A Center Transversal Theorem for Hyperplanes and Applications to Graph Drawing

Abstract

Motivated by an open problem from graph drawing, we study several partitioning problems for line and hyperplane arrangements. We prove a ham-sandwich cut theorem: given two sets of n lines in ℝ², there is a line ℓ such that in both line sets, for both halfplanes delimited by ℓ, there are $\sqrt{n}$ lines which pairwise intersect in that halfplane, and this bound is tight; a centerpoint theorem: for any set of n lines there is a point such that for any halfplane containing that point there are $\sqrt{n/3}$ of the lines which pairwise intersect in that halfplane. We generalize those results in higher dimension and obtain a center transversal theorem, a same-type lemma, and a positive portion Erdős–Szekeres theorem for hyperplane arrangements. This is done by formulating a generalization of the center transversal theorem which applies to set functions that are much more general than measures. Back to graph drawing (and in the plane), we completely solve the open problem that motivated our search: there is no set of n labeled lines that are universal for all n-vertex labeled planar graphs. In contrast, the main result by Pach and Toth (J. Graph Theory 46(1):39–47, 2004), has, as an easy consequence, that every set of n (unlabeled) lines is universal for all n-vertex (unlabeled) planar graphs.

1 Introduction

Consider a mapping of the vertices of a graph to distinct points in the plane and represent each edge by the closed line segment between its endpoints. Such a graph representation is a (straight-line) drawing if the only vertices that each edge intersects are its own endpoints. A crossing in a drawing is a pair of edges that intersect at some point other than a common endpoint. A drawing is crossing-free if it has no crossings.

One main focus in graph drawing is finding methods to produce drawings or crossing-free drawings for a given graph with various restrictions on the position of the vertices of the graph in the plane. For instance, there is plethora of work where vertices are required to be placed on integer grid points or on parallel lines in two or three dimensions.

Given a set R of n regions in the plane and an n-vertex graph G, consider a class of graph drawing problems where G needs to be drawn crossing-free by placing each vertex of G in one region of R. If such a drawing exists, then R is said to support G. The problems studied in the literature distinguish between two scenarios: in one, each vertex of the graph is prescribed its specific region (that is, the vertices and the regions are labeled); in the other, each vertex is free to be assigned to any of the n regions (that is, the vertices are unlabeled).

When regions are points in the plane, Rosenstiehl and Tarjan [19] asked if there exists a set of n points that support all n-vertex unlabeled planar graphs. This question is answered in the negative by De Fraysseix [8, 9]. On the contrary, every set of n points in general position supports all n-vertex unlabeled outerplanar graphs, as proved by Gritzmann et al. [13] and recapitulated in Lemma 14.7 in the text by Agarwal and Pach [1]. If the drawings are not restricted to be straight-line, then every set of labeled points supports every labeled planar graph, as shown by [18]. However, Ω(n) bends per edge may be necessary in any such crossing-free drawing.

When regions are labeled lines in the plane, Estrella-Balderrama et al. [11] showed that for every n≥6, there is no set of n parallel lines in the plane that support all labeled n-vertex planar graphs. The authors moreover characterized a (sub)class of n-vertex planar graphs that are supported by every set of n-parallel lines, for every labeling of the graphs in the class. That class is mainly comprised of several special families of trees. Dujmović et al. [7] showed that no set of n lines that all intersect in one common point supports all n-vertex labeled planar graphs. Moreover, they show that for every n large enough, there is a set of n lines in general position that does not support all labeled n-vertex planar graphs. They leave as the main open problem the question of whether, for every n large enough, there exists a universal set of n lines in the plane, that is, one that supports all labeled n-vertex planar graphs. In Sect. 5, as our main graph drawing result, we answer that question in the negative. The main result by Pach and Toth [17] on monotone drawings, has, as an easy consequence, that in the unlabeled case, every set of n-lines supports every n-vertex unlabeled planar graph. As a side note, we give an alternative and direct proof of that fact. The result illustrates the sharp contrast with the labeled case.

While the positive result is proved using little of the geometry in the arrangement, the non-existence of universal line sets required extraction of some (bad) substructure from any line arrangement. This prompted us to study several structural and partitioning problems for line and hyperplane arrangements.

Hyperplane Arrangements

Partitioning problems are central to our understanding of discrete and computational geometry, and while much work has focused on partitioning point sets, probability distributions or measures, much less is understood for sets of lines in ℝ² or hyperplanes in ℝ^d. This is partially due to the fact that a line (or a hyperplane), being infinite, cannot be contained in any bounded region, or even in a halfplane (except if the boundary of the halfplane is parallel to the given line). Previous work (such as cuttings [5, 15] or equipartitions [14]) has focused on identifying and bounding the number of lines/hyperplanes intersecting a set of regions. Other work [6] focused on partitioning the vertices of the arrangements rather than the lines themselves. Those results have found numerous applications. Our graph drawing problem motivates a different approach.

