4 The Constrained Empty-Rectangle Delaunay Graph

# Constrained Generalized Delaunay Graphs Are Plane Spanners1

## Abstract

We look at generalized Delaunay graphs in the constrained setting by introducing line segments which the edges of the graph are not allowed to cross. Given an arbitrary convex shape , a constrained Delaunay graph is constructed by adding an edge between two vertices and if and only if there exists a homothet of with and on its boundary that does not contain any other vertices visible to and . We show that, regardless of the convex shape used to construct the constrained Delaunay graph, there exists a constant (that depends on ) such that it is a plane -spanner. Furthermore, we reduce the upper bound on the spanning ratio for the special case where the empty convex shape is an arbitrary rectangle to , where and are the length of the long and short side of the rectangle.

## 1 Introduction

A geometric graph is a graph whose vertices are points in the plane and whose edges are line segments between pairs of vertices. A graph is called plane if no two edges intersect properly. Every edge is weighted by the Euclidean distance between its endpoints. The distance between two vertices and in , denoted by , or simply when is clear from the context, is defined as the sum of the weights of the edges along the shortest path between and in . A subgraph of is a -spanner of (for ) if for each pair of vertices and , . The smallest value for which is a -spanner is the spanning ratio or stretch factor of . The graph is referred to as the underlying graph of . The spanning properties of various geometric graphs have been studied extensively in the literature (see [7, 15] for an overview of the topic).

Most of the research has focused on constructing spanners where the underlying graph is the complete Euclidean geometric graph. We study this problem in a more general setting with the introduction of line segment constraints. Specifically, let be a set of points in the plane and let be a set of line segments with endpoints in , with no two line segments intersecting properly. The line segments of are called constraints. Two vertices and can see each other or are visible to each other if and only if either the line segment does not properly intersect any constraint or is itself a constraint. If two vertices and can see each other, the line segment is a visibility edge. The visibility graph of with respect to a set of constraints , denoted , has as vertex set and all visibility edges as edge set. In other words, it is the complete graph on minus all edges that properly intersect one or more constraints in .

This setting has been studied extensively within the context of motion planning amid obstacles. Clarkson [10] was one of the first to study this problem and showed how to construct a linear-sized -spanner of . Subsequently, Das [11] showed how to construct a spanner of with constant spanning ratio and constant degree. Bose and Keil [5] showed that the Constrained Delaunay Triangulation is a -spanner of . The constrained Delaunay graph where the empty convex shape is an equilateral triangle was shown to be a 2-spanner of  [4]. We look at the constrained generalized Delaunay graph, where the empty convex shape can be any convex polygon.

In the unconstrained setting, it is known that generalized Delaunay graphs are spanners [2], regardless of the convex shape used to construct it. These bounds are very general, but unfortunately not tight. In special cases, better bounds are known. For example, when the empty convex shape is a circle, Dobkin et al. [13] showed that the spanning ratio is at most . Improving on this, Keil and Gutwin [14] reduced the spanning ratio to . Recently, Xia showed that the spanning ratio is at most 1.998 [17]. On the other hand, Bose et al. [3] showed a lower bound of 1.58, which is greater than , which was conjectured to be the tight spanning ratio up to that point. Later, Xia and Zhang [18] improved this to 1.59.

Chew [9] showed that if an equilateral triangle is used instead, the spanning ratio is 2 and this ratio is tight. In the case of squares, Chew [8] showed that the spanning ratio is at most . This was later improved by Bonichon et al. [1], who showed a tight spanning ratio of .

In this paper, we show that the constrained generalized Delaunay graph is a spanner whose spanning ratio depends solely on the properties of the empty convex shape used to create it: We show that satisfies the -diamond property and the visible-pair -spanner property (defined in Section 3.2), which implies that it is a -spanner for:

 t=⎧⎪⎨⎪⎩2κC⋅max(3sin(αC/2),κC), \emph{if G is a triangulation}2κ2C⋅max(3sin(αC/2),κC), \emph{otherwise.}

This proof is not a straightforward adaptation from the work by Bose et al. [2] due to the presence of constraints. For example, showing that a region contains no vertices that are visible to some specific vertex requires more work that showing that this same region contains no vertices, since we allow vertices in the region that are not visible to . Induction cannot be applied in a straightforward manner as in the unconstrained case because not all pairs of vertices are visible to each other. Moreover, we prove a slightly stronger result, where constraints are not necessarily edges of the graph. We elaborate on this point in more detail in Section 2.

Though the spanning proof is very general, since it holds for arbitrary convex shapes, its implied spanning ratio is far from tight. To improve on this, we also consider the special case where the empty convex shape is a rectangle and show that it has spanning ratio at most , where and are the length of the long and short side of .

