    # Integer and Mixed Integer Tverberg Numbers

We show that the Tverberg number of Z^2 is 4m-3. We also improve the upper bounds for the Tverberg numbers of Z^3 and Z^j ×R^k.

## Authors

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## Introduction

Consider points in and a positive integer . If , the points can always be partitioned into subsets whose convex hulls contain a common point. This is the celebrated theorem of Tverberg [Tve66], which has been the topic of many generalizations and variations since it was first proved in 1966 [BS18, DLGMM19]. In this paper we focus on new versions of Tverberg-type theorems where some of the coordinates of the points are restricted to discrete subsets of a Euclidean space. The associated discrete Tverberg numbers are much harder to compute than their classical real-version counterparts (see for instance the complexity discussion of [Onn91]).

We begin our work remembering the following unpublished Tverberg-type result of Doignon. Consider points with coordinates in and a positive integer . If , then the points can be partitioned into subsets whose convex hulls contain a common point in . According to Eckhoff [Eck00] this result was stated by Doignon in a conference.

A partition of points where the intersection of the convex hulls contains at least one lattice point is called an integer -Tverberg partition and such a common point is an integer Tverberg point for that partition. Regarding the case , the integer -Tverberg partitions are called integer Radon partitions. Any configuration of at least six points in admits an integer Radon partition. This was proved by Doignon in his PhD thesis [Doi75] and later discovered independently by Onn [Onn91]. All these values for are optimal as shown by following examples. The -point configuration , exhibited by Onn in the cited paper, has no Radon partition. To address the optimality when , consider the set (According to Eckhoff [Eck00], this set was proposed by Doignon during the aforementioned conference.) This set has

points and a moment of reflection might convince the reader that it has no integer

-Tverberg partition.

More generally, one can define the Tverberg number for any subset of and an integer as the smallest positive integer with the following property: Any multiset of points in admits a partition into subsets with

 (m⋂i=1conv(Ai))∩S≠∅.

(Here, by “partition of a multiset”, we mean that each element of a multiset is contained in a number of submultisets so that the sum of its multiplicities in the is equal to its multiplicity in .) If no such number exists, we say that . Note that Doignon’s theorem, together with the discussion that follows, allows us to say

 Tv(Z2,m)={6if m=2% ,4m−3otherwise.

### Our contributions

Our first main result generalizes Doignon’s theorem. We determine the exact -Tverberg number (when is at least three) for any discrete subset of , as considered in [DLHRS17]. Before stating this result we recall the Helly number of a discrete subset of as the smallest positive integer with the following property: Suppose is a finite family of convex sets in , and that intersects in at least one point for every subfamily of having at most members. Then intersects in at least one point. If no such integer exists, we say that .

Then we have the following theorem. (The theorem is stated for with finite Helly number, as any with has for all  [Lev51].)

###### Theorem 1.

Suppose is a discrete subset of with . If , then

 Tv(S,m)=H(S)(m−1)+1.

Regarding the case , if , then

 Tv(S,2)=H(S)+1,

and if , then

 H(S)+1≤Tv(S,2)≤H(S)+2,

and both values are possible.

In particular we present a proof of Doignon’s theorem, the special case of Theorem 1 where .

0.1in0.0in Remark: Theorem 1 shows that -Tverberg numbers of planar sets are very closely related to -Helly numbers (see [Ave13, AGS17] and the references there). However, for the case , the bounds on given above cannot be improved. For example, and both have Helly number four, but , while the pigeonhole principle implies that .

Our second main result improves the upper bound on the integer Tverberg numbers for the three-dimensional case .

###### Theorem 2.

The following inequality holds for all :

 Tv(Z3,m)≤24m−31.

Our third main result is an inequality that will be used to derive improved bounds on -Tverberg numbers when is a product of a Euclidean space with some subset of a Euclidean space.

