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2016 The Contact Number Problem in Two and Three Dimensions
Foerster, Melanie
Foerster, M. (2016). The Contact Number Problem in Two and Three Dimensions (Unpublished master's thesis). University of Calgary, Calgary, AB. doi:10.11575/PRISM/28036 http://hdl.handle.net/11023/3302 master thesis
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The Contact Number Problem in Two and Three Dimensions
by
Melanie Foerster
A THESIS SUBMITTED TO THE FACULTY OF GRADUATE STUDIES IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE
GRADUATE PROGRAM IN MATHEMATICS AND STATISTICS
CALGARY, ALBERTA September, 2016
c Melanie Foerster 2016 Abstract
Given a packing of balls in two or three dimensions we want to maximize the number of contacts in the packing. This is referred to as the contact number problem. A packing of balls is totally separable if every two balls can be separated by a hyperplane that is disjoint from the interior of each ball. In this thesis we study the contact number problem for packings of congruent and incongruent balls in two and three dimensions, and we consider the totally separable version of each problem.
ii Acknowledgements
This thesis would not have been possible without the support of many important individuals. Most notably, I would like to extend sincere thanks to Dr. K´arolyBezdek. You have been an incredible, and invaluable, support both personally and academically over the last two years. It has been an honour to work with you. I will be forever grateful for your generosity. Thank you to my friends from coast to coast, especially my officemates Muhammad Khan, Michael Oliwa, and Trevor Song. This journey was made more successful with your help and encouragement. Special thanks to Dr. Robert Dawson and Dr. Dorette Pronk for igniting my love of research while I was studying in Halifax. Your support, and our stimulating mathematical conversations, have encouraged me throughout my studies.
iii Table of Contents
Abstract ii
Acknowledgements iii
Table of Contents iv
List of Figures v
List of Key Symbols vi
1 Introduction 1 1.1 Kissing Number ...... 1 1.2 Contact Number ...... 2 1.3 Motivation from Material Science ...... 3 1.4 Notation ...... 5
2 The Contact Number Problem in Two Dimensions 6 2.1 The Contact Number Problem for Congruent Disks ...... 6 2.1.1 Harborth’s Theorem ...... 6 2.1.2 Harborth’s Theorem for Totally Separable Packings ...... 10 2.1.3 Digital Packings ...... 15 2.2 The Contact Number Problem for Incongruent Disks ...... 16 2.2.1 Koebe-Andreev-Thurston Representation Theorem ...... 16 2.2.2 Totally Separable Incongruent Packings ...... 23
3 The Contact Number Problem in Three Dimensions 28 3.1 The Contact Number Problem for Congruent Balls ...... 28 3.1.1 Voronoi Diagrams ...... 28 3.1.2 The Isoperimetric Inequality ...... 29 3.1.3 Congruent Ball Packings ...... 32 3.1.4 Totally Separable Congruent Ball Packings ...... 42 3.2 The Contact Number Problem for Incongruent Balls ...... 49 3.2.1 Kuperberg-Schramm Theorem ...... 50 3.2.2 Totally Separable Incongruent Ball Packings ...... 57
Bibliography 58
iv List of Figures
1.1 Kissing number ...... 2 1.2 Contact graph ...... 3 1.3 Totally separable packing ...... 4
2.1 Contact graph ...... 7 2.2 Contact graph after polygon deletion ...... 9 2.3 Congruent packing ...... 11 2.4 Contact graph of a totally separable packing ...... 12 2.5 Contact graph of a totally separable packing after polygon deletion . . . . . 13 2.6 Totally separable congruent digital packing ...... 15 2.7 Incongruent packing ...... 16 2.8 Wooden triangles ...... 18 2.9 Simplex in R3 ...... 19 2.10 Wooden triangle vertex converging to π ...... 20 2.11 Decomposition of a polygon by lines ...... 24 2.12 Decomposition of a polygon into convex polygons ...... 24 2.13 Totally separable incongruent packing in a triangle ...... 25 2.14 Totally separable incongruent packing ...... 26 2.15 Totally separable incongruent packing of 7 disks ...... 26 2.16 Totally separable incongruent packing of 10 disks ...... 27
3.1 Voronoi diagrams ...... 29 3.2 Triangle 4o1pq ...... 33 3.3 Plane reflections ...... 34 3.4 Voronoi cells for a packing ...... 36 3.5 Surface volume contribution ...... 40 3.6 Voronoi cells for a totally separable packing ...... 45 3.7 Overlapping expanded balls ...... 48 3.8 Intersection of S(B) and S(C)...... 52 3.9 Definition of width ...... 53 3.10 Width in terms of θ and ρr(B) ...... 53
v List of Key Symbols
Symbol or abbreviation Definition bxc Lower integer of x |E(G)| Number of edges in the graph G G\H Graph H deleted from the graph G bd(x) Boundary of x vol3(x) 3-dimensional volume of x svol2(x) 2-dimensional surface volume of x
vi Chapter 1
Introduction
Discrete geometry is the study of finite sets of points, lines, balls, convex sets, and convex polytopes. In discrete geometry, questions are focused on combinatorial properties. Such questions include: How many regions can a set of lines partition the plane into? How many times does the minimum distance occur in a set of points in the plane? Some problems date back hundreds of years, and many problems are motivated by other areas of mathematics and science; techniques used in discrete geometry are thus of great use to other disciplines [24].
1.1 Kissing Number
In order to discuss the contact number problem we first must describe the kissing number problem. The kissing number, denoted k(d), is the number of non-overlapping unit balls
(of unit radius) that can touch a given unit ball in the d-dimensional Euclidean space Ed. The kissing number problem asks for the largest number of unit balls that can touch a given unit ball without overlap. The problem arose from a 1694 disagreement between Newton and Gregory regarding the number of 3-dimensional unit balls that could touch a central unit ball without overlap [8]. Newton thought the answer was 12, whereas Gregory believed the answer was 13. The question was answered many years later when Newton was proved
1 correct [34].
Figure 1.1: The kissing number in two dimensions, k(2) = 6.
The known values of k(d) are k(2) = 6 (trivial, see Figure 1.1), k(3) = 12 ([34]), k(4) = 24 ([27]), k(8) = 240 ([28]), and k(24) = 196560 ([28]). The kissing number problem is closely related to the more general problem of finding bounds for optimal ball packings.
1.2 Contact Number
A finite packing of balls in Ed is a finite family of non-overlapping balls in Ed. In this thesis, we only consider finite packings, so we will use “packing” to refer to a “finite packing”.
The contact graph of a packing of balls in Ed is a graph whose vertices correspond to the packing elements and whose two vertices are adjacent if and only if the corresponding packing elements touch each other. Figure 1.2 is an example of a packing with its corresponding contact graph. The number of edges in a contact graph is the contact number of the packing. The contact number problem asks for the largest possible contact number c(n) of a contact graph. That is, the maximum number of edges that a contact graph of n non-overlapping balls can have in Ed. The contact number problem is equivalent to Erd˝os’srepeated shortest distance problem, which asks for the largest number of repeated shortest distances among n points in Ed. The planar case was first raised by Erd˝osin 1946 [11]. Reutter conjectured an answer in 1972,
2 Figure 1.2: A packing with its corresponding contact graph. which was proved by Harborth in 1974 [16]. The more general forms of the problem were popularized by Erd˝osand Ulam. As a close combinatorial relative, it is natural to investigate the maximum contact number of totally separable packings of balls in Ed. The notion of totally separable packings is an extension of separable domains, first introduced by Erd˝os.Total separability was introduced in [12] by G. Fejes T´othand L. Fejes T´oth. A packing of balls in Ed is totally separable if every two balls can be separated by a hyperplane in Ed such that it is disjoint from the interior of each ball in the packing. An example of a totally separable packing in the plane is shown in Figure 1.3. The two dimensional version of the contact number problem for totally separable packings will be explored in Sections 2.1.2 and 2.2.2, and in three dimensions in Sections 3.1.4 and 3.2.2.
1.3 Motivation from Material Science
As previously mentioned, discrete geometric techniques are of great use in other disciplines. In particular, the contact number problem has many important applications. Consider balls that are impenetrable and have short-range attractive forces. Packings of these balls are excellent models for materials such as powders, gels, and glasses [18]. The particles in these
3 Figure 1.3: A totally separable packing.
materials are like hard balls that self-assemble due to their attractive forces. This process of self-assembly is of great interest to material scientists, chemists, statistical physicists, and biologists. Of particular interest are colloids, which consist of particles dispersed in a fluid and are kept suspended by thermal interactions [22]. Colloids occur in many different types of matter including glue, milk, and paint. Controlled colloid formation is used to understand self-assembly. Colloidal particles assemble in a way that minimizes the potential energy of the cluster. Two such particles only exert force on each other once they are infinitesimally close, at which point there is strong attraction between them. As a result, the particles stick together, resist moving apart, but also resist moving closer [3, 18]. Thus, colloidal particles experiencing attraction can be considered as being in contact with one another. Since the potential energy of a colloidal cluster at low temperatures is inversely proportional to the number of contacts between the particles [3, 19, 20], the particles are very likely to assemble in packings that maximize the contact number. This has resulted in work by material scientists on the contact number problem [3, 20].