An arrangement L of n lines in ℝ² is composed of vertices V(L) (all pairwise intersections between lines of L), edges connecting these vertices, and half-lines. If we omit the half-lines, we are left with a finite graph which can be contained in a bounded region of the plane, in particular, it is contained in the convex hull CH(V(L)) of the vertices of the arrangement. Therefore, a natural way of evaluating the portion of an arrangement contained in a given convex region C is to find the largest subset L′ of lines of L such that the arrangement of L′ (without the half-lines) is contained in C, or equivalently, such that all pairwise intersections of lines in L′ lie in C.

It is not hard to show that, in any arrangement of n lines, a line ℓ can be found such that for both closed halfplanes bounded by ℓ there are at least $\sqrt{n}$ lines which pairwise intersect in that halfplane. This provides the analogue of a bisecting line for point sets. In Sect. 3.1, we show that any two line arrangements can be bisected simultaneously in this manner, thus proving a ham-sandwich theorem for line arrangements. We also prove a centerpoint theorem: for any arrangement of n lines, there is a point q such that for any halfplane containing q, there are at least $\sqrt{n/3}$ lines of the arrangement that pairwise intersect in that halfplane. In Sect. 3.2 we generalize these notions to higher dimensions and prove a center transversal theorem: for any k and d, there is a growing function Q such that for any k sets A ₁,…,A _k of hyperplanes in ℝ^d, there is a (k−1)-flat π such that for any halfspace h containing π there is a subset $A'_{i}$ of Q(|A _i |) hyperplanes from each set A _i such that any d hyperplanes of $A'_{i}$ intersect in h. The bound Q we find is related to Ramsey numbers for hypergraphs.

Ham-sandwich theorems have a number of natural consequences. In Sect. 2 we show a same-type lemma for hyperplane arrangements: informally, for any k arrangements A ₁,…,A _k of hyperplanes in general position (no d+1 share a point) and that are large enough, we can find a large subset of hyperplanes $A'_{i}$ from each set A _i such that the convex hulls $\mathit{CH}(A'_{i})$ of the vertices in the arrangements $A'_{i}$ are well-separated, that is, no hyperplane hits d+1 of them. In the plane, we also show a positive portion Erdős–Szekeres theorem: for any integers k and c there is an integer N such that any set of N lines in general position contains k subsets A ₁,…,A _k of c lines each such that the vertices of each arrangement A _i can be separated from those of all the others by a line.

All the results above would be relatively easy to prove if the set function we were computing—the maximum subset of hyperplanes that have all d-wise intersections in a given region—was a measure. Unfortunately it is not. However, in Sect. 2, we identify basic properties much weaker than those of measures which, if satisfied by a set function, guarantee a central-transversal theorem to be true.

2 Center Transversal Theorem

The center transversal theorem is a generalization of both the ham-sandwich cut theorem and the centerpoint theorem discovered independently by Dol’nikov [10], and Živaljević and Vrećica [21]. The version of Dol’nikov is defined for a class of set functions that is more general than measures. Let ${\mathcal{H}}$ be the set of all open halfspaces in ℝ^d and let ${\mathcal{G}}$ be a family of subsets of ℝ^d closed under union operations and that contains ${\mathcal{H}}$. A charge μ is a finite set function that is defined for all set $X\in {\mathcal{G}}$, and that is monotone (μ(X)≤μ(Y) whenever X⊆Y) and subadditive (μ(X∪Y)≤μ(X)+μ(Y)). A charge μ is concentrated on a set X if for every halfspace $h\in {\mathcal{H}}$ s.t. h∩X=∅, μ(h)=0. Dol’nikov shows¹:

Theorem 1

(Center transversal theorem [10])

For arbitrary k charges μ _i , i=1,…,k, defined on ${\mathcal{G}}$ and concentrated on bounded sets, there exists a (k−1)-flat π such that

for every open halfspace $h\in {\mathcal{H}}$ containing π.

A careful reading of the proof of this theorem reveals that its statement can be generalized, and the assumptions on μ _i weakened. We first notice that the subadditive property is only used in the proof for taking the union of a finite number of halfspaces from ${\mathcal{H}}$. Therefore, define μ to be ${\mathcal{H}}$ -subadditive if

for any finite set $H\subset {\mathcal{H}}$ of halfspaces.

Next, notice that in order for the proof to go through, the set function μ need not be real-valued. Recall [4] that a totally ordered unital magma (M,⊕,≤,e) is a totally ordered set M endowed with a binary operator ⊕ such that M is closed under ⊕ operations, ⊕ has neutral element e (i.e., x⊕e=x=e⊕x) and is monotone (i.e., a⊕c≤b⊕c and c⊕a≤c⊕b whenever a≤b). Further, for all x∈M and c∈ℕ, define the cth multiple of x as $cx := \bigoplus^{c} x := \underbrace{x\oplus(x \oplus( \ldots\oplus x)\ldots)}_{c\ \mathrm{times}}$.

Then, it suffices that μ take values over M, and use e as the 0 used in the definition of a concentrated set function above. It is then the addition operator ⊕ which is to be used in the definition of the subadditive (or ${\mathcal{H}}$-subadditive) property and in the proof of the theorem. Thus, just by reading the proof of Dol’nikov under this new light we have:

Theorem 2

Let μ _i , i=1,…,k be k set functions defined on ${\mathcal{G}}$ and taking values in a totally ordered unital magma (M,⊕,≤,e). If the functions μ _i are monotone ${\mathcal{H}}$-subadditive and concentrated on bounded sets, there exists a (k−1)-flat π such that

for every open halfspace $h\in {\mathcal{H}}$ containing π.

