A Center Transversal Theorem for Hyperplanes and Applications to Graph Drawing^{1}^{1}1The preliminar version of this paper has appeared in the Proceedings of the th annual ACM symposium on Computational geometry, SoCG, 2011.
Abstract
Motivated by an open problem from graph drawing, we study several partitioning problems for line and hyperplane arrangements. We prove a hamsandwich cut theorem: given two sets of lines in , there is a line such that in both line sets, for both halfplanes delimited by , there are lines which pairwise intersect in that halfplane, and this bound is tight; a centerpoint theorem: for any set of lines there is a point such that for any halfplane containing that point there are of the lines which pairwise intersect in that halfplane. We generalize those results in higher dimension and obtain a center transversal theorem, a sametype lemma, and a positive portion ErdősSzekeres 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 labelled lines that are universal for all vertex labelled planar graphs. As a contrast, the main result by Pach and Toth in [J. of Graph Theory, 2004], has, as an easy consequence, that every set of (unlabelled) lines is universal for all vertex (unlabelled) 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 (straightline) 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 crossingfree if it has no crossings.
One main focus in graph drawing is finding methods to produce drawings or crossingfree 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 or –dimensions.
Given a set of regions in the plane and an vertex graph , consider a class of graph drawing problems where needs to be drawn crossingfree by placing each vertex of in one region of . If such a drawing exists, then is said to support . 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 labelled); in the other, each vertex is free to be assigned to any of the regions (that is, the vertices are unlabelled).
When regions are points in the plane, Rosenstiehl and Tarjan [RT86] asked if there exists a set of points that support all vertex unlabelled planar graphs. This question is answered in the negative by De Fraysseix [dFPP88, dFPP90]. On the contrary, every set of points in general position supports all vertex unlabelled outerplanar graphs, as proved by Gritzmann et al.[GMPP91] and recapitulated in Lemma in the text by Agarwal and Pach [AP95]. If the drawings are not restricted to be straightline, then every set of labelled points supports every labelled planar graph, as shown by [PW01]. However bends per edge may be necessary in any such crossingfree drawing.
When regions are labelled lines in the plane, EstrellaBalderrama et al.[EBFK09] showed that for every , there is no set of parallel lines in the plane that support all labelled vertex planar graphs. The authors moreover characterized a (sub)class of vertex planar graphs that are supported by every set of parallel lines, for every labelling of the graphs in the class. That class is mainly comprised of several special families of trees. Dujmović et al.[DEK10] showed that no set of lines that all intersect in one common point supports all vertex labelled planar graphs. Moreover, they show that for every large enough, there is a set of lines in general position that does not support all labelled vertex planar graphs. They leave as the main open problem the question of whether, for every large enough, there exists a universal set of lines in the plane, that is, one that supports all labelled vertex planar graphs. In Section 5, as our main graph drawing result, we answer that question in the negative. The main result by Pach and Toth [PT04] on monotone drawings, has, as an easy consequence, that in the unlabelled case, every set of lines supports every vertex unlabelled planar graph. As a side note, we give an alternative and direct proof of that fact. The result illustrates the sharp contrast with the labelled case.
While the positive result is proved using little of the geometry in the arrangement, the nonexistence 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 many works have focused on partitioning point sets, probability distributions or measures, much less is understood for sets of lines in or hyperplanes in . This is partially due to the fact that a line (or a hyperplane), being infinite, can’t 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 works (such as cuttings [CF90, Mat90] or equipartitions [LS03]) have focused on identifying, and bounding the number of lines/hyperplanes intersecting a set of regions. Others [CSSS89] 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 of lines in is composed of vertices (all pairwise intersections between lines of ), edges connecting these vertices, and halflines. If we omit the halflines, 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 of the vertices of the arrangement. Therefore, a natural way of evaluating the portion of an arrangement contained in a given convex region is to find the largest subset of lines of such that the arrangement of (without the halflines) is contained in , or equivalently, such that all pairwise intersections of lines in lie in .