4 1.4 Notation
As we are only examining the contact number problem in two and three dimensions, we will use “disk” to refer to the (closed) packing elements of the two-dimensional case, and “ball” to refer to the (closed) packing elements of the three-dimensional case. Note that we do not use the terms “circle” and “sphere” as they refer to just the boundary of the disk and ball, respectively. Congruent disks and balls are of unit radius. Disks and balls will be denoted by boldface uppercase letters. Families of bodies will be denoted by script letters. Points will be denoted by boldface lowercase letters. We use the standard volume measurement [33] for three-dimensional volume, which will be denoted by vol3(·), and for the surface volume, which will be denoted by svol2(·).
5 Chapter 2
The Contact Number Problem in Two Dimensions
2.1 The Contact Number Problem for Congruent Disks
In this section we consider the contact number problem for congruent disks in the plane. We start by proving the formula for the contact number of a packing of congruent disks, first given by Harborth [16]. Then we prove the formula for the contact number of a packing of totally separable congruent disks, given by Bezdek et al. [6].
2.1.1 Harborth’s Theorem
As with many problems in discrete geometry, the contact number problem originated from Erd˝os. The contact number problem is equivalent to Erd˝os’srepeated shortest distance problem, stated in 1946, that asks for the largest number of repeated shortest distances among n points in Ed [11]. In 1972 Reutter conjectured the contact number of a packing of congruent disks in the plane. Harborth proved Reutter’s conjecture in 1974 [16], which is stated in the following theorem.
Theorem 2.1 (H. Harborth, [16]). Let c(n) denote the maximum number of times the
6 minimum distance can occur among a set of n points in the plane. Then
√ c(n) = b3n − 12n − 3c. (2.1)
A proof of this theorem is given in [30]. We reproduce it here, with necessary details added.
Proof. Let Pn be a packing of n unit disks in the plane. Construct a graph Gn from Pn as follows. Take vertices to be the centres of the disks. Connect two vertices if and only if
the corresponding disks touch each other. Assume Gn has the maximum number of possible
edges. That is, |E(Gn)| = c(n). Every vertex in Gn is adjacent to at least two other vertices
because Gn has c(n) edges. As a result, Gn is two-connected, meaning it remains connected after any vertex is deleted (along with the adjacent edges).
Figure 2.1: A contact graph with b = 10 and b2 = 1, b3 = 5, b4 = 3, b5 = 1.
The outer face of Gn is bounded by a simple closed polygon P . Let b be the number of
vertices of P . Let bd be the number of vertices of P that have degree d in Gn. Since the
degree of every vertex of Gn is at least two, and any vertex of degree greater than five is in
the interior of Gn, then b = b2 + b3 + b4 + b5. See Figure 2.1 for an example of such a contact graph. The internal angle of P at a vertex of degree d is at least (d − 1)π/3 since the degree of each vertex is at least two. When P is a triangle we have equality. The sum of the internal
7 angles is (b−2)π, as is easily seen for any polygon on b vertices. From these angles we obtain
π ((2 − 1)b + (3 − 1)b + (4 − 1)b + (5 − 1)b ) ≤ (b − 2)π 2 3 4 5 3
b2 + 2b3 + 3b4 + 4b5 ≤ 3b − 6. (2.2)
Let fi be the number of internal faces of Gn with i sides. Recall Euler’s polyhedral formula [13] for a graph G is |V (G)| − |E(G)| + |F (G)| = 2. If we disregard the outer face then this becomes |V (G)| − |E(G)| + |F (G)| = 1. Since we are only considering the internal faces of Gn, this implies
n − c(n) + (f3 + f4 + ...) = 1. (2.3)
When we add up the number of sides of every internal face, each edge of P is counted once, and all the other edges of Gn are counted twice. Since faces with i sides generate ifi edges,
b + 2 (c(n) − b)) = 3f3 + 4f4 + ...
2c(n) − b ≥ 3(f3 + f4 + ...)
−b ≥ 3(f3 + f4 + ...) − 2c(n). (2.4)
Combining (2.3) and (2.4),
n − b ≥ 3(f3 + f4 + ...) − 2c(n) + 1 + c(n) − (f3 + f4 + ...)
= 2(f3 + f4 + ...) − c(n) + 1
= 2 (1 − n + c(n)) − c(n) + 1
= c(n) − 2n + 3. (2.5)
Assume that the theorem holds for all integers less than n. Delete from Gn the vertices of P and all edges adjacent to them. Figure 2.2 is the graph that remains after deleting these vertices and edges from the graph in Figure 2.1.
8 Figure 2.2: The graph that remains after deleting the outer polygon and all adjacent edges from the graph in Figure 2.1.
Each vertex in P of degree j has j − 2 adjacent edges in Gn\P , so
c(n) − b − (b3 + 2b4 + 3b5) ≤ c(n − b). (2.6)
Recall that b = b2 + b3 + b4 + b5. Then (2.6) implies
c(n) ≤ c(n − b) + (b2 + b3 + b4 + b5) + (b3 + 2b4 + 3b5)
= c(n − b) + b2 + 2b3 + 3b4 + 4b5
= c(n − b) + 3b − 6 by (2.2)
= 3(n − b) − p12(n − b) − 3 + 3b − 6 by inductive assumption
= 3n − 6 − p12 (c(n) − 2n + 3) − 3 by (2.5)
= 3n − 6 − p12 (c(n) − 2n) + 33. (2.7)
Now, we modify (2.7) as follows.
−c(n) + 3n − 6 ≥ p12 (c(n) − 2n) + 33
(−c(n) + 3(n − 2))2 ≥ 12 (c(n) − 2n) + 33
c(n)2 − 6c(n)(n − 2) + 9n2 − 36n + 36 ≥ 12c(n) − 24n + 33
c(n)2 − 6nc(n) + 9n2 − 12n + 3 ≥ 0
The roots of c(n)2 − 6nc(n) + 9n2 − 12n + 3
9 are
6n ± p(6n)2 − 4(9n2 − 12n + 3) c(n) = √ 2 36n2 − 36n2 + 48n − 12 = 3n ± 2 √ = 3n ± 12n − 3.
Recall that c(n) is the maximum number of times the minimum distance can occur among √ a set of n points in the plane. Since c(n) < 3n then c(n) ≤ 3n − 12n − 3, and since c(n) must be a natural number then
√ c(n) ≤ b3n − 12n − 3c.
Remark 1. To prove the sharpness of the above contact number formula Harborth stated a formula for n balls [16]. Harborth expresses n as
n = 3s2 + 3s + 1 + (s + 1)i + j, where s ≥ 0, 0 ≤ j ≤ s, 0 ≤ i ≤ 5, and s represents the radius of the regular hexagonal arrangement of disks, i represents the number of additional hexagonal sides comprised of disks, and j represents the number of additional individual disks. See Figure 2.3 for an example of such an extremal packing.
This completes the proof of Theorem 2.1.
2.1.2 Harborth’s Theorem for Totally Separable Packings
A packing of disks is totally separable if every two disks can be separated by a line that is disjoint from the interior of each disk in the packing. The concept of total separability was first introduced by G. Fejes T´othand L. Fejes T´othin 1973 [12]. The contact number
10 Figure 2.3: A packing of 23 (s = 2, i = 1, j = 1) congruent disks with its contact graph. The dark grey disks correspond to s = 2, the light grey disks correspond to i = 1, and the white disk corresponds to j = 1. problem for a packing of totally separable disks was studied by Bezdek et al. [6].
Theorem 2.2 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). Let c(n) denote the maximum contact number of a totally separable packing of unit disks. Then
√ c(n) = b2n − 2 nc.
This theorem and its proof are given in [6]. We reproduce the proof here with necessary details added.
Proof. Let Pn be a totally separable packing of n unit disks in the plane which has the largest number of touching pairs of all totally separable packings of n unit disks in the plane, namely c(n). Construct the graph Gn from Pn as follows. Take the vertices to be the centres of the
11 disks. Connect two vertices if and only if the corresponding disks touch each other. Assume
Gn has the maximum number of edges. That is, |E(Gn)| = c(n). Since Gn has c(n) edges,
every vertex in Gn is adjacent to at least two other vertices. As a result, Gn is two connected (that is, it remains connected after any vertex is deleted).
The outer face of Gn is bounded by a simple closed polygon P . Let b denote the number
of vertices of P , and let bd denote the number of vertices of P with degree d in Gn. The
minimum degree of a vertex of P is two and the maximum degree is four, thus b = b2 +b3 +b4. See Figure 2.4 for an example of such a contact graph.
Figure 2.4: A contact graph with b = 12 and b2 = 5, b3 = 6, b4 = 1.
Since Pn is totally separable, then the internal angle of P at a vertex of degree d is at least (d − 1)π/2. When P is a square we have equality. The sum of the internal angles is (b − 2)π since P is a b-sided polygon. From these angles we obtain the following inequality.
π ((2 − 1)b + (3 − 1)b + (4 − 1)b ) ≤ (b − 2)π 2 3 4 2
b2 + 2b3 + 3b4 ≤ 2b − 4 (2.8)
Let fi denote the number of internal faces of Gn with i sides. Since Pn is totally separable, then i ≥ 4. Euler’s polyhedral formula implies
n − c(n) + (f4 + f5 + ...) = 1. (2.9)
When we add up the number of sides of the internal faces of Gn each edge of P is counted
12 once, and all other edges of Gn are counted twice. Since faces with i sides generate ifi edges,
b + 2 (c(n) − b) = 4f4 + 5f5 + ...
2c(n) − b ≥ 4(f4 + f5 + ...)
