holds for the
2-dimensional case. For any complex-valued, Lebesgue integrable function f(x) with
x ∈ R2, by Pinsky [34] its Fourier transform is the complex-valued function f(y) defined
25
by the integral
f(y) = F(f)(y) :=
∫
R2
f(x) · e−2πi〈x,y〉dx, for any y ∈ R2. (2–8)
Let L be a 2-dimensional lattice, L0 ≺ L, and P a set of coset representatives of the
quotient group L/L0. Let P be the union of the Voronoi cells of the lattice points of P
and A the area of a Voronoi cell of the lattice L. In the finite analogue, we replace R2 in
Equation 2–8 by P . The corresponding discretized version of Equation 2–8 is the following
equation.
fP (y) :=∑p∈P
f(p) · e−2πi〈p,y〉 · A, for any y ∈ R2. (2–9)
Now let Q be a set of coset representatives of the quotient group L∗0/L∗ and Q
be the union of the Voronoi cells of the lattice points of Q. Since P is a set of coset
representatives of L/L0, P tiles R2 by translations by the sublattice L0. Similarly Q tiles
R2 by translations by the sublattice L∗ of L∗0. We claim that fP is periodic in the sense
that fP (y) = fP (y + s0) for any s0 ∈ L∗. Since s0 ∈ L∗ and p ∈ L, we have 〈p, s0〉 ∈ Z.
Then 〈p,y + s0〉 = 〈p,y〉 + 〈p, s0〉 follows that e−2πi〈p,y+s0〉 = e−2πi〈p,y〉. Hence, by
Equation 2–9, fP (y + s0) = fP (y). Therefore fP (y) is periodic. By the definition of Q,
Q is the set of the sampled points of Q in the lattice L∗0. In Equation 2–9, the expression∑
p∈P f(p) · e−2πi〈p,y〉 is the DFT as defined in Equation 2–5 or 2–7.
26
CHAPTER 3DFT ON SOME PREVIOUSLY STUDIED HEXAGONAL ARRAYS
The 2-dimensional generalized balanced ternary, denoted GBT 2, has been used in
Zapata and Ritter [48] through Gibson and Lucas [17] for the processing of hexagonally
sampled images. The nth level of the GBT 2, denoted GBT 2(n), consists of 7n hexagons.
For any n ∈ N , the GBT 2(n + 1) is the union of 7 translated copies of the GBT 2(n)
as shown in Figure 3-1. The GBT 2 is perhaps the most studied example because of the
elegant method of indexing its cells (lattice points) described in Gibson and Lucas [17],
Kitto [21], and Middleton and Sivaswamy [30]. A periodicity matrix for the GBT 2(n)
−5 0 5−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 0 5−5
−4
−3
−2
−1
0
1
2
3
4
5
−10 −5 0 5 10
−10
−5
0
5
10
Figure 3-1. The first three levels of the GBT 2 aggregates.
is Bn where B =
2 1
−1 3
. The DFT on the GBT 2(n) and its fast algorithms have
been successfully developed in Zapata and Ritter [48], and Middleton and Sivaswamy [29].
Some deficiencies of the GBT 2 have been pointed out by the people working in the Pyxis
Innovation Inc. (the shared document of Pyxis Innovation Inc., 2003). For example, the
27
number of hexagons of the GBT 2(n + 1) is 7 times that of the GBT 2(n). A factor less
than 7 would be better for image zooming in from one level to the next. When n is large,
the region occupied by the GBT 2(n) is quite different from the regions occupied by usual
images and hence it is not convenient to apply the DFT on usual images.
The DFT on arrays of rectangular shape of a hexagonal lattice (shown in Figure 3-2 (a))
was utilized in Ehrhardt [12], and Fitz and Green [14]. For any such array, the sampling
matrix of the lattice L is
1 −12
0√
32
and the periodicity matrix is of the form M =
n1n2
2
0 n2
, where n1 and n2 are integers with n2 divisible by 2. Hence the sampling
matrix in the frequency domain is V =
1n1
0
0 2√3n2
, from which it follows that the
sampling grid in the frequency domain is rectangular. Thus it is convenient to output
frequencies on a monitor that uses a rectangular grid. Let V1 and V2 be the two columns
of V . Since both n1 and n2 are integers, the length of V1 is not equal to that of V2 for
any integers n1 and n2. Hence the sampling lattice in the frequency domain is not square.
Some fast algorithms for computing such a DFT were analyzed in Ehrhardt [12].
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(a) (b)
Figure 3-2. Arrays studied in Ehrhardt [12], and Sun and Yao [40]. (a) Hexagonalsampling of a rectangular region used in [12] and Fitz and Green [14]. (b) Thearray used in Sun and Yao [40] consisting of 3n2 lattice points, where n = 9.
28
In Sun and Yao [40], a fast algorithm was developed for computing the DFT on an
array which consists of 3n2 lattice points of a hexagonal lattice. Figure 3-2 (b) shows this
array for n = 9 where the top row consists of n lattice points but the bottom row consists
of n + 1 lattice points. A periodicity matrix for this array is M =
2n n
n 2n
.
In Anterrieu et al. [2], the hexagonal array shown in Figure 3-3 (a) was studied.
Because the hexagonal array in Figure 3-3 (a) does not tile the underlying hexagonal
lattice by translations by any sublattice, the DFT on such an array cannot be formulated
using the method in Chapter 2. Hence the authors in Anterrieu et al. [2] remove half of
the boundary points to apply a 2-dimensional standard DFT. After half of the boundary
points of the array in Figure 3-3 (a) are omitted in a certain way, we get the array shown
in Figure 3-3 (b) or Figure 3-3 (c), which consists of n2 lattice points. The arrays in
Figure 3-3 (b) or Figure 3-3 (c) do tile the underlying hexagonal lattice by translations by
a sublattice. A periodicity matrix for the array in either Figure 3-3 (b) or Figure 3-3 (c) is
M =
n n
0 n
. Since M is also a periodicity matrix of the rhombus shaped array shown
in Figure 3-3 (d), the DFT on each of the arrays in Figure 3-3 (b) and Figure 3-3 (c) is
the same as the DFT on that rhombus shaped array and hence can be computed using the
fast algorithms for a 2-dimensional standard DFT.
For any array in Figure 3-2 (b) or Figure 3-3 (b), the number of lattice points on its
top row is different from that on the bottom row. Hence the shape of those arrays is not
a regular hexagon. In the next chapter, we will consider hexagonal arrays whose 6 sides
have the same number of lattice points and have the same centroid and symmetric axes as
a regular hexagon.
29
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(c) (d)
Figure 3-3. Arrays studied in Anterrieu et al. [2]. (a) An array shown in Figure 3 onPage 2533 of Anterrieu et al. [2]. (b) The array obtained from the array inFigure (a) by omitting the boundary points on the top row and the two uppersides. (c) The array obtained from the array in Figure (a) by omitting one ofits two consecutive boundary points. (d) The periodic extension from thearray in Figure (c) to an array of a rhombus shape.
30
CHAPTER 4REGULAR HEXAGONAL STRUCTURE AND ITS TWO SPECIAL TYPES
4.1 Regular Hexagonal Structures
In algebraic terms, a regular hexagonal structure (RHS) is an array structure such
that each array in the structure is a set of coset representatives of the quotient group of
two hexagonal lattices. Let V be a sampling matrix of a hexagonal lattice L with two
columns v1, v2 satisfying ‖v1‖ = ‖v2‖ and such that the angle between v1 and v2 is π3.
If V
r
s
is an arbitrary point of L, then V
r − s
r
is just V
r
s
rotated by an
angle of π/3. Hence, a hexagonal sublattice of L has a sampling matrix of the form V M,
where M =
r r − s
s r
with r, s ∈ Z.
If we take r = 2n and s = n (r = n and s = 0) in the matrix M above, then we
get a periodicity matrix for the array in Sun and Yao [40] (Anterrieu et al. [2]). Hence the
arrays in Sun and Yao [40] or Anterrieu et al. [2] constitute a RHS. In the following, we
study arrays in the type A and type B RHS which are closely related to those in Sun and
Yao [40], and Anterrieu et al. [2].
4.2 Type A Regular Hexagonal Structure
Throughout this research, we let vA1 =
1
0
, vA
2 =
−1
2√
32
, and
LA =n1(v
A1 ) + n2(v
A2 ) : n1, n2 ∈ Z
.
The lattice LA is called the type A hexagonal lattice. In this section, we consider the type
A RHS whose algebraic definition is provided in the following. For any n ∈ N, let
<An =
j(vA
1 ) + k(vA2 ) : j, k ∈ Z, |j| ≤ n, |k| ≤ n, |j − k| ≤ n
.
We call <An the nth level of the type A RHS.
31
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
(−2, −2)
(−2, −1)
(−2, 0)
(−1, −2)
(−1, −1)
(−1, 0)
(−1, 1)
(0, −2)
(0, −1)
(0, 0)
(0, 1)
(0, 2)
(1, −1)
(1, 0)
(1, 1)
(1, 2)
(2, 0)
(2, 1)
(2, 2)
−4 −3 −2 −1 0 1 2 3 4
−3
−2
−1
0
1
2
3
(−3, −3)
(−3, −2)
(−3, −1)
(−3, 0)
(−2, −3)
(−2, −2)
(−2, −1)
(−2, 0)
(−2, 1)
(−1, −3)
(−1, −2)
(−1, −1)
(−1, 0)
(−1, 1)
(−1, 2)
(0, −3)
(0, −2)
(0, −1)
(0, 0)
(0, 1)
(0, 2)
(0, 3)
(1, −2)
(1, −1)
(1, 0)
(1, 1)
(1, 2)
(1, 3)
(2, −1)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(3, 0)
(3, 1)
(3, 2)
(3, 3)
(a) (b)
Figure 4-1. Arrays of the type A RHS. (a) The lattice points of <A2 . (b) The lattice points
of <A3 . The coordinates inside each hexagon are for the basis
vA
1 ,vA2
.
The lattice points and corresponding Voronoi cells of <A2 and <A
3 are shown in
Figure 4-1 (a) and Figure 4-1 (b), respectively. To show <An is a set of coset representatives
of the quotient group of two hexagonal lattices, we need the following four lemmas.
Lemma 4.2.1. For any n ∈ N, the nth level of the regular hexagonal structure <An consists
of 3n2 + 3n + 1 lattice points.
Proof. By the definition of the type A RHS,
<An =
j(vA
1 ) + k(vA2 ) : j, k ∈ Z, |j| ≤ n, |k| ≤ n, |j − k| ≤ n
.
If j = 0, then the inequality |j − k| ≤ n becomes |k| ≤ n and, hence, k can take 2n + 1
values. If j > 0, then the common solution of the two inequalities |k| ≤ n and |j − k| ≤ n
is j − n ≤ k ≤ n and hence k takes 2n − j + 1 values. The situation for j < 0 is similar.
Hence, the total number of those hexagons is (2n+1)+2n∑
j=1
(2n− j +1) = 3n2 +3n+1.
Lemma 4.2.2. If a,b ∈ <An , then a± b ∈ <A
2n.
Proof. Let a = a1(vA1 ) + a2(v
A2 ) ∈ <A
n ,b = b1(vA1 ) + b2(v
A2 ) ∈ <A
n , and c = a − b. By the
definition of the type A RHS, |aj| ≤ n for j = 1, 2 and |a2 − a1| ≤ n. For the same reason,
|bj| ≤ n for j = 1, 2 and |b2 − b1| ≤ n. If c =
g1
g2
, then |gj| = |aj − bj| ≤ |aj| + |bj| ≤
32
n + n = 2n for j = 1, 2, and |g2 − g1| = |(a2 − b2) − (a1 − b1)| = |(a2 − a1) − (b2 − b1)| ≤|a2 − a1| + |b2 − b1| ≤ n + n = 2n. Hence c ∈ <A
2n. Obviously b ∈ <An implies −b ∈ <A
n by
the definition of the type A RHS. Thus a− b = a + (−b) ∈ <A2n.
Lemma 4.2.3. The maximal distance of the lattice points of <An from the origin 0 is n.
Proof. It follows directly from the fact that all lattice points of <An are enclosed in the
regular hexagon with vertices ±n(vA1 ),±n(vA
2 ), and ±n(vA1 − vA
2 ).
For any n ∈ N, let V A1,n = (2n + 1)vA
1 + (n + 1)vA2 , V A
2,n = n(vA1 ) + (2n + 1)vA
2 , and
LAn =
n1(V A
1,n) + n2(V A2,n) : n1, n2 ∈ Z
.
Lemma 4.2.4. LAn
⋂<A2n = 0.
Proof. By definition, V A1,n = (2n + 1)vA
1 + (n + 1)vA2 and V A
2,n = n(vA1 ) + (2n + 1)vA
2 .
It follows that V A1,n 6∈ <A
2n since 2n + 1 > 2n. Similarly V A2,n 6∈ <A
2n. Let V A3,n = V A
1,n −V A
2,n = (n + 1)vA1 − n(vA
2 ) = l1(vA1 ) + l2(v
A2 ) where l1 = n + 1 and l2 = −n. Since
|l1 − l2| = 2n + 1 > 2n, V A3,n 6∈ <A
2n. By direct computation, we have |V A1,n| = |V A
2,n| and
the angle between V A1,n and V A
2,n is π3. Since LA
n is a hexagonal lattice generated by V A1,n and
V A1,n, the set
±V A
1,n,±V A2,n,±V A
3,n
contains the only 6 closest nonzero lattice points of LA
n
from the origin 0. For any other nonzero lattice point of LAn , its distance from the origin
0 is at least |V A1,n + V A
2,n| = |(3n + 1)vA1 + (3n + 2)vA
2 | =√
9n2 + 9n + 3 > 2n. However,
by Lemma 4.2.3, the maximal distance of the lattice points of <A2n from 0 is 2n. Hence
LAn
⋂<A2n = 0.
To prove the next theorem, we need the following lemma which follows trivially from
Lemma 1 of Section I.2.2 of Cassels [7].
Lemma 4.2.5. Let L be a lattice with a sampling matrix V . If M is a periodicity matrix
and L0 is the sublattice of the lattice L such that V M is a sampling matrix of L0, then the
order of the quotient group L/L0 is the absolute value of the determinant of the matrix M.
33
In the following, we let MAn =
2n + 1 n
n + 1 2n + 1
. Since the lattice LA
n is generated
by the two vectors V A1,n and V A
2,n where V A1,n = (2n + 1)vA
1 + (n + 1)vA2 and V A
2,n =
n(vA1 ) + (2n + 1)vA
2 , the matrix MAn is a sampling matrix of the lattice LA
n and hence will
be used very often.
Theorem 4.2.6. For any n ∈ N, <An is a set of coset representatives of the quotient group
LA/LAn .
Proof. Let V A be the matrix whose two columns are vA1 and vA
2 , and V An the matrix
whose two columns are V A1,n and V A
2,n. Since V A1,n = (2n + 1)vA
1 + (n + 1)vA2 and V A
2,n =
n(vA1 ) + (2n + 1)vA
2 , we have V An = (V A)(MA
n ). By Lemma 4.2.5 the order of the quotient
group LA/LAn is the determinant of MA
n . It follows that the order of the quotient group
LA/LAn is 3n2 + 3n + 1 since it is the determinant of MA
n . By Lemma 4.2.1, the order of
<An is also 3n2 + 3n + 1. Hence, to show <A
n is a set of coset representatives of LA/LAn , it
suffices to show that no two distinct lattice points of <An are congruent modulo LA
n . For
any a,b ∈ <An with a 6= b, we have a− b ∈ <A
2n by Lemma 4.2.2. Suppose a− b ∈ LAn . By
Lemma 4.2.4, LAn
⋂<A2n = 0. It follows that a− b ∈ LA
n
⋂<A2n = 0 and hence a = b which
leads to a contradiction. Therefore <An is a set of coset representatives of LA/LA
n .
By Theorem 4.2.6, the DFT on <An is invertible. Now we consider its frequency
domain. In the proof of Theorem 4.2.6, we have V An = (V A)(MA
n ) where V A is a sampling
matrix of the lattice LA and V An is a sampling matrix of the lattice LA
n . Let V An =
[(V An )T ]−1. Since V A
n is a sampling matrix of the lattice LAn , V A
n is a sampling matrix of
the dual lattice (LAn )∗ of LA
n by Lemma 2.3.2.
Let vAn,1 and vA
n,2 be the two column vectors of the matrix V An . It is easy to check
that vAn,1 = 1√
3(3n2+3n+1)
√
3(2n + 1)
1
, vA
n,2 = 1√3(3n2+3n+1)
−√3(n + 1)
3n + 1
,
34
‖vAn,1‖ = ‖vA
n,2‖ and the angle between them is 2π3
. For any n ∈ N, let
<An =
j(vA
n,1) + k(vAn,2) : j, k ∈ Z, |j| ≤ n, |k| ≤ n, |j − k| ≤ n
where vAn,1 and vA
n,2 are defined as the above.
Theorem 4.2.7. For any n ∈ N, <An is a set of coset representatives of the quotient group
(LAn )∗/(LA)∗.
Proof. Let us use the same notation as before. From the proof of Theorem 4.2.6, we have
V An = (V A)(MA
n ). It follows that [(V An )T ]−1 = [(V A)T ]−1[(MA
n )T ]−1. Hence [(V A)T ]−1 =
[(V An )T ]−1(MA
n )T . Since V A is a sampling matrix of the lattice LA, [(V A)T ]−1 is a sampling
matrix of the dual lattice (LA)∗. For a similar reason, [(V An )T ]−1 is a sampling matrix of
the dual lattice (LAn )∗. Since (MA
n )T =
2n + 1 n + 1
n 2n + 1
, by a proof similar to the
proof of Theorem 4.2.6 it can be shown that <An is a set of coset representatives of the
quotient group (LAn )∗/(LA)∗.
From Theorems 4.2.6 and 4.2.7, and by the definitions of the type A RHS and <An ,
it is easy to deduce that the geometric shape of the support in the frequency domain of
the DFT on <An is similar to that of the support in the spatial domain. Throughout the
remainder of this research, we let
C(A, n) =
j
k
∈ Z2 : j(vA
1 ) + k(vA2 ) ∈ <A
n
.
By the definition of <An , obviously we have
C(A, n) =
j
k
∈ Z2 : |j| ≤ n, |k| ≤ n, |j − k| ≤ n
.
This notation will be used to write the DFT on <An in terms of a periodicity matrix.
Recall in Chapter 2 that, for a periodicity matrix M, a set of coset representatives of
35
the quotient group Zd/MZd is also called a set of coset representatives associated with
the periodicity matrix M. Employing the change of bases as in Section 2.3, the following
corollary follows from Theorems 4.2.6 and 4.2.7.
Corollary 4.2.8. For any n ∈ N, C(A, n) is a set of coset representatives associated with
both MAn and (MA
n )T .
4.3 Type B Regular Hexagonal Structure
Throughout this research, we let vB1 =
√3
2
12
, vB
2 =
−√
32
12
and
LB =n1(v
B1 ) + n2(v
B2 ) : n1, n2 ∈ Z
.
The lattice LB is called the type B hexagonal lattice. In this section, we consider the type
B RHS which is defined below. Subsequent discussion will show that the indexing of the
type B RHS is more complicated than that of the type A RHS. To make the definition of
the type B RHS easier, we first give the definition of a half type B RHS. For any s, d ∈ Z,
the notation mod(s, d) denotes the remainder when s is divided by d. For any integer
n > 0, let
<Bn,h =
x(vB
1 ) + (s− x)vB2 : x, s ∈ Z, 0 ≤ s ≤ 3n,−n + r + 2q ≤ x ≤ n + q
where r = mod(s, 3), q = s−r3
, vB1 =
√3
2
12
and vB
2 =
−√
32
12
. The set <B
n,h is called
a half type B RHS at level n. For any integer n > 0, we define <Bn = <B
n,h
⋃−<Bn,h where
−<Bn,h =
−u : u ∈ <Bn,h
. We call <B
n the nth level of the type B RHS.
Remark: Let x(vB1 ) + (s − x)vB
2 ∈ <Bn . By the definitions of <B
n,h and <Bn , if s > 0,
then x(vB1 ) + (s − x)vB
2 ∈ <Bn,h; if s < 0, then −(x(vB
1 ) + (s − x)vB2 ) ∈ <B
n,h; and if s = 0,
then x(vB1 ) + (s− x)vB
2 ∈ <Bn,h
⋂(−<B
n,h). Hence, for any given s satisfying −3n ≤ s ≤ 3n,
s corresponds to a horizontal layer of hexagons of <Bn as shown in Figure 4-2. <B
n has a
total of 3n + 1 layers of hexagons. The lattice points and corresponding Voronoi cells of
<B2 and <B
3 are shown in Figure 4-2 (a) and Figure 4-2 (b), respectively. The middle layer
36
in each figure corresponds to s = 0 which means that the sum of the two components of
each lattice point of that layer is 0. The layers above the middle layer correspond to s > 0,
and the layers below the middle layer correspond to s < 0.
In the following, we are going to show that <Bn is a set of coset representatives of
the quotient group of two lattices. To make its proof more concise, we need the following
preliminary results.
Lemma 4.3.1. For any integer n > 0, <Bn consists of 9n2 + 3n + 1 lattice points.
Proof. By the previous remark, <Bn has a total of 3n + 1 layers of hexagons. The middle
layer of <Bn contains 2n + 1 lattice points. Any upper layer with s = 3q for some positive
integer q ≤ n has (n + q) − (−n + 2q) + 1 = 2n − q + 1 hexagons. Any upper layer with
s = 3q + 1 for some positive integer q ≤ n − 1 has (n + q) − (−n + 1 + 2q) + 1 = 2n − q
hexagons. Any upper layer with s = 3q + 2 for some positive integer q ≤ n − 1 has
(n + q) − (−n + 2 + 2q) + 1 = 2n − q − 1 hexagons. By the definition of <Bn , the number
of hexagons down the middle layer is the same as that of hexagons above the middle layer.
Hence the total number of those hexagons is
(2n + 1) + 2n∑
q=1
(2n− q + 1) + 2n−1∑q=0
(2n− q + 2n− q − 1) = 9n2 + 3n + 1. (4–1)
Proposition 4.3.2. If ak ∈ <Bn,h for k = 1, 2, then a1 + a2 ∈ <B
2n,h.
Proof. By definition of <Bn,h, there exist xk, sk ∈ Z such that ak = xk(v
B1 ) + (sk − xk)v
B2
with 0 ≤ sk ≤ 3n and −n + rk + 2qk ≤ xk ≤ n + qk where rk = mod(sk, 3), qk = sk−rk
3. Let
x = x1 + x2 and s = s1 + s2. Then
a1 + a2 = x(vB1 ) + (s− x)vB
2 . (4–2)
Since 0 ≤ sk ≤ 3n for k = 1, 2, we have 0 ≤ s = s1 + s2 ≤ 6n. The inequality
−n + rk + 2qk ≤ xk ≤ n + qk for k = 1, 2 implies that −2n + (r1 + r2) + 2(q1 + q2) ≤
37
−4 −3 −2 −1 0 1 2 3 4
−3
−2
−1
0
1
2
3
(−2, 2) (−1, 1) (0, 0) (1, −1) (2, −2)
(−1, 2)
(1, −2)
(0, 1)
(0, −1)
(1, 0)
(−1, 0)
(2, −1)
(−2, 1)
(0, 2)
(0, −2)
(1, 1)
(−1, −1)
(2, 0)
(−2, 0)
(0, 3)
(0, −3)
(1, 2)
(−1, −2)
(2, 1)
(−2, −1)
(3, 0)
(−3, 0)
(1, 3)
(−1, −3)
(2, 2)
(−2, −2)
(3, 1)
(−3, −1)
(2, 3)
(−2, −3)
(3, 2)
(−3, −2)
(2, 4)
(−2, −4)
(3, 3)
(−3, −3)
(4, 2)
(−4, −2)
(a)
−6 −4 −2 0 2 4 6−5
−4
−3
−2
−1
0
1
2
3
4
5
(−3, 3) (−2, 2) (−1, 1) (0, 0) (1, −1) (2, −2) (3, −3)
(−2, 3)
(2, −3)
(−1, 2)
(1, −2)
(0, 1)
(0, −1)
(1, 0)
(−1, 0)
(2, −1)
(−2, 1)
(3, −2)
(−3, 2)
(−1, 3)
(1, −3)
(0, 2)
(0, −2)
(1, 1)
(−1, −1)
(2, 0)
(−2, 0)
(3, −1)
(−3, 1)
(−1, 4)
(1, −4)
(0, 3)
(0, −3)
(1, 2)
(−1, −2)
(2, 1)
(−2, −1)
(3, 0)
(−3, 0)
(4, −1)
(−4, 1)
(0, 4)
(0, −4)
(1, 3)
(−1, −3)
(2, 2)
(−2, −2)
(3, 1)
(−3, −1)
(4, 0)
(−4, 0)
(1, 4)
(−1, −4)
(2, 3)
(−2, −3)
(3, 2)
(−3, −2)
(4, 1)
(−4, −1)
(1, 5)
(−1, −5)
(2, 4)
(−2, −4)
(3, 3)
(−3, −3)
(4, 2)
(−4, −2)
(5, 1)
(−5, −1)
(2, 5)
(−2, −5)
(3, 4)
(−3, −4)
(4, 3)
(−4, −3)
(5, 2)
(−5, −2)
(3, 5)
(−3, −5)
(4, 4)
(−4, −4)
(5, 3)
(−5, −3)
(3, 6)
(−3, −6)
(4, 5)
(−4, −5)
(5, 4)
(−5, −4)
(6, 3)
(−6, −3)
(b)
Figure 4-2. Arrays of the type B RHS. (a) The lattice points of <B2 . (b) The lattice points
of <B3 . The coordinates inside each hexagon are for the basis
vB
1 ,vB2
.
x1 + x2 ≤ 2n + (q1 + q2). If r1 + r2 < 3 let s = s1 + s2, q = q1 + q2 and r = r1 + r2. Then
−2n + r + 2q ≤ x ≤ 2n + q and hence a1 + a2 ∈ <B2n,h by Equation 4–2. If r1 + r2 ≥ 3, then
r1 + r2 = 3 + r for some 0 ≤ r < 3. The equation sk = 3qk + rk for k = 1, 2 implies that
s1 + s2 = 3(q1 + q2)+ r1 + r2 = 3(q1 + q2)+3+ r = 3(q1 + q2 +1)+ r. Now let q = q1 + q2 +1.
38
The inequality xk ≤ n + qk for k = 1, 2 implies that x = x1 + x2 ≤ 2n + (q1 + q2) ≤ 2n + q.
On the other hand −n + rk + 2qk ≤ xk for k = 1, 2 implies that x = x1 + x2 ≥ −2n + (r1 +
r2)+ 2(q1 + q2) = −2n+(3+ r)+2(q1 + q2) = −2n+(1+ r)+2(q1 + q2 +1) ≥ −2n+ r +2q.
Hence we have
− 2n + r + 2q ≤ x ≤ 2n + q. (4–3)
Since s = s1 + s2 = (3q1 + r1) + (3q2 + r2) = 3(q1 + q2) + (r1 + r2) = 3(q1 + q2) + 3 + r =
3(q1 + q2 + 1) + r = 3q + r, we have r = mod(s, 3) and q = s−r3
. Hence, by Equation 4–2
and Inequality 4–3, a1 + a2 ∈ <Bn,h ⊆ <B
n .
Proposition 4.3.3. Let ak = xk(vB1 ) + (sk − xk)v
B2 ∈ <B
n,h for k = 1, 2. If s1 ≥ s2, then
a1 − a2 ∈ <B2n,h.
Proof. If s1 = s2, then a1 − a2 = (x1 − x2)vB1 + (0− (x1 − x2))v
B2 . Let x = x1 − x2. Since
ak ∈ <Bn,h for k = 1, 2, we have −3n ≤ xk ≤ 3n. Hence −6n ≤ x = x1 − x2 ≤ 6n and thus
a1 − a2 lies on the middle layer of <B2n,h since s1 − s2 = 0. If s1 > s2, let rk = mod(sk, 3)
and qk = sk−rk
3. It then follows from the definition of <B
n,h that −n+rk +2qk ≤ xk ≤ n+qk.
Hence, x = x1 − x2 ≤ (n + q1) − (−n + r2 + 2q2) = 2n − r2 + q1 − 2q2 and x = x1 − x2 ≥(−n + r1 + 2q1)− (n + q2) = −2n + r1 + 2q1 − q2. Thus we have
− 2n + r1 + 2q1 − q2 ≤ x = x1 − x2 ≤ 2n− r2 + q1 − 2q2. (4–4)
If r1 ≥ r2, let q = q1 − q2 and r = r1 − r2. Then Inequality 4–4 implies that
−2n + r + 2q ≤ x ≤ 2n + q and, by the definition of <Bn,h, we have a1 − a2 ∈ <B
2n,h . If
r1 < r2, then r2 > r1 ≥ 0 and hence r2 ≥ 1. Let q = q1 − q2 − 1 and r = 3 + r1 − r2.
Then 2n − r2 + q1 − 2q2 ≤ 2n − r2 + q1 − q2 ≤ 2n − 1 + q1 − q2 = 2n + q, and
−2n + r1 + 2q1 − q2 ≥ −2n + r1 + 2q1 − q2 − (q2 − 2 − r2) = −2n + x + 2q. Obviously
s1 − s2 = 3(q1 − q2) + (r1 − r2) = 3(q1 − q2 − 1) + (3 + r1 − r2) = 3q + r. Then by
Inequality 4–4 it follows that −2n + r + 2q ≤ x ≤ 2n + q. Hence, by the definition of <Bn,h,
we have a1 − a2 ∈ <B2n,h.
39
Proposition 4.3.4. If ak ∈ <Bn for k = 1, 2, then a1 ± a2 ∈ <B
2n.
Proof. Since ak ∈ <Bn , either ak ∈ <B
n,h or −ak ∈ <Bn,h. If ak ∈ <B
n,h for k = 1, 2, then from
Proposition 4.3.2 we have that a1 + a2 ∈ <B2n,h ⊆ <B
2n. If −ak ∈ <Bn,h for k = 1, 2, then
−a1 − a2 ∈ <B2n,h and hence, by the definition of <B
n , we have a1 + a2 ∈ <B2n. Similarly if
just one of a1 or a2 is in <Bn,h, then by Proposition 4.3.3 a1 + a2 ∈ <B
2n. Since a2 ∈ <Bn , we
have −a2 ∈ <Bn . Hence a1 − a2 = a1 + (−a2) ∈ <B
2n.
Lemma 4.3.5. The maximal distance of the lattice points of <Bn from the origin 0 is
√3n.
Proof. Without loss of generality, we only consider the lattice points of <Bn,h. For any
g ∈ <Bn,h, there exist x, s ∈ Z with 0 ≤ s ≤ 3n,−n + r + 2q ≤ x ≤ n + q, where
r = mod(s, 3), q = s−r3
, such that g = x(vB1 ) + (s − x)vB
2 . Let ψ(g) be the square
of the distance of g from the origin 0, and let κ = maxψ(g) : g ∈ <B
n,h
. Since
g = xvB1 +(s−x)vB
2 , we have ψ(g) = x2 +(s−x)2−x(s−x) = 3x2−3sx+s2. By replacing
s by 3q + r we have ψ(g) = 3x2 − 3x(3q + r) + (3q + r)2, which is a convex function of
x. Hence the maximum of the function ψ(g) can only be reached at the two end points of
x. Since −n + r + 2q ≤ x ≤ n + q and the values of the function ψ(g) at two end points
x = −n + r + 2q and x = n + q are the same, we have
κ ≤ max3(n + q)2 − 3(n + q)(3q + r) + (3q + r)2 : 0 ≤ 3q + r ≤ 3n, 0 ≤ r ≤ 2
.
Let α(q, r) = 3(n + q)2− 3(n + q)(3q + r) + (3q + r)2. Obviously α(q, r) = 3n2− 3n(q + r) +
3q2 + 3qr + r2 by simplification. If r = 0, then 0 ≤ q ≤ n and α(q, 0) = 3n2 − 3nq + 3q2
which is a convex function of q. Hence max α(q, 0) : q ∈ [0, n] = 3n2. If r = 1, then
0 ≤ q ≤ n − 1 and α(q, 1) = 3n2 − 3n(q + 1) + 3q2 + 3q + 1 which is also a convex
function of q. Therefore max α(q, 1) : q ∈ [0, n− 1] = 3n2 − 3n + 1 ≤ 3n2. If r = 2, then
0 ≤ q ≤ n − 1 and α(q, 1) = 3n2 − 3n(q + 2) + 3q2 + 6q + 4 which is a convex function
of q. Hence max α(q, 2) : q ∈ [0, n− 1] = 3n2 − 3n + 1 ≤ 3n2. Thus κ ≤ 3n2. Let
40
g = n(vB1 ) + 2n(vB
2 ). It is easy to check that g ∈ <Bn,h and that ψ(g) = 3n2. Therefore
κ = 3n2.
Let V B be the matrix whose two columns are vB1 and vB
2 . For any integer n > 0, let
MBn =
3n + 1 3n
1 3n + 1
, V B
n = (V B)(MBn ), and LB
n be the lattice generated by the two
column vectors of V Bn . The following lemma will be used in Theorem 4.3.7.
Lemma 4.3.6. LBn
⋂<B2n = 0.
Proof. Let b1 and b2 be the two column vectors of the matrix V Bn . Since V B
n =
(V B)(MBn ) = (V B)
3n + 1 3n
1 3n + 1
, we have b1 = (3n + 1)vB
1 + vB2 and
b2 = 3n(vB1 ) + (3n + 1)vB
2 . Then b1 = x(vB1 ) + (s − x)vB
2 where x = 3n + 1
and s = 3n + 2. Let r = mod(s, 3), q = s−r3
. It follows that r = 2 and q = n.
Hence −2n + r + 2q = −2n + 2 + 2n = 2 and 2n + q = 2n + n = 3n. Since
x = 3n + 1 6∈ [−2n + r + 2q, 2n + q], we have b1 6∈ <B2n,h by the definition of <B
n,h.