###### Theorem 3.

Let be a subset of a Euclidean space. Then for all positive integers and all , we have

 Tv(S′×Rk,m)≤Tv(S′,Tv(Rk,m)).

For example, choosing of the form leads to the “mixed integer” case. Then Theorem 3 implies that for all positive integers and all , we have

 Tv(Zj×Rk,m)≤Tv(Zj,Tv(Rk,m)).

Moreover, we will use Theorem 3 to obtain the following bound:

 (1) 2j(m−1)(k+1)+1≤Tv(Zj×Rk,m)≤j2j(m−1)(k+1)+1.

Our fourth main result is a generalization of Pach’s positive-fraction selection lemma [Pac98] (see [KKP15] for related bounds). Here is Pach’s result: Given an integer , there exists a constant such that for any set of points in , there exists a point , and disjoint subsets of , say , such that for all and the simplex defined by every transversal of contains . (By “transversal”, we mean a set containing exactly one element from each .)

Unfortunately the point need not be an integer point; furthermore, the proof uses the so-called “second selection lemma” that currently does not exist for integer points (see Pach [Pac98] and Matoušek [Mat02, Chapter 9]). In Section 4, we strengthen the above theorem, such that, as a consequence, the theorem now extends to the integer case—indeed, to any scenario where one has points of high half-space depth in the following sense:

Given a finite set of points in and a point , we say that is of half-space depth with respect to if any half-space containing contains at least points of (when the context is clear, we will simply say that is of depth ). Then here is our theorem.

###### Theorem 4.

For any integer and real number , there exists a constant such that the following holds. For any set of points in and any point of half-space depth at least , there exist disjoint subsets of , say , such that

• for , and

• every simplex defined by a transversal of contains .

0.1in0.0in Remark: Our proof yields a constant whose value is exponential in the dimension .

Note that the existence of integer points of high half-space depth (Lemma 1) together with Theorem 4 implies the following integer version of the positive-fraction selection lemma.

###### Corollary 1.

Let be a set of points in . Then there exists a point , and disjoint subsets of , say , such that for all , and the simplex defined by every transversal of contains .

0.1in0.0in Remark: In particular, this implies that belongs to many distinct Tverberg partitions—at least distinct Tverberg partitions, with each such Tverberg partition containing sets.

### Related Results and Organization of the paper

The problem of computing the Tverberg number for with seems to be challenging. It has been identified as an interesting problem since the 1970’s [GS79] and yet the following inequalities are almost all that is known about this problem: for the general case, De Loera et al. [DLHRS17] proved

 (2) 2d(m−1)+1≤Tv(Zd,m)≤d2d(m−1)+1,for d≥1 and m≥2.

Two special cases get better bounds:

 (3) Tv(Z3,2)≤17and5⋅2d−2+1≤Tv(Zd,2)for d≥1.

The left-hand side inequality is due to Bezdek and Blokhuis [BB03] and the right-hand side was proved by Doignon in his PhD thesis (and rediscovered by Onn).

Previously established bounds for the “mixed integer” case include the bounds for the Radon number (-Tverberg number) found by Averkov and Weismantel [AW12].

 2j(k+1)+1≤Tv(Zj×Rk,2)≤(j+k)2j(k+1)−j−k+2.

Later, De Loera et al. [DLHRS17] gave the following general bound for all mixed integer Tverberg numbers:

 Tv(Zj×Rk,m)≤(j+k)2j(m−1)(k+1)+1.

Note that (1) above is a simultaneous improvement of both of these.

Previous bounds and related work on more general -Tverberg numbers can also be found in [DLHRS17], including the following bound for any discrete :

 Tv(S,m)≤H(S)(m−1)d+1.

The following lemma about integer points of high half-space depth is used throughout the paper. See [BO16] for a proof and related results.

###### Lemma 1.

Consider a multiset of points in . If (counting multiplicities), then there is a point of half-space depth in .

