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SYLVESTER-GALLAI RESULTS AND OTHER CONTRIBUTIONS TO COMBINATORIAL AND COMPUTATIONAL GEOMETRY Jonathan Lenchner June 2008

SYLVESTER-GALLAI RESULTS AND OTHER CONTRIBUTIONS TO COMBINATORIAL AND COMPUTATIONAL GEOMETRY

DISSERTATION Submitted in Partial Fulfillment of the Requirements for the Degree of DOCTOR OF PHILOSOPHY (Mathematics) at the POLYTECHNIC UNIVERSITY by Jonathan Lenchner June 2008

Approved:

Department Head Copy No. Date

Approved by the Guidance Committee: Major: Mathematics

J´anos Pach Professor of Mathematics, Courant Institute and City University of New York

Deane Yang Professor of Mathematics, Polytechnic University Minor: Computer Science

Boris Aronov Professor of Computer Science, Polytechnic University

Herv´e Br¨onnimann (Research Supervisor) Associate Professor of Computer Science, Polytechnic University

Microfilm or other copies of this dissertation are obtainable from

UNIVERSITY MICROFILMS 300 N. Zeeb Road Ann Arbor, Michigan 48106

Vita Jonathan Lenchner received his B.A. summa cum laude, with highest distinction in mathematics from Dartmouth College in 1981. He then received a Certificate of Advanced Study in mathematics from Cambridge University as a Winston Churchill Scholar in 1982. Jonathan began his career in industry as a research analyst studying and modeling options in the precious metals markets with the Mocatta Metals Corporation in New York City. From there he took a position as research assistant to the chief operating officer and director of computing with A-Mark Precious Metals in Beverly Hills, California. In the years since, Jonathan has worked in hydrogeological consulting, telecommunications, and information technology. He has also done freelance software development, and created shareware and commercial software for the Macintosh computer. Jonathan has spent the last fourteen years at IBM, and the last eight years at the IBM T.J. Watson Research Center in Hawthorne, New York. In the mid to late nineties, before joining Research, Jonathan was the development lead for IBM’s foray into e-commerce. He currently manages a research team that specializes in knowledge management and discovery for services. He has worked in just-in-time supply chain management, e-commerce, esupport, human computer interaction, information retrieval, machine learning and data mining. Jonathan has always had a strong academic bend. Over the years he has taken classes in many academic subjects at a number of different universities. He has taught probability and discrete mathematics at the college level. Several years ago, his group at IBM Research teamed up to teach a course on data mining at Polytechnic University. Jonathan was once an active chess player and U.S. Chess master. Jonathan has always iv

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been interested in computer games and wrote his first chess program while a sophomore in college. The program played at a level just below that of the top programs of the time. Recently Jonathan teamed up with colleagues from IBM to create a much stronger, Javabased chess program. The program has the special capability of being able to describe in detail what it is thinking about during the course of a game. Jonathan also enjoys the piano and is a competitive long distance swimmer, finishing 43rd at the 2007 US Masters National 10K Open Water Swimming Championships. Jonathan lives in North Salem, New York, with his wife of eighteen years, Denise, and their three daughters, Rachel (15), Madeline (13) and Tova (10).

To my mother, Florence Lenchner (1923-1998), of blessed memory.

Acknowledgements I wish to acknowledge the following people: Herv´e Br¨onnimann, my advisor, for all his help and encouragement and for involving me in many academic and personal adventures. Noteworthy amongst these adventures were two swims across the Hudson River to raise money for Multiple Sclerosis. (Herv´e has actually done the swim three times.) Sue Whitesides, for inspiring me when I was an undergraduate student at Dartmouth, and for being the reason I took an interest in computational geometry in the first place. There are some people who have an uncanny way of describing even the most complicated arguments and making them seem simple. Sue is such a person. I thank Sue also for the invitations to her Barbados workshops. Boris Aronov for becoming interested in my work, for his many valuable suggestions which have significantly improved the quality of this thesis, and for assuring that I never take myself or my problems too seriously. Deane Yang for graciously pushing me from differential geometry to computational geometry when it became apparent that my tastes and mathematical strengths were more on the combinatorial side of things. Joe Mitchell, Estie Arkin and the other “carrot-heads” from Barbados for inspiring working sessions in Barbados and once back. I thank Joe and Estie in particular for hosting me at Stony Brook and for teaching me so much by their example. Other esteemed mathematicians and computer scientists who have welcomed me into their “club” including Erik Demaine, Branko Gr¨ unbaum, J´anos Pach and Ricky Pollack (J´anos, especially for agreeing to serve as my external Ph.D. committee member).

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My father, without whose nudging I probably would never have gotten this thesis completed. Rachel, Maddy and Tova for putting up with a missing dad for too long. I hope that the time and energy that I have put into this thesis has some positive impact on them in later life. Denise for threatening to kick me out if I didn’t get this thesis done by February, 2007, for not following through when I didn’t, and for putting up with my manic mood swings during the last year or so. Finally, I acknowledge my mother, who suffered a lot because of the personality defects that kept me from following through on the small amount of mathematical promise that I possessed some twenty-odd years ago. I gratefully acknowledge the financial support I have received from National Science Foundation Grant CCR-0133599 (the NSF Career Grant of my advisor, Herv´e Br¨onnimann) and from the IBM Corporation. Jonathan Lenchner, New York, February, 2008

AN ABSTRACT SYLVESTER-GALLAI RESULTS AND OTHER CONTRIBUTIONS TO COMBINATORIAL AND COMPUTATIONAL GEOMETRY by Jonathan Lenchner Advisor: Herv´ e Br¨ onnimann, Ph.D. Submitted in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy (Mathematics) June 2008 The first two chapters of this thesis concern Sylvester’s problem and its relatives. The first chapter extends the well known Theorem of Csima and Sawyer which states that an arrangement of n lines not all passing through a common point, which is not the Kelly-Moser arrangement of 7 lines and 3 ordinary points, must have at least

6n 13

ordinary points. We

show that besides the McKee arrangement of 13 lines and 6 ordinary points, there cannot be additional arrangements with

6n 13

ordinary points if n is odd.

Chapter Two concerns a strong form of the dual of Sylvester’s problem in the affine plane. The main result of the chapter is that in an arrangement of n lines in the affine plane, not all of which are parallel, and not all of which pass through a common point, there must be at least

2n−3 7

affine ordinary points as long as n 6= 6.

Chapter Three describes the relationship of the affine Sylvester problem considered in Chapter Two to the problem of finding arrangements with a maximum number of wedges. In this chapter we prove results about the maximum number of wedges and extend these results to give results concerning the minimum complexity of the outer layer. Chapter Four concerns collections of points and a notion of data depth called opposite quadrant depth. The main result of this chapter states that, given n points in the plane, there must be a point in the set of opposite quadrant depth n8 , and that, moreover, there are point sets for which there is no point with opposite quadrant depth greater than

n 8

+ O(1).

x

Chapter Five treats the generic problem of finding the minimum cost way of covering points (clients) by disks, given suitable cost functions, different possible placements of the clients, and constraints on how one may place the disks. In some cases exact solutions are obtained, and in others (1 + ²) or constant factor approximations are achieved. In a couple of instances NP-hardness results for the exact solution are given.

Contents Vita

iv

Acknowledgements

vii

Introduction

1

1 Contributions to the Classical Sylvester-Gallai Theory

4

1.1

Four proofs of Sylvester’s Theorem . . . . . . . . . . . . . . . . . . . . . . .

5

1.2

Finding an ordinary line in time O(n log n) . . . . . . . . . . . . . . . . . .

8

1.3

The n/2 Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.4

Proofs of the Kelly-Moser/Kelly-Rottenberg and Csima-Sawyer Theorems .

13

1.4.1

Preliminary Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

1.4.2

Sylvester-critical arrangements must contain lines of type (2, 0) . . .

19

1.4.3

The Kelly-Moser/Kelly-Rottenberg

Theorems . . . . . . . . . . .

21

1.4.4

The Saturation Property . . . . . . . . . . . . . . . . . . . . . . . . .

22

1.4.5

The Csima-Sawyer Theorem . . . . . . . . . . . . . . . . . . . . . . .

24

1.5

A Unified View of the n = 7 and n = 13 Examples . . . . . . . . . . . . . .

28

1.6

Uniqueness of the McKee arrangement for n = 13 . . . . . . . . . . . . . . .

31

1.7

Sharpening the Csima-Sawyer Theorem . . . . . . . . . . . . . . . . . . . .

43

1.8

Ordinary point bounds for small n . . . . . . . . . . . . . . . . . . . . . . .

74

1.9

Additional useful results . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

82

1.9.1

n/2 miscellany . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

82

1.9.2

Saturation versus epsilon excess . . . . . . . . . . . . . . . . . . . . .

85

1.9.3

Ordinary points in polygons . . . . . . . . . . . . . . . . . . . . . . .

86

1.10 Counting the number of arrangements and computer-assisted searches . . .

92

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3n 7

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2 The Affine or Sharp Dual of Sylvester’s Problem

97

2.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

2.2

Establishing the Sharp Dual Theorem . . . . . . . . . . . . . . . . . . . . .

98

2.3

The (5n + 6)/39 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99

2.4

Improving the bound to n/6 . . . . . . . . . . . . . . . . . . . . . . . . . . .

100

2.5

Improving the bound to n/4 − O(1) . . . . . . . . . . . . . . . . . . . . . .

102

2.6

The final push to 2n/7 − O(1) . . . . . . . . . . . . . . . . . . . . . . . . . .

107

2.7

Finding (finite) ordinary points in time O(n log n) . . . . . . . . . . . . . .

110

2.8

Ordinary point bounds for small n in the affine setting . . . . . . . . . . . .

112

2.9

Sylvester’s theorem in other spaces . . . . . . . . . . . . . . . . . . . . . . .

113

3 Wedges and Outer Layer Complexity

118

3.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

118

3.2

Euler’s Relation: Connection to Wedges . . . . . . . . . . . . . . . . . . . .

118

3.3

The Theory of Wedges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

120

3.4

Complexity of the Outer Layer . . . . . . . . . . . . . . . . . . . . . . . . .

127

3.4.1

The generic setup

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

127

3.4.2

Maximum complexity . . . . . . . . . . . . . . . . . . . . . . . . . .

128

3.4.3

Minimum complexity . . . . . . . . . . . . . . . . . . . . . . . . . . .

129

4 Opposite Quadrant Depth

141

4.1

Lower Bound Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

142

4.2

Establishing Hard Lower Bounds . . . . . . . . . . . . . . . . . . . . . . . .

146

4.3

Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

147

4.4

Recent developments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

148

5 Coverage of Point Sets by Disks

150

5.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

150

5.2

Scenario 1: Server Locations Restricted to a Discrete Set . . . . . . . . . . .

153

5.2.1

Exact solution to the 1D and 1.5D problems . . . . . . . . . . . . .

153

5.2.2

Approximating the one-dimensional discrete problem with linear cost 157

5.2.3

Toward the discrete 2-dimensional case . . . . . . . . . . . . . . . . .

162

Scenario 2: Server Locations Restricted to a Line . . . . . . . . . . . . . . .

162

5.3

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5.4

5.3.1

Servers along a fixed horizontal line . . . . . . . . . . . . . . . . . .

162

5.3.2

Finding the best axis-parallel line . . . . . . . . . . . . . . . . . . . .

169

5.3.3

Approximating the best line - any orientation . . . . . . . . . . . . .

173

Minimum-Cost Covering Tours . . . . . . . . . . . . . . . . . . . . . . . . .

176

5.4.1

A hardness result . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177

5.4.2

The case C ≤ 4: The exact solution is a single circle . . . . . . . . .

178

Bibliography

180

List of Figures 1.1

A closest (point, line) pair (p, l) as in the Kelly proof. . . . . . . . . . . . .

1.2

A connecting line l of smallest absolute angle with the connecting lines through p1 (on the line at infinity). . . . . . . . . . . . . . . . . . . . . . . .

1.3

5 7

A point configuration S where p0 is the point of minimal x-coordinate, r is a ray through p0 not passing through any additional point of S and l is the line through two or more points of S which intersects r at minimal positive distance from p0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.4

If the line l is not ordinary then it contains three points q0 , q1 , q2 . A line through p0 and one of q0 , q1 , q2 is necessarily ordinary. . . . . . . . . . . . .

1.5

10

B¨or¨oczky’s even n example with n = 12 points and 6 ordinary lines (dashed lines). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.6

9

11

Kelly-Moser example with 7 lines and 3 ordinary points. The third ordinary point is the horizontal point at infinity. In this and all future figures hollow dots indicate ordinary points. . . . . . . . . . . . . . . . . . . . . . . . . . .

12

1.7

McKee arrangement of 13 lines and 6 ordinary points. . . . . . . . . . . . .

12

1.8

An example of a line l with two attached ordinary points. The point p is attached through the finite shaded triangle and the point q is attached through the shaded infinite triangle. . . . . . . . . . . . . . . . . . . . . . .

1.9

14

An illustration of the setup and conclusion of the Three Clause Lemma. The triangle T in the Lemma is the finite triangle 4(p, q, r) in the figure. The hollow vertex is the ordinary point attached to l in T guaranteed by the Lemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiv

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1.10 Given a pseudo-triangle T which is the finite triangle determined by the pseudolines l, l0 , l00 , and a chosen reference pseudoline l, the generic placement of the first three vertices x1 , x2 , x3 in T . Given a minimal cycle x1 ≺ ... ≺ xr ≺ x1 the point xr may be in one of the four regions A, B, C or D. . . . .

16

1.11 Three pseudolines l, m and n partitioning the projective plane into four pseudo-triangular regions. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18

1.12 The pseudoline l with ordinary point p and additional line m through p. r is one of the neighbors of p along m and n is an additional line through r. . .

19

1.13 A line l of type (2, 0) with an ordinary point p. k is the second line through p and p0 is the closest vertex to p in one direction. . . . . . . . . . . . . . . 1.14 Having concluded that T1 is a cell, and so

p0

23

cannot be ordinary we draw a

third line k 00 through p0 intersecting l at the vertex q. . . . . . . . . . . . . . 1.15 An ordinary point p on the (2, 0) line l. k is the second line through p and

23

p0

is a neighboring vertex to p on one side of l. m, n are the two lines through p0 counting p as an attachment and o is a hypothetical fourth line through p0 . 24 1.16 Two lines l, k of type (2, 0) meeting at an ordinary point p. By the (2, 0) Saturation Lemma (Lemma 12) p is saturated and using a projective transformation we may assume that lines m, m0 and lines n, n0 are parallel pairs, meeting respectively at m∞ , n∞ . . . . . . . . . . . . . . . . . . . . . . . . .

25

1.17 Two lines l, k of type (2, 0) meeting at the saturated ordinary point p. We consider the case where there are necessarily two adjacent (4+)-gons amongst P, Q, R, S, e.g. R, S in the cyclical clockwise ordering. Since R is a (4+)-gon there is a closest vertex d0 adjacent to d along [d, n∞ ] and a line meeting d0 which forms an edge of R next to [d, d0 ]. We call this line h. Analogously, since S is a (4+)-gon there is a closest vertex a0 adjacent to a along [a, m∞ ] and a line (not drawn) meeting a0 which forms an edge of S next to [a, a0 ]. We call the second line i. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26

1.18 A common view of the (a) Kelly-Moser and (b) McKee examples. . . . . . .

29

1.19 Glued regular pentagons and 7-gons, together with their corresponding points at infinity. A regular n-gon has internal angle (n − 2)π/n, with π/n slices as shown. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

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1.20 Glued regular 9-gons, together with the “productive” points at infinity i.e. points which, when added, kill off more ordinary lines than they create. Points at infinity are labeled according to the angle they make with the horizontal. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

1.21 Glued squares, together with the single “productive” point at infinity - yielding an example with 8 points and 4 ordinary (dashed) lines. . . . . . . . . .

32

1.22 Glued hexagons, together together with the 6 “productive” point at infinity - yielding an example with 17 points and 13 ordinary lines. The ordinary lines are enumerated in the text. . . . . . . . . . . . . . . . . . . . . . . . .

33

1.23 B¨or¨oczky examples of 10 lines with 5 ordinary points and 12 lines and 6 ordinary points, both of which are saturated. . . . . . . . . . . . . . . . . .

33

1.24 An arrangement of 12 lines with 6 ordinary points which is not a B¨or¨ oczky arrangement and not saturated. The third from top and bottom ordinary points are not completely surrounded by triangular cells. The arrangement is just the McKee arrangement with the central vertical line omitted. . . . .

34

1.25 Suppose that ordinary points p and q reside on (2, 0) lines adjacent to the 5-crossing u. If there are no additional ordinary points adjacent to u through (2, 0) lines then p and q each absorb a unit of charge from u. p and q are also said to be associated with the 5-crossing u. . . . . . . . . . . . . . . . . . .

35

1.26 p is associated to the cell Q: An edge of the (quadrangular) cell Q is also the edge of a triangular cell whose opposite vertex p is an ordinary point, and one of the lines through p is a (2, 0) line. In such cases we distribute at least 1 4

units of charge to p from Q. . . . . . . . . . . . . . . . . . . . . . . . . . .

36

1.27 An edge of the (quadrangular) cell Q is also the edge of a triangular cell whose opposite vertex p is an ordinary point, and one of the lines through p is a (2, 0) line. A line through p cannot be a defining line of Q since then one of the lines that count p as an attachment (k in the diagram) must enter Q, but Q is assumed to be a cell. . . . . . . . . . . . . . . . . . . . . . . . . . .

37

1.28 6 hypothetical ordinary points along (2, 0) lines surrounding a 6-crossing. By saturation, every other line will have 4 attachments. Line n is such an example. 38

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1.29 For k = 5 since there cannot be ordinary points on adjacent lines so there is only one configuration of 4 ordinary point surrounding the 5-crossing, and this one configuration, by saturation, gives rise to a line, n, with 4 attached points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

38

1.30 The case of a hypothetical 4-crossing u with ordinary points adjacent to u on two separate lines l, o of type (2, 0). . . . . . . . . . . . . . . . . . . . . .

38

1.31 An ordinary point p on the (2, 0) line l surrounded by two adjacent (4+)-gons. 39 1.32 A hypothetical quadrangular cell R giving

1 4

unit of charge to four surround-

ing ordinary points, each on (2, 0) lines. Since l is a (2, 0) line, by the CsimaSawyer Lemma, k must be a (2, 0) line.

. . . . . . . . . . . . . . . . . . . .

40

1.33 The three (2, 0) lines l, k, m, each crossing at a distinct 4-crossing (indicated by a solid dot), with ordinary points as indicated with hollow dots. The ordinary points drawn closest to the 4-crossings are meant to be adjacent vertices to the respective 4-crossings. . . . . . . . . . . . . . . . . . . . . . .

41

1.34 The three lines u, v, w passing through 4-crossings q, r, p respectively, each pass through opposite ordinary points b, f, c, again in the same order. The ordinary point c, though a neighboring vertex to r is not necessarily located “close” to r. We consider first the case where w passes through z, the intersection point of u and v. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

42

1.35 The three lines u, v, w passing through 4-crossings q, r, p and opposite ordinary points b, f, c respectively intersecting in the only possible manner. . . .

43

1.36 The three lines u, v, w passing through 4-crossings q, r, p and opposite ordinary points b, f, c respectively intersecting in the only possible manner, with all vertices on the lines l, k, m accounted for.

. . . . . . . . . . . . . . . . .

44

1.37 The only possible configuration of three ordinary points along the “spokes” surrounding a 5-crossing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46

1.38 Filling out the configuration of three ordinary points surrounding a 5-crossing; the line k, which already counts p, q, r as attached points must have an additional ordinary or attached point inside the triangle T . . . . . . . . . . . .

47

1.39 The cases of four ordinary points along (2, 0) lines surrounding a 6-crossing. In either case the line k violates the No Three Closest Attachments Principle. 47

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1.40 The case of four ordinary points p1 , ..., p4 along (2, 0) lines surrounding a 6-crossing, with an ordinary point along every third spoke. There must be either an ordinary or attached point associated with the line k in (x, y) as explained in the text. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48

1.41 The case of four ordinary points p1 , ..., p4 along (2, 0) lines surrounding a 6-crossing, with an ordinary point along every third spoke. There will be a fourth ordinary or attached point associated with k unless the second (pseudo-)line through p4 intersects k in (x, y). . . . . . . . . . . . . . . . . .

49

1.42 An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. . . .

50

1.43 An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. The case where the cell P is a quadrangle. . . . . . . . . . . . . . . . . . . . . .

51

1.44 An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. The case where the cell P is a pentagon. . . . . . . . . . . . . . . . . . . . . . .

52

1.45 An ordinary point p on a (2, 0) line l surrounded by 3-crossings with three (4+)-gons, P, Q, R surrounding its adjacent triangles. . . . . . . . . . . . . .

53

1.46 An ordinary point p on a (2, 0) line l surrounded by 3-crossings with three (4+)-gons, P, Q, R surrounding p’s adjacent triangles. We assume there is a second ordinary point q on l adjacent to u and consider whether there can be ordinary points r, s absorbing charge from P . In the diagram we assume P is a quadrangle but the argument applies more generally. . . . . . . . . .

54

1.47 An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two (4+)gons, P, Q surrounding its adjacent triangles. We consider first the sub-case where l has its second ordinary point q adjacent to u absorbing charge from P and Q and ask whether the vertices r and s can absorb charge from P . .

55

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1.48 An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two (4+)-gons, P, Q surrounding its adjacent triangles. We consider the sub-case where l does not have a second ordinary point adjacent to u absorbing charge from P and Q, and ask now whether both vertices r and s can absorb charge from Q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

56

1.49 An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two quadrangles, P, Q surrounding its adjacent triangles on the same side of l. We ask which of the possible . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

1.50 Two ordinary points on a (2, 0) line l separated by a 4-crossing in an arrangement without epsilon excess. . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

1.51 We consider the case where B is not a triangular cell.. . . . . . . . . . . . .

58

1.52 The first case where P is not a triangular cell. The dashed lines indicate cells we know to be triangular. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

58

1.53 The second case where P is not a triangular cell. The dashed lines indicate cells we know to be triangular. . . . . . . . . . . . . . . . . . . . . . . . . .

59

1.54 An essential (2, 0) line l with essential crossing u. . . . . . . . . . . . . . . .

60

1.55 An essential (2, 0) line l with essential crossing u and two ordinary points sandwiched around the essential crossing u with all other vertices along l being 3-crossings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61

1.56 The essential (2, 0) line l from Figure 1.55 redrawn with u taken to be the horizontal point at infinity and n ∩ m taken to be the vertical point at infinity. 61 1.57 The essential (2, 0) line l from Figure 1.56 with the addition of lines ψ1 = l(vu , ξR ).ψ2 = l(uu , ξL ), ψ3 = l(vd , ηR ), ψ4 = l(ud , ηL ). . . . . . . . . . . . . .

62

1.58 An essential (2, 0) line l with essential 4-crossing u together with another 4-crossing v. Assume first that one of the closest vertices to v along k is not ordinary - call this vertex q, and consider the additional lines m, n through q. 64 1.59 An essential (2, 0) line l with essential 4-crossing u together with another 4-crossing v. k is known to also be a (2, 0) line with its two closest vertices to v each ordinary. We show that v is essential. . . . . . . . . . . . . . . . .

65

xx

1.60 Three essential (2, 0) lines l, k, k 0 with l∩k and k∩k 0 both essential 4-crossings. g ∩ l ∩ h is taken to be the horizontal point at infinity and k ∩ k 0 is taken to be the vertical point at infinity. If k 0 is not the line at infinity, it is one of the lines drawn passing through l. If g ∩ l ∩ h it is one of the lines through g 0 , k, h0 . Note that the lines g 0 , h0 actually fan out to the right and left. . . .

69

1.61 The arrangement considered in Figure 1.60 with the addition of the line η1 meeting the closest vertex, b, along ξ1 above g and crossing g to the right of k ∩ g. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

72

1.62 The arrangement considered in Figure 1.61 where both η3 = ξ2 and η2 = ξ3 . The two lines η2 and η3 would then intersect twice unless x = y. . . . . . .

73

1.63 The arrangement considered in earlier figures where η1 6= ξ4 , η2 6= ξ3 , and that η1 intersects g 0 at z and η2 intersects h0 at z 0 . Although not drawn, η3 = ξ2 and η4 = ξ1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

1.64 B¨or¨oczky examples with n = 4k+1 points (left) and n = 4k+3 points (right). Note the missing point at infinity (and center-point) in the right-hand example. 75 1.65 A 5-crossing with two surrounding ordinary points on two lines of type (2, 0) crossing at the 5-crossing. The line n have four attached points. . . . . . . .

78

1.66 Two (2, 0) lines with their ordinary points surrounding a 5-crossing. The line n necessarily has 4 attachments. . . . . . . . . . . . . . . . . . . . . . . . .

81

1.67 A 5-crossing together with two (2, 0) lines, one with its ordinary points surrounding the 5-crossing and the other with only one of its ordinary points adjacent to the 5-crossing. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81

1.68 Filling Figure 1.67 out a bit with known edges due to saturation. The edges [w, u] and [v, x] must actually be part of the same line. . . . . . . . . . . . .

82

1.69 A 4-crossing u with ordinary points p and q adjacent to u along (2, 0) lines l and k respectively. w is the closest vertex to u along h in the direction opposite the ordinary points p, q. . . . . . . . . . . . . . . . . . . . . . . . .

83

1.70 A sub-arrangement prohibited by Same Side Lemma (Lemma 36). l and k are (2, 0) lines so p and q are attached to h which also passes through u. Moreover, p and q are locally attached on the same side of h. . . . . . . . .

84

xxi

1.71 A (2, 0) line l with a hypothetical 5-crossing u sandwiched between ordinary points p and q. x is the closest vertex to u on k in the direction “upward” from l and likewise for y and h. For a (5 + j)-crossing place the additional j lines between k and h and apply the same argument as in the text. . . . . .

84

1.72 We consider the finite triangle 4(p, q, r) in the statement of Theorem 39. Since line h does not intersect the interior of 4(p, q, r) it is not considered. Amongst the remaining lines, the Theorem concludes that there are at least 2 ordinary points in 4(p, q, r) - indicated here with hollow circles. . . . . .

87

1.73 The finite triangle T = 4(p, q, r) with external faces F1 , ..., F7 and external edges E1 , ..., E8 as described in the proof of Theorem 39. . . . . . . . . . . . 1.74 Tight examples for Theorem 41.

88

. . . . . . . . . . . . . . . . . . . . . . . . P 1.75 A non-convex quadrangle. In order for the relation j≥2 jtj = E + Eext to

90

hold we must declare that the line l have two external edges. . . . . . . . .

91

1.76 Examples showing that (a) the conclusion of Theorem 40 does not hold for quadrangles, and (b) that there are hexagons without ordinary points. . . .

92

1.77 A wiring diagram for a 5 line pseudoline arrangement. . . . . . . . . . . . .

93

1.78 Two equivalent wiring diagrams. We can obtain an isomorphic arrangement of pseudolines from a given wiring diagram by interchanging the order of a pair of crossings as long as they do not share a pseudoline in common. . . .

93

1.79 Two vertices v1 , v2 through the line l1 with lines {lik }N k=1 crossing at v1 and lines {ljl }M l=1 crossing at v2 . Let x = liN ∩ lj1 and y = li1 ∩ ljM . l1 has an attached point via v1 and v2 iff x is ordinary and the triangle 4(v1 , v2 , x) lying projectively above l is a cell, or y is ordinary and the triangle 4(v1 , v2 , y) lying projectively below l1 is a cell. . . . . . . . . . . . . . . . . . . . . . . . 2.1

Three lines l, k and h, no two parallel to one another, with intersection points u = l ∩ k, v = h ∩ k, w = l ∩ h. . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2

99

The points p, q and r are all closest vertices to l on one side, but only p and q are closest and most extreme to one side. . . . . . . . . . . . . . . . . . .

2.4

99

The three lines l, k, h of Figure 2.1 together with lines through u parallel to h, through v parallel to l and through w parallel to k. . . . . . . . . . . . .

2.3

96

101

Examples showing that the conclusion of Theorem 45 is tight for n = 6, 7, 8. 102

xxii

2.5

The four four-line examples with each line having the All-to-One-Side Property.103

2.6

The four distinct affine three line arrangements. Any lines shown which do not intersect in the figure are assumed to be parallel. . . . . . . . . . . . . .

2.7

104

An arrangement of 10 affine pseudolines, three of which, l1 , l2 and l3 are of type (0, 1). In this context, an arrangement of affine pseudolines is a family of curves, each of which extends to infinity with a well-defined limiting slope which is the same from each end, and such that any two lines with different slopes cross, while pseudolines with equal slopes do not intersect. . . . . . .

2.8

106

A hypothetical arrangement with two lines, l, k of affine type (1, 0) with common ordinary point. We consider the case where two consecutive projective cells surrounding the triangular cells about p, Q and R say, are (4+)-gons. .

2.9

108

The hypothetical arrangement with two lines, l, k of affine type (1, 0) sharing an ordinary point, two consecutive projective cells surrounding the triangular cells about p, Q and R which are (4+)-gons, and a line extending the third edge (e3 ) in counter-clockwise order around Q from e1 = [b, d] . . . . . . . .

109

2.10 If the line l is not ordinary then it contains three points q0 , q1 , q2 - this is the case where two of the points q0 , q1 are to one side of the ray r. The line through p0 and q1 is seen to be ordinary. . . . . . . . . . . . . . . . . . . . .

111

2.11 Sylvester’s Theorem is false on the flat torus. . . . . . . . . . . . . . . . . .

114

2.12 Sylvester’s Theorem is false on the flat Klein Bottle. . . . . . . . . . . . . .

114

2.13 The same argument applied to the real projective plane does not lead to the same conclusion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

114

2.14 The Sylvester-violating flat torus example with points in one more row and column compared to Figure 2.11. . . . . . . . . . . . . . . . . . . . . . . . .

117

3.1

Example with n + 3 wedges. . . . . . . . . . . . . . . . . . . . . . . . . . . .

120

3.2

Essentially equivalent variations in the case of n = 9 lines and 12 wedges. .

120

3.3

4n/3 examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121

3.4

n wedges with no two lines parallel. . . . . . . . . . . . . . . . . . . . . . . .

121

3.5

4 or more lines with just 3 wedges. . . . . . . . . . . . . . . . . . . . . . . .

122

3.6

Inner wedges and the difficulty detecting them. . . . . . . . . . . . . . . . .

124

3.7

b4n/3c cases (a) n = 3, ρ2 = 4, (b) n = 4, ρ2 = 5 and (c) n = 5, ρ2 = 6. . . .

126

xxiii

3.8

Adding a wedge at a time in the cases n ≡ 1, 2 (mod 3) beginning with a regular hexagon and n = 9 lines with

3.9

4n 3

= 12 wedges. . . . . . . . . . . . .

126

An arrangement of 6 lines with outer layer complexity 16. There are two pairs of parallel lines. The bounding segments of the outer layer are highlighted.

128

3.10 An arrangement of 3 lines with the addition of a (dashed) line `. In the classical definition of the complexity of the zone of `, the two sides of each line crossing ` are counted separately since they bound adjacent faces of the arrangement. On the other hand, in our definition of complexity, for each line crossing ` we count the segment above ` and the segment below ` as distinct. Hence the two definitions are equivalent and under either definition the complexity of the above arrangement is 10. . . . . . . . . . . . . . . . . 3.11 An arrangement with 4 lines and complexity 10
0 minimal, or there is no line whatsoever through two or more points of S intersecting r above p0 . In the latter case we consider the oppositely oriented ray r = {p0 − λˆ : λ ≥ 0} and find the line through two or more points which intersects this ray at a point of minimal λ > 0. See Figure 1.3. In either case, the points on this line come in consecutive order in a polar ordering of points around p0 . By polar ordering of points pi ∈ S \ {p0 } we mean the ordering by angle between the rays ri = {p0 + λ(pi − p0 ) : λ ≥ 0} and ray i = {λˆı : λ ≥ 0} where ˆı denotes the unit vector in the x-direction. If our assertion about consecutive polar ordering were not true then there would be a point p∗ ∈ / l which came between q0 , q1 ∈ l in the polar ordering of points around p0 . Suppose q0 to be the point along l, amongst q0 and q1 , which is closer to r. Consider the case were r = {p0 + λj : λ ≥ 0} - the other case being symmetrical. If p∗ were above l, then l(q0 , p∗ ) would intersect r at a point closer to p0 than l ∩ r, and if p∗ were below l, then l(p∗ , q1 ) would intersect r at a point closer to p0 than l ∩ r.

10

Figure 1.4: If the line l is not ordinary then it contains three points q0 , q1 , q2 . A line through p0 and one of q0 , q1 , q2 is necessarily ordinary. Since the points about p0 can be ordered and tested in sequential pairs in O(n log n) time, such a line l, meeting r at minimal distance from p0 can be found in O(n log n) time. We show next that either l is ordinary or a line through p0 and one of the points on l is ordinary. Suppose the line l were not ordinary and contained the three points q0 , q1 , q2 as illustrated in Figure 1.4. If now l(p0 , q1 ) is ordinary we are done. If not, l(p0 , q1 ) contains another point; call it q. If q were located in the finite segment [p0 , q1 ] then l(q, q2 ) would have a smaller λ than l. On the other hand if q were in the unbounded segment of l(p0 , q1 ) (i.e. beyond q on the ray − p−→ q ) then l(q , q) would have had smaller λ than l. Hence, if l 1

0 1

0

is not ordinary then l(p0 , q1 ) must be. At CCCG 2007, Mukhopadhyay and his student Eugene Greene presented a new algorithm [62] for finding an ordinary line in time O(n log n). The big open question is whether one can do better. Olivier Devillers has suggested that perhaps there is a randomized algorithm which, for any input, could find an ordinary line in O(n) expected time.

1.3

The n/2 Conjecture

Much work has gone into obtaining lower bounds on the number of ordinary lines in a collection of n points satisfying the hypothesis of Sylvester’s Theorem. In 1951 Motzkin √ [60] found an argument giving a bound of O( n). That same year, Dirac [23] and Motzkin [60] separately conjectured that there must be at least n/2 ordinary lines. For even n there is a family of examples due to B¨or¨ oczky (as cited in [18]) with n points and precisely n/2 ordinary lines1 . These examples are most easily constructed and verified in the real projective plane. One starts with the vertices of a regular n/2-gon and then adds the n/2 points at infinity determined by the line through any pair of vertices. Each vertex of the 1

Examples which were presumably known to Dirac and Motzkin.

11

Figure 1.5: B¨or¨oczky’s even n example with n = 12 points and 6 ordinary lines (dashed lines). original regular n/2-gon then determines an ordinary line (a dashed line in the figure) with precisely one of the n/2 points at infinity. Note that there is nothing special about this example residing in RP2 ; the points and connecting lines can easily enough be rotated into R2 . In 1957 W. Moser [59] found a clever argument to show that if n is even, then the number of ordinary lines is greater than

n+11 6 .

Consider the dual problem, where by the

Yproj equation (1.1.8), we have t2 ≥ 3 +

X (j − 3)tj .

(1.3.1)

j≥4

Now the number of lines containing some point which is not a 3-crossing is at most X 2t2 + 4t4 + 5t5 + ... ≤ 2t2 + 4( (j − 3)tj )

(1.3.2)

j≥4

If now t2 ≤

n+11 6

≤ 2t2 + 4(t2 − 3)

(1.3.3)

≤ 6t2 − 12.

(1.3.4)

then 6t2 − 12 ≤ n − 1 and so at least one line contains only 3-crossings

and therefore n is odd. Thus if n is even t2 >

n+11 6 .

QED.

In 1958 Kelly and W. Moser [47] found an example of 7 points with just 3 ordinary lines. We shall often need to refer to the dual of this example which we refer to as the Kelly-Moser arrangement. See Figure 1.6. They also showed that a set of n not all collinear points must admit at least 3n/7 ordinary lines. In 1968 McKee [18] found an example of 13 points with just 6 ordinary lines. The dual of this configuration is given in Figure 1.7.

12

Figure 1.6: Kelly-Moser example with 7 lines and 3 ordinary points. The third ordinary point is the horizontal point at infinity. In this and all future figures hollow dots indicate ordinary points.

Figure 1.7: McKee arrangement of 13 lines and 6 ordinary points. In 1981 Hansen [41] presented a habilitation thesis claiming to show that there must be at least n/2 ordinary lines as long as n 6= 7, 13. However, in 1993 Csima and Sawyer [19] found a flaw in Hansen’s thesis, but in the same paper showed that except for the KellyMoser 7-point example there must be at least 6n/13 ordinary lines. The Dirac-Motzkin conjecture thus stands as follows: Conjecture 1. (Dirac-Motzkin) For n not all collinear points, n 6= 7, 13, there must be at least n/2 ordinary lines.

13

1.4

Proofs of the Kelly-Moser/Kelly-Rottenberg and CsimaSawyer Theorems

1.4.1

Preliminary Lemmas

Henceforth we shall focus on the dual setting, in which we are given a family of n lines in the projective plane – or more generally, a family of pseudolines. A family of pseudolines in the projective plane RP2 is a collection of at least two simple closed curves, each pair of which has exactly one point in common and at that point the two pseudolines properly cross. The notion of pseudolines was introduced by Levi in 1926 [54]. Since simple closed curves in RP2 are either the images of circles or lines under self-homeomorphisms of RP2 , it follows by the intersection property that any pseudoline l is the image, under self-homeomorphism of the space, of a line and that l does not divide the space into distinct connected components. In general we shall speak of arrangements of pseudolines, just like we have spoken of arrangements of lines, by which we mean not only the family of pseudolines but also the vertices, edges and faces (or cells) induced by the intersecting pseudolines. For more on pseudolines see [40]. For purposes of rendering pseudoline arrangements as drawings in the affine plane we make the additional assumption that a pseudoline is either the line at infinity or contains a unique point at infinity. This allows us to draw any pseudoline other than the line at infinity with a unique “slope” as the pseudoline tends to infinity. When we mention the “slope” of a pseudoline we mean it in this sense. Definition 1. An arrangement of n pseudolines is said to be Sylvester-critical if it has fewer than n/2 ordinary points. As noted in the previous section, only two Sylvester-critical arrangements are known, for n = 7 (the Kelly-Moser example) and n = 13 (the McKee example). Let A be an arrangement of n pseudolines. In general we shall be searching for Sylvestercritical arrangements. For the remainder of this chapter we assume that every pseudoline l ∈ A has at least three vertices. If l ∈ A had just two vertices and if m were the number of pseudolines passing through one of the two vertices then each of the m pseudolines through the one vertex would intersect each of the n − m − 1 pseudolines through the other vertex in an ordinary point and it is easy to see that there would have to be at least n − 1 ordinary points. The following lemmas are due to Kelly and Moser [47] for lines and Kelly and Rottenberg [48] for pseudolines. We provide the proofs given by Felsner in [33] slightly restated and

14

Figure 1.8: An example of a line l with two attached ordinary points. The point p is attached through the finite shaded triangle and the point q is attached through the shaded infinite triangle. ultimately generalized to pseudolines. Definition 2. Say that an ordinary point p is attached to a pseudoline l, not containing p, if l together with the two pseudolines crossing at p form a (possibly infinite) triangular cell of the arrangement. See Figure 1.8. Lemma 2 (Four Attachment Lemma). In any arrangement of pseudolines, an ordinary point can be attached to at most 4 pseudolines. Proof. An ordinary point is contained in 2 crossing pseudolines, and hence is a vertex of at most 4 faces. It can therefore be attached to at most 4 pseudolines. Definition 3. A line is said to be of type (i, j) if it contains i ordinary points and has j ordinary points attached to it. A line is of type (i, j+) if it contains i ordinary points and has j or more ordinary points attached to it. Lemma 3 ((0,3+)-Lemma). If a line l of an arrangement A contains no ordinary points, then there are at least 3 ordinary points attached to l. Lemma 4 ((1,2+)-Lemma). If a line l of an arrangement A contains a single ordinary point, then the line l has at least 2 ordinary points attached to it. In order to prove Lemmas 3 and 4, we use the following (a slight variation of which can be found in [33]):

15

Figure 1.9: An illustration of the setup and conclusion of the Three Clause Lemma. The triangle T in the Lemma is the finite triangle 4(p, q, r) in the figure. The hollow vertex is the ordinary point attached to l in T guaranteed by the Lemma. Lemma 5 (Three Clause Lemma for Lines). Let T be a triangle formed by three lines of an arrangement A. Let l be one of the defining lines of T , [p, q] the edge of T on l, and (p, q) the open subinterval of [p, q] not containing p, q. If (i) T is not a cell of the arrangement, (ii) (p, q) contains no ordinary points, and (iii) every line intersecting the interior of T also intersects [p, q], then there exists an ordinary point x attached to l through some triangle contained in T . Figure 1.9 illustrates the setup and conclusion of this lemma. The triangle T is the (finite) triangle 4(p, q, r) in the Figure. Proof. If necessary, rotate the arrangement so that T is a finite triangle. Let x be the vertex of A in T , not on l, which has the smallest distance to l. If x is ordinary then by the assumptions of the Lemma it is attached to l via a triangle in T and we are done. On the other hand suppose there are three lines l1 , l2 , l3 intersecting in x, and let v1 , v2 , v3 be their respective intersection points with l. By assumption, all the vi lie in [p, q] and we assume v2 lies between v1 and v3 . Since v2 is not ordinary there is a line m 6= l2 entering T at v2 . The intersection of m and l1 or m and l3 is of smaller distance to l than x, a contradiction. To extend the Three Clause Lemma to pseudolines, we need the following definitions: Definition 4. Given an arrangement A of pseudolines, a pseudo-triangle is any region bounded by three pseudolines.