Also s = 3n + 2 > 0 implies that b1 6∈ −<B2n,h. Therefore b1 6∈ <B
2n. Similarly
b2 = x2(vB1 ) + (s2 − x2)v
B2 where x2 = 3n and s2 = 6n + 1. Since s2 = 6n + 1 > 3(2n),
we have b2 6∈ <B2n,h by the definition of <B
n,h. It follows from s2 > 0 that b2 6∈ −<B2n,h.
Therefore b2 6∈ <B2n. Let g = b2 − b1. Then g = −vB
1 + 3n(vB2 ). It follows that
g = x3(vB1 ) + (s3 − x3)v
B2 where x3 = −1 and s3 = 3n− 1. Let r3 = mod(s3, 3), q3 = s3−r3
3.
Then we have r3 = 2 and q3 = n− 1. Hence −2n + r3 + 2q3 = −2n + 2 + 2(n− 1) = 0 and
2n+q3 = 2n+n−1 = 3n−1. Since x3 = −1 6∈ [−2n+r+2q, 2n+q], we have g 6∈ <B2n,h. Of
course, since s3 = 3n− 1 > 0 where n is a positive integer, g 6∈ −<B2n,h. Therefore g 6∈ <B
2n.
By the definition of <Bn , <B
2n is symmetric with respect to the origin. Hence we have also
that −b1,−b2,−g 6∈ <B2n.
Since LBn is a hexagonal lattice generated by b1,b2 and the angle between b1 and
b2 is π3, the lattice points in the set ±b1,±b2,±(b2 − b1) are the only 6 closest nonzero
lattice points from the origin 0. Since b1 = (3n + 1)vB1 + vB
2 and the angle between vB1
41
and vB2 is 2π
3, the norm of b1 is
√9n2 + 3n + 1. It follows that the distance of any other
nonzero lattice point of LBn from the origin 0 is at least
√3(9n2 + 3n + 1) > 5n as shown
in Figure 4-3. However by Lemma 4.3.5, the maximal distance of the lattice points of <B2n
from the origin 0 is√
3(2n) < 4n. Therefore LBn
⋂<B2n = 0.
−25 −20 −15 −10 −5 0 5 10 15 20 25−25
−20
−15
−10
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0
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Figure 4-3. The lattice points of <B2n (red) and LB (blue) where n = 6.
Theorem 4.3.7. For any n ∈ N, <Bn is a set of coset representatives of the quotient group
LB/LBn .
Proof. Since V B is a sampling matrix of the lattice LB, V Bn a sampling matrix of the
lattice LBn , and since V B
n = (V B)(MBn ), by Lemma 4.2.5, the order of the quotient
group LB/LBn equals the determinant of MB
n . It follows that the order of the quotient
group LB/LBn is 9n2 + 3n + 1 since it is the determinant of MB
n . Hence, to show <Bn is
a set of coset representatives of LB/LBn , it suffices to show that no two distinct lattice
points of <Bn are congruent modulo LB
n . For any a,b ∈ <Bn , we have a − b ∈ <B
2n by
Proposition 4.3.4. Suppose a − b ∈ LBn . By Lemma 4.3.6, we have LB
n
⋂<B2n = 0. Hence
a− b ∈ LBn
⋂<B2n = 0. It follows that a = b. Therefore <B
n is a set of coset representatives
of LB/LBn .
42
It follows from Theorem 4.3.7 that the DFT on <Bn is invertible. Next we consider
the frequency domain. In the proof of Theorem 4.3.7, we have V Bn = (V B)(MB
n ), where
V B is a sampling matrix of the lattice LB and V Bn is a sampling matrix of the lattice LB
n .
Let V Bn = [(V B
n )T ]−1. Since V Bn is a sampling matrix of the lattice LB
n , V Bn is a sampling
matrix of the dual lattice (LBn )∗ of LB
n by Lemma 2.3.2. Let vBn,1 and vB
n,2 be the two
column vectors of the matrix V Bn . Similar to the definition of <B
n,h and the definition of <Bn
in the spatial domain, we have the following two definitions in the frequency domain. For
any integer n > 0, let
<Bn,h =
x(vB
n,1) + (s− x)vBn,2 : x, s ∈ Z, 0 ≤ s ≤ 3n,−n + r + 2q ≤ x ≤ n + q
where r = mod(s, 3), q = s−r3
, vBn,1 and vB
n,2 are defined as the above. For any integer
n > 0, let <Bn = <B
n,h
⋃−<Bn,h where −<B
n,h =−u : u ∈ <B
n,h
.
Theorem 4.3.8. For any n ∈ N, <Bn is a set of coset representatives of the quotient group
(LBn )∗/(LB)∗.
Proof. The proof is similar to that of Theorem 4.2.7 and 4.3.7, and hence omitted.
From Theorems 4.3.7 and 4.3.8, and by the definitions of <Bn,h and <B
n,h, it is easy to
deduce that the geometric shape of the support in the frequency domain of the DFT on
<Bn is similar to that of the support in the spatial domain. Throughout this research, we
also let
C(B, n) =
j
k
∈ Z2 : j(vB
1 ) + k(vB2 ) ∈ <B
n
.
It will help to write the DFT on <Bn in terms of a periodicity matrix. By the change of
bases as in Section 2.3, the following corollary follows from Theorems 4.3.7 and 4.3.8.
Corollary 4.3.9. For any n ∈ N, C(B, n) is a set of coset representatives associated with
both MBn and (MB
n )T .
43
4.4 Relation Between the Type A and B RHS and Some Previously StudiedArrays
Consider the hexagonal array shown in Figure 4-4 (a) consisting of 48 = 3 · 42
lattice points. If Γn denotes the nth level of this structure, then Γn consists of 3n2 lattice
points and corresponds to the hexagonal structure investigated in Sun and Yao [40] (also
Figure 3-2 (b)). If we delete the points marked in red, then the number of lattice points
reduces to 37 = 3 · 42 − 3 · 4 + 1. The reduced array corresponds to the array <A3 shown in
Figure 4-4 (b). More generally, deleting the lattice points on the bottom and the lower two
sides of Γn reduces the 3n2 lattice points to 3n2 − 3n + 1 = 3(n− 1)2 + 3(n− 1) + 1 lattice
points. Hence, the remaining lattice points are exactly the lattice points of <An−1.
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
a) (b)
Figure 4-4. Comparison of a previously studied array structure with the type A RHS. (a)The lattice points of Γ4. (b) The lattice points of <A
3 .
Similarly, consider the hexagonal array shown in Figure 4-5 (a) consisting of 100 = n2
lattice points where n = 3k + 1 and k = 3. If Υk denotes the kth level of this structure,
then Υn consists of (3k + 1)2 lattice points and corresponds to the hexagonal structure
investigated in Anterrieu et al. [2] (also Figure 3-3 (b)). If we delete the points marked
in red, then the number of lattice points reduces to 91 = 9 · 32 + 3 · 3 + 1. The reduced
array corresponds to the array <B3 shown in Figure 4-5 (b). More generally, deleting the
lattice points on the bottom and the lower two sides of Υk reduces the n2 lattice points to
(3k + 1)2− 3k = 9k2 + 3k + 1 lattice points. Hence, the remaining lattice points are exactly
44
the lattice points of <Bk . Also by comparing Figure 4-2 with Figure 3-3 (c), we can see
that if we add 3k lattice points along 6 sides of <Bk , then the expanded array is the kind of
array shown in Figure 3-3 (c).
−6 −4 −2 0 2 4 6
−5
−4
−3
−2
−1
0
1
2
3
4
5
−6 −4 −2 0 2 4 6
−5
−4
−3
−2
−1
0
1
2
3
4
5
a) (b)
Figure 4-5. Comparison of a previously studied array structure with the type B RHS. (a)The lattice points of Υ3. (b) The lattice points of <B
3 .
45
CHAPTER 5FAST ALGORITHMS FOR COMPUTING THE DFT ON THE TWO SPECIAL TYPES
OF THE RHS
5.1 Convert the DFT on a Tile of a Tiling by Translations by a Sublattice toa Standard DFT
In this section we emphasize the correspondence between the domain of computing
the original DFT on a lattice and that of computing the standard DFT generally
presented in textbooks. Recall that Equation 2–1 for computing the DFT on a tile
of a lattice is reduced to Equation 2–7. The use of the Smith normal form to convert
Equation 2–7 to a standard DFT appeared previously in Problem 20, Chapter 2, of
Dudgeon and Mersereau [10] and in Guessoum and Mersereau [19]. In this research, by
a unimodular matrix we mean a square matrix with integer entries and determinant +1
or −1. Applying the Smith normal form to the periodicity matrix M in Equation 2–7,
we can obtain unimodular matrices E and F such that M = EDF for some diagonal
periodicity matrix D =
d1 0
0 d2
, where dj ≥ 0 for j = 1, 2, and d1 divides d2. Let
gk,l = E
k
l
, and hk,l = F T
k
l
. Also, let G = gk,l : 0 ≤ k < d1, 0 ≤ l < d2
and H = hk,l : 0 ≤ k < d1, 0 ≤ l < d2. As in Section 2.3, any function f defined on a
set of coset representatives associated with M can be extended periodically to the whole
lattice Z2, where the association is defined in Section 2.3. Then we have the following
proposition:
Proposition 5.1.1. Let G,H, M and f be as the above. Then G and H are sets of coset
representatives associated with M and MT , respectively. Furthermore, the DFT f of f
satisfies
f(hs,t) =
d1∑
k=0
d2∑
l=0
f(gk,l) · e−2πi( skd1
+ tld2
)for all hs,t ∈ H. (5–1)
46
Proof. Let E, F and D be as the above and let Γ = εk,l : 0 ≤ k < d1, 0 ≤ l < d2 where
εk,l =
k
l
. Obviously Γ is a set of coset representatives associated with D. We claim
that G and H are sets of coset representatives associated with M and MT , respectively.
If there exist gk1,l1 ,gk2,l2 ∈ G such that gk1,l1 − gk2,l2 = Mz for some z ∈ Z2, then
E(εk1,l1−εk2,l2) = Mz = EDFz and hence εk1,l1−εk2,l2 = D(Fz). Since εk1,l1 , εk2,l2 ∈ Γ and
Γ is a set of coset representatives associated with D, it follows that εk1,l1 = εk2,l2 . Thus G
is a set of coset representatives associated with M. Similarly we can show that H is a set
of coset representatives associated with MT . By Equation 2–7 we have
f(hs,t) =
d1∑
k=0
d2∑
l=0
f(gk,l) · e−2πi(hs,t)T M−1gk,l for all hs,t ∈ H. (5–2)
Since (hs,t)TM−1gk,l = (F T εs,t)
T (EDF )−1Eεk,l = εs,tTD−1εk,l = sk
d1+ tl
d2, we have
f(hs,t) =
d1∑
k=0
d2∑
l=0
f(gk,l) · e−2πi( skd1
+ tld2
)for all hs,t ∈ H. (5–3)
Thus, f can be computed using algorithms of the standard DFT.
Corollary 5.1.2. Let G, H, and f be the same as in Proposition 5.1.1, and M =
r −s
s r − s
be a periodicity matrix such that r and s are relatively prime. Then there
exist unimodular matrices E and F such that M = EDF where D =
1 0
0 N
and
N = |det(M)|. Hence the DFT f of f can be converted to a 1-dimensional standard DFT.
Proof. Without loss of generality we assume that det(M) ≥ 0. Since r and s are relatively
prime, there exist integers x and y such that xr + ys = 1. It follows from xr + ys = 1
that
x y
−s r
r −s
s r − s
1 sx− (r − s)y
0 1
=
1 0
0 N
and that the matrix
47
x y
−s r
is unimodular. Let E =
x y
−s r
−1
and F =
1 −sx + (r − s)y
0 1
.
Then M = EDF . Hence by Proposition 5.1.1 the DFT f of f can be converted to a
1-dimensional standard DFT .
5.2 Fast Algorithms for the DFT and its Inverse on the Type A RHS
Let f be a complex function defined on <An . By Equation 2–5, its DFT f satisfies the
following equation:
f(q) =∑
p∈<An
f(p) · e−2πi〈p,q〉, for all q ∈ <An . (5–4)
As in Section 2.3, f can be extended periodically to the whole lattice LA. It follows from
Equation 2–5 that, for any set of coset representatives P of LA/LAn, the set <A
n can be
replaced by P . Similarly, the set <An can be replaced by a set of coset representatives Q of
the quotient group (LAn)∗/(LA)∗.
To simplify our notation, let f and f be two functions defined on Z2 such that
f(m) = f((V A)m) and f(m) = f(V Am) for any m ∈ Z2, where V A = [(V AMAn )T ]−1.
Recall in Chapter 4 that
C(A, n) =
j
k
∈ Z2 : j(vA
1 ) + k(vA2 ) ∈ <A
n
=
j
k
∈ Z2 : |j| ≤ n, |k| ≤ n, |j − k| ≤ n
.
(5–5)
By Corollary 4.2.8, C(A, n) is a set of coset representatives associated with both MAn and
(MAn )T , where the matrix MA
n =
2n + 1 n
n + 1 2n + 1
is defined in Section 4.2. Hence, by
Equation 2–7, the DFT of f satisfies
f(k) =∑
m∈C(A,n)
f(m) · e−2πikT (MAn )−1m for all k ∈ C(A, n). (5–6)
48
For any integer n > 0 let P1 =
0
l
: 0 ≤ l ≤ 3n2 + 3n
and P2 =
2l
l
: 0 ≤ l ≤ 3n2 + 3n
. The following proposition shows that the DFT of f
on C(A, n) can be converted into a standard DFT.
Proposition 5.2.1. For any integer n > 0, each of the sets P1 and P2 is a set of
coset representatives associated with both MAn and (MA
n )T . Furthermore, f(
2k
k
) =
∑3n2+3nm=0 f(
0
m
) · e −2πilm
3n2+3n+1 for all 0 ≤ k ≤ 3n2 + 3n.
Proof. By the Smith normal form of MAn , we have MA
n = E(DAn )F where E =
1 0
−(3n + 1) 1
,DA
n =
1 0
0 3n2 + 3n + 1
, and F =
2n + 1 n
2 1
. By
computing the matrix products E
0
k
and F T
0
k
, we have P1 = E
0
k
: 0 ≤
k ≤ 3n2 +3n and P2 = F T
0
k
: 0 ≤ k ≤ 3n2 +3n. It follows from Proposition 5.1.1
that P1 is a set of coset representatives associated with MAn , P2 is a set of coset
representatives associated with (MAn )T , and f(
2k
k
) =
∑3n2+3nm=0 f(
0
m
) · e −2πilm
3n2+3n+1 ,
for all k satisfying 0 ≤ k ≤ 3n2 + 3n.
By the Smith normal form of (MAn )T , we have (MA
n )T = E(DAn )F , where E =
1 0
3n + 2 1
and F =
2n + 1 n + 1
−2 −1
are unimodular. It now follows from
Proposition 5.1.1 that the set P1 is a set of coset representatives associated with (MAn )T
and the set P2 is a set of coset representatives associated with MAn .
The next Corollary follows from Proposition 5.2.1 and the use of change of bases.
49
Corollary 5.2.2. For any n ∈ N, each of the setsk(vA
2 ) : 0 ≤ k ≤ 3n2 + 3n
and2k(vA
1 ) + k(vA2 ) : 0 ≤ k ≤ 3n2 + 3n
is a set of coset representatives of the
quotient group LA/LAn . Similarly each of the sets
k(vA
n,2) : 0 ≤ k ≤ 3n2 + 3n
and
2k(vAn,1) + k(vA
n,2) : 0 ≤ k ≤ 3n2 + 3n
is a set of coset representatives of the quotient
group (LAn )∗/(LA)∗.
Because f is originally defined on C(A, n), to apply Proposition 5.2.1 to get the values
of f on C(A, n), we need an efficient correspondence between C(A, n) and P1 in the spatial
domain, and an efficient correspondence between C(A, n) and P2 in the frequency domain.
(a) The correspondence between C(A, n) and P1 in the spatial domain
Since each of C(A, n) and P1 is a set of coset representatives associated with MAn ,
there exists a bijection ϕ between them such that a ≡ ϕ(a) mod MAn for any a ∈ C(A, n).
The bijection ϕ is called the correspondence between C(A, n) and P1 in the spatial
domain. In the following, we give an efficient method to find the correspondence ϕ.
Let a =
u
v
∈ C(A, n). By the definition of C(A, n), we have |u| ≤ n, |v| ≤ n and
|u− v| ≤ n. We consider the following three cases for u.
Case 1: u = 0. Because u = 0, we have a =
0
v
. If 0 ≤ v ≤ n, since
P1 =
0
k
: 0 ≤ k ≤ 3n2 + 3n
, we have a ∈ P1. Hence ϕ(a) = a. If −n ≤ v < 0,
then
0
3n2 + 3n + v + 1
∈ P1. Let b =
0
3n2 + 3n + v + 1
. Because b − a =
0
3n2 + 3n + 1
and MA
n =
2n + 1 n
n + 1 2n + 1
, we have b − a = MA
n
−n
2n + 1
.
Hence a ≡ b mod MAn . It follows that ϕ(a) = b =
0
3n2 + 3n + v + 1
.
50
Case 2: −n ≤ u < 0. Let b =
0
b2
where b2 = (3n+1)(u+n)+v +2n+1. Then
b− a =
−u
(3n + 1)(u + n) + 2n + 1
= MA
n
−(u + n)
2(u + n) + 1
. Hence a ≡ b mod MA
n .
Because −n ≤ u < 0 and −n ≤ v < u+n, we have b2 ≥ (3n+1)(−n+n)−n+2n+1 = n+1
and b2 ≤ (3n+1)(−1+n)+(−1+n)+2n+1 = 3n2 +n−1 ≤ 3n2 +2n. Since b =
0
b2
,
it follows that b ∈ P1. Thus ϕ(a) = b =
0
(3n + 1)(u + n) + v + 2n + 1
.
Case 3: 0 < u ≤ n. Let b =
0
b2
where b2 = (3n + 1)u + v. Then
b − a =
−u
(3n + 1)u
= MA
n
−u
2u
. Hence a ≡ b mod MA
n . Since 1 ≤ u ≤ n
and u − n ≤ v ≤ n, we have b2 = (3n + 1)u + v ≥ (3n + 1)1 + (1 − n) = 2n + 2
and b2 = (3n + 1)u + v ≤ (3n + 1)n + n = 3n2 + 2n. It follows that b ∈ P1. Thus
ϕ(a) = b =
0
(3n + 1)u + v
.
(b) The correspondence between P2 and C(A, n) in the frequency domain
By Proposition 5.2.1 and Corollary 4.2.8, each of the sets P2 and C(A, n) is a set of
coset representatives associated with the matrix (MAn )T . Hence there exists a bijection
χ between P2 and C(A, n) such that g ≡ χ(g) mod (MAn )T for any g ∈ P2. Such a
bijection is called the correspondence between P2 and C(A, n) in the frequency domain.
In the following, we show an efficient method to find the bijection χ. To make expressions
simpler, we assume that n is odd, and let n = 2l + 1 for some integer l > 0. When n is
even, the correspondence can be similarly shown. The method consists of the following two
steps. In Step 1, we find the correspondence between P2 and a set U which will be defined
in Step 1. In Step 2, we find the correspondence between U and C(A, n).
51
Step 1: The correspondence between P2 and U .
For any h ∈ Z satisfying 0 ≤ h ≤ l, let D0,h = h
2
1
. Also for any k ∈ Z satisfying
1 ≤ k ≤ l, let Ak,h = h
2
1
+
1
k
for any h ∈ Z satisfying −(l + 1) ≤ h ≤ l,
and let Bk,h = h
2
1
+
0
k
for any h ∈ Z satisfying −l ≤ h ≤ l. Furthermore
let D0 = D0,h : h ∈ Z, 0 ≤ h ≤ l ,Ak = Ak,h : h ∈ Z,−(l + 1) ≤ h ≤ l, Bk =
Bk,h : h ∈ Z,−l ≤ h ≤ l, and let U = D0
⋃(⋃3l+2
k=1 Ak)⋃
(⋃3l+2
k=1 Bk). Then U contains
l + 1 + (3l + 2)(4l + 3) = n−12
+ 1 + 3n+12
(2n + 1) = 3n2 + 3n + 1 lattice points each of
which will be shown to be congruent to an element of C(A, n) with respect to the modulo
relation associated with the matrix (MAn )T . When n = 3, the lattice points enclosed by the
blue dotted polygon in Figure 5-1 are exactly the lattice points of U . In the following, we
show an efficient method to find a bijection Φ between P2 and U such that g ≡ Φ(g) mod
(MAn )T for any g ∈ P2.
Since g ∈ P2, there exists m ∈ Z satisfying 0 ≤ m ≤ 3n2 + 3n such that g =
2m
m
.
To find Φ(g), we consider the following two cases.
Case 1: If 0 ≤ m ≤ l, then g ∈ D0 ⊂ U . Hence Φ(g) = g.
Case 2: If l < m ≤ 3n2 + 3n, since l + (4l + 3)(3l + 2) = 3n2 + 3n, there
exists k ∈ Z satisfying 1 ≤ k ≤ 3l + 2 and t ∈ Z satisfying 1 ≤ t ≤ 4l + 3 such that
m = l + (4l + 3)(k − 1) + t.
When 1 ≤ t ≤ 2l + 2, let h = t − l − 2. Then t = h + l + 2. Since m =
l + (4l + 3)(k − 1) + t, it follows that m = 2l + 2 + (4l + 3)(k − 1) + h. Hence
g − Ak,h =
2m
m
−
2h + 1
h + k
=
4l + 4 + (8l + 6)(k − 1) + 2h
2l + 2 + (4l + 3)(k − 1) + h
−
52
2h + 1
h + k
=
4l + 3 + 2(4l + 3)(k − 1)
2l + 2 + (4l + 3)(k − 1)− k
=
(4l + 3)(2k − 1)
(2l + 1)(2k − 1)
=
4l + 3 2l + 2
2l + 1 4l + 3
2k − 1
0
=
2n + 1 n + 1
n 2n + 1
2k − 1
0
. Thus g ≡ Ak,h
mod (MAn )T . Since Ak,h ∈ U , it follows that Φ(g) = Ak,h.
When 2l + 3 ≤ t ≤ 4l + 3, let h = t − 3l − 3. Then t = h + 3l + 3. Since
m = l + (4l + 3)(k − 1) + t, it follows that m = (4l + 3)(k − 1) + h + 4l + 3 = k(4l + 3) + h.
Hence g − Bk,h =
2m
m
−
2h
h + k
=
2k(4l + 3) + 2h
k(4l + 3) + h
−
2h
h + k
=
2k(4l + 3)
k(4l + 2)
=
4l + 3 2l + 2
2l + 1 4l + 3
2k
0
=
2n + 1 n + 1
n 2n + 1
2k
0
. Thus
g ≡ Bk,h mod (MAn )T . Since Ak,h ∈ U , it follows that Φ(g) = Bk,h.
We have shown that, with respect to the modulo relation associated with the matrix
(MAn )T , the elements of P2 correspond sequentially to the elements of D0, A1, B1, A2, B2,
..., A3l+2, and B3l+2.
Step 2: The correspondence between U and C(A, n) in the frequency
domain.
Recall in Step 1 that U = D0
⋃(⋃3l+2
k=1 Ak)⋃
(⋃3l+2
k=1 Bk). In this step, we show an
efficient method to find a bijection Ψ between U and C(A, n) such that u ≡ Ψ(u) mod
(MAn )T for any u ∈ U .
By the definitions of D0 and C(A, n), we have D0 ⊆ C(A, n). It follows that Ψ(u) = u
for any u ∈ D0.
For any u ∈ Ak, by the definition of Ak, there is h ∈ Z satisfying −l − 1 ≤ h ≤ l such
that u = Ak,h where Ak,h =
2h + 1
h + k
. We consider the following three cases for k to
find Ψ(u).
53
−8 −6 −4 −2 0 2 4 6 8
−8
−6
−4
−2
0
2
4
6
8
(−3,−3)
(−3,−2)
(−3,−1)
(−3,0)
(−2,−3)
(−2,−2)
(−2,−1)
(−2,0)
(−2,1)
(−1,−3)
(−1,−2)
(−1,−1)
(−1,0)
(−1,1)
(−1,2)
(0,−3)
(0,−2)
(0,−1)
(0,0)
(0,1)
(0,2)
(0,3)
(1,−2)
(1,−1)
(1,0)
(1,1)
(1,2)
(1,3)
(2,−1)
(2,0)
(2,1)
(2,2)
(2,3)
(3,0)
(3,1)
(3,2)
(3,3)
(4,0)
(4,1)
(4,2)
(4,3)
(5,0)
(5,1)
(5,2)
(5,3)
(5,4)
(6,0)
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(7,0)
(7,1)
(7,2)
(7,3)
(7,4)
(7,5)
(7,6)
(8,1)
(8,2)
(8,3)
(8,4)
(8,5)
(8,6)
(9,2)
(9,3)
(9,4)
(9,5)
(9,6)
(10,3)
(10,4)
(10,5)
(10,6)
(1,4)
(1,5)
(1,6)
(1,7)
(2,4)
(2,5)
(2,6)
(2,7)
(2,8)
(3,4)
(3,5)
(3,6)
(3,7)
(3,8)
(3,9)
(4,4)
(4,5)
(4,6)
(4,7)
(4,8)
(4,9)
(4,10)
(5,5)
(5,6)
(5,7)
(5,8)
(5,9)
(5,10)
(6,6)
(6,7)
(6,8)
(6,9)
(6,10)
(7,7)
(7,8)
(7,9)
(7,10)
(−6,1)
(−6,2)
(−6,3)
(−6,4)
(−5,1)
(−5,2)
(−5,3)
(−5,4)
(−5,5)
(−4,1)
(−4,2)
(−4,3)
(−4,4)
(−4,5)
(−4,6)
(−3,1)
(−3,2)
(−3,3)
(−3,4)
(−3,5)
(−3,6)
(−3,7)
(−2,2)
(−2,3)
(−2,4)
(−2,5)
(−2,6)
(−2,7)
(−1,3)
(−1,4)
(−1,5)
(−1,6)
(−1,7)
(0,4)
(0,5)
(0,6)
(0,7)
(0,−7)
(0,−6)
(0,−5)
(0,−4)
(1,−7)
(1,−6)
(1,−5)
(1,−4)
(1,−3)
(2,−7)
(2,−6)
(2,−5)
(2,−4)
(2,−3)
(2,−2)
(3,−7)
(3,−6)
(3,−5)
(3,−4)
(3,−3)
(3,−2)
(3,−1)
(4,−6)
(4,−5)
(4,−4)
(4,−3)
(4,−2)
(4,−1)
(5,−5)
(5,−4)
(5,−3)
(5,−2)
(5,−1)
(6,−4)
(6,−3)
(6,−2)
(6,−1)
(−7,−10)
(−7,−9)
(−7,−8)
(−7,−7)
(−6,−10)
(−6,−9)
(−6,−8)
(−6,−7)
(−6,−6)
(−5,−10)
(−5,−9)
(−5,−8)
(−5,−7)
(−5,−6)
(−5,−5)
(−4,−10)
(−4,−9)
(−4,−8)
(−4,−7)
(−4,−6)
(−4,−5)
(−4,−4)
(−3,−9)
(−3,−8)
(−3,−7)
(−3,−6)
(−3,−5)
(−3,−4)
(−2,−8)
(−2,−7)
(−2,−6)
(−2,−5)
(−2,−4)
(−1,−7)
(−1,−6)
(−1,−5)
(−1,−4)
(−10,−6)
(−10,−5)
(−10,−4)
(−10,−3)
(−9,−6)
(−9,−5)
(−9,−4)
(−9,−3)
(−9,−2)
(−8,−6)
(−8,−5)
(−8,−4)
(−8,−3)
(−8,−2)
(−8,−1)
(−7,−6)
(−7,−5)
(−7,−4)
(−7,−3)
(−7,−2)
(−7,−1)
(−7,0)
(−6,−5)
(−6,−4)
(−6,−3)
(−6,−2)
(−6,−1)
(−6,0)
(−5,−4)
(−5,−3)
(−5,−2)
(−5,−1)
(−5,0)
(−4,−3)
(−4,−2)
(−4,−1)
(−4,0)
..
..
..
..
.
..
..
..
..
..
..
.
..
..
..
..
..
.
..
..
..
..
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..
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Figure 5-1. The lattice points enclosed by the blue dashed polygon form the set of coset
representatives U of (LA3 )∗/(LA)∗.
Case 1: 1 ≤ k ≤ l + 1. Let a = 2h + 1 and b = h + k. Since −l − 1 ≤ h ≤ l, we have
−2l − 1 ≤ a ≤ 2l + 1. Because n = 2l + 1, it follows that −n ≤ a ≤ n. Hence |a| ≤ n.
Since −l − 1 ≤ h ≤ l and 1 ≤ k ≤ l + 1, we have −l ≤ b ≤ 2l + 1 = n. Since −l > −n, it
follows that −n ≤ b ≤ n. To ensure Ak,h ∈ C(A, n), we also need to show that |a− b| ≤ n.
Since −l − 1 ≤ h ≤ l and 1 ≤ k ≤ l + 1, we have h + 1 − k ≤ l + 1 − 1 = l < n and
h + 1− k ≥ (−l − 1) + 1− (l + 1) = −2l − 1 = −n. Since a− b = h + 1− k, it follows that
|a− b| ≤ n. Therefore Ak,h ∈ C(A, n). It follows that Ψ(u) = Ψ(Ak,h) = Ak,h.
Case 2: l + 2 ≤ k ≤ 2l + 1. Since h satisfies −(l + 1) ≤ h ≤ l, we consider the
following three sub-cases.
54
Sub-case 1: If −(l + 1) ≤ h ≤ k − 2l − 3, let a = 2h + n + 1, b = h + k − n − 1
and u = u −
−n
n + 1
. Since u = Ak,h =
2h + 1
h + k
, we have u =
2h + 1
h + k
−
−n
n + 1
=
2h + n + 1
h + k − n− 1
=
a
b
.
We claim that u ∈ C(A, n). Since l + 2 ≤ k ≤ 2l + 1 and −(l + 1) ≤ h ≤ k − 2l − 3,
we have a = 2h + n + 1 ≤ 2(k − 2l − 3) + n + 1 ≤ 2(2l + 1 − 2l − 3) + n + 1 = n− 3 < n
and a ≥ −2(l + 1) + n + 1 = −2(l + 1) + 2l + 1 + 1 = 0 > −n. Hence |a| < n.
For the same reason, we have b = h + k − n − 1 ≤ (k − 2l − 3) + k − n − 1 =
2k − 2l − n − 4 ≤ 2(2l + 1) − 2l − n − 4 = 2l − n − 2 = 2l − (2l + 1) − 2 = −3 < n,
and b ≥ −(l + 1) + (l + 2) − n − 1 = −n. Hence |b| ≤ n. Also we have a − b =
(2h+n+1)−(h+k−n−1) = h+2n+2−k ≤ (k−2l−3)+2n+2−k = 2n−2l−1 = 2n−n = n,
and a−b = h+2n+2−k ≥ −(l+1)+2n+2−(2l+1) = 2n−3l = 2(2l+1)−3l = l+2 > 0 > −n.
Hence |a− b| ≤ n. Therefore u ∈ C(A, n).
Because u − u =
−n
n + 1
=
2n + 1 n + 1
n 2n + 1
−1
1
, we have u ≡ u mod
(MAn )T . It follows that Ψ(u) = u.
Sub-case 2: If k−2l−2 ≤ h ≤ 2l−k+1, we claim that u ∈ C(A, n). Let a = 2h+1
and b = h + k. Since u = Ak,h =
2h + 1
h + k
=
a
b
, we need to show that |a| ≤ n,
|b| ≤ n, and |a − b| ≤ n. Because l + 2 ≤ k ≤ 2l + 1 and k − 2l − 2 ≤ h ≤ 2l − k + 1, we
have 2h + 1 ≤ 2(2l − k + 1) + 1 = 4l − 2k + 3 ≤ 4l − 2(l + 2) = 2l − 2 < 2l + 1 = n, and
2h+1 ≥ 2(k−2l−2)+1 = 2k−4l−3 ≥ 2(l +2)−4l−3 = −2l+1 > −2l−1 = −n. Hence
|2h+1| < n, i.e., |a| < n. For the same reason, we have h+k ≤ (2l−k+1)+k = 2l+1 = n,
and h+k ≥ (k−2l−2)+k = 2k−2l−2 ≥ 2(l+2)−2l−2 = 2 > −n. Hence |h+k| < n, i.e.,
|b| < n. We also have h+1−k ≤ (2l−k+1)+1−k = 2l+2−2k ≤ 2l+2−2(l+2) = −2 < n,
and h + 1 − k ≥ (k − 2l − 2) + 1 − k = −2l − 1 = −n. Hence |h + 1 − k| ≤ n. Since
55
a − b = h + 1 − k, it follows that |a − b| ≤ n. Therefore u ∈ C(A, n). It follows that
Ψ(u) = u.
Sub-case 3: If 2l − k + 2 ≤ h ≤ l, let a = 2h − n, b = h + k − 2n − 1 and
u = u −
n + 1
2n + 1
. Since u = Ak,h =
2h + 1
h + k
, we have u =
2h + 1
h + k
−
n + 1
2n + 1
=
2h− n
h + k − 2n− 1
=
a
b
.
We claim that u ∈ C(A, n). Since l + 2 ≤ k ≤ 2l + 1 and 2l − k + 2 ≤ h ≤ l, we
have a = 2h − n ≤ 2l − n < 2l < n and a ≥ 2(2l − k + 2) − n = 4l − 2k + 4 − n ≥4l − 2(2l + 1) + 4 − n = 2 − n > −n. Thus |a| < n. For the same reason, we have
b = h + k − 2n − 1 ≤ l + (2l + 1) − 2n − 1 = 3l − 2n < 3n − 2n = n, and
b ≥ (2l − k + 2) + k − 2n − 1 = 2l + 1 − 2n = n − 2n = −n. Thus |b| ≤ n. Also we have
a−b = (2h−n)−(h+k−2n−1) = h+n+1−k ≤ l+n+1−k ≤ l+n+1−(l+2) = n−1 < n,
and a−b = h+n+1−k ≥ (2l−k+2)+n+1−k = 2l+n−2k+3 = 2l+n−2(2l+1)+3 =
n− 2l + 1 > −2l > −2l − 1 = −n. Thus |a− b| ≤ n. Therefore u ∈ C(A, n).
