Problems and Results in
Discrete and Computational Geometry
A dissertation submitted to the
Graduate School
of the University of Cincinnati
in partial fulfillment of the
requirements for the degree of
Doctor of Philosophy
In the School of Electronic and Computing Systems
of the College of Engineering and Applied Sciences
By
Justin W. Smith
M.S. University of Cincinnati
December 2010
Committee Chair: George B. Purdy, Ph.D.
Abstract
Let S be a set of n points in R3 , no three collinear and not all coplanar. If
at most n − k are coplanar and n is sufficiently large, the total number of
planes determined is at least 1 + k n−k
− k2 n−k
. For similar conditions
2
2
and sufficiently large n, (inspired by the work of P. D. T. A. Elliott in [1]) we
also show that the number of spheres determined by n points is at least 1 +
orchard
n−1
−t3
(n−1), and this bound is best possible under its hypothesis. (By
3
torchard
(n), we are denoting the maximum number of three-point lines attainable
3
by a configuration of n points, no four collinear, in the plane, i.e., the classic
Orchard Problem.) New lower bounds are also given for both lines and circles.
We demonstrate an infinite family of pseudoline arrangements each with no
member incident to more than 19 (4n − 10) points of intersection, where n is the
number of pseudolines in the arrangement. We also prove a generalization of
the Weak Dirac that holds for more general incidence structures.
Acknowledgments
The past several years have been very eventful (including, most notably, the
birth of my son). There were times that were very difficult, and times that
were not so difficult. Through it all, there have been many individuals that
have supported me with direction, motivation, and hope. I am very thankful
for their support.
My adviser, Prof. George B. Purdy, has been a source of knowledge and
inspiration throughout my research. He has that wonderful ability to find problems that are both interesting and solvable. I would also like to thank Ben Lund
for his assistance on various results, particularly, the results relating to the Dirac
conjectures. (Most notably, the Strong Dirac counter-example in Section 4.1.2
is due to him.) I greatly appreciate them both for assistance and friendship over
the past several years.
My wife has provided me with support in every possible way. She always
keeps me grounded, and helps me see the world from a different perspective. I
am grateful for her love and patience through all of these challenging times.
There are many people whom also should be listed here, including others in
my family and various faculty and staff members at the University of Cincinnati.
They each have played an valuable role in my life or education. I am blessed
to have such great people in my life, and I am grateful for their help along the
way.
ii
Contents
1 Introduction
1
1.1
Lines, Circles, Planes and Sphere . . . . . . . . . . . . . . . . . .
1
1.2
Pseudolines and Dirac’s Conjectures . . . . . . . . . . . . . . . .
3
1.2.1
Strong Dirac . . . . . . . . . . . . . . . . . . . . . . . . .
3
Weak Dirac . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
2 Background
6
2.1
A Question Of J.J. Sylvester . . . . . . . . . . . . . . . . . . . .
6
2.2
Duality and the Projective Plane . . . . . . . . . . . . . . . . . .
7
2.2.1
Polar Dual . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2.2.2
Parabolic Dual . . . . . . . . . . . . . . . . . . . . . . . .
12
Melchior’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.3.1
Euler’s Polyhedral Formula . . . . . . . . . . . . . . . . .
13
2.3.2
Melchior’s Proof of the Inequality . . . . . . . . . . . . . .
15
2.4
How Many? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
2.5
Da Silva and Fukuda’s Conjecture . . . . . . . . . . . . . . . . .
18
2.6
Pseudolines and Other Curves . . . . . . . . . . . . . . . . . . . .
21
2.3
3 Planes and Spheres
3.1
Lines Determined by Points in R2 . . . . . . . . . . . . . . . . . .
25
Circles Determined by Points in R2 . . . . . . . . . . . . .
28
Planes Determined by Points in R3 . . . . . . . . . . . . . . . . .
31
3.1.1
3.2
25
iii
3.3
3.2.1
A Derivation from Melchior’s Inequality . . . . . . . . . .
32
3.2.2
Planes Incident to Exactly Three Points . . . . . . . . . .
34
3.2.3
Total Number of Determined Planes . . . . . . . . . . . .
35
3.2.4
Planes Determined by At Most Four Points . . . . . . . .
38
3.2.5
Corollaries and Conjectures . . . . . . . . . . . . . . . . .
41
Spheres Determined by Points in R3 . . . . . . . . . . . . . . . .
43
3.3.1
Spheres Incident To Exactly Four Points . . . . . . . . . .
44
3.3.2
The Orchard Problem . . . . . . . . . . . . . . . . . . . .
45
3.3.3
Total Number of Spheres Determined . . . . . . . . . . .
49
4 Dirac’s Conjectures
4.1
4.2
55
Strong Dirac . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
4.1.1
Wedges and Eppstein’s observation . . . . . . . . . . . . .
55
4.1.2
Pseudoline counterexample to Strong Dirac . . . . . . . .
59
Weak Dirac . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
4.2.1
α-Curve combinatorics . . . . . . . . . . . . . . . . . . . .
62
4.2.2
A conditional Weak Dirac for α-curve combinatorics . . .
64
5 Conclusions and Future Work
69
5.1
Planes and Spheres . . . . . . . . . . . . . . . . . . . . . . . . . .
69
5.2
Pseudolines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
iv
List of Figures
1.1
The dual of Felsner’s arrangement with 6k + 7 = 31 lines (including the line at infinity) and no line incident to more than
3k + 2 = 14 points of intersection. . . . . . . . . . . . . . . . . .
2.1
A perpendicular from point pk to line pi pj is shorter than the
perpendicular from pi to line pj pk . . . . . . . . . . . . . . . . . .
2.2
7
The Kelly-Moser configuration has seven points that determine
only three ordinary lines, and thus, t2 (7) = 3. . . . . . . . . . . .
2.3
4
17
McKee’s configuration showing thirteen points determining only
six ordinary lines. . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.4
Known values of t2 (n) originally published in [2]. . . . . . . . . .
18
2.5
A Böröczky configuration of ten points determining five ordinary
lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6
19
A counter example to Da Silva and Fukuda’s conjecture. Originally published in [3]. . . . . . . . . . . . . . . . . . . . . . . . .
20
2.7
Pappus’s theorem requires the points x, y and z to be collinear. .
22
2.8
An arrangement of pseudolines that is not realizable over any field. 23
4.1
A single wedge from Felsner’s arrangement. . . . . . . . . . . . .
4.2
A set of m beams all intersection at a point P , on a side of the
wedge. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3
56
57
A set of m beams all intersection at a point P , on a side of the
wedge. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
58
4.4
The wedge for j = 1, the base case for our induction. . . . . . . .
4.5
The arrangement for j = 1, containing 3(6j + 2) + 1 = 25 pseu-
4.6
60
dolines, each of which incident to at most 10 vertices. . . . . . .
61
The wedge for j = 2. . . . . . . . . . . . . . . . . . . . . . . . . .
61
vi
Chapter 1
Introduction
1.1
Lines, Circles, Planes and Sphere
The problem of maximizing the number of three-point lines goes back as far
as 1821 when several such problems appeared in [4] in a section called “Trees
planted in rows.” The first question of the section essentially asks (but in a
rhyming verse) how to plant nine trees such that they form ten rows of three.
That section also includes several variations on this question. Many years later,
Sylvester demonstrated a configuration of n points in the plane, lying on a
cubic curve, that determines ∼
1 2
8n
three-point lines [5, pp. 106-107]. (More
specifically, Sylvester’s configuration was answering his own question of how to
plant 81 trees to form 800 rows of three!) See [6] and [7, pp. 315–318] for a
complete history of the Orchard Problem.
In Section 3.3.2, we discuss an unexpected equivalence to the Orchard Problem that we found while researching a lower bound for the number of spheres
determined by a set of points. This lower bound, best possible under its hy orchard
pothesis, is 1 + n−1
− t3
(n − 1), where torchard
(n) is the upper bound on
3
3
the number of three-point lines determined by n points, no four collinear, in the
plane. This equivalence was discovered after realizing that a similar subtractive term exists for circles (but was overlooked in [1]). We must first, however,
1
present several lower bounds on determined planes which, by using methods
inspired by P. D. T. A. Elliott, provide the tools necessary to obtain our results
for spheres.
Motzkin’s 1951 paper, [8], is often cited for his conjecture (also conjectured
by Dirac in [9]) that among any set of 2n noncollinear points in the plane there
exist n ordinary (i.e., two-point) lines. Included in that paper, [8], are several
results concerning planes (or hyperplanes) determined by points in three, or
higher, dimensional space.
Using two counter-examples, i.e., the Desargues configuration and a set of
points all lying on two skew lines in R3 , Motzkin demonstrated that a threepoint plane need not exist among a finite set of non-coplanar points. So Motzkin
defined the proper analog to an “ordinary line” to be an “ordinary plane”, i.e.,
a plane in which all but one of the incident points may lie on a line. Similarly,
an “ordinary hyperplane” in d-space is defined to be one in which all but one
of its points are incident to a (d − 2)-flat (i.e., an affine subspace of dimension
d − 2). S. Hansen later proved that an ordinary hyperplane necessarily exists
among a finite set of points in (d > 3)-space [10]. Hansen also found a new
infinite family of configurations in R3 with no three-point planes [11].
Given a set of n points in R3 , let m be the number of planes determined,
and let t be the number of lines. Purdy conjectured, under various conditions,
that m − t + n > 2 and m > t. Extending this to Rd , it was conjectured by
Purdy that under certain conditions the number of hyperplanes determined is
at least the number of (d − 2)-flats. (See Section 3.2.5 for discussion of these
conjectures.)
Compared to the work done concerning points in the plane, relatively little
has been done to extend Motzkin’s results concerning planes in R3 . Several
results in this area were obtained by Erdős and Purdy in [12], with which the
present paper is related. In [12], they prove that given a finite set of n > 552
points, not all coplanar and no three collinear, there are at least n−1
+1
2
planes determined. (This result verified m − t + n > 2 for that case.) They also
2
proved that given such a point set there exist 12 n2 − cn determined planes (for
a suitable c) incident to at most four points. Chapter 3 improves upon both of
these results.
1.2
Pseudolines and Dirac’s Conjectures
Given an arrangement L of lines in the plane, let r(L) be the maximum number
of vertices of L on any line of L. Let r∗ (n) be the minimum of r(L) over
all arrangements of n lines that don’t all share a common point. In 1951,
Dirac (working in the dual context) conjectured a lower bound on r∗ (n). This
conjecture is commonly stated as follows.
Conjecture 1.2.1 (Strong Dirac). For all sufficiently large n, there exists a
constant c such that r∗ (n) ≥ n/2 − c.
A set of unbounded curves in the plane is a set of pseudolines if each pair
of curves intersects in exactly one point, and at that point cross. In Section
4.1.2, we demonstrate an infinite family of pseudoline arrangements each with
no member incident to more that 4n/9 − 10/9 vertices. This implies that,
if the Strong Dirac is true, proving it will require methods that utilize the
“straightness” of lines. This is surprising since most results known for lines in
the plane also hold for pseudolines.
In 1961, Erdős proposed a weaker version of this conjecture. It was proved
independently in 1983 by both Beck[13] and Szemerédi and Trotter[14].
Theorem 1.2.2 (Weak Dirac). r∗ (n) = Ω(n).
In Section 4.2.2, we demonstrate that the Weak Dirac holds for a variety of
incidence structures.
1.2.1
Strong Dirac
To be specific, Dirac conjectured in 1951 that among any set of n non-collinear
points, P , there must exist a point incident to at least d n2 e lines spanned by P .[9]
3
This bound can be attained for odd n when the points lie on two intersecting
lines. Typically, Dirac’s original conjecture is stated in a slightly weaker form
(i.e., the “Strong Dirac”).
1
In [15], Akiyama et al. show that the b n2 c bound (i.e., the Strong Dirac
conjecture with c = 0) can be attained for all sufficiently large n except those of
the form 12k + 11 (which they left as an open problem). However, there exists
a family of configurations, with an arbitrarily large number of points, for which
the conjecture is false for c = 0. This infinite family of counterexamples is due
to Felsner and contains 6k + 7 points with none incident to more than 3k + 2
spanned lines when k is even, and 3k + 3 when k is odd. [7, p. 313] The dual
form for this family is demonstrated in Figure 1.1.
∞
Figure 1.1: The dual of Felsner’s arrangement with 6k + 7 = 31 lines (including
the line at infinity) and no line incident to more than 3k + 2 = 14 points of
intersection.
In Section 4.1.2, we provide a counter-example showing the Strong Dirac
does not apply to pseudolines. The counter-example demonstrates a family of
1 This conjecture is sometimes referred to as the Dirac-Motzkin conjecture, see for example
the first paragraph of [13]. However, that label more often refers to the conjecture that an
arrangement of n non-collinear points is spanned by at least n
ordinary lines.
2
4
arrangements containing 18k + 7 pseudolines with none incident to more than
8k + 2 vertices.
1.3
Weak Dirac
In the original paper, Dirac showed that in any arrangement of n non-collinear
√
points there exists a point incident to at least n spanned lines. Dirac’s original
proof was essentially combinatorial. A generalization of his proof is included in
Section 4.2.1.
With little progress having been made toward proving the original conjecture, in 1961 Erdős published (among a multitude of other conjectures) a weakened version of Dirac’s original conjecture. [16, p. 245] This version of the
conjecture was first proved in 1983, nearly simultaneously, in two well-known
papers by Beck [13]; and Szemerédi and Trotter [14].