16

Figure 1.10: Given a pseudo-triangle T which is the finite triangle determined by the pseudolines l, l0 , l00 , and a chosen reference pseudoline l, the generic placement of the first three vertices x1 , x2 , x3 in T . Given a minimal cycle x1 ≺ ... ≺ xr ≺ x1 the point xr may be in one of the four regions A, B, C or D. Definition 5. Given an arrangement A of pseudolines, a finite pseudo-triangle T and a chosen pseudoline l forming one of the edges e of T , suppose v and v 0 are two vertices of A that lie in T . Write v ≺ v 0 if there is a pseudoline k that meets e with v, v 0 ∈ k and v lies between e and v 0 . Lemma 6. Given an arrangement A of pseudolines, a finite pseudo-triangle T and a chosen pseudoline l forming one of the edges e of T , then the transitive closure of the ≺ relation described in Definition 5 gives rise to a partial ordering of the vertices contained in T . Proof. For the transitive closure, ≺c , of the ≺ relation not to be a partial ordering there would have to be a cycle among the vertices with respect to the ≺ relation. Suppose x1 ≺ ... ≺ xr ≺ x1 were such a cycle, with r minimal. x1 ≺ x2 ≺ x1 is impossible since if x1 ≺ x2 by virtue of a pseudoline k, and x2 ≺ x1 by virtue of a second pseudoline k 0 , then k and k 0 would have two points in common, an impossibility. Thus r ≥ 3 and so, generically, we have the situation in Figure 1.10. We cannot now have x3 ≺ x1 since for a pseudoline to pass through x1 then x3 and then through l it would have to cross either m or n again, in either case an impossibility. Hence r > 3. Let us now consider where the vertex xr may lie. Suppose first that xr ∈ n. If xr ≺ x3 then we have a cycle x3 ≺ x4 ... ≺ xr ≺ x3 which is shorter than our assumed minimal cycle. On the other hand, if x3 ≺ xr then we have the trivial cycle x1 ≺ x2 ≺ xr ≺ x1 which is again too short. It follows that xr ∈ / n. Next suppose that xr ∈ m. If xr ≺ x2 then we have

17

a cycle x2 ≺ x3 ≺ ... ≺ xr ≺ x2 which is too short. On the other hand if x2 ≺ xr then, since xr ≺ x1 , there is a pseudoline passing through xr and x1 in addition to m which is impossible. It follows that xr ∈ / m. We next consider the regions A, B, C and D depicted in Figure 1.10 in turn. If xr ∈ A then l(xr , x1 ) crosses n between x2 and l at a point x∗ . But then there is a cycle xr ≺ x∗ ≺ x3 ≺ ... ≺ xr which is too short. Hence xr ∈ / A. If xr ∈ B with xr ≺ x1 it is easy to see that l(x1 , xr ) must either cross m twice or n twice, in either case an impossibility. Hence xr ∈ / B. If xr ∈ C with xr ≺ x1 then again there exists x∗ ∈ n, with x∗ between l and x2 so we get the same short cycle and hence same contradiction as in the case xr ∈ A. Hence xr ∈ / C. It follows that xr ∈ D. We ask next where xk−1 may be. It cannot be x3 since then l(x3 , xr ), which goes through l, then x3 , then xr , would cross m or n twice. Now that we have established that xk−1 6= x3 , the arguments showing that xk−1 ∈ / m and xk−1 ∈ / n are precisely the same as for xr . See above. If xk−1 ∈ A then l(xk−1 , xr ) passes through n between x2 and l, say at a point x∗ and as before we have the short cycle x∗ ≺ x3 ≺ ... ≺ xk−1 ≺ x∗ . The case xk−1 ∈ C is argued analogously; again l(xk−1 , xr ) passes through n between x2 and l and we use this point to obtain a short cycle. Finally, if xk−1 ∈ B then l(xr , xk−1 ) passes either twice through m or twice through n. It follows that we must have xk−1 ∈ D. Analogously we argue that xk−j ∈ D for 2 ≤ j ≤ k − 3. But in particular this means that x3 ∈ D, a contradiction. The lemma follows. Lemma 7 (Three Clause Lemma for Pseudolines). The Three Clause Lemma remains true for arrangements of pseudolines as well as arrangements of lines. Proof. We utilize the same notation as in the Three Clause Lemma for Lines (Lemma 5). As in the proof of that lemma, if necessary, rotate the arrangement so that T becomes a finite pseudo-triangle. Let ≺ denote the “closer to” relation with respect to l in T , in other words the transitive closure of the former ≺ relation. Let v ∗ be a minimal element (vertex) with respect to ≺. If v ∗ were not ordinary, then the same argument as in the Three Clause Lemma for Lines shows that there would be a vertex closer to l than v ∗ . Hence v ∗ is ordinary, and by virtue of having no closer vertex to l along its two defining pseudolines, it is attached to l. In recent work, Perlstein and Pinchasi [67] have shown that families of “pseudo-parabolas” admit a partial ordering as in Lemma 6 and this allowed them to achieve an interesting Sylvester-Gallai like result for these objects, namely that t2 + p2 ≥

n 5

− 1, where t2 = the

number of ordinary points and p2 = the number of two-edged faces (also called digons).

18

x

T1

T2 p

l

q

T1

m

T2

n

Figure 1.11: Three pseudolines l, m and n partitioning the projective plane into four pseudotriangular regions. The authors define a family of “pseudo-parabolas” to be any family of x-monotone curves in the plane such that any pair intersect properly (i.e. cross) at precisely two points. We are now in a position to prove the (0, 3+) and (1, 2+) Lemmas. Proof of the (0, 3+) Lemma (Lemma 3). Let l be a pseudoline of A without ordinary points and let x be a vertex of A not on l. Without loss of generality we can assume that l is not the line at infinity and that, moreover, x is not on the line at infinity. We can define a new ≺∗ relation with respect to l, the line at infinity, l∞ , and x. Say that two vertices v0 , v1 ∈ A are such that v0 ≺∗ v1 if v0 and v1 lie in the same half-plane between l∞ and l as x, and there is a pseudoline k which passes through l∞ , v1 , v0 and l in that order. Just like ≺C was seen to be a partial ordering, the transitive closure, ≺∗c of ≺∗ can be shown to be free of cycles and hence a partial ordering.2 Moreover, ≺∗c and ≺c agree for triangles contained in the half-plane between l∞ and l containing x. With this ≺∗c relation, the usual distance argument shows that the vertex x ∈ A closest to l is an ordinary point attached to l. Now let m and n be the lines (pseudolines) crossing at x and let p, q be their intersection points with l. l, m and n collectively partition the projective plane into four (pseduo-)triangular regions as illustrated in Figure 1.11. Let T1 and T2 be the regions indicated in Figure 1.11. Both pseudo-triangles have an edge which is an actual edge of the arrangement, [q, x] for T1 and [p, x] for T2 . Let v be a third vertex on l. All (pseudo-)lines crossing l at v intersect the interior of T1 and T2 . Since we assume all (pseudo-)lines have, and therefore l has, at least three vertices, neither T1 nor T2 are cells. Hence the Three Clause Lemma applies and we may find attached points to l in both T1 and T2 , making three in total, including x. 2

Looking back at the proof of Lemma 6, and in particular Figure 1.10, note that we can partition the half-plane above l into regions A, B, C and D just like we partitioned the finite pseudo-triangle determined by l, l0 and l00 , and argue exactly as in that proof.

19

Figure 1.12: The pseudoline l with ordinary point p and additional line m through p. r is one of the neighbors of p along m and n is an additional line through r. Proof of the (1, 2+) Lemma (Lemma 4). Let p be the ordinary point on l and let r be one of the neighbors of p on the other (pseudo-)line, m, through p. Let n be a second (pseudo-)line through r and let q be the point of intersection of n and l. The (pseudo-)line l is partitioned by p and q as depicted in Figure 1.12. We denote by [p, q] the segment of l contained in the cell T1 and denote by [q, p] the segment along l outside T1 . If both intervals contain additional vertices then we consider the two (pseudo-)triangular regions T1 and T2 . Both regions have a side [p, r] which is an edge in the arrangement. Hence, by the Three Clause Lemma one finds ordinary points attached to l in T1 and T2 . On the other hand if all the vertices along l fall into one interval, then if they are in [p, q] we can apply the Three Clause Lemma to T1 and T3 and if they are in [q, p] we can apply the Three Clause Lemma to T2 and T4 . Hence in all cases we can find two ordinary points attached to l and the lemma is proved.

1.4.2

Sylvester-critical arrangements must contain lines of type (2, 0)

The ensuing discussion, up through the end of this chapter, applies more generally to pseudolines, though we use the term “line” for brevity. All arrangements live in the real projective plane, RP2 . Theorem 8. A Sylvester-critical arrangement must contain lines of type (2, 0). Proof. Suppose we had a Sylvester-critical arrangement of n lines without lines of type (2, 0). Then by virtue of the (0, 3+) and (1, 2+) Lemmas, all lines are of type (0, 3+),

20

(1, 2+), (2, 1+), (3, 0+), (4, 0+) and so on. Let k0 denote the number of lines of type (0, 3+), k1 the number of lines of type (1, 2+) and so forth. Additionally, let 3 + ²0 denote the average number of attached points for lines of type (0, 3+), 2 + ²1 denote the average number of attached points for lines of type (1, 2+) and so forth, where ²i ≥ 0, ∀i ≥ 0. Now, if we denote by K the number of ordinary points, then by virtue of the fact that we have a Sylvester-critical arrangement, we have X

iki = 2K < n

(1.4.1)

i≥1

since the left hand side counts each ordinary point twice, once for each line containing it. Then, since

X

ki = n

(1.4.2)

i≥0

we obtain k0 > k2 +

X

(i − 1)ki .

(1.4.3)

i≥3

On the other hand, if we let M denote the total number of ordinary point-line attachments, and we count the second coordinates of the (i, j) tuples, then by virtue of the Four Attachment Lemma we have X

M = (3 + ²0 )k0 + (2 + ²1 )k1 + (1 + ²2 )k2 +

²i ki ≤ 4K = 2

i≥3

so 3k0 + 2k1 + k2 ≤ 2

X

X

iki

(1.4.4)

i≥1

iki .

(1.4.5)

i≥1

Thus 3k0 ≤ 3k2 + 2

X

iki

(1.4.6)

i≥3

or k0 ≤ k2 +

X2 i≥3

3

iki

which contradicts Equation 1.4.3 since i − 1 ≥ 32 i for i ≥ 3.

(1.4.7)

21

1.4.3

The Kelly-Moser/Kelly-Rottenberg

3n 7

Theorems

We use the (0, 3+) and (1, 2+) Lemmas, together with the Four Attachment Lemma, and the same basic argument as above to now give a slightly more transparent proof of the Kelly-Moser Theorem than is usually given. In what follows, we write k2,0 for the number of lines of type (2, 0) and k2,1 for the number of lines of type (2, 1+). The following (pseudoline version) is actually due to Kelly and Rottenberg [48]. The version for lines is due to Kelly and Moser [47]. Theorem 9. In any arrangement of n pseudolines which do not all pass through a common point, there must be at least 3n/7 ordinary points. Proof. Suppose the Theorem were false; then there would be an arrangement A with k1 + 2k2,0 + 2k2,1 +

X

6n . 7

(1.4.8)

ki = n

(1.4.9)

iki
k1 + 2k2,0 + 2k2,1 + iki 7

(1.4.10)

X 7 iki ) ki > (k1 + 2k2,0 + 2k2,1 + 6

(1.4.11)

i≥3

or k0 + k1 + k2,0 + k2,1 +

X i≥3

so that

i≥3

i≥3

X 7 1 4 4 k0 > k1 + k2,0 + k2,1 + ( i − 1)ki . 6 3 3 6

(1.4.12)

i≥3

On the other hand, as in the proof of the previous theorem, we have (3 + ²0 )k0 + (2 + ²1 )k1 + (1 + ²2,1 )k2,1 +

X i≥3

²i ki ≤ 2k1 + 4k2,0 + 4k2,1 +

X

2iki (1.4.13)

i≥3

where (1 + ²2,1 ) denotes the average number of attached points for lines of type (2, 1+). Since for all i, ²i ≥ 0 and ²2,1 ≥ 0, the above inequality remains true if we remove the terms

22

involving ²i or ²2,1 . Hence we have X 2i 4 k0 ≤ k2,0 + k2,1 + ki . 3 3

(1.4.14)

i≥3

Since

7i 6

−1 >

2i 3

for i ≥ 3, equations 1.4.12 and 1.4.14 cannot simultaneously hold, and so

the theorem follows. Furthermore, for there to be precisely we must have k0 =

1.4.4

4 3 k2,0

3n 7

ordinary points, we see that

and ki = 0 otherwise (including k2,1 = 0).

The Saturation Property

In this and subsequent sections of this chapter we shall confine our attention to the case of lines, though all of the definitions and results apply equally well to pseudolines. We find the constant reference to pseudolines and pseudo-triangles unnecessarily cumbersome. We remind the reader of our assumption from Section 1.4.1, holding for the entirety of this chapter, that all lines have at least three vertices. By the Four Attachment Lemma an ordinary point can have at most 4 lines counting it as an attached point. We introduce the following important notion: Definition 6. An ordinary point p is saturated (in attachments) if exactly 4 lines count p as an attached point. An arrangement is saturated (in attachments) if all its ordinary points are saturated in attachments. The following is obvious from the definition: Lemma 10. An ordinary point is saturated iff it is surrounded by triangular cells of the arrangement. Lemma 11. A saturated ordinary point cannot have an ordinary point as a neighboring vertex. Proof. Let p be a saturated ordinary point with adjacent vertex v. The edge between v and p is shared by two of p’s surrounding triangular cells and thus v is at least a 3-crossing. The following lemma will be used frequently in what follows: Lemma 12 ((2,0) Saturation Lemma). The ordinary points on a line of type (2, 0) are necessarily saturated. Proof. Let l be a line of type (2, 0) in an arrangement A and let p be one of the two ordinary points on l. Let k be the second line through p and let p0 be the nearest vertex to p along

23

k on one side of l. Consider a second line k 0 through p0 as in Figure 1.13. By the Three Clause Lemma, since l has just 2 ordinary points and no attached points, either triangle T1 or T2 is a cell. Without loss of generality, suppose it is T1 . p0 cannot be ordinary since

Figure 1.13: A line l of type (2, 0) with an ordinary point p. k is the second line through p and p0 is the closest vertex to p in one direction. otherwise it would be attached to l through T1 . Hence let k 00 be a third line through p0 which intersects l at a vertex q. See Figure 1.14. Applying the Three Clause Lemma to the

Figure 1.14: Having concluded that T1 is a cell, and so p0 cannot be ordinary we draw a third line k 00 through p0 intersecting l at the vertex q. 4(p, p0 , q) which contains T1 we conclude that the segment (p, q) on this side must contain an ordinary point. On the other hand, the segment (p, q) on the other side, cannot contain an ordinary point, so the triangle 4(p, p0 , q) which is labelled T2 in the Figure (but which is smaller than the T2 from the prior figure) must be a cell. Thus p is surrounded by triangles “above” l. Considering the closest vertex to p along k “below” l allows us to similarly conclude that p is surrounded by triangles from below as well, and hence p is saturated.

24

Figure 1.15: An ordinary point p on the (2, 0) line l. k is the second line through p and p0 is a neighboring vertex to p on one side of l. m, n are the two lines through p0 counting p as an attachment and o is a hypothetical fourth line through p0 . Figure 21 from the article by Csima and Sawyer [19] shows that the ordinary points on a line of type (3, 0) need NOT be saturated. Also important is the following: Lemma 13 (Top 3-Crossing Lemma). Let p be an ordinary point on line l of type (2, 0) and let k be the other line through p. Then the closest vertex to p along k (in either direction) is necessarily a 3-crossing. Proof. Let p0 be a closest vertex to p along k as in the statement of the Lemma. Since p is saturated, let m and n be additional lines through p0 as depicted in Figure 1.15 and let o be a hypothetical fourth line through p0 intersecting l at a vertex q. Consider the two triangles 4(p, p0 , q). Neither is a cell, so by the Three Clause Lemma l must contain an ordinary point in (p, q) in each of the triangles. However, l has just 2 ordinary points, a contradiction. The Lemma follows.

1.4.5

The Csima-Sawyer Theorem

The key lemma which Csima and Sawyer [19] added to the mix is the following: Lemma 14 (Csima-Sawyer Lemma). Let A be an arrangement of n lines such that two lines of type (2, 0) intersect in an ordinary point. Then A is the Kelly-Moser arrangement (Figure 1.6). Proof. Suppose we have two (2, 0) lines l, k intersecting at a common ordinary point p. By the (2, 0) Saturation Lemma (Lemma 12), p is saturated. By the Top 3-Crossing Lemma (Lemma 13) all neighboring vertices to p are 3-crossings. See Figure 1.16. Without loss of generality suppose mkm0 , nkn0 . Let us identify by P the cell adjacent to 4(a, b, p) along

25

Figure 1.16: Two lines l, k of type (2, 0) meeting at an ordinary point p. By the (2, 0) Saturation Lemma (Lemma 12) p is saturated and using a projective transformation we may assume that lines m, m0 and lines n, n0 are parallel pairs, meeting respectively at m∞ , n∞ . [a, b], and analogously identify cells Q, R and S. Since the vertices a, b, c, d are all 3-crossings either some two consecutive cells amongst P, Q, R, S in the clockwise ordering are (4+)-gons or two opposite cells amongst P, Q, R, S - say Q, S - are actually triangles with a common vertex at, in this case m∞ . However, if Q, S are triangular cells with a common vertex at m∞ , then there can be no additional vertices along m other than a, b, m∞ while all crossings at a, b have been identified. It follows that all additional lines pass through m∞ and so are either the line at infinity or lines parallel to m. Lines parallel to m can’t pass through Q, however, and if such lines intersected R or P , they would each create additional ordinary points on k - hence there can be at most one such line. But if the one line passed through R say, it would create an ordinary point on n attached to k, which is impossible since k cannot have attached points. It follows that there is at most one additional line - the line at infinity. Indeed, since we have accounted for only one of the ordinary points on l, k there must be this additional line - giving us the Kelly-Moser arrangement. The only remaining case is that there are two adjacent (4+)-gons amongst P, Q, R, S, e.g. R, S. There is then a first vertex d0 adjacent to d along [d, n∞ ] and a line meeting d0 which forms an edge of R next to [d, d0 ]. Call this line h and suppose h meets l at a vertex v. See Figure 1.17. Since d0 cannot be attached to l, and a line cannot cut through R since it is a cell, there is a line through [d0 , v) that cuts through the interior of 4(d, d0 , v) and so we may apply the Three Clause Lemma to conclude that l has an ordinary point in (d, v). Now consider the (4+)-gon S. There is a first vertex a0 adjacent to a along [a, m∞ ] and a

26

Figure 1.17: Two lines l, k of type (2, 0) meeting at the saturated ordinary point p. We consider the case where there are necessarily two adjacent (4+)-gons amongst P, Q, R, S, e.g. R, S in the cyclical clockwise ordering. Since R is a (4+)-gon there is a closest vertex d0 adjacent to d along [d, n∞ ] and a line meeting d0 which forms an edge of R next to [d, d0 ]. We call this line h. Analogously, since S is a (4+)-gon there is a closest vertex a0 adjacent to a along [a, m∞ ] and a line (not drawn) meeting a0 which forms an edge of S next to [a, a0 ]. We call the second line i. line meeting a0 which forms an edge of S next to [a, a0 ] - call this line i. By virtue of the fact that i cannot pass through R, and hence through (d, d0 ) and h cannot pass through S, and hence through (a, a0 ), i must intersect l outside (c, v) in a point v 0 . Otherwise, unless i = h and v = v 0 (which is possible), the lines i and h would cross twice. Again the point a0 cannot be attached so there is a line through [a0 , v 0 ) that cuts the interior of 4(a, a0 , v 0 ) and so we may again apply the Three Clause Lemma to conclude that l has a third ordinary point in (a, v 0 ), contradicting the fact that l is a (2, 0) line. It follows that our arrangement is the Kelly-Moser. We are now in a position to use the same kind of double counting argument we used to prove the Kelly-Moser result to prove the 6n/13 result of Csima and Sawyer. Theorem 15 (Csima-Sawyer, 1993). An arrangement of n pseudolines, not all passing through a common point, which is not the Kelly-Moser arrangement (Figure 1.6), must contain at least 6n/13 ordinary points. Proof. We argue very much as in the proof of Theorem 9 and adopt the same notation. Since each pseudoline of type (2, 0) intersects two pseudolines of type other than (2, 0) in

27

its two ordinary points we have 2k2,0 ≤ k1 + 2k2,1 +

X

iki .

(1.4.15)

i≥3

The left hand side of Equation 1.4.15 counts the number of ordinary points on pseudolines of type (2, 0) and the right hand side counts the number of ordinary points on all other pseudolines. As in the proof of the Kelly-Moser (Kelly-Rottenberg) Theorem, we argue by contradiction. We assume we have an arrangement A with fewer than 6n/13 ordinary points, and such that no two lines of type (2, 0) intersect in an ordinary point. Then k1 + 2k2,0 + 2k2,1 +

X

iki


i≥3

we have k0 >

(1.4.17)

X 13 (k1 + 2k2,0 + 2k2,1 + iki ) 12

(1.4.18)

i≥3

X 13 7 7 1 k1 + k2,0 + k2,1 + ( i − 1)ki . 12 6 6 12

(1.4.19)

i≥3

Equation 1.4.14 from the Kelly-Moser (Kelly-Rottenberg) argument remains true, so we copy that equation again below: X 2i 4 k0 ≤ k2,0 + k2,1 + ki . 3 3

(1.4.20)

i≥3

Now, rewriting equation 1.4.15 in terms of k1 gives k1 ≥ 2k2,0 − 2k2,1 −

X

iki

(1.4.21)

i≥3

which when substituted into 1.4.19 gives: k0 >

X X 13 1 7 7 (2k2,0 − 2k2,1 − iki ) + k2,0 + k2,1 + ( i − 1)ki , 12 6 6 12 i≥3

i≥3

(1.4.22)

28

or

X 4 k0 > k2,0 + k2,1 + (i − 1)ki . 3

(1.4.23)

i≥3

Since i − 1 ≥

2i 3

for i ≥ 3, equations 1.4.20 and 1.4.23 cannot simultaneously hold and the

Theorem is established. Using the same idea as in the above proof we have the following: Lemma 16. In an arrangement of lines which is not the Kelly-Moser arrangement, a saturated ordinary point p is either adjacent to a (4+)-crossing or two of p’s surrounding and adjacent triangular cells are adjacent to distinct (4+)-gons.

1.5

A Unified View of the n = 7 and n = 13 Examples

There are various ways to view the Kelly-Moser and McKee counterexamples to the Dirac-Motzkin n/2 conjecture. The literature of this subject considers these examples to be “sporadic,” or not related. However, they can be seen to come from a common construction. We work in RP2 . For the Kelly-Moser example, start with the vertices of two equal size equilateral triangles, glue the triangles together along one edge, and add a center point. This gives a configuration with 5 points and 4 ordinary lines. Now greedily add points at infinity with the objective of reducing the ratio of ordinary lines to points, stopping when adding an additional point would create more ordinary lines than it would “kill off.” Since the 4 original ordinary lines formed 2 pairs of parallel lines, we greedily add two points, one to kill each pair of parallel lines. See Figure 1.18(a). For the McKee example, start with the vertices of two equal sized regular pentagons, glue the pentagons together along an edge, add a center point, and again greedily add points at infinity with the objective of reducing the ratio of ordinary lines to points. We start with 9 points and 12 ordinary lines. There are two sets of 4 ordinary lines all sharing a common direction (slope), so we greedily add points at infinity to remove these. In the process we create two new finite ordinary lines, and one additional ordinary line at infinity. Adding vertical and horizontal points at infinity saturates this example, yielding McKee’s configuration of 13 points and 6 ordinary lines as shown in Figure 1.18(b). We may try the same procedure with equal sized regular 7-gons. Glued regular pentagons and 7-gons are shown, together with their corresponding points at infinity, in Figure 1.19. In the case of glued 7-gons, we start with 13. To count the ordinary lines, note that there are two ordinary lines containing points a and g, 6 ordinary lines containing points c, d, e, i, j and

29

(a)

(b)

Figure 1.18: A common view of the (a) Kelly-Moser and (b) McKee examples. k, and 8 ordinary lines containing points b, f, h and l - for a total of

4+36+32 2

= 36 ordinary

lines in total. There are two sets of 6 ordinary lines sharing the respective directions r∞ , l∞ , so adding these points at infinity adds three ordinary lines [`(m, l∞ ), `(m, r∞ ) and `(l∞ , r∞ )] while killing off 12, leaving a configuration with 15 points and 27 ordinary lines. Adding vertical and horizontal points at infinity is the next most productive thing to do, adding four ordinary lines [`(a, v∞ ), `(g, v∞ ), `(d, h∞ ) and `(j, h∞ )] while killing off 9, one of which was `(l∞ , r∞ ) - thus leaving a configuration of 17 points and 22 ordinary lines. Unfortunately there is nothing very productive to do at this stage. For example, adding a point at infinity, z∞ , to kill off the ordinary lines `(e, f ) and `(k, l) would kill off just two ordinary lines since {d, g, i} and {c, a, j} are already collinear, while creating three new ordinary lines: `(h, z∞ ), `(m, z∞ ) and `(b, z∞ ). The example therefore becomes satiated with 21 points and 26 ordinary lines. Pairs of regular 9 and odd n-gons for n > 9 only fair worse. We illustrate the case of glued 9-gons, the analysis of which is a bit different from that of glued 7-gons. See Figure 1.20. In this case, there are quite a few “productive” points at infinity that can be added, one for each of the directions determined by an edge of the 9-gon - yielding a configuration of 27 points. However, in the end we are left with considerably more than

n 2

ordinary lines

(in fact well over n ordinary lines). To see this note first that there are 5 ordinary lines through m and e. The ordinary lines through m are `(m, h), `(m, g), `(m, b), `(m, c) and `(m, 0∞ ). To see that `(m, h) and `(m, g) pass between vertices i and j note simply that `(i, j) is parallel to `(g, h) and that `(h, j) is vertical. Moreover, by virtue of the fact that `(m, h) and `(m, g) pass between i and j, their slopes do not correspond to any of the points at infinity. Analogous reasoning allows us to conclude that n, l, d and f each have three ordinary lines through them. For example, the ordinary lines through l are `(l, h), `(l, g) and `(l, b). All the ordinary lines so far listed and associated with the glued 9-gons are

30

Figure 1.19: Glued regular pentagons and 7-gons, together with their corresponding points at infinity. A regular n-gon has internal angle (n − 2)π/n, with π/n slices as shown. distinct, giving already 22 ordinary lines. For larger m, we can again add productive points at infinity to kill off ordinary lines determined by pairs of points in either of the m-gons, but there become more and more ordinary lines determined by pairs of points, one from the bottom m-gon and one from the top m-gon. To see this a bit more concretely, observe that for a line to pass through a point in the bottom (top) m-gon and two points on the top (resp. bottom) m-gon, the line must actually pass through two points in the bottom (resp. top) m-gon since a line through two points in any of the of the m-gons, other than the center point, makes an angle of

kπ m

with the horizontal, for integral k. But consider points, for example, on the bottom

quarter of the bottom m-gon. As m gets large, the “windows” between successive points on the top quarter of the bottom m-gon become relatively larger in the sense of allowing bottom-quarter points to “see” more points in the upper m-gon. In our analysis we have been biased towards considering just odd m-gons because the Kelly-Moser example arose from gluing triangles and the McKee example arose from gluing pentagons. For completeness note that we can glue together squares in the same fashion and get 8 points with 4 ordinary lines - see Figure 1.21 - to obtain an example which is

31

Figure 1.20: Glued regular 9-gons, together with the “productive” points at infinity - i.e. points which, when added, kill off more ordinary lines than they create. Points at infinity are labeled according to the angle they make with the horizontal. equivalent to the 8 point B¨or¨oczky example3 . Gluing regular hexagons gives an example with 17 ordinary points and 13 ordinary lines. See Figure 1.22. The ordinary lines are `(k, π/2∞ ), `(k, 5π/6∞ ), `(k, 2π/3∞ ), `(k, π/3∞ ), `(k, π/6∞ ), ,`(i, b), `(i, 5π/6∞ ), `(h, e), `(h, π/6∞ ), `(c, j), `(c, π/6∞ ), `(d, g), and `(d, 5π/6∞ ). For octagons and higher we run into the same problem as for odd m-gons (in fact this problem begins to manifest itself already at m = 6) where we begin to get many ordinary lines between points in the lower and upper m-gons.

1.6

Uniqueness of the McKee arrangement for n = 13

In this section we seek to prove that the McKee arrangement is unique amongst 13 line arrangements in having just 6 ordinary points. Before getting to the uniqueness theorem, we establish a few useful facts. 3 Being more formal, we can say that two point configurations are isomorphic iff their dual line arrangements are isomorphic.

32

Figure 1.21: Glued squares, together with the single “productive” point at infinity - yielding an example with 8 points and 4 ordinary (dashed) lines. Lemma 17. An arrangement of n lines with 6n/13 or fewer ordinary points must be saturated in attachments. Proof. The Kelly-Moser arrangement of Figure 1.6 has 3 lines of type (2, 0) and 4 lines of type (0, 3) and is therefore saturated. We may thus focus on arrangements with n lines and precisely 6n/13 ordinary points. In this case, equation 1.4.20 stands, but equation 1.4.23 is not a strict inequality and so both

X 2i 4 ki k0 ≤ k2,0 + k2,1 + 3 3

(1.6.1)

X 4 k0 ≥ k2,0 + k2,1 + (i − 1)ki 3

(1.6.2)

i≥3

and

i≥3

can be satisfied as long as 4 k0 = k2,0 + k2,1 + 2k3 , 3

ki = 0, ∀i > 3

(1.6.3)

and k1 arbitrary. The number of ordinary points is then 1 3 k2,0 + k2,1 + k1 + k3 2 2

(1.6.4)

and the number of lines times points attached is at least 4 3k0 + 2k1 + k2,1 = 3( k2,0 + k2,1 + 2k3 ) + 2k1 + k2,1 = 4k2,0 + 4k2,1 + 2k1 + 6k3 3 or at least 4 times the number of ordinary points.

(1.6.5)

33

Figure 1.22: Glued hexagons, together together with the 6 “productive” point at infinity yielding an example with 17 points and 13 ordinary lines. The ordinary lines are enumerated in the text. It is worth noting that the B¨or¨ oczky examples of n lines with n/2 ordinary points are all saturated. For n ≡ 2 (mod 4) the B¨or¨ oczky arrangements have n lines of type (1, 2) and for n ≡ 0 (mod 4) the B¨or¨oczky arrangements have n/4 lines of type (2, 0), n/2 lines of type (1, 2) and n/4 lines of type (0, 4). See Figure 1.23. There is, however, an arrangement of 12 lines with 6 ordinary points that is not due to B¨or¨ oczky and which is not saturated. This example is obtained by removing a line from the McKee 13 line arrangement (Figure 1.7). See Figure 1.24. Hansen has conjectured [41] that this example together with the B¨or¨ oczky arrangements are the only arrangements with exactly n/2 ordinary points.

Figure 1.23: B¨or¨oczky examples of 10 lines with 5 ordinary points and 12 lines and 6 ordinary points, both of which are saturated.

34

Figure 1.24: An arrangement of 12 lines with 6 ordinary points which is not a B¨or¨ oczky arrangement and not saturated. The third from top and bottom ordinary points are not completely surrounded by triangular cells. The arrangement is just the McKee arrangement with the central vertical line omitted. There are also examples from the theory of maximal wedges (Chapter 3) of arrangements which are saturated but do not have a minimal number of ordinary points. The second arrangement in Figure 3.3 is an example of 15 lines with 20 ordinary points (there are 10 pairs of parallel lines) which is saturated. We note that for an arrangement to have exactly

6n 13

ordinary points both equations

1.6.1 and 1.6.2 must be equalities. In particular, for equation 1.6.1 to be an equality we must have ²i = 0 for all i in equation 1.4.13. Thus Lemma 18. In an arrangement with

6n 13

ordinary points all lines have the minimum number

of attached points. In other words all lines with 0 ordinary points have 3 attachments, all lines with 1 ordinary point have 2 attachments, no line with 2 ordinary points has more than 1 attachment, and lines with 3 ordinary points have no attachments. Furthermore, there are no lines with more than 3 ordinary points. Lemma 19. An arrangement of 13 lines and 6 ordinary points must have 3 lines of type (2, 0). Proof. By equation 1.6.3 we have 4 k0 = k2,0 + k2,1 + 2k3 , 3

ki = 0, ∀i > 3

(1.6.6)

and by Theorem 8, k2,0 > 0. Since k0 must be integral we necessarily have k2,0 ≡ 0 (mod

35

Figure 1.25: Suppose that ordinary points p and q reside on (2, 0) lines adjacent to the 5-crossing u. If there are no additional ordinary points adjacent to u through (2, 0) lines then p and q each absorb a unit of charge from u. p and q are also said to be associated with the 5-crossing u. 3). By the Csima-Sawyer Lemma (Lemma 14), no two (2, 0) lines share an ordinary point, so that 6 ≥ 2k2,0 and thus k2,0 = 3. Charging Scheme In order to prove that the McKee arrangement is the only arrangement of 13 lines with exactly 6 ordinary points, we shall apply a “charging” argument. We use the familiar Yproj equation (1.1.8): t2 = 3 +

X X (j − 3)tj + (k − 3)pk . j≥4

We refer to the quantity

(1.6.7)

k≥4

X X (k − 3)pk (j − 3)tj + j≥4

(1.6.8)

k≥4

as the “charge” of the arrangement. As noted earlier, the first sum gives the “excess crossings” and the second sum the “polygonal excess” of the arrangement. We define a procedure for attributing charge in an unambiguous way to the ordinary points in arrangements where all ordinary points lie on (2, 0) lines (such as Sylvester critical 13 line arrangements, by Lemma 19). When we attribute charge to an ordinary point p we also refer to the point p as “absorbing” charge. First of all, if we have a (4+)-crossing we spread the charge associated with the crossing to ordinary points which are adjacent to the (4+)-crossing along (2, 0) lines. Specifically, if we have a (3 + i)-crossing u and there are j > 0 ordinary points adjacent to u along one or more (2, 0) lines then we attribute

i j

units of charge to each of these ordinary points.

Additionally, we say that each of the j adjacent ordinary points are associated to the (4+)-crossing u. See Figure 1.25. Next, suppose we have a (4+)-gon Q. If an ordinary point p along a (2, 0) line is the vertex of a triangular cell whose opposite edge is also an edge of Q, we say that p

36

Figure 1.26: p is associated to the cell Q: An edge of the (quadrangular) cell Q is also the edge of a triangular cell whose opposite vertex p is an ordinary point, and one of the lines through p is a (2, 0) line. In such cases we distribute at least 14 units of charge to p from Q. is associated to Q, and we attribute some of the charge in Q to p. See Figure 1.26. If there are l > 0 such ordinary points associated with the (3 + k)-gon Q then we attribute

k l

units of charge to each ordinary point. Since p is saturated (Lemma 20), we pause to note that neither line through p can be one of the defining lines of Q. See Figure 1.27 and the accompanying description. If we then sum up the charge associated with all ordinary points, which are either adjacent to a (4+)-crossing via a (2, 0) line or adjacent to a (4+)-gon via a surrounding triangular cell, the number cannot exceed the total charge of the arrangement, or t2 − 3. Lemma 20 ( 6n 13 Charging Lemma). In an arrangement of n lines with exactly

6n 13

ordinary

points: (i) An ordinary point p on a (2, 0) line l can always absorb at least

1 2

unit of charge.

(ii) An ordinary point p on a (2, 0) line l can always absorb strictly more than

1 2

unit of

charge unless p together with the other ordinary point q on l are each adjacent to the same 4-crossing on l. Proof. Let p be an ordinary point on some (2, 0) line l. The proof of the lemma breaks into several cases: Case 1: p is adjacent along l to a (5+)-crossing. Case 2: p is adjacent along l to a 4-crossing, but the 4-crossing does not have a second ordinary point adjacent to it along l. Case 3: p is adjacent along l only to 3-crossings. Case 4: p and the second ordinary point on l surround a 4-crossing.

37

Figure 1.27: An edge of the (quadrangular) cell Q is also the edge of a triangular cell whose opposite vertex p is an ordinary point, and one of the lines through p is a (2, 0) line. A line through p cannot be a defining line of Q since then one of the lines that count p as an attachment (k in the diagram) must enter Q, but Q is assumed to be a cell. Since p is saturated it cannot be adjacent to a 2-crossing (an ordinary point). Hence the above four cases exhaust all possibilities. Case 1: The ordinary point p is adjacent along l to a k-crossing u where k ≥ 5. Observe, by saturation, that the largest number of associated ordinary points we can have around u is k, i.e. at a position along every other “spoke” emanating from u. However, if k is odd and every other spoke contains an associated ordinary point, then every line through u would actually contain a spoke with an ordinary point, and hence be a (2, 0) line. However, if k were such a (2, 0) line then one of the spokes associated with k would have an ordinary point and one would not. The spoke without an ordinary point would have two adjacent spokes with ordinary points which, by virtue of residing on (2, 0) lines would necessarily be saturated and attached to k - an impossibility. On the other hand, if k is even and there were an associated ordinary point on every other spoke then every other line (the lines without associated ordinary points) would have 4 attached points (see Figure 1.28) - which is impossible in an arrangement of

6n 13

ordinary points by virtue of Lemma 18. It

follows that we have less than k ordinary points adjacent to u along (2, 0) lines, and so each ordinary point can absorb at least

k−3 k−1

units of charge. Now

k−3 k−1

>

1 2

as long as k > 5, but

looking at Figure 1.29, we readily conclude that an arrangement with 4 saturated ordinary points around a 5-crossing would again have a line with 4 attachments. Hence, in all cases, an ordinary point next to (5+)-crossing can absorb more than half a unit of charge. Case 2: The ordinary point p is adjacent to a 4-crossing u along the (2, 0) line l but there is no ordinary point adjacent to u on the other side of l (see Figure 1.30). If there

38

Figure 1.28: 6 hypothetical ordinary points along (2, 0) lines surrounding a 6-crossing. By saturation, every other line will have 4 attachments. Line n is such an example.

Figure 1.29: For k = 5 since there cannot be ordinary points on adjacent lines so there is only one configuration of 4 ordinary point surrounding the 5-crossing, and this one configuration, by saturation, gives rise to a line, n, with 4 attached points. is no additional ordinary point around u on a (2, 0) line then p would absorb a full unit of charge from u, so without loss of generality assume there is an ordinary point q on a (2, 0) line o as drawn. q necessarily lies at the intersection of the lines o and m in the Figure, since otherwise it could not be saturated. Note that p absorbs at least

1 3

unit of charge from u

since there are at most three ordinary points along (2, 0) lines adjacent to u. If either of the cells marked P and Q are (4+)-gons then p would absorb at least an additional charge through these polygons and so absorb more than

1 2

1 4

unit of

unit of charge in total. Hence we

Figure 1.30: The case of a hypothetical 4-crossing u with ordinary points adjacent to u on two separate lines l, o of type (2, 0).

39

Figure 1.31: An ordinary point p on the (2, 0) line l surrounded by two adjacent (4+)-gons. may assume that P and Q are triangular cells. In the case of Q this means that m, m0 meet to form the triangular cell. But, by saturation, m, m0 must meet to the “bottom right” of q to form a triangular cell. Hence these points where m and m0 meet are one and the same, and we call the point x. We conclude that a, b, q and x are the only vertices on the line m. However x and b are 3-crossings by the Top 3-Crossing Lemma (Lemma 13) and a must be a 3-crossing or p would absorb at least than

1 2

1 4

unit of charge from it and again absorb greater

unit of charge in total. In sum then, m would have only 7 crossings which is clearly

too few. Hence, if p is adjacent to a 4-crossing along a (2, 0) line l, but l does not have both of its ordinary points adjacent to u, then p can absorb more than

1 2

unit of charge.