−b ≥ 4(f4 + f5 + ...) − 2c(n). (2.10)
Combining (2.9) and(2.10),
n − b ≥ 4(f4 + f5 + ...) − 2c(n) + 1 − (f4 + f5 + ...) + c(n)
= 3(f4 + f5 + ...) − c(n) + 1
= 3 (1 + c(n) − n) − c(n) + 1
= 2c(n) − 3n + 4. (2.11)
Assume that the theorem holds for all integers smaller than n. Delete from Gn the vertices of P and all edges adjacent to them. Figure 2.5 is the graph that remains after deleting these vertices and edges from the graph in Figure 2.4.
Figure 2.5: The graph that remains after deleting the outer polygon and all adjacent edges from the graph in Figure 2.4.
Each vertex in P of degree j is adjacent to j − 2 edges in Gn\P , so
c(n) − b − (b3 + 2b4) ≤ c(n − b). (2.12)
13 Recall that b = b2 + b3 + b4. Then (2.12) implies
c(n) ≤ c(n − b) + (b2 + b3 + b4) + (b3 + 2b4)
= c(n − b) + b2 + 2b3 + 3b4
= c(n − b) + 2b − 4 by (2.8) √ = 2(n − b) − 2 n − b + 2b − 4 by inductive assumption √ = 2n − 4 − 2 n − b
= 2n − 4 − 2p2c(n) − 3n + 4. by (2.11) (2.13)
Now, we modify (2.13) as follows.
−c(n) + 2n − 4 ≥ 2p2c(n) − 3n + 4
(−c(n) + 2(n − 2))2 ≥ 4 (2c(n) − 3n + 4)
c(n)2 − 4c(n)(n − 2) + 4n2 − 16n + 16 ≥ 8c(n) − 12n + 16
c(n)2 − 4nc(n) + 4n2 − 4n ≥ 0
The roots of c(n)2 − 4nc(n) + 4n2 − 4n are
4n ± p(4n)2 − 4(4n2 − 4n) c(n) = √ 2 16n2 − 16n2 + 16n = 2n ± 2 √ = 2n ± 2 n. (2.14)
Recall that c(n) is the maximum contact number of a totally separable packing of n disks in √ the plane. Since c(n) < 2n, then (2.14) implies that c(n) ≤ 2n − 2 n. Since c(n) must be a
14 natural number then √ c(n) = b2n − 2 nc, the sharpness of which is determined using digital packings (see Section 2.1.3). This com- pletes the proof of Theorem 2.2.
Figure 2.6: A totally separable packing of congruent disks, also a digital packing, with its underlying contact graph.
2.1.3 Digital Packings
It is important to note the relationship that totally separable packings of congruent disks have to digital packings. Digital packings are a special case of totally separable packings which maximize the number of contacts. Finding all extremal arrangements is a difficult problem. The interested reader is directed to the paper of Harary and Harborth for the details [15], specifically for the proof of sharpness of the formula given in Theorem 2.2. Figure 2.6 is an example of a digital packing in two dimensions. The extension to three-dimensional digital packings holds, and was made by Alonso and Cerf [2].
15 2.2 The Contact Number Problem for Incongruent Disks
In this section we consider the contact number problem for incongruent disks in the plane. We start by proving the (sharp) upper bound for the contact number of a packing of incongruent disks, given by Koebe, Andreev, and Thurston [30]. Then we give an upper bound for the contact number of a packing of totally separable incongruent disks, which has previously not been documented.
2.2.1 Koebe-Andreev-Thurston Representation Theorem
In 1936, Koebe proved that a finite graph is a contact graph of a disk packing if and only if it is planar. It was later rediscovered (independently) by Andreev and Thurston [30], and as a result all three are typically credited. This theorem is stated, and proved, below. An example of a packing of incongruent disks with its contact graph is given in Figure 2.7.
Figure 2.7: A packing of incongruent disks with its underlying contact graph.
Theorem 2.3 (P. Koebe, E. Andreev, and W. Thurston, [30]). Given any planar graph G with vertex set V (G) = {v1, v2,..., vn} and edge set E(G), we can find a packing of n (not
necessarily congruent) disks C = {C1, C2,..., Cn} in the plane with the property that Ci and
Cj touch each other if and only if vivj ∈ E(G) for 1 ≤ i ≤ n.
16 We follow the structure of the proof given in [30]. However, as that proof has multiple sections left without explanation, we use Thurston’s proof, as given by Marden and Rodin in [23], to fill in the gaps.
Proof. Note that it is sufficient to prove Theorem 2.3 for maximal planar graphs. If a planar graph G has any non-triangular faces, add a vertex in each such face, and connect it to all the vertices of the corresponding face. Assume this new graph can be represented by disks and satisfies Theorem 2.3. Now, erase the disks corresponding to the added vertices. The graph obtained is a good representation of the original contact graph.
Let G be a triangulated graph with vertex set V (G) = {v1, v2,..., vn}, edge set E(G), and face set F (G) including the exterior triangle. Recall Euler’s polyhedral formula [13] for a graph G is |V (G)| − |E(G)| + |F (G)| = 2. Since G is a triangulated graph, every face is a triangle. Thus, summing the number of edges of each face (including the exterior triangle) counts each edge twice. That is, 3|F (G)| = 2|E(G)|. Using Euler’s polyhedral formula, we have
3|F (G)| = 2 (|V (G)| + |F (G)| − 2)
= 2|V (G)| + 2|F (G)| − 4
|F (G)| = 2|V (G)| − 4
= 2n − 4. (2.15)
Let r = (r1, r2, . . . , rn) be a vector of n positive real numbers with r1 + r2 + ... + rn = 1. For each face vivjvk of G, consider a triangle cut out of wood whose vertices vi, vj, vk are the centres of three mutually tangent disks of radii ri, rj, and rk respectively. Try to glue the wooden triangles together along their edges in the same way they are connected in G. Let
σ(vi) be the sum of the angles at vi of all the wooden triangles that have vi as a vertex.
The best case is if these triangles fit together perfectly. In this case, for every vertex vi of G not in the exterior triangle, σr(vi) = 2π. Thus, we obtain a good representation of G
17 Figure 2.8: A representation of seven wooden triangles.
by disks of radii r1, r2, . . . , rn. If the triangles do not fit together perfectly, (2.15) implies that
n X σr(vi) = |F (G)|π = (2n − 4)π. i=1
It is important to note that the exterior triangle is not a wooden triangle. Figure 2.8 represents the ideal case in which the wooden triangles fit together perfectly. This also demonstrates the fact that the exterior triangle itself is not a wooden triangle but simply composed of many wooden triangles. Each of these wooden triangles, in addition to the external triangle, contribute π to the above sum.
Let ∆ ⊆ Rn denote the relatively open (n − 1)-dimensional simplex defined by
( n )
X ∆ := r = (r1, r2, . . . , rn) ri > 0 for all i, and ri = 1 , i=1 shown in R3 in Figure 2.9. Let
( n )
X H := x = (x1, x2, . . . , xn) xi = (2n − 4)π . i=1
18 Figure 2.9: The simplex ∆ in R3.
Let f be the continuous function f : ∆ → H, where
f(r) = (σr(v1), σr(v2), . . . , σr(vn)) .
Without loss of generality, we can assume that v1, v2, and v3 are the vertices of the exterior triangle of G. It suffices to show that, for example,
2π 2π 2π x∗ = , , , 2π, 2π, . . . , 2π (2.16) 3 3 3 is in the image of f. This is easy to see, simply by summing the terms of x∗ we have
2π 3 + (n − 3)2π = 2π + 2πn − 6π = (2n − 4)π. 3
Claim 1 (J. Pach and P. Agarwal, [30]). The function f : ∆ → H is a one-to-one mapping.
Proof. Let r and r0 be two distinct points of ∆, and let I be the set of indices i for which
0 ri < ri. It is clear that I 6= ∅ and I 6= {1, 2, . . . , n}.
Consider a wooden triangle vivjvk whose vertices are the centres of three mutually touch- ing disks of radii ri, rj, and rk respectively. Increase ri and decrease, or leave unchanged, rj and rk so that the three corresponding disks remain tangent. Then the angle of the trian- gle vivjvk at the vertex vi decreases. Similarly, increase ri and rj, and decrease, or leave
19 unchanged, rk. Then the sum of the angles at vertex vi and vj will decrease. Thus,
X X σr(vi) > σr0 (vi), (2.17) i∈I i∈I which implies that f(r) 6= f(r0), and completes the proof of the claim.
Figure 2.10: As the radius of the middle circle decreases, the angle at the corresponding wooden triangle vertex converges to π.
Let δ = (δ1, δ2, . . . , δn) be a boundary point of ∆, and let I denote the set of all indices with δi = 0. If r tends to δ then in each wooden triangle that has at least one vertex in
{vi|i ∈ I} the sum of the angles at such a vertex tends to π. This fact is demonstrated for a single vertex in Figure 2.10. The details for multiple vertices are complex, and can be found in Section 6 of [23].
