Classifying Convex Bodies by their Contact and Intersection Graphs

02/05/2019 ∙ by Anders Aamand, et al. ∙ 0

Suppose that A is a convex body in the plane and that A_1,...,A_n are translates of A. Such translates give rise to an intersection graph of A, G=(V,E), with vertices V={1,...,n} and edges E={uv| A_u∩ A_v≠∅}. The subgraph G'=(V, E') satisfying that E'⊂ E is the set of edges uv for which the interiors of A_u and A_v are disjoint is a unit distance graph of A. If furthermore G'=G, i.e., if the interiors of A_u and A_v are disjoint whenever u≠ v, then G is a contact graph of A. In this paper we study which pairs of convex bodies have the same contact, unit distance, or intersection graphs. We say that two convex bodies A and B are equivalent if there exists a linear transformation B' of B such that for any slope, the longest line segments with that slope contained in A and B', respectively, are equally long. For a broad class of convex bodies, including all strictly convex bodies and linear transformations of regular polygons, we show that the contact graphs of A and B are the same if and only if A and B are equivalent. We prove the same statement for unit distance and intersection graphs.

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1 Introduction

Consider a convex body , i.e., a convex, compact region of the plane with non-empty interior, and let be a set of translates of . Then gives rise to an intersection graph , where and , and a unit distance graph , where if and only if and and have disjoint interiors. In the special case that (i.e., the convex bodies of have pairwise disjoint interiors), we say that is a contact graph (also known as a touch graph or tangency graph). Thus, defines three classes of graphs, namely the intersection graphs , the unit distance graphs , and the contact graphs of translates of .

The study of intersection graphs has been an active research area in discrete and computational geometry for the past three decades. Numerous papers consider the problem of solving classical graph problems efficiently on various classes of geometric intersection graphs. From a practical point of view, the research is often motivated by the applicability of intersection graphs when modeling wireless communication networks and facility location problems. If a station is located at some point in the plane and is able to transmit to and receive from all other stations within some distance then the stations can be represented as disks in such a way that two stations can communicate if and only if their disks overlap.

Meanwhile, the study of contact graphs of translates of a convex body has older roots. It is closely related to the packings of such a body, which has a very long and rich history in mathematics going back (at least) to the seventeenth century, where research on the packings of circles of varying and constant radii was conducted and Kepler famously conjectured upon a 3-dimensional counterpart of such problems, the packing of spheres. An important notion in this area is that of the Hadwiger number of a body , which is the maximum possible number of pairwise interior-disjoint translates of that each touch but do not overlap . The Hadwiger number of is thus the maximum degree of a contact graph of translates of . In the plane, the Hadwiger number is for parallelograms and for all other convex bodies. We refer the reader to the books and surveys by László and Gábor Fejes Tóth [8, 9] and Böröczky [1].

Another noteworthy result on contact graphs is the Circle Packing Theorem (also known as the Koebe–Andreev–Thurston Theorem): A graph is simple and planar if and only if it is the contact graph of some set of circular disks in the plane (the radii of which need not be equal). The result was proven by Koebe in 1935 [16] (see [10] for a streamlined, elementary proof). Schramm [21] generalized the circle packing theorem by showing that if a smooth planar convex body is assigned to each vertex in a planar graph, then the graph can be realized as a contact graph where each vertex is represented by a homothet (i.e., a scaled translation) of its assigned body.

In this paper we investigate the question of when two convex bodies and give rise to the same classes of graphs. We restrict ourselves to convex bodies that have the URTC property (unique regular triangle constructibility). This is the property that given two interior disjoint translates of that touch, there are exactly two ways to place a third translate such that is interior disjoint from and , but touches both. Convex bodies with the URTC property include all linear transformations of regular polygons except squares and all strictly convex bodies [11].

The main result of the paper is summarized in the following theorem.

Theorem 1.

Let and be convex bodies with the URTC property. Then each of the identities , , and holds if and only if the following condition is satisfied: there is a linear transformation of such that for any slope, the longest segments contained in and , respectively, with that slope are equally long.