## 2 Preliminaries

Throughout this paper, we fix a convex shape . We assume without loss of generality that the origin lies in the interior of . A homothet of is obtained by scaling with respect to the origin, followed by a translation. Thus, a homothet of can be written as

 x+λC={x+λz:z∈C},

for some scaling factor and some point in the interior of after translation. We refer to as the center of the homothet .

For a given set of vertices and a set of constraints , the constrained generalized Delaunay graph is usually defined as follows. Given any two visible vertices and , let be any homothet of with and on its boundary. The constrained generalized Delaunay graph contains an edge between and if and only if there exists is a constraint or there exists a such that there are no vertices of in the interior of visible to both and . We assume that no four vertices lie on the boundary of any homothet of .

Now, slightly modify this definition such that there is an edge between and if and only if there exists a such that there are no vertices of in the interior of visible to both and . Note that this modified definition implies that constraints are not necessarily edges of the graph, since constraints may not necessarily adhere to the visibility property. Indeed, our modified graph is always a subgraph of the constrained generalized Delaunay graph. Therefore, any result proven on our modified graph also holds for the graph that includes all the constraints. As such, we prove the stronger result on our modified graph. For simplicity, in the remainder of the paper, when we refer to the constrained generalized Delaunay graph, we mean our modified subgraph of the constrained generalized Delaunay graph.

### 2.1 Auxiliary Lemmas

Next, we present three auxiliary lemmas that are needed to prove our main results. First, we reformulate a lemma that appears in [16].

###### Lemma 1

Let be a closed convex curve in the plane. The intersection of two distinct homothets of is the union of two sets, each of which is either a segment, a single point, or empty.

Though the following lemma (see also Figure 1) was applied to constrained -graphs in [4], the property holds for any visibility graph. We say that a region contains a vertex if lies in the interior or on the boundary of . We call a region empty if it does not contain any vertex of in its interior. We also note that we distinguish between vertices and points. A point is any point in , while a vertex is part of the input.

###### Lemma 2

Let , , and be three arbitrary points in the plane such that and are visibility edges and is not the endpoint of a constraint intersecting the interior of triangle . Then there exists a convex chain of visibility edges from to in triangle , such that the polygon defined by , and the convex chain is empty and does not contain any constraints.

Let and be two vertices that can see each other and let be a convex polygon with and on its boundary. Extend to half-lines with source all constraints and edges that have as an endpoint and intersect (see Figure 2a). Define the clockwise neighbor of to be the half-line that minimizes the strictly positive clockwise angle with and define the counterclockwise neighbor of to be the half-line that minimizes the strictly positive counterclockwise angle with . We define the cone that contains to be the region between the clockwise and counterclockwise neighbor of . Finally, let , the region of that contains with respect to , be the intersection of and (see Figure 2b).

###### Lemma 3

Let and be two vertices that can see each other and let be any convex polygon with and on its boundary. If there is a vertex in (other than and ) that is visible to , then there is a vertex (other than and ) that is visible to both and and such that triangle is empty.

Proof. We have two visibility edges, namely and . Since lies in , is not the endpoint of a constraint such that and lie on opposite sides of the line through this constraint. Hence, we can apply Lemma 2 and we obtain a convex chain of visibility edges from to and the polygon defined by , and the convex chain is empty and does not contain any constraints. Furthermore, since the convex chain is contained in triangle , which in turn is contained in , every vertex along the convex chain is contained in (see Figure 3).

Let be the neighbor of along this convex chain. Hence, is visible to and contained in . Furthermore, can see , since the line segment is contained in the polygon defined by , and the convex chain, which is empty and does not contain any constraints. This also implies that triangle is empty.

## 3 The Constrained Generalized Delaunay Graph

Before we show that every constrained generalized Delaunay graph is a spanner, we first show that they are plane.

### 3.1 Planarity

In order to show that the constrained generalized Delaunay graph is plane, we first observe that no edge of the graph can contain a vertex, as this vertex would lie in and be visible to both endpoints of the edge.

###### Observation 4

Let be an edge of the constrained generalized Delaunay graph. The line segment does not contain any vertices other than and .

###### Lemma 5

The constrained generalized Delaunay graph is plane.

Proof. We prove this by contradiction, so assume that there exist two edges and that intersect properly, i.e. not at their endpoints. It follows from Observation 4 that neither nor lies on and that neither nor lies on , so the edges intersect properly. Since is contained in and is contained in , the boundaries of and intersect or one of and contains the other.