3 Center Transversal Theorem for Arrangements

Let A be an arrangement of n hyperplanes in ℝ^d. We write V(A) for the set of all vertices (intersection points between any d hyperplanes) of A and CH(A)=CH(V(A)) for the convex hull of those points. In the arguments that follow, by abuse of language, we will write A and mean V(A) or CH(A). For example, we say that the arrangement A is above hyperplane h when all points in V(A) are above h. More generally for a region Q in ℝ^d, we say that the arrangement A does not intersect Q if CH(A) does not intersect Q. We say that the k arrangements A ₁,A ₂,…,A _k are disjoint if their convex hulls do not intersect. They are separable if they are disjoint and no hyperplane intersects d+1 of them simultaneously.

Let ${\mathcal{H}}$ be the set of all open halfspaces in ℝ^d and let ${\mathcal{G}}$ be a family of subsets of ℝ^d closed under union operations and that contains ${\mathcal{H}}$. For any set $S\in {\mathcal{G}}$, let μ _A (S) be the maximum number of hyperplanes of A that have all their vertices inside of S, that is,

In particular, μ _A (ℝ^d)=μ _A (CH(A))=n and μ _A (∅)=d−1.

3.1 Lines in ℝ²

We start with the planar case. Thus, A is a set of lines in ℝ², and ${\mathcal{H}}$ is the set of all open halfplanes. Recall the Erdős–Szekeres theorem [12].

Theorem 3

(Erdős–Szekeres)

For all integers r, s, any sequence of n>(r−1)(s−1) numbers contains either a non-increasing subsequence of length r or an increasing subsequence of length s.

We show:

Lemma 4

For any two sets $S_{1}\in {\mathcal{H}}$ and $S_{2} \in {\mathcal{G}}$,

Proof

Let ℓ be a line defining two open halfplanes ℓ ⁺ and ℓ ⁻ such that S ₁=ℓ ⁻ and let $S_{2}' = S_{2}\setminus\ell^{-}$. Rotate and translate the plane so that ℓ is the (vertical) y axis, and ℓ ⁺ contains all points with positive x coordinate. Let A′ be a maximum cardinality subset of A such that V(A′)⊆S ₁∪S ₂. Let l ₁,…,l _|A′| be the lines in A′ ordered by increasing order of their slopes, and let Y=(y ₁,…,y _|A′|) be the y coordinates of the intersections of the lines l _i with line ℓ, in the same order. For any set A ₁⊆A′ such that the y _i values of the lines in A ₁ form an increasing subsequence in Y, notice that V(A ₁)⊆S ₁. Likewise, for any set A ₂⊆A′ that forms a non-increasing subsequence in Y, we have $V(A_{2})\subseteq S_{2}'$. Any such set A ₁ is of size |A ₁|≤μ _A (S ₁) and any such set A ₂ is of size $|A_{2}|\leq\mu_{A}(S_{2}')\leq\mu_{A}(S_{2})$.

Therefore, Y has no non-decreasing subsequence of length μ _A (S ₁)+1 and no non-increasing subsequence of length μ _A (S ₂)+1, and so by Theorem 3, μ _A (S ₁∪S ₂)=|A′|=|Y|≤μ _A (S ₁)μ _A (S ₂). □

Corollary 5

The set function μ _A takes values in the totally ordered unital magma $({\mathbb{R}}_{0}^{+},\cdot,\leq, 1)$; it is monotone and ${\mathcal{H}}$-subadditive.

We thus apply the generalized center transversal theorem with k=2 to obtain a ham-sandwich cut theorem:

Theorem 6

For any arrangements A ₁ and A ₂ of lines in ℝ², there exist a line ℓ bounding closed halfplanes ℓ ⁺ and ℓ ⁻ and sets $A_{i}^{\sigma}$, i∈1,2, σ∈+,− such that $A_{i}^{\sigma}\subseteq A_{i}, |A_{i}^{\sigma}|\geq|A_{i}|^{1/2}$, and $V(A_{i}^{\sigma})\in\ell^{\sigma}$.

Note that this statement is similar to the result of Aronov et al. [2] on mutually avoiding sets. Specifically, two sets A and B of points in the plane are mutually avoiding if no line through a pair of points in A intersects the convex hull of B, and vice versa. Note that, on the other hand, our notion of separability for lines is equivalent to the following definition in the dual. Two sets A and B of points in the plane are separable if there exists a point x such that all the lines through pairs of points in A are above x and all the lines through pairs of points in B are below x or vice versa. Aronov et al. show in Theorem 1 of [2] that any two sets A ₁ and A ₂ of points contains two subsets $A_{i}'\subseteq A_{i}$, $|A_{i}'|\geq|A_{i}/12|^{1/2}$, i∈{1,2} that are mutually avoiding. That this bound is tight, up to a constant, was proved by Valtr [20]. In the dual, Theorem 6 states that for any two sets A ₁ and A ₂ of points in ℝ², there exist a point ℓ and sets $A_{i}^{\sigma}$, i∈1,2, σ∈+,− such that $A_{i}^{\sigma}\subseteq A_{i}, |A_{i}^{\sigma}|\geq|A_{i}|^{1/2}$, and all lines through pairs of points in $A_{i}^{+}$ are above ℓ and all lines through pairs of points in $A_{i}^{-}$ are below ℓ. While similar, neither the two results nor the two notions of mutually avoiding and separable are equivalent. It is not difficult to show that no result/notion immediately implies the other. Moreover, neither our proof of Theorem 6 nor the proof of Theorem 1 in [2] give two sets that are, at the same time, mutually avoiding and separable.