Many results concerning points and lines do not depend on the straightness
of lines; these include the Szemerédi-Trotter incidence bound [14] and Csima
and Sawyer’s bound on the number of ordinary points.[17] The Weak Dirac is
among these, being derivable from the Szemerédi-Trotter incidence bound. As
we will demonstrate in Section 4.2.2, it can also be derived from considerably
weaker incidence bounds.
5
Chapter 2
Background
2.1
A Question Of J.J. Sylvester
In 1893, J.J. Sylvester posed the following problem [18],
Prove that it is not possible to arrange any finite number of real
points so that a right line through every two of them shall pass
through a third, unless they all lie in the same right line.
A simpler way might be to ask, “Does every set of non-collinear points necessarily determine an ordinary line (i.e. a line passing through exactly two of
them)?”.
Sylvester’s problem remained without a solution (and was perhaps even forgotten) until it was raised again independently by Erdős in the 1930s. In 1943,
Erdős published this question in American Mathematical Monthly [19], and a solution appeared soon after in [20]. The first solution to this problem is commonly
attributed to T. Gallai, thus it is now called the Sylvester-Gallai Theorem.1
Although several proofs of the Sylvester-Gallai Theorem are known, one
credited to L.M. Kelly (and published by Coxeter in [22]) is considered the
most elegant. His proof is as follows.
1 Erdős was sure that Sylvester had a proof of his conjecture, but it (apparently) was never
published.[21]
6
Figure 2.1: A perpendicular from point pk to line pi pj is shorter than the
perpendicular from pi to line pj pk .
Proof. Assume we have a noncollinear configuration of points, P, such that any
line determined by two points also passes through a third. Assume the points
of P are numbered p1 , p2 , . . . , pn . There exists a point, pi , and a line, pj pk ,
such that the point-line pair (pi , pj pk ) determine the smallest nonzero distance
along a perpendicular among all point-line pairs of the configuration. Let q be
the foot of the perpendicular from pi to pj pk . Let pm be a third point point on
line pj pk . (See Figure 2.1.) Among the three (or more) points on the line, i.e.
pj , pk , and pm , two must lie to one side of q. Assume they are ordered on the
line pj pk qpm (q might coincide with pk ). The point-line pair (pk , pi pj ) forms a
shorter perpendicular distance, and thus a contradiction.
2.2
Duality and the Projective Plane
Consider the similar problem, “Does every arrangement of lines, no two being
parallel and not all passing through a common point, necessarily determine an
ordinary intersection (i.e. an intersection of exactly two lines)?”. The answer is
“yes”. In fact, this problem is equivalent to that of Sylvester. The equivalence
of these two questions is due to a duality that exists between points and lines
7
in a projective plane.
Euclid’s Elements describes plane geometry in terms of constructions (e.g.,
triangles, squares or etc.) attainable via compass and straight-edge. Essential
to many of the constructions is the concept of a right angle. Consider a line l0
and two lines which cross it, l1 and l2 . By the last of Euclid’s five postulates,
called the “parallel postulate”, these crossing lines, l1 and l2 , will intersect on
whichever side of l0 having interior angles summing to less than two right angles,
i.e., 180◦ . If l1 and l2 intersect at all, one of the two pairs of interior angles
must sum to less than 180◦ by Proposition 17 of Book One in Euclid’s Elements.
Thus, by constructing right angles at two distinct points on a line, one creates
two parallel lines, i.e., two lines which do not intersect. However, to obtain a
“duality” between points and lines, parallel lines cannot be allowed.
Although Euclid’s five postulates are quite natural, perhaps to the point
of seeming essential, one can also construct interesting alternative geometries
using less restrictive sets of postulates (or equivalently, “axioms”). Considered
most general among the alternative geometries is Projective Geometry. In this
geometry, there is no concept of measure, of angles or distance, or “betweeness”
among three points on a line. In contrast to Euclidean, Projective Geometry
asks “How much remains if we discard the compass . . . , and use the straightedge alone?” [23, p. 1].
A common conceptual model given for the projective plane (i.e., two-dimensional
Projective Geometry) is that of a sphere in R3 . (See Felsner’s description [24,
pp 70-72].) For this model, one translates an arrangement of point or lines on a
plane to an arrangement of points or great circles (resp.) on a sphere in a way
that preserves their incidences.
Consider the plane, π, as lying tangent to a sphere in R3 centered at the
origin. Points on the plane can be mapped in the obvious way to vectors (i.e.,
one-dimensional linear subspaces) in R3 , and, likewise, lines can be mapped to
two-dimensional linear subspaces. With this set of one- and two-dimensional
subspaces in mind, consider their intersection with the sphere centered at the
8
origin: the original points on the plane become points on the sphere and lines
in the plane π become great circles on the sphere.
By identifying antipodal points (i.e., considering them equivalent) on the
sphere, we have a complete representation of the points and lines on the plane.
Furthermore, there exists a unique great circle that lies in a plane parallel to π.
(Thus, it does not correspond to any line on π.) This unique great circle is the
so-called “line at infinity”. With some inspection, one can verify that lines that
are parallel on the plane correspond to great circles on the sphere that intersect
at the line at infinity.
Consider what happens when one rotates the sphere along with its arrangement of great circles. After rotation, the great circles can be reprojected onto π
(or any other plane) to form another line arrangement. This new arrangement
of lines is, in some sense, equivalent to the original; its incidence structure has
not been affected. In fact, one could produce any number of these equivalent
line arrangements by rotating the sphere and re-projecting. Such an operation
that preserves the incidence structure (as well as cross-ratios2 ) of the original
lines is called a “projective transformation” or “homography”.
Because of this, for an arrangement of lines in the plane, one may designate
one (or none) of them as the line at infinity and produce an equivalent arrangement. Thus, equivalence between two arrangements of lines is determined
by the incidence relationships within each arrangement, and not the angles or
distances between lines. In projective geometry (and more specifically combinatorial geometry), it’s incidence that is of primary concern.
There exists a “Principle of Duality” by which one may interchange points
and lines while preserving their analogous incidence relationships. Such an interchange is called a “duality mapping”. (Several such mappings are discussed in
the subsequent subsections.) Of course, one must also exchange other concepts
for their analog, e.g., “concurrent lines” for “collinear points”. For example,
from the claim “any two points determine a connecting line,” one may derive
2 Although cross-ratios are not pertinent to this dissertation, more information can be
found about them in, e.g., [25, Ch. 4.3]
9
the claim “any two lines determine an intersection point.”
Since the Principle of Duality does not hold for the Euclidean plane (because
incidence is not the only concern in such a context), one might wonder how this
concept might be used to reduce an algorithmic problem concerning points to
the analogous problem concerning lines. In order for our claims to be made
evident, we must further describe how to implement a duality mapping (or
transform).
A duality transform is a function which provides a mapping between points
and lines (in R2 ) such that their incidence relationships are preserved. For
example, let D be a duality transform. For any three collinear points p1 , p2 ,
and p3 , the lines D(p1 ), D(p2 ) and D(p3 ) must be concurrent (i.e., intersect at
a common point). Conversely, any three concurrent lines should transform into
three collinear points.
Again, let D be a duality transform. The following two properties are assumed (as axioms). In the subsections below, specific duality transforms will be
shown to have these two properties.
Axiom 2.2.1. D = D−1 , i.e. D is its own inverse.
Axiom 2.2.2. Point D(L) lies on line D(p) iff point p lies on line L.
Following from these two axioms is the following theorem:
Theorem 2.2.3. Points p1 and p2 determine line L if and only if lines D(p1 )
and D(p2 ) intersect at point D(L).
Proof. Assume points p1 and p2 determine line L, but lines D(p1 ) and D(p2 ) do
not intersect at point D(L). In this case at least one of the following is true:
• Point D(L) does not lie on D(p1 ).
• Point D(L) does not lie on D(p2 ).
By Axiom 2.2.2, this contradicts our original assumption.
The reverse implication follows from Axiom 2.2.1.
10
One should note that the duality transforms discussed here are not bijective
functions (i.e., one-to-one correspondences). In fact, such a bijective function
cannot exist for R2 . To see why this is true, for sake of contradiction, assume
that D is a bijective duality transform for R2 , and let l1 and l2 be two distinct
parallel lines in R2 . The two points D(l1 ) and D(12 ) must be distinct points
that, subsequently, determine a line l3 . Since D must preserve incidence, D(l3 )
should be a point in R2 incident to both l1 and l2 , but no such point exists
and, thus, a contradiction. The reader will see below how these limitations are
overcome in practice.
The next two sections will demonstrate two commonly used duality transforms. See [23, p. 15] for an axiomatization of Projective Geometry. For a
discussion of duality in the projective plane see [23, pp. 24-32]. See [26, pp.
201-205] and [27, pp. 12-15] for further discussion of duality transforms.
2.2.1
Polar Dual
Perhaps the first duality transform in common use is called the polar dual. It
maps point (a, b) to line ax + by = 13 and vice versa.
One can quickly see that this transform is only a partial function of lines
to points, since not every line can be expressed by an equation of the form
ax + by = 1 (i.e the lines which pass through the origin, e.g. x + y = 0).
The lines, which are not mapped, would conceptually correspond to points on
the line at infinity and require special treatment. This is consistent with our
representation of points by ordered pairs of numbers, since points on the line
at infinity also cannot be represented by a pair of real numbers. In practice,
a simple translation would be used so that no line in an arrangement passes
through the origin.
From the definition, it is apparent that this mapping possesses the property
required by Axiom 2.2.1. The following is proof that it also has the property
required by Axiom 2.2.2.
3 The
line ax + by = −1 is also commonly used.
11
Proof. Let P be the point at (a, b) which lies on a line L given by the equation
cx + dy = 1. Since P lies on L, we know that ca + db = 1.
By definition, D(P ) is the line given by equation ax + by = 1, and D(L) is
the point at (c, d). Obviously ca + db = ac + bd = 1, thus, we can see that point
D(L) lies on line D(P ).
One can see that the converse is also true, completing the proof.
2.2.2
Parabolic Dual
Within Computational Geometry, a more common duality transformation is
the “parabolic dual” 4 . It maps the point at (a, b) to the line given by equation
y = 2ax − b and vice versa.
Just as with the polar dual, this is only a partial function since vertical lines
(e.g. x=2) cannot be represented in the form required. Likewise, these (vertical)
lines correspond through duality with points on the line at infinity. Of course
for any given (finite) arrangement of lines, vertical lines can be detected. A
subsequent rotation can then be used to create a different arrangement without
vertical lines that preserves the incidence relationships.
From the definition, it is apparent that this mapping possesses the property
required by Axiom 2.2.1. The following is proof that it also has the property
required by Axiom 2.2.2.
Proof. Let P be the point at (a, b) which lies on a line L given by the equation
y = 2cx − d. Since P lies on L, we know that b = 2ca − d.
By definition, D(P ) is the line given by equation y = 2ax − b, and D(L) is
the point at (c, d). Since we know that b = 2ca − d, and thus d = 2ac − b, we
can see that point D(L) lies on line D(P ).
One can see that the converse is also true, completing the proof.
4 The author has not found the term “parabolic dual” in the literature, although it seems
to follow naturally from the description given by O’Rourke in [26, p 202].
12
2.3
Melchior’s Inequality
Prior to Gallai’s proof in [20] (and even before Erdős had published his question
in American Mathematical Monthly), an inequality was published in Deutsche
Mathematik that provides proof of the Sylvester-Gallai Theorem [28]. Let S
be an arrangement of lines. Let tk be the number of points through which k
lines of S intersect. Assuming |S| > 3 and not all lines pass through the same
intersection point, the following (i.e., Melchior’s Inequality) holds:
X
(k − 3)tk 6 −3
k>2
This immediately shows that there are at least three “ordinary” intersection
P
points, i.e., by rearranging the terms one can derive t2 > 3 + k>4 (k − 3)tk . It
further demonstrates that for every point incident to four (or more) lines, there
exists at least one other point incident to exactly two.
2.3.1
Euler’s Polyhedral Formula
In order to prove Melchior’s inequality, one may start from Euler’s polyhedral
formula. This formula demonstrates an invariant relationship between the vertices, edges, and faces determined by a plane graph. (A plane graph is the
drawing of a graph in the plane such that each vertex is distinct and no two
edges touch. By definition such a graph, capable of being drawn this way, is a
planar graph.) Let V be the set of vertices, E the edges, and F the faces (or
planar regions) determined by the graph. Euler’s polyhedral formula says that
|V | − |E| + |F | = 2. No fewer than 19 proofs exist for this formula [29]. See [30,
p 65] and [24, p 4] for two (distinct) simple proofs.
A similar invariant exists for the vertices, edges and faces determined by an
arrangement of lines in the projective plane. Any example quickly shows that
the analogous formula for an arrangement of lines is instead:
|V | − |E| + |F | = 1
13
An informal proof can be seen by returning to the spherical model for the
projective plane (see Section 2.2).
Let A be an arrangement of lines. Let S be an arrangement of great circles
on a sphere that would (each) project onto the lines of A. We may now construct
a graph. Consider the intersection of any two great circles to be a vertex. Two
vertices are adjacent if they they are adjacent intersection points on a great
circle. (A graph drawn on a sphere is equivalent to one drawn on the plane.
This can be seen by “stretching” an arbitrary face of the drawing out onto the
plane.) Since we have constructed a plane graph, its constituents are correlated
by Euler’s formula, |V | − |E| + |F | = 2. Now return to our original arrangement
of lines. Each intersection of lines corresponds to two intersections (at antipodal
points) on the sphere. Likewise, the faces and “edges” are counted twice. Thus,
2(|V | − |E| + |F |) = 2, or equivalently |V | − |E| + |F | = 1.
