Dense Point Sets with Many Halving Lines

A planar point set of n points is called γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma $$\end{document}-dense if the ratio of the largest and smallest distances among the points is at most γn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma \sqrt{n}$$\end{document}. We construct a dense set of n points in the plane with neΩ(logn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ne^{\Omega ({\sqrt{\log n}})}$$\end{document} halving lines. This improves the bound Ω(nlogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega (n\log n)$$\end{document} of Edelsbrunner et al. (Discrete Comput Geom 17(3):243–255, 1997). Our construction can be generalized to higher dimensions, for any d we construct a dense point set of n points in Rd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^d$$\end{document} with nd-1eΩ(logn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n^{d-1}e^{\Omega ({\sqrt{\log n}})}$$\end{document} halving hyperplanes. Our lower bounds are asymptotically the same as the best known lower bounds for general point sets.


Introduction
Let P be a set of n points in the plane in general position, that is, no three of them are on a line. A line, determined by two points of P, is a halving line if it has exactly (n −2)/2 points of P on both sides. Let f (n) denote the maximum number of halving lines that a set of n points can have. It is a challenging unsolved problem to determine f (n).
The first bounds are due to Lovász [8], and Erdős et al. [6]. They established the upper bound O(n 3/2 ), and the lower bound (n log n) (see also [5] for a different lower bound construction). Despite great interest in this problem, there was no progress until the very small improvement due to Pach et al. [10]. They improved the upper bound to O(n 3/2 / log * n). The iterated logarithm of n, log * n, is the number of times the log function must be iteratively applied before the result is at most 1.
The best known upper bound is O(n 4/3 ), due to Dey [3]. The lower bound has been improved by Tóth [13] to ne ( √ log n) . Nivasch [9] simplified the construction of Tóth. Suppose that γ > 0. A planar point set P of n points is called γ -dense if the ratio of the largest and smallest distances determined by P is at most γ √ n. There exist arbitrarily large γ -dense point sets if and only if see [4,7]. Dense point sets are important in the analysis of some geometric algorithms, as they can be considered "typical" for some practical applications, like computer graphics.
Recall that f (n) is the maximum number of halving lines of a (not necessarily dense) set of n points in the plane. Edelsbrunner et al. [4] showed, that a γ -dense point set can have at most O(γ √ n f (3γ √ n)) halving lines. Agarwal and Sharir [1] combined it with the result of Dey [3] and obtained that for any fixed γ , a γ -dense point set can have at most O(n 7/6 ) halving lines.
On the other hand, Edelsbrunner et al. [4] constructed 2-dense point sets with (n log n) halving lines. Note that at that time (n log n) was the best known lower bound for general point sets as well. In this note we give a better construction. Our lower bound is again asymptotically the same as the best known lower bound for general point sets and it answers a question of Nivasch [9].
Our construction can be generalized to higher dimensions. Let d ≥ 2 and let P be a set of n points in the d-dimensional space, R d , in general position, that is, no d + 1 of them are on a hyperplane. A hyperplane, determined by d points of P, is a halving hyperplane if it has exactly (n − d)/2 points of P on both sides. Let f d (n) denote the maximum number of halving hyperplanes that a set of n points in R d can have. So f 2 (n) is the same as f (n) above. The best upper bounds are f 3 (n) ≤ O(n 5/2 ) [12], f 4 (n) ≤ O(n 4−1/18 ) [11], and for d > 4, f d (n) ≤ O(n d−ε d ) where ε d = 1/(2d) d [2]. The planar lower bound construction can be "lifted" to higher dimensions and it gives the lower bound n d−1 e ( √ log n) for f d (n) [13]. A set of n points in R d is called γ -dense if the ratio of the largest and smallest distances determined by P is at most γ d √ n. Just like in the planar case, our lower bounds are asymptotically the same as the best known lower bounds for general point sets. The planar construction is divided into three steps, presented in the next three sections. The first step is the bulk of the result, and it is based on the ideas of Nivasch [9]. However, we have to be more careful to keep the distances under control. In the last section we show how to generalize it to higher dimensions.

First Construction
Definition 2.1 (a) For any point set S in the plane, if points p, q ∈ S determine a halving line, then the segment pq is called a halving segment. (b) A geometric graph is a graph drawn in the plane with straight line segments as edges.
First, we construct point sets S with n points and ne ( √ log n) halving lines, such that the diameter of S is 1 and the smallest distance is (n −8 ). Then, using several, slightly modified copies of S, we increase the smallest distance to (n −1/2 ) while the number of halving lines remains asymptotically the same.

