A Lyapunov function for Glauber dynamics on lattice triangulations

A Lyapunov function for Glauber dynamics on lattice triangulations

Alexandre Stauffer University of Bath, Bath, UK; a.stauffer@bath.ac.uk. Supported in part by a Marie Curie Career Integration Grant PCIG13-GA-2013-618588 DSRELIS.
Abstract

We study random triangulations of the integer points , where each triangulation has probability measure with denoting the sum of the length of the edges in . Such triangulations are called lattice triangulations. We construct a height function on lattice triangulations and prove that, in the whole subcritical regime , the function behaves as a Lyapunov function with respect to Glauber dynamics; that is, the function is a supermartingale. We show the applicability of the above result by establishing several features of lattice triangulations, such as tightness of local measures, exponential tail of edge lengths, crossings of small triangles, and decay of correlations in thin rectangles. These are the first results on lattice triangulations that are valid in the whole subcritical regime . In a very recent work with Caputo, Martinelli and Sinclair, we apply this Lyapunov function to establish tight bounds on the mixing time of Glauber dynamics in thin rectangles that hold for all . The Lyapunov function result here holds in great generality; it holds for triangulations of general lattice polygons (instead of the square) and also in the presence of arbitrary constraint edges.

Keywords and phrases. Lattice triangulations, Glauber dynamics, Lyapunov function.
MSC 2010 subject classifications. Primary 60J10; Secondary 60K35, 52C20, 05C10.

1 Introduction

Consider the set of integer points in the plane. A triangulation of is a maximal collection of edges (straight line segments) such that each edge has its endpoints in and, aside from its endpoints, intersects no other edge of and no point of . Figure 1 illustrates a triangulation with .

Figure 1: A lattice triangulation.

Our goal is to study properties of random triangulations. Let be the set of all triangulations of . It is known that, for any triangulation , every triangle in has area exactly , and the set of midpoints of the edges of does not depend on . In particular, this set is

which is the set of half-integer points in excluding . This allows us to regard random lattice triangulations as a spin system since a lattice triangulation can be seen as a collection of variables , where denotes the edge (representing the spin) of the midpoint in . However, many challenges arise when trying to make use of the vast literature on spin systems to study lattice triangulations. For example, lattice triangulations have unbounded dependences as long edges affect far away regions, the interaction between the spins depends on the triangulation, and some useful properties in the study of spin systems do not hold, one example being the FKG inequality, see Appendix B.

There is a natural Markov chain (or Glauber dynamics) over , where transitions are given by flips of uniformly random edges [12, 18]. More precisely, if is the current state of the Markov chain, a transition consists of picking a non-boundary edge of uniformly at random, and if the two triangles containing in form a strictly convex quadrilateral (in which case they actually form a parallelogram), then with probability we remove and replace it by the opposite diagonal of that quadrilateral. Otherwise, the Markov chain stays put; see Figure 2.

Figure 2: A flip of edge in a triangulation. In the triangulation on the left, the flippable edges are marked in blue, while unflippable edges are in black.

The graph on induced by the edge-flipping operation above is usually referred to as the flip graph, and is known to be connected [15]. In addition, since the transition matrix is symmetric and aperiodic, this Markov chain converges to the uniform distribution on . Very little is currently known regarding the dynamic properties of this Markov chain, in particular no non-trivial bound on its mixing time (the time until the Markov chain is close enough to its stationary distribution) is known.

In [6] we introduced a real parameter and considered weighted triangulations: each triangulation has weight , where denotes the norm of the edge . Adapting the Markov chain above using the so-called heat-bath dynamics gives a Markov chain whose stationary distribution is proportional to the weights. Simulation suggests that this Markov chain has intriguing behavior, undergoing a phase transition at ; see Figure 3.

Figure 3: lattice triangulations produced by the edge-flipping Markov chain with (left) and (right). The triangulation in Figure 1 was obtained by the edge-flipping Markov chain with .

It is believed that for any , which we call the subcritical regime, regions far from one another in the triangulation evolve roughly independently. This suggests the presence of decay of correlations and small mixing time. On the other hand, for , which we call the supercritical regime, the Markov chain faces a rigidity phenomenon: long edges give rise to rigid regions of aligned edges. This suggests the presence of “bottlenecks” in the Markov chain, giving rise to exponential mixing time. Finally, in the case, which is the case of uniformly random triangulations, relatively long edges appear but simulation suggests that the regions of aligned edges are not as rigid as in the supercritical regime.