We first show that this implies that , , , or . If one of and contains the other, this holds trivially. If the two homothets intersect and either or , we are done, so assume that neither nor lies in . Lemma 1 states that the boundaries of and intersect each other at most twice. These intersections split the boundary of into two parts: one that is contained in and one that is not. Since and , and lie on the arc of that is not contained in (see Figure 4). However, intersects , since otherwise cannot intersect . Let and be the two intersections of with the boundary of (if the boundary of is parallel to , and are the two endpoints of the interval of this intersection). We note that and split into two parts, one of which is contained in , and that and cannot lie on the same part. In particular, one of and lies on the part that is contained in , proving that , or .

In the remainder of the proof, we assume without loss of generality that (see Figure 4). Let be the intersection of and . Hence, can see both and . Also, is not the endpoint of a constraint intersecting the interior of triangle . Therefore, it follows from Lemma 2 that there exists a convex chain of visibility edges from to . Let be the neighbor of along this convex chain. Since is part of the convex chain, which is contained in , which in turn is contained in , it follows that is a vertex visible to contained in . Furthermore, since the polygon defined by , and the convex chain does not contain any constraints, lies in . Thus, it follows from Lemma 3 that there exists a vertex in that is visible to both and , contradicting that is an edge of the constrained generalized Delaunay graph.

### 3.2 Spanning Ratio

Let and be two distinct points on the boundary of . These two points split into two parts. For each of these parts, there exists an isosceles triangle with base such that the third vertex lies on that part of . We denote the base angles of these two triangles by and . We define as follows:

 αC=min{max(αx,y,α′x,y):x,y∈∂C,x≠y}. (1)

Given a graph and an angle , we say that an edge of satisfies the -diamond property, when at least one of the two isosceles triangles with base and base angle does not contain any vertex visible to both and . A graph satisfies the -diamond property when all of its edges satisfy this property [12]. Some examples of are the following: When is a circle, , when is a rectangle where and are the length of its long and short side, , and when is an equilateral triangle, .

###### Lemma 6

Let be any convex polygon. The constrained generalized Delaunay graph satisfies the -diamond property.

Proof. Let be any edge of the constrained generalized Delaunay graph. Since is an edge, there exists a such that does not contain any vertices that are visible to both and . The vertices and split the boundary of into two parts and each of these parts defines an isosceles triangle with base . Let and be the base angles of these two isosceles triangles and assume without loss of generality that (see Figure 5). Let be the third vertex of the isosceles triangle having base angle .

Translate and scale such that it corresponds to . This transformation does not affect the angles and . Hence, since and both lie on the boundary of , the pair is one of the pairs considered when determining in Equation 1. Hence, since , it follows that . Let be the third vertex of the isosceles triangle having base and base angle that lies on the same side of as triangle (see Figure 5). Since , triangle is contained in triangle . By convexity of , is contained in . Hence, since does not contain any vertices visible to both and , triangle does not contain any vertices visible to both and either. Hence, satisfies the -diamond property.

For the next property, let be a point in the interior of and let and be two distinct points on , such that , , and are collinear. Again, and split into two parts. Let and denote the length of these two parts. We define as follows:

 κC,O=max{max(ℓx,y,ℓ′x,y)|xy|:x,y∈∂C,x≠y,and x,y,and O are collinear}.

We note that the constrained generalized Delaunay graph does not depend on the location of inside , as the presence of any edge is defined in terms of , which does not depend on the location of . Therefore, we define as follows:

 κC=min{κC,O:O is in the interior of C}.

Throughout the remainder of this section, we assume that is picked such that . Some examples of are the following: When is a circle, with being the center of , when is a rectangle where and are the length of its long and short side, with being the center of , and when is an equilateral triangle, with being the center of .

Given a constrained generalized Delaunay graph , let and be two vertices on the boundary of a face of the constrained generalized Delaunay graph, such that can see and the line segment does not intersect the exterior of . If for every such pair and on every face , there exists a path in of length at most , then satisfies the visible-pair -spanner property. We show that the constrained generalized Delaunay graph satisfies the visible-pair -spanner property. However, before we do this, we bound the length of the union of the boundary of a sequence of homothets that have their centers on a line.

Let a set of vertices be given, such that all vertices lie on one side of the line through and . For ease of exposition, assume the line through and is the -axis and all vertices lie on or above this line. We consider only point sets for which there exists , a set of homothets of , such that the center of each homothet lies on the -axis, has and on its boundary, and no contains any vertices other than and . Let be the boundary of above the -axis and let be the part of the boundary of between and that lies above the -axis.

###### Lemma 7

Let be the homothet with and on its boundary and its center on the -axis. It holds that

 k∑i=1|∂(vi,vi+1)|≤|∂C(v1,vk+1)|.

Proof. We prove the lemma by induction on , the number of homothets. If , is the same as , so the lemma holds.