Definition 2.2
We say that a set of points forms an arithmetic progression, if its points are on a horizontal line and consecutive points are at the same distance. Its size is the number of points it contains, its step is the distance between the consecutive points, its width is the distance between the first and last points.    For i ≥ 1, let

Suppose again that we already have
, and replace p with the plain points p (k) ∈ S o i , 0 ≤ k ≤ a i − 1. We call these points the children of p, and p is the parent of the new points. For Again, we call these points the children of b, and b is the parent of the new points.
For each halving segment Add the segments qp (k) , 0 ≤ k ≤ a i − 1, and qb (k) , 0 ≤ k ≤ a i , to H o i . See Fig. 1. We say that these new segments in H o i are the children of pb ∈ H o i−1 , and pb is the parent of the new segments and q.
We also say that the point q is the child of b and b is the parent of q. Finally, we say that the point q is assigned to segment pb.
Extend the relations "parent" and "child" to their transitive closure and call the resulting relation "ancestor" and "descendant", respectively.
A halving segment s ∈ H o i is the k-th ancestor (k-th descendant) of s ∈ H o i , if it is an ancestor (descendant) and i = i + k (i = i − k). Similarly, a point p ∈ S o i is the k-th ancestor (k-th descendant) of p ∈ S o i , if it is an ancestor (descendant) and We show that for any halving segment s in H o i , all of its descendant segments are in a very narrow strip of s, in a very strong sense, while other points, that is, those points which are not endpoints of descendant segments, will never get into that strip. See Fig. 3.

Claim 1 Let i ≥ 0 and s ∈ H o
i . For every r ∈ Desc(s), the extensionr of r lies in Then there is a sequence s = s 0 , s 1 , . . . , s j = r such that for every s r y = 1 Let Move s 1 horizontally to the right by ε o i+1 /4, let s 1 = b 1 p 1 be the resulting segment and lets 1 =p 1b 1 be the extension of s 1 . Then b 1 is exactly the midpoint of s. The point p 1 has the same y-coordinate as either p or b, and their (horizontal) distance is at most . We can show by the same argument that for any 1

be a child of s i . Then q i is below the line of s i if and only if q i+1 is below the line of s i+1 .
Proof First we show part (a) of the claim. Let j ≥ i, q ∈ S o j that is not an endpoint of s or any descendant of s.
Similarly, there is a sequence s 0 , s 1 , . . . , s i = s such that for every k, 0 ≤ k ≤ i, s k ∈ H o k , and for every k, 0 ≤ k < j, s k is a parent of s k+1 .
Consequently, q / ∈ S 2α (s). Here we used that o ≥ j.
Case 2 Suppose that q is plain. The argument will be a more complicated version of the previous one. There is a sequence Case 2.2 Suppose now that l < i. If s l−1 = s l−1 , then we can proceed exactly as in Case 2.1. Let m ≤ i be the smallest number with the property that s m = s m , and do the same calculation.
So we can assume that s l−1 = s l−1 . In this case b l is an endpoint of s l . For simplicity denote b l also as p l . By the assumption, p i is not an endpoint of s i . Let m be the smallest number such that p m is not the endpoint of s m . Now we do the same calculation again.
Let be the horizontal line through p m , . . . , p j . Let t be the intersection of and the line of s m . We have |t , so it follows that q / ∈ S 2α (s i ). This concludes the proof of part (a). Part (b) follows directly from the calculations, we only have to observe that taking a child of the point q, it cannot "jump" to the other side of the strip S 2α (s).
We are ready to prove that the segments in Similarly, P i = P (1) i ∪ P (2) i ∪ P (3) i where P (1) i is the set of (plain) children of points of P i−1 \ {p i−1 }, P Let s i be a child of s i−1 . It follows from Claim 2 that if p ∈ P i−1 is on the left (resp. right) side of the line of s i−1 , then any descendant p ∈ P i of p is on the left (resp. right) side of s i . Since each p ∈ P i−1 is replaced by the same number, a i of plain points, it follows that s i is halving P (1) i . By a similar argument, s i is halving P (2) i . And by the locally halving property of s i , it is halving the set P (3) i . So, s i is halving P i . Since s i−1 is halving B i−1 , and each bold point has exactly 2a i−1 + 1 bold children, it follows again from Claim 2 that s i is halving the set B