Our contribution.

In this paper we construct a height function on lattice triangulations: given a triangulation , the function attributes a positive real value to each midpoint in . Our main result (Theorem 2.3) establishes that this function behaves as a Lyapunov function with respect to Glauber dynamics. This means that the value of the function at any midpoint behaves like a supermartingale. Theorem 2.3 holds for any , and gives the first result on the dynamics of lattice triangulations that are valid in the whole subcritical regime. A crucial feature of Theorem 2.3 is that it holds in great generality: also in the case of triangulations of general lattice polygons111The set of vertices does not need to be the square, but can be the set of integer points inside any, even non-convex, lattice polygon (a polygon whose vertices are points of ). and in the presence of arbitrary constraint edge222Triangulations where some given set of edges are forced to be present, see Section 2 for precise definitions..

The definition of the height function is a bit involved, so we defer it, as well as the statement of our main result (Theorem 2.3), to Section 2. In particular, the height function is defined in terms of a novel type of geometric crossings, and uses a new partition on the edges of a triangulation in terms of what we call regions of influence. We believe these two new concepts (which we introduce and analyze in Sections  3 and 4) are of independent interest. The proof of Theorem 2.3 is given in Section 5.

Our main result has a large range of consequences and applications, which we discuss in Sections 68. For example, we apply it to establish that the length of an edge of a triangulation has an exponential tail (Corollary 7.4), that local measures are tight (Theorem 7.1), and that there are crossings of triangles of constant size (Theorem 7.5). In the particular case of triangulations of rectangles, where is a fixed integer independent of , our technique yields the existence of local limits (Theorem 8.4), decay of correlations (Theorem 8.3), and recurrence of random walks on the induced graph (Corollary 8.7). In a very recent work with Caputo, Martinelli and Sinclair [5], we apply Theorem 2.3, as well as a number of other results from this paper, to establish tight bounds on the mixing time of triangulations.

Motivation and previous works.

Lattice triangulations have appeared in a broad range of contexts. A number of beautiful combinatorial arguments have been recently developed to estimate the number of lattice triangulations [2, 12, 18]. For example, a very elegant argument by Anclin [2] shows that the cardinality of is at most . Despite not being the best known upper bound, Anclin’s argument is quite general and applies to lattice triangulations of general lattice polygons; we state and use it later, see Lemma 2.1. The best known bounds on are  [12] and  [18] for some positive constant .

Lattice triangulations have also been studied in other areas. For example, in algebraic geometry, they play a key role in the famous construction of plane algebraic curves by Viro [17], which has connections with Hilbert’s Sixteenth problem, and also appeared in the theory of discriminants [10] and toric varieties [7]. Lattice triangulations have also been studied in the contexts of discretization of random surface models [9] and two-dimensional quantum gravity [16]. Several other applications of triangulations are discussed in [8].

Much less is known about random triangulations. The only result on the mixing time of the above edge-flipping Markov chain is [6]. There we showed that, for any , the mixing time is at least for some constant . We also showed that, for all sufficiently small , the mixing time is of order , and a random triangulation has decay of correlations. Extending these results to the whole subcritical regime turned out to be quite challenging, especially since similar results for other spin systems make use of fundamental properties that do not hold on lattice triangulations. This led us to look for new geometric properties of lattice triangulations and to develop new techniques.

Concurrently to [6], a similar model of random lattice triangulations has independently appeared in the statistical physics literature [13, 14]. Following [6], a similar model has been introduced to study the mixing time of random rectangular dissections and dyadic tilings [4].

2 Notation and statement of main result

Unless stated otherwise, henceforth we let be any subset of such that contains all points of that lie inside some lattice polygon, including the vertices of the polygon. A lattice polygon is defined as a polygon whose vertex set only contains points of , and whose edges do not intersect one another (aside from their endpoints) and do not contain points of in their interior, see Figure 4(a). (In a first reading the reader can consider as mentioned in Section 1.) Let be the set of triangulations of , and be the set of midpoints of edges of a triangulation of .

Figure 4: (a) A lattice polygon (black edges): the black points form the induced set and the white points belong to . (b) Two constraint edges (blue edges): the green and red half-integer points form the set of midpoints , the black and blue edges form the boundary condition , and the red points form the midpoints of the boundary condition. (c,d) Two triangulations consistent with the boundary condition of part (b).