20 Let F (I) denote the set of faces of G with at least one vertex in {vi|i ∈ I}. Then,
X lim σr(vi) = |F (I)|π. (2.18) r→δ i∈I
P Note that we express the sum as a limit because we cannot explicitly express i∈I σr(vi). For any fixed r ∈ ∆ and for any nonempty proper subset I ⊂ {1, 2, . . . , n} there exists a point δ in bd(∆), the boundary of ∆, such that δi = 0 for all i ∈ I, and δi > ri for all i 6∈ I. P If we move r along a line towards δ then by (2.17), i∈I σr(vi) will increase. Combined with (2.18) we obtain that X σr(vi) < |F (I)|π. i∈I That is, the image of f : ∆ → H lies in the (n − 1)-dimensional relatively open bounded convex polytope P ∗ determined by
n X X xi = (2n − 4)π and xi < |F (I)|π, (2.19) i=1 i∈I
where I varies over all nonempty proper subsets of {1, 2, . . . , n}. Since f : ∆ → P ∗ is one-to-one and all accumulation points of f(r), as r tends to bd(∆), lie on bd(P ∗) then we obtain the following claim.
Claim 2 (A. Marden and B. Rodin, Section 7 [23]). The function f : ∆ → P ∗ is a surjective mapping.
Proof. This proof uses elementary topology. The reader is directed to [26] for details on these concepts. Since f : ∆ → P ∗ is continuous and one-to-one, then by Invariance of the Domain, f is a homeomorphism. Note that f(r) → bd(P ∗) as r → bd(∆). Recall from topology that if X and Y are Hausdorff spaces and X is compact and connected, and Y is connected, then if g : X → Y is continuous and open, then g is surjective. Using the one point compactifications, it follows that f : ∆ → P ∗ is surjective.
21 ∗ ∗ ∗ ∗ To complete the proof of Theorem 2.3 we need to show that the point x = (x1, x2, . . . , xn) defined in (2.16) belongs to the polytope P ∗. Clearly,
n X ∗ xi = (2n − 4)π. i=1
Let I be a nonempty proper subset of {1, 2, . . . , n}. If |I| = n − 1 or |I| = n − 2 then every face of G has at least one vertex belonging to {vi|i ∈ I}. That is, |F (I)| = |F (G)| = 2n − 4. In these cases, X 4π x∗ ≤ 2π(n − 3) + < (2n − 4)π = |F (I)|π. i 3 i∈I Finally, the following claim completes the proof of Theorem 2.3.
Claim 3 (A. Marden and B. Rodin, Section 8 [23]). For any subset I ⊆ {1, 2, . . . , n} with 1 ≤ |I| ≤ n − 3, G has more than 2|I| faces that have at least one vertex belonging to
{vi|i ∈ I}.
Proof. The details of the proof are nontrivial and go beyond the scope of this thesis. We direct the reader to Section 8 of [23] for a proof of this claim.
For any such I,
X ∗ xi ≤ 2π|I| < |F (I)|π. i∈I Thus, x∗ satisfies the relations defining P ∗. That is, x∗ ∈ P ∗ = f(∆) as required, which completes the proof of Theorem 2.3.
Remark 2. Recall that for a triangulated graph G, 3|F (G)| = 2|E(G)|. Euler’s polyhedral
22 formula then implies that, for a triangulated graph,
2 |V (G)| − |E(G)| + |E(G)| = 2 3 1 − |E(G)| = 2 − |V (G)| 3 |E(G)| = 3|V (G)| − 6.
Of all planar graphs on n vertices, triangulated graphs have the most number of vertices. Thus, Theorem 2.3 implies that the contact number for a packing of n incongruent disks in the plane is at most 3n − 6, and this is a sharp bound.
2.2.2 Totally Separable Incongruent Packings
Having considered the contact number problem for congruent disks, incongruent disks, and totally separable congruent disks, it is natural to now consider the contact number problem for a packing of totally separable incongruent disks. To obtain an upper bound for the contact number of a totally separable packing of incongruent disks, we will use the following lemma of G. Fejes T´othand L. Fejes T´oth.
Lemma 2.4 (G. Fejes T´othand L. Fejes T´oth, [12]). Decompose a convex polygon Π hav- ing k + 4 sides (k ≥ −1) by a finite number of straight lines into convex polygons. Let
P1,P2,...,Pn be some of these polygons. Then Π can be decomposed into convex polygons
Q1,Q2,...,Qn having numbers of sides s1, s2, . . . , sn such that Pi ⊂ Qi and
s1 + s2 + ... + sn ≤ 4n + k.
Figures 2.11 and 2.12 give an example of the decomposition of a polygon using Lemma 2.4.
Proof. The proof follows from induction on n. For n = 1 the statement is trivial. Suppose the lemma is true for all m > n. We will prove the lemma for m = n.
23 Figure 2.11: The decomposition of the polygon Π by the dashed lines.
Figure 2.12: Note Qi contains Pi from Figure 2.11.
Let l be a line decomposing Π such that on both sides of l there is at least one of the
polygons P1,P2,...,Pm. The line l decomposes Π into two polygons Π1 and Π2. Let Π1 and Π2 have k1 + 4 and k2 + 4 sides, respectively, such that k1 + k1 ≤ k. Let m1 and m2 be the number of polygons P1,P2,...,Pm that are contained in Π1 and Π2, respectively. Since m1 < n and m2 < n, we can apply the lemma to Π1 and Π2 and the polygons P1,P2,...,Pm
contained in them. The polygons Q1,Q2,...,Qm obtained satisfy the lemma for m = n.
24 Corollary 2.5. Let c(n) denote the contact number of a totally separable packing of incon- gruent disks. Then c(n) ≤ 2n − 2.
Proof. Without loss of generality, we can assume that a totally separable packing of incon- gruent disks is contained within a triangle. See Figure 2.13 for such a packing, with Pi indicating each polygon that contains a disk. Then
2c(n) < s1 + s2 + ... + sn
≤ 4n − 1 by Lemma 2.4
c(n) ≤ 2n − 2 since the contact number is a natural number.
Figure 2.13: A packing formed using the method of G. Fejes T´othand L. Fejes T´oth[12].
In the paper [12] in which Lemma 2.4 is given, G. Fejes T´othand L. Fejes T´othcon- jecture that the arrangement of disks given in Figure 2.14 is the densest totally separable arrangement of arbitrary disks. Figures 2.15 and 2.16 give this arrangement for 7 and 10 incongruent disks, respectively. Note that the contact number of each, 9 and 14 respectively, is one more than the contact number for the same number of an arrangement of totally sep-
25 arable congruent disks. See Figure 2.6 for the totally separable arrangement of 10 congruent disks.
Figure 2.14: A totally separable packing of 28 incongruent disks with 46 contacts.
Remark 3. It is important to note that the configuration given in Figure 2.14 (with six small disks surrounding one disk three times as large) may not be the optimal configuration for a totally separable packing of incongruent disks. This configuration is used here because it is conjectured to be the densest totally separable arrangement of arbitrary disks [12]. Indeed, the densest arrangement of congruent disks is the configuration which gives the maximum number of contacts [30].
Figure 2.15: A totally separable packing of 7 incongruent disks with 9 contacts.
26 Figure 2.16: A totally separable packing of 10 incongruent disks with 14 contacts.
27 Chapter 3
The Contact Number Problem in Three Dimensions
3.1 The Contact Number Problem for Congruent Balls
In this section we consider the contact number problem in three dimensions for congruent balls. We first introduce Voronoi diagrams and the isoperimetric inequality, which will be used in the proofs of the main theorems of this section. The first main theorem gives an upper bound for the contact number of a packing of congruent balls, given by Bezdek and Reid [5]. The second main theorem gives an upper bound for the contact number of a packing of totally separable congruent balls, given by Bezdek et al. [6].
3.1.1 Voronoi Diagrams
Voronoi diagrams are an important construction in discrete geometry and are a powerful tool used to solve computational problems. As a result, Voronoi diagrams are of great interest outside of mathematics, particularly in computer science [4], but also in biology, physiology, chemistry, physics, crystallography, meteorology, and geography [32]. Voronoi diagrams date back to the 17th century. Descartes proposed that the solar system
28 was comprised of convex regions with a star at the centre of each region [32]. Dirichlet and Voronoi later formalized this concept, also known as Dirichlet tessellations. A Voronoi diagram is a division of space based on the nearest neighbour property given a set of generating points called sites. Each point in a Voronoi cell is associated with a region of space closest to a site. Every side of a Voronoi cell is the side of another cell, and each vertex of a Voronoi cell is the vertex of another cell. Thus, a Voronoi diagram is a partition of space into convex polytopes. Note that some cells are necessarily unbounded. These are the cells generated by the sites on the boundary of the convex hull of the generating sites because there are points arbitrarily far away but still closest to those sites. See Figure 3.1 for two examples of a Voronoi diagram.
Figure 3.1: Voronoi diagrams for seven and nine sites in the plane, respectively.
Three dimensional Voronoi diagrams are used in the proofs of Theorems 3.2 and 3.11 below. For more details on Voronoi diagrams, the interested reader can refer to [4, 10, 13, 32].
3.1.2 The Isoperimetric Inequality
The isoperimetric problem asks for the largest possible area that can be enclosed within a given perimeter. The problem dates back to, and was solved by, the Greeks [7] who proved that a circle has a larger area than any polygon with the same perimeter. For an in-depth history of this problem, the interested reader is directed to [7]. The isoperimetric problem gives way to the isoperimetric inequality.