Case 3: p is adjacent along the (2, 0) line l only to 3-crossings. By Lemma 16 there are two adjacent triangular cells around p, each adjacent to (different) (4+)-gons. If one of these (4+)-gons is actually a k-gon for k ≥ 5 then p will absorb at least of charge from the k-gon and at least absorb more than

1 2

1 4

k−3 k

>

1 4

units

unit of charge from the other polygon, and so

unit of charge in total. Hence we may assume that both (4+)-gons

are actually quadrangles. Thus in Figure 1.31 some two adjacent cells amongst P, Q, R, S are quadrangles and the others are triangles. If, for example, P, R were quadrangles and p could absorb just

1 4

unit of charge from each then there would have to be some ordinary

point to the right of p along l to absorb charge from R and a third ordinary point to the left of p along l to absorb charge from P - which is impossible since l is a (2, 0) line. Hence we may assume that R, S are quadrangular cells and that P, Q are triangular cells. But now, for R to just give

1 4

unit of charge to p it must have three other ordinary points, all

on (2, 0) lines, also pulling

1 4

unit of charge from it, as drawn in Figure 1.32. It follows

by the Csima-Sawyer Lemma (Lemma 14) that the line k in the Figure is a (2, 0) line. But by assumption, P is a triangular cell so the triangle marked T in the Figure, with edges on k, l and one edge in common with P satisfies the conditions of the Three Clause Lemma (Lemma 5) and so we would conclude that k either has another ordinary point or an attached point, neither of which is possible. It follows that R can give more than

1 4

unit

40

Figure 1.32: A hypothetical quadrangular cell R giving 14 unit of charge to four surrounding ordinary points, each on (2, 0) lines. Since l is a (2, 0) line, by the Csima-Sawyer Lemma, k must be a (2, 0) line. of charge to p and so in total p can absorb more than

1 2

unit of charge in this case.

Case 4: p and the second ordinary point q on l surround a 4-crossing. If there were four ordinary points along (2, 0) lines through u then we would have lines through u with four attachments, which is impossible. Hence there are at most three ordinary points along (2, 0) lines through u. If there were three such ordinary points, then the reasoning from case 2 shows that p and q can each absorb more than

1 2

unit of charge. If there are just two

ordinary points p and q on l which are the sole ordinary points adjacent to u through (2, 0) lines, then by the charging rules p and q each absorb

1 2

unit of charge from u, and possibly

more from elsewhere. The Lemma is thus established. Theorem 21. The McKee arrangement (Figure 1.7) is the unique arrangement of 13 lines with 6 ordinary points. Proof. Suppose we have an arrangement A of 13 lines and 6 ordinary points. By virtue of equation 1.6.3, the fact that k2,0 must be integral, k2,0 > 0 (Theorem 8), and the fact that two lines of type (2, 0) cannot intersect in an ordinary point (the Csima-Sawyer Lemma Lemma 14), A must have exactly 3 lines of type (2, 0) and consequently all ordinary points reside on these lines. Since t2 = 6 there are 3 units of charge in the arrangement, and by the

6n 13

Charging

Lemma (Lemma 20) there must be a 4-crossing adjacent to and between each of the ordinary points on each of the (2, 0) lines, and moreover if any other (2, 0) line intersects in the 4crossing, such a line has no ordinary points which are neighbors to the 4-crossing. It follows

41

k

l a p b

f e

q

r c

m d

Figure 1.33: The three (2, 0) lines l, k, m, each crossing at a distinct 4-crossing (indicated by a solid dot), with ordinary points as indicated with hollow dots. The ordinary points drawn closest to the 4-crossings are meant to be adjacent vertices to the respective 4-crossings. that t4 = 3, and hence by the Yproj equation (equation 1.1.8), ti = 0, for all i > 4 and, moreover, all cells in the arrangement are triangles. By parity, if we count all crossings on a (2, 0) line, we know that each (2, 0) line must contain another 4-crossing. Only two (2, 0) lines can cross at a 4-crossing since two of the crossing lines have attached points. It follows that if l1 , l2 , l3 are the (2, 0) lines with 4crossings u1 , u2 , u3 such that li ’s ordinary points surround ui then if l1 also contains u2 it must be that l2 contains u3 (l2 cannot contain u1 since then l1 would intersect l2 at both u1 , u2 ) and l3 contains u1 . We then have a distribution of (2, 0) lines and ordinary points as depicted in Figure 1.33. We next show that the two remaining lines through each of the 4 crossings (which in Figure 1.33 are labelled p, q and r) cut through opposite pairs of triangles, e.g. that of the two remaining lines through p, one cuts through the triangles “above” and “below” p and one cuts through the triangles to the “left” and “right” of p. By symmetry, it suffices to show that there is an additional line through p cutting into the top and bottom triangles. Thus consider the vertex b, which is necessarily saturated. Since it is the closest vertex to p along l, since k cannot have attached points, we conclude that there must be a line through p cutting the triangles above and below p and counting b as an attachment. In this way we see that there is a line crossing each of the vertices p, q and r cutting each pair of opposite triangles (6 lines in all). Now consider the second line through b. This line meets the line k in a vertex, call it z. If z lies at or above f in the finite interval [f, p) then we can apply the Three Clause Lemma to show that k has an attached point a contradiction. In the same way we get a contradiction using the Three Clause Lemma if

42

k

w

l a

u

p b

f q e

z

r c

d

m v

Figure 1.34: The three lines u, v, w passing through 4-crossings q, r, p respectively, each pass through opposite ordinary points b, f, c, again in the same order. The ordinary point c, though a neighboring vertex to r is not necessarily located “close” to r. We consider first the case where w passes through z, the intersection point of u and v. z were to lie at or below e in the infinite interval [e, p). The line through b therefore must pass through q. Analogously, the line through r passes through f and the line through p passes through c. We label these three lines u, v and w and first prove that they cannot all intersect in a common point. Suppose, on the contrary that they do, as depicted and described in Figure 1.34. Since f is attached to u, the segment [f, z] contains no vertices. Therefore we may apply the Three Clause Lemma to triangle 4(f, p, z) to conclude that it must be a cell. However this would give p three surrounding ordinary points along (2, 0) lines, a possibility we have ruled out. It follows that w does not pass through z. Since w cannot create a vertex in [f, z], the arrangement thus far must look like Figure 1.35. Since p is not adjacent to any ordinary points along k there must be an additional 3-crossing vertex on k between p and f and on the infinite edge between p and e, and similarly for the lines l and m with respect to the 4-crossings r, q respectively. We thus have the situation in Figure 1.36 which now has all vertices on the lines l, k, m accounted for (as one can verify by counting 12 crossings per line). The new vertices interior to the finite triangle 4(p, q, r) along k, l, m we label i1 , i2 , i3 and those external to the finite triangle we label e1 , e2 , e3 respectively. Now since the ordinary point b is saturated, there is a line passing from i2 to x (and not through (x, b) since b is attached to w). Once through x the line has crossed both w and m so must intersect k in the finite interval [p, q]. However, the only unaccounted for crossing in this interval is through i1 (the additional line through q, not drawn, passes through a and the additional line through p, not drawn, passes through d).

43

Figure 1.35: The three lines u, v, w passing through 4-crossings q, r, p and opposite ordinary points b, f, c respectively intersecting in the only possible manner. Similarly we conclude that there is a line passing through i3 , y, i2 and one passing through i1 , z, i3 . We have thus accounted for all intersections inside the finite triangle 4(p, q, r). Since the choice of finite triangle is arbitrary and all triangles are completely symmetrical, we can do the same for any of the other 3 triangles with the same vertices. In all cases, we find that we have no choice about how lines should intersect. By looking at just the finite 4(p, q, r) we note that we have actually not accounted for one of the thirteen lines (which we would account for when examining any of the other three triangles). The unaccounted-for line passes through e2 and e3 and is analogous to the line l(i2 , y, i3 ) which entirely misses the triangle 4(p, q, r) to the left of the vertex i1 . While one can inspect any of the triangles determined by the three 4-crossings in the McKee arrangement (Figure 1.7) to establish graph theoretic isomorphism with the arrangement we have been led to, the established uniqueness of this arrangement, in itself, also establishes this isomorphism. The proof is therefore complete.

1.7

Sharpening the Csima-Sawyer Theorem

A closer look at the proof of Theorem 8 leads to the following sharpened result: Theorem 22. A Sylvester-critical arrangement must contain at least 3 lines of type (2, 0).

44

w

l e2

a

k e1

u

p b i1 x i2 f

z

y i3

q

r

e3

d

m

c v

e

Figure 1.36: The three lines u, v, w passing through 4-crossings q, r, p and opposite ordinary points b, f, c respectively intersecting in the only possible manner, with all vertices on the lines l, k, m accounted for. Proof. As usual write k2,0 for the number of lines of type (2, 0) and k2,1 for the number of lines of type (2, 1+) and suppose k2,0 ≤ 2. With this additional notation, equation 1.4.3 becomes k0 > k2,0 + k2,1 +

X (i − 1)ki

(1.7.1)

i≥3

and equation 1.4.5 becomes 3k0 + 2k1 + k2,1 ≤ 2(k1 + 2k2,0 + 2k2,1 +

X

iki )

(1.7.2)

i≥3

so

2X 4 k0 ≤ k2,0 + k2,1 + iki . 3 3

(1.7.3)

i≥3

But for equations 1.7.1 and 1.7.3 to hold simultaneously, since k0 is an integer, there must be an integer N such that k2,0 + k2,1 +

X 2X 4 iki . (i − 1)ki < N ≤ k2,0 + k2,1 + 3 3

However, 43 k2,0 − k2,0 ≤

2 3

while

(1.7.4)

i≥3

i≥3

P

i≥3 (i

− 1)ki ≥

2 3

P

i≥3 iki

so the right hand side of 1.7.4

is at most its left hand side plus 23 , while the left hand side is integral. Hence there is no way an integer can fit between the two sides and so equation 1.7.4 is impossible and the Theorem follows.

45

Lemma 17 can actually be seen to follow from the (2, 0) Saturation Lemma (Lemma 12) by virtue of the following: Lemma 23. In an arrangement with

6n 13

(or fewer) ordinary points all ordinary points lie

on lines of type (2, 0). Proof. Suppose we have an arrangement A as in the statement of the Lemma. We may assume A has

6n 13

ordinary points, since otherwise A is the Kelly-Moser arrangement which

has all ordinary points on its (2, 0) lines (see Figure 1.6). Rewriting equation 1.6.3 in terms of k2,0 gives

4 k2,0 = k0 − k2,1 − 2k3 , ki = 0, ∀i > 3 3

(1.7.5)

n = k0 + k1 + k2,0 + k2,1 + k3

(1.7.6)

6n 1 3 = k1 + k2,0 + k2,1 + k3 . 13 2 2

(1.7.7)

Of course,

and counting ordinary points

Now, rewriting 1.7.6 in terms of k0 and substituting into 1.7.5 gives: 1 k2,0 = n − k1 − 2k2,0 − 2k2,1 − 3k3 . 3

(1.7.8)

Twice the right hand side of 1.7.7 exactly meshes with 1.7.8 to give k2,0 =

3n 13

(1.7.9)

and since no two ordinary points lie on a single (2, 0) line (the Csima-Sawyer Lemma Lemma 14), this accounts for all

6n 13

ordinary points.

In the spirit of Lemma 18 we make the following definition: Definition 7. We say that an arrangement has no epsilon excess if all lines with 0 ordinary points have 3 attachments, all lines with 1 ordinary point have 2 attachments, no line with 2 ordinary points has more than 1 attachment, and all lines with 3 or more ordinary points have no attachments. Theorem 24 (Strong Charging Lemma). In an arrangement of n lines without epsilon excess, which is not the Kelly-Moser arrangement, an ordinary point p on a (2, 0) line l can absorb at least 1 unit of charge unless p together with the other ordinary point q on l are

46

each adjacent to the same 4-crossing on l, in which case, as in the they can each absorb at least

1 2

6n 13

Charging Lemma,

unit of charge.

Proof. The proof proceeds by examining the same four exhaustive cases as we examined in the proof of the

6n 13

Charging Lemma (Lemma 20).

Case 1: The ordinary point p is adjacent to a k-crossing u where k ≥ 5. The cases k = 5, 6 must be handled separately and then we handle the case k ≥ 7 with a general argument. If k = 5 we show that there can be at most two ordinary points adjacent to u along (2, 0) lines and hence each ordinary point can absorb one unit of charge from u. Suppose on the other hand there were three such surrounding ordinary points. By saturation, in the clockwise ordering of the five lines about u we cannot have two consecutive (2, 0) lines unless neither has an ordinary point adjacent to u. If we label the half-lines or “spokes” emanating from u s1 , ..., s10 , one sees that we cannot have an ordinary point adjacent to u at every third spoke, since e.g. if s1 , s4 , s7 are such spokes, then the line containing s7 would be adjacent to the line containing s1 . Moreover, we cannot have an ordinary point adjacent to u in every other spoke, e.g. along s1 , s3 , s5 because the lines containing s1 and s5 are adjacent. The only remaining possibility is that pictured in Figure 1.37. By saturation, the additional lines through p, r must meet at a vertex x along k.

Figure 1.37: The only possible configuration of three ordinary points along the “spokes” surrounding a 5-crossing. Moreover, by the Top 3-Crossing Lemma (Lemma 13), x is a 3-crossing so the additional lines through p and r (the non-spokes) are the only additional lines through x. Thus the line which is not a spoke through p must count r as an attachment and the line which is not a spoke through r must count p as an attachment. See Figure 1.38. Note that k already counts p, q and r as attachments. However, by the Three Clause Lemma (Lemma 5) applied to the triangle T we conclude that k either contains an additional attached or ordinary point, an impossibility by the assumption of no epsilon excess. It follows that a 5-crossing can be surrounded by at most two ordinary points along (2, 0) lines under the assumptions of the Lemma. We stop to note a general principle. By the preceding argument, in any arrangement

47

Figure 1.38: Filling out the configuration of three ordinary points surrounding a 5-crossing; the line k, which already counts p, q, r as attached points must have an additional ordinary or attached point inside the triangle T . without epsilon excess, and given a vertex u, we cannot have a line passing through u which has three attached points such that the three attached points all reside on (2, 0) lines with the attached points each adjacent vertices to u along their respective (2, 0) lines. Let us call this observation the No Three Closest Attachments Principle. Now consider the case of a 6-crossing u where we need show that there cannot be more than three surrounding ordinary points along (2, 0) lines. Suppose there were in fact four such surrounding ordinary points. Then either there is an ordinary point along every third spoke or there are ordinary points along some two spokes si , si+2 . But the only way to have four ordinary points around u if we have ordinary points adjacent to u along si , si+2 would be to have one of the two situations depicted in Figure 1.39. But in either case the line

Figure 1.39: The cases of four ordinary points along (2, 0) lines surrounding a 6-crossing. In either case the line k violates the No Three Closest Attachments Principle. k would violate the No Three Closest Attachments Principle. For the case of a 6-crossing we are thus left with the sub-case of four ordinary points, one along every third spoke, as in Figure 1.40. We claim that there is necessarily either an ordinary point in (x, y) or an

48

Figure 1.40: The case of four ordinary points p1 , ..., p4 along (2, 0) lines surrounding a 6crossing, with an ordinary point along every third spoke. There must be either an ordinary or attached point associated with the line k in (x, y) as explained in the text. ordinary point attached the line k on the P1 side in [x, y]. Consider the cell marked P1 in Figure 1.40 which has two edges [x, u], [w, u] along consecutive spokes. If P1 is a triangular cell then the existence of an ordinary or attached point as claimed follows by applying the Three Clause Lemma (Lemma 5) to 4(x, w, y). On the other hand, if P1 is a (4+)-gon then consider the three edges of P1 , e1 = [x, u] plus the next two edges e2 , e3 in the clockwise ordering. The lines extending these three edges form a unique triangle T which shares just the edge e2 with P1 , and moreover, note that the line extending e3 necessarily intersects the line extending e1 in (x, y). Applying the Three Clause Lemma to T allows us to again conclude that there is an ordinary or attached point associated with k as earlier claimed. If we consider now the second line through p4 and its intersection point, call it z, on k, we would obtain a fourth ordinary or attached point associated with k unless z ∈ (x, y). See Figure 1.41. But now we shift attention to the line l, where by the same reasoning there is an ordinary or attached point associated with l in [x0 , y 0 ]. The second line through p2 will then create a fourth ordinary or attached point associated with l unless this second line through p2 intersects l inside (x0 , y 0 ). However m necessarily intersects this second line in (y, w) since it cannot pass through [u, w] and must escape 4(y, w, u), and if this second line were to intersect l in [x0 , y 0 ] it would have to pass a second time through m, an impossibility. It follows that in an arrangement without epsilon-excess, a 6-crossing can be surrounded by at most 3 ordinary points along (2, 0) lines and hence in all such cases the ordinary points around the 6-crossing can each absorb at least one unit of charge from the 6-crossing. We now consider the case where u is a k-crossing for k ≥ 7 and we have k − 2 ordinary points along (2, 0) lines emanating from u. If we can prove there cannot be k − 2 such ordinary points then each of the at most k − 3 ordinary points will each absorb at least

49

Figure 1.41: The case of four ordinary points p1 , ..., p4 along (2, 0) lines surrounding a 6crossing, with an ordinary point along every third spoke. There will be a fourth ordinary or attached point associated with k unless the second (pseudo-)line through p4 intersects k in (x, y). k−3 k−3

= 1 unit of charge from the k-crossing and we will be done with the generic case of an

ordinary point along a (2, 0) line adjacent to a (5+)-crossing. Since 3(k − 2) ≥ 2k if k ≥ 7 it is not possible to have an ordinary point at every third spoke. Thus we must have some two ordinary points at spokes si , si+2 for some i. Ruling out the spokes on lines adjacent to the lines containing si , si+2 leaves 2k − 8 remaining spokes. Now we cannot have an ordinary point on the other ends of the lines containing si , si+2 by the No Three Closest Attachments Principle, so we are down to 2k − 10 spokes. These remaining spokes are in two groups of k − 5 consecutive spokes, on k − 5 adjacent lines. If k is odd there are an even number of spokes in each group so utilizing at most every other spoke in each group gives a total of k − 5 + 2 = k − 3 ordinary points. On the other hand if k is even then there are an odd number ≥ 3 of spokes in each group. However, if we label the spokes in each group s1 , ..., sk−5 and the corresponding spokes in the other group s01 , ..., s0k−5 then if s1 and s3 have ordinary points then s01 and s03 cannot also have ordinary points since then the line containing s2 and s02 would have four attached points4 . It follows again that at most k − 5 + 2 = k − 3 spokes have closest ordinary points. Hence we have handled the case of (7+)-crossings and so in all cases in which a (5+)-crossing is beside an ordinary point on a (2, 0) line, the ordinary point can absorb a full unit of charge. Case 2: An ordinary point p on a (2, 0) line l is beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. By the Top 3-Crossing Lemma (Lemma 13), the second line through q must count p as an attachment and the second line through p must count q as an attachment. See Figure 1.42. k (or l) 4

In fact neither spoke can have a closest ordinary point by the No Three Closest Attachment Principle.

50

Figure 1.42: An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. cannot now have a second ordinary point adjacent to u by the No 3 Closest Attachment Principle. We show that p must be able to absorb

1 2

unit of charge in addition to the

1 2

unit

of charge it can absorb from u. By symmetry the same will hold for q. By saturation of q, lines m, m0 meet to form the bottom right cell about q. Now if the vertex v is a 4-crossing then p can absorb at least

1 2

unit of charge from it for a total of at least one unit of absorbed

charge. Alternatively the lines n, n0 must meet to to form a triangular cell Q to the upper left of p (since m, m0 cannot so meet to the bottom left of p, and if neither so met, p would absorb at least

1 4

unit of charge from each for a total absorption of at least 12 ). We then

consider the (4+)-gon P in Figure 1.43 and first suppose P to be a quadrangle. Suppose some of P ’s charge is given to an ordinary point s as shown in the Figure. Then the line m has one ordinary point and two attached points. However, we may apply the Three Clause Lemma (Lemma 5) to the triangle Q to assure it has an additional ordinary or attached point, an impossibility. On the other hand if P gives charge to r, then the same reasoning applies to the line m0 which already has three attached points, and would have an additional ordinary or attached point again by applying the Three Clause Lemma to Q. Hence if P is a quadrangle, p can absorb at least

1 2

unit of charge from it. The case where P is a pentagon

is just as easy. See Figure 1.44. In this case it suffices to rule out absorption through just one of the edges of P (though we can do better). If r and s were ordinary points absorbing

51

Figure 1.43: An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. The case where the cell P is a quadrangle. charge from P then m0 would have three attachments. Then applying the Three Clause Lemma to 4(a, b, c) would give us a fourth ordinary or attached point associated with m0 , a contradiction. If P were a k-gon for k ≥ 6, then, without any additional care we can say that p absorbs at least

k−3 k



1 2

unit of charge and so in all cases p (and by symmetry q)

absorbs at least one unit of charge in total if it is one of two caddy-corner ordinary points adjacent to a 4-crossing u along (2, 0) lines. Case 3: The ordinary point p on a (2, 0) line l is surrounded by 3-crossings. Since our arrangement is not the Kelly-Moser, by Lemma 16 it follows that there are at least two (4+)-gons adjacent to triangular cells surrounding p. Suppose, to start, that there are actually three such surrounding (4+)-gons. In other words we have the situation in Figure 1.45. Moreover, within this case suppose first that there is an ordinary point q on l adjacent to u (necessarily) absorbing charge from P and Q. See Figure 1.46. We will see that the argument is more general but for simplicity suppose P is a quadrangle and ask whether there can be ordinary points r and/or s, as drawn, absorbing charge from P . The lines through r and s which pass through the vertex of P opposite u may be one and the same line or two lines - the argument is the same in either case. If r is ordinary and absorbs charge from P then the line m has three attached points, and we can apply the Three Clause Lemma

52

Figure 1.44: An ordinary point p on a (2, 0) line l beside a 4-crossing u which has a second (2, 0) line k through it and an ordinary point q adjacent to u along k. The case where the cell P is a pentagon. (Lemma 5) to the triangle with edges on lines j, m, m0 containing R since R is a (4+)-gon giving m epsilon excess. On the other hand if s is ordinary, absorbing charge from P , then we can apply the Three Clause Lemma to triangle T in the Figure since then s will have a line counting it as an attachment on the “upper right” which will cut into T - and we thereby get a third ordinary or attached point associated with l - again an impossibility. If P is a pentagon the same basic argument applies, with the only exception that there is not a single line through r and s. It follows that if P, Q, R are all (4+)-gons and there is a second ordinary point q next to u along l, then p can absorb a total of at least one unit of charge. So suppose there is no second ordinary point adjacent to u along l. Referring to Figure 1.46 again, note that there must be ordinary points r, s absorbing charge from P and P must be a quadrangle. Note that r is not necessarily attached to m0 as drawn in the Figure since the line connecting r and s may not be part of the arrangement. Nonetheless m0 has attached points s and p. Now let m ∩ m0 = m∗ . Then applying the Three Clause Lemma to the triangle 4(x, y, m∗ ) containing r and then to the triangle 4(v, w, m∗ ) containing R shows that m0 has a total of 4 ordinary plus attached points, an impossibility. Hence there must be exactly two (4+)-gons surrounding p. Given that there must be exactly two (4+)-gons, P, Q around p, suppose first that P

53

Figure 1.45: An ordinary point p on a (2, 0) line l surrounded by 3-crossings with three (4+)-gons, P, Q, R surrounding its adjacent triangles. and Q are on opposite sides of l. See Figure 1.47. In the first sub-case we suppose that the second ordinary point q along l is adjacent to u absorbing charge from P and Q and we ask whether vertices r and s can absorb charge from P , again initially assuming P to be a quadrangle. First if s absorbs charge from P then we consider the familiar triangle T in the Figure and note that since s is saturated, there is a line counting it as an attachment form the top right which cuts into T so the Three Clause Lemma (Lemma 5) applies to give l a third associated ordinary or attached point. On the other hand if r absorbs charge from P then the line m has three attached points. Ignore the line k for a moment and consider the triangle T 0 . T 0 is defined to be the triangle adjacent to Q with defining lines m and the lines determined by the next two edges of Q in the clockwise ordering, following the vertex v. Since k cuts this triangle, the Three Clause Lemma applies and gives m a fourth ordinary or attached point - a contradiction. If P were a pentagon there could still not be an ordinary point s absorbing charge from it, so in all cases p can absorb from P , and by symmetry, also

1 2

1 2

unit of charge

unit of charge from Q. Hence, with two (4+)-gons on

opposite sides of l and the second ordinary point q of l beside u, p can absorb a full unit of charge. So next, assume there is not an ordinary point on the other side of u to absorb charge from P and Q. We may then assume that Q is a quadrangle. If p does not absorb

1 2

unit

of charge from Q then we must have the configuration in Figure 1.48 where, as earlier, we note that r, s may not actually lie on the same line. We focus on the line n which passes through s, z (but not necessarily r) and x = m ∩ n - noting that it must be of type (2, 0) since by the Csima-Sawyer Lemma k cannot be of type (2, 0). Applying the Three Clause Lemma first to the finite 4(x, y, z), and then to T - which is not a cell since s is saturated,

54

Figure 1.46: An ordinary point p on a (2, 0) line l surrounded by 3-crossings with three (4+)-gons, P, Q, R surrounding p’s adjacent triangles. We assume there is a second ordinary point q on l adjacent to u and consider whether there can be ordinary points r, s absorbing charge from P . In the diagram we assume P is a quadrangle but the argument applies more generally. and so the bottom right line counting s as an attachment cuts into T - we conclude that n has two additional ordinary or attached points, which is impossible given that it is a (2, 0) line. Hence p can absorb half a unit of charge from Q and, by symmetry, half a unit of charge from P . Thus we are left just with the case where p has surrounding (4+)-gons to one side of l, as depicted in Figure 1.49. But this is precisely the situation we saw in the proof of the Csima-Sawyer Lemma where examination of triangles 4(w, t, x) and 4(y, v, z) yields two additional ordinary or attached points along l - an impossibility. We have thus completed the case where p is surrounded by 3-crossings. Case 4: p and the second ordinary point q on l surround a 4-crossing. We repeat essentially the same comments we made in Case 4 of the

6n 13

Charging Lemma. If there were

four ordinary points along (2, 0) lines through u then we would have lines through u with four attachments, which is impossible. Hence there are at most three ordinary points along (2, 0) lines through u. If there were three such ordinary points, then the reasoning from case 2 shows that p and q can each absorb one unit of charge. If there are just two ordinary points p and q on l which are the sole ordinary points adjacent to u through (2, 0) lines, then by the charging rules p and q each absorb

1 2

unit of charge from u, and possibly more

from elsewhere. The proof of the Theorem is finally complete. Note that the argument we made regarding the impossibility of a 6-crossing having two (2, 0) lines passing through it with each of their ordinary points adjacent to the 6-crossing

55

Figure 1.47: An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two (4+)gons, P, Q surrounding its adjacent triangles. We consider first the sub-case where l has its second ordinary point q adjacent to u absorbing charge from P and Q and ask whether the vertices r and s can absorb charge from P . can be made more generally. We can add as many lines as we want through the crossing in addition to the (2, 0) lines and those lines adjacent to the (2, 0) lines through the crossing. Refer to figures 1.40 and 1.41. Lemma 25. In an arrangement without epsilon excess, a crossing (i.e. a vertex) cannot have two (2, 0) lines passing through it, both of whose ordinary points surround the crossing. We can actually refine the Strong Charging Lemma by virtue of the following surprising result: Lemma 26. In an arrangement without epsilon excess, if the two ordinary points p and q of a (2, 0) line l are each adjacent to the same 4-crossing on l, then p and q must together absorb either exactly one unit of charge or at least two units of charge. Proof. We have the generic situation in Figure 1.50 and show first that the cells A, B, C, D must be triangles. Suppose for example that B were not a triangle. Then consider the vertex v shared by B and D and the next vertex, call it w in the counter-clockwise ordering of vertices in B. The line through w which contains the edge of B after [v, w] in the counterclockwise ordering of edges intersects l at a vertex x. See Figure 1.51. It is possible that the vertex x lies at or to the “left” of the vertex y but in any case, we consider the triangle 4(v, w, x) which shares an edge, but no interior with B. A line cannot pass through w without passing through the interior of 4(v, w, x) because it would then pass through the

56

Figure 1.48: An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two (4+)gons, P, Q surrounding its adjacent triangles. We consider the sub-case where l does not have a second ordinary point adjacent to u absorbing charge from P and Q, and ask now whether both vertices r and s can absorb charge from Q. interior of B. Hence we may apply the Three Clause Lemma (Lemma 5) to conclude that l has an additional ordinary or attached point in [v, x], contradicting the fact that l is a (2, 0) line. Hence, by symmetry, A, B, C, D are all triangular cells. Next we consider the case where one of the other four cells adjacent to the triangular cells about p (or q) are not triangles. To begin, we consider the case where exactly one of these other four cells adjacent to the triangular cells about p (or q) are not triangles. See Figures 1.52 and 1.53. Let us call the lines i, j, m and n “outside lines” since in a sense they are the outside lines counting p and q as attachments. There are two possibilities. Either two outside lines, e.g. the lines i and j meet at a point not on k as in Figure 1.52, or the two lines i and j intersect at a point not on k, but there are one or more additional lines cutting through the cell P as in Figure 1.53, each additional line contributing an edge to the cell. Consider the case of Figure 1.52 first. Note that there cannot be an ordinary point absorbing charge from P through [u, x] since Q is a cell. Also there cannot be an ordinary point o absorbing charge from P through [w, x] since one of the lines through o would have to enter Q. Finally, suppose there is an ordinary point o absorbing charge from P through [v, w]. One of the lines through o would have to be l(v, p). Moreover, o, residing on a (2, 0) lines must be saturated by the (2, 0) Saturation Lemma (Lemma 12). Hence o is attached to the line h and the second line through o meets h at m ∩ h. Thus far then, h has three attachments. But now consider the triangle 4(u, y, m ∩ h) that is unbounded to the top right in the Figure. Applying the Three Clause Lemma to this triangle gives h a fourth

57

Figure 1.49: An ordinary point p on a (2, 0) line l surrounded by 3-crossings with two quadrangles, P, Q surrounding its adjacent triangles on the same side of l. We ask which of the possible .

Figure 1.50: Two ordinary points on a (2, 0) line l separated by a 4-crossing in an arrangement without epsilon excess. ordinary or attached point - violating the assumption of no epsilon excess. Hence, in the case of Figure 1.52, p can absorb a full unit of charge from P . Next consider the case of Figure 1.53. Again, charge cannot be absorbed form P through [u, x], and the same argument using the Three Clause Lemma and no epsilon excess shows that charge cannot be absorbed from P through [v, w]. Suppose then there is an ordinary point o absorbing charge from P through [w, x]. There are two cases, either o resides on the line i in (w, x) or o resides inside the triangle 4(w, x, z). Suppose first that o lies on the line i. Then since i counts q as an attached point, it must be that l(o, w) is a (2, 0) line. Now l(o, w) intersects the line h at a point h∗ and applying the Three Clause Lemma to the triangle 4(v, w, h∗ ) adjacent to P , we conclude that l(o, w) must contain an ordinary point in (h∗ , w). Since this ordinary point is saturated, it is not hard to see that it must reside at

58

A

B

y p

C

u

q

w v

x

l

D

Figure 1.51: We consider the case where B is not a triangular cell..

Figure 1.52: The first case where P is not a triangular cell. The dashed lines indicate cells we know to be triangular. l(v, p) ∩ l(o, w). Moreover, by saturation again, we must have h∗ = h ∩ m and so we apply the Three Clause Lemma as before to he triangle 4(u, y, h∗ ) that is unbounded to the top right in Figure 1.53 to see that the line h has epsilon excess. Thus o cannot reside on i and so must reside inside the triangle 4(w, x, z). In this case, the line l(o, w) cannot be a (2, 0) line, by the Top 3-Crossing Lemma (Lemma 13), since the vertex x is a 4-crossing. Hence the line l(o, x) is a (2, 0) line and intersects the line g at a point g ∗ and applying the Three Clause Lemma to the triangle 4(x, r, g ∗ ) adjacent to Q, we conclude that l(o, x) must contain an ordinary point in (x, g ∗ ). As before, since this ordinary point is saturated, it is not hard to see that it must reside at l(r, q) ∩ l(o, x). Moreover, by saturation again, we must have g ∗ = g ∩ n and so we apply the Three Clause Lemma to he triangle 4(u, s, g ∗ ) that is unbounded to the top left in Figure 1.53 to see that the line g has epsilon excess, having three attached points, and one more by the final Three Clause Lemma argument. Therefore the Lemma is proved in the case where precisely one of the four cells adjacent to the triangular cells about p or q is a quadrangle.

59

Figure 1.53: The second case where P is not a triangular cell. The dashed lines indicate cells we know to be triangular. Finally, if P were a (5+)-gon, then still P could not have ordinary points absorbing charge from it along the edges adjacent to [v, u], so p could draw a full unit of charge. Hence in all cases where there is a single non-triangular cell adjacent to the triangular cells about p, p can absorb a full unit of charge from this cell and the Lemma is established. The argument in the case where P is one of two (or more) non-triangular cells adjacent to the triangular cells surrounding p and q is no different - still P cannot have ordinary points absorbing charge from it along the edges adjacent to [v, u], so p can absorb at least half a unit of charge from P , and analogously p or q could absorb at least a half unit of charge from any such additional non-triangular cell. The Lemma follows. Corollary 27. In an arrangement of n lines and

6n 13

ordinary points there must be at least

3 lines of type (2, 0), each of whose ordinary points are surrounded by triangles which are themselves surrounded by triangles. Proof. By Lemma 23 all ordinary points lie on (2, 0) lines. By the Strong Charging Lemma (Theorem 24), Lemma 26 and the fact that the total charge in any arrangement must be 3 less than the number of ordinary points (equations 1.6.7 and 1.6.8) we must have three (2, 0) lines with ordinary points together absorbing one unit of charge, which by the proof of the preceding lemma are necessarily ordinary points surrounded by triangles which are themselves surrounded by triangles. For future use, we make the following definitions:

60

Definition 8. (i) A 4-crossing u is essential if u lies on a (2, 0) line l with ordinary points p and q surrounding u and such that each of the triangular cells surrounding p and q are themselves surrounded by triangular cells. (ii) A (2, 0) line l is essential if its ordinary points p and q each absorb just

1 2

unit of charge. (Equivalently l is essential iff it contains

an essential 4-crossing - however, in a moment we are going to extend the charging rules and this statement will no longer be true.) The following, unfortunately, had to be downgraded from a lemma to a conjecture, because I found a flaw in the argument at the eleventh hour: Conjecture 2. In a saturated arrangement without epsilon excess an essential (2, 0) line must contain a second (4+)-crossing. What we know about this conjecture: Let A be a saturated arrangement without epsilon excess and let l ∈ A be an essential (2, 0) line without an additional (4+)-crossing. By virtue of l being essential we know we have a sub-arrangement as depicted in Figure 1.54 - and we contend that we necessarily have a

Figure 1.54: An essential (2, 0) line l with essential crossing u. sub-arrangement of the general form depicted in Figure 1.55. We draw the lines g and h as pseudolines for simpler rendering. Let us consider any one of the lines intersecting l but not passing through p, u or q. Call such a line m and let us consider the closest vertex v0 to l along m that is projectively above l. There is only one line m0 which could intersect m at v0 in such a way that the triangle formed by l, m and m0 lying above l could conceivably form a triangular cell. This line m0 necessarily intersects l at a neighboring vertex along l to l ∩ m. If any of the other lines intersecting l at vertices other than u passed through v0 we would be able to use a Three Clause Lemma (Lemma 5) argument to conclude that l had

61

Figure 1.55: An essential (2, 0) line l with essential crossing u and two ordinary points sandwiched around the essential crossing u with all other vertices along l being 3-crossings. a third ordinary or attached point. But if m ∩ m0 were ordinary then it would be attached to l - which is impossible; hence a third line must pass through v0 and the third line must pass through u. Clearly k cannot intersect in v0 since then k would intersect either g or h twice. Of course both g and h cannot intersect at v0 since they already intersect at u. If just g intersects at v0 with g ∩ m ordinary then g ∩ m0 would also be ordinary giving g epsilon excess. It follows that m, m0 and one of g and h intersect v0 , and so the general picture in Figure 1.55 is correct and all vertices along g and h are 3-crossings except for u. Hence, by the no epsilon excess assumption, g and h must be (0, 3) lines. Let us redraw the arrangement, as in Figure 1.56, with u taken to be the horizontal point at infinity and

Figure 1.56: The essential (2, 0) line l from Figure 1.55 redrawn with u taken to be the horizontal point at infinity and n ∩ m taken to be the vertical point at infinity.

62

n ∩ m taken to be the vertical point at infinity. The line k (not drawn) is conceivably the line at infinity, though we have not proved this to be necessarily the case. Consider now the closest vertex vu on m above g and the closest vertex vd on m below h. Label the analogous vertices uu and ud on n. Glancing at Figure 1.55 we see that k cannot cross at any of these vertices since, e.g., the line denoted by γ intersects n above g before k does. Moreover, note that, for example, the vertex ud must be affinely below h since if ud were actually affine above uu then their would have to be no vertices along n affinely below h, which is impossible since f ∩ n is such a vertex. Analogous arguments apply to confirm the affine placement of the vertices vu , vd and uu with respect to the lines g and h. The line through vu which crosses g furthest to the right along g, say at a point ξR , induces a third attached point to g in the finite interval [m ∩ g, ξR ]. Analogously, the line through uu which crosses g furthest to the left along g, at a point ξL , induces an attached point to g in the finite interval [ξL , n ∩ g]. g will thus have a total of four attached points unless ξR is further to the right than ξL . If we denote by ηL , ηR the analogous vertices along h we must have ηR further to the right along h than ηL . See Figure 1.57. Put

Figure 1.57: The essential (2, 0) line l from Figure 1.56 with the addition of lines ψ1 = l(vu , ξR ).ψ2 = l(uu , ξL ), ψ3 = l(vd , ηR ), ψ4 = l(ud , ηL ). ψ1 = l(vu , ξR ).ψ2 = l(uu , ξL ), ψ3 = l(vd , ηR ), ψ4 = l(ud , ηL ). Then note that, for example, ψ1 must be one of the lines passing through g, l and h sloping from upper left to lower right (by “sloping from upper left to lower right” we just mean this in a relative sense: of the two lines meeting g at a given vertex, between m ∩ g and n ∩ g, one line slopes relatively speaking, from top left to bottom right - it is such a line that we are referring to). If ψ1 were a line sloping in the opposite direction, then if ψ10 were the other line through ξR , since ψ10 cannot intersect m affinely between g and vu , ψ1 and ψ10 would necessarily intersect twice.

63

It follows then that either (1) ψ4 intersects g at a vertex to the left of ξR or (2) ψ3 intersects g at a vertex to the right of ξL . Suppose for the moment that ξL is not the vertex just to the right of m ∩ g along g, and that analogously, ξR is not the vertex just to the left of n ∩ g along g, ηL is not the vertex just to the right of m ∩ h along h, and ηR is not the vertex just to the left of n ∩ h along h. Then in case (1) ψ1 and ψ4 intersect twice while in case (2) ψ2 and ψ3 intersect twice, in either case a contradiction. It follows that in at least one case, say in the case of ψ1 , that the intersection point ξR = ψ1 ∩ g is at the vertex just to the left of n ∩ g along g. Let us call this case, where there is a line from a closest vertex of m (or n) of the type vu , vd , uu , ud to the vertex along g or h which is adjacent to n (or m) on g or h a “long extender.” Using this terminology, our rather weak conclusion is that one of the lines ψ1 , ψ2 , ψ3 or ψ4 must be a long extender. In order to obtain our next result we need to make the following small extension to the charging rules. Extension to the Charging Rules: If a (2, 0) lines l contains a k-crossing u for k ≥ 4, u is not adjacent to either of the ordinary points on l and there are no (2, 0) lines in addition to l passing through u, then spread the k − 3 units of charge associated with u evenly between the ordinary points on l - in other words, give them each

k−3 2

additional units of

charge. With this addition to the charging rules it is conceivable that a (2, 0) line l contains an essential 4-crossing without itself being essential since the ordinary points on l may absorb charge from a (4+)-crossing on l which is not adjacent to them. Lemma 28. An essential (2, 0) line with essential 4-crossing u cannot contain an additional (4+)-crossing v unless v is also essential. Proof. Suppose first that v is a 4-crossing and that the “middle” line of the three lines in addition to l passing through v is k, as depicted in Figure 1.58. Suppose there is at most one ordinary point adjacent to v along k. Let q be the vertex on k adjacent to v which is not ordinary and let m, n be two lines passing through q in addition to k. Then we can apply the Three Clause Lemma to the triangles formed by l, m and [v, q], and l, n and [v, q], to get a third ordinary or attached point associated with l unless both m and n pass between the two ordinary points on l. However, since there is just one vertex, u, between the ordinary points, and m and n already intersect at q, this is impossible. It follows that the two closest vertices to v along k are ordinary and that the lines passing through these closest vertices in addition to k both pass through u.