The paper is organized as follows. In Section 1, we prove Theorem  1 using a somewhat similar strategy to Birch’s proof of the planar case of the original Tverberg theorem [Bir59]. In Section 2, we prove Theorem 2 using techniques reminiscent of those in [DLHRS17]. In Section 3, we prove Theorem 3 (adapting an approach by Mulzer and Werner [MW13, Lemma 2.3]) and collect some consequences of the main theorems presented above, including (1). Finally, in Section 4, we prove Theorem 4 by proving a new lemma and adapting the methods of Pach in [Pac98].

## 1. Tverberg Numbers over Discrete Subsets of R2: Proof of Theorem 1

We start with the proof of the special case because it nicely illustrates the techniques of the more general proof of Theorem 1.

### 1.1. Proof of the special case S=Z2

The theorem will follow easily from the following two lemmas, the first covering the case and the second the case .

###### Lemma 2.

Consider a multiset of points in with and . If is a point of depth , then there is an -Tverberg partition of with as Tverberg point.

###### Lemma 3.

Consider a multiset of points in with . If is a point of depth two, then there is a Radon partition of with as Tverberg point.

###### Proof of special case S=Z2..

Consider a multiset of at least points in . By Lemma 1, has an integer point of depth . If is an element of with multiplicity , then take the singletons as of the sets in the Tverberg partition. Then is a point of depth of the remaining points. If , we are done, and if , the point is in the convex hull of the remaining points and we take them to be the last set in the desired partition. If , according to Lemma 2, there is an -Tverberg partition of the remaining points with as Tverberg point. There is thus an -Tverberg partition of with as Tverberg point. The case is treated similarly with the help of Lemma 3 in place of Lemma 2. ∎

###### Proof of Lemma 2.

Since is not in , up to a radial projection, we can assume that the points of are arranged in a circle around . Define and to be respectively the quotient and the remainder of the Euclidean division of by . Define moreover to be .

Suppose first that is a point of depth . In such a case, we arbitrarily select a first point in , and label clockwise the points with elements in according to the following pattern:

 1,2,…,m,1,2,…,e,1,2,…,m,1,2,…,e,1,2,…m,1,2,…k,

where . Note that we have . Each half-plane delimited by a line passing through contains at least consecutive points in this pattern and thus has at least one point with each of the different labels. Partitioning the points so that each subset consists of all points with a fixed label, we therefore obtain an -Tverberg partition with as Tverberg point.

Suppose now that is not a point of depth . There is thus a closed half-plane , delimited by a line passing through , with . The complementary closed half-plane to , which we denote by , is such that . Define to be . Since , we have . Denote the points in by , where the indices are increasing when we move clockwise. We label with from to , and then label with from to . We then continue labeling the points of , still moving clockwise, using labels See Figure 1 for an illustration of the labeling scheme.

The labeling pattern is such that any sequence of consecutive points either has all labels, or contains the two consecutive points and . Let us prove that any closed half-plane delimited by a line passing through contains at least one point with each label. Once this is proved, the conclusion will be immediate by taking as subsets of points those with same labels, as above.

If such an does not simultaneously contain and , then contains at least one point with each label. Consider thus a closed half-plane delimited by a line passing through and containing and . Note that according to Farkas’ lemma ([Sch03] Theorem 5.3) , cannot be separated from and by a line passing through , since they are all in . This means that either contains , or contains . In any case, contains a point with each label. ∎

###### Proof of Lemma 3.

As before, we assume that the points in are arranged on a circle centered at . If is even, it clearly suffices to label the points in order, alternating between 1 and 2. We may therefore assume that

is odd, and thus

. If is a point of depth three, it suffices to label the points alternating labels between 1 and 2, except with two consecutive points labeled 1. If is odd but is not a point of depth three, then and there is a half-plane containing with . The complementary half-plane has and we follow a similar strategy as in the second half of Lemma 2. Namely, we denote the points in by , where the indices are increasing when we move clockwise. Then we label with 2, with 1, with 1, and with 2. We continue this pattern for , labeling with 1 if is odd, and with 2 if is even. For the remaining points in we continue labeling clockwise, alternating between the labels 1 and 2.