29 The isoperimetric inequality is given by
4πA ≤ L2 (3.1)
where A is the area of a closed curve of length L. Following [29], we will show that equality holds if and only if the given curve is a circle. Let C be a simple closed smooth parametric curve. The arc length of C, a well-known result in calculus [1], is given by
b Z s dx2 dy 2 L = + dt. (3.2) dt dt a
The area enclosed by C can be expressed as a line integral
Z Z b dx A = − ydx = − y dt. C a dt
It can be shown [29] that we must choose a parameter to ensure the square root is eliminated from (3.2) and the best choice is t = (2πL)s. Then,
2π 2π Z " # Z dx2 dy 2 ds2 + dt = dt dt dt dt 0 0
L2 = 2π
30 and
2π Z " # dx2 dy 2 dx L2 − 4πA = 2π + + 2y dt dt dt dt 0 2π 2π Z Z " # dx 2 dy 2 = 2π + y dt + 2π − y2 dt. (3.3) dy dt 0 0
Since the first term of (3.3) is clearly non-negative, then we must show that
Z 2π dy 2 Z 2π dt ≥ y2dt. 0 dt 0
Lemma 3.1 (Wirtinger’s Inequality, [31]). Let F : [0, π] → R be a smooth function such that F (0) = F (π) = 0. Then
Z π dF 2 Z π dt ≥ F (t)2dt 0 dt 0
and equality holds if and only if F (t) = D sin t for all t ∈ [0, π], where D is a constant.
It follows from Wirtinger’s Inequality (Lemma 3.1) that
Z 2π dy 2 dt ≥ 0. 0 dt
Thus, both terms of (3.3) are non-negative, so
4πA ≤ L2.
Wirtinger’s Inequality (Lemma 3.1) also implies that C must be a circle for the case of equality since equality holds only if both sides of (3.3) are zero. The isoperimetric inequality has many applications in mathematical physics, analysis,
31 and geometry [29]. For a comprehensive list of works related to the isoperimetric inequality, the interested reader is directed to Notes for Section 6.2 (number 3) in [33]. The isoperimetric inequality is used in the proofs of Theorems 3.2 and 3.11 below.
3.1.3 Congruent Ball Packings
The contact number problem for packings of congruent balls in three dimensions has been studied in depth by Bezdek and Reid in [5]. In this section we will prove the following theorem of Bezdek and Reid which gives an upper bound for the contact number of a packing of unit balls.
Theorem 3.2 (K. Bezdek and S. Reid, [5]). Let c(n) be the contact number of an arbitrary
packing of n ≥ 2 unit balls in E3. Then
2 c(n) < 6n − 0.926n 3 .
In order to prove this theorem we use of two theorems of Hales without proof. The first theorem of Hales (Theorem 3.3) concerns the kissing number problem that caused a debate between Newton and Gregory [8]. The second of Hales’s theorems we will use (Theorem 3.6) gives an upper bound on the ratio of the volume of a unit ball to the volume of its corresponding Voronoi cell.
3 First, let B be the unit ball centred at the origin o of E . Let P := {c1 + B, c2 +
3 B,..., cn + B} denote the packing of n unit balls with centres c1, c2,..., cn in E that has the largest contact number c(n) of all packings of n unit balls in E3. Note that P might not be uniquely determined up to congruence, in which case P is any extremal packing. Let rˆ := 1.58731. This choice ofr ˆ will be explained later.
Theorem 3.3 (T. Hales, [14]). Let B1, B2,..., B14 be elements of a packing of unit balls in
3 E . Assume the balls B2, B3,..., B13 each touch B1. Then the distance between the centres
of B1 and B14 is at least 2.52.
32 This theorem is an extension of the kissing number problem, which asks for the maximum number of unit balls that can touch a given unit ball. There was a great debate between Newton and Gregory regarding this problem. Newton believed the solution was 12, whereas Gregory maintained the solution was 13. Many years later, Newton was proved correct by Sch¨utteand van der Waerden [34]. Note that the arrangement of the 12 balls is not unique. In fact, the centres of the 12 balls can be at the vertices of a regular icosahedron such that none of the 12 balls touch each other [9].
Let oi be the centre of the unit ball Bi, 1 ≤ i ≤ 13. Assume that B1 is tangent to
B2, B3,..., B13 at the points tj ∈ bd(Bj) ∩ bd(B1), 2 ≤ j ≤ 13.
Let α be the angle opposite to the equal sides of the isosceles triangle 4o1pq, as seen
in Figure 3.2, with d(o1, p) = 2 and d(p, q) = d(o1, q) =r ˆ, where d(x, y) is the Euclidean distance between x and y. It is easy to see that
1 cos α = (3.4) rˆ
π with α < 3 .
Figure 3.2: The triangle 4o1pq.
Lemma 3.4 (K. Bezdek and S. Reid, [5]). Let T be the convex hull of the points t2, t3,..., t13. Then the radius of the circumscribed circle of each face of the convex polyhedron T is less than sin α.
Proof. Let F be a face of T with vertices tj, where j ∈ IF ⊂ {2, 3,..., 13}, and let cF be
33 the centre of the circumscribed circle of F . The triangle 4o1cF tj is a right triangle with a
right angle at cF and an acute angle at o1 of measure βF for all j ∈ IF ⊂ {2, 3,..., 13}. We
need to show that βF < α. We will prove this by contradiction. Assume that βF ≥ α. Then
π π either βF > 3 or α ≤ βF ≤ 3 . π First we consider the case of βF > 3 . Reflect the point o1 about the plane of F and let 0 o1 be the new point.
0 Figure 3.3: The reflections about F and o1.
0 0 The triangle 4o1o1oj is a right triangle with right angle at o1 and an acute angle βF at
0 00 o1 for all j ∈ IF ⊂ {2, 3,..., 13}. Then reflect the point o1 about the point o1 and let o1 be the new point obtained, see Figure 3.3.
00 00 π 00 Let B1 be the unit ball centred at o1. Since βF > 3 then d(o1, o1) < 2 so we can translate 00 00 000 B1 along the line o1o1 away from the point o1 such that it is tangent to B1. Let B1 be the
000 unit ball at this new position. However, this implies that B1 is tangent to B1 in addition
to the original 12 unit balls B2, B3,..., B13. This contradicts the solution of Sch¨utteand van der Waerden which resolved the Newton and Gregory argument [34], concluding that the kissing number in three dimensions is 12.
π 0 00 00 Now, we consider the case of α ≤ βF ≤ 3 . Using the definitions of o1, o1, B1, the π 00 inequality βF ≤ 3 implies that the 14 unit balls B1, B1, B2, B3,..., B13 form a packing in
34 3 E . The inequality α ≤ βF implies that
00 d(o1, o1) ≤ 4 cos α 4 = by (3.4) rˆ = 2.51998 ...
< 2.52. (3.5)
00 00 Note that o1 is the centre of the unit ball B1 and o1 is the centre of the unit ball B1, and recall Theorem 3.3 of Hales. Inequality (3.5) contradicts Theorem 3.3. This completes the proof of Lemma 3.4.
Theorem 3.5 (K. Bezdek and S. Reid, [5]). Let B1, B2,..., B13 be elements of a packing 3 ˆ of unit balls in E . Assume the balls B2, B3,..., B13 each touch B1. Let Bi be the ball ˆ ˆ concentric to Bi having radius rˆ, 1 ≤ i ≤ 13. Then the boundary bd(B1) of B1 is covered ˆ ˆ ˆ by B2, B3,..., B13. That is, 13 ˆ [ ˆ bd(B1) ⊂ Bj. j=2 ˆ Proof. By projecting the faces F of T from the centre point o1 onto the sphere bd(B1), we ˆ ˆ obtain a tiling of bd(B1) into spherically convex polygons F . Thus, it is sufficient to show that if F is an arbitrary face of T with vertices tj, where j ∈ IF ⊂ {2, 3,..., 13}, then its ˆ ˆ ˆ central projection F ⊂ bd(B1) is covered by the balls Bj, where j ∈ IF ⊂ {2, 3,..., 13}. In order to achieve this it is sufficient to prove that the projection cˆF of cF from the centre point ˆ ˆ o1 onto the sphere bd(B1) is covered by each of the balls Bj, where j ∈ IF ⊂ {2, 3,..., 13}.
Let the angle at o1 in the triangle 4o1ojcˆF be denoted by βF . Lemma 3.4 implies that
βF < α. Since d(o1, oj) = 2 and d(o1, cˆF ) =r ˆ, then comparing the triangle 4o1ojcˆF to the original triangle 4o1pq, we see that d(oj, cˆF ) < rˆ holds for all j ∈ IF ⊂ {2, 3,..., 13}. This completes the proof of Theorem 3.5.
n Take the union ∪i=1(ci +r ˆB) of the balls c1 +r ˆB, c2 +r ˆB,..., cn +r ˆB of radiir ˆ centred
35 3 at the points c1, c2,..., cn in E . Continuing the proof of Theorem 3.2, we will use the following theorem of Hales.
Theorem 3.6 (T. Hales, [14]). Let F be an arbitrary (finite or infinite) family of non-
overlapping unit balls in E3. Let B be the unit ball belonging to F centred at the origin o of E3. Let P be the Voronoi cell of F that corresponds to B. Let Q be a regular dodecahedron √ √ π that circumscribes B (note the circumradius is 3 tan 5 = 1.2584 ...). Let r := 2, and
let rB be the ball of radius r centred at the origin o. Let vol3(·) denote the 3-dimensional volume of the given set. Then
vol (B) vol (B) vol (B) 3 ≤ 3 ≤ 3 < 0.7547. vol3(P) vol3(P ∩ rB) vol3(Q)
Figure 3.4: Voronoi cells for a packing of balls.