We now provide a more formal inductive proof (similar to the one provided
by Felsner in [24, p 73]).
Lemma 2.3.1. Given an arrangement of lines in the projective plane, let V be
the set of vertices, E the edges, and F the faces determined by the arrangement.
The following invariant must hold:
|V | − |E| + |F | = 1
Proof. Let S be an arrangement of lines in the projective plane. Given |S| = 2,
we have |V | = 1, |E| = 2 and |F | = 2 (i.e. edges and faces “wrap around” in
the projective plane).
Assume it is true for |S| = n. Let l be a line not in S, and let S 0 = S ∪ {l} so
that |S 0 | = n+1. Let V 0 , E 0 , and F 0 be the vertices, edges and faces, respectively,
of S 0 . By adding l to S additional vertices, edges, and faces are created. For
each edge created by l in E 0 , a face of F is split into two. For each vertex created
by l in V 0 , an edge E is split into two. Therefore, the value of |V | − |E| + |F | is
not affected by the addition of l, i.e., |V 0 | − |E 0 | + |F 0 | = |V | − |E| + |F | = 1.
14
2.3.2
Melchior’s Proof of the Inequality
One can find proof of Melchior’s inequality in the original article, [28]. The
article is in German and might not be readily accessible to the reader. Hence,
a proof similar to that of Melchior is provided below.
Theorem 2.3.2 (Melchior’s Inequality). Let tk be the number of intersection
points through which k lines pass. Given an arrangement of lines, not all concurrent, in the projective plane, the following inequality holds:
X
(k − 3) · tk 6 −3
k>2
Proof. Let fk be the number of faces determined by a set of exactly k edges from
P
the arrangement, and let F = k>3 fk be the total number of faces. (Note that
when an arrangement of lines is not all concurrent each face has at least three
sides.) Let tk be the number of points (i.e., intersections) at which k lines cross.
P
Let E be the number of edges, and V = k>2 tk be the number of vertices. We
start with the following identity:
2·
X
k · tk = 2E =
k>2
X
k · fk
k>3
By combining this with Lemma 2.3.1 (i.e., F =
P
k>3
fk = E + 1 − V ), we get
the inequality:
2E =
X
k · fk > 3F = 3E + 3(1 − V )
k>3
Thus,
2E − 3F = 3(V − 1) − E > 0
From this we can derive Melchior’s inequality:
3(V − 1) − E = 3V − E − 3
=3·
X
k>2
tk −
X
k>2
15
k · tk − 3 =
X
(3 − k) · tk − 3 > 0
k>2
Therefore,
X
(k − 3) · tk 6 −3
k>2
This inequality will be leveraged in Section 3.2.1 to obtain an analogous
bound for planes in R3 .
2.4
How Many?
A natural extension to Sylvester’s problem is to ask how many ordinary lines
must be determined by a set of n non-collinear points. Let t2 (n) be the minimum
number of ordinary lines determined over all sets of n non-collinear points in
the plane. From Kelly’s proof (in Section 2.1) we know t2 (n) > 1, and from
Melchior’s inequality (in Section 2.3) we know t2 (n) > 3. Is it possible that, as
n gets larger, t2 (n) never surpasses some constant? Erdős and de Bruin asked a
similar question in [31]. More specifically, they asked whether lim t2 (n) = ∞.
n→∞
√
In 1951, Motzkin confirmed this by showing that t2 (n) > n ([8]) 5 . The same
year, in [9] Dirac conjectured that t2 (n) > b n2 c for all n.
Only two specific cases are known for which n non-collinear points determine less than d n2 e ordinary lines. The first is the “Kelly-Moser configuration”,
published in [33], of seven points that determine only three ordinary lines. In
the same paper, Kelly and Moser prove t2 (n) >
3n
7 .
Their configuration thus
demonstrated that “in a certain sense, this is a best possible bound”[33]. See
Figure 2.2.
The other known configuration for which t2 (n) 6 d n2 e, due to McKee and
published in [2], consists of thirteen points that determine only six ordinary
lines. Eight of the points are on the vertices of two adjacent pentagons, with
the midpoint of the shared side containing an additional point. The remaining
four points are on the line at infinity. See Figure 2.3.
5 This
1951 paper of T.S. Motzkin introduced the term ordinary line (see [32]).
16
Figure 2.2: The Kelly-Moser configuration has seven points that determine only
three ordinary lines, and thus, t2 (7) = 3.
Figure 2.3: McKee’s configuration showing thirteen points determining only six
ordinary lines.
17
n
t2 (n)
3
3
4
3
5
4
6
3
7
3
8
4
9
6
10
5
11
6
12
6
13
6
Figure 2.4: Known values of t2 (n) originally published in [2].
In 1981 Hansen published, in [34], for his habilitation an erroneous proof
that t2 (n) >
n
2.
Reportedly, his proof was difficult to read and dubious from
the time of its publication. Csima and Sawyer made a careful study of Hansen’s
proof. They discovered his error, and ultimately improved the bound published
by Kelly and Moser. Csima and Sawyer demonstrated in [17] that, with the
exception of n = 7, t2 (n) >
6n
13 .
This bound is again “best possible”, in a
certain sense, because of McKee’s configuration of thirteen points.6 For n 6 13,
all values of t2 (n) are known and were published by Crowe and McKee in [2].
A table of these values is provided in Figure 2.4.
Asymptotically, Dirac’s conjectured bound cannot be improved. An infinite
family of configurations, due to Károly Böröczky and published in [2], is known
which demonstrates t2 (2m) 6 m for all m. Consider m points at the vertices of
a regular m-gon. The m
2 pairs of points determine only m distinct directions.
Thus, also select the m points on the line at infinity corresponding to these m
directions. Such a configuration determines only m ordinary lines, i.e., the line
connecting each vertex, v, to the point on the line at infinity determined by it’s
neighboring vertices. See Figure 2.5 for the configuration for m = 5.
2.5
Da Silva and Fukuda’s Conjecture
Let P be a set of n points, and L be a set of m lines in E2 . The set L is said to
isolate P if every point in P lies in a distinct region of E2 \ L.
In [35] Da Silva and Fukuda study minimal arrangements of lines which isolate a point set, i.e., min(P ) (|L0 | ∈ {L : L isolates P }). They demonstrate a
method to compute such a minimal arrangement. When the resulting arrangement is allowed to instead consist of “pseudolines” their method runs in time
6 The author finds it interesting that both proven linear bounds on t have a closely related
2
configurations demonstrating the bound to be “best possible”.
18
Figure 2.5: A Böröczky configuration of ten points determining five ordinary
lines.
O(|P | log(|P |)). (See Section 2.6 for more information on pseudolines.)
In the conclusion of their paper ([35]), they made the following conjecture.
Conjecture 2.5.1. Let S be a set of n point in R2 . Let l be a line that
contains no point of S and partitions S into two subset, B and R, such that
S1 6 S2 6 d n2 e. There exist points p1 ∈ B and p2 ∈ R such that the line
connecting them passes through no other point from S.
Unfortunately, this conjecture was shown to be false for |S| = 9. Using
oriented matroids, Finschi and Fukuda enumerated all possible arrangement of
few points in search of a counter-example ([3]). The conjecture holds for point
sets containing eight or fewer points. However, for nine points they found a
counter-example. See Figure 2.6. Because of the combinatorial explosion (e.g.,
15 296 266 possible “abstract order types” for nine points) 7 , it would likely
be difficult to extend their search much further (beyond nine points). The
7 The author does not know whether a bounded growth rate for the number of abstract
order types is known.
19
Figure 2.6: A counter example to Da Silva and Fukuda’s conjecture. Originally
published in [3].
conjecture is still open for larger point sets.
One might consider the partitioning of the point set by a line, as done by
Finschi and Fukuda, as establishing a “coloring” for the point set; That is, one
side might be, e.g, blue and the other side red. This leads one to the concept of
“bichromatic” point sets. Conjectures concerning bichromatic sets were made
as early as 1965 when Ron Graham asked whether a bichromatic arrangement of
lines (i.e., the dual context) necessarily determines a “monochromatic” intersection point. Chakerian later, in [36], published a proof that a “monochromatic”
intersection point is necessarily determined.
Da Silva and Fukuda’s conjectured (stronger) version of the Sylvester-Gallai
theorem provided motivation Pach and Pinchasi to further study bichromatic
point sets [37]. Their primary results related to how many lines, in proportion
to the total, pass through few points (e.g. 6 6) and at least one of each color.
The work of Pach and Pinchasi, in turn, motivated the author and his adviser
to attempt to improve their bounds. These results are covered in Chapter ??.
20
2.6
Pseudolines and Other Curves
There are many results concerning points and lines that do not depend on the
straightness of lines. This is the case for several well-known results including the
Szemerédi-Trotter incidence bound [14] and Csima and Sawyer’s bound on the
number of ordinary points.[17] Such results rely on the topology of the plane,
and are (typically) consequences of Euler’s polyhedral formula along with the
combinatorial properties of points and lines.
The term “Pseudogerade”, translated “pseudoline”, was coined by Levi for
the title of his 1926 paper, “Die Teilung der projektiven Ebene durch Gerage
oder Pseudogerade”.[38] According to Grünbaum [39, p. 40], Levi observed in
that paper that an arrangement of (non-straight) pseudolines could violate the
theorem of Pappus (or Desargues’) while remaining faithful to other important
qualities held by an arrangement of lines. Such arrangements that would violate
geometric laws were their pseudolines “made straight” are called non-stretchable
and are NP-hard to recognize.[40] Interestingly, as a consequence of the topological representation theorem of Folkman and Lawrence [41] arrangements of
pseudolines can be shown to be equivalent to rank 3 oriented matroids. (See
[42].)
An arrangement of pseudolines is defined as a set of simple closed curves in
the projective plane in which
• every pair share exactly one point,
• and at that point cross.
When examining small arrangement, one might suppose that allowing the
pseudolines to “bend” does not do much to generalize their structure. In fact,
any pseudoline arrangement with eight or fewer lines are combinatorially equivalent to an arrangement of (straight) lines.[43, p. 134] The large disparity between
line and pseudoline arrangements becomes apparent as their size increases. It
is known that for sufficiently large n that the number of combinatorially distinct line arrangement is bounded above by 2O(n log n) . However, the number of
21
x
y
z
Figure 2.7: Pappus’s theorem requires the points x, y and z to be collinear.
2
distinct pseudoline arrangements is known to be bound below by 2Ω(n ) . [43,
p. 135] (An upper bound for the number of distinct pseudoline arrangements
n
2
was found to be 3( 2 ) ≈ 20.792n by Knuth in [44, p. 39], and recently improved
2
to 20.657n in [45].) Thus, the proportion of pseudolines arrangement that are
stretchable quickly goes to zero and n gets large.
The (combinatorial) difference between lines and pseudolines can be seen in
an arrangement with as few as nine lines. An easily seen demonstration of this
difference is taken directly from Pappus’s8 theorem, which requires that among
a set of eight lines, a collinearity exists among certain intersections. See Figure
2.7. (Pappus’s theorem holds for a projective plane over any field, [25, Chpt.
5]. Thus, the non-“stretchability” of the pseudoline arrangement in the figure
is a somewhat fundamental to the geometry of the projective plane.)
By weakening the first condition for pseudolines (above), allowing curves to
cross more than once, an obvious generalization is obtained. It was this generalization (“curves”) for which Pach and Sharir, in [46], proved an upper bound
on the number of incidences between curves and their points of intersection.
They classified an arrangement of curves as having “α degrees of freedom” and
8 Some
sources, e.g, [25], spell this as “Pappos”.
22
Figure 2.8: An arrangement of pseudolines that is not realizable over any field.
“multiplicity-type β”. To clarify, a set of curves S satisfies these conditions if:
1. for any α points there are at most β curves of S incident to all of them,
and
2. any pair of curves from S intersect in at most β points.
Let P be a set of points in the plane. Let n = |S| and m = |P |. In [46],
Pach and Sharir proved the following incidence bound, which is a generalization
of the Szemerédi-Trotter incidence bound.
Theorem 2.6.1 ([46]). The number of incidences between S and P is at most
c(α, β) mα/(2α−1) n(2α−2)/(2α−1) + m + n
where c(α, β) is a positive constant depending only on α and β.
Let Pk ⊆ P be a set of points each incident to at least k curves. By substituting m = |Pk | and noting that the number of incidences must be greater than
mk, one finds the following corollary.
23
Corollary 2.6.2. The size of the set Pk is at most
b
c(α, β)
where b =
n
n2
+
kb
k
,
2α−1
α−1 .
This incidence bound will be used in Section 4.2 to prove a theorem analogous
to the weak–Dirac for arrangements of curves.
Like so many other results in this field, their proof of this incidence bound
relies upon Euler’s Polyhedral Formula, and not the algebraic nature of these
curves. Thus, it’s possible that a stronger incidence bound holds for these types
of arrangements.
24
Chapter 3
Planes and Spheres
3.1
Lines Determined by Points in R2
This section contains a new result for lines in the plane. Its inclusion was motivated, in part, because it demonstrates in a more familiar setting the method
of proof that will be used again for planes in 3-space.
In this section, we will consider a set of n points in R2 . Let tk be the
number of lines incident to exactly k of these points. Let t be the total number
P
of determined lines, i.e., t = i>2 ti
In [47], Erdős and Purdy proved the following lemma. For the convenience
of the reader, we provide a proof.
Lemma 3.1.1. If r points are on a line, l, and s points are not on l, then
t2 > rs − s(s − 1).