It is not hard to show that, in any arrangement of lines, a line can be found such that for both closed halfplanes bounded by there are at least lines which pairwise intersect in that halfplane. This provides the analogue of a bisecting line for point sets. In Section 3.1, we show that any two line arrangements can be bisected simultaneously in this manner, thus proving a hamsandwich theorem for line arrangements. We also prove a centerpoint theorem: for any arrangement of lines, there is a point such that for any halfplane containing , there are at least lines of the arrangement that pairwise intersect in that halfplane. In Section 3.2 we generalize these notions to higher dimensions and prove a center transversal theorem: for any and , there is a growing function such that for any sets of hyperplanes in , there is a flat such that for any halfspace containing there is a subset of hyperplanes from each set such that any hyperplanes of intersect in . The bound we find is related to Ramsey numbers for hypergraphs.
Hamsandwich theorems have a number of natural consequences. In Section 2 we show a sametype lemma for hyperplane arrangements: informally, for any arrangements of hyperplanes in general position (no share a point) and that are large enough, we can find a large subset of hyperplanes from each set such that the convex hulls of the vertices in the arrangements are wellseparated, that is, no hyperplane hits of them. In the plane, we also show a positive portion ErdősSzekeres theorem: for any integers and there is an integer such that any set of lines in general position contains subsets of lines each such that the vertices of each arrangement 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 wise intersections in a given region – was a measure. Unfortunately it is not. However, in Section 2, we identify basic properties much weaker than those of measures which, if satisfied by a set function, guarantee a centraltransversal theorem to be true.
2 Center transversal theorem
The center transversal theorem is a generalization of both the hamsandwich cut theorem, and the centerpoint theorem discovered independently by Dol’nikov [Dol92], and Živaljević and Vrećica [ŽV90]. The version of Dol’nikov is defined for a class of set functions that is more general than measures. Let be the set of all open halfspaces in and let be a family of subsets of closed under union operations and that contains . A charge is a finite set function that is defined for all set , and that is monotone ( whenever ) and subadditive (). A charge is concentrated on a set if for every halfspace s.t. , . Dol’nikov shows^{4}^{4}4Dol’nikov actually shows a slightly more general theorem that allows for nonconcentrated charges. For the sake of simplicity we only discuss the simplified version even though our generalizations extend to the stronger original result.:
Theorem 1 (Center transversal theorem [Dol92]).
For arbitrary charges , , defined on and concentrated on bounded sets, there exists a flat such that
for every open halfspace containing .
A careful reading of the proof of this theorem reveals that its statement can be generalized, and the assumptions on 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 . Therefore, define to be subadditive if
for any finite set of halfspaces.
Next, notice that in order for the proof to go through, the set function need not be realvalued. Recall [Bou07] that a totally ordered unital magma is a totally ordered set endowed with a binary operator such that is closed under operations, has neutral element (i.e., ) and is monotone (i.e., and whenever ). Further, for all and , define the multiple of as .
Then, it suffices that take values over , and use 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 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 , be set functions defined on and taking values in a totally ordered unital magma . If the functions are monotone subadditive and concentrated on bounded sets, there exists a flat such that
for every open halfspace containing .
3 Center transversal theorem for arrangements
Let be an arrangement of hyperplanes in . We write for the set of all vertices (intersection points between any hyperplanes) of and for the convex hull of those points. In the arguments that follow, by abuse of language, we will write and mean or . For example, we say that the arrangement is above hyperplane when all points in are above . More generally for a region in , we say that the arrangement does not intersect if does not intersect . We say that the arrangements are disjoint if their convex hulls do not intersect. They are separable if they are disjoint and no hyperplane intersects of them simultaneously.
Let be the set of all open halfspaces in and let be a family of subsets of closed under union operations and that contains . For any set , let be the maximum number of hyperplanes of that have all their vertices inside of , that is,
In particular, and .