64

Figure 1.58: An essential (2, 0) line l with essential 4-crossing u together with another 4crossing v. Assume first that one of the closest vertices to v along k is not ordinary - call this vertex q, and consider the additional lines m, n through q. In order for the line l to be essential, by the Extension to the Charging Rules, it must be the case that one of the lines g, k or h is a (2, 0) line. By symmetry it suffices to suppose that h were a (2, 0) line. Let q be the ordinary point adjacent to v along k “below” l and let n be the second line through q which, as we know, also passes through u. n ∩ h cannot be the closest vertex to v below l along h or q would be attached to h - an impossibility since h is assumed to be of type (2, 0). So let r be the closest vertex to v along h below l. r clearly cannot be ordinary, because by (2, 0)-saturation (Lemma 12) it would create an attachment for l. Therefore there are two additional lines passing through r - but unless one of these lines passes through u, we would necessarily have a third ordinary or attached point associated with l by the Three Clause Lemma (Lemma 5). But if one of the lines through r passed through u then this same line would pass twice through n, an impossibility. Hence g and h cannot be (2, 0) lines. It follows that k must be a (2, 0) line. It remains to show that v cannot be a (5+)-crossing and then, given that v is a 4-crossing on a (2, 0) line, that it must be essential. Suppose that v is a (5+)-crossing. For any “non-outside” line amongst the lines through v not including l (analogs of the line k in Figure 1.58), we conclude, as earlier, that both closest vertices must be ordinary, with the second line through these vertices passing through u. But then none of these ordinary points can be saturated, so none can lie on (2, 0) lines. Further, the same argument as we used at the end of the preceding paragraph shows that the outside lines (the analogs of g and h in Figure 1.58) cannot be of type (2, 0). Thus if v is a (5+)-crossing, none of the lines through v are (2, 0) lines and, by the Extension to the Charging Rules, l could use the crossing to absorb more charge - contrary to the assumption

65

that l is essential. Hence v cannot be a (5+)-crossing. Given that v is necessarily a 4-crossing we now show that v is essential. Since k is a (2, 0) line q is saturated. Refer to Figure 1.59. We must show that the triangles surrounding

Figure 1.59: An essential (2, 0) line l with essential 4-crossing u together with another 4crossing v. k is known to also be a (2, 0) line with its two closest vertices to v each ordinary. We show that v is essential. q (equivalently p) are themselves surrounded by triangles. Note that the line n crosses l at a vertex other than u since n crosses i and i passes through u. Let x = h ∩ n and y = n ∩ l. Then if the triangle 4(v, x, y) that resides “below” l were not a triangular cell we could apply the Three Clause Lemma to conclude that l has a third ordinary or attached point. Hence the triangles about p and q are themselves surrounded by triangles on the side closest to l. As far as the triangles facing away from l, consider the triangle bounded by the lines n, m and h which is adjacent to the triangular cell to the bottom right of q. If this triangle is not a cell, the line m is cut by another line o at some vertex which is as close as possible to a = m ∩ n. If we put b = m ∩ o and c = o ∩ k then we can apply the Three Clause Lemma to 4(a, b, c) to conclude that k has a third ordinary or attached point. Hence v is necessarily essential and the Lemma is established. In a hypothetical arrangement with Lemma 29. In an arrangement with

6n 13

6n 13

ordinary points we can say something stronger.

ordinary points a (2, 0) line l whose ordinary points

surround an essential 4-crossing must itself be essential. Proof. Suppose l is a (2, 0) line, as in the statement of the Lemma, with essential 4-crossing u sandwiched between its ordinary points. Then if l contains no further (4+)-crossings it

66

is automatically essential, so suppose l contains a second (4+)-crossing v. Arguing as in the proof of the previous lemma, v must be a 4-crossing where, say, lines g, k, h cross as in Figure 1.59, and moreover, such that the middle crossing line k has two closest vertices to v, call them p and q, which are ordinary. By Lemma 23 all ordinary points lie on lines of type (2, 0) so if k is not a (2, 0) line the second lines through p and q are both of type (2, 0), but since both lines pass through u, at least one has an attachment, an impossibility. It follows that k is a (2, 0) line and so l is essential. If Conjecture 2 could be proved, a simple argument using Lemma 29 could now be given to show that there does not exist arrangements of n lines with 6n/13 ordinary points for even n. However, we shall have to leave this as a conjecture: Conjecture 3. There does not exist an arrangement of n lines with 6n/13 ordinary points for any even n. Definition 9. Say an ordinary point p is associated to a line l if either p is contained in l or p is attached to l. This definition allows several somewhat shorter computations than we have made. For example, to notice that an arrangement with no (2, 0) lines must contain at least

n 2

ordinary

points, we count the number of point-line associations, first from the perspective of the lines, then from the perspective of the points. On the one hand, each ordinary point is associated with at most 6 lines, and on the other, if there are no (2, 0) lines, then each line is associated with at least 3 ordinary points, hence if m is the number of ordinary points, 3 + ² is the average number of ordinary points associated with the various lines, then we have 6m ≥ (3 + ²)n or m ≥ n2 . Lemma 30. (i) A Sylvester critical arrangement with exactly three lines of type (2, 0) necessarily has zero epsilon excess and has k ordinary points and 2k + 1 lines for some k. (ii) A Sylvester critical arrangement with an even number of lines must have at least six (2, 0) lines, and finally (iii) a Sylvester critical arrangement with an even number of lines and exactly six (2, 0) lines must have zero epsilon excess. Proof. (i) With k2,0 = 3 we have 6m ≥ 2k2,0 + 3(n − k2,0 ) n−1 m ≥ 2

(1.7.10) (1.7.11)

67

If n is even then of course m ≥ Otherwise,

n−1 2

=

d n2 e

n−1 2

means the arrangement is not Sylvester critical.

− 1 and (i) is established.

For (ii), from 1.7.10, we have m≥

n k2,0 − 2 6

(1.7.12)

so by the same reasoning, if n is even, in order for the arrangement, to be Sylvester critical we must have k2,0 ≥ 6. For (iii) with k2,0 = 6 and n even, we have 6m ≥ 2k2,0 + (3 + ²)(n − 6) n m ≥ − 1 + ²(n − 6) 2 so ² = 0.

In fact, we can generalize the previous lemma: Lemma 31. Let A be an arrangement of n0 + 2λ lines with {0, 1, 2, ...}. Then A has at least has exactly

3n0 13

3n0 13

6n0 13

+ λ ordinary points, for λ ∈

lines of type (2, 0). Moreover, if such an arrangement

lines of type (2, 0) it must have zero epsilon excess.

Proof. We have 6m = 2k2,0 + (3 + ²)(n − k2,0 ) 6(

6n0 13

+ λ) = 2k2,0 + 3n0 + 6λ − 3k2,0 + ²(n0 + 2λ − k2,0 ) k2,0 =

3n0 + ²(n0 + 2λ − k2,0 ) 13

which establishes both assertions. It is worth noting that every conceivable Sylvester-critical arrangement would have n0

+ 2λ lines and

6n0 13

+ λ ordinary points for some non-negative integer λ. To see this it

suffice to show that any rational number p where

6 13

≤p
13, if η1 = ξ4 and η4 = ξ1 then η1 and η4 intersect twice, a contradiction. Analogously, and again under the assumption n > 13, if η3 = ξ2 and η2 = ξ3 then η2 and η3 intersect twice. V. WLOG we may assume that η3 = ξ2 and η4 = ξ1 . We argue that either ψ1 passes through the first vertex along g 0 above g or ψ2 passes through the first vertex along h0 below h. WLOG we assume the former. We consider the line η1 and note that b = η1 ∩ ξ1 cannot be the first vertex along η1 to the left of g 0 . Call the first vertex to the left of g 0 along η1 , vw . To avoid a Three Clause Lemma problem there must be at least one line through vw which intersects g 0 above g 0 ∩ η1 , however such a line, after passing through vw would have to pass through g, but there are no available vertices for it to pass through, finally yielding a contradiction.

70

Proof. We proceed according to the above outline. I. Suppose there were an arrangement A of n lines with

6n 13

or fewer ordinary points for n

odd. By the Csima-Sawyer Theorem it would necessarily be the case that n is a multiple of 13 and there are exactly

6n 13

ordinary points. By the Strong Charging Lemma (Lemma 24)

coupled with the fact that the total charge in an arrangement must be less than the number of ordinary points, there must be an essential (2, 0) line l ∈ A, say, with ordinary points p and q surrounding an essential crossing v. By crossing parity, l must contain an additional (4+)-crossing u in addition to v. By Lemma 28 u is essential, and then by Lemma 29 the (2, 0) line k containing u between its ordinary points must also be essential. Applying the same reasoning again to the line k we conclude that A must contain a third essential (2, 0) line k 0 with k and k 0 meeting at an essential 4-crossing. So l has ordinary points p and q separated by essential 4-crossing v and a second essential 4-crossing u, which it shares with k and let r and s be the ordinary points surrounding u on k. To avoid giving l additional ordinary or attached points (by virtue of the Three Clause Lemma) we know that the lines through r, s besides k pass through v counting p and q as attached points. If l contained an additional (4+)-crossing it would necessarily be essential by Lemma 28. However, if r0 , s0 were the surrounding ordinary points to this additional essential crossing, then there would necessarily also be lines through r0 , s0 passing through v and counting p and q as attached points. The lines would therefore have to be the same lines as those through r and s and so would each have a total of four ordinary plus attached points, violating the assumption of no epsilon excess. It follows that l contains two ordinary points, two essential 4-crossings and that all remaining crossings are 3-crossings. We now let g ∩ l ∩ h be the horizontal point at infinity and k ∩ k 0 be the vertical point at infinity, as in the proof outline and as in Figure 1.60. It is a routine argument to see that the arrangement has the generic form of this Figure insofar as all vertices along the lines g, h (and hence g 0 , h0 are 3-crossings except for one ordinary point and one 4-crossing. By symmetry, it suffices to make the argument for the lines g and h. Of the lines crossing l at 3-crossings there are lines which cross each vertex with relatively larger slopes (“right slanting” lines) and those with relatively smaller slopes (“left slanting” lines). Consider now the vertex just to the left of u along l and the left slanting line through this vertex, labeled ξ1 . Consider the first vertex along ξ1 above l. Note that the fourth line through g ∩ l ∩ h cannot meet ξ1 at this closest vertex since g ∩ ξ1 is definitely even closer. Moreover, if ξ1 were met at this closest vertex by any line other than g and ψ1 then we could apply the Three Clause Lemma to obtain an attached point to l contradicting the fact that l is a (2, 0) line. If g ∩ ξ1 were ordinary then g would have four associated points, and if ξ1 ∩ ψ1

71

were ordinary then it would be attached to l - in either case a contradiction. The same reasoning applies to the first vertex along ξ2 below l and then to successive vertices formed by pairs of left and right slanting lines moving westward. We therefore conclude that in the respects claimed, the picture in Figure 1.60 is accurate. II. Next notice that since there are three lines parallel to k, either ξ1 or ξ3 must contain a finite vertex above g unless ψ1 and ψ3 are both parallel to k. But then the infinite vertices formed by the line at infinity (a line in the arrangement) and ξ3 and ξ1 will give rise to excess attached points associated with g by the Three Clause Lemma unless (i) there are no vertices along g between ξ3 ∩g and infinity to the right and no vertices along g between ξ1 ∩g and infinity to the left and (ii) ξ2 kξ3 and ξ1 kξ4 . In this case, if we count lines crossing along g we then get 12 and so we obtain just the 13 line Csima-Sawyer arrangement, contrary to the assumption that n > 13. WLOG then suppose it is ξ1 that has finite vertices above g and consider the closest such vertex along ξ1 above g. Observe that to avoid a fourth associated point to g there must be a line meeting this closest vertex which crosses g to the right of k ∩ g and is not the fourth line meeting g ∩ l ∩ h. It follows that there are vertices above g on ξ3 and so there must be a line meeting the closest vertex above g along ξ3 which crosses g to the left of g ∩ k. Analogous statements apply to the lines crossing ξ2 and ξ4 at their closest respective vertices below h. III. Let us start by considering the line, which we call η1 , meeting the closest vertex, b, along ξ1 above g and crossing g to the right of k ∩ g. See Figure 1.61. If there are several lines through b crossing g to the right of k ∩ g let η1 be the line crossing g at the left-most of these vertices. Let us suppose initially that η1 6= ξ4 so that η1 ∩ g is at or to the right of ψ3 ∩ g. Let us establish first that ψ3 passes through α3 . If ψ3 does not pass through α3 then the third line through α3 in addition to h0 and ξ2 could not pass through g at a vertex to the right of ψ3 ∩ g by the Three Clause Lemma. Further, the third line through α3 could not pass through g at a vertex to the left of ψ1 ∩ g because such a line would necessarily pass through η1 twice. Thus the third line through α3 would have to be ψ1 . But then the third line in addition to ξ4 and g 0 through α1 would intersect ψ1 twice, a contradiction. Hence ψ3 passes through α3 . But now if η1 = ψ3 then the third line through α1 besides ξ4 and g 0 must be ψ1 and the third line through g 0 ∩ η1 besides g 0 and η1 will have no way of exiting the quadrangle Q(b, η1 ∩ g 0 , g 0 ∩ g, ψ1 ∩ g). Hence η1 6= ψ3 . Let us now switch attention to ξ3 . Note that ξ3 has vertices above g since η1 ∩ ξ3 is such a vertex. So consider the closest vertex, b0 , along ξ3 above g, and the line η3 meeting b0 which intersects g to the left of r. n3 cannot intersect g to the left of ψ1 ∩ g since it would

72

Figure 1.61: The arrangement considered in Figure 1.60 with the addition of the line η1 meeting the closest vertex, b, along ξ1 above g and crossing g to the right of k ∩ g. necessarily cross η1 twice. Moreover, we cannot have η3 = ψ1 for the same reason we could not have η1 = ψ3 . Hence η3 = ξ2 . The same argument also holds below h; in other words if η2 is the line crossing ξ2 at its first vertex below h and intersecting h to the right of, but as close as possible to s, and η4 is the analogous line with respect to ξ4 , then either η2 = ξ3 or η4 = ξ1 . IV. Thus we have that above g, either η1 = ξ4 or η3 = ξ2 , and below h, either η2 = ξ3 or η4 = ξ1 . But now if, say η3 = ξ2 and η2 = ξ3 then we would have the situation in Figure 1.62 which would lead to a contradiction since the two lines have two points of intersection x and y unless actually x = y. But note that if x = y then the only vertices along η3 or ξ3 would be those given in Figure 1.62 so that all additional lines would have to pass through x. If n = 13 then x would be a 3-crossing and everything would actually work out fine5 . However, if n > 13 then x would be an (n − 10)-crossing which of course for n ≡ 0 (mod 13) and n odd is very large. Moreover, the essential (2, 0) line k 0 would have to pass through x, which is impossible by Lemma 28. 5 Indeed, pushing this line of reasoning further we could obtain an alternative proof that the McKee arrangement is the unique arrangement with 6 ordinary points amongst 13 line arrangements.

73

Figure 1.62: The arrangement considered in Figure 1.61 where both η3 = ξ2 and η2 = ξ3 . The two lines η2 and η3 would then intersect twice unless x = y. V. Without loss of generality, let us assume that η1 6= ξ4 , η2 6= ξ3 (and so η3 = ξ2 and η4 = ξ1 ), and that η1 intersects g 0 at z and η2 intersects h0 at z 0 . See Figure 1.63. It is easy to see that the finite open intervals (x, z) and (x0 , z 0 ) do not contain ordinary points. For example, suppose (x, z) contained an ordinary point, call it e. The second line through e besides g 0 , call it φ, passes through the essential 4-crossing on k which is at infinity so is parallel to k. But this means that φ intersects η1 between g 0 and k, violating the crisscrossing nature of the lines through g 0 , k, h0 (where all vertices in the strip between g 0 and h0 are contained on one of these lines). Now consider the lines ψ1 and ψ2 - we show that either ψ1 passes through α1 or ψ2 passes through α2 . Just like in the preceding argument, if ψ1 crossed g 0 above g at a vertex α then there could not be an ordinary point in the finite (x, α) so if α1 6= α then g 0 would have a fourth associated point by the Three Clause Lemma. Thus if ψ1 does not pass through α1 , it must cross g 0 (and h0 ) below h. Analogously, if ψ2 does not pass through α2 it must cross h0 (and g 0 ) above g. But if both ψ1 crosses h0 below h and ψ2 crosses g 0 above g then ψ1 and ψ2 would cross twice. Hence either ψ1 passes through α1 , or ψ2 passes through α2 . We assume the former.

74

Figure 1.63: The arrangement considered in earlier figures where η1 6= ξ4 , η2 = 6 ξ3 , and that η1 intersects g 0 at z and η2 intersects h0 at z 0 . Although not drawn, η3 = ξ2 and η4 = ξ1 . Now note that the line ξ4 cannot pass through b since, by virtue of the fact that η4 = ξ1 , this would imply that the lines ξ4 and ξ1 intersect twice. Thus, let zw be the vertex of η1 just to the west of z. Any line passing through zw must cross g 0 in the finite (x, z) since there is no place else for such a line to go given that ψ1 passes through α1 . But then the line through zw crossing g 0 at a southern-most point zs induces a fourth associated point to g 0 via an application of the Three Clause Lemma to 4(zw , z, zs ), a contradiction. The Theorem is therefore established.

1.8

Ordinary point bounds for small n

Given precisely n lines in the (projective) plane, not all lines passing through a common point, a certain amount of work has gone into computing the lower bound m(n) on the number of ordinary points for particular small values of n. Values have been obtained by Crowe and McKee [18] and Brakke [8] as far back as 1968 and 1972, to which, by virtue of the Csima-Sawyer Theorem (Theorem 15) and the known

n 2

examples of B¨or¨ oczky (see

section 1.3), we can add values for n = 20 and n = 24. Very recently, Xiaomin Chen, in his

75

Ph.D. thesis [11], determined the value m(15) = 9 using a computer search. n

3

4

5

6

7

8

9

10

11

12

13

14

15

16

m(n)

3

3

4

3

3

4

6

5

6

6

6

7

9

8

17

18

20

21

22

23

24

n m(n)

9

19

10

11

12

The values of m(n) could conceivably be different for lines and pseudolines, though the known values in the above table are valid for both. The best known asymptotic examples for odd n are also due to B¨or¨ oczky (again as reported in [18]). The bounds are considerably worse then in the even n case, and are most easily verified in the dual setting where we have points and count ordinary lines. There are two separate cases, when n ≡ 1 (mod 4) and n ≡ 3 (mod 4). For the case n = 4k + 1, one starts with a regular 2k-gon and the 4k points of Figure 1.5, and then adds the point at the center of the regular 2k-gon. See the left hand example in Figure 1.64. Note that none of

Figure 1.64: B¨or¨ oczky examples with n = 4k + 1 points (left) and n = 4k + 3 points (right). Note the missing point at infinity (and center-point) in the right-hand example. the ordinary lines from the even n case went through the center of the k-gon, so these lines are still ordinary. In addition, half of the lines determined by the newly added center and the points at infinity are ordinary. Therefore the 4k + 1 points determine 3k ordinary lines. For the case n = 4k + 3 we start with a regular 2k + 2-gon, add the points at infinity just as in the other examples, but this time do not add the center point, and furthermore, delete a point at infinity which is not determined by an edge of the 2k + 2-gon. Two of the ordinary lines associated with the vertices of the 2k + 2-gon are no longer ordinary, by virtue of the removal of the single point at infinity. In addition, the removal of the point at

76

infinity adds k ordinary lines corresponding to k distinct pairs of vertices determining lines with the same direction as the missing point at infinity. In total then, the 4k + 3 points give rise to 3k ordinary lines. The B¨or¨oczky examples with either n = 4k + 1 or n = 4k + 3, and 3k ordinary lines give the best possible bounds for n = 9, 11 and 15. This optimality requires a proof in the case n = 9 (Crowe and McKee [18]) and follows by Chen’s computer enumeration [11] for n = 15. The B¨or¨oczky example alone suffices for n = 11 by the Csima-Sawyer Theorem. Though the following Lemma is a consequence of Chen’s computer-aided verification that m(15) = 9, it is nice to see that the result can be obtained by a simple analytical argument. I actually obtained this result, along with the next one, quite early in my investigations (i.e. before developing the Strong Charging Lemma (Lemma 24)). Lemma 33. There does not exist an arrangement of 15 lines with 7 ordinary points. Proof. Suppose we had an arrangement with 15 lines and just 7 ordinary points. Then the Yproj Equation (1.1.8) implies that t4 + 2t5 + 3t6 + 4t7 ≤ 4.

(1.8.1)

Now note that if we multi-count crossings, we have the fundamental relation6 µ ¶ Xµ ¶ n i = ti 2 2

(1.8.2)

98 = 3t3 + 6t4 + 10t5 + 15t6 + 21t7 .

(1.8.3)

i≥2

so that

Since the coefficients of all terms on the right hand side are divisible by 3 except the t5 term, if t5 = 0, 1 we get equations that are unsolvable modulo 3, so the only possibility is that t5 = 2 and by 1.8.1 t4 = t6 = t7 = 0. By the Yproj Equation (1.1.8) we also have pk = 0 for k > 3. In other words the arrangement contains all triangles and therefore is saturated. By Lemma 16 then each ordinary point is adjacent to a (4+)-crossing, and since there are only 5-crossings in this category, each ordinary point is adjacent to a 5-crossing. Since each line is crossed by an even number of lines, and there are only 2-crossings, 3-crossings and 5-crossings, there can be no line with 1 or any odd number of ordinary points. By Theorem 22 there are at least three lines of type (2, 0) and by the Csima-Sawyer Lemma (Lemma 14) exactly three lines of type (2, 0). We now characterize precisely what 6

as also used in [18].

77

other types of lines there may be. For this computation we use three basic equations. First, counting the total number of lines we have k0 + k2,0 + k2,1 + k4 + k6 = 15 or k0 + k2,1 + k4 + k6 = 12.

(1.8.4)

Second, counting ordinary points, we have 2k2,0 + 2k2,1 + 4k4 + 6k6 = 14 so k2,1 + 2k4 + 3k6 = 4.

(1.8.5)

Finally, because the number of attached points can be no more than 4 times the number of ordinary points, we have (using the (0, 3+) Lemma) 3k0 + k2,1 ≤ 28.

(1.8.6)

By 1.8.5 if k6 = 1 then k4 = 0, k2,1 = 1 so by 1.8.4 k0 = 10 contradicting 1.8.6. Hence k6 = 0. Now in 1.8.5 suppose k4 = 2, k2,1 = 0. Then by 1.8.4 k0 = 10 again contradicting 1.8.6. Finally, in 1.8.5 suppose k4 = 1, k2,1 = 2. Then by 1.8.4 k0 = 9 once again contradicting 1.8.6. So we necessarily have k4 = 0 and so k2,1 = 4 and k0 = 8. The number of attached points is then 4(1 + ²2,1 ) + 8(3 + ²0 ) or 28 + ²2,1 + ²0 whence ²0 = ²2,1 = 0 meaning that we must have 4 lines of type (2, 1) and 8 lines of type (0, 3). Using this information about the available lines, we show that there can be at most 3 ordinary points adjacent to each 5-crossing. Suppose we had 4 ordinary points adjacent to a given 5-crossing. Note that since two (2, 0) lines cannot intersect in an ordinary point, 6 of the 7 ordinary points reside on (2, 0) lines. Further, by the Top 3-Crossing Lemma (Lemma 13) if a 5-crossing is adjacent to an ordinary point that resides on a (2, 0) line the 5-crossing must be adjacent to the ordinary point along the (2, 0) line. Hence of the 4 ordinary points adjacent to the 5-crossing, at least three must be adjacent along (2, 0) lines. It is then immediate, if we are to avoid placing an ordinary point along a “spoke” in such a way that it would be attached to a (2, 0) line that actually there are two ordinary points on the same (2, 0) line surrounding the 5-crossing, and then two additional ordinary points, also on a (2, 0) line surrounding the same 5-crossing, as in Figure 1.65. But now the line n will have 4 attached points - which is impossible since we earlier concluded that we had

78

Figure 1.65: A 5-crossing with two surrounding ordinary points on two lines of type (2, 0) crossing at the 5-crossing. The line n have four attached points. only lines of types (2, 0), (2, 1) and (3, 0). It follows that each 5-crossing is surrounded by at most 3 ordinary points. However, there are 7 ordinary points and each ordinary point must be adjacent to a 5-crossing, so there are not enough 5-crossings to go around and hence there can be no arrangement of 15 lines and just 7 ordinary points, as claimed. Lemma 34. There does not exist an arrangement of 17 lines with 8 ordinary points. Proof. Suppose we had an arrangement with 17 lines and 8 ordinary points. Then the Yproj Equation (1.1.8) implies that t4 + 2t5 + 3t6 + 4t7 + 5t8 ≤ 5.

(1.8.7)

Next, multi-counting crossings we obtain µ ¶ Xµ ¶ n i = ti 2 2

(1.8.8)

128 = 3t3 + 6t4 + 10t5 + 15t6 + 21t7 + 28t8 .

(1.8.9)

i≥2

so

This latter equation we will consider modulo 3. If t8 = 1 then the arrangement contains only triangles and so by virtue of Lemma 16 each ordinary point is adjacent to the 8-crossing and so there are 4 lines, each with 2 ordinary points crossing at the 8-crossing (in fact they would all be (2, 0) lines). However, these lines, having 2 ordinary points, an 8-crossing and all the rest 3-crossings, would have an odd number of total crossings which is impossible. Hence t8 = 0. Next, since all remaining terms on the right hand side, with the exception of the t5 term, are divisible by 3 we conclude that t5 6= 0 and t5 6= 1. It follows that t5 = 2, t6 = t7 = 0, and t4 = 0 or t4 = 1.

79

We next show that there cannot be any excess attachments in the sense of there being an ²i > 0 from equation 1.4.13. In particular this will show that there are are no lines with 4 attachments. In the process we also show that our hypothetical arrangement of 17 lines and 8 ordinary points is necessarily saturated. First of all, k0 + k1 + k2,0 + k2,1 +

X i≥3

so k0 =

ki =

X 17 iki ) (k1 + 2k2,0 + 2k2,1 + 16

(1.8.10)

i≥3

X 17 1 9 9 k1 + k2,0 + k2,1 + ( i − 1)ki . 16 8 8 16

(1.8.11)

i≥3

Now, from the Csima-Sawyer Lemma (Lemma 14), we know that the number of ordinary points on (2, 0) lines is no greater than the number of ordinary points on all other lines, hence k1 ≥ 2k2,0 − 2k2,1 −

X

iki

(1.8.12)

i≥3

which we then substitute into equation 1.8.11 to give k0 ≥

X X 17 1 9 9 (2k2,0 − 2k2,1 − iki ) + k2,0 + k2,1 + ( i − 1)ki 16 8 8 16 i≥3

or

(1.8.13)

i≥3

X 5 (i − 1)ki . k0 ≥ k2,0 + k2,1 + 4

(1.8.14)

i≥3

Now, from equation 1.4.13 one obtains equation 1.4.14 which we copy below X 2i 4 ki . k0 ≤ k2,0 + k2,1 + 3 3

(1.8.15)

i≥3

However, if actually ²i > 0 for some ²i then we actually have X 2i 4 k0 ≤ k2,0 + k2,1 + ki − 1. 3 3

(1.8.16)

i≥3

We show that equations 1.8.14 and 1.8.16 are incompatible and that equations 1.8.14 and 1.8.15 together imply that the arrangement is saturated. By the Csima-Sawyer Lemma (Lemma 14) our hypothetical arrangement can have at most 4 (2, 0) lines. It follows that the difference between the first term on the right-hand

80

side of 1.8.16, 43 k2,0 , and the first term on the right hand side of 1.8.14, 54 k2,0 is at most 31 . P P Since i≥3 (i − 1)ki is term for term greater than or equal to i≥3 2i 3 ki it follows that the right hand side of 1.8.16 is less than the right hand side of 1.8.14 and so the two inequalities cannot simultaneously hold. Hence ²i = 0, ∀i and in particular there are no lines with 4 attachments. Now consider the two equations 1.8.14 and 1.8.15. If k2,0 = 4 then all ordinary points would lie on (2, 0) lines and by the (2, 0) Saturation Lemma (Lemma 12) the arrangement would be saturated. Hence assume that k2,0 = 3 - the only other possibility by Lemma 22. Equations 1.8.14 and 1.8.15 become X 15 + k2,1 + (i − 1)ki , 4 i≥3 X 2i ki . ≤ 4 + k2,1 + 3

k0 ≥ k0

i≥3

But since all terms except the

15 4

in the first equation are integers, the first equation is

equivalent to k0 ≥ 4 + k2,1 +

X (i − 1)ki

(1.8.17)

i≥3

and we therefore necessarily have 4 k0 = k2,0 + k2,1 + 2k3 , ki = 0 ∀i ≥ 4. 3

(1.8.18)

But as in the proof of Lemma 17, this implies that the number of ordinary points is 1 3 k2,0 + k2,1 + k1 + k3 2 2

(1.8.19)

and the number of lines times points attached is 4 3k0 + 2k1 + k2,1 = 3( k2,0 + k2,1 + 2k3 ) + 2k1 + k2,1 = 4k2,0 + 4k2,1 + 2k1 + 6k3 (1.8.20) 3 or 4 times the number of ordinary points, thus establishing saturation. Since we have two 5-crossings, equation 1.1.4 says that we have at most 1 (4+)-gon and by Lemma 16, since all ordinary points are saturated, it follows that each ordinary point is adjacent to a (4+)-crossing. We next establish that there is a 4-crossing. We have at least 3 (2, 0) lines and no two of them can pass through both 5-crossings. Moreover, the ordinary points on a (2, 0)

81

line can only be adjacent to a (4+)-crossing along the (2, 0) line by the Top 3-Crossing Lemma (Lemma 13). We cannot have two (2, 0) lines with their ordinary points each surrounding the same 5-crossing since, as seen in Figure 1.66, by saturation, this would

Figure 1.66: Two (2, 0) lines with their ordinary points surrounding a 5-crossing. The line n necessarily has 4 attachments. imply the existence a line with 4 attachments (the line n in the Figure). Hence of the three known (2, 0) lines, one must have its ordinary points surrounding one of the 5-crossings, while another has its ordinary points surrounding the other 5-crossing, while the third has one of its ordinary points adjacent to one of the 5-crossings and its other ordinary point adjacent to the other 5-crossing. We thus have two 5-crossings as picture in Figure 1.67. Now we know that all the lines passing through these 5-crossings contain an even number of

Figure 1.67: A 5-crossing together with two (2, 0) lines, one with its ordinary points surrounding the 5-crossing and the other with only one of its ordinary points adjacent to the 5-crossing. ordinary points. Moreover, only one of the lines, other than the (2, 0) lines m, n have fewer than 2 attachments. Hence, by virtue of the fact that there is no epsilon excess (²i = 0 ∀i), we conclude that there are at most two additional lines with an even number > 0 ordinary points, and in fact that these lines, if they exist are (2, 1) lines. With the three (2, 0) lines and the as many as two (2, 1) lines, we have multi-counted for just 10 of the 16 ordinary points. Hence there must be some lines with an odd number of ordinary points, and hence, by parity, a 4-crossing. It follows that the arrangement must have all triangles. However, filling out Figure 1.67 with known edges we obtain Figure 1.68. By the Top 3-Crossing Lemma (Lemma 13) both u and v are 3-crossings and since the arrangement

82

Figure 1.68: Filling Figure 1.67 out a bit with known edges due to saturation. The edges [w, u] and [v, x] must actually be part of the same line. consists entirely of triangles, the edges [w, u] and [v, x] actually are part of the same line, call it l. But now l will have to cross k in two places to form all triangles unless one of w, x is not a 3-crossing or the one crossing point of l and k is the only remaining crossing of k. But counting k’s existing crossings this would mean that l ∩ k is actually a 9-crossing which is impossible since we know that t9 = 0. We are left with the case that either w or x is not a 3-crossing. We have, however, declared that n is the line with two 5-crossings, so w is not a 5-crossing and it cannot be a 4-crossing by parity. Hence it must be that x is actually a 5-crossing. But if x is a 5-crossing then it has an ordinary point to the “right” of it along n as well, and so the ordinary points on any of the three (2, 0) lines surround one of the 5-crossings. It follows that there are two around one and the same 5-crossing, which would imply that there is a line with 4 attachments, which there cannot be, a contradiction, proving the Theorem.

1.9 1.9.1

Additional useful results n/2 miscellany

I collect here results that seem useful for making progress towards the n/2 conjecture. In order not to let the length and time duration of this dissertation extend to infinity, I have decided to leave them as loose ends. Lemma 35. In an arrangement without epsilon excess a 4-crossing u cannot have two (2, 0) lines l, k, each with ordinary points adjacent to u. Proof. By virtue of the Top 3-Crossing Lemma (Lemma 13) the lines l, k, 4-crossing u and respective ordinary points p, q must be configured as in Figure 1.69. (Refer back to Case 2

83

in the proof of the Strong Charging Lemma, Theorem 24.) Consider the closest vertex, w,

Figure 1.69: A 4-crossing u with ordinary points p and q adjacent to u along (2, 0) lines l and k respectively. w is the closest vertex to u along h in the direction opposite the ordinary points p, q. to u along the line h in the direction opposite the ordinary points. Clearly w 6= v since there are additional vertices on h besides u and v (e.g. m ∩ h). An additional line, j, through w intersects i at a vertex z. Apply the Three Clause Lemma to the two triangles 4(u, w, z) which share the finite edge [u, w]7 to obtain two additional ordinary or attached points for the line i in addition to the two already drawn - giving the arrangement epsilon excess. In fact the proof of the above lemma can be used to give a generalization of the No Three Closest Attachment Principle. In order to state this generalization most succinctly, we introduce the following definition. Definition 10. Suppose we have three lines h, k and l all passing through a common vertex u. Moreover, let p be a closest vertex to u along l and let q be a closest vertex to u along k. See Figure 1.70. In the projective plane lines do not actually have two sides, however, say that p and q are locally on the same side of h if when we slide h in parallel until we reach the first of p or q (say p), then we slide h through a portion of that segment [u, q] on k which is free of additional vertices. Lemma 36 (Same Side Lemma). In an arrangement without epsilon excess, given a crossing u, two lines l, k cannot pass through u with respective ordinary points p, q adjacent 7

Finite as per the rendering in Figure 1.69.

84

to u if there is a third line h through u which counts both p and q as attachments locally on the same side of h. Note that the lines l and k need not be (2, 0) lines. Figure 1.70 gives an example of a sub-arrangement of 5 lines with 2 ordinary points which is a prohibited by the lemma.

Figure 1.70: A sub-arrangement prohibited by Same Side Lemma (Lemma 36). l and k are (2, 0) lines so p and q are attached to h which also passes through u. Moreover, p and q are locally attached on the same side of h. Proof. The proof is essentially the same as the proof of Lemma 35. In Lemma 35 we picked the closest vertex to u along the line h in the opposite direction form p and q. In the Same Side Lemma we can pick any line between l and k in place of h and do the same. The No Three Closest Attachment Principle is an immediate consequence of the Same Side Lemma since if a line l which passes through a crossing u, has three attachments, each of which are adjacent to u, then two of these must be locally on the same side of l. Lemma 37. In an arrangement without epsilon excess a (2, 0) line cannot have its two ordinary points surrounding a (5+)-crossing. Proof. We consider the case of a 5-crossing. See Figure 1.71. The argument for a (6+)-

Figure 1.71: A (2, 0) line l with a hypothetical 5-crossing u sandwiched between ordinary points p and q. x is the closest vertex to u on k in the direction “upward” from l and likewise for y and h. For a (5 + j)-crossing place the additional j lines between k and h and apply the same argument as in the text. crossing is essentially the same. In the Figure x and y are the vertices closest to u along k

85

and h in the “above” l direction. We contend that there is some line through x or y that does not pass through either u, s or t (for example, the line g in the drawing). Suppose x is ordinary. If the additional line through x passes through s, then x is attached to l(s, w) and we can apply the Same Side Lemma (Lemma 36) to l(s, w) with respect to x and q to obtain epsilon excess. If the second line through x passes through t then l(x, t) must pass through y, with y being ordinary and the Same Side Lemma applied to l(z, t) and p, y establishes epsilon excess. Hence we can assume that neither x nor y is ordinary and so at least one of the lines through these points does not pass through s or t as claimed. Let us call the line not passing through these top crossings g, as in the Figure, and consider the triangle T . Since any line passing into T through both the left and right edges would pass through the interior of the edge [x, u] or into one of the triangular cells about p, all lines through the interior of T pass through the bottom edge. Moreover, any line not passing through the interior of T but passing through the apex of T opposite the edge along l, must similarly pass through the interior of the edge [x, y] or into a triangular cell about p. The Three Clause Lemma (Lemma 5) thus implies that l, and hence the arrangement, has epsilon excess. The supposed arrangement with a (5+)-gon T is thus not possible and the Lemma is established.

1.9.2

Saturation versus epsilon excess

It would be nice to better clarify the relationship between saturation and epsilon-excess. Epsilon-excess plus what additional assumptions imply saturation, and correspondingly, saturation plus what additional assumptions imply no epsilon-excess? The B¨or¨oczky examples of n lines and

n 2

ordinary points (Figure 1.23) are all saturated.

In the examples with n ≡ 2 (mod 4) all lines are of type (1, 2) and so these examples have no epsilon-excess. In case n ≡ 0 (mod 4) there are lines of type (0, 4) so these arrangements have epsilon excess. Note that the non-B¨or¨oczky arrangement of 12 lines and 6 ordinary points (Figure 1.24) has no epsilon excess - in fact it has 2 lines of type (2, 0), 8 lines of type (1, 2) and 2 lines of type (0, 3), so that no epsilon excess does not imply saturation (or, for example, saturation of ordinary points on lines of type (1, 2)). There are several stronger notions of “no epsilon excess” which are possible. One can demand that all lines with two ordinary points have no attachments, or that there be no lines with four or more ordinary points. Both of these conditions are met in hypothetical arrangements of

6n 13

ordinary points by

Lemma 18. Moreover, we can extend the notion of attachment to include what Hansen

86

refers to in his thesis [41] as “semi-attached” ordinary points. If a line l contributes an edge to a cell with k edges, and j of the k − 2 vertices of the k-cell, which are not on l, are ordinary, we can attribute

j k−2

units of attachment to l due to these ordinary points. Then,

in the non-B¨or¨oczky example of Figure 1.24 two of the (1, 2) lines would pick up half an attached point from each of the two quadrangular cells and be reclassified as type (1, 3). If we use this new definition of attachments and types in the (−, −) sense, then a stricter still definition of “no epsilon excess” emerges - e.g. as one in which lines with no ordinary points have 2 attached points, those with one ordinary point have 2 attachments, and those with 2 or more ordinary points have no attachments, and attachments are taken in the stricter sense. It is not difficult to see that this stricter definition of “no epsilon excess” implies saturation. Indeed, if an ordinary point p is not saturated then p is the vertex of some k-cell for some k > 3 and p fractionally contributes attachments in excess of the (0, 3+)-Lemma and (1, 2+)-Lemma guarantees on each of the k − 2 lines contributing edges of the k-cell other than the two lines through p. Suppose we allow for fractional attachments and say that an arrangement has Υ units of fractional epsilon excess if amongst those lines of type (xi , yi ) with xi + yi > 2 we P put Υ = i xi + yi − 3. Then, the following becomes rather self-evident: Lemma 38. An arrangement A is Sylvester critical iff A has more lines of type (2, 0) than fractional epsilon excess, i.e. iff k2,0 > Υ. Proof. In the case of fractional epsilon excess an ordinary point always has a total of four lines counting it as an attachment (some lines perhaps only counting it fractionally). Thus, 6m = 2k2,0 + 3(n − k2,0 ) + Υ

(1.9.1)

so 6m ≥ 3n iff Υ ≥ k2,0 which is an equivalent statement to that of the Lemma.

1.9.3

Ordinary points in polygons

Melchior’s argument (Proof #2 from section 1.1) showed that the Sylvester-Gallai Theorem is a consequence of Euler’s formula regarding the facial structure of line or pseudoline arrangements in the projective plane. Euler’s formula also holds within any polygon determined by the lines of an arrangement. However, at least in its most obvious form, there is no Sylvester-Gallai Theorem for the lines intersecting in generic polygons. In this section we formulate and prove variants of the Sylvester-Gallai Theorem suitable for lines intersecting in simple polygons with up to five edges.

87

Figure 1.72: We consider the finite triangle 4(p, q, r) in the statement of Theorem 39. Since line h does not intersect the interior of 4(p, q, r) it is not considered. Amongst the remaining lines, the Theorem concludes that there are at least 2 ordinary points in 4(p, q, r) - indicated here with hollow circles. Theorem 39. Consider any triangle formed by the lines of an arrangement and the vertices formed inside or on the boundary of the triangle by the defining lines of the triangle together with the lines which intersect the interior of the triangle. Amongst these vertices and lines only, there must be at least 2 ordinary points. To better understand the statement of the Theorem, see Figure 1.72. Proof. Let T be an arbitrary triangle formed by the lines of an arrangement. In any polygon Euler’s relation holds and says that V −E+F =1

(1.9.2)

where V, E and F , respectively, refer to the vertices, edges and faces of the arrangement lying inside or on the boundary of T . We would like to use 1.9.2 to tell us something about the familiar expression YT =

X

(3 − j)tj +

j≥2

X (3 − k)pk

(1.9.3)

k≥3

where tj denotes the number of j-crossings in T and pk denotes the number of k-gons in T . We still have

X j≥2

tj = V,

X k≥3

pk = F.

