Points with large quadrant-depth

Given a set P of points in the plane we are interested in points that are 'deep' in the set in the sense that they have two opposite quadrants both containing many points of P. We deal with the extremal version of this problem. A pair (a, b) of numbers is admissible if every point set P contains a point p ∈ P that determines a pair (Q,Qop) of opposite quadrants, such that Q contains at least an a-fraction and Qop contains at least a b-fraction of the points of P. We provide a complete description of the set F of all admissible pairs (a, b). This amounts to identifying three line segments and a point on the boundary of F. In higher dimensions we study the maximal a, such that (a, a) is opposite orthant admissible. We show that 1/(2γ) d a d 1/γ for γ = 22d--1 2d--1. Finally we deal with a variant of the problem where the opposite pairs of orthants need not be determined by a point in P. Again we are interested in values a, such that all subsets P in ℜd admit a pair (O,Oop) of opposite orthants both containing at least an a-fraction of the points. The maximal such value is a = 1/2d. Generalizations of the problem are also discussed.

Finally we deal with a variant of the problem where the opposite pairs of orthants need not be determined by a point in P . Again we are interested in values a, such that all subsets P in R d admit a pair (O, O op ) of opposite orthants both containing at least an a-fraction of the points. The maximum such value is a = 1/2 d . Generalizations of the problem are also discussed.

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A point p = (p 1 , p 2 ) in the plane defines four quadrants Q 1 (p), Q 2 (p), Q 3 (p), and Q 4 (p) centered at p, each being the intersection of two halfspaces defined by one horizontal and one vertical line through p. As usual the quadrants are numbered in counterclockwise order starting from Q 1 = {(x 1 , x 2 ) ∈ R 2 : x 1 ≥ p 1 and x 2 ≥ p 2 }. There are two pairs of opposite quadrants (Q 1 , Q 3 ) and (Q 2 , Q 4 ). We write Q op to denote the quadrant opposite to Q, e.g., Q op 3 = Q 1 . In the extended abstract of this article, which appeared in the 2010 SOCG Conference Proceedings [ABF+], we have considered the quadrants to be open. However, our results hold for closed quadrants as well. The closed quadrants have the advantage that we need no assumption about general position.
For a given set P of points we ask for points p ∈ P that are 'deep' in the set in the sense that p has two opposite quadrants both containing many points of P . A pair (Q, Q op ) of opposite quadrants centered at a point p ∈ P is called (a, b)-admissible if |Q∩P | ≥ a(n−1) and |Q op ∩P | ≥ b(n−1). We call a pair (a, b) admissible if every finite point set P contains a point with two opposite quadrants that are (a, b)-admissible. Thus, (a, b) is not admissible if there exists a point set P in which no point has two opposite quadrants that are (a, b)admissible. Indeed it follows from our proofs that if (a, b) is not admissible, then there exists a number N , such that for all n ≥ N there is an n-element set P in which no point has two opposite quadrants that are (a, b)-admissible.
Brönnimann, Lenchner, and Pach [BLP] define the notion of opposite-quadrant depth for point sets in the plane as the maximum a, such that (a, a) is admissible. They prove that every set of points in the plane has opposite-quadrant depth at least 1 8 . We give a new and simpler proof of this result below, Theorem 1.1.
In Section 2 we provide a complete description of the set F of all admissible pairs. The shape of F turns out to be surprisingly complicated (see Figure 3).
In Section 3 we ask for the maximum a, such that (a, a) is admissible in higher dimensions. In dimension d we obtain upper and lower bounds that differ by a factor of 2. In Section 4 we discuss further generalizations.
The notion of opposite-quadrant depth, resp. opposite-orthant depth, is related to centerpoints and some measures of statistical depth, such as hyperplane depth. We refer to [Ede] for information on centerpoints and to [LPS] for statistical depth. Brönnimann et al. [BLP] also mention a connection with conflict-free colorings. Related notions of depth have been studied e.g. in [BPZ1] and [BPZ2].
As a warm up and for the purpose of introducing some convenient notation we now reprove the main result from [BLP].
(1) Any set P of n points in the plane has opposite-quadrant depth at least 1 8 .