1.1 Other Related Work

Several papers have compared classes of intersection graphs of various geometric objects, see for instance [2, 4, 5, 12, 17]. Most of the results are inclusions between classes of intersection graphs of one-dimensional objects such as line segments and curves.

A survey by Swanepoel [22] summarizes results on minimum distance graphs and unit distance graphs in normed spaces, including bounds on the minimum/maximum degree, maximum number of edges, chromatic number, and independence number.

Perepelitsa [20] studied unit disk intersection graphs in normed planar spaces and showed that they are -bounded in any such space. Kim et al. [13] improved Perepelitsa’s bound. For other work on intersection graphs of translates of a fixed convex body, see [6, 7, 14, 15].

In the area of computational geometry, Müller et al. [19] gave sharp upper and lower bounds on the size of an integer grid used to represent an intersection graph of translates of a convex polygon with corners at rational coordinates. Their results imply that for any convex polygon with rational corners, the problem of recognizing intersection graphs of translates of is in NP. On the contrary, it is open whether recognition of unit disk graphs in the Euclidean plane is in NP. Indeed, the problem is -complete (and thus in PSPACE), and using integers to represent the center coordinates and radii of the disks in some graphs requires exponentially many bits [3, 18].

1.2 Preliminaries

We begin by defining some basic geometric concepts and terminology.

1.2.1 Convex Bodies and Graphs as Point Sets

For a subset of the plane we denote by the interior of . We say that is a convex body if is compact, convex, and has non-empty interior. We say that is symmetric if whenever , then . It is well-known that if is a symmetric convex body, then the map defined by

is a norm. Moreover and .

It follows from these properties that for translates and it holds that if and only if and if and only if . This means that when studying contact, unit distance, and intersection graphs of a symmetric convex body , we can shift viewpoint from translates of to point sets in and their -distances: If is a set of points we define and to be the graphs with vertex set and edge sets and , respectively. Moreover, if for all distinct points it holds that , we say that is compatible with and define to be the graph with vertex set and edge set . Then , and , respectively, are isomorphic to the intersection, unit distance, and contact graph of realized by the translates . When studying contact, unit distance, and intersection graphs of a symmetric, convex body we will view them as being induced by point sets rather than by translates of .

1.2.2 The URTC Property

We say that a (not necessarily symmetric) convex body in the plane has the URTC property if the following holds: For any two interior disjoint translates of , call them and , satisfying that

, there exists precisely two vectors

such that for , but . If is symmetric, this amounts to saying that for any two points with , the set has size two. Gehér [11] proved that a symmetric convex body has the URTC property if and only if the boundary does not contain a line segment of length more than  in the -norm.

1.2.3 Drawing of a Graph

A drawing of a graph as an intersection graph of a convex body is a point set and a set of straight line segments such that is isomorphic to and is exactly the line segments between the points which are connected by an edge in . We define a drawing of a graph as a contact and unit distance graph similarly.

1.2.4 Notation

For a norm on and a line segment with endpoints and we will often write instead of . Also, if is a symmetric convex body and , we define .

1.3 Structure of the Paper

In Section 2, we establish the sufficiency of the condition of Theorem 1, which is relatively straightforward.

In Section 3 we show how to reduce Theorem 1 to the case where the convex bodies are symmetric. For contact graphs, we then prove the following more general version of the necessity of the condition of Theorem 1 in Section 4.

Theorem 2.

Let and be symmetric convex bodies with the URTC property such that is not a linear transformation of . There exists a graph such that for all and all subgraphs , is not isomorphic to . In particular .

As we will also discuss in Section 4 the same result holds if is replaced by for everywhere in the theorem above. The proof is identical.

In Section 5 we prove the following result which combined with Theorem 2 yields the necessity of the condition of Theorem 1 for intersection graphs.

Theorem 3.

Let and be symmetric convex bodies. If there exists a graph such that for all and all subgraphs , is not isomorphic to , then .

This result holds for general symmetric convex bodies. An improvement of Theorem 2 to general symmetric convex bodies (not necessarily having the URTC property) would thus yield a version of Theorem 1 that also holds for general convex bodies.