(1.9.4)

88

Figure 1.73: The finite triangle T = 4(p, q, r) with external faces F1 , ..., F7 and external edges E1 , ..., E8 as described in the proof of Theorem 39. However edges along the border of T are not shared by two faces so we don’t have X

kpk = 2E

k≥3

and similarly, since edges don’t extend outward from the borders of T we don’t have X

jtj = E.

j≥2

However, let us declare that we have a single “external face” wherever there is an edge lying along the border of T and similarly say that every line which is either a defining line of T or intersects the interior of T has a single “external edge” on the outside of T . See Figure 1.73. If we let

then we have

Fext = number of External Faces

(1.9.5)

Eext = number of External Edges

(1.9.6)

X

kpk + Fext = 2E,

(1.9.7)

k≥3

X j≥2

jtj = E + Eext .

(1.9.8)

89

Let us also define Excess Crossings =

X (j − 3)tj . j≥4

We claim that Eext ≤ Fext + Excess Crossings + 1.

(1.9.9)

To see this, start with a triangle S consisting just of its three defining lines. We have Eext = 3, Fext = 3, Excess Crossings = 0, t2 = 3. Adding a line to S always increments Eext by 1. When we place a line creating a crossing along the border of S (i.e. a 2-crossing) we increment Fext by 1. Thus the only way we can place a line and increase the number of 2-crossings is by either (1) creating 2 new 2-crossings, or (2) creating 1 new 2-crossing and adding an excess crossing. In both cases we add 2 to the right hand side of 1.9.9 while adding just 1 to the left hand side. Placing a line to add just 1 to the left hand side of 1.9.9 but nothing to the right hand side uses up 2 2-crossings. Since we start with just 3 2-crossings and any time we increase the number of 2-crossings we add a net of 1 to the right hand side of 1.9.9 the claim follows. Substituting 1.9.2, 1.9.4, 1.9.7, 1.9.8 and 1.9.9 into 1.9.3 gives YT

= 3V − (E + Eext ) + 3F − (2E − Fext ) = 3 + Fext − Eext ≥ 2 − Excess Crossings

i.e.

X X (3 − j)tj + (3 − k)pk ≥ 2 − Excess Crossings j≥2

(1.9.10)

k≥3

so that t2 +

X (3 − k)pk ≥ 2

(1.9.11)

k≥3

whence t2 ≥ 2. The same sort of argument can be used to prove the following, which is ostensibly not

90

Figure 1.74: Tight examples for Theorem 41. a corollary: Theorem 40. If a triangle T as considered in Theorem 39 has only one (3+)-crossing amongst its defining vertices and we consider just the lines of T together with lines intersecting the interior of T . then T must contain an ordinary point in addition to the two at its defining vertices. Proof. By the assumptions of the Theorem, only one of the defining 2-crossings in the original triangle S from the proof of Theorem 39 can be used. Hence the analog of equation 1.9.9 is

so

Eext ≤ Fext + Excess Crossings

(1.9.12)

X X (3 − k)pk ≥ 3 − Excess Crossings (3 − j)tj +

(1.9.13)

j≥2

k≥3

and the result follows. Henceforth we refer to polygons determined by the lines of an arrangement simply as “polygons in arrangements” and the ordinary points contained in these polygons, formed just by the defining lines of the polygons or lines intersecting the polygon’s interior, as “ordinary point in polygons.” Conjecture 4. A triangle in an arrangement without ordinary points on its boundary must have at least three ordinary points in its interior. Theorem 41. A quadrangle or pentagon P in an arrangement must contain at least one ordinary point. See the Figure 1.74 for tight examples. Proof. In the case of quadrangles or pentagons, there is the possibility of non-convexity, so

91

Figure 1.75: A non-convex quadrangle. In order for the relation hold we must declare that the line l have two external edges.

P

j≥2 jtj

= E + Eext to

we must alter our definition of an external edge slightly in order that the relation X

jtj = E + Eext

(1.9.14)

j≥2

can still hold. See Figure 1.75. We say that a line l determines k external edges if the portion of l that lies outside of P forms k + 1 distinct segments (k projective segments). With this definition, relation 1.9.14 is easily seen to hold. Then, given a (not necessarily convex) quadrangle or pentagon, we start off with either 4 or 5 2-crossings, which is enough to lay down two lines without creating additional external polygons or creating any excess crossings, the analog of equation 1.9.9 is

so

Eext ≤ Fext + Excess Crossings + 2

(1.9.15)

X X (3 − k)pk ≥ 1 − Excess Crossings (3 − j)tj +

(1.9.16)

j≥2

k≥3

and the result follows. The conclusion of Theorem 40 does not hold for 4-gons and higher. Similarly, there are 6-gons (and clearly 2n-gons for all n > 3) without any ordinary points. See Figure 1.76. The methods we have developed do not seem powerful enough to decide whether or not there are ordinary points in odd n-gons for n ≥ 7. However, we hazard the following: Conjecture 5. Any odd n-gon for n ≥ 7 must contain at least one ordinary point.

92

Figure 1.76: Examples showing that (a) the conclusion of Theorem 40 does not hold for quadrangles, and (b) that there are hexagons without ordinary points. Questions regarding lower bounds on the number of ordinary points in polygons as a function of the number of lines intersecting the interior of the polygon may also be pursued.

1.10

Counting the number of arrangements and computerassisted searches

Given that we have taken some time to argue about the feasibility of Sylvester-critical arrangements for n not too much beyond 13, it is interesting to pause to consider bounds on the total number of arrangements for a given number n, to get a sense for whether such computations would be possible using a naive computer search. I touch on this issue in just cursory fashion. The real question of course is whether you could go much further than I have, with a sophisticated computer search. I will describe some of the ways in which a sophisticated computer search might be implemented at the end of this section. In 1980 Goodman [37] came up with a nice way of compactly visualizing pseudoline arrangements using so-called wiring diagrams. A sample wiring diagram, taken from [32], is given in Figure 1.77. We consider here just simple pseudoline arrangements and their wiring diagrams. By a simple arrangement we mean one where all intersection points are ordinary. Every simple pseudoline arrangement is isomorphic to a wiring diagram W such that we start with n horizontal wires, and at distinct x-coordinates, precisely two wires cross. In the course of the traversal from the left side of the diagram to right side, ever pair of wires crosses exactly once, and at the end of the diagram on the right, when all interchanges are made, the order of the wires is exactly reversed from what it was at the start. By convention,

93

Figure 1.77: A wiring diagram for a 5 line pseudoline arrangement.

Figure 1.78: Two equivalent wiring diagrams. We can obtain an isomorphic arrangement of pseudolines from a given wiring diagram by interchanging the order of a pair of crossings as long as they do not share a pseudoline in common. we initially label the wires from 1 to n with 1 at the bottom. When we get to the extreme right, the wires are numbered from bottom to top in the order n, n − 1, ..., 1. Counting the initial ordering and each of the successive orderings following an interchange, we get a ¡ ¢ total of n2 + 1 distinct permutations or orderings of the wires. Following Goodman and ¡ ¢ Pollack [38, 39] call the sequence of n2 + 1 permutations a simple allowable sequence. For non-simple arrangements, consecutive permutations are allowed to differ by the reversal of an arbitrarily long sequence. One does not allow the the simultaneous flipping of several non-intersecting contiguous blocks. It follows by the so-called Sweeping Lemma (Lemma 6.1 of [33]) that an arrangement of pseudolines can be transformed into a wiring diagram. However, many allowable sequences can correspond to the same pseudoline arrangement. See Figure 1.78. The number of simple allowable sequences, a number usually denoted by An , has been computed exactly by Stanley [74]. Bn is typically used to denote the number of simple pseudoline arrangements. Using the sharp version of the Zone Theorem obtained by Bern et al. [5] (i.e. with constant 5.5), Knuth [50] showed that 2

n2 − 5n 6 2

2

≤ Bn ≤ 20.7194n .

2

Subsequently Felsner [32] improved the upper bound to Bn ≤ 20.6974n using the notion of a replace matrix. A replace matrix is a binary n x n matrix M with the two properties: n X

mij = n − i (i = 1, ..., n),

j=1

mij ≥ mji (i < j).

94

Felsner showed that every simple arrangement of pseudolines corresponds to a unique n x 2

n replace matrix and then used a probabilistic argument to get his 20.6974n upper bound. Goodman and Pollack [39] have shown that the number of simple arrangements of lines is (for large enough n) much smaller, with an upper bound of O(2n log n ). Felsner [32] computed B10 = 18, 410, 581, 880 exactly using a recursive program. I do not know of any bounds of this kind for the number of general arrangements (i.e. not just simple arrangements). We thus see that it is not practical to do a completely naive search/enumeration of pseudolines arrangements in the hope of finding a Sylvester-critical arrangement, or satisfying ourselves that there is none such for given n. In his thesis Chen [11] used a computer enumeration to establish that m(15) = 9. In what follows we wish merely to ascertain whether or not there is a Sylvester-critical arrangement of n pseudolines for given, relatively small, n. Chen’s enumeration program generates allowable sequences, noting that ordinary points occur between successive permutations iff a pair of adjacent numbers are interchanged. He starts with the identity permutation (1 2 · · · n) and then systematically generates all allowable sequences starting with this permutation, noting that one can reverse a contiguous block in a subsequent permutation iff the numbers in the block are in increasing order. For the m(15) computation, the enumeration can stop once it finds that an allowable sequence has 9 ordinary points, since the B¨or¨ oczky example for n ≡ 3 (mod 4) already gives this bound. To remove degeneracies Chen also makes use of the observation from Figure 1.78, generalized to arbitrary arrangements, that if we have two disjoint blocks B1 = [i · · · j], B2 = [i0 · · · j 0 ] each of which is reversed at some point in an allowable sequence, then we can always assume that i < i0 . Finally, he uses the fact that there must be an ordinary point to begin the enumeration with the assumption that the second permutation is (2 1 3 4 5 · · · n). To Chen’s observations we can also add the following: Although, if we are given an allowable sequence it is not easy to detect (4+)-gons, we can trivially detect (4+)-crossings, and so make use of the Yproj Equation (1.1.8) in the form X (i − 3)ti ≤ 3 + t2 .

(1.10.1)

i≥4

Also, as observed in the section on ordinary point bounds for small n (Section 1.8), µ ¶ Xµ ¶ n i = ti 2 2 i≥2

(1.10.2)

95

must hold mod 3, which yields µ ¶ n = t2 + 10t5 + 28t8 + · · · (mod 3). 2

(1.10.3)

Now the coefficients of each of the terms on the right are of the form µ ¶ 2 + 3n (2 + 3n)(1 + 3n) n(n + 1) = =1+9 ≡ 1 (mod 3) 2 2 2

(1.10.4)

Hence we can rewrite 1.10.3 as µ ¶ X n = t2+3i (mod 3). 2

(1.10.5)

i≥0

and utilize this fact as well. If we are just interested in ruling out Sylvester-critical arrangements for a given n, rather than in computing m(n), we can also utilize the fact that Sylvester-critical arrangements must contain (at least three) lines of type (2, 0) (Theorem 22). Thus if we declare line 1 (henceforth l1 ) to be of type (2, 0), we avoid generating allowable sequences where l1 contains other than two ordinary points, and moreover, we stop generating allowable sequences down a recursive branch of the code where l1 is discovered to have an attached point. To show that such a strategy is feasible, we need show that it is not prohibitively expensive to detect attached points. Any two projectively adjacent vertices have the potential to give rise to an attachment. Let v1 , v2 be two such adjacent vertices along l1 . Let the horizontal sweep of the arrangement described by the allowable sequence determine the up/down, above/below directions in the arrangement with the assumption that the arrangement has no horizontal or vertical lines. Suppose that {lik }N k=1 are the lines crossing at v1 in addition to l1 and that {ljl }M l=1 are the lines crossing at v2 in addition to l1 . Using Chen’s enumeration conventions, the increasing subsequences (1 i1 · · · iN ), (1 j1 · · · jM ) are reversed respectively at v1 and v2 . Now if x = liN ∩ lj1 and y = li1 ∩ ljM , then l1 has an attached point via v1 and v2 iff x is ordinary and the triangle 4(v1 , v2 , x) lying projectively above l is a cell, or y is ordinary and the triangle 4(v1 , v2 , y) lying projectively below l1 is a cell. See Figure 1.79. Now y cannot lie horizontally at or to the left of v2 because then ljM would start out below l1 to the left of the first vertex (and we have assumed there are no vertical lines). Analogously x cannot lie horizontally at or to the right of v1 because then liN would start out below l1 . Thus, if y is ordinary it will be attached to l1 iff there is no line which intersects ljM horizontally between v2 and y. Analogously, if x is ordinary it is attached to

96

Figure 1.79: Two vertices v1 , v2 through the line l1 with lines {lik }N k=1 crossing at v1 and lines {ljl }M crossing at v . Let x = l ∩ l and y = l ∩ l . l has an attached point via 2 iN j1 i1 jM 1 l=1 v1 and v2 iff x is ordinary and the triangle 4(v1 , v2 , x) lying projectively above l is a cell, or y is ordinary and the triangle 4(v1 , v2 , y) lying projectively below l1 is a cell. l1 iff there is no line which intersects liN horizontally between x and v1 . The intersection x = liN ∩ lj1 occurs when there is a transposition involving iN and j1 . x is ordinary iff the transposition involves just two indices. If x is ordinary, it is attached to l iff there is no transposition involving iN between the transposition that occurs at x and that which occurs at v1 . Equivalently, if x is ordinary it is attached to l iff iN has the same set of indices above it (below it) in the permutations after the crossing at x and before the crossing at v1 . Analogously, if y = li1 ∩ ljM is ordinary, it is attached to l iff there is no transposition involving jM horizontally between v2 and y (or, equivalently, jM has the same set of indices above it after the crossing at v2 and before the crossing at y). We can either start with the assumption that the first two permutations are (1 2 3 · · · n), (2 1 3 4 5 · · · n) and l1 is (2, 0), or instead assume that both l1 , l2 are (2, 0) and that the first permutation involves the interchange of l1 , l2 and at least one other line (by the CsimaSawyer Lemma - Lemma 14 l1 ∩ l2 cannot be ordinary). Which set of assumptions, when coupled with equations 1.10.1 and 1.10.5 is more constraining would need to be determined from practice. I leave a more careful treatment of these algorithmic enumeration issues for future work.

Chapter 2

The Affine or Sharp Dual of Sylvester’s Problem 2.1

Introduction

In 2004 I began considering the following variant of the Sylvester problem: In an arrangement of n lines in the affine plane, not all of which are parallel and not all of which pass through a common point, must there be a (finite) ordinary point? In my initial papers on this subject I chose the word “Euclidean” instead of the word “affine” in the description of this problem. Both terms imply that we do not consider points at infinity, and moreover, it is no longer the case that any two lines intersect since we may have parallel lines. The word “Euclidean” implies a choice of underlying metric (the “Euclidean” or square root of the sum of the squares metric) and since none of the proof methods in this chapter actually require the use of a metric, I now prefer the term “affine” which is metric-agnostic. In Chapter 1 I mentioned that Sylvester’s problem had a dual in the projective plane, but in fact it also has a dual in the affine plane. The traditional point-connecting line version of Sylvester’s problem asks: Given a finite set of not all collinear points in the (affine) plane, must there be a connecting line between pairs of points, which contains precisely two of the points? However we can rotate such a point set so that none of the points share a common x-coordinate and then apply the incidence-preserving duality transformation (u, v) 7→ {(x, y) : y = ux − v}1 to get a set of lines, not all of which pass through a common point and no two of which are parallel. The lines have an ordinary point of intersection in the affine plane iff the points have an ordinary connecting line. Thus an equivalent 1

See [21] for details on this transformation.

97

98

formulation of Sylvester’s problem in the affine plane is: Given an arrangement of lines, not all of which pass through a common point and no two of which are parallel, must there be an ordinary point? The question of whether such an ordinary point of intersection also exists if one mandates that not all the lines pass through a common point and not all the lines are parallel has weaker conditions then the original problem, and the positive resolution is stronger. Hence we refer the new problem as the Sharp Dual of Sylvester’s Problem. In Chapter 3 I point out that a positive answer to the sharp dual problem does not follow from Euler’s formula. However, we can adapt our proof of the more classical Sylvester Theorem (Proof #4 from section 1.1) to obtain a positive answer in the sharp dual case. A careful double counting argument using the Csima-Sawyer Theorem (Theorem 15) enables us to show that in fact there must be at least

5n+6 39

ordinary points. Finer observations enable

one to push the bound successively further. Although it would be possible to present just the final bound, I have chosen to present the succession of ideas, and the impact of each on the bound.

2.2

Establishing the Sharp Dual Theorem

Let us make the following definition which will simplify the statement of our results: Definition 11. An arrangement of n lines in the affine plane is said to be trivial if either all lines are parallel or all lines pass through a common point. An arrangement is said to be non-trivial otherwise. Theorem 42 (Sharp Dual of Sylvester’s Theorem). In a non-trivial arrangement of n lines in the affine plane there must be at least one (finite) ordinary point Proof. Suppose to the contrary we have a non-trivial arrangement without ordinary points. Consider again proof #4 from section 1.9.3 of the Classical Sylvester Theorem. As long as our arrangement has three lines intersecting at finite vertices u, v and w with one of the vertices having a third line intersecting the interior of the finite triangle 4(u, v, w), the rest of the argument carries over. If the arrangement does not have three lines intersecting at three finite vertices then necessarily all lines in the arrangement are parallel to one or the other of two lines, in which case all vertices are ordinary. Hence we may assume we have three lines l, k and h, no two parallel to one another, as in Figure 2.1, forming the finite triangle 4(u, v, w). There is a third line n through u, which if not parallel to h would induce a finite triangle amongst the vertices of {m, k, h} (or {l, m, h}) with the line l (or k) cutting through u into the

99

Figure 2.1: Three lines l, k and h, no two parallel to one another, with intersection points u = l ∩ k, v = h ∩ k, w = l ∩ h.

Figure 2.2: The three lines l, k, h of Figure 2.1 together with lines through u parallel to h, through v parallel to l and through w parallel to k. interior of the triangle. Thus n must be parallel to h and analogously the third line through v must be parallel to l and the third line through w must be parallel to k, yielding the nested finite triangles in Figure 2.2. But now applying the same reasoning to the vertices u0 , v 0 , w0 we necessarily have a line through u0 parallel to l(v 0 , w0 ), a line through v 0 parallel to l(u0 , w0 ) and a line through w0 parallel to l(u0 , v 0 ). Continuing in this fashion we see that the arrangement would necessarily have an infinite number of lines, contrary to assumption, and proving the Theorem.

2.3

The (5n + 6)/39 Result

Theorem 43. In a non-trivial arrangement of n lines in the affine plane, there must be at least (5n + 6)/39 (finite) ordinary points. Proof. We consider the problem embedded in the real projective plane, where the CsimaSawyer Theorem (Theorem 15) says that there must be at least

6n 13

ordinary points except

if n = 7 and the arrangement is the Kelly-Moser arrangement (see Figure 1.6). We handle this n = 7 case at the very end of the argument. 5n+6 n 2 If our result were false then more than d 6n 13 − 39 e = d 3 − 13 e of these ordinary points

would have to lie on the line at infinity. In other words there would have to be at least 2 d n3 − 13 e pairs of parallel lines. To this arrangement add the line at infinity. This “kills off”

100

the at least d n3 −

2 13 e

ordinary points and creates at most bn −

2n 3

+

4 13 c

= b n3 +

4 13 c

new

ordinary points. By Csima-Sawyer applied to the new arrangement (as long as n 6= 6, a case we cover in the next paragraph) we have at least d 6(n+1) 13 e ordinary points. But then there must have

− been at least d 6(n+1) 13

n 3



4 13 e

= d 5n+6 39 e finite ordinary points earlier, contradicting our

initial assumption. The result is thus proved. If n = 6 the theorem claims that there is at least one finite ordinary point. By the KellyMoser Theorem we know that there are at least three total (projective) ordinary points. If all such points were on the line at infinity the implication would be that we have three pairs of parallel lines. Adding a seventh line at infinity would yield a projective arrangement without ordinary points, a clear impossibility. Finally, let us consider the case n = 7. If the arrangement is not the Kelly-Moser arrangement with some two or three ordinary points at infinity, we would be done by the prior argument. However, looking at Figure 1.6, in the Kelly-Moser arrangement we see that each pair of ordinary points is already on a line of the arrangement; hence for two points of the arrangement to be at infinity, one of the lines in the arrangement would have to be at infinity - in which case we would be in the already handled n = 6 case. The theorem follows.

2.4

Improving the bound to n/6

As in the projective case we will make use of arguments about attached points. We will just consider finite attached points, but allow for the fact that finite points may be attached via infinite triangles, as in the case of the point q back in Figure 1.8. Definition 12. Say that an arrangement has affine type (i, j) if it has i finite ordinary points and j finite attached points. The Four Attachment Lemma (Lemma 2) of course still holds. Our central lemma is the following: Lemma 44 (Affine (0,1+) Lemma). Let A be a non-trivial affine arrangement of n lines. Then if a line l ∈ A does not contain an ordinary point, it must have at least one (finite) ordinary point attached to it. Proof. If all the finite vertices are on a single line, then all but that line must be parallel, and all vertices are ordinary. There is thus no line without finite ordinary points.

101

Figure 2.3: The points p, q and r are all closest vertices to l on one side, but only p and q are closest and most extreme to one side. Otherwise, let l ∈ A be a line without finite ordinary points and let x be the closest vertex to l on one side of l and the most extreme to one side (a side which, for convenience, we shall refer to as the “right” side) if there are several such vertices. See Figure 2.3. To see that this notion is inherently non-metric, note that we can pick x to be any vertex which is “closest” to l in the sense that x ∈ / l and any line k through x contains no vertices which are affinely between x and l. If k and l are parallel, then there should be no third parallel line between k and l and, moreover, there should be no further vertex on k either to the “right” or to the “left” of x. We argue that in this case x must be ordinary. If x is indeed ordinary then the triangle defined by l and the two lines through x must be a cell of the arrangement (possibly infinite if one of those lines is parallel to l) and so x is attached to l. If x is not ordinary, then there are at least three lines through x, let us call them l1 , l2 and l3 , with l3 possibly parallel to l, and l2 intersecting l between l1 and l3 (or to the right of l1 if l3 is parallel). Then the intersection y of l2 and l must be non-ordinary, yet any line through it must intersect l1 or l3 in a point that is closer to l than x, or to the right of x on l3 if l3 is parallel to l, in either case a contradiction. Theorem 45. Let A be a non-trivial affine arrangement of n lines. Then A has at least n/6 (finite) ordinary points. Proof. Let ki denote the number of lines of A containing exactly i finite ordinary points, and suppose that there are fewer than n/6 finite ordinary points in total. Then we have X

iki


i≥0

so k0 >

X

3iki

(2.4.3)

i≥1

X (3i − 1)ki .

(2.4.4)

i≥1

But also there are at most 4 lines counting a given ordinary point as an attachment, so that if 1 + ²0 denotes the average number of finite attached points for lines with no finite ordinary points (Lemma 44), and ²i denotes the average number of finite attached points for lines with i ≥ 1 finite ordinary points, we have, for ²i ≥ 0, ∀i ≥ 0, (1 + ²0 )k0 +

X

² i ki ≤ 2

i≥1

so that k0 ≤

X

X

iki

(2.4.5)

i≥1

2iki .

(2.4.6)

i≥1

But 3i − 1 ≥ 2i for i ≥ 1, so equations (2.4.4) and (2.4.6) cannot simultaneously hold and the theorem follows. Examples showing that the d n6 e bound is tight for n = 6, 7, 8 are given in Figure 2.4.

2.5

Improving the bound to n/4 − O(1)

For future reference, let us refer to the six line arrangement which is the left-most arrangement in Figure 2.4 as the Little B¨ or¨ oczky arrangement, since, projectively speaking, it is the smallest example of those due to B¨or¨ oczky with n lines and

n 2

ordinary points.

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Lemma 46 (Sharp Affine (0,1+) Lemma). Let A be an affine arrangement which is not the Little B¨ or¨ oczky arrangement, then at most three lines of A can have no ordinary point and just one attached point. All other lines without ordinary points must have two or more attached points. Proof. Consider any line l ∈ A with no ordinary points. If l has only one attached point then the vertices of A not contained in l are either all to one side of l or the other - otherwise we could argue as in the proof of the Affine (0, 1+) Lemma (Lemma 44) to conclude that a closest vertex to l on either side is an attached point. Call this property of having all vertices to one side or another of a given line, the All-to-One-Side Property. We will show that if four lines in an arrangement have the All-to-One-Side Property then at least one of the lines must not be a (0, 1) line. Suppose we have just a four line arrangement in which all lines have the All-to-One-Side Property. We proceed with a case analysis to show that the arrangement must be one of those in Figure 2.5.

Figure 2.5: The four four-line examples with each line having the All-to-One-Side Property. We begin the case analysis by observing that there are just four distinct three-line affine arrangements - those depicted in Figure 2.6. To prove that the only four-line arrangements with the All-to-One-Side Property are those in Figure 2.5 it suffices to show that if we add a line to any of the arrangements in Figure 2.6, preserving the All-to-One-Side Property, we necessarily obtain one of the arrangements in Figure 2.5. To this end, consider first the sub-arrangement (a) from Figure 2.6. If the fourth line l4 is not parallel to l1 , l2 or l3 and does not pass through l1 ∩ l2 ∩ l3 then if we label the

104

Figure 2.6: The four distinct affine three line arrangements. Any lines shown which do not intersect in the figure are assumed to be parallel. vertices on l4 in left to right order p1 , p2 , p3 with p2 ∈ li , 1 ≤ i ≤ 3, then p1 lies to one side of li and p3 lies to the other side of li and so li does not have the All to One Side Property. If l4 is parallel to one of the existing lines then we have arrangement (b) in Figure 2.5 and if l4 passes through l1 ∩ l2 ∩ l3 then we have arrangement (a) in Figure 2.5. Next consider sub-arrangement (b) from Figure 2.6. If l4 is not parallel to any of the existing lines and does not pass through l1 ∩ l3 or l2 ∩ l3 then again we label the vertices of l4 in left to right order p1 , p2 , p3 and arrive at a contradiction as in the previous case. On the other hand if l4 passes through l1 ∩ l3 or l2 ∩ l3 we have arrangement (b) in Figure 2.5 while if l4 kl3 we have arrangement (c) in Figure 2.5. If, however, l4 is parallel to l1 and l2 , then one of the lines l1 , l2 , l4 lies vertically between the other two, and so also contains vertices on both sides so does not have the All to One Side Property. Thus if we have three lines forming either sub-arrangement (a) or (b) from Figure 2.6 then the fourth line necessarily yields an arrangement in Figure 2.5. Next consider sub-arrangement (c) from Figure 2.6. Consider the addition of l4 . l4 must intersect one of the lines l1 , l2 or l3 - suppose it is l1 . It is easy to see that l4 can intersect l1 only at l1 ∩ l3 or l1 ∩ l2 . If l4 passes through l1 ∩ l3 say, then if l4 is not parallel to l2 then it intersects l2 at a third vertex. Whichever line intersects l2 at the middle vertex will not have the All to One Side Property. It follows that adding a fourth line l4 to (c) in Figure 2.6 requires that l4 pass through an existing vertex and be parallel to the one line not passing through that vertex - thus yielding sub-arrangement (b) in Figure 2.5.

105

Finally consider sub-arrangement (d) from Figure 2.6. If any one of the three lines contains additional vertices, then they all do, and so l2 will not have the All to One Side Property. Thus adding a fourth line to sub-arrangement (d) in Figure 2.6 can only lead to arrangement (d) in Figure 2.5. Thus in all cases if we have four lines with the All to One Side Property, the four lines must form one of the arrangements in Figure 2.5. We examine next whether an arrangement can have any of these sets of lines as a sub-arrangement, with the lines maintaining their All-to-One-Side Property and also each of the four lines being of type (1, 0). It is easy to see that adding an additional line not through the common intersection point in arrangement (a) forces one of the pre-existing lines to lose the All-to-One-Side Property, and analogously for arrangement (d), if we add a line not parallel to the existing lines. Thus only cases (b) and (c) are interesting. First consider arrangement (b) from Figure 2.5. Note that two of the existing lines currently have ordinary points, so there must be additional lines. Label the top horizontal line in the Figure l3 , the bottom horizontal line l4 and the diagonal line ld . The vertex ld ∩l3 cannot be ordinary so consider a third line k passing through this point. k must intersect l4 . However, if k ∩ l4 is to the left of ld ∩ l4 then the vertical line does not have the All to One Side Property while if k ∩ l4 is to the right of ld ∩ l4 then ld does not have the All to One Side Property. Hence arrangement (b) in Figure 2.5 is not possible. Finally consider arrangement (c) from Figure 2.5. Lines can only be added to this arrangement, preserving the All to One Side Property for the first four lines, by adding lines parallel to and between a pair of existing parallel lines, or connecting a pair of diagonally opposite vertices. Adding additional parallel lines, however, creates ordinary points - so there cannot be any of these. On the other hand, the arrangement as drawn has four ordinary points, so the lines through pairs of diagonally opposite vertices must be added, yielding the six line Little B¨or¨oczky arrangement. The Lemma is established. It is possible that Lemma 46 can be improved - e.g. it may be true that as long as the arrangement is not that the Little B¨or¨ oczky arrangement, then there can be at most two lines of affine type (0, 1). However, in order to prove such a result one would have to utilize some properties of the straightness of lines, as the example in Figure 2.7 shows. Theorem 47. Let A be a non-trivial affine arrangement of n lines. Then A has at least n 3 4−8

finite ordinary points, unless n = 6 in which case A may have just one ordinary point.

Proof. Let p be the fraction of lines with one ordinary point and no attachments and m the number of (finite) ordinary points. Then, since each (1, 0) line contributes a half an

106

Figure 2.7: An arrangement of 10 affine pseudolines, three of which, l1 , l2 and l3 are of type (0, 1). In this context, an arrangement of affine pseudolines is a family of curves, each of which extends to infinity with a well-defined limiting slope which is the same from each end, and such that any two lines with different slopes cross, while pseudolines with equal slopes do not intersect. ordinary point, we have m≥

pn , 2

(2.5.1)

and, under the assumption that n > 6, counting either ordinary or attached point-line associations, we have 6m ≥ 3 + pn + 2(n − (pn + 3)), i.e., m≥

(2 − p)n 1 − . 6 2

(2.5.2)

(2.5.3)

From 2.5.1, −pn ≥ −2m so substituting that into 2.5.3 gives 2n − 2m − 3 , 6 8m ≥ 2n − 3, m ≥

(2.5.4) (2.5.5)

and therefore m ≥

n 3 − . 4 8

(2.5.6)

107

2.6

The final push to 2n/7 − O(1)

We can push the n/4 result even further by virtue of the following striking analog of the Csima-Sawyer Lemma (Lemma 14). Lemma 48. Two lines of affine type (1, 0) cannot intersect in their ordinary point unless the arrangement consists just of two intersecting lines or is the six line arrangement in Figure 2.4. Proof. The proof is similar to the Csima-Sawyer Lemma but with a couple of extra nuances. Suppose the two (1, 0) lines l, k intersect at their ordinary point p. Since there are more than just these two lines, there is another vertex, q, on one of the lines, say on k. q is not ordinary so there are at least two lines passing through q in addition to k. By the Three Clause Lemma (Lemma 5) the two lines must intersect l at the closest (finite) vertices to p on either side. Applying the same reasoning to these two closest vertices to p on l, we see that p is saturated in finite triangular cells and also that each vertex around p is a 3-crossing. Now consider the projective cells surrounding the triangular cells about p. Ordering these cells clockwise, either there are two opposite pairs of projective triangular cells or two consecutive projective (4+)-gons. If there are opposite pairs of projective triangular cells, then the triangles meet at a possibly infinite vertex z. Any additional line would have to pass through z (in the projective sense if z is infinite) and so create a finite ordinary point on k or l which is impossible. Hence we have just a six line arrangement. If z were finite it would be attached to k, l which is impossible since both are (1, 0) lines. Hence z must be at infinity. If l1 , l2 are the two lines meeting at z and l3 , l4 are the other line contributing edges to the triangular cells about p, then since there are no additional lines, l3 , l4 similarly cannot intersect at a finite point and so we have the Little B¨or¨ oczky arrangement. On the other hand, suppose that there are two consecutive projective (4+)-gons surrounding triangular cells about p; see Figure 2.8. There are two cases: either (1) there is a line crossing into and forming an edge of Q in the northeast quadrant, or there is a line crossing into and forming an edge of R in the southeast quadrant or (2) there are no such lines. In case (1) suppose such a crossing line and edge exists for Q. Then if e1 = [b, d], there are two successive edges, going either clockwise around Q with e2 lying on l(a, b) and the line extending e3 intersecting k at a finite point north of b, or going counterclockwise around Q with e2 lying on l(c, d) and the line extending e3 intersecting l at a finite point east of d. The argument is the same in either case, so suppose we have the latter situation and the line extending e3 intersects l at a finite point g as depicted in Figure 2.9. Then,

108

Figure 2.8: A hypothetical arrangement with two lines, l, k of affine type (1, 0) with common ordinary point. We consider the case where two consecutive projective cells surrounding the triangular cells about p, Q and R say, are (4+)-gons. since a line cannot pass through f into Q, we can apply the Three Clause Lemma (Lemma 5) to conclude that l must have a finite ordinary or attached point in [d, g], contrary to assumption. Thus, revert to considering Figure 2.8 and in particular, consideration of case (2) where there are no lines crossing into and creating an edge of Q in the northeast quadrant, and also no lines crossing into and forming an edge of R in the southeast quadrant. Note that a line of affine type (1, 0) must either have an infinite ordinary point or two infinite attached points by the (1, 2+) Lemma (Lemma 4). By our assumption about Q and R, there cannot be a line parallel to l, so l cannot contain an infinite ordinary point. We show that l cannot even have a single infinite attached point. Again by the assumptions on Q and R, there can be no lines in the arrangement intersecting l to the east of d. Thus any pair of parallel lines giving rise to an infinite ordinary point attached to l must intersect l at or to the west of a. However any pair of parallel lines forming an attached point to the north of l must not be cut to the north by either l(a, c) or l(b, d). But unless the parallel lines are parallel to or identical with l(a, c) and l(b, d), they will cut into R, which is impossible. They cannot be parallel with l(a, c) and l(b, d) since then they would not meet at an ordinary point. Further, if the parallel lines were actually equal to l(a, c) and l(b, d) we would have our the Little B¨or¨oczky arrangement - but actually in this case Q and R would not be (4+)-gons. The argument is clearly the same for a pair of parallel lines forming an

109

Figure 2.9: The hypothetical arrangement with two lines, l, k of affine type (1, 0) sharing an ordinary point, two consecutive projective cells surrounding the triangular cells about p, Q and R which are (4+)-gons, and a line extending the third edge (e3 ) in counter-clockwise order around Q from e1 = [b, d] . attached point to the south of l. It follows that in case (2), l cannot have either an infinite ordinary or attached point, and so this case is ruled out and the Lemma is established. Theorem 49. Let A be a non-trivial affine arrangement of n lines. Then A has at least 2n−3 7

(finite) ordinary points, unless n = 6 in which case A may have just one ordinary

point. Proof. Let p be the fraction of lines with one ordinary point and no attachments and m the number of (finite) ordinary points. Then, by virtue of Lemma 48, we have m ≥ pn,

(2.6.1)

and, under the assumption that n 6= 6, counting both ordinary point-line and attached point-line associations, and applying the Four Attachment Lemma (Lemma 2) together with the Sharp Affine (0,1+) Lemma (Lemma 46) we have 6m ≥ 3 + pn + 2(n − (pn + 3)),

(2.6.2)

where 3 is the maximum number of (0, 1) lines, pn is the number of (1, 0) lines and n−(pn+3)

110

is the number of lines with at least a total of two ordinary plus attached points. The right hand side of (2.6.2) is equal to 2n − pn − 3, so that together with −pn ≥ −m from (2.6.1), we obtain m≥

2n − 3 . 7

In projective arrangements, since the choice of which is the line at infinity is completely arbitrary, we have the following immediate corollary: Corollary 50. Let A be a projective arrangement of n lines which is not the six line arrangement in Figure 2.4, and which does not have all lines passing through a common point. Then there are at least

2n−3 7

ordinary points off any line which is not part of the

arrangement. Slightly less obvious is the following: Corollary 51. Let A be a projective arrangement of n lines, no n − 1 of which pass through a common point. Also, assume that A is not a seven line arrangement having the six line arrangement in Figure 2.4 as a sub-arrangement. Then there are at least

2n−5 7

ordinary

points off any line in the arrangement. Proof. Let l ∈ A and consider the arrangement with l removed. By the assumption about no n − 1 of the lines passing through a common point and A \ {l} not being our favorite six line arrangement, we may apply the previous Corollary to conclude that there are at least 2(n−1)−3 7

2.7

=

2n−5 7

ordinary points off of l; points which are ordinary with or without l.

Finding (finite) ordinary points in time O(n log n)

In this section we extend the Mukhopadhyay et al. [61] argument for finding an ordinary point amongst an arrangement of projective lines, described in section 1.2, to find a finite ordinary point in a non-trivial affine arrangement of lines, also in time O(n log n). As with the projective problem, we work in the dual formulation. Upon dualizing the lines to points via the standard transformation, {(x, y) : y = mx + b} 7→ (m, −b), we must find an ordinary line which is not vertical since only these dualize back to finite ordinary points. First find a point which does not share an x-coordinate with any other point in the configuration. If there is no such point then add the vertical point at infinity. The addition of this point creates no ordinary lines. Rotate the configuration with the addition of the vertical point

111

r

q0

q2

q1

l

p0

Figure 2.10: If the line l is not ordinary then it contains three points q0 , q1 , q2 - this is the case where two of the points q0 , q1 are to one side of the ray r. The line through p0 and q1 is seen to be ordinary. at infinity so that all points are finite. Apply the algorithm for the projective problem to find an ordinary line in time O(n log n). Otherwise, if there is a point, call it p0 , which does not share an x-coordinate with any other point, consider the ray, defined as in the projective problem, r = {p0 + λˆ : λ ≥ 0}. Either the point configuration is a near pencil with n − 1 points along a vertical line, plus p0 , not on the line, or there are two points p, q such that l(p, q) passes through r coming as close as possible to p0 without passing through p0 . If the arrangement is a near pencil, then any of the n − 1 collinear points together with p0 determine an ordinary line. Otherwise we run through the same basic argument as in the projective case but with a couple of differences. We are not able to choose p0 to have minimal x-coordinate, so the points p, q and subsequent points in the argument may be on either side of r. Furthermore, we need to verify at each stage that the found ordinary lines are not vertical. Although the argument is much the same as in the projective case, we run through it in detail for completeness. Either l(p, q) is ordinary or p, q are two of three points which we shall refer to as q0 , q1 , q2 , which we label in order of increasing x-coordinate. There are two cases, one where all of q0 , q1 , q2 are to one side of r, and another where two of the three points, say q0 , q1 are to one side, and q2 is on the other side. Consider the case where q0 , q1 , q2 are all to one side (say the right side) of r first. As in the projective case, we consider the line l(p0 , q1 ) exactly as in the former Figure 1.4. The only new possibility is that there can be a point q to the left of p0 on l(p0 , q1 ). However, in this case the line l(q, q0 ) would have crossed r closer to p0 than did l, a contradiction. In the second case q0 , q1 are to one side, and q2 is on the other side of the ray r, as depicted in Figure 2.10. In this case the line l(p0 , q1 ) must be ordinary. Suppose there were a third point q on l(p0 , q1 ). q can be in (∞, p0 ), (p0 , q1 ) or (q1 , ∞). If q ∈ (∞, p0 ) then l(q, q0 ) crosses r closer to p0 than does l. If q ∈ (p0 , q1 ) then l(q2 , q) crosses r closer to p0 than does l. Finally if q ∈ (q1 , ∞) then l(q0 , q) crosses r closer to p0 than does

112

l. l(p0 , q1 ) is not vertical by choice of p0 . The running time of this algorithm is essentially the same as that for the projective case (O(n log n)), since the only difference is that at the end of the algorithm we see if there are some three points on l and if so, our ordinary line is not always the line through p0 and the centerpoint along l, but rather determined by an additional condition, namely how many points are to either side of r.