(2) If P is in convex position, then it has opposite-quadrant depth at least 1 4 .
Before starting with the proof let us introduce the following convenient notation. Given a set P of n points, the weight of a subset A of the plane is ω(A) = |A∩P | n−1 . In terms of weights, a pair (Q, Q op ) of opposite quadrants at a point p ∈ P is (a, b)-admissible if and only if ω(Q) ≥ a and ω(Q op ) ≥ b.
In many cases we will choose a subset P ′ of P of some specified weight. When we choose a set P ′ ⊆ P of weight a we mean that P ′ contains exactly ⌊a(n − 1)⌋ points from P . This way, the weight of P ′ may be less than a, but the addition of any point would result in a weight which is at least a. The additional point will correspond to that point in P that determines an admissible pair of quadrants. Theorem 1.1 is an easy consequence of the proof of the following lemma.
Proof. Given the point set P , choose the sets P L , P R , P B , and P T of weight 1 4 each consisting of the first points in P from the left, right, bottom, and top, respectively (see Figure 1). That is, we choose the ⌊ 1 4 (n − 1)⌋ points from P with the smallest x 1 -values, the largest x 1 -values, the smallest x 2 -values, and the largest x 2 -values, respectively.
It follows that P ′ = P \ (P L ∪ P R ∪ P B ∪ P T ) = ∅ and we claim that every point in P ′ has the desired property. Let p be such a point and assume that min ω(Q 1 (p)), ω(Q 3 (p)) = s = ω(Q 1 (p)). Since P T is contained in Q 1 (p) ∪ Q 2 (p) it follows that ω(Q 2 (p)) ≥ 1 4 − s. Considering P R we obtain ω(Q 4 (p)) ≥ 1 4 − s. Consequently min ω(Q 1 (p)), ω(Q 3 (p)) = s and min ω(Q 2 (p)), ω(Q 4 (p)) ≥ 1 4 − s, which proves the lemma. Proof of Theorem 1.1. For part one of the theorem it is enough to observe that either s or 1 4 − s is at least 1 8 . For the second part note that if P is in convex position one of the four quadrants of p is empty. Therefore, one of the two minima in the lemma is zero and the other minimum is at least 1 4 . Figure 2: Opposite-quadrant depth at most 1 8 .
It is easy to see that the second part of Theorem 1.1 is best possible by taking P to be the set of vertices of a regular n-gon. In [BLP] it is shown that the first part of Theorem 1.1 is best possible for arbitrarily large values of n of the form n = 4 · 3 k . The example in Figure 2 shows a simple construction that proves that the first part of Theorem 1.1 cannot be improved, i.e., we show that for every ε > 0 there exists N (ε), such that for all n ≥ N (ε) there is an n-element set P with opposite-quadrant depth less than 1 8 + ε. Fix ε > 0 and n ≥ 17 8ε + 1. To describe the example we identify a point set in the plane with the induced dominance order, that is, we say (x 1 , x 2 ) (y 1 , y 2 ) if x 1 ≤ y 1 and x 2 ≤ y 2 . Based on this order we can talk about chains and antichains of a point set. The example of Figure 2 consists of eight chains each of either ⌊ n 8 ⌋ or ⌈ n 8 ⌉ points. In the figure the chains are represented by gray segments. Now n ≥ 17 8ε + 1 is equivalent to ( 1 8 + ε)(n − 1) ≥ n 8 + 2, which is strictly more than one point together with the points in one chain. Therefore, if a quadrant at p contains, besides p, no more than one chain, then its weight is less than 1 8 + ε. The eight chains come in two groups of four chains each. One group is the set of weight 1 2 consisting of the first points in P from the left (and the bottom) and the other group of weight 1 2 consists of the first points in P from the right (and the top). We consider two cases: First for p ∈ P look at the pair (Q 2 (p), Q 4 (p)). The chains in each group are arranged in such a way that Q 2 (p) and Q 4 (p) both contain p and an integral number of chains. But there is no point p with Q 2 (p) and Q 4 (p) each containing two chains.