Note that the bound in Theorem 6 is tight: assume n is the square of an integer. Construct the first line arrangement A ₁ with $\sqrt{n}$ pencils of $\sqrt{n}$ lines each, centered at points with coordinates (−1/2,i) for $i=1,\ldots,\sqrt{n}$, and the slopes of the lines in pencil i are distinct values in $[1/2-(i-1)/\sqrt{n},1/2-i/\sqrt{n}]$. Thus all intersections other than the pencil centers have x coordinates greater than 1/2. The line x=0 delimits two halfplanes in which $\mu_{A_{1}}(x\leq0) = \sqrt{n}$ since any set of more than $\sqrt{n}$ lines have lines from different pencils which intersect on the right of x=0, and $\mu_{A_{1}}(x\geq0) = \sqrt{n}$ since any set of more than $\sqrt{n}$ lines has two lines in the same pencil which intersect left of x=0. Since $\mu_{A_{1}}$ is monotone, no vertical line can improve this bound on both sides. Perturb the lines so that no two intersection points have the same x coordinate. For A ₂, build a copy of A ₁ translated down, far enough so that no line through two vertices of A ₁ intersects CH(A ₂) and conversely. Therefore any line not combinatorially equivalent to a vertical line (with respect to the vertices of A ₁ and A ₂) does not intersect one of the arrangements and so there is no better cut than x=0.

Applying the generalized center transversal theorem with k=1 gives a centerpoint theorem with a bound of |A|^1/3. A slightly more careful analysis improves that bound.

Theorem 7

For any arrangement A of lines in ℝ², there exists a point q such that for every halfplane h containing q there is a set $A'\subseteq A, |A'|\geq\sqrt{|A|/3}$, such that V(A′)∈h.

Proof

Let H be the set of halfplanes h such that $\mu_{A}(h)< z = \sqrt{|A|/3}$. The halfspace depth δ(q) is the minimum value of μ _A (h) for any halfspace containing q. Therefore, the region of depth ≥z is the intersection of the complements $\overline{h}$ of the halfplanes h∈H. If there is no point of depth ≥z then the intersection of the complements of halfplanes in H is empty, and so (by Helly’s Theorem) there must be three halfplanes h ₁, h ₂, and h ₃ in H such that the intersection of their complements $\overline{h_{1}}\cap \overline{h_{2}}\cap\overline{h_{3}}$ is empty. But then, there is at least one point q∈h ₁∩h ₂∩h ₃. Let $h'_{i}$ be the translated halfplanes h _i with point q on the boundary. Since $h'_{i}\subseteq h_{i}$, $\mu_{A}(h'_{i})\leq\mu_{A}(h_{i})<z$. The point q and the three halfplanes through it are witness that there is no point of depth ≥z.

The three lines bounding those three halfplanes divide the plane into 6 regions. Every line misses one of the three regions $h'_{1}\cap\overline{h'_{2}}\cap \overline{h'_{3}}$, $\overline{h'_{1}}\cap h'_{2}\cap\overline{h'_{3}}$, and $\overline {h'_{1}}\cap\overline{h'_{2}}\cap h'_{3}$. Classify the lines in A depending on the first region it misses, clockwise. The largest class A′ contains ≥|A|/3 lines. Assume without loss of generality that all lines in A′ miss $h'_{1}\cap\overline{h'_{2}}\cap\overline{h'_{3}}$, then all intersections between lines of A′ are in $h'_{2}\cup h'_{3}$. By Lemma 4,

a contradiction. □

3.2 Hyperplanes in ℝ^d

We first briefly review a bichromatic version of Ramsey’s theorem for hypergraphs.

Theorem 8

For all p,a,b∈ℕ, there is a natural number R=R _p (a,b) such that for any set S of size R and any 2-colouring $c:{S \choose p} \rightarrow \{1,2\}$ of all subsets of S of size p, there is either a set A of size a such that all p-tuples in $A \choose p$ have colour 1 or a set B of size b such that all p-tuples in $B\choose p$ have colour 2.

Lemma 9

For any two sets $S_{1}\in {\mathcal{H}}$ and $S_{2} \in {\mathcal{G}}$,

Proof

Let h be a hyperplane defining two open halfplanes h ⁺ and h ⁻ such that S ₁=h ⁻ and let $S_{2}' = S_{2}\setminus h^{-}$. Let A′ be a maximum cardinality subset of A such that V(A)⊆S ₁∪S ₂. Colour every subset of d hyperplanes in A′ with colour 1 if their intersection point is in h ⁻ and with colour 2 otherwise.