2.8

Ordinary point bounds for small n in the affine setting

For non-trivial arrangements of n lines in the affine plane, we can consider the analog of the function m(n) which we considered in Chapter 1 (Section 1.8). In the affine setting, we denote the corresponding function by m∗ (n) and it denotes, for a given number n, the minimum number of finite ordinary points amongst all affine arrangements with precisely n lines. We shall not take this analysis very far, but at least establish the following values: n

3

4

5

6

7

8

9

10

m∗ (n)

2

2

2

1

2

2

3

3

11

12

13 4

In order to keep this section as brief as possible we borrow some results from Chapter 3. In particular, we use the affine expression (equation 3.2.3) X X (3 − k)ρk = n + 3, (3 − j)τj + j≥2

(2.8.1)

k≥2

where τj is the number of affine j-crossings and ρk is the number of affine k-gons - together with Theorem 61 which asserts that ρ2 ≤ b4n/3c. For n = 3, 4, 5, b4n/3c = n + 1. Since τ2 and ρ2 are the only positive terms on the left hand side of Equation 2.8.1 we have τ2 + ρ2 ≥ n + 3,

(2.8.2)

so together with the fact that for n = 3, 4, 5 we have ρ2 ≤ b4n/3c = n + 1 we get that for these values m∗ (n) ≥ 2. Examples establishing the fact that actually m∗ (n) = 2 in each of these cases are provided in Figure 3.7. The fact that m∗ (n) has lower bounds as indicated in the table for n = 6, 7, 8, 9, 10 and 13 follows from Theorem 49. The examples of Figure 2.4 establish these as exact values of m∗ (n) for n = 6, 7 and 8. Figure 3.1 is an example of 9 lines and 3 finite ordinary points so m∗ (9) = 3 is established. To see that m∗ (10) = 3, start with the B¨or¨ oczky example of 10 lines with 5 ordinary points (Figure 1.23), take a line through any two of the ordinary points - note that such a line

113

is not part of the arrangement - and send this line to infinity via a projective transformation. All lines in the original arrangement remain finite, and there are just 3 finite ordinary points. Analogously, to establish that m∗ (13) = 4, start with the McKee example (Figure 1.7) and take any two ordinary points not currently lying on a line in the arrangement, and take these points to infinity by a projective transformation, giving an isomorphic arrangement of 13 (finite) lines with 4 finite ordinary points.

2.9

Sylvester’s theorem in other spaces

The Sylvester problem and its dual can be considered in other spaces besides the Euclidean or affine plane, and the real projective plane. We consider first the hyperbolic plane, H2 . There are various Euclidean models of the hyperbolic plane, but for our purposes, the easiest is the so-called projective disk model. In this model the points of H2 are the points of the open unit disk, and the lines (geodesics) are the chords through the disk. Given a set of not-all-collinear points in the open unit disk, we can apply Sylvester’s Theorem in R2 to find a line containing just two of the points, which when restricted to the open unit disk provides a proper ordinary line in H2 . Sylvester’s Theorem thus holds in H2 . The dual of Sylvester’s Theorem, stated in the affine plane, says that given an arrangement such that any two lines intersect at a point there must be an ordinary point of intersection. This dual statement (though not a dual in the hyperbolic sense) remains true in H2 . We can extend the open chords in H2 to lines in R2 where the dual of Sylvester’s Theorem guarantees an ordinary point, say at the intersection of lines l and k. But the corresponding chords, chord(l) and chord(k) are assumed to already intersect in H2 - so this point of intersection is ordinary in H2 as well. Sylvester’s Theorem is false on the torus. The “lines” of this manifold are the geodesics, or curves which are everywhere, locally, shortest paths. We only consider the case of the “flat” torus. A “flat” manifold is a manifold in which every neighborhood is isometric to a piece of Euclidean space (so angles are preserved as well as distances). The flat torus can actually be isometrically embedded into R4 . We model the flat torus in the usual topological way as a square with its oriented edges identified. Because of flatness, this model, when unrolled to form its universal cover, is locally an isometrically accurate model. An illustration showing that Sylvester’s Theorem is false on the flat torus is given in Figure 2.11. Patches isometrically equivalent to a piece of R2 may be taken to include the nearest neighbors of any of the marked points. Any line (geodesic) through two points passes through an additional point.

114

Figure 2.11: Sylvester’s Theorem is false on the flat torus. Sylvester’s Theorem is likewise false on the flat Klein Bottle. The flat Klein Bottle cannot be isometrically embedded in R4 but can be so embedded in R5 . With the usual identification of edges, the same example as before for the flat torus, is also a counterexample to Sylvester’s Theorem for the flat Klein Bottle. See Figure 2.12. Note that the line

Figure 2.12: Sylvester’s Theorem is false on the flat Klein Bottle. (geodesic) containing the points in the second row also contains the points in the third row. Thus, the line through the points in the second column intersects the line containing all points in the second and third row twice! Given that Sylvester’s Theorem is false on both the flat torus and the flat Klein Bottle, it is important to see why the same argument does not lead to the same conclusion in the projective plane, which can be modeled in analogous fashion. See Figure 2.13. Note that

Figure 2.13: The same argument applied to the real projective plane does not lead to the same conclusion. the line through points A and B does not pass through any of the additional points. Sylvester’s Theorem has also been considered in two and higher complex dimensions where the problem takes on a decidedly different character. In C2 “lines” are one-complexdimensional affine spaces of the form {a + bz : z ∈ C} for fixed a, b ∈ C2 (or more generally,

115

in Cn where a, b ∈ Cn ). Thus complex lines are two real-dimensional affine subspaces of C2 (or Cn ) thought of as a four-dimensional (or 2n-dimensional) real vector space, though not all two real-dimensional affine subspaces are complex lines. See [51] for a more complete discussion. It has been known for quite some time2 that there are cubic curves in C2 , with 9 inflection points such that any line through two of the inflection points passes through another. Thus Sylvester’s Theorem is false in C2 . In 1966 J. P. Serre [71] asked whether there were such Sylvester-violating point configurations in C3 that spanned all three dimensions - and in 1986, Kelly [49], applying deep results from algebraic geometry, found the answer to be no. In fact, the only such point configurations in Cn for any n ≥ 2 must be coplanar. The key lemma, due to Hirzebruch [43] associates with each arrangement of lines in the complex plane an arrangement of algebraic surfaces, calculates their Chern numbers and applies an inequality due to Miyaoka [58] and Yau [76]. Subsequently, Elkies et al. [27] gave an elementary proof of the Kelly result, not relying on the Hirzebruch lemma. Using the same elementary techniques, Elkies et al. [27] then considered higher-dimensional quarternionic spaces, and showed that any point configuration violating Sylvester’s Theorem would have to be three-dimensional in the sense of spanning no more than a three-quarternionic-dimensional flat. It is not known, however, whether such Sylvester-violating quarternionic configurations actually exist. While the cubic example in C2 shows that Sylvester’s Theorem is false in that space, the Hirzebruch inequality immediately implies a somewhat weaker result, namely that given a finite set of points in C2 there is always a line through at most three of the points. No elementary proof of this result is known [9]. However, Solymosi and Swanepoel have recently shown [73], that the weaker fact, that there is always a line through at most 5 points, follows from very elementary reasoning and that the kissing number for circles in the real (or Argand) plane is 6. They push the same elementary reasoning even further to get that in the quarternionic plane there is always a (quarternionic) line through at most 24 points - this time exploiting the fact that the kissing number for balls in R4 is 24. In 2004, Chv´atal [14] considered the Sylvester problem in finite metric spaces using a notion of metric betweenness. Given a metric space (V, ρ) where ρ is the metric, following Menger [57], Chv´atal defined the betweenness relation B(ρ) for triples of distinct points u, v, w ∈ V by (u, v, w) ∈ B ↔ ρ(u, w) = ρ(u, v) + ρ(v, w). 2

Perhaps even to Sylvester, as noted in [49].

(2.9.1)

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Writing [uvw] if (u, v, w) ∈ B, one can define a line containing points a, b as {a} ∪ {b} ∪ {x : [xab]} ∪ {x : [axb]} ∪ {x : [abx]}.

(2.9.2)

However, as shown in [14], Sylvester’s Theorem is false with such a definition, and Chv´ atal considers instead a certain recursive closure of the definition of a line given by 2.9.2. The easiest way to understand the recursive closure is to start with a, b ∈ V and consider the line determined by a and b to be the return value of the following program: LineFromPoints(a, b) : S ← {a, b} while some triple of points T has |T ∩ S| = 2 and T ∈ B S ←S∪T return S

With this new definition, Chv´ atal conjectured, and his student, Xiaomin Chen [12], ultimately proved that Sylvester’s Theorem holds in arbitrary finite metric spaces where the notion of line is connected in this way with a betweenness relation. To conclude this section it is worthwhile to look back at the examples where Sylvester’s Theorem did not hold and observe that the notion of a line in these cases is not connected with a betweenness relation in the sense of Menger and Chv´ atal and hence satisfy ourselves that we do not have counter-examples to the Chv´ atal-Chen Theorem. In order to view these examples as finite metric spaces we must view the chosen point sets as the entire space. Infinitesimally (i.e. in the neighborhood of any patch) these spaces do satisfy the Menger and Chv´atal betweenness relation - but to extend the infinitesimal betweenness relations to the entire space (and so to the point set), via the recursive closure, would require that we consider an infinite number of points. We consider just the flat torus example from Figure 2.11. If we take the distance between horizontally or vertically adjacent points in √ this example to be unity, then all points are either distance 1 or distance 2 from one another. Thus there are no points between any two others in the Menger-Chv´ atal sense, so that if we were to derive lines from this example using betweenness, we would obtain a line for every pair of points. To get around this problem one is tempted to try to just add more points to each row and column, as in Figure 2.14. Again we assume neighboring horizontal and vertical points are unit distance from one another. Then we observe that although there is a toroidal line through points a1, c2, a3 and c4, no one of these points lies between some other two in the Menger-Chv´ atal sense, and so these points do not lie on a

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Figure 2.14: The Sylvester-violating flat torus example with points in one more row and column compared to Figure 2.11. single Chv´atal-type line.

Chapter 3

Wedges and Outer Layer Complexity 3.1

Introduction

As an extension of my investigation of affine line arrangements where not all lines intersect at a common point and not all lines are parallel I was led to consider bounds on the number of faces (cells) whose boundary consists just of two unbounded edges, or rays. We call such an unbounded face a wedge. In the following section we explain the connection between ordinary points and wedges and motivate our investigation. The next section develops the theory of wedges and culminates with a proof of a d4n/3e bound on the maximum number of wedges. Finally we use results from the theory of wedges to obtain lower bounds on the complexity of the outer layer.

3.2

Euler’s Relation: Connection to Wedges

Melchior’s proof of Sylvester’s Theorem (the second proof in Section 1.1) shows that the existence of projective ordinary points is a relatively simple consequence of Euler’s relation in the projective plane. However, we now show that the same is not the case for the Sharp Dual of Sylvester’s Theorem (see section 2.1). Recall (Definition 11) that an affine arrangement is said to be trivial if either all lines are parallel or all lines pass through a common point and non-trivial otherwise. Begin then with a non-trivial arrangement of n lines in the affine plane and transfer the arrangement

118

119

to an arrangement of circles on the sphere, all of which pass through the south pole. To do this, imagine the plane sitting on top of the sphere and project points from the plane to the sphere by stereographic projection through the south pole. Finally join up all circles by adding the south pole itself. If we let V, E and F denote, respectively, the number of vertices, edges and faces in the induced spherical arrangement, then Euler’s relation (on the sphere) says that V − E + F = 2.

(3.2.1)

and, as usual, putting tj = number of vertices where j lines cross, pk = number of faces surrounded by k edges, we obtain Ysph =

X

(3 − j)tj +

j≥2

X (3 − k)pk = 3(V − E + F ) = 6.

(3.2.2)

k≥2

Now, returning to the affine plane, the point at the south pole has n lines crossing, so if we exclude this point from the vertex set, and denote by τj and ρk the analogs of tj and pk , then in the affine plane, we obtain X X (3 − k)ρk = n + 3. (3 − j)τj + j≥2

(3.2.3)

k≥2

In equation (3.2.3) only the τ2 and ρ2 terms are positive and they both have coefficient 1. τ2 is the number of vertices where two lines cross, and ρ2 is the number of wedges. Upper bounds on the number of wedges therefore immediately imply lower bounds on the number of ordinary points. In particular, if we knew that there could be no more than n + 2 wedges, relation (3.2.3) would immediately give a positive answer to the Sharp Dual of Sylvester’s Problem. However, there are line arrangements with n + 3 wedges. Figure 3.1 gives an arrangement of 9 lines with 12 wedges. This arrangement has a family of “essentially equivalent variations” with respect to the property of having n + 3 wedges as the examples of Figure 3.2 illustrate. The essential ingredients are two oppositely oriented similar triangles, with coinciding medians which each cut pairs of opposite wedges. One might be tempted to conjecture that these are the only line arrangements with n+3 wedges. However, Figure 3.3 gives an example with 16 lines and 12 wedges and another with 20 wedges and 15 lines. We describe a process for generating arrangements with 4n/3

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Figure 3.1: Example with n + 3 wedges.

Figure 3.2: Essentially equivalent variations in the case of n = 9 lines and 12 wedges. wedges for any n ≥ 6 and n a multiple of 3 in Theorem 3.3.

3.3

The Theory of Wedges

Lemma 52. In any non-trivial affine arrangement of n lines without parallel lines there are at most n wedges. Proof. Order the lines cyclically by slope l1 , ..., ln . One obtains n adjacency pairs (l1 , l2 ), ...,(ln−1 , ln ), (ln , l1 ). A wedge must be formed between some such pair, on one unbounded end or the other (what we shall refer to as either the “top” or “bottom,” where the choice of “top” is arbitrary). Since no two lines are parallel, each line contains at least two points of intersection. Hence, if (li , li+1 ) corresponds to a wedge on “top,” it cannot correspond to a wedge on “bottom” (and vice versa). The lemma follows. It is very easy to achieve an arrangement of n lines, no two parallel, with n wedges, as the example in Figure 3.4 shows. Another example is provided by extending the edges of a regular n-gon for n odd. If we label the lines which extend the edges in counter-clockwise order as we traverse the n-gon, starting with an arbitrary l1 , as l1 , ..., ln , then (l1 , ldn/2e ) is a wedge. Continuing around cyclically we obtain n wedges. While there are other convex n-gons whose extended edges yield n wedges, many do not.

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Figure 3.3: 4n/3 examples.

Figure 3.4: n wedges with no two lines parallel. Lemma 53. An arrangement of n lines with n + k wedges must contain at least k pairs of parallel lines in k distinct directions. Proof. Adding a line to an arrangement can add at most two wedges: one at the “top” of the line and one at the “bottom.” To see this, suppose that adding a line lk could add more than one wedge at the “top.” Far from all points of intersection, in the direction of “top,” lk lies between two lines, l1 and l2 say. For lk to add more than one wedge at the “top” there must not have been a wedge between l1 and l2 before placing lk , and there must be a pair of such faces after placing lk . But for there to have been no wedge earlier means that there had to be a line lj intersecting both l1 and l2 in the direction of “top” after the point where l1 and l2 intersect (i.e closer to “top”). But then clearly the newly placed line lk can create a wedge with l1 or l2 , but not both. Adding a third parallel line between two existing parallel lines clearly cannot contribute any wedges. Placing n − j lines, no two of which are parallel creates at most n − j wedges by Lemma 52, and then placing the j remaining lines, each parallel to a distinct line amongst the initial n − j, creates at most 2j additional wedges for n + k wedges in total. The lemma follows. Figure 3.5 is an example of 4 lines with 3 wedges, and is thus an example of n lines with fewer than n wedges. It is also an example of a convex n-gon whose edges do not extend to form n wedges. In fact one can produce examples with 3 wedges and arbitrarily many lines. In Figure 3.5, we may either add many lines parallel to the line l, all placed between

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points p and r, or if we prefer an example with no two lines parallel, add lines through p which pass into the regions U and B, but do not pass through the point q.

Figure 3.5: 4 or more lines with just 3 wedges. Lemma 54. Let L be a line arrangement and let C denote the extreme points of the convex hull of the the vertices of L. Then L must contain at least one wedge for each element of C. Proof. Observe that the elements of C are themselves vertices of L. Let v ∈ C. Any line intersecting in v contributes an edge extending from v to infinity. Since there are at least two lines intersecting at v there is at least one wedge per extreme vertex. Corollary 55. In any arrangement of n lines, not all of which are parallel, there must be at least 3 wedges. Proof. If all the intersection points of the collection of lines L lie on one line then we have a family of parallel lines cutting a single line. In this case there are 4 wedges. Otherwise some three intersection points form a triangle and Lemma 54 applies. As a corollary we have the following classical theorem due to Levi [54]: Corollary 56. In any arrangement of n lines in the projective plane, the number of triangular cells is at least n. Proof. First observe that every line is incident to at least 3 triangles. To see this, consider an arbitrary line l which we may take to be the line at infinity. The triangles incident to l are the wedges amongst the remaining lines. Hence by Corollary 55 l has at least 3 incident triangles. Since there are 3 lines per triangle, we can allocate 1/3 of a triangle to each of its component lines and conclude that there must be at least n triangles in total.

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In fact we can say the same thing for pseudolines since Lemma 54 and Corollary 55 are a consequence of the crossing properties of lines and not their straightness. (This observation is also due to Levi [54].) Corollary 57. In any arrangement of n pseudolines in the projective plane, the number of triangular cells is at least n. Obtaining a lower bound on triangular cells in affine arrangements, where one does not consider the unbounded cells, is a harder problem. In 1889 Roberts [68] claimed to have a proof showing that a simple affine arrangement of n lines must have n − 2 triangles. However, the proof was shown to be flawed, but finally proved in 1979 by Shannon [72]1 . Interestingly, although an arrangement of n lines must have n − 2 triangles, an arrangement of n pseudolines my have as few as d2n/3e triangles. See [33] for details. The following was first proved by Ching and Lee in [13]. Corollary 58. Given a collection of n lines it is possible to find the convex hull of the set of intersection points in time O(n log n). Proof. Without loss of generality, assume no line has infinite slope. Sort the lines by increasing slope, and in case of ties, by increasing y-intercept, labelling the sorted sequence l1 , ..., ln . Follow this with a sequence l10 , ..., ln0 which is the same sequence, but this time sorted by increasing slope, and in case of ties, by decreasing y-intercept. Consider in turn the intersection of pairs 0 (l1 , l2 ), ..., (ln−1 , ln ), (ln , l10 ), (l10 , l20 ), ..., (ln−1 , ln0 ), (ln0 , l1 ).

The intersection point giving rise to each wedge appears in this list. Lemma 54 implies that the set of intersection points of line pairs in this list includes the extreme points of the convex hull of intersection points of all pairs. Compute the convex hull of this set of up to 2n intersection points, not a priori knowing which are the extreme points, using any of the known O(n log n) algorithms. Once we have found the convex hull it is easy enough to generate a list of all intersection points on the hull, still in O(n log n) time. For each point found in the final convex hull above, keep track of the “first” and “last” lines intersecting at that point in the cyclical ordering l1 , .., ln , l10 , ..., ln0 , l1 . In other words if l2 , l3 , l4 intersect at a point p, then l2 is first 1

Despite the flawed proof, this result is often referred to as the Roberts Theorem. Shannon proved a stronger result, namely that any non-trivial arrangement of n hyperplanes in Rd contains at least n − d simplicial d-dimensional cells.

124

and l4 last. If ln0 , l1 , l2 intersect at p, then ln0 is first and l2 last. If l, k ∈ {l1 , .., ln , l10 , ..., ln0 } are the first and last lines crossing at extreme convex hull point pi and ˆl, kˆ ∈ {l1 , .., ln , l0 , ..., l0 } 1

n

are the first and last lines crossing at the next extreme convex hull point pi+j , where points are ordered consistently with the line ordering, then we must just gather the intersection points of each line between k, ˆl (in the cyclical ordering), and the line segment connecting pi , pi+j . The collection of all such points, together with the original extreme hull points, is a complete hull point enumeration. It is also possible to enumerate all wedges in O(n log n) time, however, not simply by enumerating the hull points. There may be “inner” wedges - wedges that emanate from a point inside the convex hull of the intersection points. The left hand diagram in Figure 3.6 provides an example. The convex hull points are marked with solid dots. The vertex

Figure 3.6: Inner wedges and the difficulty detecting them. of the inner wedge is marked with a hollow dot on the left. The difficulty is detecting the difference between the diagram on the left, which has an inner wedge, and the diagram on the right, which does not. Note that we can fit arbitrarily many inner wedges next to the wedge on the left, thus showing that there can be Ω(n) inner wedges. In fact it is not clear that one can tell whether an arbitrary face, given by (li , li+1 ) in the cyclical ordering, is a wedge in less than O(n) time. Nonetheless, finding wedges is related to computing the envelope of a collection of lines. Recall that the envelope of a collection of lines is defined as the polygon whose boundary consists of the bounded edges of all the unbounded faces of the associated arrangement. Keil’s algorithm [46] for computing the envelope of a collection of lines actually pieces together the envelope by completely delineating all unbounded faces. As a result we have:

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Theorem 59 (Keil). Given a collection of n lines in the plane, it is possible to completely describe all unbounded faces in the associated arrangement in time O(n log n). Corollary 60. Given a collection of n lines in the plane, it is possible to identify all wedges in time O(n log n). The main result of this section is the following: Theorem 61. In an arrangement of n lines, not all of which pass through a common point, there are at most b4n/3c wedges. Furthermore, this bound is tight. Proof. We begin by establishing the b4n/3c upper bound. Given n lines, in order for there to be n + k wedges, by Lemma 53, requires k independent pairs of parallel lines. But between the k pairs of parallel lines there can be no wedge. In other words, of the 2n pairs 0 (l1 , l2 ), . . . , (ln−1 , ln ), (ln , l10 ), . . . , (ln−1 , ln0 ), (ln0 , l1 ),

2k of the pairs cannot be wedges. It follows that n + k ≤ 2n − 2k, i.e. k ≤ n/3. So the number of wedges is at most b4n/3c. It remains to establish tightness of the bound. To this end we give explicit examples showing the bound being attained. We first give examples for the case when n ≥ 6 is a multiple of 3. Start with a convex 2n/3-gon with opposite sides parallel (for example a regular 2n/3-gon) and extend the edges to infinity2 . Label the extended edges (lines) in the order in which they are encountered in a clockwise traversal of the edges of the polygon, starting with an arbitrary edge: l1 ,l2 ,...,l2n/3 . Then the wedges are formed by (l1 , ln/3 ), (l2 , ln/3+1 ), etc. By virtue of the fact that we started with a convex polygon with opposite sides parallel, the line that bisects the wedge (l1 , ln/3 ) also bisects the wedge formed by the pair of corresponding parallel lines, i.e. (ln/3+1 , l2n/3 ). There is one bisector line for each such pair of wedges. Since there were 2n/3 original lines with 2n/3 wedges, adding n/3 bisector lines adds 2n/3 additional wedges, yielding a total of n lines with 4n/3 wedges. The case n ≥ 6 where n is a multiple of 3 is thus established. Figure 3.7 establishes the cases n = 3, 4, and 5. It remains to consider the cases of n = m + 1 and n = m + 2 where m ≥ 6 with c = m a multiple of 3. Note that b 4(m+1) 3

4m 3

+ 1 and b 4(m+2) c = 3

4m 3

+ 2 so we need to

show how to add a line at a time to an example like those of Figures 3.1 through 3.3 while also adding a single wedge. For this purpose, find a circle big enough so that it contains 2

It is also possible to do this construction starting with two identically oriented regular n/3-gons with a common center point, of arbitrary relative sizes, with one n/3-gon rotated by π/(n/3) degrees.

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Figure 3.7: b4n/3c cases (a) n = 3, ρ2 = 4, (b) n = 4, ρ2 = 5 and (c) n = 5, ρ2 = 6. all intersection points. Traverse the circle clockwise beginning at its intersection with any pair of parallel lines. Label the lines as they are encountered l1 , l2 ,...,ln ,l2 ,l1 ,l3 ,l5 ,l4 ,l6 ,...,ln . See Figure 3.8. We encounter a pair of parallel lines, followed by a splitting line, a pair

Figure 3.8: Adding a wedge at a time in the cases n ≡ 1, 2 (mod 3) beginning with a regular hexagon and n = 9 lines with 4n 3 = 12 wedges. of parallel lines, followed by a splitting line, and so forth, until we run through all lines. After reaching ln we encounter the parallel lines in reverse order, but otherwise the overall order is unchanged. Now suppose we add a new line ln+1 that splits the wedge (l2 , l3 ), passing through l2 ∩ l3 . Since this new line has no parallel, the new ordering will be l1 ,l2 ,ln+1 ,l3 ...,ln ,l2 ,l1 ,ln+1 ,l3 ,l5 ,l4 ,l6 ,...,ln . Previously (l1 , l3 ) was a wedge. Now (l1 , ln+1 ) is a

127

wedge since ln+1 plainly intersects l1 “after” l3 does, and l1 ∩ l3 was a previous last point of intersection along l1 . Since we have also added a wedge between the original (l2 , l3 ) the net result of adding the splitter ln+1 is that we have added a wedge. Similarly we can split the wedge (l3 , l4 ) with a line ln+2 yielding the sequence l1 ,l2 ,ln+1 ,l3 ,ln+2 ,l4 ,...,ln ,l2 ,l1 , ln+1 ,l3 ,ln+2 ,l5 ,l4 ,l6 ,...,ln , and thereby adding another wedge. We have thus established the tightness of the b4n/3c bound and hence the theorem follows. Given a suitable random distribution on the space of lines, it is interesting to now consider the expected number of wedges in an arrangement. If we could ignore the possibility of there being inner wedges, the problem would reduce to that of determining the expected number of extreme points of the convex hull of the vertices of the arrangement, since (i) almost surely the vertex of a wedge does not happen to lie somewhere in the middle of a supporting line since (a) if the supporting line were part of the arrangement then three lines would meet at this vertex, which almost surely doesn’t happen and (b) if the supporting line were not part of the arrangement then we would have three collinear vertices not all along an existing line, which also almost surely doesn’t happen and (ii) we can assume there is no more than one wedge at an extreme point, because again, almost surely, no three lines meet at a common vertex. It is an interesting fact that the expected number of extreme points in the convex hull of the points of intersection of lines chosen randomly as duals from a uniform distribution of points in the unit square [0, 1]2 or unit disk {z : kzk ≤ 1} is actually bounded by a constant for all n, where n is the number of lines. See [36] and [22] respectively. As n → ∞ intuition suggests that it becomes more and more unlikely that a randomly chosen internal unbounded face is a wedge, but this does not necessarily imply that the number of internal wedges goes to zero, or is even bounded.

3.4 3.4.1

Complexity of the Outer Layer The generic setup

For a given number n we consider all non-trivial arrangements of n lines in the affine plane without general position assumptions (i.e. lines may be parallel and more than two lines may pass through a single point) and ask: Amongst all such arrangements, what is the maximum and minimum number of line segments which may bound the outer layer? By outer layer we mean the collection of all unbounded cells. We refer to the number

128

of segments bounding such unbounded cells as the outer layer complexity of the arrangement. Note that we count the segments belonging to two unbounded cells only once. For example, the outer layer complexity of the arrangement in Figure 3.9 is 16.

Figure 3.9: An arrangement of 6 lines with outer layer complexity 16. There are two pairs of parallel lines. The bounding segments of the outer layer are highlighted.

3.4.2

Maximum complexity

In the case of maximum complexity, if we introduce the usual general position assumption (no two lines parallel, no three lines passing through a common point), the problem reduces to that of finding the maximum complexity of the zone of a line3 . To see this, consider the problem temporarily in the projective plane, add a line at infinity, and then rotate the arrangement, containing the line at infinity, again to the affine plane where all lines are finite. The complexity of the outer layer in the old arrangement is the complexity of the zone of the line rotated in from infinity in the new arrangement - though our definition of complexity is somewhat different from the classical one. The classical definition of the complexity of the zone of a line ` is the total number of edges of faces in the arrangement (minus the line `), whose faces intersect `. See Figure 3.10 to appreciate the distinction. When considering the complexity of the zone, our definition of complexity differs from the classical one by how we count the segments which cross the line `. Although the perspective is slightly different, in either case, each line (other than `) contributes precisely 3

As classically stated, i.e. with general position assumptions; see [5], [10], [25], and [64].

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Figure 3.10: An arrangement of 3 lines with the addition of a (dashed) line `. In the classical definition of the complexity of the zone of `, the two sides of each line crossing ` are counted separately since they bound adjacent faces of the arrangement. On the other hand, in our definition of complexity, for each line crossing ` we count the segment above ` and the segment below ` as distinct. Hence the two definitions are equivalent and under either definition the complexity of the above arrangement is 10. two segments which intersect `. Thus the answer to our maximum complexity question, for lines in general position, is answered by the theorems giving best possible constants in the Zone Theorem. In particular, in 1985 Chazelle et al. [10] and in 1986, Edelsbrunner et al. [25]4 showed that the Zone complexity is at most 6n and then in 1991, Bern et al. [5] showed the complexity to be at most 5.5n and gave examples to show that this bound was tight up to an additive constant.

3.4.3

Minimum complexity

In any non-trivial arrangement of n lines, each line must contribute at least its two unbounded segments to the complexity. We may then consider the 2n “slots” between successive unbounded edges as we walk, say, counterclockwise around the outer layer. Any time we have a non-wedge, there must be at least another edge contributing to the complexity. Thus, using Theorem 61, the overall complexity must be at least 2n + (2n − b4n/3c) or at least 8n/3. The examples in Figure 3.9 and the right hand drawing in Figure 3.3 are therefore tight minimum complexity examples. The question remains whether the lower 4

This paper has a famous flawed proof of the higher dimensional Zone Theorem which says that for an arrangement of n hyperplanes in Rd the total number of faces bounding the cells intersected by another hyperplane is at most O(nd−1 ). A correct proof was given in [26]. The two dimensional proof, establishing the constant 6 mentioned in the text, is however correct.

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bound of 8n/3 is correct for all n. In fact for very small n it is trivially not, as the example in Figure 3.11 illustrates. In our analysis above, specifically in the statement “Any time we

Figure 3.11: An arrangement with 4 lines and complexity 10


4n 3

4n 3

= 12 wedges but

8n 3 .

We begin with two identical regular n/3-gons, positioned one on top of the other, then rotate one of them about the common center (the n/3-gon which we shall call “the second one”) by π/(n/3) radians. By virtue of the fact that n ≡ 3 (mod 6), the n/3-gons have

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an odd number of edges. The generic construction to get edges of the two n/3-gons to infinity, yielding a total of

2n 3

4n 3

wedges is now to extend all

wedges, and then add

n 3

wedge

bisectors that each cut a pair of oppositely oriented wedges in two, giving rise to n lines with 4n 3

wedges. This construction works regardless of whether n/3 is even or odd. By virtue

of the fact that n/3 is odd, however, the original set of wedges, before adding the bisector lines, each come from lines extending edges in the same n/3-gon. Indeed, if we number the associated lines in the same order as the associated edges in one of the polygons, travelling say in counter-clockwise order, l1 , ..., lm where m = n/3 then (l1 , ldm/2e ) forms a wedge, (l2 , ldm/2e+1 ) forms a wedge, and so forth. Furthermore, the wedges formed by the extended edges of the second n/3-gon will appear between successive wedges formed by extended edges of the first n/3-gon. See Figure 3.12 for the case n/3 = 7.

Figure 3.12: Two concentric regular 7-gons, one offset from the other by π/7 radians of rotation. The lines determined by the edges of the solid 7-gon are labelled l1 , ..., l7 . The wedges formed by lines extending the edges of the solid 7-gon are interleaved with wedges formed by lines extending the edges of the dashed 7-gon. Now if we continue in this way and add the n/3 lines that split opposite facing wedges, we will obtain an arrangement of 4n/3 wedges but between one pair of wedges and the next will be an unbounded quadrilateral. We note however, that we can shrink one of the offset n/3-gons and do better. In Figure 3.12 we note that the intersection point of the dashed lines lies on a line with the center of the two 7-gons and the intersection point of l4 and l2 . Thus we may shrink the dashed 7-gon about its center such that the intersection point of the dashed lines meets l4 ∩ l2 . See Figure 3.13. By symmetry we will in this way have unbounded triangles between successive pairs of wedges and so have total outer layer

132

Figure 3.13: Two concentric regular 7-gons after shrinking one to obtain unbounded triangles between successive wedges. There will be successive pairs of wedges after adding splitting lines. complexity 8n/3. Although Figure 3.13 is typical of the case n odd, n ≥ 7, the cases n = 3, 5 are slightly different. In the case n = 7 we note that there is one solid edge between the edges determining lines forming the wedges, which sandwich the wedge determined by the dashed lines (in Figure 3.12, the line determined by this edge would be l3 ). For n ≥ 7 there will be

n−5 2

edges between sandwiching wedges. In the case n = 5 there are no such

intermediate solid edges, so the dashed pentagon, in this case, would have to shrink so that the dashed lines meet in a middle vertex. The right hand drawing in Figure 3.14 illustrates this point. In the case n = 3 the two solid wedges sandwiching the dashed wedge share a common edge, so in this case the dashed triangle must be shrunk so that its vertices lie on the solid triangle, as the left hand drawing in Figure 3.14 illustrates.

Figure 3.14: Nested triangles and pentagons after shrinking one (the triangle and pentagon, respectively, formed by the dashed lines) to obtain unbounded triangles between successive wedges. Lines splitting opposite pairs of wedges have not been drawn. Now that we have examples with outer layer complexity exactly 8n/3 for n ≡ 3 (mod 6), we proceed to add lines one at a time to any such example while keeping the outer layer complexity to

8m 3

+ d19k/6e, where k is the number of lines added, and 1 ≤ k ≤ 5. The

133

process is the same for n/3 = 3, 5, 7, etc. so to keep things uncluttered we illustrate for the case of n/3 = 3 in Figure 3.15. We begin with two nested triangles and 3 splitting lines

Figure 3.15: Adding one line at a time l1 , l2 , ... (dashed lines) to an arrangement of complexity 8m 3 . l1 adds 4 to the total complexity and successive lines l2 , ..., l5 each add 3 to the complexity yielding total complexity of 8m 3 + d19k/6e for 1 ≤ k ≤ 5. for a total of 9 (solid) lines. We place the first additional line (l1 in Figure 3.15) so that it splits any wedge. If we say that this line splits a wedge at its “top” then at its “bottom” it will cut through an unbounded triangle (forming an unbounded quadrilateral) and also contribute an unbounded edge at the very “bottom,” contributing in total, 3 edges to the outer layer complexity. In addition, the edge the line cuts through at the bottom, which previously separated a wedge from an unbounded triangle, now contributes 2 edges to the outer layer. Hence the net addition is 4 edges to the outer layer complexity. If we now add additional lines splitting the same wedge (i.e. in Figure 3.15, the addition of l2 cutting the wedge formed by l and l0 ) each line will contribute a net addition of 3 edges to the outer layer complexity. The net result is a complexity of

8m 3

+ 3k + d k6 e =

8m 3

+ d 19k 6 e for m, k

defined as in the statement of the theorem. Although the maximum complexity question for lines in general position is answered by the best bounds for the Zone Theorem, the analogous question for minimum complexity is still interesting. By Lemma 52 we know that an arrangement of n lines, no two of which are parallel, has at most n wedges. Therefore a natural question to ask is whether the right lower bound on complexity for lines in general position is 3n (2n for the unbounded edges and n more for the non-wedges). If we do not allow parallel lines but do allow more than

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two lines to pass through a common point, the 3n bound is attained by a family of n − 1 lines passing through a common point plus one additional line. However, as we shall see, the 3n bound cannot be attained for lines in general position (i.e. if we add the constraint that no three lines pass through a common point) for any n > 3. We begin with the following slight sharpening of Lemma 52: Lemma 63. In any affine arrangement of n ≥ 3 lines in general position there are at most n wedges if n is odd, and at most n − 1 wedges if n is even, and these bounds can be realized for every n. Proof. With an apology to the careful reader, we repeat the argument from Lemma 52: Consider the sequence of unbounded edges and the associated lines as we traverse them travelling along a circle very far from the intersection points of any of the lines. Label the associated lines as we encounter them l1 , l2 , ..., ln . Since there are no parallel lines we next encounter the lines a second time in precisely the same order as the first time. Since there are at least 3 lines, if there is wedge between a cyclically ordered pair (li , li+1 ) at one end, there cannot be a wedge between the same pair at the opposite end. It follows that there are at most n wedges. Now suppose there is a wedge at the “top” of (l1 , l2 ). By virtue of the fact that no three lines pass through a common point, there cannot be a wedge at the “top” of (ln , l1 ) or (l2 , l3 ). For there to be n wedges then there must be wedge in every other such “slot,” i.e. at (l1 , l2 ), (l3 , l4 ), etc. If n is even, say n = 2N , then the slots are (l1 , l2 ), (l3 , l4 ), ..., (l2N −1 , l2N ) but then the next slot is the “bottom” of (l1 , l2 ). However, the “top” of (l1 , l2 ) already has a wedge, so the “bottom” cannot have one. Thus if n is even there cannot be a wedge in every other slot, and so there cannot be n wedges. An example of n lines in general position with n wedges is given by extending the lines of a regular n-gon, for n odd. See Figures 3.16 and the right hand drawing in Figure 3.17. Each of the lines in such an arrangement contributes to two wedges, so removing any line gives an arrangement of n − 1 lines with n − 2 wedges. Now for an arrangement of n > 3 lines in general position to have complexity 3n, there would need to be n wedges, and hence a wedge in every other “slot.” There are more than 3 lines, so suppose we have wedges at positions (l1 , l2 ) and (l3 , l4 ). The (l2 , l3 ) slot must contain edges from both l1 and l4 and so we see that each non-wedge contains at least 2 additional edges and hence an arrangement with a wedge in every other slot must have total complexity at least 4n. Interestingly, if we remove a line from the arrangement in Figure 3.16 we obtain an arrangement of 4 lines and 3 wedges with complexity just 14. Moreover,

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Figure 3.16: A regular pentagon with lines extended, giving rise to 5 wedges. we may start with the arrangement obtained by extending the edges of a regular 7-gon, remove two lines and obtain an arrangement of 5 lines with complexity 18, showing that for a given n the arrangement of minimum complexity, under the general position assumption, is not necessarily an arrangement with the maximum number of wedges. See Figure 3.17. Can we do asymptotically better than 4n? An interesting observation about complexity 4n

Figure 3.17: A regular 7-gon with lines extended, and the same arrangement with two lines removed. The arrangement on the right has 5 lines and complexity 18. is that on average we have exactly one extra edge in each of the 2n slots between unbounded edges. Since if two wedges are separated by only one slot the one slot must have at least four edges, we know we cannot do better than a packing that has a wedge, followed by two unbounded triangles, followed by a wedge, followed by two unbounded triangles, and so on. However, if we start in this fashion with a wedge followed by two unbounded triangles, followed by another wedge, we necessarily have the sub-arrangement pictured in Figure 3.18. The only lines extending to the right of line l1 must be l2 , l3 , and l4 - in other words, in an arrangement in general position featuring, consecutively, a wedge, followed by two unbounded triangles and then another wedge, there cannot be any other lines! In fact the same argument follows if we had any sequence consisting of a wedge, some m consecutive

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l1

l2

l3

l4

Figure 3.18: A wedge followed by two unbounded triangles, followed by another wedge. unbounded triangles, followed by a wedge. There would have to be one “outer” line common to both wedges, and all additional lines would have to contribute to the unbounded edges of the unbounded triangles sandwiched between the wedges, there making m + 2 lines in total, counting the line common to the two wedges. We ask next what the minimum complexity may be to the “left” of the “outer” line (i.e. the line l1 in Figure 3.18) in an arrangement like we have been considering, of m + 2 lines, where to the “right” of the outer line we have a wedge followed by m unbounded triangles, followed by another wedge. We seek to show that the average slot must contain at least one unbounded edge. If this were not the case, then first of all we would have to have another wedge to the left of the “outer” line, and furthermore, since there cannot be a sequence where the set of adjacent triangles is shorter than m - since this would require an arrangement of total size less than m + 2, we must have a sequence of triangles of exactly length m, whence we would necessarily have two wedges side by side, which is impossible since then three lines would pass through a point, violating the general position assumption. Since the average slot on the left of the “outer” edge contains at least one extra unbounded edge, and the the slots on the right each contain an unbounded edge except for the two wedges, we have minimum total complexity 4n − 2. We next demonstrate that the 4n − 2 bound is in fact realizable for every n > 3 using some illustrations. We start with the arrangement in Figure 3.18 which has complexity 4n − 2 and add lines sequentially that contribute an unbounded triangle to the right of l1 and an unbounded triangle to the left (top) of l2 without disturbing anything else, in this way keeping the overall complexity always at 4n − 2. See Figure 3.19. Inductively line lm+1 is added so that it intersects l1 below any other intersection points and remains to the left of lm until crossing l3 at which point it crosses lm−1 , lm−2 , ..., l4 and finally l2 . In this way there remain two wedges and all triangles to the right of l1 , and an unbounded 4-gon, a

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l1

l2

l3

l4 l6

l5

Figure 3.19: Adding lines l5 , l6 , ...: the 4n − 2 construction. wedge, and a set of equally many (m − 1) triangles to the left of l1 , thus keeping complexity 4n − 2. We thus have proven: Theorem 64. Any affine arrangement of n > 3 lines in general position has outer layer complexity at least 4n − 2 and this bound is achievable for every such n. So we see that the minimum outer layer complexity for arrangements in general position is 4n − 2 while by the classical maximum complexity results for the zone of a line, the maximum complexity is 5.5n + O(1). An interesting further question is what the expected complexity is, given a suitable distribution on the space of lines, e.g. lines chosen uniformly as duals from the set of points in the unit square or circle. As noted earlier, the expected number of extreme points in the convex hull of the intersection points of such random line arrangements is bounded by a constant for all n. These facts, together with consideration of the 4n − 2 argument and Figure 3.19 might lead one to conjecture that: (1) Any arrangement of n lines in general position such that there are only 3 extreme points in the convex hull of the intersection points, has outer layer complexity 4n − 2. (2) Any arrangement of n lines in general position such that there are only k extreme points in the convex hull of the intersection points, has outer layer complexity at most 4n + f (k) = 4n + O(1). However, neither of these conjectures actually hold. Figure 3.20 gives an example of 6 lines with 3 extreme points in the convex hull of the intersection points, with outer layer complexity 24 - hence providing a counterexample to the first conjecture. Furthermore, looking at Figure 3.21. we see that we can begin with Figure 3.20 and add a line at a time in such a way that we keep just 3 vertices in the convex hull, and with

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Figure 3.20: An example of 6 lines with 3 extreme points in the convex hull of the intersection points, with outer layer complexity 24 and so the first conjecture is false. each line addition we add 5 edges to the outer layer complexity, thus yielding outer layer complexity 5n + O(1). Questions of maximum and minimum complexity can also be asked in higher dimensions, where, for example in three dimensions, complexity is measured in terms of the number of bounding faces. More generally the outer layer complexity of an arrangements of hyperplanes in d-dimensional affine space is the number of (d − 1)-dimensional faces bounding the unbounded cells5 . When considering minimality questions, we rule out the two “trivial” arrangements (1) where all hyperplanes intersect in a common (d−2)-dimension flat and (2) where all hyperplanes are parallel. With these preliminaries, at Snowbird, 2006, I identified the following problems: Problem 1: In an arbitrary affine arrangement of n lines, what is the maximum outer layer complexity? In an affine arrangement of n hyperplanes in dimension d what is the maximum outer layer complexity? As noted earlier in this chapter, the answer to the question of the maximum complexity of the outer layer, for lines in general position, is answered by the theorems giving best possible constants in the Zone Theorem. In particular, Bern, et al. [5] showed the maximum complexity to be at most b5.5nc − 4, and that this answer is tight up to an additive constant. In dimension d the complexity is O(nd−1 ) though the best known constant, due 5

Or if we want to get fancy, the number of k-dimensional faces for any 0 ≤ k ≤ d − 1. This distinction is only important for dimension d ≥ 3.