Finally, for every p ∈ P either Q 1 (p) or Q 3 (p) contains no more than one chain. Hence every point has two adjacent quadrants with weight less than 1 8 + ε.

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Recall that a pair (a, b) ∈ [0, 1] 2 is called admissible if every finite point set P in the plane contains a point p ∈ P , such that there is a quadrant Q centered at p with ω(Q) ≥ a and In this section we provide a full characterization of the set F of all admissible pairs. We begin with some easy observations.
• F is symmetric in the sense that if (a, b) ∈ F, then also (b, a) ∈ F.
• F is also monotone decreasing, that is, if (a, b) ∈ F and (0, 0) (a ′ , b ′ ) (a, b) then (a ′ , b ′ ) ∈ F. Figure 3 depicts the half of the set F where a ≥ b, the other half is obtained by reflection about the diagonal line a = b. In our analysis, we will determine the part of the boundary of F shown in the figure. That is, we will always assume that a ≥ b.
Theorem 1.1 shows that ( 1 8 , 1 8 ) ∈ F and the example from Figure 2 implies that if both a and b are greater than 1 8 , then (a, b) / ∈ F.
Though the next proposition does not really contribute to the boundary of F it provides a good first approximation to the set.
Proof. First note that we only have to prove that every pair (a, b) with 3a + 5b = 1 is in F. The proposition then follows for all pairs (0, 0) (a ′ , b ′ ) (a, b).
So given a set P of points and a pair (a, b) with 3a + 5b = 1, choose the set A of weight 2a consisting of the first points in P from the left and the set Z of weight a + b consisting of the first points in P from the right. Consider the horizontal median point p in the strip S between the sets A and Z. From the assumption it follows that ω(S) ≥ 4b (see Figure 4).   From ω(A) ≥ 2a it follows that one of the left quadrants of p has weight at least a (w.l.o.g. the upper one Q 2 (p)). If ω(Q 4 (p)) ≥ b we are done. Hence, we may assume that ω(Q 4 (p)) < b. From the weight of Z it then follows that ω(Q 1 (p)) ≥ a. The weight of points below p in S is at least 2b, therefore, ω(Q 4 (p)) < b implies that ω(Q 3 (p)) ≥ b. We have thus found an appropriate pair of opposite quadrants.
The next observation implies that ( 1 2 , 0) belongs to the boundary of F.
Proof. Let P be a set of n points in the plane and let p ∈ P be the point of P with the largest x 1 -coordinate. All points of P \ {p} are contained in the second and third quadrant of p. Hence, one of these two quadrants has weight at least 1 2 . This shows that ( 1 2 , 0) belongs to F.
To see that (a, 0) does not belong to F for any a > 1 2 , consider a set P of n points evenly distributed on a circle, or equivalently the vertices of a regular n-gon. It is left to the reader to verify that no point in P has a quadrant of weight greater than 1 2 .
We now get to the concavity at ( 1 5 , 1 10 ) on the boundary of F and the two segments bounding F that meet in this point. Proof. Given a set P of points and a pair (a, b), choose two vertical lines, such that the set A of points to the left of both lines has weight a + b, the set S in the strip between the lines has weight 4b and the set Z of points to the right of the two lines has the remaining weight a + b. Consider the horizontal median point p of the middle set S (see Figure 5).
One of the quadrants of p has weight at least 1 4 . Without loss of generality we assume that this is true for Q 1 (p). The restriction 1 4 ≥ a implies ω(Q 1 (p)) ≥ a. If ω(Q 3 (p)) ≥ b we are done. Hence, we may assume that ω(Q 3 (p)) < b. From the weight of A it then follows that ω(Q 2 (p)) ≥ a. Similarly from the weight of points below p in S it follows that ω(Q 4 (p)) ≥ b. Hence, (Q 2 , Q 4 ) is an appropriate pair of opposite quadrants.
Proof. In the proof we will need that 2a ≥ 3b, which follows from 4a + 2b = 1 and 3 16 ≤ a. Given a set P of points and a pair (a, b), choose two vertical lines, such that the set A of points to the left of both lines has weight 2a − b, the set S in the strip between the lines has weight 4b and the set Z of points to the right of the two lines has the remaining weight 2a − b. Consider the horizontal median point p of the middle set S (see Figure 6).