Because u − u =
n + 1
2n + 1
=
2n + 1 n + 1
n 2n + 1
0
1
, we have u ≡ u mod
(MAn )T . It follows that Ψ(u) = u.
Case 3: 2l + 2 ≤ k ≤ 3l + 2. Since h satisfies −(l + 1) ≤ h ≤ l, we consider the
following two sub-cases.
Sub-case 1: If −(l + 1) ≤ h ≤ −1, let a = 2h + n + 1, b = h + k − n − 1 and
u = u −
−n
n + 1
. Since u = Ak,h =
2h + 1
h + k
, we have u =
2h + 1
h + k
−
−n
n + 1
=
2h + n + 1
h + k − n− 1
=
a
b
.
We claim that u ∈ C(A, n). Since 2l + 2 ≤ k ≤ 3l + 2 and −(l + 1) ≤ h ≤ −1, we have
a = 2h+n+1 ≤ −2+n+1 < n and a ≥ −2(l+1)+n+1 = −2(l+1)+2l+1+1 = 0 > −n.
Hence |a| < n. For the same reason, we have b = h+k−n−1 ≤ −1+k−n−1 = k−n−2 ≤
56
(3l +2)−n−2 = 3l−n = 3l− (2l +1) = l−1 < l < n, and b ≥ −(l +1)+(2l +2)−n−1 =
l − n ≥ −n. Hence |b| ≤ n. Also we have a − b = (2h + n + 1) − (h + k − n − 1) =
h+2n+2−k ≤ −1+2n+2−k = 2n+1−k ≤ 2n+1−(2l+2) = 2n−2l−1 = 2n−n = n, and
a−b = h+2n+2−k ≥ −(l+1)+2n+2−(3l+2) = 2n−4l−1 = 2(2l+1)−4l−1 = 1 > −n.
Hence |a− b| ≤ n. Therefore u ∈ C(A, n).
Because u − u =
−n
n + 1
=
2n + 1 n + 1
n 2n + 1
−1
1
, we have u ≡ u mod
(MAn )T . It follows that Ψ(u) = u.
Sub-case 2: If 0 ≤ h ≤ l, let a = 2h−n, b = h+k−2n−1 and u = u−
n + 1
2n + 1
.
Because u = Ak,h =
2h + 1
h + k
, we have u =
2h + 1
h + k
−
n + 1
2n + 1
=
2h− n
h + k − 2n− 1
=
a
b
.
We claim that u ∈ C(A, n). Since 2l + 2 ≤ k ≤ 3l + 2 and 0 ≤ h ≤ l, we have
a = 2h− n ≤ 2l− n < 2l < n and a ≥ 0− n = −n. Hence |a| < n. For the same reason, we
have b = h + k− 2n− 1 ≤ l + (3l + 2)− 2n− 1 = 4l− 2n + 1 = 4l− 2(2l + 1) + 1 = −1 < n,
and b = h+k−2n−1 ≥ 0+k−2n−1 ≥ 2l +2−2n−1 = 2l−2n+1 = 2l−2(2l +1)+1 =
−2l − 1 = −n. Hence |b| ≤ n. Also we have a − b = (2h − n) − (h + k − 2n − 1) =
h + n + 1 − k ≤ l + n + 1 − k ≤ l + n + 1 − (2l + 2) = n − l − 1 < n, and
a− b = h + n + 1− k ≥ 0 + n + 1− k = 2l + 2− k ≥ 2l + 2− (3l + 2) = −l > −n. Hence
|a− b| ≤ n. Therefore u ∈ C(A, n).
Because u − u =
n + 1
2n + 1
=
2n + 1 n + 1
n 2n + 1
0
1
, we have u ≡ u mod
(MAn )T . It follows that Ψ(u) = u.
We have shown that Ψ(u) is either u, or u −
−n
n + 1
, or u −
n + 1
2n + 1
. For
any k ∈ Z satisfying 1 ≤ k ≤ 3l + 2 and h ∈ Z satisfying −l ≤ h ≤ l, by the definition
57
of Ak,h and Bk,h, we have Ak,h − Bk,h =
1
0
. As shown in Figure 5-1, the method for
finding Ψ(Ak,h) can be modified easily to find Ψ(Bk,h). Finally let χ be the composition of
Φ and Ψ. Then χ is the required correspondence between P2 and C(A, n) in the frequency
domain.
5.3 Fast Algorithms for the DFT and its Inverse on the Type B RHS
In this section, we show an efficient method to convert the DFT and its inverse on the
type B RHS to a 1-dimensional standard DFT and a 1-dimensional standard inverse DFT,
respectively. It turns out that the DFT on <Bn can be similarly computed to the DFT on
<An . For any complex function f defined on <B
n , by Equation 2–5, its DFT f satisfies the
following equation:
f(q) =∑
p∈<Bn
f(p) · e−2πi〈p,q〉, for all q ∈ <Bn . (5–7)
As in the previous section, the set <Bn can be replaced by any set of coset representatives
P of LB/LBn , and <B
n can be replaced by a set of coset representatives Q of the quotient
group (LBn )∗/(LB)∗. Let f be a function defined on Z2 such that f(m) = f((V B)m) for any
m ∈ Z2. By Corollary 4.3.9, C(B, n) is a set of coset representatives associated with both
MBn and (MB
n )T , where the matrix MBn =
3n + 1 3n
1 3n + 1
is defined in Section 4.3.
Hence, by Equation 2–7, the DFT of f satisfies
f(k) =∑
m∈C(B,n)
f(m) · e−2πikT (MBn )−1m, for all k ∈ C(B, n). (5–8)
The following proposition shows that the DFT of f on C(B, n) can be similarly
converted to a standard DFT.
Proposition 5.3.1. For any integer n > 0, let S1 =
0
l
: 0 ≤ l ≤ 9n2 + 3n
and S2 =
l
l
: 0 ≤ l ≤ 9n2 + 3n
. Then the set S1 is a set of coset representatives
58
associated with MBn and the set S2 is a set of coset representatives associated with (MB
n )T .
Furthermore, f(
l
l
) =
∑9n2+3nm=0 f(
0
m
) · e −2πilm
9n2+3n+1 for all l satisfying 0 ≤ l ≤
9n2 + 3n.
Proof. By the Smith normal form of MBn we have MB
n = E(DBn )F , where E =
1 0
−3n 1
,DB
n =
1 0
0 9n2 + 3n + 1
, and F =
3n + 1 3n
1 1
. Obviously
S1 = E
0
l
: 0 ≤ l ≤ 9n2 + 3n and S2 = F T
0
l
: 0 ≤ l ≤ 9n2 + 3n.
Hence, it now follows from Proposition 5.1.1 that S1 is a set of coset representatives
associated with MBn , S2 is a set of coset representatives associated with (MB
n )T , and
f(
l
l
) =
∑9n2+3nm=0 f(
0
m
) · e −2πilm
9n2+3n+1 for all l satisfying 0 ≤ l ≤ 9n2 + 3n.
The correspondence between C(B, n) and S1 in the spatial domain can be established
in a likewise fashion.
5.4 Computational Complexity and Cooley-Tukey Factorization for the DFTon the Type A and Type B RHS
For the nth level of the type A RHS <An , let N be its size. By Lemma 4.2.1, N =
3n2 + 3n + 1. From the results of Section 5.2, the computational complexity of the DFT on
<An is the complexity of the correspondences (from C(A, n) to P1, and from P2 to C(A, n))
plus the computational complexity of the 1-dimensional standard Fourier transform
whose input size is N . The computational complexity of those two correspondences is
obviously O(N) as shown in the algorithms. It was shown in Frigo and Johnson [15] that
the computational complexity for the 1-dimensional standard Fourier transform of an
arbitrary input size N is O(N log N) although the sizes that are products of small factors
are transformed most efficiently. Thus the computational complexity of the DFT on <An is
still O(N log N). Similarly the computational complexity of the DFT on <Bn is O(N log N)
where N = 9n2 + 3n + 1.
59
Let f be a complex function defined on <An . By Equation 2–5, its DFT f satisfies the
following equation:
f(q) =∑
p∈<An
f(p) · e−2πi〈p,q〉, for all q ∈ <An . (5–9)
We generated a function f defined on <A298 using a random function with values ranging
from 0 to 100. The set <A298 consists of N = 267307 > 5122 = 262144 lattice points.
Although the number 267307 is prime, on a 2.8 GHz PC, the Matlab computational time
of the DFT of f is less than four seconds using the method provided in this chapter.
According to Moler [31], Version 6 of Matlab uses the Fast Fourier Transform in the West
(FFTW) whose webpage is www.fftw.org. As shown in Frigo and Johnson [15], FFTW
computes the DFT of arbitrary input size and utilizes Rader’s algorithm (Rader [35]) for
prime sizes. The Rader’s algorithm computes the DFT of size N by way of two DFTs and
one inverse DFT of size N − 1. For the DFT on <An , its size is N = 3n2 + 3n + 1. Since
N − 1 = 3n(n + 1), when n > 2, the product n(n + 1) has at least three non trivial integer
factors and hence N − 1 has at least four non trivial integer factors. Therefore, when N
is prime, the Rader’s algorithm can be applied efficiently for computing the DFT on <An
using the Cooley-Tukey algorithm for two DFTs and one inverse DFT of size N − 1.
When the determinant of the matrix MAn can be factored, we can also apply Theorem
2.6 (the Cooley-Tukey factorization of matrix form) in Zapata and Ritter [48]. Let us recall
that theorem in the following form:
Theorem 5.4.1. Let N be a d-dimensional periodicity matrix and IN a set of coset
representatives associated with N. Let f : IN → C. Suppose there exist periodicity
matrices P and Q and there exist sets of coset representatives IP and IQ associated with
P and Q respectively such that N = PQ, |det(P)| > 1, |det(Q)| > 1, IP ⊆ IN and
IQ ⊆ IN . Let INT , IPT , IQT be a set of coset representatives associated with NT , PT and
QT respectively such that IPT ⊆ INT and IQT ⊆ INT . Then the DFT f of f is given by
f(QT m + l) =∑p∈IP
C(p, l) · e−2πi lTN−1p · e−2πi mT P−1p, (5–10)
60
for all l ∈ IQT , m ∈ IPT , where C(p, l) =∑
q∈IQf(Pq + p) · e−2πi lT Q−1q.
For our case, the determinant of the matrix MAn is 3n2 + 3n + 1. Suppose that there
exist integers 1 < a, b ∈ Z such that 3n2+3n+1 = ab. Let TAn =
1 0
−3n− 1 3n2 + 3n + 1
and F =
2n + 1 n
2 1
. Since MA
n = TAn F and F is unimodular, MA
n and TAn have the
same periods.
Since 3n2 + 3n + 1 = ab, we have TAn =
1 0
−3n− 1 ab
= GH where G =
1 0
−3n− 1 a
and H =
1 0
0 b
. It is easy to check that P =
0
j
: 0 ≤ j < a
is a set of coset representatives associated with G and that Q =
0
k
: 0 ≤ k < b
is
a set of coset representatives associated with H. Let E = Gq + p : p ∈ P, q ∈ Q. Then
E is a set of coset representatives associated with TAn . By direct computation, we have
E =
1 0
−3n− 1 a
0
k
+
0
j
: 0 ≤ j < a, 0 ≤ k < b
=
0
j + ak
: 0 ≤ j < a, 0 ≤ k < b
=
0
l
: 0 ≤ l ≤ 3n2 + 3n
.
(5–11)
It is easy to verify that all conditions of Theorem 5.4.1 are satisfied.
61
CHAPTER 6PYXIS STRUCTURE
6.1 Definition and Labeling of the Pyxis Structure
6.1.1 The Definition of the Pyxis Structure
The Pyxis structure P was proposed in Peterson [33] and mentioned in Chapter 1. To
give a recursive definition based on vector additions, we need the following notation. First
recall from Chapter 4 that
vA1 =
1
0
,vA
2 =
−1
2√
32
,vB
1 =
√3
2
12
,vB
2 =
−√
32
12
,
LA =n1(v
A1 ) + n2(v
A2 ) : n1, n2 ∈ Z
and LB =
n1(v
B1 ) + n2(v
B2 ) : n1, n2 ∈ Z
.
Let ρ1 be a fixed positive number. For any n satisfying 1 < n ∈ N, let ρn = ( 1√3)n−1ρ1,
and let v2n−1,k = ρ2n−1vAk and v2n,k = ρ2nv
Bk for k ∈ 1, 2. Now we define a sequence of
lattices by Ln = n1vn,1 + n2vn,2 : n1, n2 ∈ Z for any n ∈ N. From this definition we can
see that, if we rotate the lattice L2n+1 by 30 about the origin and multiply the resulting
lattice by√
3, then we get the lattice L2n. The following lemma shows the inclusion
relation of those lattices.
To give a recursive definition of the Pyxis structure, for any n ∈ N, we let βn,1 =
vn,1 + vn,2, βn,2 = vn,2, βn,3 = −vn,1, βn,j = −βn,j−3 for j = 4, 5, 6, and βn =
βn,j : j = 1, 2, ..., 6. Figure 6-1 shows the generators of the lattices L1 and L2 as
well as the elements of β1 and β2. Let βn = βn ∪ 0 where 0 =
0
0
. Recall
from Chapter 2 that, for any lattice L and ∅ 6= X,Y ⊆ L, X + Y denotes the set
x + y ∈ L : x ∈ X, y ∈ Y .Let P (0) = 0, P (1) = β1, and for any integer n > 1 let
P (n) = P (n− 1)⋃
(P (n− 2) + βn).
62
−15 −10 −5 0 5 10 15
−15
−10
−5
0
5
10
15
β2,1
β2,2
β2,3
β2,4
β2,5
β2,6
β1,2 β
1,1
β1,3
β1,4
β1,5
β1,6
. (0,0)
Figure 6-1. The generators of the lattices L1 and L2, and the lattice points contained in β1
and β2. In this figure, the two red vectors are v1,1 and v1,2, the two dashedgreen vectors are v2,1 and v2,2, β1 consists of the six black * points, and β2
consists of the six blue o points.
For any integer n ≥ 0, the set P (n) is called the Pyxis structure at level n. We will show in
Theorem 6.1.4 that P (n − 1) and P (n − 2) + βn are disjoint. The next lemma shows the
containment relationship among P (n) and Ln for any n ∈ N.
Lemma 6.1.1. For any n ∈ N, we have Ln ⊂ Ln+1 and P (n) ⊂ Ln.
Proof. First we prove Ln ⊂ Ln+1. Because the proof when n is even can be similarly
done, we assume that n is odd, i.e., n = 2k − 1 for some k ∈ N. Since vA1 =
1
0
,
vA2 =
−1
2√
32
, vB
1 =
√3
2
12
and vB
2 =
−√
32
12
, we have vA
1 = 1√3(vB
1 − vB2 )
and vA2 = 1√
3(vB
1 + 2vB2 ). Then for any n1, n2 ∈ Z, we have n1v2k−1,1 + n2v2k−1,2 =
n1ρ2k−1vA1 + n2ρ2k−1v
A2 = ρ2k−1(n1v
A1 + n2v
A2 ) = 1√
3ρ2k−1(n1(v
B1 − vB
2 ) + n2(vB1 + 2vB
2 )) =
ρ2k((n1 + n2)vB1 + (2n2 − n1)v
B2 ) = (n1 + n2)v2k,1 + (2n2 − n1)v2k,2 ∈ L2k = Ln+1. Thus
Ln ⊆ Ln+1. Obviously the length of v2k−1,1 and v2k−1,2 is ρ2k−1 and the angle between
v2k−1,1 and v2k−1,2 is 2π3
. It follows that the length of any vector in the set L2k−1 is at
63
least ρ2k−1 which is bigger than the length of v2k,1. Hence v2k,1 6∈ L2k−1 and L2k 6= L2k−1.
Therefore L2k−1 ⊂ L2k, i.e., Ln ⊂ Ln+1.
By the definition of βn, we have βn ⊂ Ln for any n ∈ N. Also by the definition
of P (1), we have P (1) = β1. It follows that P (1) ⊆ L1. Obviously P (0) = 0 ⊂ L0.
Using induction, assume that P (k) ⊆ Lk for any k satisfying 0 ≤ k < n. By the
definition of P (n), for any n ≥ 2, we have P (n) = P (n − 1)⋃
(P (n − 2) + βn). By the
assumption of this induction, we have P (n − 2) ⊆ Ln−2 and P (n − 1) ⊆ Ln−1. By the
result of the previous paragraph, we have Ln−2 ⊆ Ln−1 ⊆ Ln. Because βn ⊂ Ln and
P (n) = P (n − 1)⋃
(P (n − 2) + βn), it follows that P (n) ⊆ Ln. Since P (n) contains only
finite number of lattice points, P (n) 6= Ln. Thus P (n) ⊂ Ln.
The first 4 levels of the Pyxis structure are shown in Figure 6-2. Notice that the area
of a cell of P (n) is three times that of a cell of P (n + 1). The factor three is a suitable
scaling factor for re-sampling an image from one resolution into another. This is one of
the advantages of the Pyxis structure over the GBT2 where, as we mentioned in Chapter
3, the scaling factor seven in GBT2 would make an image change rapidly if we re-sample
it into the grid of the previous level. The following two lemmas will be used to prove
Theorem 6.1.4, which provides some properties of the Pyxis structure.
Lemma 6.1.2. If n ≥ 2 and f : Ln−2 × βn → Ln−2 + βn is the map defined by
f(a,b) = a + b for any a ∈ Ln−2 and b ∈ βn, then f is a bijection.
Proof. Obviously f is onto. We will show that f is also one to one. If f(a,b) = f(a, b) for
some (a,b), (a, b) ∈ Ln−2×βn, then a+b = a+ b. Hence b− b = a−a. Since a, a ∈ Ln−2
and Ln−2 is a lattice, we have a− a ∈ Ln−2. It follows that b− b ∈ Ln−2. Hence there are
l1, l2 ∈ Z such that
b− b = l1vn−2,1 + l2vn−2,2. (6–1)
64
0 6
12
3
4 5
06
01
02
03
04
05
00 60
1020
30
40 50
006
001002
003
004 005
606
601602
603
604 605
106
101102
103
104 105
206
201202
203
204 205
306
301302
303
304 305
406
401402
403
404 405
506
501502
503
504 505
060
010
020
030
040
050
000 600
100200
300
400 500
0606
0601
0602
0603
0604
0605
0106
0101
0102
0103
0104
0105
0206
0201
0202
0203
0204
0205
0306
0301
0302
0303
0304
0305
0406
0401
0402
0403
0404
0405
0506
0501
0502
0503
0504
0505
0006
0001
0002
0003
0004
0005
6006
6001
6002
6003
6004
6005
1006
1001
1002
1003
1004
1005
2006
2001
2002
2003
2004
2005
3006
3001
3002
3003
3004
3005
4006
4001
4002
4003
4004
4005
5006
5001
5002
5003
5004
5005
0060
00100020
0030
0040 0050
6060
60106020
6030
6040 6050
1060
10101020
1030
1040 1050
2060
20102020
2030
2040 2050
3060
30103020
3030
3040 3050
4060
40104020
4030
4040 4050
5060
50105020
5030
5040 5050
0600
0100
0200
0300
0400
0500
0000 6000
10002000
3000
4000 5000
Figure 6-2. The first 4 levels of the Pyxis structure where P (1) consists of 7 bluehexagons, P (2) consists of 13 red hexagons, P (3) consists of 55 greenhexagons, and P (4) consists of 133 black hexagons. In this figure, the blackvector is the sum of the two blue vectors. It follows that the coordinates of thelattice point labeled 0205 is the sum of the coordinates of the two latticepoints labeled 0030 and 0001, respectively.
Let D = (1, 1), (0, 1), (−1, 0), (−1,−1), (0,−1), (1, 0). It follows easily from the definition
of βn that
βn = n1vn,1 + n2vn,2 : (n1, n2) ∈ D .
Since b, b ∈ βn, there exist (n1, n2), (n1, n2) ∈ D such that b = n1vn,1 + n2vn,2 and
b = n1vn,1 + n2vn,2. It follows that
b− b = (n1 − n1)vn,1 + (n2 − n2)vn,2. (6–2)
By combining Equations 6–1 and 6–2, we have
(n1 − n1)vn,1 + (n2 − n2)vn,2 = l1vn−2,1 + l2vn−2,2.
65
Because vn−2,j = 3vn,j for j = 1, 2, it follows that
(n1 − n1)vn,1 + (n2 − n2)vn,2 = 3(l1vn,1 + l2vn,2).
Hence n1 − n1 = 3l1. However (n1, n2), (n1, n2) ∈ D implies that |n1 − n1| ≤ 2. Thus
n1 − n1 = 0. Similarly n2 − n2 = 0. Hence b = b and a = a. Therefore f is one to one.
Lemma 6.1.3. For any n satisfying 0 ≤ n ∈ Z, we have Ln
⋂βn+1 = ∅.
Proof. We will just give a proof when n is even since the proof is similar when n is odd.
Suppose that βn+1
⋂Ln 6= ∅. Since βn+1 = n1vn+1,1 + n2vn+1,2 : (n1, n2) ∈ D where
D = (1, 1), (0, 1), (−1, 0), (−1,−1), (0,−1), (1, 0), there exist (n1, n2) ∈ D and l1, l2 ∈ Zsuch that
n1vn+1,1 + n2vn+1,2 = l1vn,1 + l2vn,2. (6–3)
Since n is assumed to be even, vn+1,k = ρn+1vAk and vn,k = ρnv
Bk for k = 1, 2. Then
Equation 6–3 becomes
n1ρn+1vA1 + n2ρn+1v
A2 = l1ρnv
B1 + l2ρnv
B2 . (6–4)
Since ρn =√
3ρn+1, it follows that n1vA1 + n2v
A2 =
√3(l1v
B1 + l2v
B2 ). Hence
n1vA1 + n2v
A2 =
√3(
l1√3(2vA
1 + vA2 ) +
l2√3(−vA
1 + vA2 )) = (2l1 − l2)v
A1 + (l1 + l2)v
A2 .
Then n1 = 2l1 − l2 and n2 = l1 + l2. Thus n1 + n2 = 3l1. However (n1, n2) ∈ D implies that
n1 + n2 equals ±1 or ±2, which contradicts that n1 + n2 = 3l1 for some l1 ∈ Z. Therefore
βn+1
⋂Ln = ∅.
Theorem 6.1.4. For any n satisfying 0 ≤ n ∈ Z, we have
1. P (n) ⊂ P (n + 1);
2. P (n + 1)⋂
(P (n) + βn+2) = ∅;3. |P (n)| = 1
5[3n+2 − (−2)n+2].
66
Proof. 1. It follows easily from the definition of the Pyxis structure that P (n) ⊆ P (n + 1)
and P0 6= P1. Since P (0) = 0, we have 0 ∈ P (n) for any n ≥ 0. Suppose that
P (n) = P (n + 1) for some n ≥ 1. Since P (n + 1) = P (n)⋃
(P (n − 1) + βn+1), we
have P (n) = P (n)⋃
(P (n − 1) + βn+1). It follows that P (n − 1) + βn+1 ⊆ P (n).
Furthermore 0 ∈ P (n − 1) implies that 0 + βn+1 ⊆ P (n − 1) + βn+1. Therefore
βn+1 ⊆ 0 + βn+1 ⊆ P (n− 1) + βn+1 ⊆ P (n) which contradicts Lemma 6.1.3.
2. Since P (n) ⊂ Ln ⊂ Ln+1 and P (n + 1) ⊂ Ln+1, if P (n + 1)⋂
(P (n) + βn+2) 6= ∅, we
have Ln+1
⋂βn+2 6= ∅ which contradicts Lemma 6.1.3.
3. The conclusion is obviously true for n = 0, 1. When n > 1, by the definition the
Pyxis structure, we have P (n) = P (n− 1)⋃
(P (n− 2) + βn). Since P (n− 1)⋂
(P (n− 2) +
βn) = ∅ by Item 2, we have |P (n)| = |P (n− 1)|+ |P (n− 2)+βn|. By Lemma 6.1.2 we have
|P (n− 2) + βn| = |P (n− 2)× βn| = 6|P (n− 2)|. Hence |P (n)| = |P (n− 1)|+ 6|P (n− 2)|.We also have initial conditions |P (0)| = 1 and |P (1)| = 7. Thus we can get |P (n)| =
15[3n+2 − (−2)n+2].
6.1.2 The Labeling of the Pyxis Structure
In this subsection, we assign to each lattice point of P (n) a unique label which
consists of a string of integers λ1λ2...λn such that 0 ≤ λi ≤ 6 for each 1 ≤ i ≤ n and either
λi = 0 or λi+1 = 0 for any i satisfying 1 ≤ i < n. Because labels of lattice points of P (n),
as opposed to their coordinates, consist of integers, such an assignment is useful for quick
data retrieval of the Pyxis structure.
Let βn,0 = 0 for any n ∈ N. Since βn = βn∪0, we have βn = βn,j : j = 0, 1, 2, ..., 6.To define the labeling, we prove that, for any a ∈ P (n), there exists uniquely determined
λi ∈ 0, 1, 2, ..., 6 for each i satisfying 1 ≤ i ≤ n such that a =∑n
i=1 βi,λi, and either
λi = 0 or λi+1 = 0 for each i satisfying 1 ≤ i < n. The next lemma will be used in the
proof of the following theorem.
Lemma 6.1.5. Let λi ∈ 0, 1, 2, ..., 6 for i = 1, 2, ..., n. If either λi = 0 or λi+1 = 0 for
any i satisfying 1 ≤ i < n, then∑n
i=1 βi,λi∈ P (n).
67
Proof. If n = 1, then this lemma is obviously true. Suppose that it is true for any n < t
for some integer t. When n = t, let a =∑t
i=1 βi,λi. If λt = 0, then βt,λt = 0. Hence
a =∑t−1
i=1 βi,λi∈ P (t−1) by the assumption of this induction. By Item 1 of Theorem 6.1.4,
we have P (t− 1) ⊂ P (t). Hence a ∈ P (t). If λt 6= 0, then λt−1 = 0 by the condition of this
lemma. Hence
a = βt,λt +t−2∑i=1
βi,λi. (6–5)
The assumption of this induction follows that∑t−2
i=1 βi,λi∈ P (t− 2). Since λt 6= 0, we have
βt,λt ∈ βt. Hence by Equation 6–5, we have a ∈ βt + P (t− 2) ⊂ P (t).
Theorem 6.1.6. For any n ∈ N and a ∈ P (n), there exists uniquely determined
λi ∈ 0, 1, 2, ..., 6 for each i = 1, 2, ..., n such that a =∑n
i=1 βi,λiand either λi = 0 or
λi+1 = 0 for each i satisfying 1 ≤ i < n.
Proof. When n = 1, it is obviously true. Suppose that it is true for any n < t for some
integer t. When n = t, by definition we have P (t) = P (t − 1)⋃
(P (t − 2) + βt). Hence
a ∈ P (t) implies that either a ∈ P (t − 1) or a ∈ P (t − 2) + βt. If a ∈ P (t − 1),
then by the assumption of this induction, a has such an expression in P (t − 1), which is
a =∑t−1
i=1 βi,λi. Let λt = 0. Then βt,λt = 0. It follows that a =
∑ti=1 βi,λi
=∑n
i=1 βi,λi,
which is the required expression of a in P (n). If a ∈ P (t − 2) + βt, then a = b + c for
some b ∈ P (t − 2) and c ∈ βt. By the assumption of this induction, b has the required
expression b =∑t−2
i=1 βi,λiin P (t − 2). Since c ∈ βt, there is λt ∈ 1, 2, ..., 6 such that
c = βt,λt . Let λt−1 = 0. Then a = b + c =∑t
i=1 βi,λi=
∑ni=1 βi,λi
is the required expression
of a in P (n).
To show the uniqueness, suppose that a has another such expression a =∑n
i=1 βi,λi,
where λi ∈ 0, 1, 2, ..., 6 for each i = 1, 2, ..., n and either λi = 0 or λi+1 = 0 for any i
satisfying 1 ≤ i < n. It follows that
n∑i=1
βi,λi=
n∑i=1
βi,λi. (6–6)
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If λn = λn, then Equation 6–6 becomes
n−1∑i=1
βi,λi=
n−1∑i=1
βi,λi. (6–7)
By Lemma 6.1.5, we have∑n−1
i=1 βi,λi∈ P (n − 1). By the assumption of this induction,
it follows from Equation 6–7 that λi = λi for any i satisfying 1 ≤ i ≤ n − 1. Thus
the expression of a is unique in this case. If λn 6= 0 but λn = 0, then it follows from
Equation 6–5 that
βn,λn =n−1∑i=1
βi,λi−
n−1∑i=1
βi,λi. (6–8)
By Lemma 6.1.1, we have βi ⊂ Ln−1 for each i = 1, 2, ..., n− 1. By Equation 6–8, it follows
that βn,λn ∈ Ln−1. Since λn 6= 0, we have βn,λn ∈ βn. Hence βn,λn ∈ βn
⋂Ln−1 = ∅
by Lemma 6.1.3, which is a contradiction. If λn = 0 but λn 6= 0, then a contradiction
can be similarly shown. If λn 6= 0 and λn 6= 0, then λn−1 = 0 and λn−1 = 0 by the
requirement of such expressions. By Lemma 6.1.5, we have∑n−2
i=1 βi,λi∈ P (n − 2)
and∑n−2
i=1 βi,λi∈ P (n − 2). Since βn,λn ∈ βn and βn,λn
∈ βn, by Lemma 6.1.2, it
follows from Equation 6–6 that∑n−2
i=1 βi,λi=
∑n−2i=1 βi,λi
and βn,λn = βn,λn. Since
∑n−2i=1 βi,λi
=∑n−2
i=1 βi,λi, by the assumption of this induction, we have λi = λi for each i
satisfying 1 ≤ i ≤ n− 2. Furthermore, since βn,λn = βn,λn, we have λn = λn. Therefore the
uniqueness is proved for all cases.
For any n ∈ N and a ∈ P (n), the unique expression a =∑n
i=1 βi,λiin Theorem 6.1.6
is called the standard expression of a in P (n). The string λ1λ2...λn of integers in that
standard expression is called the label of a in P (n). If Va denotes the Voronoi cell of
a ∈ Ln, then the string λ1λ2...λn is also called the label of Va. The label of each lattice
point of P (n) also serves as the label of the Voronoi cell of that lattice point. It is easy to
verify that, in Figure 6-2 of the previous subsection, the center hexagon of P (1) is labeled
0, and other six hexagons of P (1) are labeled 1,2,3,4,5,6 going counter-clockwise. For any
69
n satisfying 0 ≤ n ∈ Z, let Λn denote the set of the labels of lattice points of P (n). The
following corollary shows the structure of Λn.
Corollary 6.1.7. For any n satisfying 2 < n ∈ Z, we have
1. Λn = λ1λ2...λn : λi ∈ 0, 1, 2, ..., 6 for each i, and λi · λi+1 = 0 for each i < n .
2. Λn = λ0j : λ ∈ Λn−2, j = 1, 2, 3, 4, 5, 6⋃ λ0 : λ ∈ Λn−1 .
Proof. 1. Since Λn is the set of the labels of lattice points of P (n), by Lemma 6.1.5 and
Theorem 6.1.6, we have
Λn = λ1λ2...λn : λi ∈ 0, 1, 2, ..., 6 for each i, and λi · λi+1 = 0 for each i < n .
2. It follows easily from Item 1 that
λ0j : λ ∈ Λn−2, j = 1, 2, 3, 4, 5, 6⋃λ0 : λ ∈ Λn−1 ⊆ Λn.
Conversely, for any λ ∈ Λn, let λ = λ1λ2...λn. If λn = 0, then λ = λ0 where
λ = λ1λ2...λn−1 ∈ Λn−1. If λn 6= 0, by Theorem 6.1.6, we have λn−1 = 0. Hence
λ = λ0j where λ = λ1λ2...λn−2 ∈ Λn−2 and j = λn ∈ 1, 2, ..., 6. Therefore Λn ⊆λ0j : λ ∈ Λn−2, j = 1, 2, 3, 4, 5, 6⋃ λ0 : λ ∈ Λn−1.
6.1.3 Addition of the Labels of the Pyxis Structure
In this subsection, we first define the sum of labels of two lattice points in P (n) to
be the label of the sum of these two lattice points which are taken as the vectors of R2.
The addition of Pyxis labels will be used in Section 6.2. Our algorithm for adding any
two labels in Λn uses the fact that 13a + 1
3b ∈ P (n + 2) for any n ∈ N and a,b ∈ P (n)
which is proved in this subsection. Tables 6-1 and 6-2 are also used in the algorithm. The
computational complexity of our algorithm is shown to be of order n.
Let a,b ∈ P (n) be labeled a1a2...an and b1b2...bn, respectively. If a + b ∈ P (n) and
the label of a + b in P (n) is c1c2...cn, then the label c1c2...cn is called the sum of the labels
a1a2...an and b1b2...bn, and we write a1a2...an ⊕ b1b2...bn = c1c2...cn in Λn. For example, the
three dashed vectors in Figure 1-2 show that 0506⊕ 2005 = 1040 in Λ4.
70
Partial addition table for P (3).⊕ 001 002 003 004 005 006 010 020 030 040 050 060
001 104 010 002 000 006 060 103 205 003 005 603 105002 010 205 020 003 000 001 206 204 306 004 006 104003 002 020 306 030 004 000 205 301 305 401 005 001004 000 003 030 401 040 005 002 306 402 406 502 006005 006 000 004 040 502 050 001 003 401 503 501 603006 060 001 000 005 050 603 104 002 004 502 604 602
Partial addition table for P (4).⊕ 0001 0002 0003 0004 0005 0006 0010 0020 0030 0040 0050 0060
0001 0104 0020 0002 0000 0006 0010 0105 0103 0205 0003 0005 06030002 0020 0205 0030 0003 0000 0001 0104 0206 0204 0306 0004 00060003 0002 0030 0306 0040 0004 0000 0001 0205 0301 0305 0401 00050004 0000 0003 0040 0401 0050 0005 0006 0002 0306 0402 0406 05020005 0006 0000 0004 0050 0502 0060 0603 0001 0003 0401 0503 05010006 0010 0001 0000 0005 0060 0603 0602 0104 0002 0004 0502 06040010 0105 0104 0001 0006 0603 0602 1040 0100 0020 0000 0060 06000020 0103 0206 0205 0002 0001 0104 0100 2050 0200 0030 0000 00100030 0205 0204 0301 0306 0003 0002 0020 0200 3060 0300 0040 00000040 0003 0306 0305 0402 0401 0004 0000 0030 0300 4010 0400 00500050 0005 0004 0401 0406 0503 0502 0060 0000 0040 0400 5020 05000060 0603 0006 0005 0502 0501 0604 0600 0010 0000 0050 0500 6030
By the plot of the lattice points of P (3) (P (4) respectively) in Figure 6-2, it is
easy to verify Tables 6-1 (6-2 respectively) which gives partial addition tables for P (3)
(P (4) respectively) and will be used in our algorithm. The following lemma gives some
properties of labels of the Pyxis structure which will be used very often for the addition of
labels.