3.1 Lines in
We start with the planar case. Thus, is a set of lines in , and is the set of all open halfplanes. Recall the ErdősSzekeres theorem [ES35]
Theorem 3 (ErdősSzekeres).
For all integers , , any sequence of numbers contains either a nonincreasing subsequence of length or an increasing subsequence of length .
We show:
Lemma 4.
For any two sets and ,
Proof.
Let be a line defining two open halfplanes and such that and let . Rotate and translate the plane so that is the (vertical) axis, and contains all points with positive coordinate. Let be a maximum cardinality subset of such that . Let be the lines in ordered by increasing order of their slopes, and let be the coordinates of the intersections of the lines with line , in the same order. For any set such that the values of the lines in form an increasing subsequence in , notice that . Likewise, for any set that forms a nonincreasing subsequence in , we have . Any such set is of size and any such set is of size .
Therefore, has no nondecreasing subsequence of length and no nonincreasing subsequence of length , and so by Theorem 3, . ∎
Corollary 5.
The set function takes values in the totally ordered unital magma ; it is monotone and subadditive.
We can thus apply the generalized center transversal theorem with to obtain a hamsandwich cut theorem:
Theorem 6.
For any arrangements and of lines in , there exists a line bounding closed halfplanes and and sets , , such that , and .
Note that this statement is similar to the result of Aronov et al.[AEG94] on mutually avoiding sets. Specifically, two sets and of points in the plane are mutually avoiding if no line through a pair of points in intersects the convex hull of , 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 and of points in the plane are separable if there exists a point such that all the lines through pairs of points in are above and all the lines through pairs of points in are below or vice versa. Aronov et al. show in Theorem of [AEG94] that any two sets and of points contains two subsets , , that are mutually avoiding. That this bound is tight, up to a constant, was proved by Valtr [Val97]. In the dual, Theorem 6 states that for any two sets and of points in , there exists a point and sets , , such that , and all lines through pairs of points in are above and all lines through pairs of points in are below . 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 in [AEG94] give two sets that are, at the same time, mutually avoiding and separable.
Note that the bound in Theorem 6 is tight: assume is the square of an integer. Construct the first line arrangement with pencils of lines each, centered at points with coordinates for , and the slopes of the lines in pencil are distinct values in . Thus all intersections other than the pencil centers have coordinates greater than . The line delimits two halfplanes in which since any set of more than lines have lines from different pencils which intersect on the right of , and since any set of more than lines has two lines in the same pencil which intersect left of . Since is monotone, no vertical line can improve this bound on both sides. Perturb the lines so that no two intersection points have the same coordinate. For , build a copy of translated down, far enough so that no line through two vertices of intersects and conversely. Therefore any line not combinatorially equivalent to a vertical line (with respect to the vertices of and ) does not intersect one of the arrangements and so there is no better cut than .
Applying the generalized center transversal theorem with gives a centerpoint theorem with a bound of . A slightly more careful analysis improves that bound.
Theorem 7.
For any arrangement of lines in , there exists a point such that for every halfplane containing there is a set , such that .
Proof.
Let be the set of halfplanes such that . The halfspace depth is the minimum value of for any halfspace containing . Therefore, the region of depth is the intersection of the complements of the halfplanes . If there is no point of depth then the intersection of the complements of halfplanes in is empty, and so (by Helly’s Theorem) there must be 3 halfplanes , , and in such that the intersection of their complements is empty. But then, there is at least one point . Let be the translated halfplanes with point on the boundary. Since , . The point and the 3 halfplanes through it are witness that there is no point of depth .
The 3 lines bounding those 3 halfplanes divide the plane into 6 regions. Every line misses one of the three regions , , and . Classify the lines in depending on the first region it misses, clockwise. The largest class contains lines. Assume without loss of generality that all lines in miss , then all intersections between lines of are in . By Lemma 4,
a contradiction. ∎
3.2 Hyperplanes in
We first briefly review a bichromatic version of Ramsey’s theorem for hypergraphs.