Proof. The result is true for s = 0 and s = 1, so suppose s > 1. We shall use
induction on s. Let p, not on l, be the last point to be added for a total of s
points off of l. Assume it’s true for s − 1. By adding point p, at most s − 1
existing lines are spoiled, and r lines are created of which at most s − 1 already
25
exist. Thus,
t2 > r(s − 1) − (s − 1)(s − 2) − (s − 1) + r − (s − 1) = rs − s(s − 1).
The following lemma is due to Elliott [1].
Lemma 3.1.2. The number of lines determined by at most three points in a
plane is at least one-half the total number of lines. More specifically,
t2 + t3 >
t+3
.
2
Proof. Beginning with Melchior’s Inequality [28],
t2 > 3 +
X
(k − 3)tk .
k>3
Adding t2 + t3 to both sides yields,
2t2 + t3 > 3 + t2 + t3 + t4 + 2t5 + . . . .
Obviously, t = t2 + t3 + t4 + .... So,
2(t2 + t3 ) > 3 + t.
The lemma follows.
We shall use one more lemma, due to Kelly and Moser [33], which is also
a consequence of Melchior’s Inequality. Let ri be the total number of points
incident to exactly i determined lines.
Lemma 3.1.3. The total number of lines is at least one-third the total number
26
of point-line incidences. That is,
3t − 3 >
X
i · ti =
i>2
X
i · ri .
i>2
Proof. Again starting from Melchior’s Inequality,
−3 >
X
(k − 3)tk = −t2 + t4 + 2t5 + 3t6 + . . . .
k>2
By adding 3t to both sides,
3t > 3 + 2t2 + 3t3 + 4t4 + 5t5 + 6t6 + . . . .
The lemma follows.
By combining Lemmas 3.1.2 and 3.1.3 we see the following,
6(t2 + t3 ) > 3(t + 3) > 12 +
X
i · ti = 12 +
i>2
X
i · ri ,
i>2
or simplified,
t2 + t3 > 2 +
1X
i · ri .
6
(3.1)
i>2
Theorem 3.1.4. Let S be a set of n points in R2 . If n > 72k 2 + 2k − 1, and
no more than n − k points are collinear, then
t2 + t3 > k(n − k) − k(k − 1).
We define the degree of a point to be the number of determined lines to
which it is incident.
Proof.
Case 1: There exist two points, p and q, of degree < 6k.
Let l be the line determined by p and q. Assume l is incident to exactly
n − x points. Each line through p, other than l, must be incident to less than 6k
27
points, otherwise the degree of point q would be too high, i.e. a contradiction.
Likewise for lines through q. Thus, k 6 x < 36k 2 = (6k)2 .
def
By Lemma 3.1.1, we know t2 > f (x) = x(n−x)−x(x−1) = −2x2 +(n+1)x.
The second derivative of f (x) is negative, i.e., f 00 (x) = −4. Therefore,
min
k6x636k2
f (x) = min{f (k), f (36k 2 )}.
One can verify that when n > 72k 2 +2k −1, f (36k 2 ) > f (k), thus the lemma
is true for this case.
Case 2: There exist at least n − 1 points of degree > 6k.
From (3.1) we know (for positive k),
t2 + t3 > 2 +
1X
i · ri > 2 +
6
i>2
1
6k(n − 1)
6
= kn − k + 2 > k(n − k) − k(k − 1).
Therefore, the lemma is true in both cases.
We should mention that one could derive a similar bound for t2 + t3 using a
result of Kelly-Moser from [33] (i.e., t > kn− 12 (3k +2)(k −1) when at most n−k
are collinear and n sufficiently large) with Lemma 3.1.2. Using the Kelly-Moser
result one would need a larger value for k but could achieve a better lower bound
for n. One could also derive a result similar to the corollary in the following
section using this method.
3.1.1
Circles Determined by Points in R2
We shall now apply Theorem 3.1.4 to produce a corollary on the number of
circles determined by at most four points (among a finite set of points) in the
plane. The proof will utilize circular inversion.
Circular inversion is a transformation of the euclidean plane, i.e., a mapping
28
of R2 → R2 . This transformation has several properties useful to the combinatorial geometer, e.g., for demonstrating the existence of circles determined
by a finite set of points. Motzkin may have been first to use circular inversion
for this purpose. More specifically, Motzkin ([8]) proved that in a finite set of
points in the plane, not all collinear and not all cocircular, each point is incident
to either a three-point line or a three-point circle. Prompted by a conjecture
of Erdős, Elliott improved upon this by demonstrating a lower bound on the
total number of circles determined by such a set of points [1]. Using a similar method, Bálint and Bálintová ([48]) further extended the results of Elliott.
For any reader unfamiliar with circular inversion, we would recommend [49, pp.
334–346] for coverage of this transformation.
P. D. T. A. Elliott’s 1967 result [1] that the number of circles is at least
n−1
is slightly wrong. The lower bound is actually
2
1+
n−1
n−1
1
−
> 1 + (n − 1)(n − 3),
2
2
2
as may be seen by taking a circle with n − 1 points on it and a point p off the
circle. It is easy to arrange the points such that p lies on b n−1
2 c three-point
lines. (We discovered this when we tried to prove that the number of spheres is
at least n−1
and discovered that there is a subtractive term derived from the
3
Orchard Problem.) Apparently, this counter-example even escaped the notice
of the well-known geometer Beniamino Segre whom Elliott cited as providing an
eight point counter-example to his result. Elliott’s proof can easily be modified
to show the correct result with the same lower bound of 394 for n. By the
way, Bálint and Bálintová also published this more accurate lower bound, i.e.,
1 + 12 (n − 1)(n − 3), in [48]. (One of the anonymous referees informed us that
the correct bound was also published by Bálint in his 1969 thesis, [50].)
We define Invp (S) to be the circular inversion of a point set S about the
point p ∈
/ S. (In this case S may be finite or infinite.) Although we will not give
a formal definition of circular inversion, we provide its basic properties (see [49]
29
for details):
• Invp is self-inverse, i.e., q = Invp (Invp (q)).
• If l is a line passing through p, then Invp (l) = l.
• If l is a line not passing through p, then Invp (l) is a circle passing though
p.
Let S be a set of n points, one of which is p. Let S 0 = Invp (S\{p}) be the
set of n − 1 points formed by circular inversion about p. We begin with the
following lemma.
Lemma 3.1.5. If at most n − k points in S are incident to any line or circle,
then in S 0 , at most (n − 1) − k = n − k − 1 will be on any line. Thus, (by
Theorem 3.1.4) when n > 72k 2 + 2k, the number of determined lines incident
to at most three points in S 0 is at least k(n − k − 1) − k(k − 1). In S 0 ∪ {p}, at
most (n − 1)/2 of these lines can be incident to p.
Corollary 3.1.6. Let S be a set of n points in R2 with at most n − k (k >
1) points on any line or circle. If n > 72k 2 + 2k, then there exist at least
1
8 (2k
− 1)(n2 − (2k + 1)n) circles determined in S.
Let cr be the number of circles incident to r > 3 points in S. We denote by
(p)
cr
the number of circles in S incident to r points, one of which is p.
Proof. This corollary is true when there exists a circle or line incident to n − 1
points (see the exceptional case for Elliot’s result above), so assume k > 2.
Let S 0 = Invp (S\{p}), for some arbitrary p ∈ S. By Lemma 3.1.5, at least
k(n − k − 1) − k(k − 1) determined lines in S 0 are incident to at most three
points. Since at most
(p)
n−1
2
of these lines are incident to p in S 0 ∪ {p}, it must be
(p)
the case that c3 + c4 > k(n − k − 1) − k(k − 1) − (n − 1)/2. By repeating this
30
argument for all n points in S, each circle is counted at most four times. Thus,
c3 + c4 >
n
4
k(n − k − 1) − k(k − 1) −
n−1
2
=
3.2
1
(2k − 1)(n2 − (2k + 1)n).
8
Planes Determined by Points in R3
Let S = {p1 , p2 , . . . , pn }, a set of n points in R3 , no three collinear and not all
coplanar.
One method for determining the number of planes determined by a set of
points in R3 is to take an arbitrary point, p ∈ S, and project from p onto a
plane π, the other points of S. The lines determined (by the projected points)
on π correlate to the planes determined by the point set.
Definition 3.2.1. Let L be the set of
n
2
lines determined by points of S. Let
π be any plane which intersects every line in L, but does not contain any point
of S. Let pi ∈ S be the point from which we project, and let li,j be the line
connecting point pi with another point pj ∈ S. The projection of point pj on
π is the point of intersection of li,j with π. Since no three points are collinear,
this projection of pj is unique with respect to pi .
Although no three points are collinear in S, there might exist three points
whose projections onto π are collinear. Also, the point from which we project
is not itself projected; so a set of n points will produce n − 1 points projected
onto π.
Definition 3.2.2. Let tk (p) be the number of lines determined on π, when proP
jecting from point p, containing exactly k (> 2) points. Let t(p) = k>2 tk (p).
Definition 3.2.3. Let mk (p) be the number of planes containing exactly k
P
(> 3) points, one of which is point p. Let m(p) = k>3 mk (p).
31
Lemma 3.2.4. For any point p ∈ S,
tk (p) = mk+1 (p).
(3.2)
Proof. Let pn ∈ S be an arbitrary point from which we project the others onto
a plane π. Let si and sj on π be the projections of distinct points pi ∈ S and
pj ∈ S, respectively. Let l be the line determined by si and sj . The points pi ,
pj and pn determine a plane, P , which contains line l. Any other point whose
projection lies on l is incident to P . Since each point in S − {pn } has a unique
projection onto π, the identity follows.
This identity, and derivations thereof, will be utilized throughout this section
to obtain our primary results.
3.2.1
A Derivation from Melchior’s Inequality
Let tk be the number of lines determined by points in a plane containing exactly
k points.
We again use Melchior’s Inequality [28],
−3 >
X
(k − 3)tk .
k>2
Applying this inequality to the projection onto π from point p1 ∈ S, along
with the identity (3.2), we see that
−3 >
X
(k − 3)tk (p1 ) =
k>2
X
(k − 4)mk (p1 ).
k>3
Since this inequality is true for whichever point we choose, we may extend
this by summing over all n points in S,
−3n >
n X
X
(k − 4)mk (pi ).
i=1 k>3
32
Let mk be the number of planes determined by points of S containing exactly
k points. Each k-plane is counted exactly k times in this summation, yielding
−3n > (−1)(3)m3 + (0)(4)m4 + (1)(5)m5 + (2)(6)m6 + . . . .
From this we get the following theorem:
Theorem 3.2.5. Let S be a set of n points in R3 , no three collinear and not all
coplanar. Let mk be the number of planes determined by points of S containing
exactly k points. It must be the case that
−3n >
X
k(k − 4)mk .
k>3
By separating the m3 term from the summation and reordering the inequality, one can see our first corollary:
Corollary 3.2.6. Given a set of n points in R3 , no three collinear and not all
coplanar, there exists at least n planes containing exactly three points. More
specifically,
m3 > n +
X k(k − 4)
k>4
3
mk .
From Theorem 3.2.5, one may also derive a bound for the number of planes
determined by at most four points. The inequality from Theorem 3.2.5 can be
rewritten,
3n 6 3m3 + 0m4 − 5m5 − 12m6 − . . . .
Let m be the total number of determined planes. By adding 5m to both
sides, we see that
5m + 3n 6 8m3 + 5m4 + 0m5 − 7m6 − . . . ,
or
5m + 3n 6 8(m3 + m4 ).
33
From this we get our next corollary:
Corollary 3.2.7. Let S be a set of n points in R3 , no three collinear and not all
coplanar. Let m be the total number of planes determined by S. The number of
planes determined by S containing at most four points is more than five-eighths
of the total number of determined planes. More specifically,
m3 + m4 >
3.2.2
1
(5m + 3n).
8
Planes Incident to Exactly Three Points
In [17], Csima and Sawyer published their well-known result that among any
set of n 6= 7 points in a plane, not all collinear, there exist >
6n
13
lines incident to
exactly two points.1 We will now use this result to obtain an analogous result
for the number of planes determined by exactly three points in 3-space.
Let S be a set of n points in R3 , no three collinear and not all coplanar. Let
p1 ∈ S be the point from which we project the others onto π.
Since the points are not all coplanar, their n − 1 projections onto π are not
all collinear. Thus, Csima and Sawyer’s result can be applied:
m3 (p1 ) = t2 (p1 ) >
6(n − 1)
.
13
By summing the above inequality for all n points, we would count each
three-point plane three times. By taking one-third of that total, we arrive at
the following:
m3
1
2n(n − 1)
4 n
= (m3 (p1 ) + m3 (p2 ) + . . . + m3 (pn )) >
=
.
3
13
13 2
Theorem 3.2.8. Let S be a set of n points in R3 , no three collinear and not all
4 n
2
coplanar. There exists at least 13
2 planes determined by exactly three points.
1 In August 2012, Ben Green and Terence Tao published on ArXiv a manuscript claiming
to prove the long conjectured t2 (n) > n
, for all sufficiently large n. Their work is still being
2
reviewed.
2 Using the Green-Tao result, this bound becomes 1 n , for sufficiently large n.
3 2
34
3.2.3
Total Number of Determined Planes
The following lemma is an extension of a result of Erdős and Purdy (called
“Lemma 2”) in [12].
Lemma 3.2.9. Let S be a set of n points in R3 , not all coplanar and no three
collinear. Let m be the total number of planes determined by S. If exactly n − k
of the points are coplanar, then
m>1+k
n−k
k
n−k
−
.
2
2
2
Proof.
Case k = 1: Let S be the set of n points. Let M be the plane containing n − 1
points, and p the point not on M . Each pair of points on M , along with p,
determine a three-point plane. The relation is true with equality.