Now we formally define the notion of boundary conditions via constraint edges. Consider a collection of edges for some such that each edge of has endpoints in and, aside from its endpoints, does not intersect other edges of or points in . We say that

a triangulation is compatible with if for all .

see Figure 4(b–d). We refer to as a boundary condition and let

For convenience, we assume that (resp., ) always contains the boundary edges (resp., the midpoints of the boundary edges) of the lattice polygon induced by ; for example, in the case of , we have that contains the edges of length connecting consecutive points on the boundary of . See Figure 4(b) for another example. Define

to be the collection of all possible sets of constraint edges with endpoints .

Consequently, for any boundary conditions , contains the boundary of the aforementioned lattice polygon induced by . Henceforth, for any , we denote by the set of midpoints of the edges in .

Given any , any set of constraint edges and any , let

with parameter and starting state .

(When the starting state is not important, we will simply denote the above Markov chain by .) Given a initial triangulation , the Markov chain evolves as follows. Pick a uniformly random midpoint . If is a constraint edge (i.e., ) or is unflippable, do nothing. Otherwise, let denote the triangulation obtained by flipping in . Then with probability , flip in , otherwise do nothing. For any edge , we denote by the length of .

The stationary measure of is denoted by , and is given by

(1)

where We omit the dependence on from to simplify the notation.

Given any midpoint , define as the set of edges of midpoint that are compatible with . In symbols,

(2)

Despite not being the best known upper bound on the number of triangulations, we mention the following upper bound due to Anclin as it holds for arbitrary boundary conditions . Anclin showed that if we order the midpoints in from top to bottom and left to right, and we construct the triangulation by sampling edges one by one following this order, then for each midpoint there are at most two edges of that are compatible with all previously sampled edges. This immediately implies the following upper bound.

Lemma 2.1 (Anclin’s bound, [2]).

Given any set of constraint edges , we have

We refer to the edges of of smallest length as the ground state edges of given . The ground state edges of are either composed of a single edge or are the two opposite unit diagonals (i.e., the diagonals of a square of side length 1). Let

(3)

be the set of ground state edges given . Also, define the set of all possible edges as

We consider that the edges in are open line segments. Hence, two edges that intersect only at their endpoints are considered to be disjoint. The ground state triangulation is defined as the triangulation with smallest total edge length. The following lemma from [6] gives that a ground state triangulation can be easily constructed by independently adding the smallest edge of each midpoint that is compatible with the boundary condition.

Lemma 2.2 (Ground State Lemma, from [6, Lemma 3.4]).

Given any boundary condition , the ground state triangulation given is unique (up to possible flips of unit diagonals), and can be constructed by placing each edge in its minimal length configuration consistent with , independent of the other edges.

The flip operation induces a natural partial order on . It is known that for any non-ground-state edge there is a unique edge such that can be obtained from via a decreasing flip; see, for example, [6, Section 2.2]. In this case we say that is the parent of . When belongs to a triangulation where can be flipped to a shorter edge (which necessarily is ), in this triangulation is the largest diagonal of a parallelogram, which is referred to as the minimal parallelogram of . Then for two distinct edges , we say that

iff there is a sequence such that
(4)

In other words, if , there is a length-increasing sequence of edges with for all , and such that for each there exists two triangulations adjacent in the flip graph satisfying for all , and . Hence, the ground state edges of midpoint are the edges such that there exists no with . We say that if either or .

Given a boundary condition , a triangulation , a midpoint and a ground state edge (whose midpoint is not necessarily ), define the set

(5)

We will show later in Proposition 4.2 that if and only if does not intersect . For any , let denote the length of . Given a parameter , define the function as

(6)

Note that, for any and , letting be the midpoint of , we obtain

If is the ground state triangulation, then for all . We can regard as a height function for triangulations: given any triangulation , can be seen as a height value to the midpoint of .

The theorem below shows that there are values of for which is a Lyapunov function. For this reason, in many parts of the paper we will refer to as the Lyapunov function. Let denote the probability measure induced by , and let be the corresponding expectation.

Theorem 2.3.

For any , there exists , and , each depending only on , for which the following holds. Let be any boundary condition, be any triangulation, and be a random triangulation obtained by applying one step of . For any , if , then

3 Partition of triangulations and trees of influence

Fix any boundary condition . Given a triangulation and a midpoint , we say that is increasing if it is a flippable edge of and after flipping we obtain a (strictly) larger edge. We could define decreasing edges in a similar way, however for technical reasons we need to include some constraint edges in the set of decreasing edges, namely the constraint edges which would be flippable and decreasing if they were not in . We do this by calling decreasing if it is not a unit diagonal and it is the largest edge of all triangles of containing . Note that if is decreasing according to the above definition and , then is flippable and after flipping we obtain a (strictly) smaller edge. For any and triangulation , define the following subsets of :

(7)

Note that for any triangulation and midpoint , we have that is flippable but does not change its length after being flipped.