2 Sufficiency of the Condition of Theorem 1

This section establishes the sufficiency of the condition stated in Theorem 1 – this is the easy direction. It is worth noting that for this direction our result holds for general convex bodies.

Essentially, we show that the classes of contact, unit distance, and intersection graphs arising from a convex body are closed under linear transformations of and under operations on maintaining the signature of which we proceed to define.

Definition 4 (Profile).

Let be a convex body. A profile through at angle is a closed line segment of maximal length which has argument and is contained in .

Definition 5 (Signature).

Let be a convex body. The signature of is the function satisfying that for every , is the length of a profile through at angle .

Lemma 6.

Let be a convex body and a vector with argument and magnitude . Then

  1. if and only if .

  2. if and only if .

Proof.

First, suppose that . Then and by convexity, the line segment from to , which has length and argument , is contained in . It follows that . If further , there exists a vector of length and argument such that . It follows that implying that , so .

Second, suppose and let be the endpoints of a profile, , through at angle such that the vector from to has argument . Then as . It follows that implying . If further , then . Let be such that and are not colinear. (Such a point exists as has non-empty interior.) The interiors of the triangles with vertices and are contained in and , respectively and their intersection is non-trivial. It follows that , as desired.

Second, the following lemma demonstrates closure of contact, unit distance, and intersection graphs under linear transformations.

Lemma 7.

Let be a compact and connected subset of the plane and be an invertible linear transformation. Then , , and .

Proof.

Suppose that is a realization of a graph as either a contact, unit distance, or intersection graph of . Then is a realization of as a contact, unit distance, or intersection graph, respectively, of . ∎

Finally, we combine the above observations to prove our sufficient condition.

Theorem 8.

Let be convex bodies. If there exists a linear transformation such that the signatures of and of are identical, i.e., , then , , and .

Proof.

Having already established Lemma 7, it suffices to show that if then , , and .

Let be translated copies of in the plane; let be vectors satisfying for every ; and let translated copies of , , be defined by for . Now, consider any pair and denote by and the argument and magnitude of the vector . By Lemma 6, if and only if , which is true if and only if . Similarly, if and only if . It follows that if is a realization of a graph as a contact, unit distance, or intersection graph then is a realization of as a contact, unit distance, or intersection graph, respectively.

3 Reducing to Symmetric Convex Bodies

In this section we show that for proving the necessity of the condition of Theorem 1 it suffices to consider only symmetric convex bodies with the URTC property. We use the well-known trick in discrete geometry of considering the “symmetrization” of a convex body , .

Lemma 9.

Let be a convex body with signature . Then is a symmetric convex body with signature .

Proof.

It is well-known and easy to check that is symmetric, bounded and convex. For the statement concerning the signature, let be given. Consider a profile through at angle . Then is a line segment of the same length and argument as and hence, . Conversely, consider a profile through at angle and let and be the endpoints of with . Then the vector has argument and magnitude , but since

and by convexity of , there is a line segment from to in with the same argument and magnitude as . It follows that . ∎

Thus, for every convex body there is a symmetric convex body with the same signature and which by Theorem 8 therefore has the same contact, unit distance, and intersection graphs as .

Proposition 10.

Let and be convex bodies such that . Then has the URTC property if and only if has the URTC property. In particular this applies when is the symmetrization of .

Proof.

Suppose that has the URTC property. Let and be translates of satisfying that but . Let and . Since ,  Lemma 6 implies that but . Again using Lemma 6 we obtain that for a vector it holds that intersects and but only at their boundary if and only if intersects and but only on at the boundary. As has the URTC property it follows that there are exactly two choices of such that intersect and but only at their boundary. Since and were arbitrary it follows that has the URTC property. The converse implication is identical. ∎

Combined with the work done in the main body of this paper (Section 4 and 5Theorem 1 follows immediately:

Proof of Theorem 1.