The labeling pattern is such that any sequence of consecutive points either has both labels, or contains the two consecutive points and . As in Lemma 2 it suffices to show that any closed half-plane delimited by a line passing through contains at least one point with each label.

If such an does not simultaneously contain and , then contains at least one point with each label. Consider thus a closed half-plane delimited by a line passing through and containing and . Note that according to Farkas’ lemma, cannot be separated from and by a line passing through , since they are all in . This means that either contains , or contains and . In any case, contains a point with each label. ∎

### 1.2. Proof of the general case

The proof of the general case is split into three lemmas addressing the lower bound, the upper bound for , and the upper bound for , respectively.

###### Lemma 4.

For any discrete set with finite Helly number , we have .

###### Proof.

It suffices to exhibit a subset , of cardinality , with the property that no point in is of half-space depth with respect to . By Lemma 2.6 in [ADLS17], there exists a set of points in in convex position with the property that . Let be the multiset given by taking each point in with multiplicity , so . No points of are in . Since was taken to be in convex position, for any point in , there exists a line such that one side of that line has at most points in . Thus cannot contain a point of half-space depth with respect to . ∎

###### Lemma 5.

For any discrete set with finite Helly number , we have whenever . For the case , we have .

###### Proof.

The proof of Lemma 5 is the same as the proof of Theorem 1. In particular, we can use Lemmas 2 and 3 as they are stated, except that we use the following result (Theorem 2 in [BO16] with

being the uniform probability measure on

) in place of Lemma 1. For any discrete discrete subset of a Euclidean space with finite Helly number , and any set with , there exists a point that is of half-space depth with respect to . ∎

###### Lemma 6.

For a discrete set with finite Helly number , we have .

###### Proof.

The case implies that consists of a single point, so the result trivially follows. If , it must be that all points in are collinear (as any set containing a non-degenerate triangle has Helly number at least 3), and thus we can take median of any set with at least points in as the desired -Tverberg point. Thus for the remainder of the proof we assume that .

Given any set of points in , there exists an -Tverberg partition, say by the classical Tverberg theorem. We denote by the convex hulls of the subsets in . As is a nonempty polygon, say , (possibly just a point or line segment) we pick an arbitrary vertex of .

It suffices to show that . We can assume that is not a vertex of any , since otherwise .

Since is a vertex of , it must be contained in a one dimensional face of at least one . Since is not a vertex of any , in fact is in the relative interior of . For to be a vertex of , it must also be in another one dimensional face, say , of some other , such that is not parallel to . Moreover, must be in the relative interior of , and we also have .

Denote by and the vertices of and respectively. We have that are the vertices of a convex quadrilateral with diagonals intersecting at , by the assumption that and are non parallel. Out of the four triangles , any three have at least one vertex in common, and therefore intersect in . Since , the four triangles therefore all intersect in . This intersection point is , the point where the diagonals of the quadrilateral intersect. ∎

## 2. Tverberg Numbers over Z3: Proof of Theorem 2

We state the following lemma without proof; it is a consequence, upon close inspection of the argument, of the proof of the main theorem in the already mentioned paper by Bezdek and Blokhuis [BB03].

###### Lemma 7.

Consider a multiset of at least points in and a point of depth in . There is a bipartition of into two subsets whose convex hulls contain .

Next, we prove the following.

###### Lemma 8.

Consider a multiset of points in with and . If is a point of depth , then there is an -Tverberg partition of with as Tverberg point.

###### Proof.

Since is not an element of , we assume without loss of generality that the points of are located on a sphere centered at , as in the proof of Theorem 1.