One of the quadrants of p has weight at least 1 4 . Without loss of generality we assume that this is true for Q 1 (p), i.e., ω(Q 1 (p)) ≥ 1 4 ≥ a. If ω(Q 3 (p)) ≥ b we are done. Hence, we may assume that ω(Q 3 (p)) < b. From the weight of points below p it then follows that ω(Q 4 (p)) ≥ b. Since the weight of A is at least 2a − b ≥ 2b it follows that ω(Q 2 (p)) ≥ b. If one of the quadrants Q 2 (p) and Q 4 (p) has weight at least a we are done. But if ω(Q 2 (p)) < a and ω(Q 4 (p)) < a the points in the union of the second, third and fourth quadrant of p would have total weight less than 2a + b which contradicts the choice of p. It remains to show that for 1 4 ≤ a ≤ 1 3 the boundary of F is a segment supported by the line 3a + 3b = 1. The following theorem shows that all pairs (a, b) on this segment are admissible. In Figure 9 we present a point set P with no point having opposite quadrants that are (a + ε, b + ε)-admissible for 3a + 3b = 1, a ≥ 1 4 , and any ε > 0. The analysis of the example is given in Proposition 2.7.
Proof. Given a set P of points and a pair (a, b), choose a vertical line, such that the set A of points to the left of the line has weight a + b. Choose another vertical line, such that the set Z of points to the right of this line has weight a + b. The set S in the strip between the lines also has weight a + b. This set S is divided vertically into a top part T , a middle part M and a bottom part B, such that each of T and B has weight 2b (see Figure 8). The vertical lines are chosen, such that the weight of M is (a + b) − 4b = a − 3b. From a ≥ 1 and 3a + 3b = 1 it follows that this weight is non-negative. To see that at least one of the quadrants of p or q has weight at least a we sum up the weights of the first and second quadrant of p and the third and fourth quadrant of q: ω(Q 1 (p)) + ω(Q 2 (p)) + ω(Q 3 (q)) + ω(Q 4 (q)) ≥ 4a.
Proof. As in the previous proof we will need that a − 3b ≥ 0, which follows from a ≥ 1 4 and 3a + 3b = 1.
The example consists of four chains/antichains each of weight 1 2 (a − 3b), a circle of weight a − 3b and 12 chains/antichains each of weight b (see Figure 9). The total weight of the set S 1 consisting of an antichain and three chains is 1 2 (a − 3b) + 3b = 1 2 (a + 3b). Since a − 3b ≥ 0 this is at most a.
Since the example is invariant under rotations of 90 degrees it is enough to show that there is no point p, such that (Q 1 (p), Q 3 (p)) is (a + ε, b + ε)-admissible for ε > 0. Since we need ω(Q 1 (p)) > a we cannot take p from S 1 or S 2 . When considering p ∈ S 4 we get strong restrictions from the requirement ω(Q 3 (p)) > b. To match this we need to have two of the antichains of size b of S 4 in Q 3 (p). Therefore Q 1 (p) can only contain the chain of size 1 2 (a − 3b) from S 4 , which yields ω(Q 1 (p)) ≤ 1 2 (a − 3b) + 1 2 (a + 3b) = a, which is not enough. Any point p from the circle C has ω(Q 1 (p) ∩ C) ≤ 1 2 ω(C) = 1 2 (a − 3b). This rules out points from the circle because for such points ω(Q 1 (p)) ≤ 1 2 (a − 3b) + 1 2 (a + 3b) = a which is not enough.
The last possibility is to take p ∈ S 3 . Such a point, however, does not match the requirement ω(Q 3 (p)) > b.
Altogether this shows that there is no point with an (a + ε, b + ε)-admissible pair of opposite quadrants.  [BLP] have considered α 3 . They claim that α 3 ≥ 1/2016, this however is based on the false assumption that every set of 9 points in R 3 has a point p with two opposite orthants each containing a point from P \ {p}. Indeed, the least n such that this holds is n = 17. This is related to the Erdős-Szekeres lemma, details can be found in the proof of Thm. 3 of [Fel]. With the correct value α 3 ≥ 2 −7 = 1/128.