Case k = 2: Let p and q be the two points not on M . The line pq intersects
M at a point r ∈
/ S. There can be at most b n−2
2 c pairs that determine a line
through r, and therefore at most b n−2
2 c pairs that determine a plane with p that
is incident to q (or vice versa). Therefore, the number of planes determined is
n−2 at least 1 + 2 n−2
− 2 .
2
Case k > 3: Following the same argument as the case for k = 2, since there
are k2 pairs of points not on M , the number of planes determined is at least
1 + k n−k
− k2 n−k
.
2
2
Before we begin the primary result of this section, we will also need the
following lemma.
Lemma 3.2.10. Let mk be the number of planes incident to exactly k points.
Let m be the total number planes determined by a point set. Given a set of n
points, no three collinear and not all coplanar,
6m > 3n +
X k k>3
35
2
mk .
Proof. Theorem 3.2.5 states the following,
−3n >
X
(k 2 − 4k)mk =
k>3
X
(k 2 − k)mk − 3 ·
k>3
X
k · mk .
(3.3)
k>3
By negating this inequality one gets,
3m3 + 0m4 − 5m5 − 12m6 − 21m7 − 32m8 . . . > 3n.
(3.4)
Similarly, one can unwind the summation from (3.3) to get,
9m3 + 12m4 + 15m5 + 18m6 + 21m7 + 24m8 + . . .
> 3n +
X
(k 2 − k)mk . (3.5)
k>3
Adding (3.4) to (3.5) produces,
12m3 + 12m4 + 10m5 + 6m6 > 6n +
X
(k 2 − k)mk .
k>3
Therefore,
12m > 12(m3 + m4 + m5 + m6 ) > 6n +
X
(k 2 − k)mk .
k>3
Dividing the inequality by two produces the lemma.
This leads us to the following theorem, which is a generalization of the KellyMoser Theorem (called “Theorem 4.1” in [33]) to three dimensions:
Theorem 3.2.11. Let S be a set of n points in R3 , no three collinear and at
def
most n − k coplanar. If n > g(k) = 54k 2 + 92 k, the total number of planes
determined by S is at least 1 + k n−k
− k2 n−k
.
2
2
def
The function, f (k) = 1 + k
n−k
2
−
36
k
2
n−k
2
, is a cubic polynomial of k
and can be rewritten as:
f (k) =
1
9
3 3 1
1
k + (1 − 5n)k 2 + n(−1 + 2n)k + 1.
4
4
4
Let c1 and c2 , where c1 < c2 , be the function’s two local extrema at
√
−1 + 5n ± 1 − n + 7n2 . For all n > 4, f (c1 ) > 0 and f (c2 ) < 0. Fur-
thermore, f 00 (c1 ) < 0 and f 00 (c2 ) > 0.
Proof of Theorem 3.2.11: Trivially, S contains
n
2
pairs of points. We define
the degree for a pair of points to be the number of determined planes to which
the pair is incident.
Case 1: More than
These (>
n
2)
n
2
pairs of points have degree < 6k.
pairs cannot form a matching, hence two pairs of low degree
(< 6k) must share a point. Assume two such pairs are {p, q} and {p, r}. Let M
be the plane determined by the points p, q, and r. Let a < 6k be the number of
planes determined by S, not including M , incident to the pair {p, q}. Likewise,
let b < 6k be the number of planes incident to the pair {p, r}. Since no three
points are collinear, any plane passing through {p, q} can share at most one
point of S, other than p, with a plane though the pair {p, r}. Therefore, at
most a · b < 36k 2 points of S are not on M . If M has exactly n − x points on
it, then k 6 x < 36k 2 .
We claim that for all n > g(k), the following two conditions are true:
◦ 36k 2 < c2 , where c2 is the second local extremum of f (k).
◦ f (36k 2 ) > f (k)
To verify the first, it would be sufficient for (5 +
√
6)n > 324k 2 + 1, and thus,
it is true for all n > 44k 2 . To verify the second, one must first consider our
formula as a function of two variables, n and k, i.e.,
def
f1 (n, k) = 1 + k
n−k
k
n−k
.
−
2
2
2
37
Considering k to be a constant,
def
f2 (n) = f1 (n, 36k 2 ) − f1 (n, k)
is a convex quadratic function of n. By solving f2 (n) = 0, one can see that for
all n > d54k 2 + 29 ke, f2 (n) is positive.
From the two properties listed above, it is obvious that for all x such that
k 6 x < 36k 2 , it is also true that f (x) > f (k). So our inequality holds in this
first case.
Case 2: At most
n
2
pairs of points have degree < 6k.
There are at least n2 − n2 = 12 n(n − 2) pairs of degree > 6k. Let Pi be the
number of pairs of points incident to exactly i planes. By Lemma 3.2.10,
X k X
1
6m >
mk =
i · Pi > n(n − 2)(6k).
2
2
k>3
i>2
Thus, for all k > 1 (and n > 4),
1
1
n−k
m > n(n − 2)k > (n − k)(n − k − 1)k + 1 = 1 + k
.
2
2
2
Thus our inequality holds in this case as well.
3.2.4
Planes Determined by At Most Four Points
Lemma 3.2.12. Let m3 be the number of planes incident to exactly three points.
Given a set of points S, no three collinear, if r points lie on a plane π, and s
do not, then m3 > s 2r − 12 rs(s − 1).
Proof. We define a plane with more than three points to be spoiled. This lemma
is obviously true for s = 0 and s = 1. We will use induction on s.
Assume s > 1. Let p be one of the s points not on π. If p is removed there
are (s − 1) 2r − 2r (s − 1)(s − 2) three-point planes. The addition of p will spoil
at most 2r (s − 1) of those planes.
38
Let q be any one of the other s − 1 points, in S and not on π. Let x ∈
/ S
be the point (possibly at ∞) at which the line pq intersects π. There can be at
most
r
2
pairs of points, in S and on π, that determine a line through x, thus
forming a four point plane with p and q.
So, the addition of p introduces 2r new planes of which at most
r
2
(s − 1)
contain four or more points. Thus,
m3 >
r
r
r
r
r
(s − 1)
−
(s − 1)(s − 2) −
(s − 1) +
−
(s − 1)
2
2
2
2
2
r
1
=s
−
rs(s − 1).
2
2
Theorem 3.2.13. Let S be a set of n points in R3 , no three collinear and at
2
def
8
most n − k coplanar. If n > g(k) = 184 + 25
k + 4k, then
n−k
k
m3 + m4 > k
− (n − k)
.
2
2
def
We now define f (k) = k
n−k
2
− (n − k)
k
2
. The function can be rewritten
in the following form:
f (k) = k 3 −
3n 2 n2
k +
k.
2
2
Let c1 and c2 , where c1 < c2 , be the two local extrema of this function at
√
√
1
3)n. We note that f (c1 ) > 0 (i.e. f (c1 ) = 363 n3 ) and f (c2 ) < 0.
6 (3 ±
Furthermore, f 00 (c1 ) < 0 and f 00 (c2 ) > 0.
For obvious reasons, the following proof is very similar to the proof for
Theorem 3.2.11 in the previous section.
Proof of Theorem 3.2.13: Trivially, S contains
n
2
pairs of points. We define
the degree for a pair of points to be the number of determined planes to which
the pair is incident.
39
Case 1: More than
These (>
n
2)
n
2
pairs of points have degree <
48
5 k.
pairs cannot form a matching, hence two pairs must share
a point. Assume two such pairs are {p, q} and {p, r}. Let M be the plane
determined by the points p, q, and r. As demonstrated in Theorem 3.2.11,
2
there can be at most ( 48
5 k) = (92 +
4
2
25 )k
points of S not on M . If M has
2
exactly n − x points on it, then ( 48
5 k) > x > k.
We claim that for all n > g(k), the following two conditions are true:
2
◦ ( 48
5 k) < c2 , where c2 is the second local extremum of f (k).
2
◦ f ( 48
> f (k)
5 k)
To verify the first, it would be sufficient for (3 +
√
3)n > 558k 2 , and thus, it is
true for all n > 118k 2 . To verify the second, one must first consider our formula
as a function of two variables, n and k, i.e.
def
f1 (n, k) = k
n−k
k
− (n − k)
.
2
2
Considering k to be a constant,
def
f2 (n) = f1
n,
48
k
5
2 !
− f1 (n, k)
is a convex quadratic function of n. By solving f2 (n) = 0, one can see that for
2
8
all n > d 184 + 25
k + 4ke (and k > 1), f2 (n) is positive.
From the two properties listed above, it is obvious that for all x such that
2
k 6 x < ( 48
5 k) , it is also true that f (x) > f (k). So our inequality holds in this
first case.
pairs of points have degree < 48
5 k.
There are at least n2 − n2 = 21 n(n − 2) pairs of degree >
Case 2: At most
n
2
48
5 k.
Let Pi be the
number of pairs incident to exactly i planes. By Corollary 3.2.7, we know that
40
48
5 (m3
+ m4 ) =
48
5
8
5
· 6 · (m3 + m4 ) > 6m. Thus, by Lemma 3.2.10
X k X
1
48
(m3 + m4 ) >
mk =
i · Pi > n(n − 2)
k.
2
2
5
i>2
k>3
Therefore, m3 + m4 > 12 n(n − 2)k > k
n−k
2
. So the theorem is true in this case
as well.
3.2.5
Corollaries and Conjectures
From Theorem 3.2.11, we have two corollaries. Both relate to conjectures found
in [32, p. 815]. In [12], Erdős and Purdy proved m > 1 + n−1
for n > 552.
2
Theorem 3.2.11 improves this by showing it to be true for n > 59 (i.e., k = 1),
which leads to the following corollary.
Corollary 3.2.14. Let S be a set of n > 59 points in R3 , no three collinear
and not all coplanar. Let m be the number of planes determined by S, and t the
number of lines. For all such point sets, m − t + n > 2.
Proof. Putting k = 1, Theorem 3.2.11 says that m > 1 + n−1
2 . Since no three
points are collinear, t = n2 = n−1
+ (n − 1). Thus, m − t + n > 2.
2
This is conjectured by Purdy to be true when n > 32 for any finite set of
points in R3 that are not all coplanar and not all on two skew lines.
Corollary 3.2.15. Let S be a set of n > 225 points in R3 , no three collinear
and no n − 1 coplanar. Let m be the number of planes determined by S, and t
the number of lines. For all such point sets, m > t.
Proof. Putting k = 2, thus n > 225, Theorem 3.2.11 shows that
n−2
n−2
n
m>2
−
>
=t
2
2
2
(for n > 9).
Purdy proved in [51] that if the points are not all coplanar and not all on
two skew lines then m > ct, for some c > 0. Erdős asked what are sufficient
41
conditions for m > t. Purdy conjectured in [32] that m > t when n is sufficiently
large, no n − 1 points are coplanar and the points do not lie on two skew lines.
This conjecture is easily seen to be false for projective geometries over finite
fields. Let q = pk , where p is any prime and k > 1. We denote by P G(d, q) the
projective d-space over GF (q), and by m
k q , we denote the Gaussian coefficient.
In P G(3, q), the number of points (same as the number of planes) is 43 q = q 3 +
q 2 +q+1, and the number of lines is 42 q = (q 2 +1)(q 2 +q+1) = q 4 +q 3 +2q 2 +q+1
[52, p. 66] [53, p. 168]. (From this one can see that n < t > m.) Since P G(3, q)
is a rank-4 matroid, the conjecture is also false for matroids.
This conjecture has also been extended by Purdy to d-dimensional space
[32, p. 815]. We define the rank of a flat (i.e., rk(f )) to be one more than
its dimension, e.g., points have rank one, lines rank two, etc. A set of flats
F = {f1 , f2 , . . . , fr } is defined to be a covering set of flats for a point set
S ⊂ Rd if every point in S is incident to a flat in F . Furthermore, the rank of a
set of flats F is defined to be the sum of the ranks of its members, i.e., rk(F ) =
rk(f1 ) + rk(f2 ) + . . . + rk(fr ). A point configuration S is irreducible if there
does not exist a covering set of r (> 2) flats with rank at most d + 1 = rk(Rd ).
Let wk be the number of determined flats of rank k. For any sufficiently
large irreducible point configuration in Rd , Purdy conjectures that wd > wd−1 .
By projection (i.e., one which preserves the number of (d − 2)-flats), it would
follow that wd > wd−1 > wd−2 > . . . > w1 , thus implying unimodality in this
case. (Ben D. Lund has found a counter-example to this conjecture in Rd for
d ≥ 4. A joint paper is currently in preparation that intends to prove a related
result for R3 , and provides a corrected variation of the conjecture for higher
dimensions.)
Seymour proved [54], in the more general context of matroids, that if no five
points are collinear then t2 > mn. This is related to a conjecture by Mason
[55] and others that the sequence of Whitney Numbers (i.e., w1 , w2 , w3 , . . .) is
log-concave, i.e., wi2 > wi−1 wi+1 for all i > 0. Purdy proved in R3 [51] that if
the points do not all lie on a plane then t2 > cmn, for some c > 0. See [32] for
42
further discussion of these and other conjectures.
3.3
Spheres Determined by Points in R3
Circular Inversion has a lesser known extension to higher dimensional space,
i.e., spherical inversion. For the convenience of the reader, we provide below
a basic definition and the relevant properties of spherical inversion. We refer
the reader to [56, pp. 83–87] for a rigorous definition and proof of the given
properties.