Lemma 6.1.8. For any n ∈ N we have the following:
1. If λ1λ2...λn ∈ Λn and k < n, then λ1λ2...λn = λ1λ2...λk00...0⊕ 00...0λk+1λk+2...λn in
Λn.
2. If 1 ≤ k ≤ n, A = λ1λ2...λk00...0 ∈ Λn, B = 00...0λk+1λk+2...λn ∈ Λn, and either
λk = 0 or λk+1 = 0, then A⊕B = λ1λ2...λn in Λn.
3. If a ∈ P (n) with label λ1λ2...λn ∈ Λn, then 13a ∈ P (n + 2) with label 00λ1λ2...λn ∈
Λn+2, and the label of a in P (n + 1) is λ1λ2...λn0 ∈ Λn+1.
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4. For any a1a2...an ∈ Λn, b1b2...bn ∈ Λn and c1c2...cn ∈ Λn, we have a1a2...an ⊕b1b2...bn = c1c2...cn in Λn if and only if 0a1a2...an ⊕ 0b1b2...bn = 0c1c2...cn in Λn+1 if
and only if a1a2...an0⊕ b1b2...bn0 = c1c2...cn0 in Λn+1.
Proof. 1. Let a =∑n
i=1 βi,λi, b =
∑ki=1 βi,λi
, and c =∑n
i=k+1 βi,λi. Since λ1λ2...λn ∈ Λn, by
Item 1 of Corollary 6.1.7, we have either λi = 0 or λi+1 = 0 for any i satisfying 1 ≤ i < n.
By Lemma 6.1.5, it follows that a,b, c ∈ P (n). By the definition of labels of the elements
in P (n), the labels of a, b and c are λ1λ2...λn, λ1λ2...λk00...0 and 00...0λk+1λk+2...λn,
respectively. Since a = b + c, we have λ1λ2...λn = λ1λ2...λk00...0 ⊕ 00...0λk+1λk+2...λn in
Λn.
2. Since A,B ∈ Λn, there exist a,b ∈ P (n) such that A and B are the labels of a
and b, respectively. Because A = λ1λ2...λk00...0 ∈ Λn, we have a =∑k
i=1 βi,λi. Similarly
B = 00...0λk+1λk+2...λn ∈ Λn implies that b =∑n
i=k+1 βi,λi. Since λ1λ2...λk00...0 ∈ Λn,
we have either λi or λi+1 for any i satisfying 1 ≤ i < k. Similarly it follows from
00...0λk+1λk+2...λn ∈ Λn that either λi or λi+1 for any i satisfying k + 1 ≤ i < n.
Furthermore, by the given condition of this Item, we have either λk = 0 or λk+1 = 0.
Hence either λi = 0 or λi+1 = 0 for any i satisfying 1 ≤ i < n. By Lemma 6.1.5, it
follows that∑n
i=1 βi,λi∈ P (n) and the label of
∑ni=1 βi,λi
in P (n) is λ1λ2...λn. Since
a + b =∑n
i=1 βi,λi, we have A⊕B = λ1λ2...λn in Λn.
3. Since a ∈ P (n) and the label of a is λ1λ2...λn, the standard expression of a is
a =∑n
i=1 βi,λi. It follows that 1
3a =
∑ni=1
13βi,λi
. Because 13βi,λi
= βi+2,λi, we have 1
3a =
∑ni=1 βi+2,λi
∈ P (n + 2). It follows that the label of 13a in P (n + 2) is 00λ1λ2...λn ∈ Λn+2.
If we let λn+1 = 0, then a =∑n+1
i=1 βi,λi. It follows that the label of a in P (n + 1) is
λ1λ2...λnλn+1, i.e., λ1λ2...λn0.
4. Since a1a2...an, b1b2...bn, c1c2...cn ∈ Λn, there exist a,b, c ∈ P (n) such that
a1a2...an, b1b2...bn and c1c2...cn are the labels of a,b and c, respectively. Then the
standard expressions of a,b and c are a =∑n
i=1 βi,ai, b =
∑ni=1 βi,bi
and c =∑n
i=1 βi,ci,
respectively.
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If a1a2...an ⊕ b1b2...bn = c1c2...cn in Λn, then a + b = c, i.e.,∑n
i=1 βi,ai+
∑ni=1 βi,bi
=∑n
i=1 βi,ci. It follows that
∑ni=1
1√3βi,ai
+∑n
i=11√3βi,bi
= 1√3
∑ni=1 βi,ci
. By rotating those
vectors by an angle of π6
as shown in Figure 6-1, it follows that∑n
i=1 βi+1,ai+
∑ni=1 βi+1,bi
=∑n
i=1 βi+1,ci. Hence 0a1a2...an ⊕ 0b1b2...bn = 0c1c2...cn in Λn+1. If 0a1a2...an ⊕ 0b1b2...bn =
0c1c2...cn in Λn+1, similarly we can show that a1a2...an ⊕ b1b2...bn = c1c2...cn in Λn.
Let an+1 = bn+1 = cn+1 = 0. Then∑n
i=1 βi,ai+
∑ni=1 βi,bi
=∑n
i=1 βi,ciif and only
if∑n+1
i=1 βi,ai+
∑n+1i=1 βi,bi
=∑n+1
i=1 βi,ci. It follows that a1a2...an ⊕ b1b2...bn = c1c2...cn if
and only if a1a2...an+1 ⊕ b1b2...bn+1 = c1c2...cn+1 if and only if a1a2...an0 ⊕ b1b2...bn0 =
c1c2...cn0.
Let L be a hexagonal lattice. For any a,b ∈ L, if their Voronoi cells share exactly one
side, then we say that a and b are next to each other in the lattice L, and each of them
is called a neighbor of the another in the lattice L. The following three lemmas will be
applied in subsequent lemmas.
Lemma 6.1.9. Let 1 < n ∈ N. If a,b ∈ βn which are next to each other in the lattice Ln,
then a + b ∈ βn−1.
Proof. It follows directly from Figure 6-1.
Lemma 6.1.10. For any n ∈ N and a,b ∈ Ln, a is next to b in the lattice Ln if and only
if b− a ∈ βn.
Proof. If a is next to b in the lattice Ln, then their Voronoi cells shares exactly one side.
After a translation of −a, it follows that the Voronoi cell of a − a shares exactly one side
with the Voronoi cell of b − a. Hence a − a, i.e., 0, is next to b − a in the lattice Ln.
Obviously βn = x ∈ Ln : x is next to 0 in the lattice Ln. Thus b − a ∈ βn. Conversely
if b − a ∈ βn, then b − a is next to 0 in the lattice Ln. After a translation of a, it follows
that a is next to b in the lattice Ln.
Lemma 6.1.11. For any a,b ∈ βn, we have:
1. If a 6= b, then either a + b = 0 or a + b ∈ βn−1
⋃βn.
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2. If a 6= ±b, then a is next to either b or −b in the lattice Ln.
Proof. 1. Obviously βn contains exactly six lattice points which are all neighbors of 0
in the lattice Ln. Since a,b ∈ βn and a 6= b, it follows that b = −a, or b is next to
either a or −a in the lattice Ln. If b = −a, then a + b = 0. If b is next to a, then by
Lemma 6.1.9, a + b ∈ βn−1. If b is next to −a in the lattice Ln, then by Lemma 6.1.10, we
have b− (−a) ∈ βn. Hence a + b ∈ βn.
2. Since a,b ∈ βn and βn contains exactly six lattice points which are all neighbors of
0 in the lattice Ln, if a 6= ±b, then a must be next to either b or −b in the lattice Ln.
The following two lemmas will be used in Theorem 6.1.14, which is one of the main
results in this subsection.
Lemma 6.1.12. For any a,b, c ∈ P (1), we have 13a + 1
3b + 1
3c ∈ P (3).
Proof. Let u = 13a + 1
3b + 1
3c and let the labels of a,b and c be a1, b1, and c1, respectively.
By applying Item 3 of Lemma 6.1.8, it follows that the labels of 13a, 1
3b and 1
3c are 00a1,
00b1, and 00c1, respectively.
If either a = 0, or b = 0, or c = 0, without loss of generality, we assume that c = 0.
It follows that u = 13a + 1
3b and the label of u is 00a1 ⊕ 00b1. By applying Table 6-1, we
have 00a1⊕ 00b1 = d10d3 for some label d10d3 ∈ Λ3. Hence the label of u is d10d3 ∈ Λ3 and
thus u ∈ P (3).
Otherwise, if a = b = c, then u = a ∈ P (1) ⊂ P (3).
Otherwise, if either a = −b, or a = −c, or b = −c, without loss of generality, we
assume a = −b. It follows that u = 13c ∈ P (3).
Otherwise, if either a = b, or a = c, or b = c, without loss of generality, we assume
that a = b. Since c 6= ±a and a, c ∈ β1, it follows that a is next to either c or −c in
L1. If a is next to c in L1, then 13a is next to 1
3c in L3. By lemma 6.1.9, it follows that
13a − 1
3c ∈ β3. Hence u = a + 1
3(c − a) ∈ P (1) + β3 ⊂ P (3). If a is next to −c in L1, then
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13a is next to −1
3c in L3. It follows that 1
3a + 1
3c ∈ β3, and 1
3a is next to 1
3a + 1
3c in L3.
Hence, by Lemma 6.1.9, u = 13a + 1
3(c + a) ∈ β2 ⊂ P (2) ⊂ P (4).
Otherwise, a,b and c are distinct and each of those three elements is next to at least
one of the other two elements in L1. Without loss of generality, we assume that a is next
to b and b is next to c in L1. It follows that a + c = b. Hence u = 13(a + b + c) = 2
3b =
b− 13b ∈ P (1) + β3 ⊂ P (3). Therefore this lemma is valid for all cases.
Lemma 6.1.13. For any a,b, c ∈ P (2), we have 13a + 1
3b + 1
3c ∈ P (4).
Proof. Let u = 13a + 1
3b + 1
3c and let the labels of a,b and c be a1a2, b1b2, and c1c2,
respectively. By applying Item 3 of Lemma 6.1.8, it follows that the labels of 13a, 1
3b
and 13c are 00a1a2, 00b1b2, and 00c1c2, respectively. Since a,b, c ∈ P (2) and P (2) =
β1
⋃β2
⋃ 0, we verify this lemma by considering the following cases.
Case 1: If a,b, c ∈ β1, then a2 = b2 = c2 = 0. By Lemma 6.1.12, we have
00a1⊕ 00b1⊕ 00c1 ∈ Λ3. By Item 4 of Lemma 6.1.8, it follows that 00a10⊕ 00b10⊕ 00c10 ∈Λ4, i.e., 00a1a2 ⊕ 00b1b2 ⊕ 00c1c2 ∈ Λ4. Hence u ∈ P (4).
Case 2: If a,b, c ∈ β2, then a1 = b1 = c1 = 0. Similar to the proof of Case 1, we
have u ∈ P (4).
Case 3: If either a = 0, or b = 0, or c = 0, without loss of generality, we assume
that c = 0. It follows that u = 13a + 1
3b and the label of u is 00a1a2 ⊕ 00b1b2. If a1 6= 0
and b1 6= 0, then by Item 1 of Corollary 6.1.7 we have a2 = b2 = 0. By applying Table 6-1,
there exists d10d3 ∈ Λ3 such that 00a1 ⊕ 00b1 = d10d3 in Λ3. By Item 4 of Lemma 6.1.8, it
follows that 00a10⊕ 00b10 = d10d30 in Λ4. If either a1 = 0 or b1 = 0, then by Table 6-2, we
have 00a1a2 ⊕ 00b1b2 in Λ4. It follows that u ∈ P (4).
Case 4: If a,b, c ∈ β1
⋃β2 and the set a,b, c⋂
β1 contains one element, without
loss of generality, we assume a ∈ β1. It follows that b, c ∈ β2. Let the labels of b and c be
0b2 and 0c2, respectively.
If b2 6= c2, then b 6= c. By Item 1 of Lemma 6.1.11, it follows that b+c ∈ P (2). Hence
either b + c ∈ β2 or b + c ∈ β1 or b + c = 0. By Item 3 of Lemma 6.1.8, it follows that
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either 13(b + c) ∈ β4 or 1
3(b + c) ∈ β3 or 1
3(b + c) = 0. Because a ∈ β1, we have 1
3a ∈ β3.
Since u = 13a + 1
3(b + c), it follows that either u ∈ β3 + β4 or u ∈ β3 + β3 or u ∈ β3 ⊂ P (4).
By Table 6-2, we have β3 + β4 ⊆ P (4). By Table 6-1, we have β3 + β3 ⊆ P (3) ⊆ P (4).
Hence u ∈ P (4).
If b2 = c2, without loss of generality, we assume that b2 = c2 = 1. Then, by Table 6-2,
the label of 13(b + c) is 0001 ⊕ 0001 = 0104 in Λ4. Since a ∈ β1, the label of a in Λ2 is
a20. Hence the label of u is 00a20 ⊕ 0104 in Λ4. By Table 6-2, there exists 0d10d3 ∈ Λ4
such that 00a20 ⊕ 0004 = 0d10d3 in Λ4 where d1 ∈ 0, 3, 4, 5. Since d1 ∈ 0, 3, 4, 5,by Table 6-1, we have 00d1 ⊕ 001 = 00t1 in Λ3 for some t1 ∈ 0, 1, 2, 6. By Item 4
of Lemma 6.1.8, it follows that 0d100 ⊕ 0100 = 0t100 in Λ4. Hence the label of u is
00a20⊕ 0104 = 0d10d3 ⊕ 0100 = 0t100⊕ 000d3 = 0t10d3 in Λ4. Thus u ∈ P (4).
Case 5: If a,b, c ∈ β1
⋃β2 and the set a,b, c⋂
β1 contains two elements, without
loss of generality, we assume that a,b ∈ β1. It follows that c ∈ β2. Hence, by Item 3 of
Lemma 6.1.8, we have 13a, 1
3b ∈ β3 and 1
3c ∈ β4.
If a 6= b, then, by Item 1 of Lemma 6.1.11, we have either 13a + 1
3b ∈ β2, or
13a + 1
3b ∈ β3, or 1
3a + 1
3b = 0. Hence either u ∈ β2 + β4 ⊂ P (2) + β4 ⊂ P (4), or
u ∈ β3 + β4 ⊂ P (4) by Table 6-2, or u ∈ β4 ⊂ P (4). Thus u ∈ P (4).
If a = b, without loss of generality, we assume that a1 = b1 = 1. Then the label
of 13a + 1
3b in P (4) is 0010 ⊕ 0010 = 1040 in Λ4. Let the label of c be 0c2. It follows
that the label of u is 1040 ⊕ 000c2 = 1000 ⊕ 0040 ⊕ 000c2 in Λ4. By Table 6-2, we
have 0040 ⊕ 000c2 = 0d10d2 in Λ4 where d1 ∈ 0, 3, 4. Hence the label of u becomes
1000 ⊕ 0d10d2 = 1000 ⊕ 0d100 ⊕ 000d2 in Λ4. Since d1 ∈ 0, 3, 4, by Table 6-2, we
have 10 ⊕ 0d1 ∈ 10, 01, 06 ⊂ Λ2. By applying Item 4 of Lemma 6.1.8 twice, it follows
that 1000 ⊕ 0d100 ∈ 1000, 0100, 0600 ⊂ Λ4. Hence the label of u belongs to the
set 1000⊕ 000d2, 0100⊕ 000d2, 0600⊕ 000d2 = 100d2, 010d2, 060d2 ⊂ Λ4. Thus
u ∈ P (4).
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The following theorem will play an important part in the algorithm for adding any
two labels.
Theorem 6.1.14. For any a,b ∈ P (n), we have 13a + 1
3b ∈ P (n + 2).
Proof. We will just give a proof when n is even since the proof for odd n is similar. Let
the labels of a and b be a1a2...an and b1b2...bn, respectively. It follows from Item 3 of
Lemma 6.1.8 that 13a, 1
3b ∈ P (n + 2), and the labels of 1
3a and 1
3b are 00a1a2...an ∈ Λn+2
and 00b1b2...bn ∈ Λn+2, respectively. Let A = 00a1a2...an and B = 00b1b2...bn. In the
following, we use induction to show that A⊕B ∈ Λn+2.
If n = 2, then this statement follows from Lemma 6.1.13. Assume this statement
is true for all integers less than n. We then have 00a3a4...an ⊕ 00b3b4...bn ∈ Λn. Let
c1c2...cn = 00a3a4...an ⊕ 00b3b4...bn in Λn. It follows that, in Λn+2, we have
00c1c2...cn = 0000a3a4...an ⊕ 0000b3b4...bn. (6–9)
By Item 1 of Lemma 6.1.8, in Λn+2, we have
A⊕B = 00a1a200...0⊕ 0000a3a4...an ⊕ 00b1b200...0⊕ 0000b3b4...bn
= 00a1a200...0⊕ 00b1b200...0⊕ 0000a3a4...an ⊕ 0000b3b4...bn.
(6–10)
By applying Equation 6–9, Equation 6–10 becomes
A⊕B = 00a1a2000...0⊕ 00b1b2000...0⊕ 00c1c2c3c4c5...cn
= 00a1a2000...0⊕ 00b1b2000...0⊕ 00c1c2000...0⊕ 0000c3c4c5...cn
= 00a1a2000...0⊕ 00b1b2000...0⊕ 00c1c2000...0⊕ 0000c3c40...0⊕ 000000c5...cn.
(6–11)
To finish the proof, we consider the following cases.
Case 1: c3 = 0. By Lemma 6.1.13, we have 00a1a2 ⊕ 00b1b2 ⊕ 00c1c2 in Λ4. Let
d1d2d3d4 = 00a1a2⊕ 00b1b2⊕ 00c1c2. By applying Item 4 of Lemma 6.1.8 for n− 2 times, it
follows that d1d2d3d4000...0 = 00a1a2000...0 ⊕ 00b1b2000...0 ⊕ 00c1c2000...0 in Λn+2. Hence
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Equation 6–11 becomes A⊕B = d1d2d3d4000...0⊕ 0000c3c4c5...cn in Λn+2. Since c3 = 0, by
Item 2 of Lemma 6.1.8, it follows that A⊕B = d1d2d3d4c3c4c5...cn in Λn+2.
Case 2: c3 6= 0. By Item 1 of Corollary 6.1.7, we have c2 = c4 = 0 in this case. In
Equation 6–11, we claim that we can assume a2 = b2 = 0 as well.
Suppose a2 6= 0. By Item 1 of Corollary 6.1.7, it follows that a1 = 0. Hence
00a1a200⊕ 0000c3c4 = 000a200⊕0000c30 in Λ6. By Table 6-1, in Λ3, we have 0a20⊕ 00c3 =
x0y in Λ3 for some x, y ∈ 0, 1, 2, ..., 6 with y 6= 0. By applying Item 4 of Lemma 6.1.8
for n − 1 times, it follows that 000a2000...0 ⊕ 0000c300...0 = 00x0y00...0 in Λn+2, i.e.,
00a1a2000...0⊕ 0000c3c40...0 = 00x0y00...0 in Λn+2. Hence Equation 6–11 becomes
A⊕B = 00x0000...0⊕ 00b1b2000...0⊕ 00c1c2000...0⊕ 0000y0c5...cn
= 00x0000...0⊕ 00b1b2000...0⊕ 00c1c2000...0⊕ 0000yc4c5...cn in Λn+2,
(6–12)
which is what we want. Similarly we can assume b2 = 0. Hence Equation 6–11 becomes
A⊕B = 00a10000...0⊕ 00b10000...0⊕ 00c10000...0⊕ 0000c3c4c5...cn in Λn+2. (6–13)
By Lemma 6.1.12 we have 00a1⊕00b1⊕00c1 in Λ3. Let xyz = 00a1⊕00b1⊕00c1 ∈ Λ3.
By applying Item 4 of Lemma 6.1.8 for n − 1 times, it follows that xyz0000...0 =
00a10000...0 ⊕ 00b10000...0 ⊕ 00c10000...0 in Λn+2. Hence Equation 6–13 becomes
A⊕ B = xyz0000...0⊕ 0000c3c4c5...cn = xyz0c3c4c5...cn in Λn+2 by Item 2 of Lemma 6.1.8.
Thus A⊕B ∈ Λn+2.
The following proposition tells whether or not a + b ∈ P (n) based on the label of
13a + 1
3b ∈ P (n + 2).
Proposition 6.1.15. For any a,b ∈ P (n), let s = 13a + 1
3b. If s1s2...snsn+1sn+2 ∈ Λn+2 is
the label of s, then a + b ∈ P (n) if and only if s1 = s2 = 0.
Proof. If a + b ∈ P (n), then by Theorem 6.1.6 there exists uniquely determined ci ∈0, 1, 2, ..., 6 for each i = 1, 2, ..., n such that a + b =
∑ni=1 βi,ci
and either ci = 0 or
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ci+1 = 0 for any i satisfying 1 ≤ i < n. It follows that 13a + 1
3b =
∑ni=1
13βi,ci
=∑n
i=1 βi+2,ci.
Since s = 13a+ 1
3b and s1s2...snsn+1sn+2 ∈ Λn+2 is the label of s, it follows that s1 = s2 = 0.
Conversely if s1 = s2 = 0, since s1s2...snsn+1sn+2 ∈ Λn+2 is the label of s, we have
s =∑n+2
i=1 βi,si=
∑n+2i=3 βi,si
=∑n+2
i=313βi−2,si
= 13
∑n+2i=3 βi−2,si
. Since s = 13a + 1
3b, it
follows that a + b =∑n+2
i=3 βi−2,si=
∑nj=1 βj,sj+2
. Because s1s2...snsn+1sn+2 ∈ Λn+2, either
si = 0 or si+1 = 0 for any i satisfying 1 ≤ i < n + 2. By Lemma 6.1.5, it follows that∑n+2
i=3 βi−2,si=
∑nj=1 βj,sj+2
∈ P (n). Thus a + b ∈ P (n).
Let n ∈ N and a,b ∈ P (n), and let a1a2...an ∈ Λn and b1b2...bn ∈ Λn be the labels
of a and b, respectively. By applying Item 3 of Lemma 6.1.8, it follows that the labels of
13a and 1
3b are 00a1a2...an ∈ Λn+2 and 00b1b2...bn ∈ Λn+2, respectively. By Theorem 6.1.14,
we have 13a + 1
3b ∈ P (n + 2). Hence there exists c1c2...cncn+1cn+2 ∈ Λn+2 such that
00a1a2...an ⊕ 00b1b2...bn = c1c2...cncn+1cn+2 in Λn+2. We call c1c2 and c3c4...cncn+1cn+2
the carry and the remainder of adding the two labels a1a2...an and b1b2...bn, respectively.
For example, by Table 6-2, we have 0001 ⊕ 0030 = 0205 in Λ4. It follows that the carry
and the remainder of adding the two labels 01 and 30 are 02 and 05, respectively. By
Proposition 6.1.15, c1c2 = 00 if and only if a + b ∈ P (n).
For any integer n > 0, we use the following algorithm to add any two labels
a1a2...an ∈ Λn and b1b2...bn ∈ Λn. From the discussion of the previous paragraph, we
need to compute 00a1a2...an ⊕ 00b1b2...bn to get the sum c1c2...cncn+1cn+2 ∈ Λn+2. Recall
that any label of the Pyxis structure has the property that, for any two consecutive digits,
at least one of them is zero. If the addition proceeds from the right digits an and bn to
the left digits a1 and b1, the idea of remainder, carry and sum of adding two digits is
similar to that of the usual addition except that, for the addition of ai and bi, we use
Table 6-1 when i is odd and use Table 6-2 when i is even. Furthermore the carry of
adding ai and bi consists of two digits and the sum of two labels obtained in the usual
way may no longer have the property that, for any two consecutive digits, at least one of
them is zero. For example, if we add two labels 10 and 05 in Λ2 using the usual digit by
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digit addition, the obtained sum 15 is no longer a legitimate Pyxis label because there
are two nonzero consecutive digits. Hence the addition of two elements in Λn is quite
complicated. However, we have successfully developed the following algorithm which
output the required sum and has the computational complexity of order n when two labels
in Λn are added.
Algorithm addLabels
Input: An integer n > 0, and two labels a1a2...an ∈ Λn and b1b2...bn ∈ Λn.
Output: The Sum = 00a1a2...an ⊕ 00b1b2...bn in Λn+2.
Step 1. If n = 1 or n = 2, then apply Table 6-1 or Table 6-2, respectively, to return
Sum = 00a1⊕ 00b1 in Λ3 or Sum = 00a1a2⊕ 00b1b2 in Λ4. Otherwise do the following Step
2 through Step 7.
Step 2. If n is odd, then let N = n + 1 and adjoin a 0 to the right end of the two
labels a1a2...an and b1b2...bn, i.e., let aN = bN = 0. Otherwise let N = n.
Step 3. Let M = N2. Also, for i = 1 to M , let aPairi = auav and bPairi = bubv
where u = 2(M − i) + 1 and v = 2(M − i) + 2.
Step 4. Let Sum be the empty set, and add aPair1 and bPair1 to get a
temporary carry denoted tempCarry and a temporary remainder denoted tempRem.
Step 5. For i = 2 to M , do the following.
• Let carrySet consist of elements of the set aPairi, bPairi, tempCarry that
are different from 00, and let carrySet1 consist of elements of carrySet whose last
digit is nonzero and let carrySet2 consist of elements of carrySet whose first digit is
nonzero. Also let kj denote the size of the set carrySetj for j = 1, 2.
1. If the first digit of tempRem is 0, let remd (means remainder) be tempRem.
(a) If k1 = 0, apply Subroutine 1 to add the elements in the set carrySet2
to obtain a carry c1c2 and a remainder r10. Let tempCarry = c1c2 and
tempRem = r10.
(b) If k1 = 1, let 0c2 denote the element in the set carrySet1.
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i. If k2 = 0, let tempCarry = 00 and tempRem = 0c2.
ii. Else if k2 = 1, let d10 denote the element in the set carrySet2. Apply
Table 6-2 to compute 000c2 ⊕ 00d10 = 0e10e2. Let tempCarry = 0e1
and tempRem = 0e2.
iii. Else, let d10 and e10 denote the two elements in the set carrySet2. By
Table 6-2, 000c2 ⊕ 00d10 = 0f10f2 for some 0f10f2 ∈ Λ4 with f2 6= 0.
Similarly, we have 000f2 ⊕ 00e10 = 0g10g2 for some 0g10g2 ∈ Λ4. By
Theorem 6.1.14, we have 0f1 ⊕ 0g1 = h1h2 ∈ Λ2. Let tempCarry = h1h2
and tempRem = 0g2.
(c) If k1 = 2, let 0c1 and 0d1 denote the two elements in the set carrySet1.
i. If k2 = 0, by Table 6-2, 000c1 ⊕ 000d1 = 0f1f2f3 for some 0f1f2f3 ∈ Λ4.
Let tempCarry = 0f1 and tempRem = f2f3.
ii. Else, let e10 denote the element in the set carrySet2.
A. If c1 = d1, by Table 6-2, we have 000c1 ⊕ 000d1 = 0f10f3 for some
0f10f3 ∈ Λ4. By Table 6-2 again, we have 000f3 ⊕ 00e10 = 0g10g3 for
some 0g10g3 ∈ Λ4. By Theorem 6.1.14, we have 0f1 ⊕ 0g1 = h1h2 for
some h1h2 ∈ Λ2. Let tempCarry = h1h2 and tempRem = 0g3.
B. Else, since c1 6= d1, by Lemma 6.1.11 we have 0c1 ⊕ 0d1 = f1f2 ∈ Λ2.
By Table 6-2, we have 00f1f2 ⊕ 00e10 = g1g2g3g4 for some g1g2g3g4 ∈Λ4. Let tempCarry = g1g2 and tempRem = g3g4.
(d) If k1 = 3, add the three elements in the set carrySet1 using the following
Subroutine 2 to obtain 0h2h3h4 ∈ Λ4. Let tempCarry = 0h2 and
tempRem = h3h4.
2. Else (now the first digit of remd is nonzero),
(a) if k1 = 0, let remd = tempRem and add the elements of the set
carrySet2 using Subroutine 1 to get a carry c1c2 and a remainder r1r2.
Let tempCarry = c1c2 and tempRem = r1r2.
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(b) Else let e10 denote tempRem and let 0f2 denote the first element of the set
carrySet1. Compute 00e1 ⊕ 0f20 using table 6-1 to get g10g3 ∈ Λ3. Let
carrySet2 = carrySet2⋃ g10.
i. If k1 = 1, let remd = g30 and add the elements of the set carrySet2
using Subroutine 1 to get a carry c1c2 and a remainder r10. Let
tempCarry = c1c2 and tempRem = r10.
ii. Else, let 0y2 denote the second element of the set carrySet1 and
compute 00g3⊕0y20 using Table 6-1 to get h10h3 ∈ Λ3. Let carrySet2 =
carrySet2⋃ h10.
A. If k1 = 2, let remd = h30 and add the elements of the set carrySet2
using Subroutine 1 to get a carry c1c2 and a remainder r10. Let
tempCarry = c1c2 and tempRem = r10.
B. Else, let 0z2 denote the third element of the set carrySet1 and
compute 00h3 ⊕ 0z20 using Table 6-1 to get l10l3 ∈ Λ3. Let
carrySet2 = carrySet2⋃ l10, remd = l30, and add the elements
of the set carrySet2 using Subroutine 1 to get a carry c1c2 and a
remainder r10. Let tempCarry = c1c2 and tempRem = r10.
• Let Sum be the concatenation of remd and Sum, and increase i.
Step 6. Let Sum be the concatenation of the tempCarry, tempRem, and Sum. If n is
odd, delete the 0 in the right side of Sum. Return Sum.
Subroutine 1: Input: A set Ω (containing at most three elements) of labels in Λ2
whose second digit is 0.
Output: The carry C = x1x2 ∈ Λ2 and remainder R = y10 ∈ Λ2 of adding labels in the
set Ω.
Let k be the size of the set Ω. Since Ω contains at most three elements, we consider
the following four cases.
• If k = 0, return C = 00 and R = 00.
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• If k = 1, let C = 00 and let R be the element in the set Ω. Return C and R.
• If k = 2, add the two elements in the set Ω using Table 6-2 to get a carry C =
x1x2 ∈ Λ2 and a remainder R = y10 ∈ Λ2, and return them.
• If k = 3, let a10, b10, and c10 denote the three elements of the set Ω.
1. If a1 = b1 = c1, then return C = a10 and R = 00.
2. Else if a1 = b1, then compute 00b1 ⊕ 00c1 using Table 6-1 to get 0r1r2 and
compute 0r1r2 ⊕ 00a1 using Table 6-1 to get d1d2d3 ∈ Λ3, and let C = d1d2 and
R = d30, and return C and R.
3. Else compute 00a1 ⊕ 00b1 using Table 6-1 to get 0r1r2 and compute 0r1r2 ⊕ 00c1
using Table 6-1 to get d1d2d3 ∈ Λ3, and let C = d1d2 and R = d30, and return C
and R.
Subroutine 2: Input: A set Ω (containing at most three elements) of labels in Λ2
whose first digit is 0.
Output: The carry C = 0x2 ∈ Λ2 and remainder R = y1y2 ∈ Λ2 of adding labels in the
set Ω.
Let k denote the size of the set Ω. Since Ω contains at most three elements, we
consider the following four cases.
• If k = 0, return C = 00 and R = 00.
• If k = 1, let C = 00 and let R denote the element in the set Ω. Return C and R.
• If k = 2, add the two elements in the set Ω using Table 6-2 to get a carry C = 0x2 ∈Λ2 and a remainder R = y1y2 ∈ Λ2, and return them.
• If k = 3, let 0a2, 0b2, and 0c2 denote the three elements of the set Ω.
1. If a2 = b2 = c2, then return C = 0a2 and R = 00.
2. Else if a2 = b2, then compute 0b2 ⊕ 0c2 using Table 6-2 to get a carry 00
and a remainder r1r2, and compute r1r2 ⊕ 0a2 using Table 6-2 to get a carry
C = 0x2 ∈ Λ2 and a remainder R = y1y2 ∈ Λ2, and return C and R.
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3. Else compute 0a2 ⊕ 0b2 to get a carry 00 and a remainder r1r2, and compute
r1r2 ⊕ 0c2 using Table 6-2 to get a carry C = 0x2 ∈ Λ2 and a remainder
R = y1y2 ∈ Λ2, and return C and R.
At each iteration in Step 5 of Algorithm addLabels, the set carrySet contains at most
three elements. It follows that the computational complexity of adding any two labels
in Λn using Algorithm addLabels is of order n. In the following, we give two examples
showing the application of Algorithm addLabels. In Example 1, for each iteration in Step
5 of the algorithm, the first digit of tempRem is 0. In Example 2, for each iteration in
Step 5 of the algorithm, the first digit of tempRem is nonzero. Hence the addition in
Example 2 is actually more complicated than that in Example 1.
Example 1: Let n = 6, A = 500104, and B = 403020.
• In Step 1, since n > 2 and n is even, go to Step 2.
• In Step 2, let N = n = 6.
• In Step 3, let M = N2
= 3. Also let aPair1 = 04, bPair1 = 20, aPair2 = 01,
bPair2 = 30, aPair3 = 50, and bPair3 = 40.