Recall that we definedr ˆ to be 1.58731. This value ofr ˆ was chosen to be slightly larger than the value of r that Hales uses in Theorem 3.6 so that we can apply this theorem to prove Theorem 3.7.
Theorem 3.7 (K. Bezdek and S. Reid, [5]). Let vol3(·) denote the 3-dimensional volume of
36 the given set. Then
nvol3(B) n < 0.7547. vol3 (∪i=1(ci +r ˆB))
n Proof. First we will partition ∪i=1(ci +r ˆB) into truncated Voronoi cells. See Figure 3.4 for the 2-dimensional representation of these cells. Let Pi be the Voronoi cell assigned to ci + B
3 for 1 ≤ i ≤ n. Note that these Voronoi cells Pi, 1 ≤ i ≤ n form a tiling of E . As a result, the truncated Voronoi cells Pi ∩ (ci +r ˆB), 1 ≤ i ≤ n generate a tiling of the non-convex √ n container ∪i=1(ci +r ˆB) for the packing P. Since r = 2 < 1.58731 =r ˆ, Theorem 3.6
applied to the truncated Voronoi cells Pi ∩ (ci +r ˆB), 1 ≤ i ≤ n completes the proof.
n The isoperimetric inequality applied to ∪i=1(ci +r ˆB) implies the following lemma. Note that svol3 (ˆrB) (4πrˆ2)3 2 = = 36π. 2 4 3 2 vol3 (ˆrB) ( 3 πrˆ )
Lemma 3.8 (K. Bezdek and S. Reid, [5]). Let svol2(·) denote the 2-dimensional surface volume of the given set. Then
n ! n !! 2 [ 3 [ 36πvol3 (ci +r ˆB) ≤ svol2 bd (ci +r ˆB) . i=1 i=1
Theorem 3.7 implies
n ! [ nvol3(B) < 0.7547vol3 (ci +r ˆB) i=1 n ! 4π [ n < 0.7547vol (c +r ˆB) 3 3 i i=1 n ! 4π n [ < vol (c +r ˆB) 3 0.7547 3 i i=1 2 n ! 4π n [ < vol2 (c +r ˆB) 3 0.7547 3 i i=1 2 n ! 4π n [ 36π < 36πvol2 (c +r ˆB) . (3.6) 3 0.7547 3 i i=1
37 Now, Lemma 3.8 and (3.6) imply
2 n !! 4π n [ 36π < svol3 bd (c +r ˆB) 3 0.7547 2 i i=1 2 n !! √ 3 3 4π n [ 36π < svol bd (c +r ˆB) 3 0.7547 2 i i=1 2 2 n !! 3 1 4 1 2 n 3 [ 3 3 3 (36) π π 2 < svol2 bd (ci +r ˆB) 3 3 (0.7547) i=1 2 n !! 4πn 3 [ 2 < svol2 bd (ci +r ˆB) . 3 (0.7547) i=1
Note that 4π 2 = 15.159805547. (0.7547) 3
Then, Theorem 3.7 and Lemma 3.8 imply the following inequality.
Corollary 3.9 (K. Bezdek and S. Reid, [5]).
2 2 15.159805n 3 < 15.15980554 . . . n 3
4π 2 3 = 2 n (0.7547) 3 n !! [ < svol2 bd (ci +r ˆB) i=1
n Next, we obtain an upper bound for svol2 (bd (∪i=1(ci +r ˆB))). Assume that ci + B ∈ P is tangent to cj + B ∈ P for all j ∈ Ti where Ti ⊂ {1, 2, . . . , n} is the family of indicies ˆ 1 ≤ j ≤ n for which d(ci, cj) = 2. Let Si := bd(ci +r ˆB) and let cˆij be the intersection of the line segment c c with Sˆ for all j ∈ T . Let C cˆ , π (respectively C (cˆ , α)) denote i j i i Sˆi ij 6 Sˆi ij ˆ ˆ π the open spherical cap of Si centred at cˆij ∈ Si with angular radius 6 (respectively α with 0 < α < π and cos α = 1 ). The family C cˆ , π , j ∈ T consists of pairwise disjoint 2 rˆ Sˆi ij 6 i ˆ open spherical caps of Si.
2 1 2 π 2 Let S := bd(B), then let uij := 2 (cj − ci) ∈ S . Let C(uij, 6 ) ⊂ S (respectively
38 2 2 π C(uij, α) ⊂ S ) denote the open spherical cap of S centred at uij with angular radius 6 (respectively α) and let Sarea(·) denote the spherical area measure on S2. Then
P svol C (cˆ , π ) P Sarea C(u , π ) j∈Ti 2 Sˆi ij 6 j∈Ti ij 6 = . (3.7) svol S C (cˆ , α) Sarea S C(u , α) 2 j∈Ti Sˆi ij j∈Ti ij
Moln´ar’sdensity bound [25] implies that
P Sarea C(u , π ) j∈Ti ij 6 < 0.89332. (3.8) Sarea S C(u , α) j∈Ti ij
Recall that P is a packing with the maximum number of contacts. Also recall that the kissing number in 3 dimensions is 12. Assume that m members of P have 12 touching neighbours and assume k members of P have at most 9 touching neighbours. Thus, there are n − m − k members of P that have either 10 or 11 touching neighbours. Without loss of generality, we can assume that 4 ≤ k ≤ n − m. Note that
π π Sarea C u , = 2π 1 − cos ij 6 6 √ ! 3 = 2π 1 − 2 and √ ! π 3 2 svol2 C ˆ cˆij, = 2π 1 − (ˆr) . Si 6 2
ˆ Theorem 3.5 implies that if ci + B has exactly 12 touching neighbours then Si ⊂
∪j∈Ti (cj +r ˆB). Using this fact, together with (3.7) and (3.8), we will obtain an upper Sn bound for svol2 (bd ( i=1 (ci +r ˆB))). See Figure 3.5 for a simplified image of the surface volume contribution.
39 Sn Figure 3.5: The surface volume the middle ball contributes to svol2 (bd ( i=1 (ci +r ˆB))) is shown in bold.
Corollary 3.10 (K. Bezdek and S. Reid, [5]).
n !! [ 24.53902 svol bd (c +r ˆB) < (n − m − k) + 24.53902k 2 i 3 i=1
Proof. Note that only the balls with fewer than 12 touching neighbours will contribute to Sn svol2 (bd ( i=1 (ci +r ˆB))). Recall that there are n − m − k balls in the packing that have 10 or 11 touching neighbours, and there are k balls in the packing that have at most 9 touching neighbours. First we consider the contribution that balls of radiusr ˆ with 10 or 11 touching neighbours Sn have to svol2 (bd ( i=1 (ci +r ˆB))). For each such ball, we will generate a lower bound for the surface volume of the spherical caps covered by other balls of radiusr ˆ. Since the unit balls
π are touching and non-overlapping, the angular measure of the spherical cap is 6 . Together with (3.8), this implies the first half of inequality (3.9). Next we consider the contribution that balls of radiusr ˆ with at most 9 touching neigh- Sn bours (and thus, with at least 3 touching neighbours) have to svol2 (bd ( i=1 (ci +r ˆB))). This is very similar to the previous case, however in this case since each ball has at least 3 touching neighbours then we multiply by 3 rather than 10. This implies the second half of inequality (3.9).
40 Therefore,
n !! [ svol2 bd (ci +r ˆB) i=1 π ! π ! 10 svol2 C ˆ cˆij, 3 svol2 C ˆ cˆij, < svol Bˆ − Si 6 (n − m − k) + svol Bˆ − Si 6 k 2 0.89332 2 0.89332
(3.9) √ √ 3 2 3 2 10 · 2π 1 − 2 rˆ 3 · 2π 1 − 2 rˆ < 4πrˆ2 − (n − m − k) + 4πrˆ2 − k 0.89332 0.89332
< (31.66164 − 23.74208) (n − m − k) + (31.66164 − 7.12262) k
= 7.91956(n − m − k) + 24.53902k 24.53902 < (n − m − k) + 24.5390k, 3
which completes the proof of Corollary 3.10.
Corollary 3.9 and Corollary 3.10 imply the following.
2 24.53902 15.159805n 3 < (n − m − k) + 24.53902k 3 3 2 15.159805n 3 < n − m − k + 3k 24.53902 2 1.85335090806n 3 < n − m − k + 3k
2 1.85335n 3 < n − m − k + 3k
2 1.85335n 3 − 3k < n − m − k (3.10)
Since the kissing number in three dimensions k(3) is 12, then 12n is clearly an upper bound for the contact number of a packing of n congruent balls. We can improve this upper bound. First recall that there are n − m − k balls in the packing with 10 or 11 touching neighbours. These balls have at least one fewer contact than the optimal case, k(3). Next recall that there are k balls in the packing with at most 9 touching neighbours. These balls have at
41 least 3 fewer contacts than k(3). Summing the kissing number for every ball counts each contact twice. Thus, the number of contacts in the packing is
1 c(n) ≤ (12n − (n − m − k) − 3k) . 2
Together with (3.10) this implies
1 c(n) ≤ (12n − (n − m − k) − 3k) 2 1 ≤ 6n − ((n − m − k) − 3k) 2 1 2 < 6n − 1.85335n 3 2 2 < 6n − 0.926675n 3
2 < 6n − 0.926n 3 .
This completes the proof of Theorem 3.2.