Theorem 8.
For all , there is a natural number such that for any set of size and any 2colouring of all subsets of of size , there is either a set of size such that all tuples in have colour 1 or a set of size such that all tuples in have colour 2.
Lemma 9.
For any two sets and ,
Proof.
Let be a hyperplane defining two open halfplanes and such that and let . Let be a maximum cardinality subset of such that . Colour every subset of hyperplanes in with colour if their intersection point is in and with colour otherwise.
For any set such that all subsets in have colour , notice that . Likewise, for any set such that all subsets in have colour , we have . Any such set is of size and any such set is of size .
Therefore, has no subset of size that has all tuples of colour 1, and no subset of size that has all tuples of colour 2, and so by Ramsey’s Theorem, . ∎
Define the operator as . The operator is increasing and closed on the set of naturals . Since for all , is a neutral element. Therefore is a totally ordered unital magma. Thus we have:
Corollary 10.
The set function takes values in the totally ordered unital magma ; it is monotone and subadditive.
Apply now the generalized center transversal theorem to obtain:
Theorem 11.
Let be sets of hyperplanes in . There exists a flat such that for every open halfspace that contains ,
The special case when gives a hamsandwich cut theorem.
Corollary 12.
Let be sets of hyperplanes in . There exists a hyperplane bounding the two closed halfspaces and and sets , , such that and .
If the arrangement has the property that no hyperplanes intersect in a common point, for any hyperplane , and so by Lemma 9, if is an open halfspace bounded by and is the corresponding closed halfspace, .
Corollary 13.
Let be sets of hyperplanes in , no of which intersect in a common point. There exists a hyperplane bounding the two open halfspaces and and sets , , such that and .
4 Sametype lemma for arrangements
Center transversal theorems, and especially the hamsandwich 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 .
A transversal of a collection of sets is a tuple where . A collection of sets has sametype transversals if all of its transversals have the same ordertype.
Note that sets have sametype transversals if and only if every of them have sametype transversals. There are several equivalent definitions for these notions.

The sets have sametype transversals if and only if they are well separated, that is, if and only if for all disjoint sets of indices , there is a hyperplane separating the sets from the sets .

Connected sets have sametype transversals if and only if there is no hyperplane intersecting simultaneously all . Sets have sametype transversals if and only if there is no hyperplane intersecting simultaneously all their convex hulls .
The sametype lemma for point sets states that there is a constant such that for any collection of finite point sets in , there are sets such that and the sets have sametype transversals. We here show a similar result for hyperplane arrangements.
A function is growing if for any value there is a such that for any .
Lemma 14.
For any integers , , and , there is a growing function such that for any collection of hyperplane arrangements , in , where no hyperplanes intersect at a common point, there are sets such that and the sets have sametype transversals.
Proof.
The proof will follow closely the structure of Matoušek [Mat02, 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 , times where only depends on and and where the size of each subset is some growing function of the previous one.
Also, it will suffice to prove the theorem for , and then apply it repeatedly for each tuple of sets. The resulting function will be , the repeated composition of , times.
So, given sets of hyperplanes in , suppose that there is an index set such that and are not separable by a hyperplane and assume without loss of generality that . Let be the ham sandwich cut hyperplane for arrangements obtained by applying Corollary 13. Then for each , each of the two open halfspaces , bounded by contains a subset such that and . Furthermore, because and by Lemma 9,
Assume without loss of generality . Then . For each , let and for each , let . Let . Then is a growing function, and .
In the worst case, we have to shrink the sets for each possible , times. Therefore for , the function in the statement of the theorem is a composition of , times, and is a growing function. ∎
In the plane, the sametype lemma readily gives a positive portion ErdősSzekeres Theorem. Recall that the ErdősSzekeres (happy ending) theorem [ES35] states that for any there is a number such that any set of points in general position in contains a subset of size which is in convex position.