Given a triangulation , we define a collection of trees whose vertices are elements of . Each tree is rooted at a midpoint in , and there will be two trees for each . We denote these trees by and . To define take one of the triangles of containing . Denote this triangle by . The tree will be defined analogously by considering the other triangle of containing . The root of is . The children of in are the midpoints of the other two edges of . Then we proceed inductively. The children of a midpoint with parent in are obtained by considering the triangle of containing but not containing . If is not the largest edge of , then has no child in ; otherwise the children of are the midpoints of the other edges of (see Figure 5 for a reference). Note that, for any two midpoints with being a child of in , we have that . This guarantees that the construction above ends. Define as a tree rooted at obtained by the union of and . We call the tree of influence of .

Figure 5: (a) A triangulation , with midpoints illustrated by gray points. (b) The tree constructed from the triangle . (c) The tree .

Although we used the term tree, it is not explicit from the construction above that , and are actually trees. However, if we orient the edges from parents to children, since parents are associated to strictly larger edges than their children, the construction above is at least guaranteed to produce a directed acyclic graph. But we have not ruled out the case that a midpoint is reached from two distinct paths from (i.e., some vertices may have two parents). Proposition 3.1 below shows that this does not happen, hence the construction described above indeed produces trees.

Given two midpoints , we will use standard terminology to say that is an ancestor (resp., descendant) of in if there exists a directed path in from to (resp., from to ) using the orientation of edges described above. We will need one more definition. Partition into squares whose edges are parallel to the axes (i.e., the faces of the square lattice). Let be the set of these squares. Given any edge , let

(8)
Proposition 3.1.

Consider any boundary condition , any triangulation , and any . The following statements hold:

  1. is a tree.

  2. For any with being an ancestor of , we have .

  3. For , we have , where are the set of leaves of , which are the vertices without children.

Proof.

Consider the set of squares . Note that partitions this set into two identical regions, which we denote by and . See Figure 6(a) for a reference.

Figure 6: (a) The set of squares , the two identical regions and , and the enlarged region . (b) The children and of decompose into three disconnected regions: , and the triangle . (c) A partition (cf. Proposition 3.4) of a triangulation of the trapezoid into the regions . The bold edges represent the edges of midpoint in , which are the roots of the trees.

Note also that and are lattice polygons. Let be all points of within distance from (including ). We obtain that contains the same points of as . Let be one of the triangles containing (say, the one intersecting ), and assume that and are the children of in . We claim that

all descendants of in are contained in . (9)

In the discussion below, refer to Figure 6(b). Since are children of , we have that and have size smaller than . This implies that the length of is at least . Let denote the vertex of the triangle that is not an endpoint of . Since the area of each triangle is equal to and has length at least , the distance between and is at most . Thus must be inside and, therefore, must be one of the vertices on the boundary of . We can use to partition into three regions: , and the triangle . Since the triangle cannot contain any integer point aside from its three vertices, we have that and are entirely contained in . Doing this construction inductively for and , we establish that the descendants of are contained in and the descendants of are contained in , which establishes (9). Since and have disjoint interior, we obtain that does not contain any cycle, proving part 1. This also gives that

(10)

For part 2, let be the squares of whose interior intersects , and let be the squares of whose interior intersects . Note that and . Also, and . Since the descendants of are contained in , we obtain part 2.

For part 3, applying (10) inductively we have

The same reasoning applies to . ∎

For any triangulation and any , define the set

In words, is the set of midpoints such that is in the tree rooted at . Note that in any tree containing , the parent of in the tree is a midpoint such that is the largest edge in the triangle containing both and .

Lemma 3.2.

For any boundary condition , any triangulation and any such that are in the same triangle and is the largest edge of this triangle, there exists exactly one tree containing both and , and is the parent of in that tree.

Proof.