For any set A ₁⊆A′ such that all subsets in $A_{1} \choose d$ have colour 1, notice that V(A ₁)⊆S ₁. Likewise, for any set A ₂⊆A′ such that all subsets in $A_{2} \choose d$ have colour 2, we have $V(A_{2})\subseteq S_{2}'$. Any such set A ₁ is of size |A ₁|≤μ _A (S ₁) and any such set A ₂ is of size $|A_{2}|\leq\mu_{A}(S_{2}')\leq\mu_{A}(S_{2})$.

Therefore, A′ has no subset of size μ _A (S ₁)+1 that has all d-tuples of colour 1, and no subset of size μ _A (S ₂)+1 that has all d-tuples of colour 2, and so by Ramsey’s Theorem, μ _A (S ₁∪S ₂)=|A′|≤R _d (μ _A (S ₁)+1,μ _A (S ₂)+1)−1. □

Define the operator ⊕ as a⊕b=R _d (a+1,b+1)−1. The operator is increasing and closed on the set ℕ_≥d−1 of naturals ≥d−1. Since R _d (d,x)=x for all x, d−1 is a neutral element. Therefore (ℕ_≥d−1,⊕,≤,d−1) is a totally ordered unital magma. Thus we have:

Corollary 10

The set function μ _A takes values in the totally ordered unital magma (ℕ_≥d−1,⊕,≤,d−1); it is monotone and ${\mathcal{H}}$-subadditive.

Apply now the generalized center transversal theorem to obtain:

Theorem 11

Let A ₁,…,A _k be k sets of hyperplanes in ℝ^d. There exists a (k−1)-flat π such that for every open halfspace h that contains π,

The special case when k=d gives a ham-sandwich cut theorem.

Corollary 12

Let A ₁,…,A _d be d sets of hyperplanes in ℝ^d. There exist a hyperplane π bounding the two closed halfspaces π ⁺ and π ⁻ and sets $A_{i}^{\sigma}\subseteq A_{i}$, σ∈{+,−}, such that $V(A_{i}^{\sigma})\subseteq\pi^{\sigma}$ and $|A_{i}^{\sigma}| \oplus|A_{i}^{\sigma}|\geq|A_{i}|$.

If the arrangement A has the property that no r+1 hyperplanes intersect in a common point, μ _A (π)≤r for any hyperplane π, and so by Lemma 9, if h is an open halfspace bounded by π and $\bar{h} = \pi\cup h$ is the corresponding closed halfspace, $\mu_{A}(\bar{h}) \leq\mu_{A}(h) \oplus r$.

Corollary 13

Let A ₁,…,A _d be d sets of hyperplanes in ℝ^d, no r+1 of which intersect in a common point. There exist a hyperplane π bounding the two open halfspaces π ⁺ and π ⁻ and sets $A_{i}^{\sigma}\subseteq A_{i}$, σ∈{+,−}, such that $V(A_{i}^{\sigma})\subseteq\pi^{\sigma}$ and $(|A_{i}^{\sigma}|\oplus|A_{i}^{\sigma}|) \oplus r \geq|A_{i}|$.

4 Same-Type Lemma for Arrangements

Center transversal theorems, and especially the ham-sandwich cut theorem, are basic tools for proving many facts in discrete geometry. We show here how the same facts can be shown for hyperplane arrangements in ℝ^d.

A transversal of a collection of sets X ₁,…,X _m is a m-tuple (x ₁,…,x _m ) where x _i ∈X _i . A collection of sets X ₁,…,X _m has same-type transversals if all of its transversals have the same order-type.

Note that m≥d+1 sets have same-type transversals if and only if every d+1 of them have same-type transversals. There are several equivalent definitions for these notions.

• 1. The sets X ₁,…,X _d+1 have same-type transversals if and only if they are well-separated, that is, if and only if for all disjoint sets of indices I,J⊆{1,…,d+1}, there is a hyperplane separating the sets X _i ,i∈I from the sets X _j ,j∈J.

• 2. Connected sets C ₁,…,C _d+1 have same-type transversals if and only if there is no hyperplane intersecting simultaneously all C _i . Sets X ₁,…,X _d+1 have same-type transversals if and only if there is no hyperplane intersecting simultaneously all their convex hulls C _i =CH(X _i ).

The same-type lemma for point sets states that there is a constant c=c(m,d) such that for any collection S ₁,…,S _m of finite point sets in ℝ^d, there are sets $S'_{i} \subseteq S_{i}$ such that $|S'_{i}|\geq c|S_{i}|$ and the sets $S'_{1},\ldots,S'_{m}$ have same-type transversals. We here show a similar result for hyperplane arrangements.

A function f is growing if for any value y ₀ there is a x ₀ such that f(x)≥y ₀ for any x≥x ₀.