Let p be any fixed point in R3 . Without loss of generality, let p be the origin
of a coordinate system. We assume, of course, that coordinates are determined
relative to some orthonormal basis. Let q, with coordinates (x, y, z), be any
other point in R3 . We define the “spherical inverse” to be (for any q 6= p):
def
Invp (q) =
q
=
||q||2
1
2
x + y2 + z2
(x, y, z)
(This inversion occurs about a sphere of unit radius centered at p. Invp (p) is
left undefined.) We apply this mapping, i.e., Inv, not only to points but also
to sets of points, either infinite or finite, intending the obvious results.
The mapping has the following properties (for any arbitrary point p):
• Invp is self-inverse, i.e., q = Invp (Invp (q)).
• If π is a plane passing through p, then Invp (π) = π.
• If π is a plane not passing through p, then Invp (π) is a sphere passing
through p.
Lemma 3.3.1. Let S be a set of n > 5 points in R3 , not all cospherical, no
n − 1 coplanar, no four cocircular and no three collinear. For any arbitrary
p ∈ S, the set Invp (S\{p}) ∪ {p} will be likewise.
Proof. For contradiction, assume q1 , q2 and q3 are three collinear points in S.
Choose the center of inversion, p, to be any other point in S. Let l be the line
43
determined by the qi . Let M be the plane incident to l and passing through
p (the center of inversion). Let N be any plane incident to l but not passing
through p. Obviously Invp (l) = Invp (M ) ∩ Invp (N ), which is the intersection
of a plane (i.e., Invp (M )) with a sphere (i.e., Invp (N )). Thus, Invp (l) is a circle
incident to these four points. By self-inversion, one can see that the converse is
also true.
The other claims are obvious.
3.3.1
Spheres Incident To Exactly Four Points
Using the lemma above and some careful counting, we now prove a lower bound
on the number of four point spheres determined by a set of points in R3 . We also
utilize a result, from an earlier section, concerning the number of three-point
planes determined by a set of points.
Theorem 3.3.2. Let S be a set of n > 5 points in R3 , not all cospherical or
coplanar, no four cocircular and no three collinear. There exist at least n3
spheres incident to exactly four points of S, where =
9
208 .
Proof. Choose any p ∈ S to be the center of inversion. Let S 0 = Invp (S\{p}),
and let S 00 = S 0 ∪ {p}.
Assume n − 1 points in S are coplanar. Let p be the point not on the plane.
Since no three points are collinear and no four cocircular, there are n−1
four
3
point spheres determined through p, and the theorem is true. We now assume
that no n − 1 points are coplanar in S.
By Lemma 3.3.1, S 00 is a set of n points such that no n−1 are coplanar. Thus,
the points of S 0 are not all coplanar (and no three collinear). From Theorem
3.2.8, we know that S 0 determines at least
2
13 (n
− 1)(n − 2) three-point planes.
There are two possible cases.
• Case 1: At most 81 (n − 1)(n − 2) of these three-point planes (determined
by points in S 0 ) pass through p.
44
2
− 18 )(n − 1)(n − 2) =
The set S thus contains at least ( 13
3
104 (n − 1)(n − 2)
four point spheres incident to point p.
• Case 2: More than 81 (n − 1)(n − 2) of these three-point planes (determined
by points in S 0 ) pass through p.
We now carefully count the
n−1
2
= 12 (n−1)(n−2) pairs of points in S 0 (as
seen from p). Each of the at least 81 (n − 1)(n − 2) planes that pass through
p account for at least three of these pairs, i.e., at least
3
8 (n
− 1)(n − 2)
pairs. This leaves at most ( 21 − 38 )(n−1)(n−2) = ( 18 )(n−1)(n−2) pairs to
form a three-point plane with p in S 00 . Hence, the set S 00 must determine
2
at least ( 13
− 18 )(n − 1)(n − 2) =
3
104 (n − 1)(n − 2)
three-point planes that
3
104 (n
− 1)(n − 2) four point
do not pass through p.
Thus, the set S (again) contains at least
spheres incident to point p.
Since the selection of p was arbitrary this argument may be repeated n
times, counting each of these spheres four times. Thus the number of four point
3
9 n
)n(n − 1)(n − 2) = 208
spheres determined by S is at least ( 14 )( 104
3 .
3.3.2
The Orchard Problem
Let t3 (S) be the number of three-point lines determined by some finite point set
S in the plane. Let torchard
(n) be the maximum of t3 (S) for all n element point
3
sets S in the plane containing no four collinear points. We refer to determining
the exact value of torchard
(n) as the “classic” Orchard Problem.
3
Consider a set of n points in three-space, no three collinear and no four
cocircular. Our ultimate aim is to prove that the minimum number of spheres
determined by these points is exactly 1 + n−1
− torchard
(n − 1).
3
3
To this end, we let S be a set of n cospherical points with no four cocircular.
We shall prove in this section that the maximum number of planes, determined
by S, that share a common point p ∈
/ S is equal to torchard
(n). For brevity, we
3
(n).
will refer to this bound as Mmax
3
45
We begin with the lemma below that will be used in the following section
(i.e., Section 3.3.3) to prove a lower bound on the total number of determined
spheres. It is this lemma that led us to the rather unexpected equivalence to
the Orchard Problem.
Lemma 3.3.3. Let S be a set of n points in R3 , exactly n − k cospherical, no
four cocircular and no three collinear. Let σ0 be the sphere to which n − k points
are incident. Let p ∈ S be any point not on σ0 . There are at least n−k
−
3
Mmax
(n − k) spheres determined by p and three points on σ0 . Furthermore,
3
each of these spheres is incident to at most 3 + k points of S.
Proof. From p, project the n−k points of σ0 onto a plane π. Let tk (p) be defined
as in Definition 3.2.2 for the n − k points being projected. Since no four points
are cocircular, tk (p) = 0 for all k > 4. So obviously, t3 (p) 6 Mmax
(n − k) 6
3
torchard
(n − k) 6 13 n−k
3
2 . All point triples on σ0 not forming a plane through
p must determine a sphere with p.
In the proof above, we assert that obviously M3max (n) 6 torchard
(n). Combi3
natorially, one can see that torchard
(n) 6 31 n2 , since no two distinct three-point
3
lines may share a pair of points. This is further improved by observing that, by
counting all pairs of points, 3t3 + t2 = n2 . Using the Csima-Sawyer result (i.e.,
t2 >
6n
13 ),
this upper bound can be lowered to b 61 n2 −
25
3
78 nc.
Improving upon Sylvester’s work, point configurations have been demonstrated by Burr, Grünbaum, and Sloane [6], and again by Füredi and Palásti’s
[57] that come very close to this theoretical bound. More specifically, these configurations have shown that torchard
(n) > b 61 n2 − 12 n + 1c.4 See [7, pp. 315–318]
3
for coverage of more recent results.
A set of n points on a sphere, no four cocircular will project onto n points
on a plane, no four collinear, so there will be at most torchard
(n) three-point
3
3 Although the same paper (apparently) proves an even better bound for torchard (n), using
3
instead the Green-Tao bound of n
for ordinary lines, the same argument gives an upper bound
2
of n(n − 2)/6 for torchard
(n).
3
4 Green and Tao have (apparently) proved that the Orchard bound is precisely 1 + bn(n −
3)/6c, for sufficiently large n.
46
lines and consequently at most that many planes through three points on the
sphere and the point of projection. However, to show these two bounds to be
equivalent we need to also prove the more difficult “converse” of this. That is,
we must prove that every orchard configuration in the plane can back project
into points on a sphere, no four cocircular. From this equivalence, we conclude
that not only is 1 + n−1
− torchard
(n − 1) the lower bound for determined
3
3
spheres (see Section 3.3.3), but that this bound is always attainable. (Note that
the projection of a circle onto a plane is a conic section, and that five points,
no three collinear, are required to determine a unique conic section [23, p. 85].)
Theorem 3.3.4. Let S be a configuration of n points, no four collinear, on
a plane π. There must exist a set S 0 of n points, all cospherical but no four
cocircular, and a point of projection p such that projecting the points from S 0
onto π produces the point set S.
Proof. Let π be the plane in R3 containing the point configuration S. We
will assign each point in S unique coordinates (x, y, 1), for some x,y ∈ R. We
further require the points of S to be rotated such that no two points share a
first coordinate.
Let w = (w1 , w2 , 1), x = (x1 , x2 , 1), y = (y1 , y2 , 1) and z = (z1 , z2 , 1) be
any four points of the configuration S. Let σp be a unit sphere, centered at
p = (p1 , p2 , 0) and tangent to π, to which the points of S 0 are incident. The
point from which we project will be the center of the sphere, i.e., p.
If any three of the four points are collinear they will not determine a conic
section, so we assume that this is not the case. We also assume that x, y and z
are labeled in a clockwise order, as seen from p.
Let a = ||w − p||, b = ||x − p||, c = ||y − p||, and d = ||z − p|| be the
euclidean distances from each of the four points on π to the center of the sphere
σp . Let ŵ =
1
a (w
− p), x̂ = 1b (x − p), ŷ = 1c (y − p), and ẑ = d1 (z − p) be unit
vectors representing four points on the sphere which project to w, x, y and z,
respectively.
47
Let mp (x, y, z) be the plane determined by points at (x̂ + p), (ŷ + p) and
(ẑ + p). The plane mp (x, y, z) (which does not contain p) intersects σp in a
circle. Let cp (x, y, z) be the circle formed by the intersection of mp (x, y, z) with
σp .
We denote by (û, v̂) the scalar product of two vectors û and v̂. We now
consider w, x, y and z to be fixed. Let c = (ŷ − x̂) × (ẑ − ŷ). Obviously, c is
a vector orthogonal to the plane mp (x, y, z). For any two points, e.g., û and v̂,
on the circle cp (x, y, z), their scalar products (as vectors relative to the center)
with c are the same, i.e., (û − p, c) = (v̂ − p, c). Therefore, ŵ is on cp (x, y, z) if
def
and only if f (p1 , p2 ) = (ŵ, c) − (x̂, c) = 0.
The function f (p1 , p2 ) can be rewritten as (Aa + Bb + Cc + Dd)/(a · b · c · d),
where A,B,C and D are constants (i.e., their values are constant relative to the
points w, x, y and z). Since we are concerned with the roots of the function f ,
we will only consider its numerator. We also fix p2 to be zero.
def
Let g0 (p1 ) = (a · b · c · d) · f (p1 , 0) = Aa + Bb + Cc + Dd, and thus,
g00 (p1 ) = p
A(w1 − p1 )
1 + (w1 − p1
+
w22
+p
B(x1 − p1 )
1 + (x1 − p1 )2 + x22
C(y1 − p1 )
D(z1 − p1 )
+p
+p
.
2
2
1 + (y1 − p1 ) + y2
1 + (z1 − p1 )2 + z22
)2
The first term of g00 (p1 ) has two conjugate poles which are the complex zeroes
of the monic quadratic 1 + (w1 − p1 )2 + w22 = p21 − 2w1 p1 + 1 + w12 + w22 . The
quadratic polynomials are all distinct, and since their roots occur in conjugate
pairs, their roots are all distinct. So, g00 (p1 ) has eight poles, implying that g0 (p1)
is not identically zero.
Let g1 , g2 , . . . , g7 be conjugates of g0 formed by reversing (in the seven possible ways remaining) the signs of its last three terms. By taking the product
of these eight functions we obtain a polynomial (i.e., with no radicals). Let
P (p1 ) = g0 (p1 ) · g1 (p1 ) · . . . · g7 (p1 ). Obviously, P (p1 ) is zero wherever g0 (p1) is
zero and furthermore, P (p1 ) is not identically zero. Thus, P (p1 ) is a polynomial
having at most a finite number of (real) roots.
48
Since there are only a finite number of four point combinations in S, there
must exists a point (p1 , 0, 1) on π to which the sphere may be tangent such that
the points of S 0 , no four cocircular, project onto the points of S. Therefore,
torchard
(n) = Mmax
(n).
3
3
3.3.3
Total Number of Spheres Determined
In this section, we shall prove the following.
Theorem 3.3.5. Let S be a set of n points in R3 , not all cospherical or coplanar, no four circular and no three collinear. If n > 883, then the number of
spheres determined by S is at least 1 + n−1
− torchard
(n − 1).5 This bound is
3
3
best possible.
We begin with the following lemma.
Lemma 3.3.6. Let S be a set of n points in R3 , at most n − k cospherical or
coplanar, no four cocircular and no three collinear. If S 0 = Invp (S\{p}) for
some p ∈ S, then S 0 is a set of n − 1 point with at most n − 1 − k coplanar.
Since spherical inversion is self-inverse, it’s also the case that S = Invp (S 0 ) ∪
{p}.
Proof. Let π be a plane incident to exactly r points in S 0 . There are two cases.
• Case: p is incident to π.
The Invp (π) is a plane containing r + 1 points (including p) in S. Since
r + 1 is at most n − k, the lemma follows.
• Case: p is not incident to π.
The Invp (π) is a sphere containing r + 1 points (including p) in S. Since
r + 1 is at most n − k, the lemma follows.
5 Using
the Green-Tao bound for torchard
(n−1), this result becomes
3
49
1
(n−1)(n2 −6n+10).
6
We now provide a lemma, analogous to the previous one, for when n − k
points are coplanar.
Lemma 3.3.7. Let S be a set of n points in R3 , exactly n − k coplanar, no
four cocircular and no three collinear. Let π be the plane to which n − k points
are incident. Let p ∈ S be any point not on π. There are at least n−k
spheres
3
determined by p and three points on π. Furthermore, each of these spheres is
incident to at most 3 + k points of S.
Proof. Since no three points are collinear and no four cocircular, each point
triple from the plane will determine a unique sphere with p.