Proof. A set of points is t-good if it contains no point determining a pair of opposite orthants each containing t + 2 or more points from P . We will prove that |P | > 2 2 d−1 (t 2 d ) implies that P is not t-good. Hence α d (t2 2 d−1 +d + 1) ≥ t + 2, which yields the bound stated in the theorem.
Let P be a t-good set. One of the orthants from each pair of opposite orthants determined by p ∈ P is small in the sense that it contains at most t points from P \ {p}.
The pattern assigned to p is a collection φ(p) of 2 d−1 sign vectors, such that • O σ (p) is small for each σ ∈ φ(p) and for each pair (O σ , Oσ) of opposite orthants determined by p either σ orσ is in φ(p).
For a given pattern φ we collect all points p ∈ P with φ(p) = φ in a set P φ . Figure 10 shows an example.
x 2 x 1 Figure 10: A 2-good set of 9 points in the plane. Small quadrants of points are indicated by gray angles. The white points may get the pattern φ = {(+, +), (+, −)} assigned.
We have partitioned the points of P according to their pattern. The upper bound on the size of any t-good set P follows from counting the possible patterns and bounding the number of points in each class P φ .
There are 2 d sign vectors paired up in 2 d−1 pairs σ,σ belonging to pairs of opposite orthants. A pattern is a selection of one sign vector from each such pair, therefore: • There are at most 2 2 d−1 different patterns.
From the definition of φ(p) it follows that v σ (p) ≤ t for all σ ∈ φ(p), hence, s(p) ≤ t 2 d−1 . Note that the score of p is the number of points in the small orthants O σ (p) with σ ∈ φ(p).
Consider two points p, q and note that q ∈ O σ (p) if and only if p ∈ Oσ(q). Suppose p and q both belong to P φ , since one of σ andσ is in φ we note that either p is counted in the score of q or q is counted in the score of p. From this we obtain: To reduce the upper bound on s(p) by one observe that for each σ ∈ φ there are points in P φ with v σ (p) < t. This yields the following bound on |P φ |: Combining the bounds for the number of patterns and the size of the classes we find that a t-good set P has at most 2 2 d−1 (t 2 d ) points.
The upper bound on the opposite-orthant depth α d presented in the following theorem is only a factor of two apart from the lower bound of Theorem 3.1. It is evident that the lower bound is not tight. In dimension 2 the upper and lower bounds yield 1 16 ≤ α 2 ≤ 1 8 . From Theorem 1.1 and the example of Figure 2 we know that α 2 = 1 8 . Indeed we suspect that in all dimensions the upper bound gives the true value of α d .
Proof. We have to construct large point sets with small opposite-orthant depth. The construction is in two steps. In the first step we build a set P 0 of 2 2 d−1 points, such that for any point p ∈ P 0 and any pair (O, O op ) of opposite orthants at p either O ∩ P 0 or O op ∩ P 0 equals {p}. In the second step we replace each point of P 0 with a carefully chosen set of t 2 d−1 points, such that the depth remains bounded by t + 1. Hence t + 2 ≥ α d (2 2 d−1 t 2 d−1 ), which yields the bound stated in the theorem.
Let σ be a sign vector andσ be the sign vector of the orthant opposite to O σ . Based on σ we define a binary relation on P , let p ∼ σ q if p ∈ O σ (q) or p ∈ Oσ(q), i.e., q ∈ O σ (p).
A set M of points in R d is monotone if there is a sign vector σ, such that p ∼ σ q for all p, q ∈ M . Equivalently, M is monotone if there is an ordering of the points so that each coordinate is increasing or decreasing in this order. Repeated application of the Erdős-Szekeres lemma implies that any n points in R d contain a monotone subset of size at least n 1 2 d−1 . This bound is best possible. A detailed construction of tight examples can be found e.g. in [Lit]. Due to this result there is a set P 0 of 2 2 d−1 points that does not contain a monotone subset of size three. Hence, for every p ∈ P 0 and every pair of opposite orthants (O, O op ) determined by p at least one of the orthants contains no point of P 0 \ {p}.