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ln

l

F

Figure 3.21: Adding one line at a time, beginning with the dashed line, ln , to Figure 3.20, in such a way as to add 5 edges to the outer layer complexity. ln+1 can be added between lines ln , l at the top while adding an edge along ln+1 to the unbounded 5-gon F , making it an unbounded 6-gon. This process can be continued to give an arrangement with just 3 extreme points in the convex hull of the intersection points and total outer layer complexity 5n + O(1). to Edelsbrunner, et al. [26], has a more complicated expression and is not known to be tight. If we don’t make a general position assumption, however, our definition of outer layer complexity differs from the intuitive notion of the complexity of the zone of a hyperplane. For simplicity, in two dimensions, if we have an arrangement with parallel lines, add a line l at infinity and rotate the arrangement so that l becomes a finite line in the affine plane, we are faced with the circumstance that vertices of the arrangement lie on the line l and we must count the edges of all faces associated with such vertices in the complexity. However, the edges emanating from the vertex are counted just once in the outer layer complexity, but twice, once for each face they bound, in the zone complexity. It appears to be unknown how removal of the general position assumption alters the conclusion of the Zone Theorem. Furthermore, since the zone of a line and the outer layer complexity differ in this case, the question of maximum outer layer complexity is an independent problem and likewise open. Problem 2: In a non-trivial affine arrangement of n hyperplanes in dimension d, not all orthogonal to a single additional hyperplane, what is the minimum outer layer complexity? Problem 3: In an affine arrangement of n hyperplanes in dimension d, in general position (i.e. any d hyperplanes intersect in exactly one point, but no d + 1 hyperplanes intersect in a point), what is the minimum outer layer complexity?

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As we have seen, in two dimensions, the minimum complexity problems are not too difficult - each line contributes two unbounded edges to the complexity and the analysis proceeds by considering how many edges may fill the 2n “slots” between successive unbounded edges as one walks, say, counter-clockwise around the outer layer. However, minimum complexity questions appear to be considerably harder in higher dimensions where there is not the simple notion of a fixed number of slots between unbounded faces. The condition that all hyperplanes not be orthogonal to a fixed additional hyperplane is needed in Problem 2 so that the problem in dimension d does not reduce to that in dimension d − 1. None of our arguments have actually required a notion of straightness. Therefore, some of the problems and results can be phrased in terms of pseudolines and pseudoplanes. For example, the 4n − 2 lower bound on complexity for lines in general position, can be restated in terms of the lower bound on the complexity of the zone of an n + 1st pseudoline amidst a collection of n + 1 pseudolines, all of which are in general position.

Chapter 4

Opposite Quadrant Depth Let S be a set of n points in the plane. Given any point p, the horizontal and vertical lines through p define four closed quadrants NW(p), NE(p), SE(p), and SW(p). We define the opposite-quadrant depth opp(p, S) of a point p with respect to S as the largest number k such that there are two opposite quadrants NW(p) and SE(p) or NE(p) and SW(p), each containing at least k elements of S. (See the concluding remarks for some motivation.) In notation: opp(p, S) = max (min (|NW(p) ∩ S|, |SE(p) ∩ S|) , min (|NE(p) ∩ S|, |SW(p) ∩ S|)) , opp(S) = max opp(p, S). p∈R2

A point p is said to be α-deep, for some α > 0, if its opposite-quadrant depth is at least αn. A (not necessarily unique) point of maximum opposite-quadrant depth is called a deepest point, and a point which is α-deep is also called an α-centerpoint. If we do not require the centerpoint to be in S, then there always exists a point that is 41 -deep at the intersection of the vertical and horizontal halving lines. In fact,

1 4

is the best general bound that can

be established on the maximum depth for any point in the plane, as shown by a set of n points uniformly distributed in a square, a disk, or even on a circle. There are also sets with deeper points, e.g., any set of n collinear points has a point of depth dn/2e. In what follows, we require the centerpoint to belong to the set S. In this case, we cannot guarantee that there always exists a 14 -deep point in S. However, we can show that S always contains a 81 -centerpoint. Theorem 65. Any set of n points in the plane has an element with opposite-quadrant depth 141

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at least bn/8c. Moreover, such a point can be found in O(n) time. The proof is inspired by the “catline” of Hubert and Rousseuw [44]. Somewhat surprisingly, Theorem 65 is essentially tight. Theorem 66. For infinitely many values of n, there are n-element sets Sn with maxp∈Sn opp(p, Sn ) = (n + 4)/8. If the points of S are in convex position, that is, if they form the vertex set of a convex polygon, we can also guarantee the existence of a 14 -centerpoint in S. In fact, this statement remains true under the weaker condition that S is an exposed set, that is, for any p ∈ S, at least one of the four quadrants NW(p), NE(p), SE(p), and SW(p) contains no element of S in its interior. With a little abuse of terminology, in this case we say that such a quadrant is empty (although it necessarily contains the point p and may perhaps contain some other boundary points that belong to S). Theorem 67. Any exposed set of n points in the plane has an element with oppositequadrant depth at least bn/4c. Moreover, such a point can be found in O(n) time. The above notions are closely related to half-space depth, multivariate regression depth, and other measures of statistical depth [55]. For instance, the half-space depth of a point p with respect to an n-element set S in d-space is the minimum number of points of S covered by a closed half-space containing p. It is well-known that one can always find a n c with respect to S [24]. In sharp contrast to the point of half-space depth at least b d+1

opposite-quadrant depth, however, now we cannot guarantee that S has an element of large half-space depth. Indeed, if the points of S are in strictly convex position, none of them has half-space depth larger than one.

4.1

Lower Bound Proofs

First, notice that having points on the same vertical or horizontal line does not create a problem, since the quadrants were defined to be closed. In fact, we can slightly perturb our original point set S by assigning to each element pi ∈ S a vector vi = (i, i2 ). Then there exists a sufficiently small ε > 0 such that the modified sets Sλ = {pi + λvi } are in general position for any λ ∈ (0, ε], in the sense that no pair of points have the same x or y coordinate. Varying λ in this small interval will not change the horizontal or vertical order of the elements of Sλ . If our theorems hold for sets in general position, we can apply them to Sε , and find an α-centerpoint pi + εvi . Then pi + λvi is α-deep with respect to Sλ for all

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Figure 4.1: First split the point set into three parts by two vertical lines not passing through any of the elements of S, such that in the left and right half planes there are precisely n/4 points and in the middle strip n/2 points. Then find a horizontal line, again not passing through any point of S, that divides the points of S into two even groups. We obtain the 3 × 2 grid of cells as given above. λ ∈ (0, ε]. Bringing back our points to their original positions in S only adds points on the boundary of the two opposite quadrants of pi . Therefore, pi is also an α-centerpoint for S. Thus, from now on we assume that S is in general position. Also, for the purposes of establishing Theorem 65 we may assume that n = |S| is divisible by 8. The proof of Theorem 65 proceeds by first splitting the set into three parts by two vertical lines, not passing through any element of S, such that in the left and right half-planes there are precisely n/4 points and the vertical strip in the middle contains n/2 points. Next, we find a horizontal line, not passing through any point of S, that divides S into two equal halves. In this way, we obtain a 3 × 2 “grid,” whose six regions will be called “cells.” See Figure 4.1 We say that a cell is heavy if it contains at least as many points as the other cell in the same column, that is, the cell below or above it. Note that if every cell in a row is heavy, then each of them contains precisely as many elements as the cell below or above it. Otherwise, there could not be n/2 points in the other row. Thus, in this case, all six cells must be heavy. Therefore, we can always pick one heavy cell per column so that not all of them belong to the same row. We distinguish two cases up to symmetries (reflections about the coordinate axes). (a) Bitonic configuration: The upper cells H1 and H3 of the first and third columns, respectively, and the lower cell H2 of the second column are all heavy. (See Figure 4.2(a).) (b) Monotone configuration: The upper cells H1 and H2 of the first two columns and the lower cell H3 of the third column are all heavy. (See Figure 4.2(b).) First we prove Theorem 65 for bitonic configurations. Let p be the highest point of the

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Figure 4.2: Proof of Theorem 65. We let H10 be the lower cell in the left column, H11 := H1 ∩ NW(p), H12 := H1 ∩ SW(p), H21 := H2 ∩ NW(p), and H22 := H2 ∩ NE(p). Case (a): H1 , H3 , and one of H21 or H22 have ≥ n/8 points. Case (b): H3 and one of NW(p) or SW(p) have at least n/8 points. If NW(p) does not, both SW(p) and H22 have at least n/8 points. lower (heavy) cell H2 of the middle column. The quadrants NW(p) and NE(p) contain the heavy cells H1 and H3 , therefore both of them have at least n/8 points of S. All the at least n/4 points in the heavy cell H2 that belong to S lie in the half-plane SW(p) ∪ SE(p). Thus, at least one of the two quadrants, SW(p) or SE(p), must contain at least n/8 points. This quadrant together with the opposite one thus each contain n/8 points, concluding the proof for bitonic configurations. Next we establish Theorem 65 for monotone configurations. We have to use a slightly different argument, because we have no information about the relative vertical order of the points of H1 and H2 . Choose p to be the lowest point in S that belongs to H2 , the upper cell in the middle column. Since SE(p) contains the heavy cell H3 , it has at least n/8 points of S. If NW(p), the quadrant opposite to SE(p), also has at least n/8 points, we are done. Otherwise, the number of points in SW(p) must exceed n/8, because NW(p) ∪ SW(p) covers the n/4 points in the first column. Moreover, if NW(p) has fewer than n/8 points, then there must be more than n/8 points in NE(p) ∩ H2 . Therefore, in this case both quadrants SW(p) and NE(p) have n/8 points. All of the above steps can be performed in linear time, first using a linear-time selection algorithm[16] to find the vertical lines with the properties described in the proof above, and the halving horizontal line, and finally performing several passes through the points to determine whether one has a bitonic or monotone configuration and to find the highest and lowest point in the appropriate cell. Hence one can find a point of depth at least n/8 from an arbitrary point set in O(n) time. This completes the proof of Theorem 65. Now we prove Theorem 67. We use the same configurations as for Theorem 65 (see

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Figure 4.3: Proof of Theorem 67 when all points belong to the convex curve shown. For each i = 1, 2, 3, we let Hi0 be the complement of Hi in column i. Case (a): if SE(p) is empty, then SW(p) contains all points in H2 and NE(p) contains all points in the last column. Case (b): if NE(p) is empty, then NW(p) contains all the ≥ n/4 points of H2 , and SE(p) ⊇ H3 ∪ H30 contains all the n/4 points of the right column; (c) if NW(p) is empty, then the situation is symmetric; and (d) if SW(p) is empty, then NW(p) contains all the n/4 points in the left column and SE(p) contains all the n/2 points in the bottom row. Figure 4.3), but in this case, without loss of generality, assume that n = |S| is divisible by 4. We want to use the assumption that every point p ∈ S is “exposed,” that is, one of the four quadrants determined by p is “empty.” We first prove Theorem 67 for bitonic configurations. As in the proof of Theorem 65, let p ∈ S be the highest point in H2 . Observe that neither NE(p) nor NW(p) can be empty since each of them contains a heavy cell (H1 or H3 ). Suppose without loss of generality that SE(p) is empty (see Figure 4.3(a)). Then p is not only the highest, but also the rightmost point in its cell. Hence, SW(p) contains all at least n/4 points in H2 , while NE(p) contains all n/4 elements of the last column. It remains to establish Theorem 67 for monotone configurations. Again, let p be the lowest element of S in H2 . Now one of the four quadrants at p must be empty. Observe

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that SE(p) covers the heavy cell H3 , so it is not empty. If N E(p) is empty, then N W (p) contains all the at least n/4 points in H2 , and SE(p) contains all the n/4 points of S in the third column, so we are done (Figure 4.3(b)). By symmetry, if N W (p) is empty, both NE(p) and SW(p) contain at least n/4 points (Figure 4.3(c)). Finally, if SW (p) is empty, then N W (p) contains all the n/4 points of S in the first column, and SE(p) contains all the n/2 points of S in the second row, so we are also done in this case (Figure 4.3(d)). The running time analysis for finding such a point p of depth at least n/4 is the same as for Theorem 65. This completes the proof of Theorem 67.

4.2

Establishing Hard Lower Bounds

Let ε be a sufficiently small positive number, let θ be the counterclockwise rotation by π 2

radians about the origin, and let S1 denote the set consisting of the following four points: T := (−2ε, 1), L := (−1, −2ε),

B := (2ε, −1),

R := (1, 2ε).

Setting T1 = {T }, L1 = {L}, B1 = {B}, R1 = {R}, we have S1 = T1 ∪ L1 ∪ B1 ∪ R1 . Now let k > 1, and suppose that Tk−1 , Lk−1 , Bk−1 , Rk−1 , and Sk−1 = Tk−1 ∪ Lk−1 ∪ Bk−1 ∪ Rk−1 have already been defined. Let Tk := Tk−1 ⊕ (εk ∆), Lk := Lk−1 ⊕ (εk θ(∆)), Bk := Bk−1 ⊕ (εk θ2 (∆)), Rk := Rk−1 ⊕ (εk θ3 (∆)), where ∆ = {L, B, R} and ⊕ stands for the Minkowski sum. Finally, define Sk := Tk ∪ Lk ∪ Bk ∪ Rk . See Fig. 4.4. Obviously, we have |Sk | = 4 · 3k−1 , and the set Sk is invariant under the rotation θ. Moreover, and this will be useful below, the difference in x- or y-coordinate of any two distinct points in Sk is at least 4εk . Clearly, opp(p, S1 ) = 1 holds for every p ∈ S1 . It is easy to show by induction on k that |NW(p) ∩ Tk | ≤

3k−1 + 1 , 2

for every p ∈ Tk . For k = 1 there is nothing to prove. Let k > 1, and suppose that we have already verified the statement for k − 1. Write p ∈ Tk in the form q ⊕ (εk r), for some

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Figure 4.4: The construction for the proof of Theorem 66, with ε = 0.25. The set S1 of four points (right), and the set S2 of 12 points has opposite-quadrant depth 2, shown with a pair of opposite quadrants with 2 and 4 points, respectively. q ∈ Tk−1 and r ∈ ∆. Using the induction hypothesis, we obtain that the interior of NW(q) contains at most

3k−2 +1 2

− 1 points of Tk−1 , each giving rise to precisely three points of Tk .

Moreover, these points are at least 4εk−1 − εk > εk away from the boundary of N W (q) in either x- or y-direction, and hence must also belong to the interior of NW(p). (Note that by construction, the boundaries of NW(q) and NW(p) are at distance at most εk in either x- or y-direction.) In addition, by the construction of ∆, the quadrant NW(p) contains at most two elements of the triple {q ⊕ (εk r) | r ∈ ∆}. That is, we have µ |NW(p) ∩ Tk | ≤ 3

¶ 3k−1 + 1 3k−2 + 1 −1 +2= , 2 2

as required. Finally, equality is attained by the point p = tk ∈ Tk defined recursively by t1 := T and tk := tk−1 ⊕ εk B. The same reasoning establishes the same bound on |NE(p) ∩ Tk |, because again N E(p) contains at most two elements of the triple {q ⊕ (εk r) | r ∈ ∆}. And again, equality is attained for p = tk , which implies that opp(tk , Sk ) =

3k−1 +1 . 2

The cases p ∈ Lk , Bk and Rk are symmetric. Hence, both bounds and the value of opp(tk , Sk ) imply that opp(Sk ) =

3k−1 + 1 |Sk | + 4 = , 2 8

for every k ≥ 1, which completes the proof of Theorem 66.

4.3

Concluding remarks

The notion of opposite-quadrant depth arises in connection with the following question related to conflict-free coloring problems [29, 42]. Take n points in the plane in general

148

position and connect two of them by an edge if the smallest axis-parallel rectangle containing them covers no other point of the set. Is it true that this graph contains a large independent set? Let f (n) denote the size of the largest independent set whose existence can always be guaranteed and suppose for a moment that there exists a point p whose opposite-quadrant depth is n/4. That is, NW(p) and SE(p), say, both contain at least n/4 points. The point p guarantees that no pair of points lying in opposite quadrants are connected by an edge. Thus, after recursively finding independents sets in the subgraphs induced by NW(p) and SE(p), their union is always an independent set in the whole graph. Thus, we √ would have f (n) ≥ 2f (n/4) − 1, implying that f (n) = Ω( n). Unfortunately, Theorem 65 yields only f (n) = Ω(n1/3 ), and, by Theorem 66, this method cannot lead to a better bound. Nevertheless, by an easy application of a lemma of Erd˝os and Szekeres, we obtain √ f (n) = Ω( n), and this bound can be slightly improved [65]. It is not known whether f (n) = o(n). The natural extension of opposite-quadrant depth to opposite-orthant depth in three dimensions seems to lead to interesting questions. Our approach does not work in this case, since the opposite-orthant depth of the intersection point of three halving planes can be zero. (For instance, place n/4 points in small neighborhoods of the points (1, 1, 1), (1, −1, −1), (−1, 1, −1) and (−1, −1, 1).) An alternate extension of opp(p, S) to three dimensions is maxσ,τ min(|S ∩ σ|, |S ∩ τ |), over all pairs σ, τ of opposite quadrants determined by a pair of axis-aligned planes. There are now six pairs of opposite quadrants. With this definition, all planar lower bounds remain valid in three dimensions, but they are not necessarily tight.

4.4

Recent developments

Rom Pinchasi and Stefan Felsner have recently found more enlightening proofs of the n/8 (arbitrary point set) and n/4 (exposed point set) bounds [63]. Their argument for the n/8 case runs as follows: Divide the points into four groups, N consisting of the top-most b n4 c points, E consisting of the right-most b n4 c points, S consisting of the bottom-most b n4 c points, and W consisting of the left-most b n4 c points. If there are remaining points choose any such point p and place axis-parallel lines through p. Then we have |N E(p)∪N W (p)| ≥

n n n n , |N W (p)∪SW (p)| ≥ , |SW (p)∪SE(p)| ≥ , |SE(p)∪N E(p)| ≥ . 4 4 4 4 (4.4.1)

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If, however, p had opposite quadrant depth
1) dependence of the cost on the radius; in fact, physically accurate simulation often requires superquadratic dependence (α > 2). A quadratic dependence (α = 2) models the total area of the served region, leading to an objective function arising in some applications. A linear dependence (α = 1) is sometimes assumed, as was done by Lev-Tov and Peleg [52], who study the base station coverage problem, minimizing the sum of radii. The linear case is important to study not only in order to simplify the problem and gain insight into the general problem, but also to address those settings in which the linear cost model naturally arises [15, 66]. For example, the model may be appropriate for a system with a narrow-angle beam whose direction can either rotate continuously or adapt to the needs of the network. Another motivation comes from robotics, in which a robot is to map or scan an environment with a laser scanner [31, 30]: For a fixed spatial resolution of the desired map, the time it takes to scan a circle corresponds to the number of points on the perimeter, i.e., is proportional to the radius. 150

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Our problem is a type of clustering problem, recently named min-size k-clustering by Bil`o et al. [6]. Clustering problems tend to be NP-hard, so most efforts are aimed at devising an approximation algorithm or a polynomial-time approximation scheme (PTAS). We also introduce a new problem, which we call minimum cost covering tour (MCCT), in which we combine the problem of finding a short tour and placing covering disks centered along the tour. The objective is to minimize a linear combination of the tour length and the transmission/covering costs. The problem arises in the autonomous robot scanning problem [31, 30], where the covering cost is linear in the radii of the disks, and the overall objective is to minimize the total time of acquisition (a linear combination of distance travelled and sum of scan radii). Another motivation is the distribution of a valuable or sensitive resource: There is a trade-off between the cost of broadcasting from a central location (thus wasting transmission or risking interception) and the cost of travelling to broadcast more locally, thereby reducing broadcast costs but incurring travel costs. Location Constraints.

In the absence of constraints on the server locations, it may be

optimal to place one server at each demand point. Thus, we generally set an upper bound, k, on the number of servers, or we restrict the possible locations of the servers. Here, we consider two cases of location constraints: (1) Servers are restricted to lie in a discrete set {t1 , . . . , tm }; or (2) Servers are constrained to lie on a line (which may be fixed, or may be selected from a family of lines as part of the optimization problem). In case servers are constrained to lie in a discrete set we call the problem itself discrete, otherwise we say that the problem is continuous. Our results.

We provide a number of new results, some improving on previous work,

some giving the first results of their kind. We consider a number of variants of the problem. On the one hand, the most general problem we would like to solve is that of bounded capacity (at most k servers), fixed additive cost per transmission, with geometric constraints on server locations (subset of discrete set, along a given line, or unconstrained), with combined tour/transmission costs (as in the MCCT problem described above), and with arbitrary nondecreasing cost function. On the other hand, such a general problem is clearly intractable by reduction to geometric TSP. Thus we study a number of special cases in order to gain some insight into the more general problem. In the discrete case studied by Lev-Tov and Peleg [52], and Bil´o et al. [6], we give improved results. For the discrete 1D problem where Y = {p1 , ..., pn } ⊂ R, Lev-Tov and Peleg gave an exact O((n + m)3 ) solution for the case α = 1. Zolotarev [77] observed that the algorithm also works for arbitrary α > 0, and more surprisingly, that the algorithm

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also works in 1.5D, i.e. where the servers must lie along the x-axis, but the clients can be anywhere in the plane. We observe that the Lev-Tov and Peleg algorithm as well as Zolotarev’s extension to 1.5D can actually be made to run in O(n2 m) time, and in 1D we give an improved algorithm that runs in time O(n2 +nm). In the 1D problem, for α = 1, Lev-Tov and Peleg also gave a simple linear-time 4-approximation, which we improve on by giving a linear-time 3-approximation and a near-linear-time 2-approximation. Refer to Zolotarev’s thesis [77] for an interesting study of the performance of the various 1D approximation algorithms under suitable distribution assumptions on the servers and clients. In the general discrete 2D case with clients Y ⊂ R2 and α ≥ 2, Bil´o et al. [6] showed the problem to be NP-Hard and gave a PTAS. In [2] we show that the problem is already NP-Hard for any superlinear cost function, i.e., f (r) = rα with α > 1. Furthermore, we generalize the problem in two new directions. If we restrict the servers to lie on a given fixed line (the “1.5D problem”), we give a dynamic programming algorithm that solves the problem exactly, in time O(n2 log n) for any Lp metric in the linear cost case, and apply Zolotarev’s solution to the 1.5D discrete problem to get an O(n4 ) algorithm in the case of superlinear cost functions. For simple approximations, an algorithm we call “Square Greedy” (SG) gives an O(n log n) 3-approximation to the square covering problem (L∞ metric) with any linear or superlinear cost function. A small variation, “Square Greedy with Growth” (SGG), gives a 2-approximation for a linear cost function, also in time O(n log n). The results are also valid for covering by Lp disks for any p, but with correspondingly coarser approximation factors. A practical example in which servers are restricted to lie along a line is that of a highway which cuts through a piece of land, and server locations which are restricted to lie along the highway. The line location problem arises when one not only needs to locate the servers, but also needs to select an optimal corridor for the placement of the highway. If the servers are restricted to lie on a horizontal line, but the location of this line may be chosen freely, then we show that the exact optimal position (with α = 1) is not computable by radicals, using an approach similar to that of Bajaj [3, 4] in addressing the unsolvability of the Fermat-Weber problem. On the positive side, we give a fully polynomialtime approximation scheme (FPTAS) requiring time O((n3 /ε) log n) if α = 1 and time O((n4 /ε) log n) if α > 1. For servers on an unrestricted line, of any slope, and α = 1, we give O(1)-approximations √ (4-approximation in O(n4 log n) time, or 8 2-approximation in O(n3 log n) time) and an FPTAS requiring time O((n5 /ε2 ) log n). We also give the first algorithmic results for the new problem, minimum cost covering

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tour (MCCT), which we introduce: Given a set Y ⊆ R2 of n clients, our goal is to determine a polygonal tour T and a set X of k disks of radii rj centered at points along T that cover P Y while minimizing the cost length(T ) + C riα . Our results are for α = 1. The ratio C represents the relative cost of touring versus transmitting. We show that MCCT is NP-hard if C is part of the input. At one extreme, if C is small then the optimum solution is a single server placed at the circumcenter of Y (we can show this to be the case for C ≤ 4). At the other extreme (if C very large), the optimum solution is a TSP among the clients. For any fixed value of C > 4, a PTAS for MCCT can be obtained using an extension of Mitchell’s m-guillotine method. See [2] for details.

5.2 5.2.1

Scenario 1: Server Locations Restricted to a Discrete Set Exact solution to the 1D and 1.5D problems

We begin by studying the following 1D and 1.5D problems: Given a set of m server locations {t1 , ..., tm }, lying along the x-axis and specified in increasing order, and n client locations {p1 , ..., pn }, also specified in increasing order along the x-axis and either located themselves on the x-axis (1D), or in the plane (1.5D), and a cost of broadcasting to client locations f (r) which is a function of the broadcast radius, we seek to determine the servers to broadcast from, and radii to broadcast to, so as to minimize the total cost of reaching all clients. For cost functions we consider separately that of linear and superlinear cost, and since constants and lower order terms are irrelevant to order of magnitude analysis we consider f (r) = rα for α = 1 or more generally for arbitrary α > 0. Whether in 1D or 1.5D we say covering by disks or circles, despite the fact that 1D “disks” are really intervals. We distinguish the case in 1.5D where the server locations are pre-specified along the x-axis from the case where we are free to pick the server locations anywhere we chose, though still along the x-axis. We refer to the former as the discrete problem, and the latter as the continuous problem. We start with the 1D problem. The Lev-Tov and Peleg algorithm for this problem [52] works by filling out a matrix Cost[i][j] which gives the minimum cost of covering the first j clients by the first i servers. The matrix is filled out one row at a time. The first row is filled out by noting that Cost[1][j] = max(dist(t1 , p1 )α , dist(t1 , pj )α ), and the remaining rows are filled out by observing that Cost[i][j] is the minimum over (i)

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Cost[i − 1][j], (ii) dist(ti , pj )α + Cost[i − 1][k − 1] where pk is the left-most client captured by the disk of radius dist(ti , pj ) about server ti , and (iii) dist(ti , pl )α + Cost[i − 1][l − 1] for all l such that 1 ≤ l ≤ j − 1 and dist(ti , pl ) > dist(ti , pj ). Because of (iii), Cost[i][j] can take O(n) time to compute, given that we have values of Cost[k][l] for 1 ≤ k ≤ i and 1 ≤ l < j. Since there are nm values to compute in total, the algorithm takes O(n2 m) time. Our 1D solution uses just a singly indexed array. We consider just “pinned” disks, which are disks with center at some server ti and radius extending out to some client pj . Clearly an optimal solution will consist entirely of pinned disks. Call the “owner” of a disk the right-most client contained in the disk. There are at most nm pinned disks, and furthermore the disks can all be stored along with their left and right-most contained client as a set of n arrays, where each array contains the disks for which client pi is owner. We can compute the disks centered at a given server, together with their right-most and left-most clients, in increasing order of their radii, by simply locating where the server is in the sorted list of clients and splitting that list at the server and merging both sublists in linear time. This takes O(n) time per server, hence O(nm) time overall. With this preprocessing the algorithm fills out the singly indexed array Cost[i]. After each iteration through the main loop, Cost[k] for 1 ≤ k ≤ i contains the minimum cost of covering clients up to pk by disks owned by clients up to pi . The pseudocode is given in Figure 1. Min1DCover : for every pinned circle C find the leftmost and rightmost points enclosed by C Cost[0] ← 0 for i ← 1 to n Cost[i] ← ∞ for each pinned circle C owned by pi pj ← leftmost point enclosed by C Cost[i] ← min{Cost[i], Cost[j − 1] + radius(C)α } k←1 while i > k & Cost[i − k] > Cost[i] Cost[i − k] ← Cost[i] k ←k+1 return Cost[n]

Figure 5.1: The 1D dynamic programming algorithm.

The critical portion of this algorithm is the back-tracking steps that take place in the while loop at the end of each iteration. Backtracking is required to maintain the fact not

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just that Cost[i] contains the minimum cost of covering clients up to pi with disks owned by clients up to pi , but also that Cost[k] likewise contains the minimum cost of covering clients up to pk with disks owned by clients up to pi for 1 ≤ k ≤ i. Without backtracking we could be in a position where Cost[i] < Cost[j] even though i > j and we later have to use this erroneous value of Cost[j]. To check that at each step i, the invariants Cost[k], 1 ≤ k ≤ i are properly maintained, we need verify that an optimal solution never contains two disks with the same owner. If C and C 0 were two disks in OPT with the same owner let C be the disk which extends furthest to the left (or either of the two disks if there is a tie). Then C contains all of the clients in C 0 so C 0 is superfluous. It follows that OPT can never contain two disks with the same owner and so Min1DCover is correct. The two steps following for each pinned circle C owned by pi get executed nm times - once for each pinned circle. However, for each of the n iterations of the outer loop, we may have to make up to n backtracking assignments. Hence the total running time is O(nm + n2 ). In 1.5D, the above algorithm cannot be made to work, as is easiest to see for α > 1 where it is possible for two disks in OPT to have the same owner. See Figure 5.2.

Figure 5.2: An example for which OPT contains two disks with the same owner (p3 ) for sufficiently large α. In 1.5D (and arbitrary α ≥ 0) the Zolotarev (Lev-Tov and Peleg) algorithm starts out filling in the first row by: Cost[1][j] = max1≤k≤j dist(t1 , pk )α = max(Cost[1][j − 1], dist(t1 , pj )α ), and then for i ≥ 1, to fill in Cost[i][j] we take the min over (i) Cost[i−1][j], (ii) dist(ti , pj )α + Cost[i − 1][k] where pk is the right-most client to the left of pj not captured by the disk of radius dist(ti , pj ) about server ti , and (iii) dist(ti , pl )α + Cost[i − 1][kl ] for all l such that 1 ≤ l ≤ j − 1 and dist(ti , pl ) > dist(ti , pj ) where pkl is the right-most client to the left of pj not captured by the disk of radius dist(ti , pl ) about server ti . There is some work required

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to show that this algorithm is correct and that it can still run in time O(n2 m), the same time as 1D Lev-Tov and Peleg. A naive implementation would require an additional O(n) time in the inner loop to determine “the right-most client to the left of pj not captured by the disk of radius dist(ti , pl ) about server ti .” To prove the correctness of Zolotarev’s algorithm we must verify that having placed a disk D centered at ti covering pj with rightmost uncovered client, pk , to the left of pj , the cost of covering all clients up to pk with servers up to ti−1 is no more than the cost of covering all clients up to pk minus any clients that may have been covered by D with servers up to ti−1 . There are two cases (i) where pk is to the right or even with ti and (ii) where pk is to the left of ti . In case (i), illustrated in Figure 5.3, any disk D0 with center to the left of ti

Figure 5.3: The disk D centered at ti covers pj with rightmost uncovered client, pk , to the left of pj . pj is not necessarily pinned to D. The case where pk lies to the right of, or is even with, ti . which contains pk either cuts D to the right of pk or wholly contains D. In either case it contains any clients covered by D to the left of pk . Hence in this case the cost of covering all clients up to pk with servers up to ti−1 is the same as the cost of covering all clients up to pk minus any clients that may have been covered by D with servers up to ti−1 . Case (ii), illustrated in Figure 5.4, is similar. If the center of the disk D0 covering pk is to the left of pk then D0 either cuts D to the right of pk , or completely contains D so contains all clients to the left of pk in D. If, on the other hand, the center of D0 is to the right of, or even with, pk (but to the left of ti ), then the same thing is true, i.e. D0 either cuts D to the right of pk , or completely contains D. For if D0 actually cut D to the left of pk then the radius of D0 would be less than the radius D meaning that D0 would also have to cut D to the right of pk which is impossible. It follows that again D0 contains all clients to the left of pk contained in D. The correctness of Zolotarev’s algorithm follows. To guarantee that Zolotarev’s algorithm runs in O(n2 m) time we must spend, on average,

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Figure 5.4: The disk D centered at ti covers pj with rightmost uncovered client, pk , to the left of pj . pj is not necessarily pinned to D. The case where pk lies to the left of ti . a constant amount of time determining the right-most client left of pj that is not covered by a given disk centered at ti covering pj . To do this, for each server, we keep a list of disks centered at the server, sorted by radius, and for each record the pinning client. For a given server (assumed to be ti ) we then process the disks for step (iii) of the algorithm in order of increasing size. We also keep a list of the clients p1 , ..., pj−1 together with the associated distances d(ti , p1 ), ..., d(ti , pj ), which we cross off right to left in one pass as we process successive disks. At each step the client furthest to the right in the list which is not crossed off will be the client we are looking for. For the first disk, with center ti and radius dist(ti , pj ) we cross off entries pj−1 , pj−2 , ... until we get to one that is further than dist(ti , pj ) from ti . We then choose the next biggest disk centered at ti , cross off additional entries (if any), etc. We thus spend a total of O(n) time processing all O(n) candidate disks in step (iii) and so the algorithm runs in the claimed (n2 m) time.

5.2.2

Approximating the one-dimensional discrete problem with linear cost

Once again we consider the one-dimensional problem with m fixed server locations X = {t1 , ..., tm } and n client locations Y = {p1 , ..., pn }, where each set is assumed to be sorted in increasing order along the x-axis. We consider approximations in the linear (α = 1) cost case. In [53] Lev-Tov and Peleg give a simple “closest center” algorithm (CC) that gives a linear-time 4-approximation. We improve to a 3-approximation in linear time, and a 2-approximation in O(m + n log m) time.

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5.2.2.1

Lev-Tov and Peleg’s closest center (CC) algorithm

The Lev-Tov and Peleg closest center (CC) algorithm simply associates with each client pi the closest server tj , and says to use a disk centered at tj of radius |tj − pi | to cover pi . Though a minor point, without a subsequent “cleaning phase” we note that the CC algorithm is not in fact a four, or even a constant factor, approximation, as one can see by considering the case of a single server t and clients {pi } with t < p1 < . . . < pn and the pi all very close to pn . After finding closest centers, one must take another pass through the sorted set of clients and servers to throw away all but the biggest disk centered at a given server. Another way to do this is to throw away clients whose closest center disk is contained in another disk with the same center point and bigger radius. Refer to this algorithm as Closest Center with Cleaning or CCC. For each pi left after cleaning, denote by tˆi ∈ {t1 , ..., tm } the closest center to pi . Further, denote by Ii the interval [tˆi , pi ] if tˆi ≤ pi or [pi , tˆi ] if pi < tˆi . We aim to prove that CCC is a 4-approximation to OPT. We require the following intermediate result, used in the Lev-Tov and Peleg proof, but which is false without cleaning: Lemma 68. In CCC, if i 6= j then Ii ∩ Ij = ∅. Proof. If tˆj ∈ Ii then tˆi would not be the closest center to pi . On the other hand if pj ∈ Ii but tˆj ∈ / Ii then pi must lie between pj and tˆj . But then, since tˆi is closer to pi then tˆj it must also be that tˆi is closer to pj thean tˆj , a contradiction. Lemma 69. CCC yields a 4-approximation to OPT in O(n + m) time. Proof. The proof of the runtime is immediate from the description of the algorithm, if the sets X and Y are sorted. Consider any disk D in OPT. Associate a disk with center tˆi and endpoint pi in CCC to D if pi ∈ D. Since all points pj must be covered by some disk in OPT, in this way each disk of CCC is associated with one or more disks of OPT. It suffices to show that the sum of the radii of disks in CCC associated with the arbitrary fixed disk D in OPT is no more than 4 times the radius of D. Of the disks Ci in CCC associated with D there are associated intervals Ii . These Ii may extend at most radius(D) to the left of D and radius(D) to the right of D, since the closest center to a client y ∈ D is at least as close as center(D). Thus the Ii must all fit in an interval that is at most 4 P P radius(D) wide. Since Ii = radius(Ci ) the lemma follows. The Lev-Tov and Peleg example of four disks that can be replaced by one in Figure 5.5 shows that the constant factor 4 is best possible for CCC.