We claim that there exist pairwise disjoint subsets of , each having in its convex hull and each being of cardinality at most . (Here “pairwise disjoint” means that each element of is present in a number of ’s that does not exceed its multiplicity in .) We proceed by contradiction. Suppose that we can find at most such subsets ’s. Then, by Carathéodory’s theorem, is not in the convex hull of the remaining points in . Therefore there is a half-space delimited by a plane containing such that . On the other hand, since each contains in its convex hull (and we can assume the are minimal with respect to containing ), we have for all . Therefore , which is a contradiction since is a point of depth in . There are thus disjoint subsets as claimed.

Let denote . Consider an arbitrary half-space delimited by a plane containing . Since for all , we have . Furthermore , so . Since is arbitrary, is a point of depth of . Also, , so Lemma 7 implies that can be partitioned into two sets whose convex hulls contain . With the subsets , we have therefore an -Tverberg partition of , with as Tverberg point. ∎

From these two lemmas we can now finish the proof of Theorem 2.

###### Proof of Theorem 2.

Consider a multiset of points in . The case is the already mentioned result by Bezdek and Blokhuis. Assume that . Applying Lemma 1, has an integer point of depth . If is an element of with multiplicity , then take the singletons as of the sets in the Tverberg partition.

If , we are done. If , the point is still in the convex hull of points in , and thus we are done. And if , the point is still a point of depth of the remaining points. Thus, we may apply Lemma 8 to get an -Tverberg partition of the remaining points, with as Tverberg point, and conclude the result. ∎

## 3. Tverberg Numbers over Q×Rk: Proof of Theorem 3

In this section, we prove Theorem 3. We adapt an approach by Mulzer and Werner [MW13, Lemma 2.3] and show how the results of our paper can be combined to improve known bounds and to determine new exact values for the Tverberg number in the mixed integer case, as well as better bounds for certain -Tverberg numbers.

###### Proof of Theorem 3.

Let . Choose a multiset in with . It suffices to prove that can be partitioned into subsets whose convex hulls contain a common point in .

Let be the projection of onto . Since , there is a partition of into submultisets whose convex hulls contain a common point in . The are the projections onto of disjoint subsets forming a partition of . For each , we can find a point projecting onto .

The points belong to . As , there exists a partition of into and a point such that for all . For each , define to be . We have, for each

 p∈conv⎛⎝⋃i∈Iℓqi⎞⎠⊆conv⎛⎝⋃i∈Iℓconv(Qi)⎞⎠=conv(Aℓ)

and the form the desired partition. ∎

Here are the new bounds and exact values we get:

1. .

2. .

3. .

4. .

5. If with finite Helly number , then

 Tv(S′×Rk,m)≤H(S′)(m−1)(k+1)+1.

The lower bound in (4) is obtained by repeated applications of Proposition 1 below. The upper bounds follow from Theorem 3, combined with the fact that , Theorem 1 for , Theorem 2, the upper bound in Equation (2), and Theorem 1 respectively.

###### Proposition 1.

Let and be two non-negative integers. Then we have

 Tv(Zj+1×Rk,m)>2Tv(Zj×Rk,m)−2.

We prove Proposition 1 by following the idea of the proof of Proposition 2.1 in [Onn91].

###### Proof of Proposition 1.

 Tv(Zj+1×Rk,m)≤2Tv(Zj×Rk,m)−2.

Choose to be a set of points in with no -Tverberg partition. Let Since has cardinality , there exists an -Tverberg partition of with . Furthermore is in . That implies either or . In either case or has an -Tverberg partition, a contradiction with our choice of . ∎

## 4. A Generalized Fraction Selection Lemma: Proof of Theorem 4

Our proof relies on the simplicial partition theorem of Matoušek, used in a similar manner as in [MR17], which states the following.

Given an integer and a parameter , there exists a constant such that for any set of points in , there exists an integer and a partition of such that

• for each , , and

• any hyperplane intersects the convex hull of less than

sets of the partition.

The constant is independent of and depends only on .

We now prove the following key lemma.