As in the proof of the previous theorem we collect sign vectors of small orthants of p ∈ P 0 in a pattern φ(p). Recall that φ(p) contains the sign vector of one of each pair of opposite orthants. We construct the set P by replacing each point p ∈ P 0 by a set Q(p) of t 2 d−1 points, such that q ∈ Q(p) and σ ∈ φ(p) implies |O σ (q) ∩ P | = |O σ (q) ∩ Q(p)| ≤ t + 1, i.e., the orthant O σ (q) is t-small for all σ ∈ φ(p).
For each p ∈ P 0 fix a 'small' box B(p) containing p, such that every choice of one point from each of these boxes yields a set with the same property as P 0 . Formally, for p, q ∈ P 0 with p ∼ σ q we require that p ′ ∼ σ q ′ for all p ′ ∈ B(p) and q ′ ∈ B(q). For the construction of Q(p) in the box B(p) it is convenient to think of B(p) as an open set. To begin with let σ 0 , σ 1 , σ 2 , . . . be an ordering of the 2 d−1 sign vectors in φ(p). Starting with S 0 = ∅ we inductively define subboxes S i for i = 1, . . . , 2 d−1 − 1 of B 0 = B(p) as follows: An example is given in Figure 11. From the construction rules it follows that for i < j and any points p i ∈ S i and p j ∈ S j we have p j = Oσ i (p i ) and consequently p i = O σ i (p j ). This shows that inserting t points into each S i such that any two of these points are in relation σ 0 yields a set Q(p) with the desired properties.
This completes the construction of a t-good set P and the proof of the upper bound on α d .

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The set F of all admissible pairs was defined as the set of all pairs (a, b) such that every set P of n points in the plane contains a point p ∈ P that determines two opposite quadrants with weights at least a and b, respectively. What if we do not require the point p to belong to the set P , and only look for a point z in the plane with the same property of two opposite quadrants determined by it. It is not hard to see that (a, b) is admissible in this model whenever, a + b ≤ 1/2. In addition we have the admissible pairs (a, 0) for all a ≤ 1. A set of points uniformly distributed on a circle C shows that this is the complete description of the set of admissible pairs. Indeed, it is not hard to check that for any point z in the plane, surrounded by C, the horizontal and vertical lines through z determine four quadrants such that the measure of the union of any two opposite ones is at exactly 1 2 . If the point z is not surrounded by C, then the measure of one quadrant is equal to 0 while the opposite quadrant has measure at least 1 2 . One way to generalize the setting is by considering two opposite quadrants determined by a vertical and a horizontal line as a diagonal in the 2-by-2 array of cells determined by these two lines. Then it is natural to consider n−1 vertical lines and n−1 horizontal lines and to find a generalized diagonal of cells in the n-by-n arrangement of cells, determined by these lines, such that each of the cells contains 'many' points of P . By a generalized diagonal we mean a set of n cells, such that no two are in the same row or in the same column (see Figure 12). Since we do not require the vertical and horizontal lines to intersect in points of the set in question, it is equivalent to consider a continuous probability measure in the plane and to look for n − 1 vertical lines and n − 1 horizontal lines and a generalized diagonal in the arrangement of the n-by-n cells, determined by these lines, such that each of the cells has 'large' measure (see Figure 12). This naturally generalizes to higher dimensions as well.
The following theorem gives a partial answer to this problem in any dimension d: Theorem 4.1. Let µ be a continuous probability measure in [0, 1] d . Let n be a positive integer and let α 1 , . . . , α n be a sequence of positive real numbers, such that n i=1 α i = 1 n d−1 . Then there exist numbers x i,j , where 1 ≤ i ≤ d and 0 ≤ j ≤ n, and d permutations π 1 , . . . , π d on {1, . . . , n} with the following properties: 1. For every 1 ≤ i ≤ d, x i,0 = 0 < x i,1 < . . . < x i,n−1 < x i,n = 1.