We have already seen the sufficiency (Theorem 8) of the condition. Theorems 2 and 3 (to be proved in the following sections) give the necessity in case and are symmetric. Suppose now that and are arbitrary convex bodies with the URTC property satisfying that for any linear transformation of , . For let . If there exists a linear transformation satisfying that then symmetri yields that . Applying Lemma 9 we obtain the contradiction that and we conclude that for no linear transformation is . Since and are symmetric and by Proposition 10 both have the URTC property, we conclude that , , and . But , , and for by Theorem 8, so the result follows. ∎

4 Necessity for Contact and Unit Distance Graphs

In this section we prove the necessity of the condition of Theorem 1 in the case of contact graphs in the setting where and are symmetric. The proof for unit distance graphs is completely identical so we will merely provide a remark justifying this claim by the end of the section. The main result of the section is slightly more general than required since we will use it in the classification of intersection graphs.

4.1 Properties of the Signature

Towards proving the main theorem of the section, we prove three lemmas regarding the behaviour of the signature of a symmetric convex body. The content of the concluding lemma is as follows: If and , are symmetric convex bodies satisfying that no linear map transforms into , then there exists a finite set of angles, and an , such that the signature of no linear transform of is -close to the signature of at every angle . This observation motivates the constructions of the section to follow.

Lemma 11.

For a symmetric convex body the signature is continuous.

Proof.

For we define the rotation matrix . We further let . It is easy to check that the mapping is a continuous map . Furthermore, since any norm on induces the same topology as that of the Euclidian norm, the map is continuous with respect to the standard topology on . Thus, the map defined by is continuous. The conclusion follows as . ∎

Lemma 12.

Let be a symmetric convex body. Let be the set of all non-singular linear maps with topology induced by the operator norm. For each the map defined by is continuous.

Proof.

If is non-singular, is also symmetric with non-empty interior and thus induces a norm on . Let be as in the proof of Lemma 11 and . Then

Note that

and so it suffices to show that the mapping is a continuous map . Now, for ,

where is the operator norm, so it suffices to show that the inversion is a continuous map . It is a standard result from the literature that on the set of invertible elements of a unital Banach algebra, , the operation of inversion, , is continuous. Since is exactly the invertible elements of the Banach algebra of linear maps , the conclusion follows. ∎

Figure 1: The two steps of the proof of Lemma 13
Lemma 13.

Let be a symmetric convex body and constants. Consider some and suppose that for every there exists (with coordinates modulo ) such that . If is sufficiently small as a function of and , for every .

Proof.

We compute all coordinates modulo . Also let be the center of .

For the first part of the argument, see the left-hand side of Figure 1. We start by letting . Then for arbitrary there exists and points such that the line segments and have arguments and , respectively, and satisfy . By convexity, the line segment is contained in and furthermore, it is easy to verify that every point on has distance at least to . Since the profile of at angle which passes through intersects , it follows that . Thus, for all , .

For the remaining argument, see the right-hand side of Figure 1. Towards our main conclusion, suppose for contradiction that there is a line segment of length and argument contained in . Let be the line segment of argument and length and note that it is contained in . By assumption there exists a boundary point of such that and has argument where . By convexity, intersects at a point . Denote by the angle and note that it is a constant depending only on and . Now, by the law of sines in ,

so for sufficiently small as a function of and , , which contradicts . ∎

Lemma 14.

Let and be symmetric convex bodies in . Suppose that for every finite set and for every , there exists a linear map satisfying that for all . Then there exists a linear map with .

Proof.

We may clearly assume that and are centered at the origin. For , let ; let ; and let be a linear map satisfying that for all .

For and denote by the open ball in the Euclidian norm with center and radius . We begin by proving that is uniformly bounded in the operator norm by showing that there exists a constant such that when is sufficiently large. To this end, let and and note that . There exists and constants such that for every , and . This implies that for every and , . Applying Lemma 13, we find that there exists an such that for every , . Thus, for so is uniformly bounded. As it moreover holds for and that , convexity of gives that for . In particular .