We will show that we can construct a path of adjacent midpoints in (i.e., midpoints of edges sharing a triangle in ) from and until the root of the tree containing both and . This path will have the property that is the parent of in the tree, for all . Assume, inductively, that we have defined with the property that for all we have that and share the same triangle, for which the largest edge is . Let be the midpoint of the largest edge in the triangle containing but not . If or is contained in only one triangle in (the later implies that as is at the boundary of the smallest lattice polygon containing ), then is the largest edge in all triangles of containing and, consequently, is the root of a tree. This gives that . Otherwise, let , and repeat this procedure. Note that , which implies that this procedure eventually ends, yielding the root of a tree containing and . It remains to show that this is the unique tree containing and . Since for each in the path , we have that is the largest edge in the triangle containing and , we obtain that cannot be a leaf in any tree and the only midpoint that can be a parent of in any tree is . This establishes that, for all , is an ancestor of in any tree containing , which implies that the root of the tree obtained by the above construction is the root of any tree containing , completing the proof. ∎

Proposition 3.3.

Given any boundary condition , any triangulation and any midpoint , the following holds:

  1. The cardinality of is either 1 or 2.

  2. If contains two midpoints, then is a leaf in both and .

  3. If , then contains two midpoints.

  4. If is such that is the largest edge in some triangle in , then contains only one midpoint.

Proof.

Lemma 3.2 implies 1 since for any there exists at least one and at most two midpoints , not necessarily distinct from , such that and are the longest edges in a triangle containing . For 2, note that the cardinality of being two implies that is not the root of a tree, and there are two midpoints such that is the parent of in one tree and is the parent of in the other tree. Therefore, has two distinct parents, one in each tree, implying that cannot be the largest edge in any triangle of ; hence cannot be the parent of any midpoint in any tree. This gives that is a leaf in all trees containing . For 3, note that if , then there are two distinct midpoints such that and are the largest edges in triangles containing . Therefore, using Lemma 3.2, we have that and are the parents of in the trees containing , implying that is contained in two trees. For 4, note that if is a triangle such that is the largest edge, then there exists at most one midpoint that can be the parent of in a tree: namely, the midpoint of the largest edge contained in a triangle with , if that midpoint exists and is different than . Therefore can belong to only one tree. ∎

For each , consider the following subset of :

Proposition 3.4.

For any boundary condition and any triangulation , the set partitions the lattice polygon with vertices in .

Proof.

Proposition 3.3 4 gives that for any triangle of , where is the largest edge of this triangle, there exists only one tree containing . Let be this tree. We have that is the parent of both and in , therefore contains the triangle , and the proof is completed. ∎

We recall the notion of the minimal parallelogram of an edge, which was introduced in [6] and appeared briefly in the paragraph preceding (4). For any edge , the minimal parallelogram of is the unique parallelogram composed of two lattice triangles for which is the longest diagonal.

Proposition 3.5.

Let be any boundary condition and be any triangulation. Let and be two triangles of in the same tree , for some . Assume that , and is an ancestor of in . Then, .

Proof.

First consider the case of being the parent of in , which gives that .

Figure 7: Illustration for the proof of Proposition 3.5 when (a) and (b).

If (see Figure 7(a)), the lemma clearly holds since

If , then we use that are part of the minimal parallelogram of . Refer to Figure 7(b). Let be the edge opposite to in the minimal paralellogram of . Note that may not belong to , and . We claim that

(11)

Using this claim, since is the smallest edge in the minimal parallelogram of , we have

and the proposition follows when is the parent of . In the general case of not being the parent of , the proposition follows by applying the above reasoning inductively along the path from to in the tree .

It remains to establish (11). If were an edge of and were flippable in (as illustrated in Figure 7(b)), then would be a decreasing edge and, by flipping , we would obtain a triangulation in which and are in the same triangle, whose largest edge is . This gives that is part of the minimal parallelogram of , as claimed. ∎

4 Crossings of ground state edges

In this section we consider a given edge and establish geometric properties of the set of edges of a triangulation that intersect ; recall the definition of from (3). In particular, given one edge intersecting , one of our main results here gives that the edges of midpoint also intersect .

We will need the following useful facts from [6]. Fix any boundary condition and any midpoint . Two edges are said to be neighbors if we can obtain from via a single flip; more formally, if there are such that and for all . It is known that the graph with vertex set and the neighborhood relation described above is a tree. This follows since, for any edge , there is at most one such that and are neighbors satisfying (in which case we see as the parent of in the tree). We consider the ground state edges of as the root of the tree, and it is possible that the tree has two neighboring roots, which are opposite unit diagonals. We will call this tree the tree induced by .