• In Step 4, let Sum be the empty set. By Table 6-2, we have aPair1 ⊕ bPair1 =
04 ⊕ 20 = 02 in Λ2. It follows that the carry and the remainder of this addition are
00 and 02, respectively. Hence tempCarry = 00 and tempRem = 02.
• In Step 5, for i = 2 to M do the following.
1. When i = 2, let carrySet consist of elements of the set
aPair2, bPair2, tempCarry
that are different from 00, i.e., let carrySet = 01, 30. Then let carrySet1 =
01 and carrySet2 = 30. It follows that k1 = k2 = 1. Since the first digit
of tempRem is 0, let remd = tempRem = 02. Since k1 = 1, let 0c2 denote
the element in the set carrySet1, i.e., let 0c2 = 01. Since k2 = 1, let d10
denote the element in the set carrySet2, i.e., let d10 = 30. Apply Table 6-2
84
to compute 000c2 ⊕ 00d10 = 0001 ⊕ 0030 = 0205, which is denoted 0e10e2.
Let tempCarry = 0e1 = 02 and tempRem = 0e2 = 05. Let Sum be the
concatenation of remd = 02 and Sum = ∅, i.e., let Sum = 02, and increase i.
2. When i = 3, let carrySet consist of elements of the set
aPair3, bPair3, tempCarry
that are different from 00, i.e., let carrySet = 50, 40, 02. Then let
carrySet1 = 02, and carrySet2 = 50, 40. It follows that k1 = 1 and k2 = 2.
Since the first digit of tempRem is 0, let remd = tempRem = 05. Since k1 = 1,
let 0c2 denote the element in the set carrySet1, i.e., let 0c2 = 02. Since k2 = 2,
let d10 and e10 denote the two elements in the set carrySet2, i.e., let d10 = 50
and e10 = 40. By Table 6-2, we have 000c2⊕ 00d10 = 0002⊕ 0050 = 0004, which
is denoted 0f10f2. Similarly, we have 000f2⊕00e10 = 0004⊕0040 = 0402, which
is denoted 0g10g2. By Table 6-2, we have 0f1 ⊕ 0g1 = 00 ⊕ 04 = 04, which is
denoted h1h2. Let tempCarry = h1h2 = 04 and tempRem = 0g2 = 02. Let Sum
be the concatenation of remd = 05 and Sum = 02, i.e., let Sum = 0502, and
increase i.
• In Step 6, let Sum be the concatenation of the tempCarry = 04, tempRem = 02,
and Sum = 0502, i.e., let Sum = 04020502.
Notice that, in Example 1, the first digit of tempRem is 0 for each iteration in Step 5
of the algorithm. In the next, we show another example where the first digit of tempRem
is nonzero for each iteration in Step 5 of the algorithm.
Example 2: Let n = 6, A = 050101, and B = 401006.
• In Step 1, since n > 2 and n is even, go to Step 2.
• In Step 2, let N = n = 6.
• In Step 3, let M = N2
= 3. Also let aPair1 = 01, bPair1 = 06, aPair2 = 01,
bPair2 = 10, aPair3 = 05, and bPair3 = 40.
85
• In Step 4, let Sum be the empty set. By Table 6-2, we have aPair1 ⊕ bPair1 =
01 ⊕ 06 = 10 in Λ2. It follows that the carry and the remainder of this addition are
00 and 10, respectively. Hence tempCarry = 00 and tempRem = 10.
• In Step 5, for i = 2 to M do the following.
1. When i = 2, let carrySet consist of elements of the set
aPair2, bPair2, tempCarry
that are different from 00, i.e., let carrySet = 01, 10. Then let carrySet1 =
01, and carrySet2 = 10. It follows that k1 = k2 = 1. Since the first
digit of tempRem is nonzero and k1 6= 0, let e10 denote tempRem and let 0f2
denote the element of the set carrySet1. It follows that e1 = 1 and f2 = 1. By
Table 6-1, we have 00e1 ⊕ 0f20 = 001 ⊕ 010 = 103, which is denoted g10g3,
i.e., g1 = 1 and g3 = 3. Let carrySet2 = carrySet2⋃ 10 = 10⋃ 10 =
10, 10. Since k1 = 1, let remd = g30 = 30 and add the elements in the set
carrySet2 = 10, 10 using Table 6-2 to get a carry c1c2 = 10 and a remainder
r10 = 40. Let tempCarry = c1c2 = 10 and tempRem = r10 = 40. Let Sum be
the concatenation of remd = 30 and Sum = ∅, i.e., let Sum = 30, and increase
i.
2. When i = 3, let carrySet consist of elements of the set
aPair3, bPair3, tempCarry
that are different from 00, i.e., let carrySet = 05, 40, 10. Then let
carrySet1 = 05, and carrySet2 = 40, 10. It follows that k1 = 1 and
k2 = 2. Since the first digit of tempRem is nonzero and k1 6= 0, let e10 denote
tempRem and let 0f2 denote the element of the set carrySet1. It follows that
e1 = 4 and f2 = 5. By Table 6-1, we have 00e1 ⊕ 0f20 = 004 ⊕ 050 = 502,
which implies that g1 = 5 and g3 = 2. Let carrySet2 = carrySet2⋃ 50 =
86
40, 10⋃ 50 = 40, 10, 50. Since k1 = 1, let remd = g30 = 20 and
add the elements in the set carrySet2 = 40, 10, 50 using Table 6-2 to get
a carry c1c2 = 00 and a remainder r10 = 50. Let tempCarry = c1c2 = 00
and tempRem = r10 = 50. Let Sum be the concatenation of remd = 20 and
Sum = 30, i.e., let Sum = 2030, and increase i.
• In Step 6, let Sum be the concatenation of the tempCarry = 00, tempRem = 50,
and Sum = 2030, i.e., let Sum = 00502030.
Theorem 6.1.16. For any integer n > 0 and two labels a1a2...an, b1b2...bn ∈ Λn, the Sum
output from Algorithm addLabels equals 00a1a2...an ⊕ 00b1b2...bn ∈ Λn+2.
Proof. By Item 4 of Lemma 6.1.8, we just need to consider the case when n is even. Also
when n ≤ 2, this theorem is obviously true. Hence we assume that n is an even integer
which is bigger than 2. Let n = 2M for some integer M > 1, A = 00a1a2...a2M ∈ Λn+2,
and B = 00b1b2...b2M ∈ Λn+2. For i = 1 to M , let Ai = 00a2(M−i)+1a2(M−i)+2...a2M−1a2M ∈Λ2i+2, and Bi = 00b2(M−i)+1b2(M−i)+2...b2M−1b2M ∈ Λ2i+2. By Theorem 6.1.14, we have
Ai ⊕ Bi ∈ Λ2i+2. Let Sum1, tempCarry1 and tempRem1 be the Sum, tempCarry and
tempRem output from Step 4 of Algorithm addLabels. For i = 1 to M , let Ti = Ai ⊕ Bi,
and let remdi, Sumi, tempCarryi, and tempRemi be the remd, Sum, tempCarry, and
tempRem output from the ith iteration of Step 5 in Algorithm addLabels.
In the following, we use induction to prove that Ti is the concatenation of tempCarryi,
tempRemi, and Sumi for any i satisfying 1 ≤ i ≤ M . When i = 1, A1 = 00a2M−1a2M
and B1 = 00b2M−1b2M . In Algorithm addLabels, obviously aPair1 = a2M−1a2M and
bPair1 = b2M−1b2M . From Step 4 of Algorithm addLabels, Sum1 = ∅ and hence
A1 ⊕ B1 is the concatenation of tempCarry1, tempRem1, and Sum1. It follows that
T1 is the concatenation of tempCarry1, tempRem1, and Sum1. Now assume that Ti−1
is the concatenation of tempCarryi−1, tempRemi−1, and Sumi−1 for some i satisfying
87
2 ≤ i ≤ M . Since
Ai = 00a2(M−i)+1a2(M−i)+2a2(M−i+1)+1a2(M−i+1)+2...a2M−1a2M
= 00a2(M−i)+1a2(M−i)+200...00⊕ 00Ai−1
(6–14)
and
Bi = 00b2(M−i)+1b2(M−i)+2b2(M−i+1)+1b2(M−i+1)+2...b2M−1b2M
= 00b2(M−i)+1b2(M−i)+200...00⊕ 00Bi−1,
(6–15)
we have
Ti = Ai ⊕Bi = 00a2(M−i)+1a2(M−i)+200...00⊕ 00b2(M−i)+1b2(M−i)+200...00⊕ 00Ai−1 ⊕ 00Bi−1
= 00a2(M−i)+1a2(M−i)+200...00⊕ 00b2(M−i)+1b2(M−i)+200...00⊕ 00Ti−1.
(6–16)
By our induction assumption, Ti−1 is the concatenation of tempCarryi−1, tempRemi−1,
and Sumi−1. Hence Equation 6–16 becomes
Ti = 00a2(M−i)+1a2(M−i)+200...00⊕ 00b2(M−i)+1b2(M−i)+200...00
⊕ 00tempCarryi−1tempRemi−1Sumi−1.
(6–17)
In Step 5 of Algorithm addLabels, let
carrySet =a2(M−i)+1a2(M−i)+2, b2(M−i)+1b2(M−i)+2, tempCarryi−1
=aPairi, bPairi, tempCarryi−1
.
(6–18)
Now we consider the following two cases.
Case 1: If the first digit of tempRemi−1 is 0, then, in the ith iteration of Step 5
of Algorithm addLabels, we let remdi = tempRemi−1. The ith iteration of Step 5 in
Algorithm addLabels computes the carry and remainder of the addition of the three
elements in the set carrySet. By Lemma 6.1.13, we have
00a2(M−i)+1a2(M−i)+2 ⊕ 00b2(M−i)+1b2(M−i)+2 ⊕ 00tempCarryi−1 = x1x2x3x4 (6–19)
88
for some x1x2x3x4 ∈ Λ4. In Algorithm addLabels, obviously aPairi = a2(M−i)+1a2(M−i)+2
and bPairi = b2(M−i)+1b2(M−i)+2. Then Equation 6–19 becomes
00aPairi ⊕ 00bPairi ⊕ 00tempCarryi−1 = x1x2x3x4. (6–20)
It follows that the carry and remainder of the addition of the three elements in the set
carrySet are x1x2 and x3x4, respectively. Hence tempCarryi = x1x2 and tempRemi =
x3x4. By Equations 6–17 and 6–19, we have
Ti = 00aPairi00...00⊕ 00bPairi00...00⊕ 00tempCarryi−1tempRemi−1Sumi−1
= 00aPairi00...00⊕ 00bPairi00...00
⊕ 00tempCarryi−100...00⊕ 0000tempRemi−1Sumi−1
= x1x2x3x400...00⊕ 0000remdiSumi−1
= x1x2x3x400...00⊕ 0000Sumi
= tempCarryitempRemi00...00⊕ 0000Sumi
= tempCarryitempRemiSumi.
(6–21)
It follows that Ti is the concatenation of tempCarryi, tempRemi, and Sumi.
Case 2: The first digit of tempRemi−1 is nonzero, i.e., the second case in Step 5 of
Algorithm addLabels. Recall that, in Algorithm addLabels, carrySet1 consist of elements
of carrySet whose last digit is nonzero and carrySet2 consist of elements of carrySet
whose first digit is nonzero. Also kj denotes the size of the set carrySetj for j = 1, 2.
If k1 = 0, similar to Case 1, we can show that Ti is the concatenation of tempCarryi,
tempRemi, and Sumi.
If k1 6= 0, we compute the sum
S = 00a2(M−i)+1a2(M−i)+200⊕ 00b2(M−i)+1b2(M−i)+200⊕ 00tempCarryi−1tempRemi−1
89
in certain procedures. By Equation 6–17, S consists of the first six digits of Ti. By
the assumption of this induction, Ti−1 ∈ Λ2i is the concatenation of tempCarryi−1,
tempRemi−1, and Sumi−1. Hence the concatenation of tempRemi−1 and Sumi−1 is a
legitimate Pyxis label. Since Ti = Ai ⊕ Bi ∈ Λ2i+2 and Sumi−1 ∈ Λ2i−4, by Item 2
of Lemma 6.1.8, it follows that S ∈ Λ6. In the ith iteration of Step 5 of Algorithm
addLabels, we apply Table 6-1 to convert the expression 00a2(M−i)+1a2(M−i)+200 ⊕00b2(M−i)+1b2(M−i)+200⊕ 00tempCarryi−1tempRemi−1 to the expression 00x000⊕ 00y000⊕00z000 ⊕ 0000r0 for some x, y, z, r ∈ 0, 1, 2, 3, 4, 5, 6. Let carrySet2 consist of nonzero
elements of the set x0, y0, z0 and add the elements in carrySet2. By Subroutine 2, the
sum of the elements in carrySet2 equals c1c2c30 for some c1c2c30 ∈ Λ4. It follows that
S = c1c2c30r0. Hence tempCarryi = c1c2, tempRemi = c30, remdi = r0, and Sumi is the
concatenation of remdi = r0 and Sumi−1. Now Equation 6–17 becomes
Ti = 00a2(M−i)+1a2(M−i)+200...00⊕ 00b2(M−i)+1b2(M−i)+200...00
⊕ 00tempCarryi−1tempRemi−1Sumi−1
= 00a2(M−i)+1a2(M−i)+200...00⊕ 00b2(M−i)+1b2(M−i)+200...00
⊕ 00tempCarryi−100...00⊕ 0000tempRemi−10...0⊕ 000000Sumi−1
= c1c2c30r00...0⊕ 000000Sumi−1
= c1c2c30000...0⊕ 0000r00...0⊕ 000000Sumi−1
= tempCarryitempRemi000...0⊕ 0000remdi0...0⊕ 000000Sumi−1
= tempCarryitempRemi000...0⊕ 0000remdiSumi−1
= tempCarryitempRemi000...0⊕ 0000Sumi
= tempCarryitempRemiSumi.
(6–22)
Hence Ti is the concatenation of tempCarryi, tempRemi, and Sumi.
By induction, Ti is the concatenation of tempCarryi, tempRemi, and Sumi for
any i satisfying 1 ≤ i ≤ M . It follows that TM is the concatenation of tempCarryM ,
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tempRemM , and SumM . The output Sum in Algorithm addLabels is also the concatenation
of tempCarryM , tempRemM , and SumM . Hence Sum = TM . Since TM = AM ⊕BM = 00a1a2...a2M−1a2M ⊕ 00b1b2...b2M−1b2M , we have Sum = 00a1a2...a2M−1a2M ⊕00b1b2...b2M−1b2M .
6.2 Pyxis P (n) Does Not Tile the Underlying Lattice by Translations by aSublattice for any n > 2
Recall from Chapter 2 that, if an array tiles its underlying lattice by translations by
a sublattice, then we can compute the DFT on it. For the nth level of the Pyxis structure
P (n), to consider its DFT, we need to know whether or not it tiles the underlying lattice
Ln by translations by a sublattice. In this section, we show some properties of the Pyxis
structure and apply them to prove that P (n) does not tile the underlying lattice by
translations by a sublattice for any n > 2. Hence the DFT on the Pyxis structure cannot
be defined using the method in Chapter 2. For any 2 < n ∈ N, the proof that P (n) does
not tile Ln when n is odd is quite different from that when n is even. Hence we deal with
them in two separate subsections.
6.2.1 Pyxis P (2n−1) Does Not Tile the Underlying Lattice by Translations bya Sublattice for any n > 1
In this subsection, we first define the concept of boundary index for P (n) and use
several properties of the boundary index of P (2n− 1) to prove that P (2n− 1) does not tile
the underlying lattice L2n−1 by translations by a sublattice for any n ≥ 2. Hence the DFT
on the Pyxis structure cannot be defined using the method in Chapter 2.
For any ∅ 6= T ⊆ L and q ∈ T , if there exists at least one neighbor b of q such
that b 6∈ T , then q is called a boundary lattice point of T , the Voronoi cell of q is called a
boundary hexagon of T . If q is a boundary lattice point of T and if there exist exactly d
neighbors that belong to L \ T , then d is called the boundary index of q in T . The number
d is also called the boundary index of the Voronoi cell of q in T . The following lemmas will
be used to prove the main result of this subsection, which is Theorem 6.2.11.
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Lemma 6.2.1. Let L be a hexagonal lattice, u ∈ L, ∅ 6= T ⊆ L, and d a positive integer.
Then q is a boundary lattice point of T having boundary index d if and only if u + q is a
boundary lattice point of u + T having the same boundary index d.
Lemma 6.2.2. For any a ∈ P (2) and b ∈ β3, we have a + b ∈ P (3). Moreover there exist
c ∈ β1 and d ∈ β3 such that a + b = c + d.
Proof. Since a ∈ P (2) ⊂ P (3) and b ∈ β3 ⊂ P (3), the label of a in P (3) is either of the
form 0i0 or i00, and the label of b in P (3) is of the form 00j where i, j ∈ 0, 1, 2, ..., 6.By Table 6-1, 0i0 ⊕ 00j = d10d3 for some d10d3 ∈ Λ3. It follows that a + b ∈ P (3). Let
c ∈ β1 and d ∈ β3 such that the labels of c and d in P (3) are d100 and 00d3, respectively.
Since 0i0⊕ 00j = d10d3 = d100⊕ 00d3, we have a + b = c + d.
Lemma 6.2.3. For any n ∈ N, a ∈ P (2n), and b ∈ β2n+1, we have a + b ∈ P (2n + 1).
Proof. When n = 1, it is followed directly from Lemma 6.2.2. Suppose it is true for all
n < t for some integer t. When n = t, P (2n) = P (2t) = P (2t − 1)⋃
(P (2t − 2) + β2t).
Hence a ∈ P (2n) implies that either a ∈ P (2t− 1) or a ∈ P (2t− 2) + β2t. If a ∈ P (2t− 1),
then a + b ∈ P (2t − 1) + β2t+1 = P (2t + 1). If a ∈ P (2t − 2) + β2t, then a = u + v
for some u ∈ P (2t − 2) and v ∈ β2t. Since v ∈ β2t and b ∈ β2t+1, it follows easily
from Lemma 6.2.2 that v + b = c + d for some c ∈ β2t−1 and d ∈ β2t+1. Hence
a + b = u + v + b = u + c + d. Since u ∈ P (2t − 2) = P (2(t − 1)) and either c ∈ β2t−1
or c = (0, 0), we have u + c ∈ P (2t − 1) by the assumption of this induction. Thus
a + b = (u + c) + d ∈ P (2t − 1) + β2t+1 ⊂ P (2t + 1). Hence this proposition is true for
n = t. Therefore it is true for all n ∈ N by induction.
Corollary 6.2.4. For any n ∈ N and q ∈ P (2n), the lattice point q is not a boundary
lattice point of P (2n + 1).
Proof. By Lemma 6.2.3, all six neighbors of q in L2n+1 are still in P (2n + 1). Hence q is
not on the boundary of P (2n + 1).
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The following lemma will be used in the proofs of Lemma 6.2.6 and some other
results.
Lemma 6.2.5. For any n ∈ N and a ∈ P (n), if the label of a is a1a2...an with an 6= 0,
then a 6∈ Ln−1.
Proof. Let a ∈ P (n) and an ∈ βn be lattice points of Ln such that their labels are
a1a2...an−200 and 00...0an, respectively. Since the label of a is a1a2...an−200, we have
a ∈ P (n − 2) ⊂ Ln−2 ⊂ Ln−1. By Item 1 of Lemma 6.1.8, we have a1a2...an =
a1a2...an−200⊕ 00...0an. It follows that a = a + an. Suppose a ∈ Ln−1. Then an = a− a ∈Ln−1 ∩ βn = ∅, which is a contradiction. Thus a 6∈ Ln−1.
Lemma 6.2.6. For any n ∈ N, if three lattice points of P (n) are mutually next to each
other in the lattice Ln, then at least one of them is in P (n− 1).
Proof. Let a,b, c be three lattice points of P (n) that are mutually next to each other in
the lattice Ln, and let A = a1a2...an ∈ Λn, B = b1b2...bn ∈ Λn and C = c1c2...cn ∈ Λn
be their labels, respectively. It is enough to show that either an = 0 or bn = 0 or cn = 0.
Suppose that an 6= 0, bn 6= 0 and cn 6= 0. Let u = b − a and v = c − a. Since a,
b and c are mutually next to each other in the lattice Ln, by Lemma 6.1.10, we have
b−a, c−a,b−c ∈ βn. It follows that u,v,u−v ∈ βn, and u and v are next to each other
in the lattice Ln. Thus βn = ±u,±v,±(u− v). Let a ∈ P (n) and an ∈ βn be lattice
points of Ln such that their labels are a1a2...an−200 and 00...0an, respectively. By Item 1
of Lemma 6.1.8, we have a1a2...an = a1a2...an−200 ⊕ 00...0an. It follows that a = a + an.
Now it follows from an ∈ βn that either an = ±u or an = ±v or an = ±(u− v).
If an = u, then an and v are next to each other in the lattice Ln. By Lemma 6.1.9,
it follows that an + v ∈ βn−1 ⊂ Ln−1. Hence an + c − a ∈ Ln−1. Since the label of a is
a1a2...an−200, we have a ∈ Pn−2 ⊂ Ln−1. It follows that an − a = −a ∈ Ln−1. Hence
c ∈ Ln−1 which contradicts Lemma 6.2.5.
If an = v, similar to the previous case, there is a contradiction.
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If an = −u, then an +u = 0, i.e. an +b−a = 0. It follows that b = a−an = a ∈ Ln−1
which also contradicts Lemma 6.2.5.
If an = u − v, then an is next to u in the lattice Ln. By Lemma 6.1.9, it follows that
an + u ∈ βn−1 ⊂ Ln−1. Hence an + b− a ∈ Ln−1. Since an− a = −a ∈ Ln−1, it follows that
b ∈ Ln−1 which also contradicts Lemma 6.2.5. For the remaining two cases, contradictions
can be similarly shown. Therefore either an = 0 or bn = 0 or cn = 0.
Lemma 6.2.7. Let n ∈ N, and a,b, c be three lattice points of P (n) that are mutually
next to each other in the lattice Ln, and let t be the centroid of the triangle with vertices
a,b, c. Then t ∈ P (n + 1).
Proof. Because a,b, c are mutually next to each other in the lattice Ln, there exist
u,v ∈ βn that are next to each other in the lattice Ln such that b = a + u and c = a + v.
By Lemma 6.2.6, we can assume that a ∈ P (n − 1). Since t is the centroid of the triangle
with vertices a,b, c, we have t = 13(a + b + c) = a + 1
3u + 1
3v. Let u = 1
3u and v = 1
3v.
Since u,v ∈ βn and they are next to each other in the lattice Ln, we have u, v ∈ βn+2 and
are next to each other in the lattice Ln+2. By Lemma 6.1.9, it follows that u + v ∈ βn+1.
Thus t = a + (u + v) ∈ P (n− 1) + βn+1 ⊂ P (n + 1).
Lemma 6.2.8. Let n ∈ N and let a,b, c be three lattice points of L2n−1 that are mutually
next to each other in the lattice L2n−1. If t = 13(a + b + c) and a 6∈ P (2n − 1), then
t 6∈ P (2n + 1).
Proof. Since a,b, c ∈ L2n−1, we have 13a, 1
3b, 1
3c ∈ L2n+1. It follows that t ∈ L2n+1.
Because a,b, c ∈ L2n−1 and they are mutually next to each other in the lattice L2n−1, by
Lemma 6.1.10, there exist u,v ∈ β2n−1 such that b = a + u, c = a + v, and u and v
are next to each other in the lattice L2n−1. Since t = 13(a + b + c), similar to the proof of
Lemma 6.2.7, we have t = a + w with w = 13(u + v) ∈ β2n. Suppose that t ∈ P (2n + 1).
Let the label of t be T = t1t2...t2n+1 ∈ Λ2n+1. Then we consider the following two cases
about t2n+1.
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Case 1: If t2n+1 6= 0, then t2n = 0. Let t and t2n+1 be two elements of P (2n + 1)
such that their labels are t1t2...t2n−100 ∈ λ2n+1 and 00...0t2n+1 ∈ Λ2n+1, respectively.
Then t = t + t2n+1. Also from the previous paragraph, we have t = a + w. Hence
t2n+1 = a + w − t. Since a ∈ L2n−1 ⊂ L2n, w ∈ β2n ⊂ L2n, and t ∈ L2n−1 ⊂ L2n, it follows
that t2n+1 ∈ L2n which contradicts Lemma 6.2.5.
Case 2: If t2n+1 = 0, then t ∈ P (2n). We claim that t2n 6= 0. Suppose t2n = 0. It
follows that t ∈ P (2n − 1) ⊂ L2n−1. Since a ∈ L2n−1 and w = t − a, we have w ∈ L2n−1.
Hence w ∈ L2n−1
⋂β2n = ∅ by Lemma 6.1.3, which is a contradiction. Thus t2n 6= 0.
It follows that t2n−1 = 0. Let t1,2n−2 ∈ P (2n − 2) be the lattice point whose label is
t1t2...t2n−2 ∈ Λ2n−2 and t2n ∈ P (2n) be the lattice point whose label is 00...0t2n ∈ Λ2n. It
follows that t = t1,2n−2 + t2n. Since t = a + w, we have a = t −w = t1,2n−2 + t2n −w. If
t2n = w, then a = t−w = t1,2n−2 ∈ P (2n−2) ⊂ P (2n−1) which contradicts the condition
of this lemma that a 6∈ P (2n− 1). Hence t2n 6= w. Since both t2n ∈ β2n and −w ∈ β2n, it
follows from Table 6-2 that t2n −w ∈ β2n
⋃β2n−1
⋃(β2n−2 + β2n).
If t2n−w ∈ β2n, then t2n−w = a−t1,2n−2 ∈ β2n
⋂L2n−1 = ∅, which is a contradiction.
If t2n − w ∈ β2n−1, then a = t1,2n−2 + t2n − w ∈ P (2n − 2) + β2n−1. By Lemma 6.2.3,
P (2n − 2) + β2n−1 ⊆ P (2n − 1). Hence a ∈ P (2n − 1) which again contradicts the
condition a 6∈ P (2n − 1). If t2n −w ∈ β2n−2 + β2n, then there are x ∈ β2n−2 and y ∈ β2n
such that t2n − w = x + y. Hence a = t1,2n−2 + t2n − w = t1,2n−2 + x + y. It follows
that y = a − t1,2n−2 − x ∈ β2n
⋂L2n−1 = ∅, which is a contradiction as well. Therefore
t 6∈ P (2n + 1).
Let ∅ 6= T ⊆ L, and let q ∈ T be a boundary lattice point of T , and Q the Voronoi
cell of q. If S is a side of Q but S is not a side of any other Voronoi cell of T , then S is
called a B-side of Q in T . The boundary lattice point q is called a C-boundary lattice
point of T if the union of all B-sides of the Voronoi cell Q is a connected set of R2.
Lemma 6.2.9. For any integer n ≥ 2 and any boundary lattice point q of P (2n− 1), q is
a C-boundary lattice point of P (2n− 1), and the boundary index of q in P (2n− 1) is either
95
1 or 3. Furthermore there exists at least one boundary lattice point of P (2n − 1) whose
boundary index is 1.
Proof. When n = 2, this lemma follows directly from the plot of the lattice points of
P (3) in Figure 6-2. Now assume this lemma is true for any n ≤ t for some integer t.
When n = t + 1, let q be any boundary lattice point of P (2t + 1). Since P (2t + 1) =
P (2t)⋃
(P (2t− 1)+β2t+1), we have either q ∈ P (2t) or q ∈ P (2t− 1)+β2t+1. If q ∈ P (2t),
then by Corollary 6.2.4, q is not a boundary lattice point of P (2t + 1) which leads to a
contradiction. Hence q ∈ P (2t − 1) + β2t+1. Let q = q0 + r for some q0 ∈ P (2t − 1) and
r ∈ β2t+1.
Suppose q0 is not a boundary lattice point of P (2t− 1). Then the six neighbors of q0
in the lattice L2t−1 are also lattice points of P (2t− 1). Let qi ∈ P (2t− 1) : i = 1, 2, ..., 6be the set consisting of the six neighbors of q0 in the lattice L2t−1 such that qi is next
to qi+1 for each i = 1, 2, 3, 4, 5. Also the Voronoi cell of qi is denoted Qi for each i =
1, 2, ..., 6. Since q0, q1, and q2 are lattice points of P (2t − 1) and mutually next to each
other, by Lemma 6.2.7, the centroid of the triangle whose vertices are q0, q1, and q2 is
in P (2t + 1). It follows that the hexagon A in Figure 6-3 is a Voronoi cell of P (2t + 1).
Similarly the hexagons B, C, D, E, and F are Voronoi cells of P (2t + 1). Each green
hexagon except A, B, C, D, E, and F is the Voronoi cell of a lattice point in the set⋃ qi + β2t+1 : i = 1, 2, ..., 6. Since qi + β2t+1 ⊂ P (2t− 1) + β2t+1 ⊂ P (2t + 1), each green
hexagon in Figure 6-3 is a Voronoi cell of P (2t + 1) which contradicts that q is a boundary
lattice point of P (2t + 1). Thus q0 is a boundary lattice point of P (2t− 1).
By the assumption of this induction, q0 is a C-boundary lattice point of P (2t − 1)
and the boundary index of q0 (in P (2t − 1)) is either 1 or 3. In the following, we assume
that the boundary index of q0 is 1 because the proof can be similarly done if the boundary
index of q0 is 3. It follows that the Voronoi cell Q0 of q0 is surrounded by five hexagons of
P (2t − 1) as shown in Figure 6-4 (a). Since q0 and its five neighbors in the lattice L2t−1
are lattice points of P (2t− 1) and since P (2t− 1) + β2t+1 ⊂ P (2t + 1), by Lemma 6.2.7, all
96
−25 −20 −15 −10 −5 0 5 10 15 20 25
−25
−20
−15
−10
−5
0
5
10
15
20
25
Q0
Q1 Q
2
Q3
Q4 Q
5
Q6
A
B
C D
E
F
Q
Figure 6-3. Voronoi cells of the lattice points of P(2t+1) which are generated from thelattice points of P(2t −1).
−25 −20 −15 −10 −5 0 5 10 15 20 25
−25
−20
−15
−10
−5
0
5
10
15
20
25
Q0
−25 −20 −15 −10 −5 0 5 10 15 20 25
−25
−20
−15
−10
−5
0
5
10
15
20
25
Q0
A
B C
D E
W
(a) (b)
Figure 6-4. The boundary hexagons of Q0 and Q. (a) The black hexagon Q0 has boundaryindex 1. (b) The boundary hexagon Q should be one of the three redhexagons.
green and red hexagons in Figure 6-4 (b) are Voronoi cells of P (2t+1). Because q = q0 +r
for some r ∈ β2t+1 and because q is a boundary lattice point of P (2t + 1), the Voronoi cell
Q of q should be one of the three red hexagons in Figure 6-4 (b). By Lemma 6.2.8, the
hexagon W in Figure 6-4 (b) is not a Voronoi cell of P (2t + 1). Hence, in Figure 6-4 (b),
the hexagon C has boundary index 1. For a similar reason, we can show that the hexagon
A has boundary index 1, and the hexagon B has boundary index 3 and the union of its
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three B-sides is a connected set of R2. Hence this lemma is true for n = t + 1. Thus this
lemma is true for any n ≥ 2.
Let L be a lattice, ∅ 6= T ⊂ L, and let p and q be two C-boundary lattice points of T .
If the union of the B-sides of the Voronoi cells of p and q is a connected set of R2, then we
say that p and q are B-connected in T .
Lemma 6.2.10. For any integer n ≥ 2, there exists w ∈ L2n−1 \ P (2n − 1) such that w
has at least four neighbors belonging to the set P (2n− 1).
Proof. By Lemma 6.2.9, there exists one boundary lattice point q0 of P (2n − 3) whose
boundary index is 1. Let Q0 be the Voronoi cell of q0. Let W be the hexagon shown
in Figure 6-4 (b). As shown in the proof of Lemma 6.2.9, W is not a Voronoi cell of
P (2n− 1), and all green and red hexagons in Figure 6-4 (b) are Voronoi cells of P (2n− 1).
Let w be a lattice point of L2n−1 such that the hexagon W is a Voronoi cell of w. Then
w 6∈ P (2n− 1) and w has at least four neighbors belonging to the set P (2n− 1).
The main result in this subsection is the following theorem.
Theorem 6.2.11. For any integer n ≥ 2, the (2n − 1)th level of the Pyxis structure does
not tile its underlying lattice L2n−1 by translations by a sublattice.
Proof. By Lemma 6.2.10, for any integer n ≥ 2, there exists w ∈ L2n−1 \ P (2n − 1)
such that w has at least four neighbors belonging to the set P (2n − 1). Suppose that
P (2n − 1) tiles L2n−1 by translations by a sublattice Ls. Then there is r ∈ Ls such
that w ∈ r + P (2n − 1). Since w 6∈ P (2n − 1), by the definition of tiling, we have
P (2n − 1)⋂
(r + P (2n − 1)) = ∅. Hence the boundary index of w in r + P (2n − 1) is at
least 4 because w has at least four neighbors belonging to P (2n − 1). By Lemma 6.2.1,
r + P (2n− 1) has the same geometry as P (2n− 1). Hence by Lemma 6.2.9, the boundary
index of w in r + P (2n − 1) is either 1 or 3. However we have shown that the boundary
index of w in r + P (2n − 1) is at least 4, which is a contradiction. Therefore P (2n − 1)
98
does not tile its underlying lattice L2n−1 by translations by a sublattice for any integer
n ≥ 2.
6.2.2 Pyxis P (2n) Does Not Tile the Underlying Lattice by Translations by aSublattice for any n > 1
In this subsection, we first show some properties of P (2n) and apply them to prove
that P (2n) does not tile the underlying lattice L2n by translations by a sublattice for any
n ≥ 2.
Lemma 6.2.12. P (2n + 2) = P (2n)⋃
(P (2n) + β2n+1)⋃
(P (2n) + β2n+2) for any n ∈ N.
Proof. By the definition of the Pyxis structure, we have P (2n + 2) = P (2n + 1)⋃
(P (2n) +
β2n+2) and P (2n+1) = P (2n)⋃
(P (2n−1)+β2n+1). By Theorem 6.1.4, P (2n−1) ⊂ P (2n).