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Figure 5.5: Four closest center disks of the same radius that can be replaced with a single disk of slightly greater radius (the dashed circle) capturing all clients, {p1 , ..., p4 } and centered at t2 or alternatively t3 . 5.2.2.2

A 3-approximation closest-center with growth (CCG) algorithm

We now describe an algorithm which also runs in linear time, but achieves an approximation factor of 3. Closest Center with Growth (CCG) Algorithm: Process the clients {p1 , ..., pn } from left to right. We will either cover pi (if not already covered) with its closest center or by extending an existing disk just enough to cover pi , whichever is cheaper. So, as we process each client, keep track of the rightmost extending disk. Let ωR denote the rightmost point of the rightmost extending disk, and let R denote the radius of this disk. (In fact the rightmost extending disk will always be the last disk placed.) If ωR is equal to, or to the right of the next client processed, pi , then pi is already covered so ignore it and proceed to the next client. If pi is not yet covered, consider the distance of pi to ωR compared with the distance of pi to its closest center tˆi . If the distance of pi to ωR is less than or equal to the distance of pi to tˆi , then grow the rightmost extending disk just enough to capture pi . Otherwise use the disk centered at tˆi of radius |pi − tˆi | to cover pi . Lemma 70. For α = 1, CCG yields a 3-approximation to OPT in O(n + m) time. Proof. Again the proof of the runtime immediately follows from the description. Consider any disk D in OPT. We attribute to each client a segment Ji as follows. If, in the execution of CCG, the client pi was not used because it had already been covered, we set Ji = ∅. If pi was captured by placing a disk centered at the closest center, tˆi to pi then set Ji = {[tˆi , pi ] if tˆi ≤ pi , [pi , tˆi ] if pi < tˆi }. On the other hand, if pi was captured by growing an existing disk with initial rightmost point ωR , let Ji denote the half-open interval through which this rightmost point moved out, i.e. Ji = (ωR , pi ]. Observe that Ji ∩ Jj = ∅ as long as i 6= j and that the sum of the lengths of the Ji equals the sum of the radii of disks in the

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Figure 5.6: An example showing that CCG cannot be better than a 3-approximation. The closest center to p1 is t1 , at distance 1 − ². The disk C1 , with center at t1 and initial radius 1 − ², then grows to pick up each of the densely packed clients p2 , ..., pn−1 , where the distance between p1 and pn−1 is again 1 − ². Finally, pn is captured by t3 . The total cost of capturing all clients is then 3 − 3² while, the single dashed disk would have sufficed at cost 1. CCG cover. As earlier, the leftmost and rightmost Ji cannot extend more than radius(D) to the left of D or radius(D) to the right of D. Let tD denote the center of D. At most one Ji corresponding to a client in D extends outward to the right from the right edge of D. If there is no such right-most interval, the we clearly have at most a 3-approximation. Thus assume there is such an interval, and call it JR . JR corresponds to a center, not growth, since it emanates from the right. Call the associated client pR . If there is a client pi to the right of tD not contained in JR then length(JR ) < radius(D) − d(tD , pi ) since otherwise in the algorithm we would have grown the disk containing pi to capture pR , rather than allow it to be captured by a center. It follows that the coverage by disks in CCG to the right of tD has sum of radii at most radius(D). The 3-approximation follows. Figure 5.6 shows that the factor 3 is actually tight. 5.2.2.3

A 2-approximation greedy growth (GG) algorithm

Finally we offer an algorithm that achieves a 2-approximation but runs in time O(m + n log m). Greedy Growth (GG) Algorithm: Start with disks centered at each server and each of radius zero. Now, amongst all clients, find the one which requires the least radial disk growth to capture it. Repeat until all clients are covered. An efficient implementation uses a priority queue to determine the client that should be captured next. One can set up the priority queue in O(m) time. Note that the priority queue will never have more than 2m

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elements, and that each pi eventually gets captured, either from the right or from the left. Each capture can be done in time O(log m) for a total running time of O(m + n log m). Lemma 71. For α = 1, GG yields a 2-approximation to OPT in O(m + n log m) time. Proof. Define intervals Ji as follows: when capturing a client pi from a server tj whose current radius (prior to capture) is rj , let Ji = (tj + rj , pi ] if pi > tj , and Ji = [pi , tj − rj ) otherwise. Our first trivial yet crucial observation is that Ji ∩ Jk = ∅ if i 6= k. Also note that the sum of the lengths of the Ji is equal to the sum of the radii in the GG cover. Consider now a fixed disk D in OPT, centered at tD , and the list of intervals Ji whose pi is inside D. We say that such Ji are associated with D. Every Ji is associated with at least one disk of OPT. As before, at most one of the Ji crosses the right edge of D. If such a Ji exists, call it JR , and define JL symmetrically. If JR exists, it cannot extend more than radius(D) to the right of D. Let λ = length(JR ). We argue, beginning in the next paragraph, that there must then be an interval of length λ in D, to the right of tD , which is free of Ji ’s. As a consequence, there are at most radius(D) worth of segments associated with D to the right of tD . Of course, this would also be true if JR did not exist. By symmetry, there would also be at most radius(D) worth of segments associated with D to the left of tD , whether JL exists or not, yielding the claimed 2-approximation. So assume JR exists. Then the algorithm successively extends JR by growth to the left up to some maximum point (possibly stopping right at pR ). Since the growth could have been induced by clients to the right of JR , that maximum point is not necessarily a client. There is, however, some client inside D that is captured last in this process. This client pi (possibly pR ) cannot be within λ of tD , since otherwise it would have been captured prior to the construction of JR . If there is no client between tD and pi we are done, since then there could be no interval Jk in between. Thus consider the client pi−1 just to the left of pi . Suppose d(pi−1 , pi ) ≥ λ. Then, if pi−1 is eventually captured from the left, we would have the region between pi−1 and pi free of Jk ’s and be done. On the other hand, if pi−1 is captured from the right, it must be captured by a server between pi−1 and pi , and that server is at least λ to the left of pi since otherwise pi would be captured by that server prior to pR . This leaves the distance from the server to pi free of Jk ’s. Hence the only case of concern is if d(pi−1 , pi ) < λ. Clearly pi−1 must not have been captured at the time when pR is captured since otherwise pi would have been captured before pR , contradicting the assumption that pi is captured by growth leftward from pR . Similarly, there cannot be a server between pi−1 and pi , since otherwise both pi−1 and pi would be captured before pR . Together with the definition of pi , this implies that pi−1 is

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captured from the left. Therefore, to the left of pi−1 , there must be one or more intervals {Jli } whose length is at least λ that are constructed before pi−1 is captured. Similarly, to the right of pi , there must be some one or more intervals {Jrj } whose length is at least λ, constructed before pi is captured. However, either the last Jli is placed before the last Jrj or vice versa. In the first case, there are no λ length obstructions left in the left-hand subproblem, so pi−1 will be covered, and with λ length obstructions remaining in the right subproblem, pi will be captured by growth rightward. The second case is symmetrical to the first. In either case we have a contradiction. To see that the factor 2 is tight, just consider servers at −2 + ², 0, and 2 − ², and clients at −1 and 1.

5.2.3

Toward the discrete 2-dimensional case

For the corresponding 2-dimensional discrete problem, where both the clients and candidate server locations are situated in R2 , not much is known. Bil`o et al. [6] showed NP-hardness for α ≥ 2 for either an unrestricted number of servers, or the number of servers restricted to a fixed number k. They also show that the 2D problem has a PTAS for any α ≥ 1, with or without a maximum number k of servers. In [2] we show that the problem is already NP-hard for α > 1. There is hope of finding some form of a closest center/greedy growth approach to obtain a near-linear constant factor approximation in the general 2-dimensional scenario, but we have not yet been able to make it work. One preliminary negative result is that the greedy growth heuristic does not work in the 2D linear cost case, as can be seen by putting servers √ √ on the integer grid [0, n] × [0, n] and clients at the midpoint of horizontal or vertical √ edges: the best solution would be a single circle of radius n, but the greedy growth (with points suitably perturbed) will select n unit circles.

5.3 5.3.1 5.3.1.1

Scenario 2: Server Locations Restricted to a Line Servers along a fixed horizontal line Exact solutions

Suppose that the servers are required to lie on a fixed horizontal line, which we take without loss of generality to be the x-axis, while the clients can lie anywhere in the plane. We call this problem the continuous 1.5D problem.

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MinSumOfRadiusCircleCover(Y ) : for every pinned circle C find the leftmost and rightmost points enclosed by C Cost[0] ← 0 for i ← 1 to n Cost[i] ← ∞ for each pinned circle C owned by pi if no points in P lie directly above C pj ← leftmost point enclosed by C Cost[i] ← min{Cost[i], Cost[j − 1] + radius(C))} return Cost[n]

Figure 5.7: The dynamic programming algorithm for linear cost function. In this section, we describe dynamic programming algorithms to compute a set of server points of minimum total cost. For notational convenience, we assume that the clients Y are indexed in left-to-right order. Without loss of generality, we also assume that all the clients lie on or above the x-axis, and that no two clients have the same x-coordinate. (If a client pi lies directly above another client pj , then any circle enclosing pi also encloses pj , so we can remove pj from Y without changing the optimal cover. A client lying below the x-axis can be moved to a symmetric position above the axis without affecting the optimal solution.) Let us call a circle C pinned if it is the leftmost smallest axis-centered circle enclosing some fixed subset of clients. Equivalently, a circle is pinned if it is the leftmost smallest circle passing through a chosen client or a chosen pair of clients. Under any Lp metric, there are at most O(n2 ) pinned circles. As long as the cost function f is non-decreasing, there is a minimum-cost cover consisting entirely of pinned circles. Linear Cost.

If the cost function f is linear (or sublinear), we easily observe that the

circles in any optimum solution must have disjoint interiors. (If two axis-centered circles of radius ri and rj intersect, they lie in a larger axis-centered circle of radius at most ri + rj .) In this case, we can give a straightforward dynamic programming algorithm that computes the optimum solution under any Lp metric. The algorithm given in Figure 5.7 finds the minimum-cost cover by disjoint pinned circles, where distance is measured using any Lp metric. We call the rightmost point enclosed by any pinned circle C the owner of C. If we use brute force to compute the extreme points enclosed by each pinned circle, and also to test whether any points lie directly above the pinned circles, such an algorithm runs

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in O(n3 ) time. With some more work, however, we can improve the running time by nearly a linear factor. This improvement is easiest in the L∞ metric, in which circles are axis-aligned squares. Each point pi is the owner of exactly i pinned squares: the unique axis-centered square with pi in the upper right corner, and for each point pj to the left of pi , the leftmost smallest axis-centered square with pi and pj on its boundary. We can easily compute all these squares, as well as the leftmost point enclosed by each one, in O(i log i) time. (To simplify the algorithm, we can actually ignore any pinned square whose owner does not lie on its right edge.) If we preprocess P into a priority search tree in O(n log n) time, we can test in O(log n) time whether any client lies directly above a given axis centered square. The overall running time is then O(n2 log n). For any other Lp metric, we can compute the extreme points enclosed by all O(n2 ) pinned Lp -circles in O(n2 ) time using the following duality transformation. If C is an Lp -circle centered at (x, 0) with radius r, let C ∗ be the point (x, r). For each client pi , let p∗i = {C ∗ |C is centered on the x-axis and pi ∈ C}, and let Y ∗ = {p∗i | pi ∈ Y }. We easily verify that each set p∗i is an infinite x-monotone curve. (Specifically, in the Euclidean metric, the dual curves are hyperbolas with asymptotes of slope ±1.) Moreover, any two dual curves p∗i and p∗j intersect exactly once; i.e., Y ∗ is a set of pseudo-lines. Thus, we can compute the arrangement of Y ∗ in O(n2 ) time. For each pinned Lp -circle C, the dual point C ∗ is either one of the clients pi or a vertex of the arrangement of dual curves Y ∗ . A circle C encloses a client pi if and only if the dual point C ∗ lies on or above the dual curve p∗i . After we compute the dual arrangement, it is straightforward to compute the leftmost and rightmost dual curves below every vertex in O(n2 ) time by depth-first search. Finally, to test efficiently whether any points lie directly above an axis-centered Lp -circle, we can use the following two-level data structure. The first level is a binary search tree over the x-coordinates of Y . Each internal node v in this tree corresponds to a canonical vertical slab Sv containing a subset pv of the clients. For each node v, we partition the x-axis into intervals by intersecting it with the Lp -furthest-point Voronoi diagram of pv , in O(|pv | log|pv |) time. To test whether any points lie above a circle, we first find a set of O(log n) disjoint canonical slabs that exactly cover the circle, and then for each slab Sv in this set, we find the furthest neighbor in pv of the center of the circle by binary search. The region above the circle is empty if and only if all O(log n) furthest neighbors are inside the circle. Finally, we can reduce the overall cost of the query from O(log2 n) to O(log n) using fractional cascading. The total preprocessing time is O(n log2 n). Theorem 72. Given n clients in the plane, we can compute in O(n2 log n) time a covering

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by circles (in any fixed Lp metric) centered on the x-axis, such that the sum of the radii is minimized. Superlinear Cost.

To solve the superlinear cost problem we adapt the discrete 1.5D

exact solution described earlier due to Zolotarev that ran in O(n2 m) time. Given a set of n clients in the continuous problem, it is easy to see that any disk in OPT will be determined by either one or two “pinning” clients which it will pass through. Taking all such choices from the n clients leads to O(n2 ) candidate server locations, which we pause to sort by increasing x coordinate in time O(n2 log n). Zolotarev’s discrete algorithm with these servers and the original n clients then yields a solution to the continuous problem in time O(n2 m) = O(n4 ) since here the number of candidate server locations m = n2 . 5.3.1.2

Fast and simple constant factor approximations

In this section we describe simple and inexpensive algorithms that achieve constant factor approximations for finding a minimum-cost cover with disks centered along a fixed horizontal line L, using any Lp metric. The main idea for the proofs of this section is to associate with a given disk D in OPT, a set of disks in the approximate solution and argue that the set of associated disks cannot be more than a given constant factor cover of D, in terms of cumulative edge length, cumulative area, and so forth. As in the previous section, the case of L∞ metric is the easiest to handle. By equivalence of all the Lp metrics, constant-factor c-approximations for squares will extend to constantfactor c0 -approximations for Lp disks. Square Greedy Cover Algorithm (SG): Process the client points in order of decreasing distance from the line L. Find the farthest point p1 from L; cover p1 with a square S1 exactly of the same height as p1 centered at the projection of p1 on L. Remove all points covered by S1 from further consideration and repeat. When two points are precisely the same distance from L, break ties arbitrarily. Obviously, SG computes a valid covering of Y by construction. We begin the analysis with a couple of simple observations. Lemma 73. In an SG covering {Si }, a box Si may not contain the center of another box Sj with i 6= j. Proof. Suppose i < j, in other words Si is placed before Sj . When Si is placed it covers all to-that-point uncovered clients within its width. Thus if the center of Sj were xj , and xj were covered by Si , then xj would have no clients above it after Si is placed - so Sj would

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not have a reason to be placed and could not be part of the cover. On the other hand, if i > j then the center of Si is outside Sj and width(Si ) ≤ width(Sj ) so Si cannot contain the center of Sj . Lemma 74. In the SG covering, any point in the plane (not necessarily a client) cannot be covered by more than two boxes. Proof. Suppose Si and Sj are two squares placed during the running of SG with i < j. By Lemma 73 neither Si nor Sj can contain each other’s midpoint. Now consider a point p ∈ Si ∩ Sj . If p were covered by a third square Sk then either Si or Sj would contain the center of Sk , or Sk would contain the center of one or the other of Si , Sj , neither of which is possible. Theorem 75. Given a set Y of n clients in the plane and any α ≥ 1, SG computes in time O(n log n) a covering of Y by axis-aligned squares centered on the x-axis whose cost is at most three times the optimal. Proof. Let Y = {p1 , . . . , pn } and consider a square S in OPT. We consider those squares {Sij } selected by SG corresponding to points {pij : pij ∈ S}, see Figure 5.8, and argue that S

Si5

Si1 Si2

Si3

Si4

L < s/2

< s/2

Figure 5.8: Squares of the SG algorithm inside a square of the optimal solution. these squares cannot have more than three times the total edge length of S. The same will then follow for all of SG and all of OPT. The argument, without modification, covers the case of cost measured in terms of the sum of edge length raised to an arbitrary positive exponent α ≥ 1. As a result of Lemma 74, at most two boxes Sij associated with points pij ∈ S processed by SG actually protrude outside of S, one on the left and one on the right. Denote by r

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the total horizontal length of these protruding parts of squares, then r ≤ s, the side length of S, since the side length of each protruding square is at most s and at most half of each square is protruding. Because of Lemma 74 the total horizontal length of all nonprotruding parts of the squares Sij is at most 2s, consequently all points covered by S in OPT are covered by a set P of squares Sij in SG whose total (horizontal) edge length j sij is at most 3s. P For exponents α > 1 observe that j sij ≤ 3s and 0 ≤ sij ≤ s for all j implies that P P α α−1 ≤ 3sα . j sij ≤ ( j sij )s To analyze the running time of the algorithm we need some more details about the data structures used: Initially, sort the points by x-coordinate and separately by distance from the line L in time O(n log n) and process the points in order of decreasing distance from L. As the point pi at distance di from L is processed, we throw away points which are within horizontal distance di from pi . This takes time O(log n + ki ) time where ki is the number of points within di from pi . Since we do this up to n times with k1 + · · · + kl = n the total running time is O(n log n). For the linear cost function, it is easy to modify the SG algorithm to get a 2-approximation algorithm. We call this algorithm Square Greedy with Growth, or SGG: Square Greedy with Growth Algorithm (SGG): Process the points as in SG. However, if capturing a point pi by a square Si would result in an overlap with already existing square Sj then, rather than placing Si , grow Sj just enough to capture pi , keeping the vertical edge furthest from pi at the same point on L. If placing Si would overlap two squares, grow the one which requires the smallest edge extension. Break ties arbitrarily. A proof somewhat similar to that of Lemma 71 shows that: Theorem 76. Given n clients in the plane, SGG computes in time O(n log n) a covering by axis-aligned squares centered on the x-axis whose cumulative edge length is at most twice the optimal. Proof. As we process points pi using SGG, attribute to each point pi a line segment si along L as follows. If processing pi resulted in the placement of a square Si centered at the projection of pi in L then attribute to pi the projection on L of a horizontal edge of Si (Case 1). If, on the other hand, processing of pi resulted in the growing of a prior square Sj to just capture pi , attribute to pi the projection on L of the portion of the horizontal edge of the expanded Sj needed to capture pi (Case 2). (This amount is at most the distance of pi to L since otherwise pi would have fallen into case 1.) We must show that the sum of the lengths of the segments is no more than twice the edge lengths of squares in OPT.

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It suffices to show that for any square S in OPT, the segments si associated with points pi ∈ S processed by SGG cannot have total edge length which exceeds twice the edge length s of S. To see this observe that the sum of the lengths of those si lying completely inside S does not exceed s since they are nonoverlapping. In addition, each of the parts of the at most two segments protruding from S can have length at most s/2, in case 1 for the same reason as in the SG algorithm, in case 2 since the total length of the segment is at most s/2. In order to make SGG efficient, we proceed as in SG. In addition, we maintain a balanced binary search tree containing the x-coordinates of the vertical sides of the squares already constructed. For each new point pi to be processed we locate its x-coordinate within this structure to obtain its neighboring squares and to decide whether case 1 or case 2 applies. This can be done in time O(log n) just as adding a new square in case 1 or updating an existing square in case 2. Removing points covered by the new or updated square is done as in SG, so that the total runtime remains O(n log n). Unlike SG, SGG is not a constant factor approximation for area. Consider n consecutive points at height 1 separated one from the next by distance of 1 + ². Processing the points left to right using SGG covers all points with one square of edge length n + (n − 1)², and so area O(n2 ), while covering all points with n overlapping squares each of edge length 2, uses total area 4n. Finally, extending these results from squares to disks in any Lp metric is not difficult. Enclosing each square in the algorithm by an Lp disk leads to an approximation factor 3 · 2α/p for SG and 2 · 21/p for SGG (which is only valid for α = 1). In particular, for L2 √ disks, SGG yields a 2 2-approximation for α = 1 and SG yields a 6-approximation for α = 2. Indeed, denote by OPTp and OPT∞ the minimum costs of a solution for Lp disks or for squares, respectively. Any square of half-edge length r is contained in an Lp disk of √ radius p 2 · r, and so a covering by squares SG∞ induces another covering SGp by Lp disks whose cost is that of SG∞ multiplied by c = 2α/p . The same relation holds between SGG∞ and SGGp . Then on the one hand OPTp ≤ SGp ≤ c · SG∞ ≤ 3c · OPT∞ . But also OPTp induces a covering by circumscribing squares of the same cost so that OPT∞ ≤ OPTp , thus OPTp ≤ SGp ≤ 3c · OPTp . The same goes for SGG when α = 1 yielding an approximation factor of 2c = 2 · 21/p instead.

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5.3.2

Finding the best axis-parallel line

When the horizontal line ` is not given but its orientation is fixed, we first prove that finding the best line, even for α = 1, is not expressible by radicals, then in this linear case give a simple approximation, and finally a PTAS. When we say that the equation for the best line is not expressible by radicals, we mean that in order to arrive at the equation for the best line one would need to solve a polynomial equation whose solution is not expressible by radicals. 5.3.2.1

A hardness result – non expressibility by radicals

Our approach is similar to the approach used by Bajaj on the non-expressibility by radicals of the solution to the Fermat-Weber problem and other geometric optimization problems [3, 4]. Theorem 77. Let c(t) =

P

i ri

denote the minimum cost of a cover whose centers lie on

the line of equation y = t. There exists a set Y of clients such that, if t0 is the value that minimizes c(t), then t0 is not expressible by radicals. The proof proceeds by exhibiting such a point set and showing by differentiating c(t) that t0 is the root of a polynomial which is not expressible by radicals. The following definitions and facts can be found in a standard abstract algebra reference; see, for example, Rotman [69]. A polynomial with rational coefficients is expressible by radicals (or solvable by radicals) if its roots can be expressed using rational numbers, the field operations, and taking kth roots. The splitting field of a polynomial f (x) over the field of rationals Q is the smallest subfield of the complex numbers containing all of the roots of f (x). The Galois group of a polynomial f (x) with respect to the coefficient field Q is the group of automorphisms of the splitting field that leave Q fixed. If the Galois group of f (x) over Q is a symmetric group on five or more elements, then f (x) is not expressible by radicals over Q. Consider the following set of points:{(3, 4), (−3, −2), (102, 2), (98, −2), (200, 2)} as shown in Figure 5.9. It is easy to see that the optimal solution must either be the line y = 2 or else a line y = t with −2 < t ≤ 2. For the line y = 2, the optimal cover consists of a √ circle at (−1, 2) of radius 20, a circle at (98, 2) of radius 4 and a point circle at (200, 2) √ of zero cost, for c(2) = 4 + 20 ≈ 8.472. For small ² > 0 the center of the first circle is given by (x + 3)2 + (4 − ²)2 = (x − 3)2 + (2 + ²)2 so that x = −1 + ², for a center √ at (−1 + ², 2 − ²) and radius 20 − 4² + 2²2 . The center of the second circle is given by (x − 98)2 + (4 − ²)2 = (x − 102)2 + ²2 , in other words the center is at (² + 98, 2 − ²) and

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Figure 5.9: A plot of the points from the proof of Theorem 77. there is a radius of

√ 16 − 8² + 2²2 . The final circle of course is located at (200, 2 − ²) with

radius ². The total cost is therefore c(2 − ²) =

p p 20 − 4² + 2²2 + 16 − 8² + 2²2 + ².

Differentiating, we obtain

−1 dc |²=0 = √ < 0 d² 5

so that the optimal solution lies in −2 < t < 2 and is given by one circle through the points (3, 4), (−3, −2), one circle through the points (102, 2), (98, −2) and a third circle just extending out to capture (200, 2). Reasoning as above, for a given −2 < t < 2, we obtain c(t) =

p p 2(t − 1)2 + 18 + 2t2 + 8 + (2 − t).

(5.3.1)

Therefore, in order to find the best horizontal line, we must minimize c(t). Setting the derivative to zero, we obtain the equation c0 (t) = p

2(t − 1)

2t +√ − 1 = 0. 2t2 + 8 2(t − 1)2 + 18

(5.3.2)

−2(t − 1)2 + 18 16 + 3/2 2 2 [2(t − 1) + 18] (2t + 8)3/2

(5.3.3)

Taking a derivative again, c00 (t) =

we see that for −2 < t < 2, c00 (t) is always positive. If we now move the −1 from Equation 5.3.2 to the right hand side and square both sides, we obtain an equation with just one term containing a square root sign. If we move this one term to the right hand side and square

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Figure 5.10: Critical points for SG as the line ` moves down from +∞. we obtain an expression free of radicals, which reduces to the following polynomial: f (t) = 1024 + 512t − 1600t2 + 1536t3 − 960t4 +368t5 − 172t6 + 28t7 − 7t8 . Using the computational system GAP [35], we compute that the Galois group of f (t) is the symmetric group S8 , so the solution to the polynomial equation 5.3.1 is not expressible by radicals. Though of course unable to compute the minimum cost solution exactly, GAP is capable of giving an approximate solution which is c(t) ≈ 8.3327196 at t ≈ 1.4024709. 5.3.2.2

Fast and simple constant-factor approximations

For appropriate α, the simple constant factor approximations for a fixed line can be extended to the case of approximations to the optimal solution on an arbitrary axis-parallel line with the same constant factors, though with an additional multiplicative factor of O(n2 ) added to the running time. In light of the fact that there is a reasonably effective FPTAS for this problem (see next section) we just sketch the construction, which applies for integer α, 0 < α ≤ 5 for the generalization of SG, and just to α = 1 for the generalization of SGG (since SGG itself just applies for he case of α = 1). Consider the case of squares, any integer 0 < α ≤ 5, and the SG approximation used for a fixed horizontal line. The key observation is that as we move a horizontal line y = t down from t = +∞, the cost of the solution arrived at by SG changes smoothly, with the exception of certain critical points. In between critical points the cost function can be expressed as an α-degree polynomial in t. The critical points are illustrated in Figure 5.10. In the case on the left, point p is precisely as far from ` as it is horizontally from q, so when ` moves further down, a new square will be required centered at the projection of q on `. In the case on the right, p and q are vertically equidistant, and again as ` moves down

172

further, a square again will appear, also centered at the projection of q on `. Since the critical points always involve 2 clients, there are O(n2 ) of them. If α > 1, between these points one gets additional critical points at the zero points of the derivative of the cost function. For α ≤ 5 these can be computed in constant time (using the real RAM model). In each of these cases we have O(n2 ) line placements to test, and at each candidate line placement we run the initial SG algorithm choosing the minimum, resulting in a total running time of O(n3 log n). If `∗ is the line which minimizes OPT over all lines, then SG over all lines is less than or equal to SG over `∗ , so SG over all line placements is also a factor 3 approximation to OPT over all line placements. The analysis of the SGG algorithm is similar. There are additional critical critical line locations due to the running of the algorithm, but still just O(n2 ) of them - and none due to critical points of the cost function, since SGG just works for the case α = 1. Once again the optimum of the pseudo-optimal solutions is found by running SGG just at the critical locations, and so SGG gives, at worst, a factor 2 approximation to OPT for edge length, and can be computed in O(n3 log n) time. The argument extending both the SG and SGG approximations for an arbitrary horizontal line from the L∞ case to the case of Lp -disks is the same as for a single horizontal line. 5.3.2.3

An FPTAS for finding the best horizontal line

We begin with the case α = 1. Let d denote the distance between the highest and lowest point. Clearly, d/2 ≤ OPT ≤ nd/2. Partition the horizontal strip of height d that covers the points into 2n/² horizontal strips each of height δ = d²/2n using 2n/² − 1 regularly spaced horizontal lines `i . For each line `i , run the exact dynamic programming algorithm and keep the best among these solutions. Consider the line `∗ that contains OPT. We can shift `∗ to the nearest `i while increasing the radius of each disk of OPT by at most δ to obtain a covering of the points by disks centered on some `i . The total increase in cost is at most δn = d²/2 ≤ ²OPT. Thus the algorithm computes a (1 + ²)-approximation in running time O((n3 /²) log n). In order to generalize this result to the case α > 1, let us write PSEUDO-OPT for the lowest cost of a solution on any of the regularly spaced horizontal lines `i , SHIFT for the result of shifting OPT to the closest of these lines, and r1 , ..., rm for the radii of the optimal

173

set of disks. For an arbitrary power α ≥ 1, we have PSEUDO-OPT ≤ SHIFT ≤ ≤

m X i=1

m X (ri + δ)α i=1 m X

riα + δα

(ri + δ)α−1

i=1

≤ OPT(1 + δα22α−1 n/d). The last line uses δ ≤ d, ri ≤ d and OPT ≥ (d/2)α . Choosing δ = εd/(α22α−1 n) gives the desired (1 + ε)-approximation. Together with the results from previous sections we have: Theorem 78. Given n clients in the plane and a fixed integer α ≥ 1, there exists an FPTAS for finding an optimally positioned horizontal line and a minimum-cost covering by disks centered on that line. It runs in time O((n3 /²) log n) for the linear cost case (α = 1) and O((n5 /²) log n) for α > 1.

5.3.3

Approximating the best line - any orientation

Finally, we give approximation results for selecting the best line whose orientation is not given. We give constant factor approximations for α ≥ 1, and an FPTAS for the linear cost case (α = 1). 5.3.3.1

Fast and simple constant-factor approximations

Given a line `, we say that a set D of disks D1 ,. . . ,Dk is `-centered if the centers of every disk Cj in D belongs to `. Lemma 79. Given a line `, an `-centered set D of k disks that cover Y , and any point p0 on `, there exist p0 ∈ Y and an `0 -centered set D0 of k disks that cover Y , where `0 is the line that joins p0 and p0 , such that the cost of D0 is at most 2α times the cost of D. Proof. We will assume without loss of generality that ` is the x-axis, p0 is the origin and that no other point in Y lies on the y-axis. The latter restriction can easily be enforced by a small perturbation. Let the coordinates of pi be (xi , yi ) and let mi denote the slope yi /xi of the line `i through p0 and pi for 1 ≤ i ≤ n. First, we reorder Y so that |m1 | ≤ · · · ≤ |mn |. In what follows we assume that x1 > 0 and y1 ≥ 0. The other cases can be treated analogously. For each disk Dj = D(tj , rj ) in D, we construct a disk Dj0 whose radius is rj0 = 2rj and center t0j is obtained from tj by rotating it around the origin counterclockwise by an

174

angle tan−1 (m1 ). The set D0 of k disks thus defined is `0 -centered, where `0 = {(x, y) ∈ R 2 | y = m1 x} and p1 ∈ `0 . To see that D0 covers Y , simply observe that d(tj , t0j ) ≤ rj for all 1 ≤ j ≤ k and apply the triangle inequality: any point in Dj must be at distance at most 2rj of t0j . The cost of this new solution is clearly at most 2α times that of D in the linear cost case. By a double appplication of this lemma, first about an arbitrary p0 yielding a point p0 = pi , then about pi yielding another p0 = pj , it is immediate that any `-centered cover of Y can be transformed into an `i,j -centered cover whose cost is increased at most four-fold, where `i,j is the line joining pi and pj . By computing (exactly or approximately) the optimal set of disks for all O(n2 ) lines defined by two different points of Y , we conclude: Theorem 80. Given n clients in the plane and a fixed α ≥ 1, in O(n4 log n) time, we can find a collinear set of disks that cover Y at cost at most 4α OP T , and for α = 1, in √ O(n3 log n) time, we can find a collinear set of disks that cover Y at cost at most 8 2OP T . 5.3.3.2

An FPTAS for finding the best line with unconstrained orientation in the linear case

We now prove that finding the best line with unconstrained orientation and a minimumcost covering with disks whose centers are on that line admits an FPTAS for the linear cost case. Theorem 81. Let Y be a set of n clients in the plane that can be covered, at linear cost (α = 1), by an optimal collinear set of disks OP T , and let ² > 0 be given. In O((n5 /²2 ) log n) time, we can find a collinear set of disks that cover Y at cost at most (1 + ²)OP T . Proof. Let R be a rectangle of minimal width h that contains Y . Using a rotating calipers approach, R can be computed in O(n log n) time. If h = 0, we can conclude that OPT = 0 and we are done. Otherwise, we can assume that R is in fact an axis-aligned rectangle, with h being its height and w being its length. We have h ≤ w and moreover √ h/2 ≤ OPT ≤ min(w/ 2, nh/2).

(5.3.4)

Let APP denote the cost of the solution computed in time O(n4 log n) according to Theorem 80. Further, let d =

²APP wn

and α = tan−1 (d).

Assume an optimal collinear solution S ∗ = {D(t∗1 ), ..., D(t∗k )} lies on the line `∗ . Although it may be obvious that `∗ passes through R we do not make this assumption. Let

175

Figure 5.11: The two lines `∗ , `ˆ separated by an angle of (at most) α. The distance of p to `ˆ is less than ∆d greater than the distance of p to `∗ and ∆d < 3w tan α. We let dˆ denote the distance from p to `ˆ and d∗ denote the distance from p to `∗ . z be a point on `∗ that is closest to the boundary of R, and let k be a closest point to z on √ √ the boundary of R (if `∗ ∩ R 6= ∅ then z = k). Since OPT ≤ w/ 2, kz − kk ≤ w/ 2, for all √ √ p ∈ P we have kz − pk ≤ kz − kk + kk − pk ≤ w/ 2 + 2w < 3w. We now distinguish two cases: π Case 1. w ≤ 2nh: For 1 ≤ i ≤ d 2α e, let `i be the line through the origin that forms an

angle iα with the x-axis.

√ 1 The number of these lines is N = O( α1 ) = O( tan−1 2², d = ). Note that for n ≥ 4 (d)

²APP APP 4OPT √ wn ≤ w4 2 ≤ 8OPT ≤ wn O( ²OPT ). Since w ≤ 2nh

wn 1/2 so that tan−1 (d) > d/21 , so that N = O(1/d) = O( ²APP )= 2

2

wn n h and h/2 ≤ OPT we get N = O( ²OPT ) = O( ²OPT ) = O( n² ).

Let OPTi denote the cost of an optimal `-collinear set of servers that cover Y , where ` is a line parallel to `i . We claim that mini OPTi ≤ (1 + 12²)OPT. To see this, observe that there is some `i such that the smallest angle βi between `i and `∗ is at most α. If we rotate `∗ by βi in the appropriate direction about z, it becomes parallel to `i . Let `ˆ denote this rotated line. When rotating `∗ we also rotate the servers ˆ t∗i . Let tˆi denote the final positions of the respective servers on `. Since for all p ∈ P , kp − zk < 3w, the distance of p to `ˆ is less than 3w tan α = 3wd = further than the distance of p to `∗ . To see this, refer to Figure 5.11. p∗ is the closest 3 ²APP n ˆ Let d∗ denote the distance of p to point to p on `∗ and pˆ is the closest point to p on `. ˆ Since d∗ + ∆d is not the shortest distance from p `∗ and dˆ denote the distance of p to `. to `ˆ it follows dˆ − d∗ < ∆d. Also, kp∗ − zk < kp − zk < 3w and tan α = kp∆d ∗ −zk . Thus ∗ dˆ − d < ∆d < 3w tan α as claimed. It follows that if we increase each of the server radii by 3 ²APP we get a solution on `ˆ n Pk P k ²APP with cost less than i=1 (ri∗ + 3 ²APP ≤ (1 + 12²)OPT. i=1 3 n n ) = OPT + Using Theorem 78 we can compute a (1+²)-approximation APPi to OPTi in O((n3 /²) log n) 1

In fact we do not have tan−1 (d) ≤ d/2 until d > 2.

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time. We then get that mini APPi < (1 + ²)(1 + 12²)OPT = (1 + 14²)OPT, which can be computed in time O((N n3 /²) log n) = O((n5 /²2 ) log n). Case 2. w > 2nh: In this case the reasoning is similar but we take a slightly different tack. let R0 be the axis-aligned rectangle obtained from R by moving the left and right sides of R inwards by nh/2 leaving R0 with width w0 = w − nh. If now a line ` had a slope of absolute value greater than M = h/w0 , i.e. greater than the aspect ratio of R0 , then ` would either intersect the top and bottom edges of the rectangle R0 or have the top-right and bottom-left corners of R0 both to the same side. Since there is at least one point of Y on the right and left edges of R, in either case, there would be a point of Y with distance larger than nh/2 to ` and, since OPT < nh/2, ` could not be an optimal line. Thus the absolute value of the slope of `∗ can be at most M . Now let µ = tan−1 (M ) and for 1 ≤ i ≤ d αµ e let `i be the line through the origin forming −1

(M ) an angle of iα with the x-axis. The number of these lines is N = O( αµ ) = O( tan ). As tan−1 (d)

earlier, tan−1 (d) > d/2, and tan−1 (M ) ≤ M for any non-negative M , so N = O(M/d) = hwn hn O( w²APP ) = O( ²OPT ). Since h/2 ≤ OPT we get N = O(n/²).

We can now argue as in Case 1 that there is some line `i such that the smallest angle βi between `i and `∗ is at most α. We then compute a (1 + ²)-approximation APPi to OPTi in O((n3 /²) log n) time and take mini APPi < (1 + 14²)OPT, which can be computed in time O((N n3 /²) log n) = O((n4 /²2 ) log n). Case 2 and hence the Theorem are thus established.

5.4

Minimum-Cost Covering Tours

We now consider the minimum cost covering tour (MCCT) problem: Given k ≥ 1 and a set Y = {p1 , . . . , pn } of n clients, determine a cover of Y by (at most) k disks centered at P X = {t1 , . . . , tk } with radii rj and a tour T visiting X, such that the cost length(T )+C riα is minimized. We refer to the tour T , together with the disks centered on X, as a covering tour of Y . More generally, the minimum cost covering tour (MCCT) problem is thus that of computing an optimal covering tour according to some objective function that combines tour length and disk radii. See Figure 5.12. Our results are for the case of linear transmission costs (α = 1). We first show a weak hardness result, then characterize the solution for C ≤ 4. A PTAS for fixed C > 4 can be found in [2].

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Figure 5.12: Example of the covering tour problem: The tour is shown in solid black, and the disks that cover the solid dots (demand points) are shown shaded.

5.4.1

A hardness result

We prove the NP-hardness of MCCT where C is also part of the input. Note that our argument does not prove the NP-hardness of MCCT where C is a fixed constant, which is the problem for which the extension of Mitchell’s m-guillotine method gives a PTAS in [2]. Note also that C appears in the run time exponent of that PTAS, and so the PTAS no longer runs in polynomial time if C is not a fixed constant. Theorem 82. MCCT with linear cost is NP-hard if the ratio C is part of the input. Proof. We show a reduction from Hamilton cycle in grid graphs. A grid graph is a finite subgraph of the two-dimensional integer grid, where nodes have an edge connecting them iff they are unit distance from one another. Hamilton cycle in grid graphs was shown to be NP-complete by Itai, Papadimitriou and Szwarcfiter in [45]. Given a set of n points on a grid, we construct an instance of MCCT in which each of the given points is a client. We set C to be larger than 2n. We claim that the grid graph has a Hamilton cycle if and only if there is a tour T visiting a set of disk centers with radii ri whose cost is at most n. Clearly a Hamilton cycle in the grid graph yields a tour of cost n with each client contained in a disk of radius 0 centered at that point. Conversely, suppose we have a tour whose cost (length plus sum of radii) is at most n. Note that no two clients can be contained in a single disk, as such a disk must have radius at least 0.5, and thus its contribution to the cost C · ri > 2n · 0.5 = n contrary to our assumption. Next we want to show that each disk in an optimal solution is centered at the client it covers. Suppose this is not the case, e.g. there is some client pj which is covered by a disk centered at tj 6= pj . Let |tj − pj | = d. Now consider an alternate feasible solution in which the tour visits tj then pj then back to tj , covering pj with a disk of radius 0. No other client is affected by this change, as the disk only covers pj . The cost of the new

178

solution is the cost of the original (optimal) solution +2d − Cd as we add 2d to the length P of the tour, but decrease C ri by Cd. Since C > 2 the new solution is better than the original optimal solution, a contradition. Hence an optimal solution to this MCCT visits every client, and moreover, since its cost is at most n and the clients lie along a grid graph, the MCCT must be along the grid graph and also be Hamiltonian.

5.4.2

The case C ≤ 4: The exact solution is a single circle

Theorem 83. MCCT in the plane with a cost function of length(T ) + C

P

ri and C ≤ 4,

has a minimum-cost solution which is to broadcast from the smallest disk containing all the clients with no tour cost. q

p

r’ t q’

p’

r

Figure 5.13: Comparison of the trip p → q → r → p to a trip back and forth along the diameter pp0 . The proof rests on the following elementary geometry lemma. Lemma 84. For three points p, q and r in the plane, the length of a trip from p to q to r and back to p is at least 4r where r is the radius of the smallest circle containing all of the points. Proof. Either (i) the smallest containing circle is determined by two of the points which are diametrically opposite each other, or (ii) the circumcircle determined by all three points, and not all three of the points lie on the same half-circle. In case (i) the lemma follows by the triangle inequality. In case (ii) there must be a point amongst the three which has no point within 90 degrees of arc length from it. To see this, place a first point on the circle. If this first point has a point within 90 degrees of arc length from itself, place such a second point. Now if the third point is within 90 degrees of either of the first two points, then all three points lie a half-circle, a contradiction. Hence we have the situation where the three points p, q r are in a configuration depicted by Figure 5.13. Consider now a trip p → q → r → p and compare it to a trip back

179

and forth along the diameter pp0 . Letting d(p, q) denote the distance from p to q, we see that d(p, q) > d(p, q 0 ) and d(q, t) > d(q, q 0 ) > d(q 0 , p0 ). The latter inequality, d(q, q 0 ) > d(q 0 , p0 ), holds since ∠p0 qq 0
d(p, q 0 ) together with d(q, t) > d(q 0 , p0 )

gives d(p, q) + d(q, t) > d(p, p0 ) and analogously d(p, r) + d(r, t) > d(p, p0 ), establishing the lemma. Proof of Theorem 83.

As usual let Y = {p1 , ..., pn } be the set of clients. Let T be a

covering tour of Y with X ⊆ T the set of disk centers and rj their radii. Further, let r(X) and r(Y ) denote the minimum radius of circles enclosing X and Y , respectively. Finally, let rmax = maxj rj . By the triangle inequality, Lemma 84 implies that the length(T ) ≥ 4r(X). Since the tour visits all the centers in X and the disks centered at X cover Y , we have r(Y ) ≤ r(X)+rmax . P By definition, the cost of T is length(T ) + C j rj , which by the observation above is at P least 4r(X) + C j rj ≥ 4r(X) + Crmax . The assumption C ≤ 4 then implies that it be at least C(r(X) + rmax ) ≥ Cr(Y ), which is the cost of covering by a single disk with a zero-length tour. Since it is at least as expensive to tour a set of points in Rd as it is to tour the projection of the points in any plane, we have: Corollary 85. MCCT with clients in Rd for and any d ≥ 1, and a cost function length(T )+ P C ri with C ≤ 4, has a minimum-cost solution which is to broadcast from the smallest disk containing all the clients with no tour cost.

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