Lemma 14

For any integers d, m, and r, there is a growing function f=f _m,d,r such that for any collection of m hyperplane arrangements A ₁,…,A _m , in ℝ^d, where no r+1 hyperplanes intersect at a common point, there are sets $A'_{i}\subseteq A_{i}$ such that $|A'_{i}|\geq f(|A_{i}|)$ and the sets $\mathit{CH}(A'_{1}),\ldots,\mathit{CH}(A'_{m})$ have same-type transversals.

Proof

The proof will follow closely the structure of Matoušek [16, Theorem 9.3.1, p. 217]. First notice that the composition of two growing functions is a growing function. The proof will show how to choose successive (nested) subsets of each set A _i , c times where c=c(m,d) only depends on m and d and where the size of each subset is some growing function of the previous one.

Also, it will suffice to prove the theorem for m=d+1, and then apply it repeatedly for each d+1 tuple of sets. The resulting function f _m,d,r will be $f_{d+1,d,r}^{m \choose{d+1}}$, the repeated composition of f _d+1,d,r , ${m \choose{d+1}}$ times.

So, given d+1 sets A ₁,…,A _d+1 of hyperplanes in ℝ^d, suppose that there is an index set I⊆{1,…,d+1} such that ⋃ _i∈I CH(A _i ) and ⋃ _i∉I CH(A _i ) are not separable by a hyperplane and assume without loss of generality that d+1∈I. Let π be the ham-sandwich cut hyperplane for arrangements A ₁,…,A _d obtained by applying Corollary 13. Then for each i∈[1,d], each of the two open halfspaces π ^σ, σ∈{+,−} bounded by π contains a subset $A^{\sigma}_{i}\subseteq A_{i}$ such that $V(A_{i}^{\sigma})\subseteq\pi^{\sigma}$ and $(|A_{i}^{\sigma}|\oplus|A_{i}^{\sigma}|) \oplus r \geq|A_{i}|$. Furthermore, because $\mu_{A_{d+1}}(\pi)\leq r$ and by Lemma 9,

Assume without loss of generality $\mu_{A_{d+1}}(\pi^{+})\geq\mu_{A_{d+1}}(\pi^{-})$. Then $\mu_{A_{d+1}}(\pi^{+})\oplus\mu_{A_{d+1}}(\pi^{+})\oplus r \geq|A_{d+1}|$. For each i∈I, let $A'_{i} = A^{+}_{i}$ and for each i∉I, let $A'_{i}=A^{-}_{i}$. Let g(x)=min{y|y⊕y⊕r≥x}. Then g is a growing function, and $|A'_{i}|\geq g(|A_{i}|)$.

In the worst case, we have to shrink the sets for each possible I, 2^d times. Therefore for m=d+1, the function f in the statement of the theorem is a composition of g, 2^d times, and is a growing function. □

In the plane, the same-type lemma readily gives a positive portion Erdős–Szekeres Theorem. Recall that the Erdős–Szekeres (happy ending) theorem [12] states that for any k there is a number ES(k) such that any set of ES(k) points in general position in ℝ² contains a subset of size k which is in convex position.

Theorem 15

For every integers k, r, and c, there is an integer N such that any arrangement A of N lines, such that no r+1 lines go through a common point, contains disjoint subsets A ₁,…,A _k with |A _i |≥c and such that every transversal of CH(A ₁),…,CH(A _k ) is in convex position.

Proof

Let m=ES(k) and let f=f _m,2,r be as in Lemma 14. Let N be such that f(⌊N/m⌋)≥c. Partition the set A of N lines into m sets A ₁,…,A _m of N/m lines arbitrarily. Apply Lemma 14 to obtain sets $A'_{1},\ldots,A'_{m}$ each of size at least c. Finally, choose one transversal (x ₁,…,x _m ) from the sets $\mathit{CH}(A'_{i})$ and apply the Erdős–Szekeres theorem to obtain a subset $x_{i_{1}},\ldots,x_{i_{k}}$ of points in convex position. Because the sets $\mathit{CH}(A'_{i})$ have the same-type property, every transversal of $\mathit{CH}(A'_{i_{1}}),\ldots,\mathit{CH}(A'_{i_{k}})$ is in convex position. □

5 Graph Drawing

Formally, a vertex labeling of a graph G=(V,E) is a bijection π:V→[n]. A set of n lines in the plane labeled from 1 to n supports G with vertex labeling π if there exists a straight-line crossing-free drawing of G where for each i∈[n], the vertex labeled i in G is mapped to a point on line i. A set L of n lines labeled from 1 to n supports an n-vertex graph G if for every vertex labeling π of G, L supports G with vertex labeling π. In this context clearly it only makes sense to talk about planar graphs. We are interested in the existence of an n-vertex line set that supports all n-vertex planar graphs, that is, in the existence of a universal set of lines for planar graphs.

Theorem 16

For some absolute constant c′ and every n≥c′, there exists no set of n lines in the plane that support all n-vertex planar graphs.

The following known result will be used in the proof of this theorem.

Lemma 17

(See [7])

Consider the planar triangulation on six vertices, denoted by G ₆, depicted on the bottom of Fig. 1. G ₆ has vertex labeling π such that the following holds for every set L of six lines labeled from 1 to 6, no two of which are parallel. For every straight-line crossing-free drawing, D, of G ₆ where for each i∈[n], the vertex labeled i in π is mapped to a point on line i in L, there is a point that is in an interior face of D and in CH(L).