Theorem 15.
For every integers , , and , there is an integer such that any arrangement of lines, such that no lines go through a common point, contains disjoint subsets with and such that every transversal of is in convex position.
Proof.
Let and let be as in Lemma 14. Let be such that . Partition the set of lines into sets of lines arbitrarily. Apply Lemma 14 to obtain sets each of size at least . Finally, choose one transversal from the sets and apply the ErdősSzekeres theorem to obtain a subset of points in convex position. Because the sets have the same type property, every transversal of is in convex position. ∎
5 Graph Drawing
Formally, a vertex labelling of a graph is a bijection . A set of lines in the plane labelled from to supports with vertex labelling if there exists a straightline crossingfree drawing of where for each , the vertex labelled in is mapped to a point on line . A set of lines labelled from to supports an vertex graph if for every vertex labelling of , supports with vertex labelling . In this context clearly it only makes sense to talk about planar graphs. We are interested in the existence of an vertex line set that supports all vertex planar graphs, that is, in the existence of a universal set of lines for planar graphs.
Theorem 16.
For some absolute constant and every , there exists no set of lines in the plane that support all vertex planar graphs.
The following known result will be used in the proof of this theorem.
Lemma 17.
[DEK10] Consider the planar triangulation on vertices, denoted by , that is depicted on the bottom of Figure 1. has vertex labelling such that the following holds for every set of lines labelled from to , no two of which are parallel. For every straightline crossingfree drawing, , of where for each , the vertex labelled in is mapped to a point on line in , there is a point that is in an interior face of and in .
Proof of Theorem 16.
Let be any set of lines, where is obtained from Theorem 15 with values , , .
[EBFK09] proved that for every , no set of parallel lines supports all vertex planar graphs. Thus if has at least lines that are pairwise parallel, then cannot support all vertex planar graphs.
[DEK10] proved that for every , no set of lines that all go through a common point supports all vertex planar graphs. Thus if has at least such lines, then cannot support all vertex planar graphs.
Thus assume that has no pairwise parallel lines and no lines that intersect in one common point. Then has a subset of 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 six sets of six lines each, such that the set is in convex position. Assume appear in that order around their common “convex hull”.
Consider an vertex graph whose subgraph is illustrated in Figure 1. has three components, , , and , each of which is a triangulation. Each of the components , , and has two vertex disjoint copies of (the vertex triangulation from Lemma 17). Map the vertices of the first copy of in to and the second copy to using the mapping equivalent to in Lemma 17. Map the vertices of the first copy of in to and the second copy to using the mapping equivalent to in Lemma 17. Map the vertices of the first copy of in to and the second copy to using the mapping equivalent to in Lemma 17. Map the remaining vertices of arbitrarily to the remaining lines of .
We now prove that does not support with such a mapping. Assume, for the sake of contradiction, that it does and consider the resulting crossingfree drawing of . In the drawing of each of , , and has a triangle as an outerface. Let , , and denote these three triangles together with their interiors in the plane.
It is simple to verify that in any crossingfree drawing of at least two of these triangles are disjoint, meaning that there is no point in the plane such that is in both of these triangles. Assume, without loss of generality, that and are disjoint. By Lemma 17, there is a point such that , and a point such that . Thus the segment is in . Similarly, by Lemma 17, there is a point such that , and a point such that . Thus the segment is in .
By Theorem 15 and our ordering of , and intersect in some point . That implies that and . That provides the desired contradiction, since and are disjoint. ∎
As a sharp contrast to Theorem 16, the following theorem shows that the situation is starkly different for unlabelled planar graphs. Namely, every set of lines supports all vertex unlabelled 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 [PT04] which states the following: given a drawing of a graph in the plane where edges of are monotone curves any pair of which cross even number of times, can be redrawn as a straightline crossingfree drawing where the coordinates of the vertices remain unchanged.