For every
Remark. Intuitively, the numbers x i,j in Theorem 4.1 define the d dimensional array of n d cells generated by the hyper-planes H i,j = {(x 1 , . . . , x d ) ∈ R d | x i = x i,j }, for all 1 ≤ i ≤ n and 1 ≤ j ≤ d. The theorem then says that there exists such an array (while the measure µ is given) that contains a generalized diagonal of n boxes with measures of at least α 1 , . . . , α n , respectively.
Proof. In fact, the proof is not much more complicated than the statement of the theorem. The proof goes by induction on d. For d = 1 simply define x 1,0 = 0 and for every 1 ≤ j ≤ n let x 1,j = x 1,j−1 +α j . In this case π 1 : [n] → [n] can be chosen to be the identity permutation.
Finally, we let x d,n = 1 and set π d (j) = n, where j is the only index between 1 and n that is not one of π −1 d (1), . . . , π −1 d (n − 1). It is straight forward to check that the numbers x i,j where 1 ≤ i ≤ d and 0 ≤ j ≤ n and π 1 , . . . , π d satisfy the requirements of the theorem.
Recall that when d = 2 and n = 2, the example where the measure µ is evenly distributed on a circle C shows that the result in Theorem 4.1 is best possible apart from the case where α i = 0 for some i. When the values α i in Theorem 4.1 are all equal, then the result is best possible in any dimension. To see this just consider the standard Lebesgue measure on [0, 1] d . It is then not very difficult to verify the following proposition: Proposition 4.2. Let λ be the standard Lebesgue measure on [0, 1] d and let n be a positive integer. Then one cannot find numbers x i,j , where 1 ≤ i ≤ d and 0 ≤ j ≤ n, such that the arrangement of n d bounded cells determined by the hyperplanes H i,j = {z : z i = x i,j } has a generalized diagonal, such that each cell of the generalized diagonal has measure strictly larger than 1/n d .
Below we give some hints for the proof but leave the details to the reader. Before that let us mention that even in the case d = 3 and n = 2 the result in Theorem 4.1 is not best possible for general values of α 1 and α 2 .
One can show the following statement is true. Let P be a finite set of points in R 3 . We assume that no two points in P have a coordinate in common. Let α 1 ≤ 1 12 and α 2 = 1 3 − 3α 1 . Then there exists a point z = (z 1 , z 2 , z 3 ) ∈ R 3 , such that the planes {x i = z i } for i = 1, 2, 3 determine two opposite orthants, one with at least α 1 n points and the other with at least α 2 n points.
Surprisingly, this statement is nearly tight when α 1 is close to 0 (and thus α 2 is close to 1 3 ). To see this consider the following construction of a set P ⊂ R 3 of n = 3m points. It has the property that for any point z and any two opposite orthants, determined by z, such that each contains at least one point of P , there are at most m = 1 3 n points of P in each of these orthants.
To construct P let S be the following set of m vectors in R 3 , very close to the origin: S = { i 100m (1, 1, 1) | i = 0, 1, . . . , m − 1}. We define A, B, and C to be the set S translated by (−1, 0, 1), (0, 1, −1), and (1, −1, 0) respectively. Finally we let P = A ∪ B ∪ C. Note that P is a set of n = 3m points in R 3 no two of which share the same x 1 , x 2 , or x 3 coordinates.
We leave it to the reader to verify that P satisfies the required properties. We also leave the challenge of giving a full characterization of such admissible pairs α 1 and α 2 .

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We gave a complete description of the set F of all admissible pairs (a, b) in the plane. This was done by identifying three line segments and the points ( 1 8 , 1 8 ) and (0, 1 2 ) from the boundary of F.
In higher dimensions we were interested in the maximum number α d , such that (α d , α d ) is admissible. We think that our upper bound on α d is tight, the lower bound leaves room for improvements. It would be interesting to get more information about the set F d of all admissible pairs (a, b) for R d .
In the relaxed setting, where the point determining a pair of opposite orthants need not belong to the point set we could determine the diagonal entry precisely, it is ( 1 2 d , 1 2 d ). This result follows from more general bounds on generalized diagonals. In this relaxed setting we have indicated some additional bounds for admissible pairs (a, b) but many questions remain.