It follows that P (2n − 1) + β2n+1 ⊆ P (2n) + β2n+1. Hence P (2n)⋃
(P (2n − 1) + β2n+1) ⊆P (2n)
⋃(P (2n) + β2n+1), i.e., P (2n + 1) ⊆ P (2n)
⋃(P (2n) + β2n+1). Hence P (2n + 1) ⊆
P (2n)⋃
(P (2n) + β2n+1)⋃
(P (2n) + β2n+2). Obviously (P (2n) + β2n+2) ⊆ P (2n)⋃
(P (2n) +
β2n+1)⋃
(P (2n) + β2n+2). Thus P (2n + 1)⋃
(P (2n) + β2n+2) ⊆ P (2n)⋃
(P (2n) +
β2n+1)⋃
(P (2n) + β2n+2), i.e., P (2n + 2) ⊆ P (2n)⋃
(P (2n) + β2n+1)⋃
(P (2n) + β2n+2). On
the other hand, it follows from Lemma 6.2.3 that (P (2n)+β2n+1) ⊆ P (2n+1) ⊆ P (2n+2).
Furthermore P (2n + 2) = P (2n + 1)⋃
(P (2n) + β2n+2) implies that P (2n) + β2n+2 ⊆P (2n + 2). Hence P (2n)
⋃(P (2n) + β2n+1)
⋃(P (2n) + β2n+2) ⊆ P (2n + 2). Therefore
P (2n + 2) = P (2n)⋃
(P (2n) + β2n+1)⋃
(P (2n) + β2n+2).
For any q ∈ P (2n), by Lemma 6.2.12, we have (q + β2n+1)⋃
(q + β2n+2) ⊂ P (2n + 2).
The 13 lattice points in the set (q + β2n+1)⋃
(q + β2n+2) and their Voronoi cells are shown
in Figure 6-5, where the six lattice points in the set q + β2n+2 surround q and the six
lattice points in the set q + β2n+1 are the vertices of the (red) Voronoi cell of q ∈ P (2n).
Lemma 6.2.13. For any n ≥ 1, λ1λ2...λ2n ∈ Λ2n, and x ∈ 1, 2, 3, 4, 5, 6, we have
λ1λ2...λ2n0⊕ 00...0x ∈ Λ2n+1.
99
−8 −6 −4 −2 0 2 4 6 8−8
−6
−4
−2
0
2
4
6
8
q
q+β2n+2,6
q+β2n+2,1
q+β2n+2,2
q+β2n+2,3
q+β2n+2,4
q+β2n+2,5
q+β2n+1,6
q+β2n+1,1
q+β2n+1,2
q+β2n+1,3
q+β2n+1,4
q+β2n+1,5
Figure 6-5. The 13 lattice points in the set (q + β2n)⋃
(q + β2n−1) and their Voronoi cells.
Proof. Let p ∈ P (2n) such that the label of p is λ1λ2...λ2n ∈ Λ2n, and let b ∈ β2n+1 such
that the label of b is 00...0x ∈ Λ2n+1. By Lemma 6.2.3, we have p + b ∈ P (2n + 1). It
follows that λ1λ2...λ2n0⊕ 00...0x ∈ Λ2n+1.
Lemma 6.2.14. Let 1 < n ∈ N. If q is a boundary lattice point of P (2n) with label
q1q2...q2n−2q2n−1q2n ∈ Λ2n, then either q2n−1 6= 0 or q2n 6= 0, and the lattice point labeled
q1q2...q2n−2 is a boundary lattice point of P (2n− 2).
Proof. If q2n−1 = q2n = 0, then q ∈ P (2n − 2). Obviously the six neighbors of q in P (2n)
are exactly the elements of the set q+β2n. By Lemma 6.2.12, P (2n) = P (2n−2)⋃
(P (2n−2)+β2n−1)
⋃(P (2n−2)+β2n). It follows that q+β2n ⊆ P (2n−2)+β2n ⊆ P (2n). Hence all
neighbors of q in P (2n) are still in P (2n), which contradicts that q is a boundary lattice
point of P (2n). Thus either q2n−1 6= 0 or q2n 6= 0.
Let q be the lattice point of P (2n − 2) labeled q1q2...q2n−2. To show q is a boundary
lattice point of P (2n − 2), it suffices to show that there exists b ∈ β2n−2 such that
q+b 6∈ P (2n−2), i.e., q1q2...q2n−3q2n−2⊕00...0j 6∈ Λ2n−2 for some j ∈ 1, 2, 3, 4, 5, 6, where
00...0j ∈ Λ2n−2 is the label of b. Since q is a boundary lattice point of P (2n) with label
q1q2...q2n−2q2n−1q2n ∈ Λ2n, there exists k ∈ 1, 2, 3, 4, 5, 6 such that q1q2...q2n−2q2n−1q2n ⊕
100
00...00k 6∈ Λ2n. By Table 6-2, we have 00q2n−1q2n ⊕ 000k = 0d2d3d4 ∈ Λ4. Hence we have
q1q2...q2n−3q2n−2q2n−1q2n ⊕ 00...000k
= q1q2...q2n−3q2n−200⊕ 00...00q2n−1q2n ⊕ 00...000k
= q1q2...q2n−3q2n−200⊕ 00...0d2d3d4.
(6–23)
Since we have shown in the previous paragraph that either q2n−1 6= 0 or q2n 6= 0, we split
the proof into two cases.
Case 1: If q2n−1 6= 0, then, by Item 1 of Corollary 6.1.7, we have q2n = 0.
Since q2n−1 6= 0 and q2n = 0, by Table 6-2, we have d3 = 0. Suppose d2 = 0.
Then, by Equation 6–23, we have q1q2...q2n−2q2n−1q2n ⊕ 00...00k = q1q2...q2n−200 ⊕00...00d4 = q1q2...q2n−20d4 ∈ Λ2n, which is a contradiction. Thus d2 6= 0. Now suppose
q1q2...q2n−3q2n−2 ⊕ 00...0d2 ∈ Λ2n−2. Let
y1y2...y2n−3y2n−2 = q1q2...q2n−3q2n−2 ⊕ 00...0d2.
Then Equation 6–23 becomes
q1q2...q2n−3q2n−2q2n−1q2n ⊕ 00...000k
= q1q2...q2n−3q2n−200⊕ 00...0d20d4
= q1q2...q2n−3q2n−200⊕ 00...0d200⊕ 00...000d4
= y1y2...y2n−3y2n−200⊕ 00...000d4
= y1y2...y2n−3y2n−20d4 ∈ Λ2n,
(6–24)
which contradicts that q is a boundary lattice point of P (2n) with label q1q2...q2n−2q2n−1q2n.
Thus q1q2...q2n−3q2n−2 ⊕ 00...0d2 6∈ Λ2n−2. Therefore the lattice point labeled q1q2...q2n−2 is
a boundary lattice point of P (2n− 2).
101
Case 2: If q2n 6= 0, then, by Item 1 of Corollary 6.1.7, we have q2n−1 = 0. Suppose
d2 = 0. Then Equation 6–23 becomes
q1q2...q2n−3q2n−2q2n−1q2n ⊕ 00...000k
= q1q2...q2n−3q2n−200⊕ 00...00d3d4
(6–25)
If d3 6= 0, then, by Item 1 of Corollary 6.1.7, we have d4 = 0. Since d3 6= 0 and d4 = 0,
by Lemma 6.2.13, we have q1q2...q2n−3q2n−200 ⊕ 00...00d3d4 ∈ Λ2n. By Equation 6–25,
it follows that q1q2...q2n−3q2n−2q2n−1q2n ⊕ 00...000k ∈ Λ2n, which contradicts that q
is a boundary lattice point of P (2n) with label q1q2...q2n−2q2n−1q2n. If d3 = 0, then
q1q2...q2n−3q2n−200 ⊕ 00...00d3d4 = q1q2...q2n−3q2n−20d4 ∈ Λ2n, which also contradicts that
q is a boundary lattice point of P (2n) with label q1q2...q2n−2q2n−1q2n. Thus d2 6= 0, which
implies that d3 = 0. Since q1q2...q2n−2q2n−1q2n ⊕ 00...00k 6∈ Λ2n, by Equation 6–23, we have
q1q2...q2n−3q2n−200 ⊕ 00...0d2d3d4 6∈ Λ2n, i.e., q1q2...q2n−3q2n−200 ⊕ 00...0d20d4 6∈ Λ2n. By
Item 2 of Lemma 6.1.8, it follows that q1q2...q2n−3q2n−2 ⊕ 00...0d2 6∈ Λ2n−2. Therefore the
lattice point labeled q1q2...q2n−2 is a boundary lattice point of P (2n− 2).
The following lemma shows some invariant properties of P (n) with respect to the
rotation of R2 by an angle of kπ3
for some k ∈ Z about the origin.
Lemma 6.2.15. Let n ∈ N, x ∈ P (2n), and α = kπ3
for some k ∈ Z. If y is the point in
R2 obtained by rotating x by an angle of α about the origin, then y ∈ P (2n). Furthermore,
x is a boundary lattice point of P (2n) with boundary index b for some b ∈ 1, 2, 3, 4, 5, 6 if
and only if y is a boundary lattice point of P (2n) with boundary index b.
Proof. Let the label of x be x1x2...x2n−1x2n ∈ Λ2n. Then x =∑2n
i=1 βi,xi. Let M =
cos(α) − sin(α)
sin(α) cos(α)
. Since M is the matrix of rotation by α = kπ
3, by the definition
of β1 and β2, it is easy to show that, for any j = 1, 2, ..., 6, we have Mβ1,j ∈ β1 and
Mβ2,j ∈ β2. For any k ∈ N, since β2k+1,j = 13k β1,j and β2k+2,j = 1
3k β2,j, we have Mβ2k+1,j =
13k Mβ1,j ∈ β2k+1 and Mβ2k+2,j = 1
3k Mβ2,j ∈ β2k+2. Obviously Mβi,0 = M0 = 0 = βi,0 for
102
any i ∈ N. Thus Mβi ⊆ βi for any i ∈ N. It follows that Mβi,xi∈ βi where βi = βi ∪ 0,
and there exists yi ∈ 0, 1, .., 6 such that βi,yi= Mβi,xi
for each i satisfying 1 ≤ i ≤ 2n.
Since ‖Mβi,xi‖ = ‖βi,xi
‖, xi 6= 0 if and only if yi 6= 0. Since x1x2...x2n−1x2n ∈ Λ2n, it
follows that y1y2...y2n−1y2n ∈ Λ2n. Because y is the point in R2 obtained by rotating x by
an angle of α about the origin, we have
y = Mx =2n∑i=1
Mβi,xi=
2n∑i=1
βi,yi. (6–26)
Since either yi = 0 or yi+1 = 0 for each i satisfying 1 ≤ i ≤ 2n − 1, by Lemma 6.1.5, it
follows that y ∈ P (2n).
Furthermore, if x is a boundary lattice point of P (2n) with boundary index b for
some b ∈ 1, 2, 3, 4, 5, 6, then there exists B ⊂ 1, 2, 3, 4, 5, 6 such that |B| = b and,
for each j ∈ 1, 2, 3, 4, 5, 6, x + β2n,j 6∈ P (2n) if and only if j ∈ B. By the result of the
previous paragraph, we have x + β2n,j ∈ P (2n) if and only if Mx + Mβ2n,j ∈ P (2n).
Hence y is a boundary lattice point of P (2n) with boundary index b. Conversely, if y is
a boundary lattice point of P (2n) with boundary index b for some b ∈ 1, 2, 3, 4, 5, 6,similarly we can show that x is also a boundary lattice point of P (2n) with boundary
index b since x = M−1y =
cos(−α) − sin(−α)
sin(−α) cos(−α)
y.
The following lemma shows that each boundary lattice point of P (2n) is a C-boundary
lattice point, and its boundary index is 1, 2, or 4.
Lemma 6.2.16. Let q be a boundary lattice point of P (2n). If the label of q is λ0j for
some λ ∈ Λ2n−2 and j ∈ 1, 2, 3, 4, 5, 6, then the boundary index of q is 1. If the label of q
is λj0 for some λ ∈ Λ2n−2 and j ∈ 1, 2, 3, 4, 5, 6, then the boundary index of q is either 2
or 4, and q is a C-boundary lattice point of P (2n).
103
Proof. If the label of q is λ0j for some λ = λ1λ2...λ2n−2 ∈ Λ2n−2 and j ∈ 1, 2, 3, 4, 5, 6,then the labels of the six neighbors of q in the lattice L2n constitute the set
λ1λ2...λ2n−20j ⊕ 00...00k : k = 1, 2, ..., 6 .
For any k ∈ 1, 2, 3, 4, 5, 6 with k 6= j, by Table 6-2, we have 000j ⊕ 000k = 00x1x2 ∈ Λ4
where x1x2 ∈ Λ2. By applying Item 4 of Lemma 6.1.8 twice, it follows that 0j ⊕ 0k =
x1x2 ∈ Λ2. Hence we have
λ1λ2...λ2n−20j ⊕ 00...00k = λ1λ2...λ2n−200⊕ 00...0x1x2. (6–27)
When x1 = 0, Equation 6–27 becomes
λ1λ2...λ2n−20j ⊕ 00...00k = λ1λ2...λ2n−200⊕ 00...00x2
= λ1λ2...λ2n−20x2 ∈ Λ2n.
(6–28)
When x2 = 0, Equation 6–27 becomes
λ1λ2...λ2n−20j ⊕ 00...00k = λ1λ2...λ2n−200⊕ 00...0x10. (6–29)
By Lemma 6.2.13, λ1λ2...λ2n−20 ⊕ 00...0x1 ∈ Λ2n−1. It follows that λ1λ2...λ2n−200 ⊕00...0x10 ∈ Λ2n. Hence by Equation 6–29, we have λ1λ2...λ2n−20j⊕ 00...00k ∈ Λ2n. Thus at
least five neighbors of q in the lattice L2n are still in P (2n). Therefore the boundary index
of q is 1.
If the label of q is λj0 for some λ = λ1λ2...λ2n−2 ∈ Λ2n−2 and j ∈ 1, 2, 3, 4, 5, 6, then
the labels of the six neighbors of q in the lattice L2n constitute the set
λ1λ2...λ2n−2j0⊕ 00...00k : k = 1, 2, ..., 6 .
For k = 1, 2, ..., 6, denote nk the neighbor of q whose label is λ1λ2...λ2n−2j0 ⊕ 00...00k.
Without loss of generality we assume that j = 1 because, by Lemma 6.2.15, the boundary
index would be the same if j ∈ 2, 3, 4, 5, 6. If k = 3, by Table 6-2, we have 00j0⊕ 000k =
104
0010⊕ 0003 = 0001. Then we have
λ1λ2...λ2n−2j0⊕ 00...00k = λ1λ2...λ2n−200⊕ 00...001
= λ1λ2...λ2n−201 ∈ Λ2n.
(6–30)
It follows that n3 ∈ P (2n). Similarly 00j0 ⊕ 0004 = 0006 implies that λ1λ2...λ2n−2j0 ⊕00...004 ∈ Λ2n. Thus n4 ∈ P (2n). Now consider n1 and n2. By Table 6-2, 00j0 ⊕ 0001 =
0010⊕0001 = 0105 and 00j0⊕0002 = 0010⊕0002 = 0104. It follows that n1 6∈ P (2n) if and
only if λ1λ2...λ2n−2j0 ⊕ 00...001 6∈ Λ2n if and only if λ1λ2...λ2n−3λ2n−200 ⊕ 00...0105 6∈ Λ2n
if and only if λ1λ2...λ2n−3λ2n−2 ⊕ 00...01 6∈ Λ2n−2 if and only if λ1λ2...λ2n−3λ2n−200 ⊕00...0104 6∈ Λ2n if and only if n2 6∈ P (2n). Similarly we can show that n5 6∈ P (2n) if and
only if n6 6∈ P (2n). Thus the boundary index of q is either 2 or 4. Obviously n5 is next to
n6, n6 is next to n1, and n1 is next to n2. It follows that q is a C-boundary lattice point of
P (2n).
Lemma 6.2.17. Let 1 ≤ n ∈ N and let q be a boundary lattice point of P (2n). If the
boundary index of q is 1, then there exist two boundary lattice points c1 and c2 of P (2n)
such that, for each j = 1, 2, cj is B-connected to q in P (2n) and the boundary index of cj
is either 2 or 4. If the boundary index of q is 2 or 4, then there exist two boundary lattice
points c1 and c2 of P (2n) such that, for each j = 1, 2, cj is B-connected to q in P (2n) and
the boundary index of cj is 1.
Proof. We prove it by induction. If n = 1, by the plot of the Voronoi cells of P (2) in
Figure 6-2, this lemma is obviously true. Assume this lemma is true for all integers less
than n. Now let q be a boundary lattice point of P (2n) and let q1q2...q2n−2q2n−1q2n ∈ Λ2n
be the label of q. By Lemma 6.2.14, we have q2n−1q2n = 0i or q2n−1q2n = i0 for some
i ∈ 1, 2, 3, 4, 5, 6.Let λ = q1q2...q2n−2 ∈ Λ2n−2 and let q be a lattice point of P (2n − 2) such that the
label of q is λ. Then the label of q is λq2n−1q2n. Since q is a boundary lattice point of
P (2n), by Lemma 6.2.14, q is a boundary lattice point of P (2n − 2). Furthermore, by
105
Lemma 6.2.12, we have P (2n) = P (2n − 2)⋃
(P (2n − 2) + β2n−1)⋃
(P (2n − 2) + β2n).
It follows that q + β2n ⊂ P (2n), and q + β2n−1 ⊂ P (2n). For any i ∈ 1, 2, 3, 4, 5, 6, by
Lemma 6.2.13, we have
q1q2...q2n−20⊕ 00...0i ∈ Λ2n−1. (6–31)
Let µi = q1q2...q2n−20 ⊕ 00...0i. When q is taken as an element of P (2n − 1), the label of
q is q1q2...q2n−20. By Equation 6–31, it follows that q + β2n−1,i ∈ P (2n − 1) and the label
of q + β2n−1,i is µi ∈ Λ2n−1. Hence, if q + β2n−1,i is taken as an element of P (2n), its label
is µi0 ∈ Λ2n. When q is taken as an element of P (2n), the label of q is q1q2...q2n−200. It
follows that the label of q + β2n,i is q1q2...q2n−200 ⊕ 00...00i. By Item 2 of Lemma 6.1.8,
we have q1q2...q2n−200 ⊕ 00...00i = q1q2...q2n−20i ∈ Λ2n. Let λ = q1q2...q2n−2. Then the
label of q + β2n,i is λ0i. By Lemma 6.2.16, the boundary index of q is 1, 2, or 4, and q is
a C-boundary lattice point of P (2n − 2). Without loss of generality, we assume that the
boundary index of q is 4 since the proofs for the other two cases can be similarly done.
Then the Voronoi cell of q in the lattice L2n−2 and the Voronoi cells of lattice points of
L2n in the set (q + β2n)⋃
(q + β2n−1) are shown in Figure 6-6, where the Voronoi cells of
lattice points in the set q + β2n are labeled µi for i = 0, 1, 2, ..., 6, and the Voronoi cells of
lattice points in the set q + β2n−1 are labeled λ0i for i = 1, 2, ..., 6. By Lemma 6.2.16, the
boundary index of q is 1, 2, or 4. Hence we consider the following cases.
If the boundary index of q is 1, then, by Lemma 6.2.16, we have q2n−1q2n = 0i for
some i ∈ 1, 2, ..., 6. Hence the label of q is λ0i. As shown in Figure 6-6, there exist two
boundary lattice points c1 and c2 of P (2n) such that, for each j = 1, 2, cj is B-connected
to q in P (2n) and the boundary index of cj is either 2 or 4.
Similarly if the boundary index of q is 2 or 4, then by Lemma 6.2.16, we have
q2n−1q2n = i0 for some i ∈ 1, 2, ..., 6. Since q2n−1 = i 6= 0, by Item 1 of Corollary 6.1.7,
we have q2n−2 = 0. By Item 2 of Lemma 6.1.8, it follows that q1q2...q2n−2q2n−1q2n =
q1q2...q2n−200 ⊕ 00...0i0 = µi0 where µi is defined in the second paragraph of this proof.
Hence the label of q is µi0 where i ∈ 1, 2, 3, 4, 5, 6. As shown in Figure 6-6, there
106
−4 −2 0 2 4 6 8
−4
−2
0
2
4
6
8
λ00
λ01 λ02
λ03 λ04
λ05
λ06
µ10 µ
20
µ30
µ40 µ
50
µ60
B1
B2
B3
B4
B02
B
01
Figure 6-6. Generation of the boundary hexagons of P (2n).
exist two boundary lattice points c1 and c2 of P (2n) such that, for each j = 1, 2, cj is
B-connected to q in P (2n) and the boundary index of cj is 1. By induction, this lemma is
true for any n ∈ N.
Lemma 6.2.18. For any n satisfying 2 ≤ n ∈ N, there exists at least one C-boundary
lattice point of P (2n) with boundary index 2.
Proof. For any i ∈ N, let Γi = 1010...10 ∈ Λ2i and, for any j ∈ 1, 2, 3, 4, 5, 6, let
Θi,j = 00...0j ∈ Λi which is the label of βi,j in P (i). We claim that Γi ⊕ Θ2i,1 6∈ Λ2i for
any i ∈ N. If i = 1, by Table 6-2, we have 0010 ⊕ 0001 = 0105. By applying Item 4 of
Lemma 6.1.8 twice, it follows that 10 ⊕ 01 6∈ Λ2. Hence Γ1 ⊕ Θ2,1 6∈ Λ2 and the claim is
true for i = 1. Assume the claim is true for any i ≤ N0 for some N0 ∈ N. In the following,
we show that the claim is also true if i = N0 + 1. Obviously
ΓN0+1 ⊕Θ2N0+2,1 = 1010...1010⊕ 0000...00001
= 1010...1000⊕ 0000...0010⊕ 0000...0001
= ΓN000⊕ 0000...0010⊕ 0000...0001.
(6–32)
107
Since 0010 ⊕ 0001 = 0105, by applying Item 4 of Lemma 6.1.8 for 2N0 − 2 times, we have
0000...0010⊕ 0000...0001 = 0000...0105 ∈ Λ2N0+2. Hence Equation 6–32 becomes
ΓN0+1 ⊕Θ2N0+2,1 = ΓN000⊕ 0000...0105. (6–33)
By the assumption, we have ΓN0 ⊕ Θ2N0,1 6∈ Λ2N0 , i.e., ΓN0 ⊕ 0000...01 6∈ Λ2N0 . By
Equation 6–33 and Item 2 of Lemma 6.1.8, it follows that ΓN0+1⊕Θ2N0+2,1 6∈ Λ2N0+2. Thus
the claim is true for any i ∈ N.
Now we prove Lemma 6.2.18. If n = 2, by the plot of Voronoi cells of P (4) as shown
in Figure 6-2, this lemma is obviously true. If n > 2, let Q = 1010...1030 ∈ Λ2n and let
q be a lattice point of P (2n) whose label is Q. By the definition of the lattice L2n and
the definition of β2n, the six neighbors of q in the lattice L2n are exactly the elements
of the set q + β2n. By the definition of labels, we have q + β2n,k ∈ P (2n) if and only if
Q⊕Θ2n,k ∈ Λ2n. By Item 1 of Lemma 6.1.8, we have Q = 1010...100000⊕ 0000...001030 =
Γn−20000⊕ 0000...001030. Hence
Q⊕Θ2n,k = Γn−20000⊕ 0000...001030⊕ 0000...00000k (6–34)
In the following, we show that Q ⊕ Θ2n,k 6∈ Λ2n for k = 1, 2, and Q ⊕ Θ2n,k ∈ Λ2n for
k = 3, 4, 5, 6.
By Table 6-2, we have 0030 ⊕ 0001 = 0205. Hence 000030 ⊕ 000001 = 000205. Since
001030 = 001000⊕ 000030, it follows that
001030⊕ 000001 = 001000⊕ 000030⊕ 000001
= 001000⊕ 000205
= 001000⊕ 000200⊕ 000005.
(6–35)
By Table 6-2, we have 0010⊕0002 = 0104. It follows that 001000⊕000200 = 010400. Then
Equation 6–35 becomes 001030⊕ 000001 = 010400⊕ 000005 = 010405. By Equation 6–34,
108
we have
Q⊕Θ2n,1 = Γn−20000⊕ 0000...010405
= Γn−20000⊕ 0000...010000⊕ 0000...000405
= Γn−20000⊕Θ2n−4,10000⊕ 0000...000405.
(6–36)
By the claim in the first paragraph, we have Γn−2 ⊕ Θ2n−4,1 6∈ Λ2(n−2). By Item 2 of
Lemma 6.1.8 and Equation 6–36, it follows that Q ⊕ Θ2n,1 6∈ Λ2n. Similarly we can show
that Q⊕Θ2n,2 6∈ Λ2n.
By Table 6-2, we have 0030 ⊕ 0003 = 0301. Since 1030 ⊕ 0003 = 1000 ⊕ 0030 ⊕ 0003,
it follows that 1030 ⊕ 0003 = 1000 ⊕ 0301. Because 10 ⊕ 03 = 01 ∈ Λ2, it follows that
1030 ⊕ 0003 = 0101 ∈ Λ4. By Equation 6–34, it follows that Q ⊕ Θ2n,3 = Γn−20000 ⊕0000...001030 ⊕ 0000...000003 = Γn−20301 ∈ Λ2n. Similarly we can show that Q ⊕ Θ2n,k ∈Λ2n for k = 4, 5, 6.
We have shown that Q⊕Θ2n,k 6∈ Λ2n for k = 1, 2, and Q⊕Θ2n,k ∈ Λ2n for k = 3, 4, 5, 6.
By the definition of the sum of two labels, it follows that q + β2n,k 6∈ P (2n) for k = 1, 2,
and q + β2n,k ∈ P (2n) for k = 3, 4, 5, 6. Thus q is a boundary lattice point of P (2n) and
the boundary index of P (2n) is 2. Furthermore, since q + β2n,1 is next to q + β2n,2 in the
lattice L2n, q is a C-boundary lattice point of P (2n).
In the following, we use the previous results to prove the main result of this
subsection.
Theorem 6.2.19. For any integer n ≥ 2, the (2n)th level of the Pyxis structure P (2n)
does not tile its underlying lattice L2n by translations by a sublattice.
Proof. By Lemma 6.2.18, there exists a C-boundary lattice point, denoted a, of P (2n)
with boundary index 2. By Lemma 6.2.17, there exist two boundary lattice points bj of
P (2n) for j = 1, 2 such that a is B-connected to bj in P (2n) and bj has boundary index 1
for each j = 1, 2. Let b3 = a and let Bj be the Voronoi cell of bj for j = 1, 2, 3. Because
b3 has boundary index 2 and is B-connected to b1 and b2 in P (2n), we can assume that
109
the Voronoi cells Bj, Bj, and Bj are shown in Figure 6-7 (a). Also, for k = 1, 2, let
uk ∈ L2n \ P (2n) be a lattice point such that uk is next to bk in the lattice L2n, and let
Uk be the Voronoi cell of uk. Since uk ∈ L2n \ P (2n) and is next to bk for k = 1, 2, the
location of Uk is as shown in Figure 6-7 (a) for k = 1, 2.
−4 −2 0 2 4 6 8
−4
−2
0
2
4
6
8
B01
B02
B1
B3
B2 U
1
U2
−4 −2 0 2 4 6 8
−4
−2
0
2
4
6
8
B01
B02
B1
B3
B2 U
1
U2 W
1
W3
W2
(a) (b)
Figure 6-7. Tiling of P (2). (a) Uj is a hexagon and next to Bj but not in P (2n) forj = 1, 2. (b) Showing where is W3 located.
Suppose that P (2n) tiles L2n by translations by a sublattice Ls. Then
⋃u + P (2n) : u ∈ Ls = L2n
and u + P (2n) = v + P (2n) whenever (u + P (2n))⋂
(v + P (2n)) 6= ∅ for u,v ∈ Ls.
Since uk ∈ L2n, it follows that there exists rk ∈ Ls such that uk ∈ rk + P (2n) for each
k = 1, 2, and either r1 + P (2n) = r2 + P (2n) or (r1 + P (2n))⋂
(r2 + P (2n)) = ∅. Since
uk ∈ rk + P (2n) and uk 6∈ P (2n), we have rk + P (2n) 6= 0 + P (2n) = P (2n). Hence
P (2n)⋂
(rk + P (2n)) = ∅ for k = 1, 2. Because either r1 + P (2n) = r2 + P (2n) or
(r1 + P (2n))⋂
(r2 + P (2n)) = ∅, we consider the following two cases.
Case 1: r1 + P (2n) = r2 + P (2n). In this case, we have uk ∈ r1 + P (2n) for each
k = 1, 2. It follows that u1 and u2 are two boundary lattice points of r1 + P (2n) and next
to each other in the lattice L2n. Since there are two lattice points b1 and b3 which are
next to u1 in the lattice L2n but not in r1 + P (2n), the boundary index of u1 in r1 + P (2n)
is at least 2 as shown in Figure 6-7 (a). For a similar reason, the boundary index of u2
110
in r1 + P (2n) is at least 2 as shown in Figure 6-7 (a). However, by Lemma 6.2.17, u1 is
B-connected to two boundary boundary lattice points in r1 + P (2n) each having boundary
index 1. It follows that the boundary index of u2 in r1 + P (2n) should be 1, which
contradicts that the boundary index of u2 in r1 + P (2n) is at least 2.
Case 2: (r1 + P (2n))⋂
(r2 + P (2n)) = ∅. Let w3 be a lattice point of L2n which
is different from b3 such that its Voronoi cell W3 is next to both U1 and U2 as shown in
Figure 6-7 (b). Since (r1 + P (2n))⋂
(r2 + P (2n)) = ∅ in this case, either w3 6∈ r1 + P (2n)
or w3 6∈ r2 + P (2n). Because U1 and U2 in Figure 6-7 (b) are symmetric about W , without
loss of generality, we assume that w3 6∈ r2 + P (2n). We claim that the boundary index of
u2 (in r2 + P (2n)) is 5. Since the boundary index of b2 (in P (2n)) is 1, the only neighbor
of b2 which does not belong to P (2n) is u2. It follows that the black hexagon sitting on
the top of B2 must be a Voronoi cell of P (2n). Hence u2 has three neighbors in P (2n)
which are b3, b2, and the one sitting on the tope of b2. Since (r2 + P (2n))⋂
P (2n) = ∅,these three neighbors are not in P (2n). Furthermore, because w3 6∈ r2 + P (2n) and
u1 ∈ (r1 +P (2n))\ (r2 +P (2n)), w3 and u1 are also neighbors of u2 which are not elements
of r2 + P (2n). Thus the boundary index of u2 (in r2 + P (2n)) is at least 5. However
by Lemma 6.2.16, the boundary index of u2 is less than or equal to 4, which leads to a
contradiction. Thus P (2n) does not tile its underlying lattice L2n by translations by a
sublattice.
111
CHAPTER 7FRACTAL DIMENSION OF THE BOUNDARY OF THE PYXIS STRUCTURE
7.1 The Limit of the Pyxis Structure
Let (R2, d) be the 2-dimensional Euclidean space, where d(x, y) denotes the distance
between x and y for any x, y ∈ R2, and let H(R2) be the set consisting of all nonempty
compact subsets of (R2, d). The next three definitions are from Barnsley [4]. For any
x ∈ R2 and B ∈ H(R2), let d(x, B) = mind(x,y) : y ∈ B. The number d(x, B)
is called the distance from the point x to the set B. For any A,B ∈ H(R2), define
d(A,B) = maxd(x, B) : x ∈ A. The number d(A,B) is called the distance from the set
A ∈ H(R2) to the set B ∈ H(R2). Also let h(A,B) = maxd(A,B), d(B, A). The number
h(A,B) is called the Hausdorff distance between A and B.
By Theorem 1 on page 37 of Barnsley [4], (H(R2), h) is a complete metric space.
Recall from Chapter 6 that
vA1 =
1
0
,vA
2 =
−1
2√
32
,vB
1 =
√3
2
12
,vB
2 =
−√
32
12
,
LA =n1(v
A1 ) + n2(v
A2 ) : n1, n2 ∈ Z
, LB =
n1(v
B1 ) + n2(v
B2 ) : n1, n2 ∈ Z
, ρ1 is
a fixed positive number, and ρn = ( 1√3)n−1ρ1 for any n satisfying 1 < n ∈ N. Also
v2n−1,k = ρ2n−1vAk and v2n,k = ρ2nv
Bk for any n satisfying 1 < n ∈ N and k ∈ 1, 2. We
have defined a sequence of lattices by Ln = n1vn,1 + n2vn,2 : n1, n2 ∈ Z, n ∈ N. It is
shown in Chapter 6 that P (n) ⊂ Ln and P (n) ⊂ P (n + 1). Now let sn denote the length
of one side of a Voronoi cell of the lattice Ln and let Pn be the union of all Voronoi cells
of the lattice points of P (n). Since Ln is a hexagonal lattice with generators vn,1 and vn,2
each having length ρn, we have sn = ρn√3. The following lemma will be used to show that
each of the two sequences, P (n) : n ∈ N and
Pn : n ∈ N
, is a Cauchy sequence in
the space (H(R2), h).
Lemma 7.1.1. For any n ∈ N, we have
1. h(P (n), P (n + 1)) ≤ ρn+1 and h(P (n), P (n + 2)) ≤ ρn+1;
112
2. h(Pn, Pn+1) ≤ (2√
3 + 1)ρn+2;
3. h(Pn, P (n)) ≤ sn = 1√3ρn.
Proof. 1. Because P (n) ⊂ P (n + 1), the distance from P (n) to P (n + 1) is 0, i.e.,
d(P (n), P (n + 1)) = 0. For any x ∈ P (n + 1), since P (n + 1) = P (n)⋃
(P (n − 1) + βn+1)
and P (n − 1) ⊂ P (n), there exist y ∈ P (n) and z ∈ βn+1 such that x = y + z. It follows
that d(x, P (n)) ≤ d(x,y) = ‖z‖ = ρn+1, where ‖z‖ denotes the norm of z. Hence
d(P (n + 1), P (n)) = maxd(x, P (n)) : x ∈ P (n + 1) ≤ ρn+1.