Claim 3 Suppose that is a line determined by two points of S i and let γ be the smaller angle determined by and the x-axis. Then either
The graph G i has n i vertices and m i edges, all of them are halving edges of S i .
We estimate now the number of points n i in S i and the number of halving lines m i in H i , that is, we prove parts (a) and (b) of Lemma 2.3. Our calculation is similar to the one by Nivasch [9].
We have n 0 = 2 and m 0 = 1. Since each halving line in H i−1 is replaced by 2a i + 1 halving lines in H i , Consequently On the other hand, Here we used that for any x > 0, e x > 1 + x. Summarizing, The number of bold points in S i−1 is m i−2 , hence the number of plain points in plain points and m i−1 bold points in S i , so Using (1) for m i , we prove by induction that Both inequalities hold trivially for i = 0. Suppose that i > 0 and 2 (i−1) 2 /2+(i−1)/2 < n i−1 . Then It follows that m i ≥ n i e ( √ log n i ) . This finishes the proof of parts (a) and (b) of Lemma 2.3. Now we prove part (c). The statement is trivial for i = 0, suppose that i > 0. Let a, b ∈ S i . First assume that y(a) = y(b). Then we have 2 −i ≤ |y(a) − y(b)| ≤ 1, so by Claim 3, Summarizing, in any case we have by inequality (2). Apply a scaling by a factor of 8/9 and the statement follows. This concludes the proof of Lemma 2.3.

Second Construction
The following statement is stronger than Lemma 2.3. It holds for all even numbers n, not just a sequence {n i }, and in part (c), instead of (n −8 i ), now we have (n −1 ), which is optimal.
On the other hand |y( p) − y(b)| ≥ 2 −i . Move p, b, and q horizontally by at most n −9 i . The horizontal distance of the line of s and the point q changed by at most d = n −9 i + 2 i+1 n −9 i . By inequality (2), n i ≥ 2 i 2 /2+i/2 , therefore, We used here that i > 10. Therefore, q remains on the same side of s after the perturbation of the points. This holds for any q = p, b, therefore, s remains a halving line.
Suppose now that i > 10 is a fixed number. Let n = n i , m = m i , S = S i , and H = H i . Let p 1 , . . . , p n be the points of S, p j = (x j , y j ). Then let p j = (x j , y j ), where x j = n 9 x j n 9 + j n 10 .
Let S = {p j | 1 ≤ j ≤ n} and let H be the segments corresponding to the segments in H . Each point of S is moved horizontally by less than n −9 , so, by Claim, the segments in H are halving segments of S . Now apply a horizontal translation so that all points of S have x-coordinates between 1 and 3. This is possible by Lemma 2.3 (c).
Let δ > 0 be a very small number. Apply the transformation (x, y) −→ (x, δ 2 · y) on S , call the resulting point set S . See Fig. 5. Clearly, halving edges remained halving edges. Since δ is very small, all halving edges are very close to the x-axis and almost parallel to it. All points of S have very small y-coordinates and their x-coordinates are between 1 and 3.
Let α denote the counterclockwise rotation about the origin by angle α and let T r denote the translation to the right by r . We have to introduce one more parameter, N . Let n 9 ≤ N ≤ n 10 be an arbitrary number.
Define the point sets S (k) + and S (k) − , called positive blocks and negative blocks, respectively, as follows. For 0 ≤ k ≤ N , let S (k) Fig. 6 The construction of S * and for 0 ≤ k ≤ N − 1, let Let S union be their union, that is, See Fig. 6.
Finally, apply a scaling of factor n 9 /(6N ) to S union and let S * be a point set obtained. We claim that S * satisfies the conditions. Recall that each block has m halving lines, n = n i and m = m i for some fixed i. The set S * contains 2N + 1 blocks, each contains n points. So |S * | = n * = 2N n + n.
Observe, that a halving line of a block has the same number of other blocks on both sides, so it is a halving line of S * as well. So, for the number of halving edges of S * , since N ≤ n 10 . This proves parts (a) and (b). By the construction, the x-coordinates of the points of S are between 1 and 3. Consequently, the x-coordinates of the points of a positive block S (k) + (resp. negative block S (k) − ) are between 0 and 3N /n 9 (resp. −3N /n 9 and 0). Therefore, the x-coordinates of the points of S union are between 3N /n 9 and −3N /n 9 . Finally, we can conclude that the x-coordinates of the points of S * are between 1/2 and −1/2.
Recall that p 1 , . . . , p n are the points of S , p j = (x j , y j ), and for every j, x j = k/n 9 + j/n 10 for some integer k. This remains true for S and still remains true if we apply a translation by l/n 9 for some integer l. If we apply now a rotation about the origin by a very small angle, the x-coordinates will change by a very small amount. So, we have the following statement: let q 1 , . . . , q n be the points of a positive block S (k) + , corresponding to the points p 1 , . . . , p n of S . Then the x-coordinate x j of q j is very close to a number of the form k/n 9 + l/n 10 for some k, l integers. Analogous statement holds for the negative blocks. So, for any two points of S union , the difference of their x-coordinates is least n −10 /2. After scaling by a factor of n 9 /(6N ) we get the final set S * , so in S * , the difference of the x-coordinates of any two points is at least n/(12N ) = ((n * ) −1 ). This proves Lemma 3.1 if n * , the number of points, is of the form (2N + 1)n, where n = n i for some i and n 9 ≤ N ≤ n 10 . For other values we have to add some extra points as follows.
Observe, that for every i > 10, (2n 10 i + 1)n i > (2n 9 i+1 + 1)n i+1 . This means, that for any large enough even n (say, n > (2n 9 10 + 1)n 10 ), there is an i > 10 and an N , n 9 i ≤ N ≤ n 10 i , such that (2N + 1)n i ≤ n ≤ (2N + 3)n i . So we can take our construction with parameters n i and N , and add at most 2n i extra points so that the conditions are still satisfied. This concludes the proof of Lemma 3.1.