Fig. 1

Illustration for the proof of Theorem 16

Proof of Theorem 16

Let L be any set of n≥c′=5N lines, where N is obtained from Theorem 15 with values k=6, c=6, r=17.

Estrella-Balderrama et al. [11] proved that for every n≥6, no set of n parallel lines supports all n-vertex planar graphs. Thus if L has at least six lines that are pairwise parallel, then L cannot support all n-vertex planar graphs.

Dujmović et al. [7] proved that for every n≥18, no set of n lines that all go through a common point supports all n-vertex planar graphs. Thus if L has at least 18 such lines, then L cannot support all n-vertex planar graphs.

Thus assume that L has no six pairwise parallel lines and no 18 lines that intersect in one common point. Then L has a subset L′ of c′/5≥N lines no two of which are parallel and no 18 of which go through one common point. Then Theorem 15 implies that we can find in L′ six sets A ₁,…,A ₆ of six lines each, such that the set {CH(A ₁),…,CH(A ₆)} is in convex position. Assume CH(A ₁),…,CH(A ₆) appear in that order around their common “convex hull”.

Consider an n-vertex graph H whose subgraph G is illustrated in Fig. 1. G∖v has three components, A, B, and C, each of which is a triangulation. Each of the components A, B, and C has two vertex disjoint copies of G ₆ (the 6-vertex triangulation from Lemma 17). Map the vertices of the first copy of G ₆ in A to A ₁ and the second copy to A ₄ using the mapping equivalent to π in Lemma 17. Map the vertices of the first copy of G ₆ in B to A ₂ and the second copy to A ₅ using the mapping equivalent to π in Lemma 17. Map the vertices of the first copy of G ₆ in C to A ₃ and the second copy to A ₆ using the mapping equivalent to π in Lemma 17. Map the remaining vertices of H arbitrarily to the remaining lines of L.

We now prove that L does not support H with such a mapping. Assume, for the sake of contradiction, that it does and consider the resulting crossing-free drawing D of H. In D the drawing of each of A, B, and C has a triangle as an outerface. Let T _A , T _B , and T _C denote these three triangles together with their interiors in the plane.

It is simple to verify that in any crossing-free drawing of G at least two of these triangles are disjoint, meaning that there is no point p in the plane such that p is in both of these triangles. Assume, without loss of generality, that T _A and T _B are disjoint. By Lemma 17, there is a point p ₁∈CH(A ₁) such that p ₁∈T _A , and a point p ₄∈CH(A ₄) such that p ₄∈T _A . Thus the segment $\overline{p_{1}p_{4}}$ is in T _A . Similarly, by Lemma 17, there is a point p ₂∈CH(A ₂) such that p ₂∈T _B , and a point p ₅∈CH(A ₅) such that p ₅∈T _B . Thus the segment $\overline{p_{2}p_{5}}$ is in T _A .

By Theorem 15 and our ordering of A ₁,…,A ₆, $\overline {p_{1}p_{4}}$ and $\overline{p_{2}p_{5}}$ intersect in some point p. That implies that p∈T _A and p∈T _B . That provides the desired contradiction, since T _A and T _B are disjoint. □

As a sharp contrast to Theorem 16, the following theorem shows that the situation is starkly different for unlabeled planar graphs. Namely, every set of n lines supports all n-vertex unlabeled planar graphs. The proof of this theorem does not use any of the tools we introduced in the previous section and is in that sense elementary. It is not difficult to verify that the theorem also follows from the main result in [17] which states the following: given a drawing of a graph G in the plane where edges of G are x-monotone curves any pair of which cross even number of times, G can be redrawn as a straight-line crossing-free drawing where the x-coordinates of the vertices remain unchanged.

Theorem 18

(See [17])

Given a set L of n lines in the plane, every planar graph has a straight-line crossing-free drawing where each vertex of G is placed on a distinct line of L. (In other words, given any set L of lines, labeled from 1 to n, and any n-vertex planar graph G there is a vertex labeling π of G such that L supports G with vertex labeling π.)

Proof

In this proof we will use canonical orderings introduced in [9] and a related structure called frame introduced in [3]. We first recall these tools. We can assume that G is an embedded edge-maximal planar graph.² Each face of G is bounded by a 3-cycle. De Fraysseix [9] proved that G has a vertex ordering σ=(v ₁,v ₂,v ₃,…,v _n ), called a canonical ordering, with the following properties. Define G _i to be the embedded subgraph of G induced by {v ₁,v ₂,…,v _i }. Let C _i be the subgraph of G induced by the edges on the boundary of the outer face of G _i . Then

• v ₁, v ₂, and v _n are the vertices on the outer face of G.

• For each i∈{3,4,…,n}, C _i is a cycle containing v ₁ v ₂.

• For each i∈{3,4,…,n}, G _i is biconnected and internally 3-connected; that is, removing any two interior vertices of G _i does not disconnect it.

• For each i∈{3,4,…,n}, v _i is a vertex of C _i with at least two neighbours in C _i−1, and these neighbours are consecutive on C _i−1.