Since is uniformly bounded we may by compactness assume that converges in operator norm to some linear map by passing to an appropriate subsequence. As moreover , it is easy to check that so in particular is non-singular. We claim that . As and are symmetric it suffices to show that . Moreover, and are both continuous by Lemma 11 so since is dense in it suffices to show that for each . To see this let and let be defined as in Lemma 12. Then is continuous so as (here we use that is non-singular). It follows that

as desired. This completes the proof. ∎

4.2 Establishing Necessity

Before proving the part of Theorem 1 concerning contact graphs we describe certain lattices which gives rise to contact graphs that can be realised in an essentially unique way. We start with the following definition.

Definition 15.

Let be a symmetric convex body with the URTC property, and the associated norm. Let be such that . We define the lattice .

Note that if has been chosen with , then using the URTC property there are precisely two vectors with . If one is the second is so regardless how we choose we obtain the same lattice. Let us describe a few properties of the lattice . Using the triangle inequality and the URTC property of it is easily verified that for distinct , with equality holding exactly if . Another useful fact is the following:

Lemma 16.

With as above it holds that . Here is the convex hull of . If in particular is another symmetric convex body for which , then for all it holds that .

Proof.

As and is convex the first inclusion is clear. For the second inclusion we note that all points on the hexagon connecting the points of in this order has by the triangle inequality and so .

For the last statement of the lemma note that if then

and similarly . ∎

Definition 17.

We say that a graph is lattice unique if and there exists an enumeration of its vertices such that

  • The vertex induced subgraph is a triangle.

  • For there exists distinct such that and both and are edges of .

Suppose that is a symmetric convex body with the URTC property, that is compatible with , and that is lattice unique. Enumerate the points of according to the definition of lattice uniqueness. Without loss of generality assume that . Then the URTC property of combined with the lattice uniqueness of gives that are uniquely determined from and and all contained in . If moreover is another convex body with the URTC property, has and is compatible with , and via the graph isomorphism , then the linear map defined by satisfies that .

Before commencing the proof of Theorem 2 let us highlight the main ideas. The most important tool is Lemma 14 according to which there exist and a finite set of directions such that for any linear tranformation of there is a direction such that and differ by at least . We will construct by describing a finite set compatible with , and defining . Now, will be a disjoint union of two sets of points, , where and will play complementary roles. The construction will be such that is a subset of a lattice and such that the corresponding induced subgraph of is lattice unique. More precisely will consist of large hexagons connected along their edges. When attempting to realize as a contact graph of the lattice uniqueness enforces that is realized as a subgraph of a lattice in essentially the same way. The remaining points of do not lie in the lattice . They constitute rigid beams in the directions from “connecting” diagonally opposite points of the hexagons of . The construction of is depicted in the left-hand side of Figure 2 and in Figure 3. When trying to reconstruct the same contact graph (or a supergraph) with beams connecting the corresponding points of , we will find that in at least one direction the beam becomes too long or too short.

Proof of Theorem 2.

We let be such that and define the lattice . We also define the infinite graph . Without loss of generality we can assume that and satisfy that , since there exists a non-singular linear transformation such that , and . Note that in this setting we can use Lemma 16 to compare to the circle of radius 1 and obtain for every .

Figure 2: Left: The points of along with the corresponding lattice unique subgraph . Right: The attachment of the beam .

As already mentioned we will construct by specifying a finite point set compatible with and define . The construction of can be divided into several sub-constructions. We start out by describing a hexagon of points for which satisfies that is lattice unique.

Construction 18 ().

For an illustration of the construction see the left-hand side of Figure 2. For we write for the distance between and in the graph , and for we define . Clearly is a lattice unique graph. Moreover, using that and satisfy it is easy to check that the points lie on a regular hexagon whose corners have distance exactly to the origin in the Euclidian norm. In particular the points has , and thus by Lemma 16.

For a given and we will construct a set of points compatible with which constitute a “beam” of argument :

Construction 19 ().

Let be the vector of argument with , and let be such that (by the URTC property we have two choices for ). For a given we define

Note that is compatible with and that is lattice unique.

For a given we want to choose as large as possible such that “fits inside” . We then wish to “attach” to