3.1.4 Totally Separable Congruent Ball Packings
The contact number problem for a totally separable packing of congruent balls in three dimensions was studied by Bezdek et al. [6]. Recall that a packing is totally separable if every two balls can be separated by a plane that is disjoint from the interior of each ball in the packing.
Theorem 3.11 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). Let c(n) be the contact number of a totally separable packing of n ≥ 2 unit balls in E3. Then
2 c(n) ≤ b3n − 1.346n 3 c.
3 Let B be the unit ball centred at the origin o of E . Let P := {c1 +B, c2 +B,..., cn +B}
3 denote the totally separable packing of n unit balls with centres c1, c2,..., cn in E that has
42 the largest contact number c(n) of all totally separable packings of n unit balls in E3. Note that P might not be uniquely determined up to congruence, in which case P is any extremal packing. In the proof of Theorem 3.11, we will use the following two lemmas without proof. Both lemmas are proved in [6].
Lemma 3.12 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). The distance between the line of √ 3 3 an arbitrary edge of the Voronoi cell Pi and its centre ci is at least 4 = 1.299 ... for any 1 ≤ i ≤ n.
Lemma 3.13 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). The distance between an arbitrary √ vertex of the Voronoi cell Pi and its centre ci is at least 2 = 1.414 ... for any 1 ≤ i ≤ n.
Let conv{·} denote the convex hull of the given set. The i-dimensional simplex Y =
3 conv{o, y1, y2,..., yi} ⊂ E with vertices y0 = o, y1, y2,..., yi is an i-dimensional or-
thoscheme if for each j, 0 ≤ j ≤ i − 1 the vector yj is orthogonal to the span span{yk −
yj|j + 1 ≤ k ≤ i} where 1 ≤ i ≤ d. That is, an orthoscheme is an i-dimensional simplex defined by i consecutive edges that are pairwise orthogonal. √ Now we dissect each of the truncated Voronoi cells Pi ∩ ci + 2B , 1 ≤ i ≤ n into 1-dimensional, 2-dimensional and 3-dimensional orthoschemes with pairwise disjoint relative interiors in the following way. √ For each x ∈ Pi ∩ ci + 2B we will assign an orthoscheme for three different cases.
Firstly, if x ∈ intPi then the orthoscheme assigned to x is the line segment conv{ci, x}.
Secondly, if x is a relative interior point of some face F of Pi then the orthoscheme assigned to x is conv{ci, f, x} where f is the orthogonal projection of ci onto the plane containing F .
Thirdly, if x is a relative interior point of some edge E of Pi with E on the face F then the orthoscheme assigned to x is conv{ci, f, e, x} where e (respectively f) is the orthogonal projection of ci onto the line containing E (respectively, onto the plane containing F ).
43 Lemma 3.14 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). Let vold(·) denote the d-dimensional
volume of the given set. Let W := conv{o, w1, w2,..., wd} be a d-dimensional orthoscheme
d d in E . Let U := conv{o, u1, u2,..., ud} be a d-dimensional simplex of E such that ||ui|| =
d(o, conv{ui, ui+1,..., ud}) for all 1 ≤ i ≤ d. If 1 ≤ ||wi|| ≤ ||ui|| holds for all 1 ≤ i ≤ d then vol U ∩ Bd vol W ∩ Bd d = d . vold(U) vold(W)
3 Let W := conv{o, w1, w2, w3} be the 3-dimensional orthoscheme with ||w1|| = 1, √ √ 3 3 ||w2|| = 4 , and ||w3|| = 2. Lemmas 3.12, 3.13, and 3.14 imply that for any 3-dimensional √ 3 orthoscheme U := conv{ci, f, e, x} of the truncated Voronoi cell Pi ∩ ci + 2B ,
3 T 3 T vol3 (U B) vol3 (W B) 3 = 3 . vold (U ) vold (W )
Since each 2-dimensional and each 1-dimensional orthoscheme of the dissection of Pi ∩ √ ci + 2B can be obtained as a limit of proper 3-dimensional orthoschemes, then we obtain the following lemma.
Lemma 3.15 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]).
√ T 4π vol3 Pi ∩ ci + 2B (ci + B) 3 √ = √ vol3 Pi ∩ ci + 2B vol3 Pi ∩ ci + 2B 3 T vol3 (W B) ≤ 3 vol3 (W ) < 0.6401.
This lemma will be used to prove the following theorem.
Theorem 3.16 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). Let vol3(·) denote the 3- dimensional volume of the given set. Then
4π n 3 √ < 0.6401. Sn vol3 i=1 ci + 3B
44 Figure 3.6: Voronoi cells for a packing of totally separable balls.
√ n Proof. First we will partition ∪i=1 ci + 3B into truncated Voronoi cells. Figure 3.6 is
an example of these Voronoi cells. Let Pi denote the Voronoi cell assigned to ci + B for
1 ≤ i ≤ n. Note that these Voronoi cells Pi, 1 ≤ i ≤ n generate a tiling of the non-convex √ √ √ n container ∪i=1 ci + 3B for the packing P. Since 2 < 3 then Lemma 3.15 applied to √ the truncated Voronoi cells Pi ∩ ci + 3B , 1 ≤ i ≤ n completes the proof. √ n The isoperimetric inequality applied to ∪i=1 ci + 3B implies the following lemma.
Lemma 3.17 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]). Let svol2(·) denote the 2-dimensional surface volume of the given set. Then
n √ ! n √ !! 2 [ 3 [ 36πvol3 ci + 3B ≤ svol2 bd ci + 3B . i=1 i=1
45 Theorem 3.16 implies
n ! 4π [ √ n < 0.6401vol c + 3B 3 3 i i=1 n ! 4π n [ √ < vol c + 3B 3 0.6401 3 i i=1 2 n ! 4π n [ √ < vol2 c + 3B 3 0.6401 3 i i=1 2 n ! 4π n [ √ 36π < 36πvol2 c + 3B . (3.11) 3 0.6401 3 i i=1
Now, Lemma 3.17 and (3.11) imply
2 n !! 4π n [ √ 36π < svol3 bd c + 3B 3 0.6401 2 i i=1 2 n !! √ 3 √ 3 4π n [ 36π < svol bd c + 3B 3 0.6401 2 i i=1 2 2 n !! √ 3 3 √ 3 4 1 2 n [ 3 3 36 π π 2 < svol2 bd ci + 3B 3 3 (0.6401) i=1 2 n !! 4πn 3 [ √ 2 < svol2 bd ci + 3B . 3 (0.6401) i=1
Note that 4π 2 = 16.9192812 .... (0.6401) 3
Then, Theorem 3.16 and Lemma 3.17 imply the following inequality.
Corollary 3.18 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]).
2 2 16.919128n 3 < 16.91912812 . . . n 3
2 4πn 3 = 2 (0.6401) 3 n !! [ √ < svol2 bd ci + 3B i=1
46 √ n Next, we obtain an upper bound for svol2 bd ∪i=1 ci + 3B . Assume that ci +B ∈
P is tangent to cj + B ∈ P for all j ∈ Ti where Ti ⊂ {1, 2, . . . , n} is the family of indices √ ˆ 1 ≤ j ≤ n for which d(ci, cj) = 2. Let Si := bd(ci + 3B) and let cˆij be the intersection of the line segment c c with Sˆ for all j ∈ T . Let C cˆ , π (respectively C (cˆ , α)) denote i j i i Sˆi ij 4 Sˆi ij ˆ ˆ π the open spherical cap of Si centred at cˆij ∈ Si with angular radius 4 (respectively α with π 1 π 0 < α < and cos α = √ ). The family C ˆ cˆ , , j ∈ T consists of pairwise disjoint 2 3 Si ij 4 i ˆ open spherical caps of Si.
2 1 2 π 2 Let S := bd(B), then let uij := 2 (cj − ci) ∈ S . Let C(uij, 4 ) ⊂ S (respectively 2 2 π C(uij, α) ⊂ S ) denote the open spherical cap of S centred at uij with angular radius 4 (respectively α) and let Sarea(·) denote the spherical area measure on S2. Then
P svol C (cˆ , π ) P Sarea C(u , π ) j∈Ti 2 Sˆi ij 4 j∈Ti ij 4 = . (3.12) svol S C (cˆ , α) Sarea S C(u , α) 2 j∈Ti Sˆi ij j∈Ti ij
Moln´ar’sdensity bound [25] implies that
√ P Sarea C(u , π ) ! j∈Ti ij 4 2 ≤ 3 1 − = 0.8786 .... (3.13) Sarea S C(u , α) 2 j∈Ti ij
Note that
π π Sarea C u , = 2π 1 − cos ij 4 4 √ ! 2 = 2π 1 − 2
and
√ ! π 2 √ 2 svol2 C ˆ cˆij, = 2π 1 − 3 Si 4 2 √ ! 2 = 6π 1 − . 2
47 Together with (3.12) and (3.13), this implies the following upper bound.
√ Figure 3.7: Two touching unit balls sharing centres with overlapping balls of radius 3.
Corollary 3.19 (K. Bezdek, B. Szalkai, and I. Szalkai, [6]).
n !! [ √ svol2 bd ci + 3B ≤ 12πn − 4πc(n) i=1
Proof. Recall that the packing has n balls and c(n) contacts. For each contact there √ n are two overlapping spherical caps which do not belong to svol2 bd ∪i=1 ci + 3B . Figure 3.7 shows two touching unit balls centred at the same points as two balls of ra- √ √ dius 3. The overlapping region of the two balls of radius 3 does not contribute to √ Sn svol2 bd i=1 ci + 3B .