Thus
h(P (n), P (n + 1)) = maxd(P (n), P (n + 1)), d(P (n + 1), P (n)) ≤ ρn+1.
Similarly d(P (n), P (n + 2)) = 0. For any x ∈ P (n + 2), since P (n + 2) =
P (n + 1)⋃
(P (n) + βn+2), we have either x ∈ P (n + 1) or x ∈ P (n) + βn+2. If
x ∈ P (n + 1), then d(x, P (n)) ≤ d(P (n + 1), P (n)) ≤ h(P (n + 1), P (n)) ≤ ρn+1. If
x ∈ P (n) + βn+2, then there exist y ∈ P (n) and z ∈ βn+2 such that x = y + z. It follows
that d(x, P (n)) ≤ ‖z‖ = ρn+2 ≤ ρn+1. Thus h(P (n), P (n + 2)) ≤ ρn+1.
2. For any x ∈ Pn, there exists y ∈ P (n) such that x lies in the Voronoi cell of y.
Since Ln is a hexagonal lattice and sn is the length of one side of a Voronoi cell of the
lattice Ln, we have d(x,y) ≤ sn. It follows that d(x, P (n)) ≤ sn. Hence d(Pn, P (n)) ≤ sn.
Because P (n) ⊂ Pn, we have d(P (n), Pn) = 0. Thus h(Pn, P (n)) ≤ sn.
3. By Item 2, we have h(Pn, P (n)) ≤ sn for any n ∈ N. By the triangle inequality, we
have
h(Pn, Pn+1) ≤ h(Pn, P (n)) + h(P (n), P (n + 1)) + h(P (n + 1), Pn+1)
≤ sn + ρn+1 + sn+1.
(7–1)
Since sn = 1√3ρn for any n ∈ N, we have sn + ρn+1 + sn+1 = 1√
3ρn + ρn+1 + 1√
3ρn+1 =
2ρn+1 + 1√3ρn+1 = (2
√3 + 1)ρn+2.
113
Proposition 7.1.2. Each of the two sequences, P (n) : n ∈ N and
Pn : n ∈ N,
converges in (H(R2), h), and these two sequences have the same limit.
Proof. By Item 1 of Lemma 7.1.1, we have h(P (n), P (n + 1)) ≤ ρn+1 for any n ∈ N. Since
ρn = ( 1√3)n−1ρ1 where ρ1 is a fixed positive number, the sequence P (n) : n ∈ N is a
Cauchy sequence in the space (H(R2), h). Because (H(R2), h) is a complete metric space,
it follows that the sequence P (n) : n ∈ N converges in H(R2).
Similarly, by Item 2 of Lemma 7.1.1, the sequence Pn : n ∈ N is a Cauchy sequence
in the space (H(R2), h). Therefore the sequence Pn : n ∈ N also converges in H(R2).
Let P be the limit of the sequence P (n) : n ∈ N. For any ε > 0, there exists
n1 ∈ N such that h(P (n),P) < 12ε for any n > n1. By Item 3 of Lemma 7.1.1, we have
h(Pn, P (n)) ≤ 1√3ρn = ( 1√
3)nρ1 for any n ∈ N. It follows that there exists n2 ∈ N such
that h(Pn, P (n)) < 12ε for any n > n2. Let n3 = max n1, n2. Then, for any n > n3, we
have h(Pn,P) < h(Pn, P (n)) + h(P (n),P) < 12ε + 1
2ε = ε. Thus P is also the limit of the
sequence
Pn : n ∈ N
.
Let P be the limit of the sequence P (n) : n ∈ N in H(R2). We call P the limit
of the Pyxis structure. The boundary of P is denoted PB and is called the boundary of the
Pyxis structure.
7.2 Fractal Dimension of the Boundary of the Pyxis Structure
To make this research more self-contained, we review one definition and one theorem
from Page 174 and Page 176 of Barnsley [4], respectively. Let A ∈ H(R2). For each ε > 0,
let N (A, ε) denote the smallest number of closed balls of radius ε > 0 needed to cover A.
If D = limε→0 ln(N (A,ε))ln(1/ε)
exists, then D is called the fractal dimension of A.
Theorem 7.2.1. Let A ∈ H(R2) and C > 0. If there exists some r with 0 < r < 1 such
that εn = Crn for any n ∈ N, and D = limn→∞ ln(N (A,εn))ln(1/εn)
exists, then A has fractal
dimension D.
To determine the fractal dimension of PB, in the following, we will show that 4n + 2 ≤N (PB, εn) ≤ 4n+1 − 4 for any n ∈ N, where εn = 3
2ρ2n+1 = 3ρ1
2(1
3)n. For any n ∈ N
114
and j = 1, 2, 4, let Bn,j be the set consisting of all boundary lattice points of P (2n) whose
boundary index is j and let bn,j be the size of the set Bn,j. The following lemma shows one
relation among these sizes.
Lemma 7.2.2. For any n ∈ N, we have bn,1 = bn,2 + bn,4.
Proof. Define a bipartite graph G with parts V1 = Bn,1 and V2 = Bn,2
⋃Bn,4 such that
e is an edge of this graph if and only if the two vertices which are end points of e are
B-connected in P (2n). Let E(G) denote the edge set of the graph G. By Lemma 6.2.17,
each vertex in the set V1 is B-connected to exactly two vertices in the set V2. It follows
that |E(G)| = 2|V1|. Also, by Lemma 6.2.17, each vertex in the set V2 is B-connected to
exactly two vertices in the set V1. It follows that |E(G)| = 2|V2|. Thus |V1| = |V2| and
hence bn,1 = bn,2 + bn,4.
−2 −1 0 1 2 3 4 5 6 7 8−1
0
1
2
3
4
5
6
7
8
9
Q R
T
−2 −1 0 1 2 3 4 5 6 7 8−1
0
1
2
3
4
5
6
7
8
9
Q R
T
A4 B
4
C4 D
2
E2
(a) (b)
Figure 7-1. The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 4, and the boundary index of R and T is 1. (b) The boundary hexagonsof P(2n+2) which have boundary index 2 or 4. Their Voronoi cells overlap theinterior of the boundary hexagon Q of P (2n), where A4, B4, and C4 haveboundary index 4, and D2 and E2 have boundary index 2.
Lemma 7.2.3. For any n ∈ N, the (2n)th level of the Pyxis structure has 4n+1 − 4
boundary hexagons. Furthermore, 4n + 2 of them have boundary index 4, 4n − 4 of them
have boundary index 2, and 2(4n)− 2 of them have boundary index 1.
115
Proof. By Lemma 7.2.2, we have the following Equation.
bn,1 = bn,2 + bn,4. (7–2)
Hence it suffices to compute bn,2 and bn,4. Let Q be a boundary hexagon of P (2n). By
Lemma 6.2.16, the boundary index of Q is 1, 2, or 4. If the boundary index of Q is 4,
then it follows from Lemma 6.2.17 that, as shown in Figure 7-1 (a), there are exactly two
boundary hexagons R and T of P (2n) such that each of R and T is B-connected to Q and
has boundary index 1. By Lemma 6.2.12, it follows that, as shown in Figure 7-1 (b), there
exist three boundary hexagons A4, B4, and C4 of P (2n + 2) with boundary index 4 and
there exist two boundary hexagons D2 and E2 of P (2n + 2) with boundary index 2 such
that each of them has interior overlapping the interior of Q. Similarly if the boundary
−15 −10 −5 0 5 10 15
−10
−5
0
5
10
15
20
Q
R
T
−15 −10 −5 0 5 10 15
−10
−5
0
5
10
15
20
Q
R
T A4
B2
C2
(a) (b)
Figure 7-2. The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 2, and the boundary index of R and T is 1. (b) The boundary hexagonsof P(2n+2) which have boundary index 2 or 4, and whose Voronoi cells overlapthe interior of the boundary hexagon Q of P (2n), where A4 has boundaryindex 4, and B2 and C2 have boundary index 2.
index of Q is 2, then as shown in Figure 7-2 (a), there are exactly two boundary hexagons
R and T of P (2n) such that each of R and T is B-connected to Q and has boundary index
1. By Lemma 6.2.12, it follows that there exists one boundary hexagon A4 with boundary
index 4 and there exist two boundary hexagons B2 and C2 of P (2n + 2) with boundary
116
index 2 such that each of them has interior overlapping the interior of Q as shown in
Figure 7-2 (b). If the boundary index of Q is 1, then it follows from Lemma 6.2.17 that,
as shown in Figure 7-3 (a), there are exactly two boundary hexagons R and T of P (2n)
−15 −10 −5 0 5 10 15
−5
0
5
10
15
20
Q
R
T
−15 −10 −5 0 5 10 15
−5
0
5
10
15
20
Q
R
T
A
B
(a) (b)
Figure 7-3. The boundary hexagons of P (2n) and P(2n+2). (a) The black hexagons Q, R,and T of P (2n) such that Q is B-connected to R and T , the boundary index ofQ is 1, and the boundary index of R and T is 2 or 4. (b) The boundaryhexagons of P(2n+2) which have boundary index 2 or 4, and whose Voronoicells overlap the interior of the boundary hexagon Q of P (2n), where A and Bhave boundary index 2.
such that each of R and T is B-connected to Q and has boundary index 2 or 4. By
Lemma 6.2.12, it follows that there exist two boundary hexagons A and B of P (2n + 2)
with boundary index 2 such that each of them has interior overlapping the interior of Q
as shown in Figure 7-3 (b). But boundary hexagons A and B are already counted from
hexagons R and T which have boundary index 2 or 4. Therefore we have the following
system of equations.
bn+1,4 = 3bn,4 + bn,2
bn+1,2 = 2bn,4 + 2bn,2.(7–3)
Let Xj =
bj,4
bj,2
for any j ∈ N, and let M =
3 1
2 2
. By System 7–3, Xn+1 = MXn
for any n ∈ N. Therefore we have
Xn+1 = Mn X1. (7–4)
117
Since there exist exactly six boundary hexagons of P (2) and each of them has
boundary index 4, we have b1,4 = 6 and b1,2 = 0. Hence X1 =
b1,4
b1,2
=
6
0
. The
eigenvalues of the matrix M are 1 and 4, and V1 =
−1
2
and V2 =
1
1
are their
associated eigenvectors. Let V =
(V1 V2
)and Λ =
1 0
0 4
. Since MV = V Λ, we
have M = V ΛV −1. Thus Mn = V ΛnV −1 =
−1 1
2 1
1 0
0 4n
(1
3)
−1 1
2 1
=
(13)
1 + 2(4n) 4n − 1
2(4n)− 2 4n + 2
. Then by Equation 7–4 we have
Xn+1 = (1
3)
1 + 2(4n) 4n − 1
2(4n)− 2 4n + 2
6
0
=
4n+1 + 2
4n+1 − 4
.
(7–5)
It follows that b2n+2,4 = 4n+1 + 2 and b2n+2,2 = 4n+1 − 4. Hence b2n+2,1 = b2n+2,2 + b2n+2,4 =
2(4n+1)− 2. Therefore b2n,4 = 4n + 2, b2n,2 = 4n − 4, and b2n,1 = 2(4n)− 2.
The following lemma will be used in the proof of Lemma 7.2.7.
Lemma 7.2.4. P2n ⊆ P2n+2 for any n ∈ N. Hence P2n : n ∈ N is an increasing
sequence of H(R2) with respect to set inclusion.
Proof. By Lemma 6.2.12, we have
P (2n + 2) = P (2n)⋃
(P (2n) + β2n+1)⋃
(P (2n) + β2n+2)
for any n ∈ N. It follows that, for any q ∈ P (2n), we have q + β2n+1
⋃q + β2n+2 ⊂ P (2n +
2). As shown in Figure 6-5 of Chapter 6, the Voronoi cell of q ∈ P (2n) is contained in the
118
union of the Voronoi cells of the lattice points of L2n+2 in the set q + β2n+1
⋃q + β2n+2.
Thus P2n ⊆ P2n+2.
To prove Lemma 7.2.7, we need the following lemma.
Lemma 7.2.5. For any k ∈ N, we have P2k ⊆ P.
Proof. By Proposition 7.1.2, P is the limit of the sequence
P2n : n ∈ N
in the space
(H(R2), h). By Theorem 1 on Page 37 through page 38 from Barnsley [4], we have
P =x ∈ R2 : there is a Cauchy sequence
xn ∈ P2n
that converges to x
. For any
x ∈ P2k, let xn = x for any n ≥ k, and let xn = 0 ∈ P2n for any n < k. By Lemma 7.2.4,
P2n : n ∈ N is an increasing sequence of H(R2) with respect to set inclusion. Since
x ∈ P2k, it follows that xn ∈ P2n for any n ≥ k. Obviously the sequence xn : n ∈ N is a
Cauchy sequence which converges to x. Hence x ∈ P . Therefore P2k ⊆ P .
Lemma 7.2.6. Let L be a hexagonal lattice, x ∈ R2, y ∈ L, and let Ny be the set
consisting of the six neighbors of y in the lattice L. If x is not contained in the Voronoi
cell of y, then d(x,y) > d(x,Ny).
Proof. We consider the following two cases.
Case 1: If there exists z ∈ Ny such that x is contained in the Voronoi cell of z, then
d(x, z) < d(x,y) since x is not contained in the Voronoi cell of y. Because z ∈ Ny, we
have d(x, z) ≥ d(x,Ny). Hence d(x,y) > d(x, z) ≥ d(x,Ny).
Case 2: If there does not exist z ∈ Ny such that x is contained in the Voronoi cell of
z, then let Ny = a,b, c,d, e, f and let HN be the hexagon with vertices in the set Ny.
Since x is not contained in the Voronoi cell of any element in the set Ny
⋃ y, the lattice
point x lies outside the hexagon HN . It follows that there is exactly one intersection point
between the line segment connecting x and y and the sides of HN . Let p denote that
intersection point.
If p ∈ Ny, then d(x,y) > d(x,p) and d(x,p) ≥ d(x,Ny). Hence d(x,y) > d(x,Ny).
119
−30 −20 −10 0 10 20 30−30
−20
−10
0
10
20
30 .
y
a b
c
d e
f
x
Figure 7-4. The distance from a point x to a given lattice point y and the distance from xto the six neighbors of y in the lattice.
If p 6∈ Ny, then p lies on exactly one side, say the side connecting the two vertices
a and b, of the hexagon HN . By symmetry, without loss of generality, we assume that
d(x, a) ≤ d(x,b) as shown in Figure 7-4. Consider the triangle with vertices a, x,
and y. The three angles of that triangle with vertices a, x, and y are denoted A, X,
and Y , respectively. Since d(x, a) ≤ d(x,b), we have Y ≤ π6. However the angle A is
bigger than the angle formed by the ray from a to y and the ray from a to b, which is
π3. Hence the angle A is bigger than the angle Y . It follows that d(x,y) > d(x, a). Since
d(x, a) ≥ d(x,Ny), we have d(x,y) > d(x,Ny).
Lemma 7.2.7. For any x ∈ PB, there exists a boundary lattice point y of P (2n) such that
d(x,y) ≤ 32ρ2n+1.
Proof. By Item 1 of Lemma 7.1.1, we have h(P (i), P (i + 2)) ≤ ρi+1 for any i ∈ N. Since Pis the limit of the sequence P (2i) : i ∈ N), we have h(P (2n),P) ≤ ∑∞
i=n h(P (2i), P (2i +
2)). Thus h(P (2n),P) ≤ ∑∞i=n ρ2i+1. If we let j = i−n, then it follows that h(P (2n),P) ≤
∑∞j=0 ρ2j+2n+1 =
∑∞j=0
13j ρ2n+1 = 3
2ρ2n+1. Hence d(P , P (2n)) ≤ 3
2ρ2n+1. Since PB ⊂
P , it follows that d(PB, P (2n)) ≤ 32ρ2n+1. For any x ∈ PB, we have d(x, P (2n)) ≤
d(PB, P (2n)) ≤ 32ρ2n+1. Because P (2n) is a finite set, there exists y ∈ P (2n) such that
d(x, P (2n)) = d(x,y). We claim that y is a boundary lattice point of P (2n). Let Ny
120
be the set consisting of the six neighbors of y in the lattice L2n. Suppose that y is not a
boundary lattice point of P (2n). Then Ny ⊂ P (2n). Since x ∈ PB and since P2n ⊆ P by
Lemma 7.2.5, it follows that x lies outside the Voronoi cell of y. Hence by Lemma 7.2.6,
d(x,y) > d(x,Ny) ≥ d(x, P (2n)), which contradicts that d(x, P (2n)) = d(x,y). Thus y is
a boundary point of P2n and d(x,y) = d(x, P (2n)) ≤ 32ρ2n+1.
To get the fractal dimension of PB, we still need the following lemmas. For any n ∈ Nand x ∈ Ln, let Vn(x) denote the Voronoi cell of x.
Lemma 7.2.8. For any n ∈ N, we have P2n ⊂ P2n−1.
Proof. For any x ∈ P2n, by the definition of P2n, there exists y ∈ P (2n) such that
x ∈ V2n(y). Since P (2n) = P (2n − 1)⋃
(P (2n − 2) + β2n), either y ∈ P (2n − 1) or
y ∈ P (2n− 2) + β2n. In the following, we consider these two cases separately.
Case 1: If y ∈ P (2n − 1), then V2n(y) ⊂ V2n−1(y) as shown in Figure 7-5 (a) where
the Voronoi cell V2n(y) has a horizontal side and the Voronoi cell V2n−1(y) has a vertical
side. Since x ∈ V2n(y), it follows that x ∈ V2n−1(y) ⊆ P2n−1.
−10 −5 0 5 10
−10
−5
0
5
10
y V
2n(y)
V2n−1
(y)
−15 −10 −5 0 5 10 15 20 25−20
−15
−10
−5
0
5
10
15
20
25
y
q
r
t
(a) (b)
Figure 7-5. The containment relation among Voronoi cells of P (2n) and P(2n+1). (a)shows that the (blue) Voronoi cell V2n(y) is contained in the (black) Voronoicell V2n−1(y), where V2n(y) has a horizontal side and V2n−1(y) has a verticalside. (b) shows that y∈ P (2n) is the centroid of the triangle with vertices q∈P(2n −1), r∈ P(2n −1), and t∈ P(2n −1). It also shows that the (blue)Voronoi cell of y∈ L2n is contained in the union of the (black) Voronoi cells ofq∈ L2n−1, r∈ L2n−1, and t∈ L2n−1.
121
Case 2: If y ∈ P (2n−2)+β2n, then there exists y0 ∈ P (2n−2) and i ∈ 1, 2, 3, 4, 5, 6such that y = y0 + β2n,i.
If i ∈ 1, 3, 5, then let q = β2n,i + β2n,2, r = β2n,i + β2n,4, and t = β2n,i + β2n,6.
Assume i = 1 because the proof for i = 3 or i = 5 is completely similar to that for i = 1.
By Table 6-2, 0001 ⊕ 0002 = 0020 ∈ Λ4. By applying Item 4 of Lemma 6.1.8 for 2n − 4
times, it follows that
00...0001⊕ 00...0002 = 00...0020 ∈ Λ2n. (7–6)
By the definition of β2n−1,2, the label of β2n−1,2 is 00...002 ∈ Λ2n−1. By Item 3 of
Lemma 6.1.8, the label of β2n−1,2 in P (2n) is 00...0020 ∈ Λ2n. By Equation 7–6, it
follows that β2n,1 + β2n,2 = β2n−1,2, i.e., q = β2n−1,2. Similarly, by Table 6-2, we have
0001 ⊕ 0004 = 0000 ∈ Λ4 and 0001 ⊕ 0006 = 0010 ∈ Λ4. Then we can show that
β2n,1 + β2n,4 = 0 and β2n,1 + β2n,6 = β2n−1,1, i.e., r = 0 and t = β2n−1,1. Hence
q, r, t ∈ β2n−1,1 and they are mutually next to each other in the lattice L2n−1. Since
q + r + t = 3β2n,i + β2n,2 + β2n,4 + β2n,6 and since β2n,2 + β2n,4 + β2n,6 = 0, we
have β2n,i = 13(q + r + t). Let q = y0 + q, r = y0 + r, and t = y0 + t. Then
13(q + r + t) = y0 + 1
3(q + r + t) = y0 + β2n,i = y. It follows that y is the centroid
of the triangle with vertices q, r, and t. Since y0 ∈ P (2n − 2) and q, r, t ∈ β2n−1,1,
by Lemma 6.2.3, we have q, r, t ∈ P (2n − 1). Because q, r, and t are mutually next
to each other in the lattice L2n−1, the three lattice points q, r, and t are also mutually
next to each other in the lattice L2n−1. Hence y is the centroid of the triangle with
vertices q ∈ L2n−1, r ∈ L2n−1, and t ∈ L2n−1 which are mutually next to each other in
the lattice L2n−1. As shown in Figure 7-5 (b), it follows that V2n(y) is contained in the
union of V2n−1(q), V2n−1(r), and V2n−1(t), where the Voronoi cells of the lattice L2n−1
have a vertical side and the Voronoi cells of the lattice L2n have a horizontal side. Hence
V2n(y) ⊆ P2n−1. Since x ∈ V2n(y), it follows that x ∈ P2n−1.
122
If i ∈ 2, 4, 6, then let q = β2n,i + β2n,1, r = β2n,i + β2n,3, and t = β2n,i + β2n,5.
Similar to the proof in the previous paragraph, we can show that x ∈ P2n−1. Thus
P2n ⊂ P2n−1.
Lemma 7.2.9. For any n ∈ N, we have P2n+1 ⊂ P2n−1.
Proof. For any x ∈ P2n+1, there exists y ∈ P (2n + 1) such that x ∈ V2n+1(y). Since
P (2n + 1) = P (2n)⋃
(P (2n − 1) + β2n+1) and P (2n) = P (2n − 1)⋃
(P (2n − 2) + β2n),
we have P (2n + 1) = P (2n − 1)⋃
(P (2n − 2) + β2n)⋃
(P (2n − 1) + β2n+1). Hence either
y ∈ P (2n− 1) or y ∈ P (2n− 2) + β2n or y ∈ P (2n− 1) + β2n+1.
Case 1: If y ∈ P (2n − 1), then V2n+1(y) ⊂ V2n−1(y) ⊆ P2n−1. It follows that
x ∈ P2n−1.
Case 2: If y ∈ P (2n − 2) + β2n, then by the proof in Case 2 of Lemma 7.2.9, there
exist three lattice points q ∈ L2n−1, r ∈ L2n−1, and t ∈ L2n−1 which are mutually next
to each other in the lattice L2n−1 such that y is the centroid of the triangle with vertices
q, r, and t. It follows that V2n(y) ⊆ P2n−1 as shown in Figure 7-5 (b). Since V2n(y)
and V2n+1(y) have the same centroid and since the length of one side of the Voronoi cell
V2n(y) is√
3 times that of the Voronoi cell V2n+1(y), we have V2n+1(y) ⊂ V2n(y). Hence
V2n+1(y) ⊂ P2n−1 and x ∈ P2n−1.
Case 3: If y ∈ P (2n − 1) + β2n+1, then there exists y0 ∈ P (2n − 1) and i ∈1, 2, 3, 4, 5, 6 such that y = y0 + β2n+1,i. Since the length of the vector β2n+1,i is
ρ2n+1 = 13ρ2n−1, as shown in Figure 7-6, we have V2n+1(y) ⊆ V2n−1(y0). Hence x ∈
V2n−1(y0) ⊆ P2n−1. Therefore P2n+1 ⊂ P2n−1.
Corollary 7.2.10. For any k ∈ N, we have P ⊆ P2k−1.
Proof. Since P is the limit of the sequence
P2n : k ≤ n ∈ N
in the space (H(R2), h).
By Theorem 1 on Page 37 through page 38 from [4], we have
P =x ∈ R2 : there is a Cauchy sequence
xn ∈ P2n : n ≥ k
that converges to x
.
123
−10 −8 −6 −4 −2 0 2 4 6 8 10
−10
−5
0
5
10
y0
y V
2n+1(y)
V2n−1
(y0)
Figure 7-6. The green Voronoi cell V2n+1(y) is contained in the black Voronoi cellV2n−1(y0).
By Lemma 7.2.8, we have P2n ⊂ P2n−1 for any n ≥ k. By Lemma 7.2.9, for any n ≥ k, we
have P2n−1 ⊆ P2k−1. Hence P2n ⊂ P2k−1 for any n ≥ k. For any n ≥ k, since xn ∈ P2n, it
follows that xn ∈ P2k−1. Because P2k−1 is a closed set of R2 and x = limn→∞ xn, it follows
that x ∈ P2k−1. Therefore P ⊆ P2k−1.
For any n ∈ N and i ∈ 1, 2, 3, 4, 5, 6, recall from Chapter 6 that Θn,i = 0000...0i ∈Λn is the label of βn,i where βn,i ∈ βn ⊆ P (n).
Lemma 7.2.11. Let n ∈ N, q be a boundary lattice point of P (2n) which is labeled
q1q2...q2n−2i0 ∈ Λ2n with i ∈ 1, 2, 3, 4, 5, 6, and q be a lattice point of P (2n + 2) which
is labeled q1q2...q2n−2i0i0 ∈ Λ2n+2. If the boundary index of q is 4, then q is a boundary
lattice point of P (2n + 2) with boundary index 4.
Proof. By Lemma 6.2.15, we only need to consider the case i = 1 because any of the other
cases can be proven by rotating the vectors q and q by an angle of kπ3
for some k ∈ N. By
Table 6-2, we have 0010⊕ 0003 = 0001 ∈ Λ4. By applying Item 4 of Lemma 6.1.8 twice, it
follows that 10⊕ 03 = 01 ∈ Λ2, i.e., i0⊕ 03 = 01 ∈ Λ2. By applying Item 4 of Lemma 6.1.8
for 2n− 2 times, i.e., by attaching 2n− 2 zeros to the left side on each term of the previous
equation, we have
00...0i0⊕ 00...003 = 00...001 ∈ Λ2n.
124
By Item 1 of Lemma 6.1.8, we have
q1q2...q2n−2i0 = q1q2...q2n−200⊕ 00...0i0.
Hence
q1q2...q2n−2i0⊕Θ2n,3 = q1q2...q2n−200⊕ 00...0i0⊕ 00...003
= q1q2...q2n−200⊕ 00...001
= q1q2...q2n−201 ∈ Λ2n.
(7–7)
Similarly it follows from 0010 ⊕ 0004 = 0006 ∈ Λ4 that q1q2...q2n−210 ⊕ Θ2n,4 =
q1q2...q2n−206 ∈ Λ2n. Since the boundary index of q is 4 and since q1q2...q2n−210 ⊕00...000i ∈ Λ2n for i = 3, 4, we have q1q2...q2n−210⊕ 00...000i 6∈ Λ2n for each i ∈ 1, 2, 5, 6.
Similar to the previous computation, we can show that
q1q2...q2n−21010⊕Θ2n+2,3 = q1q2...q2n−21001 ∈ Λ2n+2
and
q1q2...q2n−21010⊕Θ2n+2,4 = q1q2...q2n−21006 ∈ Λ2n+2.
In the following, we show that q1q2...q2n−21010⊕Θ2n+2,j 6∈ Λ2n+2 for each j ∈ 1, 2, 5, 6.By Table 6-2, we have 0010⊕ 0001 = 0105 ∈ Λ4. By Item 4 of Lemma 6.1.8 for 2n− 2
times, it follows that 00...00i0⊕ 00...0001 = 00...0105 ∈ Λ2n+2. Hence
q1q2...q2n−2i0i0⊕Θ2n+2,1 = q1q2...q2n−21010⊕ 00...00001
= q1q2...q2n−21000⊕ 00...00010⊕ 00...00001
= q1q2...q2n−21000⊕ 00...00105.
(7–8)
Since q1q2...q2n−210 ⊕ 00...001 = q1q2...q2n−210 ⊕ Θ2n,1 6∈ Λ2n, by Item 4 and Item 2 of
Lemma 6.1.8, we have q1q2...q2n−21000 ⊕ 00...00105 6∈ Λ2n+2. By Equation 7–8, it follows
that q1q2...q2n−2i0i0⊕Θ2n+2,1 6∈ Λ2n+2.
125
By Table 6-2, we have 0010 ⊕ 0002 = 0104 ∈ Λ4. Hence we can show that
q1q2...q2n−2i0i0⊕Θ2n+2,2 6∈ Λ2n+2 using a similar proof as in the previous paragraph.
Now consider q1q2...q2n−2i0i0 ⊕ Θ2n+2,5. By Table 6-2, we have 0010 ⊕ 0005 = 0603 ∈Λ4. By applying Item 4 of Lemma 6.1.8 for 2n − 2 times, it follows that 00...0010 ⊕00...0005 = 00...0603 ∈ Λ2n+2. Hence
q1q2...q2n−2i0i0⊕Θ2n+2,5 = q1q2...q2n−21010⊕ 00...00005
= q1q2...q2n−21000⊕ 00...00010⊕ 00...00005
= q1q2...q2n−21000⊕ 00...00603.
(7–9)
Since q1q2...q2n−210 ⊕ 00...006 = q1q2...q2n−210 ⊕ Θ2n,6 6∈ Λ2n, by Item 4 and Item 2 of
Lemma 6.1.8, we have q1q2...q2n−21000 ⊕ 00...00603 6∈ Λ2n+2. By Equation 7–9, it follows
that q1q2...q2n−2i0i0 ⊕ Θ2n+2,5 6∈ Λ2n+2. Similarly we can show that q1q2...q2n−2i0i0 ⊕Θ2n+2,6 6∈ Λ2n+2.
We have shown that q1q2...q2n−2i0i0⊕Θ2n+2,j ∈ Λ2n+2 for j = 3, 4, and q1q2...q2n−2i0i0⊕Θ2n+2,j 6∈ Λ2n+2 for any j ∈ 1, 2, 5, 6. Since q1q2...q2n−2i0i0 is the label of q, it follows
that the boundary index of q is 4.
Lemma 7.2.12. If q is a boundary lattice point of P (2n) and q ∈ L2n−1, then q is also a
boundary lattice point of P (2n− 1).
Proof. Let q1q2...q2n ∈ Λ2n be the label of q. We claim that q2n = 0. Suppose q2n 6= 0.
Then q2n = i for some i ∈ 1, 2, 3, 4, 5, 6. Recall that the lattice point labeled 00...0i ∈ Λ2n
is β2n,i which is an element of β2n. Let r be a lattice point of P (2n − 1) which is labeled
q1q2...q2n−1. By Item 1 of Lemma 6.1.8, we have q1q2...q2n−1q2n = q1q2...q2n−10 ⊕ 00...0q2n.
It follows that q = r + β2n,i. Hence β2n,i = q− r. Because q ∈ L2n−1 and r ∈ P (2n− 1) ⊂L2n−1, it follows that q − r ∈ L2n−1, i.e., β2n,i ∈ L2n−1. Since β2n,i ∈ β2n, it follows that
β2n,i ∈ L2n−1
⋂β2n = ∅ by Lemma 6.1.3, which is a contradiction. Thus q2n = 0. It follows
that q ∈ P (2n− 1).
126
Now we show that q is also a boundary lattice point of P (2n− 1). Suppose q is not a
boundary lattice point of P (2n − 1). Then q + β2n−1 ⊆ P (2n − 1). Let β2n−1,7 = β2n−1,1,
and let uj = q + β2n−1,j for j = 1, 2, ..., 7. Also let tj = 13(q + uj + uj+1) for each
j ∈ 1, 2, 3, 4, 5, 6. Obviously
tj = q +1
3(β2n−1,j + β2n−1,j+1) = q +
1
3β2n−2,j = q + β2n,j.
For each j ∈ 1, 2, 3, 4, 5, 6, since q, uj, and uj+1 are three lattice points of P (2n − 1)
which are next to each other in the lattice L2n−1, by Lemma 6.2.7, we have tj ∈ P (2n),
i.e., q + β2n,j ∈ P (2n). Thus q + β2n ⊆ P (2n). It follows that q is not a boundary lattice
point of P (2n), which contradicts the given condition of this Lemma. Thus q is also a
boundary lattice point of P (2n− 1).
Lemma 7.2.13. Let 1 < n ∈ N and let q be a lattice point of P (2n) whose label is
q1q2...q2n ∈ Λ2n and q be a boundary lattice point of P (2n− 2) whose label is q1q2...q2n−2 ∈Λ2n−2. If the boundary index of q is 1 and if q is a boundary lattice point of P (2n), then
the boundary index of q is 1.
Proof. Since q is a boundary lattice point of P (2n) and the label of q is q1q2...q2n−2, by
Lemma 6.2.14, either q2n−3 6= 0 or q2n−2 6= 0. Because the boundary index of q is 1, by
Lemma 6.2.16, q2n−3 = 0 and q2n−2 6= 0. Because q1q2...q2n ∈ Λ2n and q2n−2 6= 0, by Item 1
of Corollary 6.1.7, we have q2n−1 = 0. Since q is a boundary lattice point of P (2n), by
Lemma 6.2.14, it follows that q2n 6= 0. Hence, by Lemma 6.2.16, the boundary index of q
is 1.
Lemma 7.2.14. Let n ∈ N, q ∈ P (2n) and let q1q2...q2n be the label of q. If q is a
boundary lattice point of P (2n) with boundary index 2 or 4, then q2i = 0 and q2i−1 6= 0 for
any i satisfying 1 ≤ i ≤ n.
Proof. If n = 1, then, by the plot of the lattice points of P (2), this lemma is obviously
true. Now assume this lemma is true for any n < k for some k ∈ N. When n = k, let
127
q be the lattice points of P (2k − 2) whose label is q1q2...q2k−2. Since q is a boundary
lattice point of P (2k), by Lemma 6.2.14, either q2k−1 6= 0 or q2k 6= 0, and q is a boundary
lattice point of P (2k − 2). If q2k 6= 0, then by Lemma 6.2.16, the boundary index of q
is 1 which contradicts the given condition that the boundary index of q is either 2 or 4.