Third Construction
Proof of Theorem 1.1 Let S * be the point set satisfying the conditions of Lemma 3.1. Let n be the number of its points and let m be the number of its halving lines. Let ε > 0 be a very small number.
Apply the transformation (x, y) −→ (x + 1/2, ε 2 · y) on S * from Lemma 3.1, and let R denote the resulting point set. The set R contains n points and m = ne ( halving lines. The x-coordinates of its points are between 1 and 2, the y-coordinates are extremely small, and the distance between any two points is (1/n).
and let See Fig. 7. The set P contains N = n(n+1) points. For every k, each halving line of R k contains the same number of other blocks on both sides, so it is a halving line of P. Therefore, the number of halving lines of P, M = (n + 1)ne ( √ log n) = N e ( √ log N ) . The minimum distance among the points in P is (1/n) = (1/ √ N ) and the diameter of P is less than 4. This finishes the proof of Theorem 1.1, if the number of points, N , is of the form n(n + 1) where n is even. If we want a construction with N even number of points where n(n + 1) < N < (n + 2)(n + 3) for some n even, we take the construction with n(n + 1) points and add N − n(n + 1) < 4n + 6 = O(N 1/2 ) extra points so that the conditions are still satisfied. One possible way to do it is that we add the vertices of a regular (N − n(n + 1))-gon, inscribed in a circle of radius 3 about the origin, so that none of the extra points are on the (original) halving lines. This concludes the proof.

Remark 4.1
The number of halving lines of P is ne √ (log 2)/11 √ log n / √ log n , while in the construction of Nivasch [9] it is ne The point set P

Remark 4.2
The point set P is 4-dense, it can be proved by the same calculation as in [4]. We omit the details.  Let ε > 0 be a very small number. We set its value later. By an application of a suitable affine transformation (flattening) we can assume in addition that (d) each point of S has y-coordinate |y| < ε 3 .