The following two lemmas place an upper bound on the intersection of two
sets. The proof for the primary theorem of this section utilizes the inclusionexclusion principle, which requires an upper bound for the size of the intersection
of two such sets.
Lemma 3.3.8. Let S be a set of n points in R3 , exactly n−k (k > 2) cospherical,
no four cocircular and no three collinear. Let σ0 be the sphere to which n − k
points are incident. Let p ∈ S and q ∈ S be any two distinct points not on σ0 .
Let σp be the set of spheres determined by p and three points from σ0 . Likewise,
let σq the set of spheres determined by q and three points from σ0 . It must be
the case that |σp ∩ σq | 6 31 n−k
2 .
Proof. The spheres in σp ∩ σq all contain at least five and at most 3 + k points.
Furthermore, no two distinct spheres in σp ∩σq can share a pair of points from σ0 ,
since that pair along with p and q determine a sphere. This reduces our problem
to determining how many distinct point triples from σ0 can there be such that
no two triples share a pair. From this, we see that |σp ∩ σq | 6 31 n−k
2 .
Lemma 3.3.9. Let S be a set of n points in R3 , exactly n − k (k > 2) coplanar,
no four cocircular and no three collinear. Let π be the plane to which n − k
points are incident. Let p ∈ S and q ∈ S be any two distinct points not on π.
Let σp be the set of spheres determined by p and three points from π. Likewise,
50
let σq the set of spheres determined by q and three points from π. It must be the
case that |σp ∩ σq | 6 13 n−k
2 .
Proof. The spheres in σp ∩ σq all contain at least five and at most 3 + k points.
Furthermore, no two distinct spheres in σp ∩σq can share a pair of points from σ0 ,
since that pair along with p and q determine a sphere. The lemma follows.
Proof of Theorem 3.3.5. We shall consider all possible cases.
• Case: Exactly n − 1 points in S are cospherical.
Follows from Lemma 3.3.3.
• Case: Exactly n − 1 points in S are coplanar.
Follows from Lemma 3.3.7.
• Case: Exactly n − 2 points in S are cospherical.
Let σ0 be the sphere incident to n − 2 points. Let p and q be the two
points not on σ0 . Let σp be the set of spheres determined by p and three
points on σ0 , and let σq be defined in similarly.
From the Inclusion-Exclusion Principle, we know |σp ∪ σq | = |σp | + |σq | −
|σp ∩ σq |. The spheres in σp ∪ σq are each incident to at most five points.
From Lemmas 3.3.3 and 3.3.8, we see that
n−2
1 n−2
1 n−2
−
−
3
3
2
3
2
n−2
n−2
n−1
>2
−
>1+
− torchard
(n − 1)
3
3
2
3
|σp ∪ σq | > 2
(for all n > 10). Note that by the Füredi-Palásti configuration [57],
torchard
(n) > b 61 n2 − 12 n + 1c.
3
• Case: Exactly n − 2 points in S are coplanar.
Let π be the plane incident to n − 2 points. Let p and q be the two points
not on π. Let σp be the set of spheres determined by p and three points
from π, and let σq be defined similarly.
51
From the Inclusion-Exclusion Principle, we know |σp ∪ σq | = |σp | + |σq | −
|σp ∩ σq |. The spheres in σp ∪ σq are each incident to at most five points.
From Lemmas 3.3.7 and 3.3.9, we see that
|σp ∪ σq | > 2
n−2
1 n−2
n−1
−
> 1+
− torchard
(n − 1)
3
3
3
2
3
(for all n > 8).
• Case: Exactly n − 3 points in S are cospherical.
Let σ0 be the sphere incident to n − 3 points. Let p, q and r be the three
points not on σ0 . Let σp be the set of spheres determined by p and three
points on σ0 , and let σq and σr be defined similarly.
Again, the Inclusion-Exclusion Principle shows that
|σp ∪ σq ∪ σr | > |σp | + |σq | + |σr | − |σp ∩ σq | − |σp ∩ σr | − |σq ∩ σr |.
The set σp ∩ σq ∩ σr contains precisely those spheres incident to six points.
Since these spheres are already subtracted, we need not alter the formula
above.
Therefore, the number of spheres incident to at most five point is at least
3
n−3
1 n−3
n−3
−
−
3
3
2
2
n−3
n−3
n−1
=3
−2
>1+
− torchard
(n − 1)
3
3
2
3
(for all n > 11).
• Case: Exactly n − 3 points in S are coplanar.
Let π be the sphere incident to n − 3 points. Let p, q and r be the three
points not on π. Let σp be the set of spheres determined by p and three
points on π, and let σq and σr be defined similarly.
52
Again, the Inclusion-Exclusion Principle shows that
|σp ∪ σq ∪ σr | > |σp | + |σq | + |σr | − |σp ∩ σq | − |σp ∩ σr | − |σq ∩ σr |.
The set σp ∩ σq ∩ σr contains precisely those spheres incident to six points.
Since these spheres are already subtracted, we need not alter the formula
above.
Therefore, the number of spheres incident to at most five point is at least
n−3
n−3
n−1
3
−
>1+
− torchard
(n − 1)
3
3
2
3
(for all n > 10).
• Case: At most n − 4 points in S lie on any plane or sphere.
Let S 0 = Invp (S\{p} for some p ∈ S. S 0 is a set of n − 1 points with at
most n − 3 on any plane.
By combining Theorem 3.2.11, Corollary 3.2.7 and Lemma 3.3.6, we know
that if n > 883 then the number of planes determined at most four points
n−5 4
in S 0 is at least 58 1 + 4 n−5
− 2 2 = 18 (10n2 − 125n + 380). We
2
need an upper bound on how many of these planes pass through p.
From the proof of Lemma 3.3.3, we know that the number of planes de
termined by points in S 0 passing through p is less than 31 n−1
2 . All other
planes determined in S 0 correspond to a sphere determined by points in
S. The number of spheres, passing through p, determined by at most five
points in S is at least
1 n−1
1
1
2
(10n − 125n + 380) −
=
(26n2 − 363n + 1132).
8
3
2
24
We can repeat this argument for all n points in S, counting each sphere at
53
most five times. Thus, the number of spheres determined by S is at least
n
5
1
24
(26n2 − 363n + 1132) =
(for all n > 34).
54
1
(26n3 − 363n2 + 1132n)
120
n−1
>1+
− torchard
(n − 1)
3
3
Chapter 4
Dirac’s Conjectures
4.1
Strong Dirac
As discussed in the introduction (Section 1.2.1), the Strong Dirac conjecture
states that among an arrangement of n lines, there must exist a line incident
to n/2 − c points of intersection, for some constant c. In this section, we show
this conjecture to be false for pseudolines. (See Section 2.6 for more information
about pseudolines.)
4.1.1
Wedges and Eppstein’s observation
A beautiful feature of Felsner’s arrangement, Figure 1.1, is its symmetry. It’s the
symmetry of a regular hexagon (i.e., the dihedral group D6 ). While studying
simplicial pseudoline arrangements (ones in which each planar face has three
sides), Eppstein observed that arrangements with dihedral symmetry can be
generated, similar to a kaleidoscope, from the contents of a single “wedge”.[29]
Figure 4.1 shows a single wedge from Felsner’s arrangement.
He noted that the entire path of a line through an arrangement can be
traced by considering that line to be “bouncing”, like a laser beam bouncing off
mirrors, from one side of the wedge to the other. (Notice that in Figure 4.1 the
beams must “retrace” their path after the third bounce.) In fact for straight55
Figure 4.1: A single wedge from Felsner’s arrangement.
line arrangements, this bouncing must follow the law of reflection: the angle of
incidence equals the angle of reflection. By applying basic trigonometry, one may
deduce for straight-line arrangements the number and locations of the bounces
as a function of the wedge angle and the beam’s initial angle of incidence.
Necessary conditions to generate a valid arrangement
To generate an arrangement from a wedge, the wedge must have an angle of π/k
for some positive integer k ≥ 2. The arrangement is produced by alternately
rotating and duplicating the wedge or its mirror image, k times each, so that
they fill the plane.
For pseudoline arrangements, the “bouncing” beams need not obey the law
of reflection. However, the beams must have two properties (inherent in the law
of reflection) that are not immediately obvious.
Lemma 4.1.1. When multiple beams meet at a point on the side of a wedge,
the order (by angle) that they enter the point, must be reverse of the order that
they leave the point.
Proof. Let w be an edge of the wedge, and P be the point at which a set of m
beams meet. Without loss of generality, we assume that w is the top of a wedge.
Label the beams entering point P clockwise from the wedge l1 , l2 , . . . , lm . Let π
be the permutation corresponding to the order the beams exiting point P . Per
56
w
l1
P
l2
...
...
lπ(m)
...
lm
...
lπ(2)
lπ(1)
Figure 4.2: A set of m beams all intersection at a point P , on a side of the
wedge.
a clockwise ordering, beam lπ(i) exits before beam lπ(j) whenever i < j. See
Figure 4.2.
We claim that for a valid arrangement to be produced π(i) = m − i + 1, for
1 ≤ i ≤ m. Assume the contrary is true for some valid arrangement. Let k be
the smallest k such that π(k) < m − k + 1. Consider the arrangement of lines
generated by this wedge, and let P 0 , lying on w0 , be one of the k points that
corresponding to the original P . Each beam that bounces at P in the wedge
corresponds to two lines through P 0 , both of which cross w0 and each other at
that point. See Figure 4.3.
Consider a circle of radius centered at P 0 and the 4m + 2 intersection
points on the circumference formed by these lines. Each of the original beams
corresponds to four points. (The radius of this circle must be small enough so
that P 0 is the only point of intersection among these lines.) A pair of lines
that cross within the circle, must form points that alternate as one progresses
around the circumference. However, beginning with a point corresponding to
lπ(k) and progressing around the circle to its match, one will either find three or
one points (depending on the direction) corresponding to lm−k+1 . The lack of
symmetry implies that there exist two lines that meet but do not cross at P 0 .
57
...
lm-k+1
...
lπ(k)
...
...
lπ(k)
...
w'
lm-k+1
...
P'
...
lπ(k)
...
lm-k+1
...
...
lm-k+1
...
lπ(k)
...
Figure 4.3: A set of m beams all intersection at a point P , on a side of the
wedge.
Theorem 4.1.2. When the wedge angle is π/k, any beam that does not pass
through the tip of the wedge must must bounce k or k − 1 times. And a beam
passing through the tip does not bounce.
Proof. Consider a set of lines k in a projective plane all passing through the
same point P . Let L be a line not passing through P . It must be the case that
L forms k points of intersection with the original k lines. In a Euclidean plane,
there can exist k − 1 iff L is parallel to one of the original k lines.
The theorem follows by letting the edges of the wedge generate the set of k
lines, and letting L be a line generated from a beam.
As with Felsner’s arrangement a beam might retrace its path after the d k2 eth
bounce. Berman, in [58], further develops Eppstein’s “kaleidoscope” method to
construct and classify many types of symmetric simplicial pseudoline arrangements.
Each beam can be broken into a sequence of segments that are split at the
locations of “bounces”; each beam has two infinite segments. We provide the
58
following lemma without proof.
Lemma 4.1.3. For a wedge to generate a valid pseudoline arrangement, the
following are necessary conditions:
• No segment intersects itself.
• No two segments intersect more than once (including end points).
• Two segments that “touch” must cross at that point.
• A finite segment must have its endpoints on opposite sides of the wedge
4.1.2
Pseudoline counterexample to Strong Dirac1
Theorem 4.1.4. For any j ∈ N+ , there exists an arrangement of n = 18j + 7
pseudolines such that no pseudoline is incident to more than 8j + 2 vertices.
Proof. We will describe the construction of a wedge for a pseudoline arrangement for arbitrary j, and show that it has the claimed number of pseudolines
and intersection property. We refer to Berman [58, Fig.11] for a proof that the
described wedge actually represents a pseudoline arrangement.
For an arbitrary j, the wedge angle will be π/(6j +2). There are four distinct
symmetry classes of pseudolines, plus the line at infinity. Two of these will be
represented by the sides of the wedge; we will call these the top and bottom
edges. Two will be represented by beams; we will call these the red and blue
beams.
Let ri be the point at which the ith bounce of the red beam occurs, counting
from infinity. Likewise, let bi be the point at which the ith bounce of the blue
beam occurs. After the beams reach the points r3j+1 or b3j+1 , respectively, the
beams “retrace” their paths. More specifically, for any j, rk = r6j+2−k and
bk = b6j+2−k .
We call r3j+1 and b3j+1 the “terminating points” for their respective beams.
Prior to reaching its terminating point, every third bounce of the blue beam
1 The
construction described in this section was found by Ben Lund.
59
coincides with a bounce of the red beam (i.e., ri = b3i for i ≤ j). The two
beams are parallel to the bottom edge before the first bounce, and both b1 and
r1 are on the top edge.
We will proceed by induction. For j = 1, the theorem holds; the arrangement
generated from this wedge contains 3(6j +2)+1 = 25 pseudolines, each of which
incident to at most 8j +2 = 10 vertices. See Figure 4.4 for the wedge, and Figure
4.5 for the associated arrangement.
Figure 4.4: The wedge for j = 1, the base case for our induction.
Assume that the theorem holds for j − 1. While maintaining for the points
each existing bounce their distances to the corner of the wedge from the previous
case, we reduce the wedge angle to π/(6j +2). In order to produce from this new
wedge a valid arrangement, we must specify how to construct {r3j−1 , r3j , r3j+1 }
and {b3j−1 , b3j , b3j+1 } for their respective beams.