Thus q2k−1 6= 0 and q2k = 0. By Lemma 7.2.13, if the boundary index of q is 1, then
the boundary index of q is also 1, which again contradicts the given condition that the
boundary index of q is either 2 or 4. Thus the boundary index of q is either 2 or 4. By
the assumption of this induction, we have q2i = 0 and q2i−1 6= 0 for any i satisfying
1 ≤ i ≤ k − 1. Thus this lemma is true when n = k. By induction, it is true for any
n ∈ N.
Recall that, for any n ∈ N, B2n,4 denotes the set consisting of all boundary lattice
points of P (2n) each having boundary index 4. For any q ∈ B2n,4, let q1q2...q2n ∈ Λ2n
be the label of q. Since q has boundary index 4, by Lemma 7.2.14, we have q2n = 0 and
q2n−1 = a for some a ∈ 1, 2, 3, 4, 5, 6. Let
Ψn(q) = q +1
2β2n−1,a.
The following lemma shows that Ψn is a map from B2n,4 to PB, where PB is the boundary
of P and defined at the beginning of this chapter.
Lemma 7.2.15. For any n ∈ N and q ∈ B2n,4, we have Ψn(q) ∈ PB.
Proof. Since q ∈ B2n,4 ⊂ P (2n), let the label of q be q1q2...q2n ∈ Λ2n. Then Ψn(q) =
q + 12β2n−1,a where a = q2n−1. Since q ∈ B2n,4, the boundary index of q is 4. By
Lemma 7.2.14, we have q2i = 0 and q2i−1 6= 0 for each i satisfying 1 ≤ i ≤ n. Because
q2n = 0, we have q ∈ P (2n − 1). Since P (2n − 1) + β2n+1 ⊂ P (2n + 1), it follows
that q + β2n+1,a ∈ P (2n − 1) + β2n+1 ⊂ P (2n + 1). By induction, it is easy to show that
q+∑k
i=0 β2n+2i+1,a ∈ P (2n+2k+1) for each k ≥ 0. Since P (2n+2k+1) ⊂ P (2n+2k+2) ⊂P2n+2k+2, we have q +
∑ki=0 β2n+2i+1,a ∈ P2n+2k+2 for each k ∈ N. By Lemma 7.2.5, we
have P2n+2k+2 ⊆ P . It follows that q +∑k
i=0 β2n+2i+1,a ∈ P for each k ∈ N.
128
Since β2n+2i+1,a = 13i β2n+1,a for each i ≥ 0, the sequence
∑ki=0 β2n+2i+1,a : k ∈ N
is
a Cauchy sequence in the space (R2, d). Hence the limit of this sequence exists. As usual,
the limit of the sequence∑k
i=0 β2n+2i+1,a : k ∈ N
is denoted∑∞
i=0 β2n+2i+1,a. Obviously
we have
∞∑i=0
β2n+2i+1,a =∞∑i=0
1
3iβ2n+1,a =
3
2β2n+1,a =
1
2β2n−1,a. (7–10)
It follows that q +∑∞
i=0 β2n+2i+1,a = q + 12β2n−1,a. Hence q + 1
2β2n−1,a is a limit point of
the sequenceq +
∑ki=0 β2n+2i+1,a : k ∈ N
. Since q +
∑ki=0 β2n+2i+1,a ∈ P for each k ∈ N
as shown in the previous paragraph and since P is a closed subset of R2, it follows that
q + 12β2n−1,a ∈ P , i.e., Ψn(q) ∈ P .
We claim that Ψn(q) ∈ PB. Suppose Ψn(q) 6∈ PB. Then there exists an open
δ-neighborhood Nδ(Ψn(q)) = y ∈ R2 : d(Ψn(q),y) < δ for some δ > 0 such that
Nδ(Ψn(q)) ⊂ P . Because δ > 0 and limn→∞ ρn = 0, there exists c ∈ N such that ρ2c+1 < δ2.
Since q ∈ B2n,4 and a = q2n−1, it follows from Lemma 7.2.11 that q +∑k
i=0 β2n+2i+1,a is a
boundary lattice point of P (2n + 2k + 2). To make the following expressions simpler, let
sk = q +∑k
i=0 β2n+2i+1,a. Because sk = q +∑k
i=0 β2n+2i+1,a ∈ P (2n + 2k + 1) as shown
in the previous paragraph and because sk is a boundary lattice point of P (2n + 2k + 2),
it follows from Lemma 7.2.12 that sk is also a boundary lattice point of P (2n + 2k + 1).
Hence there exists l ∈ 1, 2, 3, 4, 5, 6 such that sk + β2n+2k+1,l 6∈ P (2n + 2k + 1). Since
sk + β2n+2k+1,l ∈ L2n+2k+1, it follows that sk + β2n+2k+1,l 6∈ P2n+2k+1. By Corollary 7.2.10,
we have P ⊆ P2n+2k+1. It follows that sk + β2n+2k+1,l 6∈ P . Since Ψn(q) = q + 12β2n−1,a, we
have
Ψn(q)− sc = (q +1
2β2n−1,a)− (q +
c∑i=0
β2n+2i+1,a) =1
2β2n−1,a −
c∑i=0
1
3iβ2n+1,a.
It follows that d(sc, Ψn(q)) = ‖Ψn(q)−sc‖ = ‖12β2n−1,a−
∑ci=0
13i β2n+1,a‖ = ‖1
2β2n−1,a− 3
2(1−
13c+1 )β2n+1,a‖ = ‖1
2β2n−1,a− 1
2(1− 1
3c+1 )β2n−1,a‖ = ‖ 12(3c+1)
β2n−1,a‖ = 12(3c+1)
ρ2n−1 = 12ρ2n+2c+1.
129
By the triangle inequality, we have
d(sc + β2n+2c+1,l, Ψn(q)) ≤ d(sc + β2n+2c+1,l, sc) + d(sc, Ψn(q)).
Since d(sc + β2n+2c+1,l, sc) = ‖β2n+2c+1,l‖ = ρ2n+2c+1 and since d(sc, Ψn(q)) = 12ρ2n+2c+1,
it follows that d(sc + β2n+2c+1,l, Ψn(q)) ≤ ρ2n+2c+1 + 12ρ2n+2c+1 < 2ρ2n+2c+1 = 2
3n ρ2c+1 ≤2ρ2c+1 < δ. Hence sc+β2n+2c+1,l ∈ Nδ(Ψn(q)) ⊆ P . This contradicts that sk+β2n+2k+1,l 6∈ Pfor any k ∈ N. Therefore Ψn(q) ∈ PB.
The next three lemmas will be applied in the proof of Lemma 7.2.20.
Lemma 7.2.16. Let n ∈ N and x,y ∈ P (2n). If both x and y are boundary lattice points
of P (2n) each having boundary index 2 or 4, then d(x,y) 6= ρ2n and d(x,y) 6= 2ρ2n.
Proof. Let x1x2...x2n−1x2n ∈ Λ2n and y1y2...y2n−1y2n ∈ Λ2n be the labels of x and y,
respectively. Since the boundary index of x is either 2 or 4, by Lemma 7.2.14, we have
x2n = 0. It follows that x ∈ P (2n− 1) ⊂ L2n−1. Similarly we have y ∈ L2n−1.
If d(x,y) = ρ2n, then x is next to y in the lattice L2n. It follows that x = y + z for
some z ∈ β2n. Since z = x − y and x,y ∈ L2n−1, we have z ∈ L2n−1
⋂β2n. However,
by Lemma 6.1.3, we have L2n−1
⋂β2n = ∅. It follows that z ∈ ∅ which is a contradiction.
Thus d(x,y) 6= ρ2n.
If d(x,y) = 2ρ2n, then ‖x− y‖ = 2ρ2n. Since L2n is a hexagonal lattice, it follows that
x − y = 2β2n,i for some i ∈ 1, 2, 3, 4, 5, 6. Hence x = y + 2β2n,i. By Table 6-2, we have
000i⊕000i = 0j0k ∈ Λ4 and k 6= 0. By applying Item 4 of Lemma 6.1.8 for 2n−4 times, it
follows that 00...000i⊕ 00...000i = 00...0j0k ∈ Λ2n. Hence β2n,i + β2n,i = β2n−2,j + β2n,k, i.e.,
2β2n,i = β2n−2,j + β2n,k. Since x = y + 2β2n,i, it follows that x = y + β2n−2,j + β2n,k. Hence
β2n,k = x − y − β2n−2,j. Because x,y ∈ L2n−1 and β2n−2,j ∈ L2n−2 ⊂ L2n−1, it follows that
β2n,k = x − y − β2n−2,j ∈ L2n−1. Since β2n,k ∈ β2n, it follows that β2n,k ∈ L2n−1
⋂β2n = ∅,
which is a contradiction. Thus d(x,y) 6= 2ρ2n.
130
Lemma 7.2.17. Let L be a hexagonal lattice with generators a and b such that ‖a‖ =
‖b‖ = r for some 0 < r ∈ R, and the angle between a and b is 2π3. If x ∈ L and
0 < ‖x‖ < r√
7, then either ‖x‖ = r or ‖x‖ = r√
3 or ‖x‖ = 2r.
Proof. Since x ∈ L and L is a hexagonal lattice with generators a and b, there exist
m,n ∈ Z such that x = ma + nb. Because ‖a‖ = ‖b‖ = r and the angle between a and b
is 2π3
, we have
‖x‖ = ‖ma + nb‖
=
√m2‖a‖2 + n2‖b‖2 + 2mn‖a‖ · ‖b‖cos(2π
3)
= r√
m2 + n2 −mn.
(7–11)
If m < 0 or n < 0, we can rotate x by 2π3
or 4π3
to obtain y = m2a + n2b so that m2 ≥ 0
and n2 ≥ 0. Without loss of generality, we assume that m,n ≥ 0. By the symmetry of m
and n in Equation 7–11, we also assume that m ≥ n.
If n = 0 or n = m, then ‖x‖ = rm. Since 0 < ‖x‖ < r√
7, it follows that 0 < m <√
7.
Because m ∈ Z, it follows that m = 1 or m = 2. Hence ‖x‖ = r or ‖x‖ = 2r.
Otherwise we have n ≥ 1 and m− n ≥ 1. Then we consider the following two cases.
Case 1: If m−n = 1, then m = n+1 and ‖x‖ = r√
n2 + n + 1. Since 0 < ‖x‖ < r√
7,
it follows that n2 + n + 1 < 7. Hence (n + 3)(n − 2) < 0. Because n > 0, it follows that
n− 2 < 0. Hence 1 ≤ n < 2. It follows that n = 1. Thus ‖x‖ = r√
3.
Case 2: If m − n ≥ 2, then m ≥ n + 2 and ‖x‖ = r√
m(m− n) + n2 ≥r√
(n + 2)2 + n2 = r√
n2 + 2n + 4. Since n ≥ 1, it follows that ‖x‖ = r√
7 which
contradicts the condition that 0 < ‖x‖ < r√
7.
Lemma 7.2.18. For any n ∈ N and x,y ∈ B2n,2
⋃B2n,4 with d(x,y) =√
3ρ2n, there exists
z ∈ P (2n) such that z is next to both x and y in the lattice L2n.
Proof. If n = 1, this lemma is true by the plot of lattice points of P (2) (Figure 6-2).
Assume this lemma is true for any n ≤ k for some k ∈ N. Now we show that it is also
131
true when n = k + 1. Let the labels of x and y be x1x2...x2kx2k+1x2k+2 ∈ Λ2k+2 and
y1y2...y2ky2k+1y2k+2 ∈ Λ2k+2, respectively. Since x,y ∈ B2k+2,2
⋃B2k+2,4, by Lemma 7.2.14,
we have x2i = y2i = 0, x2i−1 6= 0, and y2i−1 6= 0 for any i satisfying 1 ≤ i ≤ k + 1.
Let q and r be lattice points of P (2k) whose labels are x1x2...x2k−1x2k ∈ Λ2k and
y1y2...y2k−1y2k ∈ Λ2k, respectively. Also let a = x2k+1 and b = y2k+1. By Item 1 of
Lemma 6.1.8, we have x1x2....x2k−1x2kx2k+1x2k+2 = x1x2...x2k−1x2k00 ⊕ 00...00x2k+1x2k+2.
Since x2k+1x2k+2 = a0, it follows that x = q + β2k+1,a. Similarly we have y = r + β2k+1,b.
Now we consider the following two cases.
Case 1: If q = r, then y = q + β2k+1,b. Hence x,y ∈ q + β2k+1. By Lemma 6.2.12,
we have q + β2k+1
⋃q + β2k+2 ∈ P (2k + 2). In Figure 6-5 of Chapter 6, the Voronoi cells
of lattice points in q + β2k+1 and q + β2k+2 are in blue and in green, respectively. Since
d(x,y) =√
3ρ2n =√
3ρ2k+2, as shown in that figure, there exists z ∈ q + β2k+2 such that z
is next to both x and y in the lattice L2k+2.
Case 2: If q 6= r, then we claim that d(q, r) =√
3ρ2k. By the triangle inequality, we
have
d(q, r) ≤ d(q,q + β2k+1,a) + d(q + β2k+1,a, r + β2k+1,b) + d(r + β2k+1,b, r)
= ‖(q + β2k+1,a)− q‖+ d(x,y) + ‖r− (r + β2k+1,b)‖
= ‖β2k+1,a‖+ d(x,y) + ‖β2k+1,b‖
= ρ2k+1 + d(x,y) + ρ2k+1
= d(x,y) + 2ρ2k+1.
(7–12)
Hence d(x,y) ≥ d(q, r) − 2ρ2k+1. Suppose d(q, r) ≥ 2ρ2k. Then d(x,y) ≥ 2ρ2k − 2ρ2k+1 =
6ρ2k+2 − 2√
3ρ2k+2 = 2√
3(√
3 − 1)ρ2k+2 >√
3ρ2k+2 which contradicts the condition that
d(x,y) =√
3ρ2n. Thus d(q, r) < 2ρ2k. Since q 6= r and since d(q, r) < 2ρ2k <√
7ρ2k,
by Lemma 7.2.17, we have either d(q, r) = ρ2k or d(q, r) =√
3ρ2k. Because x,y ∈B2k+2,2
⋃B2k+2,4, by Lemma 6.2.14, q and r are boundary lattice points of P (2k). Since
132
x2k = 0, by Lemma 6.2.16, the boundary index of q is either 2 or 4. So is the boundary
index of r. Hence, by Lemma 7.2.16, d(q, r) 6= ρ2k. Thus d(q, r) =√
3ρ2k.
−25 −20 −15 −10 −5 0 5−15
−10
−5
0
5
10
15
Q R
S
−25 −20 −15 −10 −5 0 5−15
−10
−5
0
5
10
15
Q R
S
X Y Z
(a) (b)
Figure 7-7. For any two hexagons of P(2k+2) each having boundary index 2 or 4 such thatthe distance between them is ρ2k+2, there exists a hexagon of P(2k+2) that isnext to both of them, where Q, R and S are hexagons of P (2k). (a) Thehexagon S is next to Q and R, and its centroid lies below the line connectingthe centroids of Q and R. (b) shows that, for any two blue hexagons ofP(2k+2) (each having boundary index 2 or 4) such that the distance betweenthem is ρ2k+2, there exists a hexagon of P(2k+2) that is next to these two bluehexagons.
Since d(q, r) =√
3ρ2k, by the induction assumption, there exists s ∈ P (2k) such
that s is next to both q and r in the lattice L2k. Because ‖r − q‖ = d(q, r) =√
3ρ2k and
r−q ∈ L2k, we have r−q = β2k−1,j for some j ∈ 1, 2, 3, 4, 5, 6. By Lemma 6.2.15, we can
rotate the vectors q, r, and s by an angle of − jπ3
or − jπ3
+ π so that the line connecting
q and r is horizontal. Hence we assume that the line connecting q and r is horizontal and
the Voronoi cells Q, R, and S of q, r, and s are as shown in Figure 7-7 (a), where the
Voronoi cells Q and R are in black and the Voronoi cell S is dashed. Since q, r, s ∈ P (2k),
by Lemma 6.2.12, it follows that all blue, green and red hexagons are Voronoi cells of
P (2k + 2). Since x = q + β2k+1,a and y = r + β2k+1,b, the Voronoi cell of x ∈ P (2k + 2)
is one of the blue hexagons overlapping Q and the Voronoi cell of y ∈ P (2k + 2) is one of
the blue hexagons overlapping R. Because d(x,y) =√
3ρ2k+2, the Voronoi cells X and Y
of x ∈ P (2k + 2) and y ∈ P (2k + 2) must be as shown in Figure 7-7 (b). Let z be a lattice
133
point of P (2k + 2) whose Voronoi cell Z is shown in Figure 7-7 (b). Then z is next to both
x and y.
To simplify the proof of Lemma 7.2.20, we need the following lemma.
Lemma 7.2.19. Let n ∈ N and x,y ∈ B2n,4. If R is the rotation of R2 by an angle of kπ3
about the origin for some k ∈ Z, then R(x), R(y) ∈ B2n,4 and d(Ψn(R(x)), Ψn(R(y))) =
d(Ψn(x), Ψn(y)).
Proof. Since x,y ∈ B2n,4, by Lemma 6.2.15, we have R(x), R(y) ∈ B2n,4.
Let x1x2...x2n−1x2n ∈ Λ2n, y1y2...y2n−1y2n ∈ Λ2n, x1x2...x2n−1x2n ∈ Λ2n and
y1y2...y2n−1y2n ∈ Λ2n be the labels of x, y, R(x) and R(y), respectively. Also let a = x2n−1,
b = y2n−1, c = x2n−1, and d = y2n−1. By the definition of Ψn, we have Ψn(x) = x+ 12β2n−1,a,
Ψn(y) = y + 12β2n−1,b, Ψn(R(x)) = R(x) + 1
2β2n−1,c, Ψn(R(y)) = R(y) + 1
2β2n−1,d.
Since x1x2...x2n−1x2n is the label of x, we have x =∑2n
i=1 βi,xi. Hence
R(x) =2n∑i=1
R(βi,xi). (7–13)
For any i satisfying 1 ≤ i < 2n, by Item 1 of Corollary 6.1.7, we have either xi = 0 or
xi+1 = 0. Hence either βi,xi= 0 or βi+1,xi+1
= 0. Since R is the rotation of R2 about
the origin, it follows that either R(βi,xi) = 0 or R(βi+1,xi+1
) = 0. For each i satisfying
1 ≤ i ≤ 2n, if βi,xi6= 0, then βi,xi
∈ βi. Because R is the rotation of R2 by an angle of
kπ3
about the origin for some integer k, it follows that R(βi,xi) ∈ βi. Thus R(βi,xi
) ∈ βi for
each i satisfying 1 ≤ i ≤ 2n. Hence Equation 7–13 is the standard expression of R(x) in
P (2n). Since x1x2...x2n−1x2n ∈ Λ2n is the label of R(x), the Equation
R(x) =2n∑i=1
βi,xi. (7–14)
is also the standard expression of R(x) in P (2n). By Theorem 6.1.6 for the unique
standard expression, it follows from Equation 7–13 and Equation 7–14 that R(βi,xi) = βi,xi
for each i satisfying 1 ≤ i ≤ 2n. Hence R(β2n−1,a) = β2n−1,c. Since Ψn(R(x)) =
134
R(x) + 12β2n−1,c, it follows that
Ψn(R(x)) = R(x) +1
2R(β2n−1,a) = R(x +
1
2β2n−1,a) = R(Ψn(x)).
For a similar reason, we have
Ψn(R(y)) = R(y) +1
2R(β2n−1,b) = R(y +
1
2β2n−1,b) = R(Ψn(y)).
Therefore
d(Ψn(R(x)), Ψn(R(y))) = d(R(Ψn(x)), R(Ψn(y))) = d(Ψn(x), Ψn(y)).
The next lemma will be used to show that Ψn is a one to one map and to obtain the
fractal dimension of PB.
Lemma 7.2.20. For any n ∈ N and x,y ∈ B2n,4 with x 6= y, we have d(Ψn(x), Ψn(y)) >
3ρ2n+1.
Proof. Let x1x2...x2n−1x2n ∈ Λ2n and y1y2...y2n−1y2n ∈ Λ2n be the labels of x and y,
respectively. Since x,y ∈ B2n,4, by Lemma 7.2.14, we have x2i−1 6= 0, y2i−1 6= 0, and
x2i = y2i = 0 for each i = 1, 2, ..., n. By the definition of Ψn, we have Ψn(x) = x + 12β2n−1,a
with a = x2n−1, and Ψn(y) = y + 12β2n−1,b with b = y2n−1. If n = 1, let q = r = 0. If
n > 1, let q and r be the lattice points of P (2n − 2) which are labeled x1x2...x2n−2 and
y1y2...y2n−2, respectively. Since x2n−1 = a and y2n−1 = b, we have x = q + β2n−1,a and
y = r + β2n−1,b. Hence we have
Ψn(x) = x +1
2β2n−1,a = q + β2n−1,a +
1
2β2n−1,a = q +
3
2β2n−1,a (7–15)
and
Ψn(y) = y +1
2β2n−1,b = r + β2n−1,b +
1
2β2n−1,b = r +
3
2β2n−1,b. (7–16)
135
Now we consider the following two cases.
Case 1: If q = r, since x 6= y, we have x2n−1 6= y2n−1, i.e., a 6= b. It follows
that β2n−1,a and β2n−1,b are two different lattice points of the lattice L2n−1. Hence
d(β2n−1,a, β2n−1,a) ≥ ρ2n−1. Since q = r, it follows that d(Ψn(x), Ψn(y)) = d(q+ 32β2n−1,a, r+
32β2n−1,b) = d(3
2β2n−1,a,
32β2n−1,b) = 3
2d(β2n−1,a, β2n−1,b) ≥ 3
2ρ2n−1 = 9
2ρ2n+1 > 3ρ2n+1.
Case 2: If d(q, r) ≥ √7ρ2n−2, then by Equations 7–15 and 7–16, we have
d(Ψn(x), Ψn(y)) = d(q +3
2β2n−1,a, r +
3
2β2n−1,b) (7–17)
By the triangle inequality and Equation 7–17, we have
d(q, r) ≤ d(q,q +3
2β2n−1,a) + d(q +
3
2β2n−1,a, r +
3
2β2n−1,b) + d(r +
3
2β2n−1,b, r)
= ‖(q +3
2β2n−1,a)− q‖+ d(Ψn(x), Ψn(y)) + ‖r− (r +
3
2β2n−1,b)‖
= ‖3
2β2n−1,a‖+ d(Ψn(x), Ψn(y)) + ‖3
2β2n−1,b‖
=3
2ρ2n−1 + d(Ψn(x), Ψn(y)) +
3
2ρ2n−1
= d(Ψn(x), Ψn(y)) + 3ρ2n−1.
(7–18)
It follows from Equation 7–18 that d(Ψn(x), Ψn(y)) ≥ d(q, r) − 3ρ2n−1. Since d(q, r) ≥√
7ρ2n−2, it follows that d(Ψn(x), Ψn(y)) ≥ √7ρ2n−2 − 3ρ2n−1 = 3
√7ρ2n − 3
√3ρ2n. Since
ρ2n =√
3ρ2n+1, it follows that d(Ψn(x), Ψn(y)) ≥ 3√
21ρ2n+1 − 9ρ2n+1 = 3(√
21− 3)ρ2n+1 >
3ρ2n+1.
Case 3: If q 6= r and d(q, r) <√
7ρ2n−2, then 0 < ‖q − r‖ <√
7ρ2n−2. By
Lemma 7.2.17, it follows that either ‖q − r‖ = ρ2n−2 or ‖q − r‖ =√
3ρ2n−2 or ‖q − r‖ =
2ρ2n−2. Hence either d(q, r) = ρ2n−2 or d(q, r) =√
3ρ2n−2 or d(q, r) = 2ρ2n−2.
Since x and y are boundary lattice points of P (2n), by Lemma 6.2.14, q and r are
boundary lattice points of P (2n − 2). Because the label of q is x10x30...x2n−30 ∈ Λ2n−2
and x2n−3 6= 0, by Lemma 6.2.16, the boundary index of q is either 2 or 4. For a similar
reason, the boundary index of r is either 2 or 4 as well. Hence, by Lemma 7.2.16, we
have d(q, r) 6= ρ2n−2 and d(q, r) 6= 2ρ2n−2. Therefore d(q, r) =√
3ρ2n−2. Since q, r ∈
136
B2n−2,2
⋃B2n−2,4, by Lemma 7.2.18, it follows that there exists s ∈ P (2n − 2) such that s
is next to both q and r in the lattice L2n−2. Because q, r ∈ P (2n − 2) ⊂ L2n−2, we have
r − q ∈ L2n−2. Since ‖r − q‖ = d(q, r) =√
3ρ2n−2, it follows that the angle between the
vector r − q and the positive x-axis is jπ3
for some j ∈ Z. After the lattice points x, y, q
and r are rotated by − jπ3
about the origin, the vector r − q is horizontal. If s lies above
the line connecting q and r, then we continue to rotate those lattice points by π about the
origin so that s lies below the line connecting q and r. By Lemma 7.2.19, for any rotation
R of R2 by kπ3
for some k ∈ Z, we have
d(Ψn(R(x)), Ψn(R(y))) = d(Ψn(x), Ψn(y)).
Hence, without loss of generality, we assume that the vector r − q is horizontal and s lies
below the line connecting q and r as shown in Figure 7-8. By Equations 7–15 and 7–16,
we have
d(Ψn(x), Ψn(y)) = d(q +3
2β2n−1,a, r +
3
2β2n−1,b)
= ‖r− q +3
2(β2n−1,b − β2n−1,a)‖.
(7–19)
By the symmetry of q and r in Equation 7–19, without loss of generality, we also assume
that q lies to the left side of r as shown in Figure 7-8. Then the two vectors r − q and
β2n−1,6 have the same direction. Because ‖r − q‖ = d(q, r) =√
3ρ2n−2 = 3ρ2n−1 =
3‖β2n−1,6‖, it follows that
r− q = 3β2n−1,6. (7–20)
Since q, r, s ∈ P (2n− 2), by Lemma 6.2.12, it follows that all blue, green and red hexagons
in Figure 7-8 are Voronoi cells of P (2n). Because x ∈ B2n,4, the boundary index of x is 4.
Referring to Figure 7-8, since x = q + β2n−1,a, it follows that a ∈ 1, 2, 3, 4. Similarly,
because y ∈ B2n,4, we have b ∈ 1, 2, 5, 6. By combining Equations 7–19 and 7–20, we
137
have
d(Ψn(x), Ψn(y)) = ‖3β2n−1,6 +3
2(β2n−1,b − β2n−1,a)‖
=3
2‖2β2n−1,6 + β2n−1,b − β2n−1,a‖.
(7–21)
−25 −20 −15 −10 −5 0 5−15
−10
−5
0
5
10
15
q
q+β2n−1,1
q+β2n−1,2
q+β2n−1,3
q+β2n−1,4
r
r+β2n−1,1
r+β2n−1,2
r+β2n−1,6
r+β2n−1,5
s
Figure 7-8. The lattice points q, r, s∈ P (2n− 2) and the lattice points in q + β2n,r + β2n, and s + β2n, where s is next to both q and r in the lattice L2n−2.
If a = b, then Equations 7–21 becomes
d(Ψn(x), Ψn(y)) =3
2‖2β2n−1,6‖ = 3‖β2n−1,6‖ = 3ρ2n−1 = 9ρ2n+1 > 3ρ2n+1.
If β2n−1,b = −β2n−1,a, then Equations 7–21 becomes
d(Ψn(x), Ψn(y)) =3
2‖2β2n−1,6 − 2β2n−1,a‖ = 3‖β2n−1,6 − β2n−1,a‖. (7–22)
Since β2n−1,6, β2n−1,a ∈ L2n−1, we have β2n−1,6 − β2n−1,a ∈ L2n−1. Because a 6= 6, β2n−1,6 −β2n−1,a 6= 0. Hence 0 6= β2n−1,6−β2n−1,a ∈ L2n−1. It follows that ‖β2n−1,6−β2n−1,a‖ ≥ ρ2n−1.
Then by Equation 7–22 we have d(Ψn(x), Ψn(y)) ≥ 3ρ2n−1 = 9ρ2n+1 > 3ρ2n+1.
If a 6= b and β2n−1,b 6= −β2n−1,a, then the two vectors β2n−1,a and β2n−1,b are not
parallel. Since ‖β2n−1,a‖ = ‖β2n−1,b‖ = ρ2n−1, it follows that ‖β2n−1,a − β2n−1,b‖ < 2ρ2n−1.
However ‖2β2n−1,6‖ = 2ρ2n−1. Hence 2β2n−1,6 6= β2n−1,a − β2n−1,b, i.e., 2β2n−1,6 + β2n−1,b −
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β2n−1,a 6= 0. Because β2n−1,a, β2n−1,b, β2n−1,6 ∈ L2n−1 and L2n−1 is a lattice, we have
2β2n−1,6 +β2n−1,b−β2n−1,a ∈ L2n−1. Thus 0 6= 2β2n−1,6 +β2n−1,b−β2n−1,a ∈ L2n−1. It follows
that
‖2β2n−1,6 + β2n−1,b − β2n−1,a‖ ≥ ρ2n−1.
Therefore, by Equation 7–21, we have d(Ψn(x), Ψn(y)) ≥ 32ρ2n−1 = 9
2ρ2n+1 > 3ρ2n+1.
The next corollary follows directly from Lemma 7.2.20.
Corollary 7.2.21. For any n ∈ N, Ψn is a one to one map from B2n,4 to PB.
Theorem 7.2.22. The fractal dimension of PB is ln 4ln 3
, where PB is the boundary of the
limit of the Pyxis structure.
Proof. For any n ∈ N, let εn = 32ρ2n+1. Since ρ2n+1 = 1
3n ρ1, we have εn = Crn where
C = 3ρ1
2and r = 1
3. By Lemma 7.2.7, PB can be covered by those closed balls with radius
εn and centered at boundary lattice points of P (2n). By Lemma 7.2.3, it follows that
N (PB, εn) ≤ 4n+1 − 4, where N (PB, εn) is the smallest number of closed balls of radius εn
needed to cover PB.
Recall that B2n,4 is the set consisting of boundary lattice points of P (2n) each having
boundary index 4. Let B2n,4 = Ψn(q) : q ∈ B2n,4. By Corollary 7.2.21, the map Ψn is
one to one. It follows that |B2n,4| = |B2n,4|. By Lemma 7.2.3, we have |B2n,4| = 4n + 2.
Hence |B2n,4| = 4n + 2. By Lemma 7.2.20, for any x,y ∈ B2n,4 with x 6= y, we have
d(Ψn(x), Ψn(y)) > 3ρ2n+1 = 2εn. Since |B2n,4| = 4n+2, it follows that N (B2n,4, εn) ≥ 4n+2,
where N (B2n,4, εn) is the smallest number of closed balls of radius εn needed to cover B2n,4.
By Lemma 7.2.15, we have B2n,4 ⊆ PB. It follows that N (B2n,4, εn) ≤ N (PB, εn). Hence
N (PB, εn) ≥ 4n + 2.
We have shown that 4n + 2 ≤ N (PB, εn) ≤ 4n+1 − 4 for any n ∈ N. It follows that
ln(4n + 2)
ln(1/εn)≤ ln(N (PB, εn))
ln(1/εn)≤ ln(4n+1 − 4)
ln(1/εn). (7–23)
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Since εn = Crn = C3n , we have
limn→∞
ln(4n + 2)
ln(1/εn)= lim
n→∞ln(4n + 2)
ln(3n/C)= lim
n→∞n ln 4 + ln(1 + 2(4−n))
n ln 3− ln C=
ln 4
ln 3.
Similarly we can show that limn→∞ln(4n+1−4)
ln(1/εn)= ln 4
ln 3. Hence, by Inequality 7–23, we have
limn→∞
ln(N (PB, εn))
ln(1/εn)=
ln 4
ln 3.
By Theorem 7.2.1, it follows that the fractal dimension of PB is ln 4ln 3
.
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CHAPTER 8SUMMARY OF THIS RESEARCH
We have provided a systematic mathematical theory for computing the DFT on
a lattice and studied the DFT on some important hexagonal array structures such as
the regular hexagonal structure and the Pyxis structure. It is shown that each array of
the type A (type B) RHS is a set of coset representatives of the quotient group of two
hexagonal lattices. Hence the DFT on such an array is amenable to the standard DFT.
We developed an efficient method for computing the DFT on these arrays and have shown
the relation between these arrays and some previously studied hexagonal arrays.
The Pyxis structure consists of a sequence of hexagonal arrays with a scheme for
labeling the lattice points of the arrays. It originated with PYXIS Innovation Inc. and
is applied in a digital earth reference model (Zheng et al. [50]). Developing algebra and
DFT for the Pyxis structure, computing the fractal dimension of the Pyxis structure, and
developing algorithms for the addition of Pyxis labels are research topics proposed by
the Pyxis Innovation Inc. We have provided a recursive definition of the Pyxis structure
and an algebraic method to label the lattice points of the Pyxis structure. Based on the
recursive definition and labeling of the Pyxis structure, we have developed an algorithm
with computational complexity of order n to add any two labels of length n. Also we
have shown that P (n), the nth level of the Pyxis structure, does not tile the underlying
hexagonal lattice by translations by a sublattice for any n > 2. Hence the DFT on P (n)
can not be evaluated using the method in Chapter 2. Furthermore, we have shown that
the limit of the Pyxis structure in the space of non empty compact subsets of the two
dimensional Euclidean space with Hausdorff metric exists, and the fractal dimension of its
boundary is ln 4ln 3
.
The methods for computing the DFT on hexagonal arrays developed in this research
can be applied to many general periodically sampled images. As shown in Wikipedia [45],
the DFT is widely employed in signal processing and related fields to analyze the
141
frequencies contained in a sampled signal, to solve partial differential equations, and
to perform other operations such as convolutions. Since hexagonal grids have some
advantages over the usual square grids in several application domains, our research has
promising applications.
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BIOGRAPHICAL SKETCH
Xiqiang Zheng was born in Jiangxi Province of China. He received his Master of
Science in mathematics from Jiangxi University in January of 1991. From 1991 to 1999,
he taught at Nanchang Institute of Aeronautical Technology, China. Before he came to
University of Florida for his PhD degree, he studied at Bowling Green State University
for one year. He will receive his PhD degree on applied mathematics in December of 2007
and his research interests include applied mathematics, image processing, and pattern
recognition.
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