Construction in the Space
First, we define two planar sets of points, block A and block B. Block A contains two parts, the important part and the unimportant part. Both parts contain m points. The important part is a translated copy of S such that all points have x-coordinate 1 ≤ x ≤ 2 and y-coordinate |y| < ε 3 . Then all halving lines of S are very close to the x-axis. The unimportant part contains the points p 1 , . . . , p m/2 and q 1 , . . . , q m/2 , where p i = (−2 + i/m, ε 2 ), q i = (−1.5 + i/m, −ε 2 ). If ε is small enough, then all halving lines of the important part will have p 1 , . . . , p m/2 on one side and q 1 , . . . , q m/2 on the other. Therefore, they are also halving lines of the whole block A. The origin O = (0, 0) is called the center, the x-axis is called the axis and the x y plane is called the plane of block A.
Block  = (0, 0) is called the center, the x-axis is called the axis and the x y plane is called the plane of block B. Note that B is symmetric about the origin. Now take a maximal symmetric (about the origin) packing of discs, of spherical radii 1/m, on the unit sphere, whose center is the origin. Let c 1 , . . . , c k , c 1 , . . . , c k be their centers, c i is the reflection of c i . Since the packing was maximal, the discs of spherical radii 2/m around c 1 , . . . , c k , c 1 , . . . , c k cover the sphere. Therefore, k = (m 2 ). On the other hand, any two centers are at distance at least 1/m (actually, almost 2/m). Suppose for simplicity that k is even. Perturb the points c i so that no three of them determine a plane through the origin and none of them is on the z-axis. Let 1 , . . . , k/2 , be the lines through the origin and c 1 , . . . , c k/2 , respectively, and let 1 , . . . , k/2 be the lines through the origin and c k/2+1 , . . . , c k , respectively. For each line i , 1 ≤ i ≤ k/2, take a block A so that its center is the origin and its axis is i . Its plane can be arbitrary through i that does not go through any other c j , j = i. For each line i , 1 ≤ i ≤ k/2, take a block B so that its center is the origin and its axis is i . Again, its plane can be arbitrary through i that does not go through any other c j . Choose now the parameter ε > 0 to be a very small number. Finally slightly perturb the points so that they are in general position. We obtain the point set P. Now we have n = (m 3 ) points. The maximum distance is at most 4, the minimum is at least 1/m = (n −1/3 ), so P is γ -dense for some γ .
Let A be a block of type A, and B a block of type B. Take two points, u, v, of the important part of block A that determine a halving line and a point w of block B. The plane determined by these three points is almost a halving plane of P. It halves any other block, since it goes almost through the origin, it also halves block A, by the choice of u and v, and since B is almost symmetric about the origin, the plane has one more points of B on one side than on the other. See Fig. 8. So, has one more or one less points of P above than below it. An easy calculation shows that we have k 2 m 2 e ( √ log n) = n 2 e ( √ log n) such planes. We can assume without loss of generality that at least half of them have one less points of P above than below. Add the point (0, 0, 2) to P. It is still dense, and now it has n 2 e ( √ log n) halving planes. This proves the result in three dimensions, if the number of points n = (m 3 ) for some even m and n is odd. In order to extend the result to all odd n, we have to add some extra points to the construction, in pairs, symmetric about the origin. We can do it so that it does not decrease the number of halving planes and the resulting set is still dense. Suppose now that d > 3. The construction is analogous, we only sketch it. We use the same blocks, A and B as before. We take a maximal symmetric (about the origin) packing of d − 1-dimensional discs, of spherical radii 1/m, on the unit sphere, whose center is the origin.
Let c 1 , . . . , c k , c 1 , . . . , c k be their centers, c i is the reflection of c i . Since the packing was maximal, k = (m d−1 ). Suppose that k is even. Let 1 , . . . , k/2 , 1 , . . . , k/2 be the lines through the origin and c 1 , . . . , c k , respectively. For each line i , 1 ≤ i ≤ k/2, take a block A so that its center is the origin and its axis is i . For each line i , 1 ≤ i ≤ k/2, take a block B so that its center is the origin and its axis is i . Finally slightly perturb the points so that they are in general position. We obtain the point set P. Now we have n = (m d ) points. The maximum distance is at most 4, the minimum is at least 1/m = (n −1/d ), so P is γ -dense for some γ .
Let A be a block of type A, and let B 1 , B 2 , . . . , B d−2 be different blocks, each of type B. Take two points, u, v, of the important part of block A that determine a halving line and for 1 ≤ i ≤ d − 2, let w i ∈ B i . The hyperplane determined by these d points is almost a halving plane of P. It halves any other block, since it goes almost through the origin, it also halves block A, by the choice of u and v, and since B i is almost symmetric about the origin, has one more points of B i on one side than on the other. An easy calculation shows that we have n d−1 e ( √ log n) such planes. Let diff( ) be the number of points of P above minus the number of points below . ("Above" and "below" are defined with respect to the d-th coordinate.) It follows, that |diff( )| ≤ d − 1 and diff( ) − d is even. So, for some x, where |x| ≤ d − 1 and x − d is even, there are still n d−1 e ( √ log n) such planes with diff( ) = x. We can assume without loss of generality that x ≤ 0. Add the points (0, 0, . . . , 0, 2 + i/n), 1 ≤ i ≤ x to P. The resulting set is still dense and now it has n 2 e ( √ log n) halving planes.