We begin by extending the red by placing r3j−1 , r3j , and r3j+1 on alternating
sides of the wedge, each slightly closer to the corner of the wedge than the
previous. This extension adds only 6 vertices to its associated red lines.
To extend the blue beam, we must cross the red beam placing b3j−1 on
the opposite side of the wedge. The subsequent point, b3j , coincides with rj .
Finally, place b3j+1 at an appropriate location on the opposite side of the edge,
slightly farther from the corner than rj+1 . This extension adds a total of eight
vertices for the associated blue lines, and two more for the red lines.
We must now consider the additional vertices formed on the sides of the
wedge (which correspond to the axes of symmetry). To one set of axes, we
60
Figure 4.5: The arrangement for j = 1, containing 3(6j+2)+1 = 25 pseudolines,
each of which incident to at most 10 vertices.
added eight vertices each; to the other, we added only six each. See Figure 4.6
for the j = 2 case, i.e., the first complete “extension”.
Figure 4.6: The wedge for j = 2.
In the resulting arrangement, there will be 18j + 7 lines with none incident
to more than 8(j − 1) + 2 + 8 = 8j + 2 vertices, completing the induction.
61
4.2
Weak Dirac
In this section, we show that incidence bounds imply analogs to the Weak Dirac
theorem for a variety of different types of arrangements.
4.2.1
α-Curve combinatorics
First, we will define exactly the class of arrangements we will consider.
Definition 4.2.1. Let G = (L, P, I) be a connected, simple, bipartite graph on
vertex sets L and P and edge set I. If each pair of elements in L is connected
to exactly α elements of P , then G is an α-curve combinatorics.
This is a generalization of a line combinatorics, as defined by Bartolo et. al.
[59]. It can be thought of as the unordered combinatorics of a set of curves in
the plane, each intersecting α times.
Definition 4.2.2. For any curve combinatorics G = (L, P, I), let tk (G) be the
number of elements of P with degree k and, let r(G) be the maximum degree
of any element in L.
Some bounds on tk and r follow from the definition of α-curve combinatorics.
The following upper bounds on tk are a generalization of Lemma 2.1 in [13].
Theorem 4.2.3. Let G = (L, P, I) be an α-curve combinatorics with |L| = n.
1. tk (G) ≤ α
n
2
/
k
2
for 2 ≤ k ≤ n;
2. tk (G) < 2αn/k for α(2n)1/2 ≤ k ≤ n.
Proof.
1. We will count triplets (l, l0 , p), with l, l0 ∈ L, p ∈ P , and (l, p), (l0 , p) ∈ I,
in two different ways.
There are n2 pairs of elements in L; each of these pairs is adjacent to α
elements in P . Thus, there are α n2 such triplets.
62
There are at least tk (G) elements of P having degree k or more; each of
these is adjacent to k2 or more pairs of lines. Thus, the number of such triplets
is at least tk (G) k2 . Solving the inequality yields the result.
2.
Let P = {p1 , p2 , ..., pt }, and let Li ⊆ L be the elements of L that are
adjacent to pi .
n =
≥
≥
t
[ Li i=1
t
X
X
i=1
1≤i<j≤t
|Li | −
kt − α
|Li ∩ Lj |
t
.
2
Assume, for contradiction, that t ≥ d2αn/ke.
(2n)1/2
2αn − α
2
n ≥
> n.
Dirac’s original proof of the Ω(n1/2 ) bound on r relied only on combinatorial
arguments [9]. The following is a generalization of his argument.
Theorem 4.2.4. Let G = (L, P, I) be a α-curve combinatorics with |L| = n
such that no set of α elements of P is adjacent to every member of L. Then
r(G) = Ω(n1/(α+1) ).
Proof. Let g be the maximum degree of any element of L. Let h be the maximum
number of elements of L adjacent to any subset Pα ⊂ P of size α.
We will show that g ≥ h. Let P 0 ⊂ P be a subset of size α with each member
adjacent to the same set L0 ⊂ L of size h. By our hypothesis, there exists a
l0 ∈ L not adjacent to every element of P 0 . By the definition of an α-curve
combinatorics, there must be α paths of length 2 between l0 and each element of
L0 . However, no two elements of L0 can both be adjacent to the same element
63
p0 ∈ (P \ P 0 ), otherwise there will be a K2,α+1 . Thus, l0 must be adjacent to at
least h elements of P , and thus g ≥ h.
Assume h = nx , for some value x. Let l ∈ L have degree g. Since l has α
paths of length 2 to each l0 ∈ L \ l, αg h ≥ n − 1, which implies g ≥ n(1−x)/α .
We’ve demonstrated that g ≥ max(nx , n(1−x)/α ) which has its minimum
value when x = 1/(α + 1). This completes the proof.
4.2.2
A conditional Weak Dirac for α-curve combinatorics
The following lemma can be used to easily prove analogs to the Weak Dirac
theorem for several specific types of arrangements.
Lemma 4.2.5. Let δ, be constants with 0 ≤ δ < /2. Let G = (L, P, I) be an
α-curve combinatorics with |L| = n. If
tk (G) = O(n2+δ /k 2+ + n/k)
for all k such that 0 ≤ k < n, then either
1. G is a Kn,α , or
2. there is an element ` ∈ L with degree Ω(n1−δ/ ).
Proof. For any pair {l0 , l00 } ∈ L × L, l0 6= l00 , let P{l0 ,l00 } ⊂ P be the elements to
which both l0 and l00 are adjacent. Note that |P{l0 ,l00 } | = α for all pairs {l0 , l00 }.
Let ld be the number of pairs {l0 , l00 } such that min(deg(p) : p ∈ P{l0 ,l00 } ) = d.
Counting pairs of elements in L two ways,
n
X
li =
i=1
n
.
2
Proposition 4.2.6. For any c < 1, there exists a v such that
n/2v
n
li < c
.
2
(δ/)
X
i=2v n
64
(4.1)
Proof. The total number of points with degree 2j (i.e., t2j ) is bounded above
by C(n2+δ /2j(2+) + n/2j ) for some constant C. In addition, an element of P
j+1 with degree 2j+1 is adjacent to at most 2 2 pairs of elements in L. Thus,
j+1
2X
li
<
i=2j
1 2j+1
t2j
α
2
< (22j+1 C/α)(n2+δ /2j(2+) + n/2j )
< (2C/α)(n2+δ /2j + n2j ).
Let Ct = 2C/α.
n/2v
blog2 nc−v
X
X
li
<
i=2v n(δ/)

j+1
2X

j=v+(δ/)blog2 nc

li 
i=2j
blog2 nc−v
X
<
Ct (n2+δ /2j + n2j )
j=v+(δ/)blog2 nc

<
Ct n2+δ
∞
X

blog(n)c−v
2−j + n
j=v+(δ/)blog2 nc
X
2j 
0
Ct n2+δ (2−(v+(δ/)blog2 nc) /(1 − 2− )) + n2 /2v−1
< Ct n2+δ 2−v n−(δ/) /(1 − 2− ) + n2 /2v−1
< Ct 2−v n2+δ n−δ /(1 − 2− ) + (1/2v−1 )n2
1
1
+
n2 .
< Ct
2(v−1) (2 − 1) 2v−1
<
Choosing a sufficiently large v ensures that the sum is less than c
n
2
, completing
the proof of Proposition 4.2.6.
Proposition 4.2.7. One of the following must hold:
1. either G contains a complete bipartite subgraph KΩ(n),α , or
2. |P | = Ω(n2−δ/ ).
Proof. Proposition 4.2.6 together with equation 4.1 immediately implies that
either
65
1.
P2v nδ/
2.
Pn
If
P2v nδ/
i=1
li = Ω(n2 ), or
i=n/2v li
i=1
= Ω(n2 ).
li = Ω(n2 ), then |P | = Ω(n2 )/(2v nδ/ ) = Ω(n2−δ/ ), and the
theorem is proved.
Pn
2
Assume
i=n/2v li = Ω(n ). We will construct induced subgraphs G ⊇
G1 ⊇ G2 ⊇ . . . ⊇ Gα , as needed, by incrementally removing nodes from G.
Select a point p1 ∈ P with deg(p1 ) ≥ n/2v . Let L1 ⊆ L be the set of nodes
adjacent to p1 . Let P1 be the set of elements in P \ {p1 } that are adjacent
to some element in L1 . Construct G1 = (L1 , P1 , I1 ) as the induced subgraph
formed by L1 and P1 . Clearly, G1 is a (α − 1)-curve combinatorics.
From Proposition 4.2.6, either |P1 | = Ω(|L1 |2−δ/ ) = Ω(n2−δ/ ), or there is
an element p2 ∈ P with deg(p2 ) ≥ n/22v . In the second case, repeat the above
procedure to get a graph G2 .
This process may be repeated until either:
• we find a subset Pj ⊆ P , j ≤ α, with |Pj | = Ω((n/2jv )2−δ/ ) = Ω(n2−δ/ ),
or
• we can construct a complete bipartite subgraph of G containing the nodes
Lα ∪ {p1 } ∪ . . . ∪ {pα }, with |L| ≥ n/2vα = Ω(n).
This completes the proof of Proposition 4.2.7.
By Proposition 4.2.7, there are two cases to consider.
First, assume |P | = Ω(n2−δ/ ). Since every point of P is adjacent to at least
one line of L, the pigeonhole principle implies that there must be a line adjacent
to Ω(n1−δ/ ) points of P , proving the theorem.
Now, assume there exists a complete bipartite subgraph G0 = (L0 , P 0 , I 0 ), G0 ⊆
G with |L0 | = Ω(n) and |P 0 | = α. If |L0 | = n, then G is a Kn,α , proving the
theorem. Otherwise, there exists an element l ∈ (L \ L0 ) not adjacent to all α
elements of P 0 . By the definition of a curve combinatorics, l must have α paths
of length 2 to every element of L0 . However, since no two elements of L0 can
66
both be incident to the same point in P \ P 0 (otherwise a forbidden subgraph
would exist), l must be adjacent to |L0 | = Ω(n) elements of P \ P 0 .
Application to Pach-Sharir curves
Pach and Sharir defined “well-behaved” curves as follows [46].
Definition 4.2.8. Let C be a set of simple curves in the plane. The set C is
said to have β degrees of freedom and multiplicity-type α, if
• for β points there are at most α curves passing through all of them, and
• any pair of curves intersect in at most α points.
These two conditions correspond, respectively, to a forbidden Kβ+1,α and a
forbidden K2,α+1 in the bipartite graph encoding the incidences between such
a set of curves and a set of points.
Pach and Sharir proved the following upper bound on the number of incidences between “well-behaved” curves and points [46].
Theorem 4.2.9. Let P be a set of m points and let C be a set of n simple
curves all lying in the plane. If C has β degrees of freedom and multiplicity-type
α, then the number of incidences between P and C is
Oα,β mβ/(2β−1) n(2β−2)/(2β−1) + m + n .
We can use this with Lemma 4.2.5 to prove the following corollary.
Corollary 4.2.10. Let C be an arrangement of n simple curves all lying in
the plane. If C has β degrees of freedom, multiplicity-type α, and every pair
of curves intersects exactly α times, then either all of the curves intersect in a
single set of α points, or
r(C) = Ωα,β (n).
Proof. Let tk = tk (C). The corollary will follow immediately from Lemma 4.2.5
and the following upper bound on the number of curves incident to k or more
67
points:
tk = O(n2 /k 2+ + n/k).
Since there are at most α
n
2
points, this inequality clearly holds for small k.
Otherwise, by Theorem 4.2.9 we have
β/(2β−1) (2β−2)/(2β−1)
ktk = O(tk
n
+ tk + n).
Thus, either tk = O(n/k) or tk = O(n2 /k 2+ ), where = 1/(β − 1).
68
Chapter 5
Conclusions and Future
Work
5.1
Planes and Spheres
In this paper, our results for point sets in R3 assumed no three points are
collinear. It would be interesting to know whether similar results could be
obtained by assuming that at most j points are collinear, for some fixed j > 3.
If one were to allow an arbitrary number of collinear points, the lower bounds
for the number of objects determined (e.g., lines, planes, etc.) become weaker.
Consider the class of configurations which have exactly n − k points collinear
for a fixed k. We note the following:
• At most
k
2
+ k(n − k) + 1 = O(n) lines are determined.
• At most
k
3
+ (n − k)
k
2
+ k = O(n) planes are determined.
• At most
k
4
+ (n − k)
k
3
+
k
2
n−k
2
= O(n2 ) spheres are determined.
It would also be interesting to know whether the results of this paper could
be matched, or improved upon, using oriented matroids.
69
5.2
Pseudolines
It is not clear whether the pseudoline arrangement constructed in Section 4.1.2
is best possible. That is, there might exist an infinite family of pseudolines
arrangements in which among a set of n pseudolines no member is incident to
(4/9−ct )n points of intersection. This naturally leads to the question of what is
the largest cp for which there does not exist an arrangement of n pseudolines, for
sufficiently large n, in which no member is incident to cp n points of intersection.
We would have preferred to have improved our “two-extremes” result in
Section 4.2.2, i.e., Proposition 4.2.7, to have shown that |P | = Ω(n2−δ/(2) ).
This would make the result more readily applicable to other incidence bounds
such as the one provided by Bourgain, Katz and Tao in [60] (i.e., “Theorem 6.2”
in their paper). However, we do not know how to prove this.
There are also other contexts in which our Weak Dirac result applies. For example, the incidence bound by Solymosi and Tao in [61] with a slightly stronger
hypothesis shows the existence of a variety incident to Ω(n1− ) points.
70
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