Given a boundary condition and a midpoint , we denote by the ground state edge of midpoint given (with an arbitrary choice among unit diagonals). Since ground state edges of distinct midpoints are all compatible with one another, we have that is a ground state triangulation. In the lemma below we use the partial order on the set , which is defined in (4), and the set of midpoints of constraint edges .

Proposition 4.1.

Given any boundary condition , any midpoint , any two edges such that , and any triangulation containing , one can obtain a triangulation containing by performing a sequence of decreasing flips from .

Proof.

Since the graph induced by is a tree, there is a unique path in this graph. We claim that there exists a sequence of decreasing flips from that produce a triangulation containing . With this the lemma follows since we can apply this claim repetitively for until we obtain a triangulation containing .

Now we establish the above claim. If is decreasing in , the claim follows since we can flip to obtain . From now on assume that is not decreasing, and let be the midpoint of . Let be the sum of the length of the edges of that cross , where the set is defined in (8). Let . We have that is a decreasing and flippable edge. Otherwise would imply that is a constraint edge, which gives that is a ground state edge, contradicting that . Let be the triangulation obtained by flipping in . By Proposition 3.12, , hence intersects . Using this and the fact that we obtain that . Repeating these steps we obtain a sequence of triangulations so that the -th triangulation in the sequence is obtained via a decreasing flip of an edge of the -th triangulation, and the value of monotonically decreases along the sequence. Since for any triangulation containing we have that , we obtain that this procedure will eventually make be a flippable and decreasing edge, establishing the claim. ∎

The lemma below gives the first property of crossings of ground state edges. We denote by the indicator function.

Proposition 4.2 (Monotonicity).

Given any boundary condition , any midpoint , any ground state edge , and any two edges such that then

Proof.

We show that if then . If does not intersect , then there is a triangulation so that and . Proposition 4.1 gives that we can perform a sequence of decreasing flips from until obtaining a triangulation with since . Since is in ground state, is not flipped in this sequence. This implies that is contained in and, consequently, cannot intersect . ∎

The following is a simple geometric lemma that we will need later.

Lemma 4.3.

In any triangle of a lattice triangulation, the largest angle is at least and the other angles are at most .

Proof.

Without loss of generality, assume that and “empty” boundary condition (i.e., contains only the unit horizontal and vertical edges at the boundary of ). The lemma will follow for arbitrary choices of and since we can choose large enough so that , which gives that the set of triangulations of with any boundary condition is contained in the set of triangulations of with empty boundary condition. Now this property clearly holds (with equality) for any ground state triangulation of . Proposition 4.1 implies that any triangulation can be obtained by a sequence of increasing flips from some ground state triangulation. Hence it suffices to show that the property in the statement of the lemma is preserved under increasing flips. Let and be two triangles sharing an edge such that is increasing. So is the smallest diagonal of the parallelogram . Let and be the two new triangles obtained after flipping . Note that the largest angles of and are larger than the largest angles of and . Moreover, the other angles of and are obtained by splitting angles of , respectively, where are not the largest angle of . ∎

The next proposition gives an upper bound on the number of small edges intersecting a given ground state edge. For any triangulation , any , and any , let

be the set edges of that intersect and have length at most . Note that is a set of edges (rather than a set of midpoints), and the midpoints of the edges in are given by . A crucial property of the lemma below is that the bounds do not depend on .

Proposition 4.4.

Given any boundary condition , any and any triangulation , all the following statements hold:

  1. If an edge of intersects , then , with strict inequality when the midpoint of is not .

  2. For any , the midpoints of are contained in the ball of radius centered at the midpoint of .

  3. There exists a universal such that for any we have that the cardinality of is at most and the cardinality of is at most .

Proof.

First we establish the lemma when is either a unit horizontal, a unit vertical or a unit diagonal. Then 1 holds trivially since any edge with the same midpoint as has length at least , and an edge with midpoint different than can only intersect if its length is larger than . Parts 2 and 3 follows since any edge of length at most that intersects must be completely contained inside a ball of radius centered at the midpoint of .

Now let be a ground state edge that is not a unit vertical, unit horizontal or unit diagonal. This means that is constrained by a constraint edge ; that is, . The proof uses the concept of excluded regions introduced in [6]. The excluded region of an edge is obtained by taking its minimal parallelogram and considering the infinite strips between both pairs of opposite sides of the parallelogram, as illustrated by the shaded area in Figure 8. The interior of the excluded region contains no point of , cf. [6, Proposition 3.3]. The endpoints of the constraint edge are in regions and , which are the two components of the complement of the excluded region of that contain an endpoint of