48 Therefore,
n !! √ √ 2 svol C cˆ , π [ 2 Sˆi ij 4 svol2 bd ci + 3B ≤ svol2( 3B)n − √ c(n) 2 i=1 3 1 − 2 √ √ 2 · 6π 1 − 2 2 2 = 4π( 3) n − √ c(n) 2 3 1 − 2 √ 2 12π 1 − 2 = 12πn − √ c(n) 2 3 1 − 2
= 12πn − 4πc(n).
Corollary 3.18 and 3.19 imply the following.
4π 2 3 12πn − 4πc(n) > 2 n (0.6401) 3
4π 2 3 −4πc(n) > 2 n − 12πn (0.6401) 3
−1 2 3 c(n) < 2 n + 3n (0.6401) 3
2 c(n) < −1.346n 3 + 3n
This completes the proof of Theorem 3.11.
3.2 The Contact Number Problem for Incongruent Balls
In this section we consider the contact number problem for incongruent balls in three di- mensions. We start by proving an upper bound for the contact number of a packing of incongruent balls, given by Kuperberg and Schramm [21]. Then we give a trivial upper bound for the contact number of a packing of totally separable incongruent balls.
49 3.2.1 Kuperberg-Schramm Theorem
The contact number problem for incongruent balls in three dimensions was studied by Ku- perberg and Schramm through the use of the average kissing number. The average kissing number, denoted K(P), of a packing P of n balls is 2c(n)/n. Recall that c(n) is the contact number of the packing, which is equivalent to the number of edges in the corresponding contact graph. In the following theorem and its proof we use the notation of G. Kuperberg and O. Schramm [21].
Theorem 3.20 (G. Kuperberg and O. Schramm, [21]). Let P be a packing of incongruent
balls in R3, and let K(P) be the average kissing number of P. Then
√ K(P) < 8 + 4 3 = 14.928 ....
Before we prove this, we will first show that K(P) ≤ 24. Let P be a packing of incongru- ent balls in R3. Let E be the set of unordered pairs of balls in P that kiss. Let r(B) be the radius of the ball B in P. By a well-known result (the solution to the Newton and Gregory argument) [34] at most 12 unit balls with disjoint interiors can touch a given unit ball. Let C be a ball with r(C) > 1, and B a ball with r(B) = 1. If C kisses B then C contains a (unique) unit ball that kisses B. As a result, B cannot kiss more than 12 balls at least as large as B. Let f be the function f : E → P that assigns to {B, C} ∈ E the smaller of the balls B and C, or either if they are the same size. Since f is at most 12 to 1, |E| ≤ 12|P|. Thus, 2|E| K(P) = ≤ 24. |P|
Let E(B) denote the set of balls that kiss B. Let ρ > 1 be a constant, which will be determined in Lemma 3.22. For each ball B, let S(B) be the concentric spherical shell with
50 radius ρr(B). For each B, C ∈ P define
area (C ∩ S(B)) a(B, C) := . (3.14) area (S(B))
Lemma 3.21. The size of the packing P is bounded below by
X |P| ≥ (a(B, C) + a(C, B)) . {B,C}∈E
Proof. Since the interiors of the balls in P are disjoint, then for any B,
X X 1 ≥ a(B, C) ≥ a(B, C). (3.15) C∈P C∈E(B)
There are |P| balls in the packing, so summing over all B,
X |P| ≥ (a(B, C) + a(C, B)) . (3.16) {B,C}∈E
√ Lemma 3.22. The optimal value of ρ is 3.
Proof. Suppose B intersects S(C) and C intersects S(B) as shown in Figure 3.8. Let b be the centre of B, and c the centre of C. Let q be a point of the spherical disk C ∩ S(B) on the relative boundary in S(B). Clearly,
d(b, c) = r(B) + r(C)
d(b, q) = ρr(B)
d(c, q) = r(C)
where d(x, y) is the Euclidean distance between x and y. Let θ = ∠cbq be the angular radius of C ∩ S(B).
51 Figure 3.8: The intersection of the balls S(B) and S(C).
By the law of cosines,
(r(B) + r(C))2 + (ρr(B))2 − r(C)2 cos θ = 2 (r(B) + r(C)) ρr(B) r(B)2 + 2r(B)r(C) + r(C)2 + ρ2r(B)2 − r(C)2 = 2 (r(B) + r(C)) ρr(B) r(B)2 + ρ2r(B)2 + 2r(B)r(C) = 2 (r(B) + r(C)) ρr(B) r(B)(r(B) + ρ2r(B) + 2r(C)) = 2 (r(B) + r(C)) ρr(B) r(B) + ρ2r(B) + 2r(C) = . (3.17) 2ρ (r(B) + r(C))
Let w be the width between two parallel planes as seen in Figure 3.9. We will express w in terms of θ and ρr(B). It is easy to see from Figure 3.10 that
w = ρr(B) − cos θρr(B)
= ρr(B) (1 − cos θ) . (3.18)
Recall that the well-known formula for the surface area of a spherical cap of a sphere of
52 Figure 3.9: The width, w, between two parallel planes.
Figure 3.10: The width between two parallel planes can be expressed in terms of θ and ρr(B).
radius R is 2πRw [17]. In our case we are interested in the spherical cap (C ∩ S(B)), seen in bold in Figures 3.8 and 3.10. So,
area (C ∩ S(B)) = 2πρr(B)w. (3.19)
We want to express this in terms of θ and area (S(B)). Equations (3.14) and (3.19) imply
area (C ∩ S(B)) = a(B, C)area (S(B))
2πρr(B)w = a(B, C)4π (ρr(B))2 w = a(B, C). 2ρr(B)
53 From (3.18), w 1 − cos θ = . 2ρr(B) 2
Thus, 1 − cos θ area (C ∩ S(B)) = area (S(B)) . (3.20) 2
Combining (3.20) and (3.14), 1 − cos θ a(B, C) = , (3.21) 2 and combining (3.17) and (3.21),
1 r(B) + ρ2r(B) + 2r(C) a(B, C) = − . 2 4ρ (r(B) + r(C))
Similarly, we can switch B and C to obtain
1 r(C) + ρ2r(C) + 2r(B) a(C, B) = − . 2 4ρ (r(C) + r(B))
Then,
r(B)(3 + ρ2) + r(C)(3 + ρ2) a(B, C) + a(C, B) = 1 − 4ρ (r(B) + r(C)) (3 + ρ2)(r(B) + r(C)) = 1 − 4ρ (r(B) + r(C)) 3 + ρ2 = 1 − . (3.22) 4ρ
Note that a(B, C) + a(C, B) does not depend on r(B) nor r(C). To maximize (3.22) we
54 must find the roots of
2ρ(4ρ) − 4(3 + ρ2) 8ρ2 − 12 − 4ρ2 = 16ρ2 16ρ2 4(ρ2 − 3) = 16ρ2 ρ2 − 3 = . 4ρ2
√ Thus, ρ = 3 to maximize (3.22).
Now we use Lemmas 3.21 and 3.22 to prove Theorem 3.20.
Proof of Theorem 3.20. Assuming S(B) intersects the ball C and S(C) intersects the ball B, and using the optimal value of ρ from Lemma 3.22,
3 + 3 a(B, C) + a(C, B) = 1 − √ 4 3 3 = 1 − √ 2 3 √ 3 = 1 − . 2
If S(B) does not intersect the ball C then a(B, C) = 0. In general,
√ 3 a(B, C) + a(C, B) ≥ 1 − . (3.23) 2
Applying Lemma 3.21 to (3.23) implies
1 |P| ≥ |E| √ 3 1 − 2 √ |P| ≥ |E| 4 + 2 3 .
This completes the proof of Theorem 3.20.
55 Let c(n) denote the contact number for a packing of incongruent balls in three dimensions. It follows from Theorem 3.20 that
1 c(n) ≤ √ n 3 1 − 2
≤ 7.464101615 . . . n.
Remark 4. Since each ball C that kisses B and intersects S(B) must have
(ρ − 1)r(B) r(C) ≥ 2 then there is a finite bound on the number of balls C that kiss B such that a(B, C) > 0. Thus, there is some constant α < 1 that depends on ρ such that
X a(B, C) ≤ α. C∈E(B)
This inequality can be used in place of (3.15) and would multiply the upper bound by a factor of α. Therefore, the upper bound on the contact number for a packing of incongruent balls can be improved by finding a good estimate of α.
56 3.2.2 Totally Separable Incongruent Ball Packings
The contact number problem for a totally separable packing of incongruent balls in three dimensions has not been studied in depth, and has no published papers that refer to it, as it is a difficult problem. We can obtain a trivial upper bound using the method of Kuperberg and Schramm [21]. The goal is for this upper bound to be improved upon.
Corollary 3.23. Let c(n) denote the contact number of a totally separable packing of balls in E3. Then c(n) ≤ 6n.
Proof. This follows the method of Kuperberg and Schramm [21]. Let P be a totally separable packing of balls in E3, and let K(P) be the average kissing number of P. Since the kissing number for a totally separable packing of balls in three dimensions is 6 then
K(P) ≤ 12.
The average kissing number is equivalent to 2|E|/n, where |E| is the number of edges of the contact graph, which is simply c(n). Hence, K(P) ≤ 12 is equivalent to c(n) ≤ 6n.
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