For example, the ordering in Fig. 2(a) is a canonical ordering of the depicted embedded graph G.

Fig. 2

Illustration for the proof of Theorem 18: (a) Canonical ordering of G, (b) Frame $\mathcal {F}$ of G

A frame $\mathcal {F}$ of G [3] is the oriented subgraph of G with vertex set $V( \mathcal {F}):=V( G)$, where:

• v ₁ v ₂ is in $E( \mathcal {F})$ and is oriented from v ₁ to v ₂.

• For each i∈{3,4,…,n} in the canonical ordering σ of G, edges pv _i and v _i p′ are in $E( \mathcal {F})$, where p and p′ are the first and the last neighbour, respectively, of v _i along the path in C _i−1 from v ₁ to v ₂ not containing edge v ₁ v ₂. Edge pv _i is oriented from p to v _i , and edge v _i p′ is oriented from v _i to p′, as illustrated in Fig. 2(b).

By definition, $\mathcal {F}$ is a directed acyclic graph with one source v ₁  and one sink v ₂. The frame $\mathcal {F}$ defines a partial order $<_{\mathcal{F}}$ on $V( \mathcal {F})$, where $v <_{\mathcal{F}}w$ whenever there is a directed path from v to w in $\mathcal {F}$.

Translate the given set L of lines so that all vertices of the arrangement of lines have negative y coordinates, and sort the lines ℓ _i ∈L according to the x coordinate b _i of the intersection of ℓ _i with the x axis. Therefore, the lines ℓ _i ∈L have equation y=a _i (x−b _i ), with b ₁<b ₂<⋯<b _n . Because all intersections among lines of L have negative coordinates, all b _i are distinct, and the values 1/a _i are sorted. Note that the slopes a _i might be positive or negative. Let $\hat{a} = \min|a_{i}|$. For any segment of slope in $[-\hat{a},\hat{a}]$ connecting two points (x _i ,y _i )∈ℓ _i and (x _j ,y _j )∈ℓ _j above the x axis (that is, y _i ,y _j >0), x _i <x _j if and only if i<j.

Construct a linear extension v _ρ(1),v _ρ(2),…,v _ρ(n) of the partial order $<_{\mathcal{F}}$ and define the bijection π:V→[n] as π(v _ρ(i))=i. That is, the vertices of G will be placed on the lines in such a way that the partial order $<_{\mathcal{F}}$ is compatible with the order determined by the values b _i of the lines.

We prove by induction that for every value $\hat{y}$ and every i≥2, it is possible to draw G _i such that v ₁ and v ₂ are placed on points (b ₁,0), (b _n ,0), and the y coordinates of all other vertices are in the horizontal slab $(0,\hat{y}]$. The base case (i=2) is obviously true.

Note that we could have formulated the induction on the slopes of the edges of G _i in the drawing. In fact those two formulations imply each other: for any value $0<s\leq\hat{a}$, there is a $\hat{y}_{s}>0$ such that any segment whose endpoints lie on distinct lines of L and have y coordinates in $[0,\hat{y}_{s}]$, the slope of the segment is in [−s,s]. This is easy to see: draw an upward cone with apex on each point (b _i ,0) and bounded by the lines of slopes s and −s through that point. Define $\hat{y}_{s}$ as the y coordinate of the lowest intersection point between any two such cones. Any segment with a slope not in [−s,s] and with its lowest point inside a cone must have its highest point inside the same cone, therefore no segment connecting two different lines inside the horizontal slab $[0,\hat{y}_{s}]$ can have such a slope.

Assume by induction that the statement is true for G _i−1. We will show how to draw G _i for a specific value $\hat{y}$. The point v _i will be placed on the point on line π(v _i ) with y coordinate $\hat{y}$. Let s ₁ and s ₂ be the slopes of the segments v ₁ v _i and v _i v ₂, and let s=max(|s ₁|,|s ₂|)/2 or $\hat{a}$, whichever is smaller. Let y ₁ be the intersection of the line of slope s through v _i and line ℓ ₁ and y ₂ the intersection of the line of slope −s through v _i and ℓ _n . Note that y ₁ and y ₂ are strictly positive. Let $\hat{y}' = \min(y_{1},y_{2},\hat{y}_{s})$. Apply the induction hypothesis to draw G _i−1 in the horizontal slab $[0,\hat{y}']$. Thus, in the drawing of G _i−1, all edges have slope at most $s\leq\hat{a}$. Then by construction, the path in C _i−1 from v ₁ to v ₂ not containing edge v ₁ v ₂ is x-monotone (that is, all its edges are oriented rightwards), and v _i is above the supporting line of each edge on that path. Therefore, v _i can see all vertices in C _i−1 and all edges adjacent to v _i can be drawn. □

We conclude this part with an intriguing 3D variant of this graph drawing problem. A graph is linkless if it has an embedding in 3D such that any two cycles of the graph are unlinked.³ These graphs form a three-dimensional analogue of the planar graphs.

Open Problem 19

Is there an arrangement of labeled planes in 3D such that any labeled linkless graph has a linkless straight-line embedding where each vertex is placed on the plane with the same label?