###### Lemma 9.

For any integer , there exists a constant such that the following holds. For any set of points in and a real number , there exists a partition , , of such that

• for each , and

• the convex hull of any transversal of contains all points in of half-space depth at least .

###### Proof.

Apply the simplicial partition theorem (Theorem 5) to with , and let the resulting partition be . Note that as for each , we have . Now partition arbitrarily each of biggest sets in into two equal parts, and let the resulting partition be . Clearly each set of this partition has size in the interval . This proves the first part. Note also that each hyperplane intersects the convex hull of at most twice as many sets, i.e., less than sets of the partition .

To see the second part, let be any point of half-space depth at least , and any transversal of . For contradiction, assume that . Then there exists a hyperplane containing in one of its two open half-spaces, say , and containing in the half-space . We will show that then there exists an index such that . But then , a contradiction to the fact that is a transversal of .

It remains to show the existence of a set such that . Towards this, we bound . Each point of lying in belongs to a set such that either

• , in which case we are done, or

• . As contains at least one point of , we must have . As argued earlier, there are less than such sets.

Thus we have

 (4) ∣∣P∩H−∣∣<2cd⋅r1−1d⋅2nr=4cd⋅n⌈(4cdα)d⌉1d≤α⋅n.

On the other hand, as has half-space depth at least and , we have , a contradiction to inequality (4). ∎

0.1in0.0in Remark: In particular, for , there exist at least -sized subsets, each of whose convex hull contains all integer points of depth at least .

###### Proof of Theorem 4..

Given the point set in and a point of half-space depth , apply Lemma 9 with and to get a partition consisting of sets, where . By discarding at most points of , we can derive a partition on the remaining points of , say , such that the ’s are equal-sized disjoint subsets of , i.e., for all . Furthermore, every transversal of contains all points in of half-space depth at least , and thus .

For each transversal of , the point lies in the convex hull of , and by Carathéodory’s theorem, there exists a -sized subset of whose convex hull also contains . By the pigeonhole principle, there must exist sets of , say the sets , such that at least

 (5) (n2r)r(rd+1)(n2r)r−(d+1)≥(n2r)(d+1)(erd+1)d+1=1(erd+1)d+1⋅d+1∏i=1|Pi|.

distinct transversals of contain .

The rest of the proof follows the one of Pach [Pac98]. In brief, we view the ’s as parts of a -partite hypergraph with vertices corresponding to points in and a hyperedge corresponding to each transversal of containing . As there are such transversals by inequality (5), we apply a weak form of the hypergraph version of Szemerédi’s regularity lemma (see [Mat92] Theorem 9.4.1) to derive the existence of constant-fraction sized subsets such that the following is true, for some constant :

for any , with for , we have the property that there exists at least one transversal of whose convex hull contains .

Then the same-type lemma ([BV98] Theorem 2) applied to gives constant-fraction sized subsets such that each transversal of has the same order type with respect to .

We can set up the parameters for the same-type lemma and the weak regularity lemma such that , for all . Then the weak regularity lemma implies that there exists at least one transversal of that contains . However, as each transversal of has the same order type, it must be that each transversal of contains . These are the required subsets.

The size of each is a constant-fraction of , say , where the constant depends on the constants in inequality (5), in the weak regularity lemma and in the same-type lemma. All of these depend only on and . ∎

## 5. Acknowledgments

This work was partially supported by NSF grant DMS-1440140, while the first and third authors were in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the Fall 2017 semester. The first and second authors were also partially supported by NSF grants DMS-1522158 and DMS-1818169. The fourth author was supported by ANR grant SAGA (ANR-14-CE25-0016). Last, but not least, we are grateful for the information and comments we received from N. Amenta, G. Averkov, A. Basu, J.P. Doignon, P. Soberón, D. Oliveros, and S. Onn. In particular we thank Pablo Soberón for noticing that the same methods used in an earlier version could be extended to yield Theorem 1.

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