Theorem 18.
[PT04] Given a set of lines in the plane, every planar graph has a straight line crossing free drawing where each vertex of is placed on a distinct line of . (In other words, given any set of lines, labelled from to , and any vertex planar graph there is a vertex labelling of such that supports with vertex labelling .)
Proof.
In this proof we will use canonical orderings introduced in [dFPP90] and a related structure called frame introduced in [BDH09]. We first recall these tools. We can assume is an embedded edge maximal planar graph.^{5}^{5}5A planar graph is edgemaximal (also called, a triangulation), if for all , the graph resulting from adding to is not planar. Each face of is bounded by a cycle. De Fraysseix [dFPP90] proved that has a vertex ordering , called a canonical ordering, with the following properties. Define to be the embedded subgraph of induced by . Let be the subgraph of induced by the edges on the boundary of the outer face of . Then

, and are the vertices on the outer face of .

For each , is a cycle containing .

For each , is biconnected and internally connected; that is, removing any two interior vertices of does not disconnect it.

For each , is a vertex of with at least two neighbours in , and these neighbours are consecutive on .
For example, the ordering in Figure 2(a) is a canonical ordering of the depicted embedded graph .
A frame of [BDH09] is the oriented subgraph of with vertex set , where:

is in and is oriented from to .

For each in the canonical ordering of , edges and are in , where and are the first and the last neighbour, respectively, of along the path in from to not containing edge . Edge is oriented from to , and edge is oriented from to , as illustrated in Figure 2(b).
By definition, is a directed acyclic graph with one source and one sink . The frame defines a partial order on , where whenever there is a directed path from to in .
Translate the given set of lines so that all vertices of the arrangement of lines have negative coordinates, and sort the lines according to the coordinate of the intersection of with the axis. Therefore, the lines have equation , with . Because all intersections among lines of have negative coordinates, all are distinct, and the values are sorted. Note that the slopes might be positive or negative. Let . For any segment of slope in connecting two points and above the axis (that is, ), if and only if .
Construct a linear extension of the partial order and define the bijection as . That is, the vertices of will be placed on the lines in such a way that the partial order is compatible with the order determined by the values of the lines.
We prove by induction that for every value and every , it is possible to draw such that and are placed on points , , and the coordinates of all other vertices are in the horizontal slab . The base case () is obviously true.
Note that we could have formulated the induction on the slopes of the edges of in the drawing. In fact those two formulations imply each other: for any value , there is a such that any segment whose endpoints lie on distinct lines of and have coordinates in , the slope of the segment is in . This is easy to see: draw an upward cone with apex on each point and bounded by the lines of slopes and through that point. Define as the coordinate of the lowest intersection point between any two such cones. Any segment with a slope not in 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 can have such a slope.
Assume by induction that the statement is true for . We will show how to draw for a specific value . The point will be placed on the point on line with coordinate . Let and be the slopes of the segments and , and let or , whichever is smaller. Let be the intersection of the line of slope through and line and the intersection of the line of slope through and . Note that and are strictly positive. Let . Apply the induction hypothesis to draw in the horizontal slab . Thus, in the drawing of , all edges have slope at most . Then by construction, the path in from to not containing edge is monotone (that is, all its edges are oriented rightwards), and is above the supporting line of each edge on that path. Therefore, can see all vertices in and all edges adjacent to 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^{6}^{6}6Two disjoint curves in 3D are unlinked if there is a continuous motion of the curves which transforms them into disjoint coplanar circles without one curve passing through the other or through itself.. These graphs form a threedimensional analogue of the planar graphs.
Open Problem 19.
Is there an arrangement of labelled planes in 3D such that any labelled linkless graph has a linkless straightline embedding where each vertex is placed on the plane with the same label?
Acknowledgements
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