If , we assume that the induction hypothesis holds for all sets of at most homothets. Since homothet does not contain any vertices other than and , it follows that none of the homothets are fully contained in the union of the other homothets.

Let be the homothet that defines the rightmost intersection with the -axis when is not part of the set of homothets. Let be the leftmost intersection of and the -axis (see Figure 6). Let and let be the part of between and above the -axis. Let be the part of between and above the -axis. To prove the lemma, we need to show that .

Let , so . Since
, it follows from the induction hypothesis that . Since the center of lies on the -axis, it follows that . We consider two cases: (a) lies to the left of , (b) lies on or to the right of .

Case (a): If lies to the left of , let be the homothet centered on the -axis with and on its boundary (see Figure 6a). Hence, it follows that . Since has and on its left boundary, it is contained in , and since it has on its right boundary, it is contained in . Hence, is contained in the intersection of and . Since the length of the boundary of this intersection above the -axis is and is convex, it follows that . Hence, we have that

 k∑i=1|∂(vi,vi+1)| = |∂(v1,vk)|+|∂(vk,vk+1)| = |∂(v1,vk)|+|∂(vk,r)|−|∂(vk,r)|+|∂Ck|−|∂(l,vk)| ≤ c⋅|v1r|−|∂(vk,r)|+c⋅|lvk+1|−|∂(l,vk)| = c⋅|v1vk+1|+c⋅|lr|−|∂(l,vk)|−|∂(vk,r)| = |∂C(v1,vk+1)|+|∂C(l,r)|−|∂(l,vk)|−|∂(vk,r)| ≤ |∂C(v1,vk+1)|.

Case (b): If lies on or to the right of (see Figure 6b), we have that

 k∑i=1|∂(vi,vi+1)| = |∂(v1,vk)|+|∂(vk,vk+1)| ≤ |∂(v1,vk)|+|∂(vk,r)|+|∂(l,vk)|+|∂(vk,vk+1)| ≤ c⋅|v1r|+c⋅|lvk+1| ≤ c⋅|v1vk+1| = |∂C(v1,vk+1)|,

completing the proof.

###### Lemma 8

The constrained generalized Delaunay graph satisfies the visible-pair -spanner property.

Proof. Let and be two vertices on the boundary of a face of the constrained generalized Delaunay graph, such that can see and the line segment does not intersect the exterior of . Assume without loss of generality that lies on the -axis. Let be the homothet of with and on its boundary and its center on . We aim to show that there exists a path between and of length at most . Since by definition is at least , showing that there exists a path between and of length at most completes the proof. If is an edge of the constrained generalized Delaunay graph, this follows from the triangle inequality, so assume this is not the case.

We grow a homothet with its center on by moving its center from to , while maintaining that lies on the boundary of (see Figure 7a). Let be the first vertex hit by that is visible to and lies in . We assume without loss of generality that lies above . Since is the first vertex satisfying these conditions, is either an edge or a constraint: Since is the first visible vertex we hit in , we have that contains no vertices visible to . Hence, there is no vertex visible to both and . Therefore, Lemma 3 implies that does not contain any vertices visible to . Finally, if is not a constraint, contains the region that is visible to both and . Hence, if is not a constraint, the region that is visible to both and does not contain any vertices and is an edge.

We continue constructing a sequence of vertices until we hit by moving the center of along towards and each time we hit a vertex , we require that it lies on the boundary of until we hit the next vertex that is visible to and is not the endpoint of a constraint that lies in the counterclockwise angle (see Figure 7b). Since is the first vertex satisfying these conditions starting from , is either an edge or a constraint. This in turn implies that these vertices all lie above , since is visible and does not intersect the exterior of .

Unfortunately, we cannot assume that there exists an edge between every pair of consecutive vertices: If is a constraint, there can be vertices visible to both and on the opposite side of the constraint. For pairs of vertices that do not form an edge, we refine the construction of the sequence between them: We start with such that it does not cross and lies on its boundary. We construct a sequence of vertices from to by moving the center of along towards , maintaining that lies on its boundary (see Figure 7c). For the first vertex we hit, we require that it is visible to and lies in .

We continue moving the center of along towards , but we now maintain that lies on the boundary of . Each time we hit a vertex , we require that it lies on the boundary of until we hit the next vertex that is visible to and is not the endpoint of a constraint that lies in the counterclockwise angle . In other words, we construct a more fine-grained sequence when consecutive vertices define a constraint and there is no edge between them. Note that we may need to repeat this process a number of times, since there need not be edges between the vertices of the finer grained sequence either. However, since the point set is finite, this process terminates.

This way, we end up with a path from to that lies above . Furthermore, since is convex, we can upper bound the length of each edge by the part of the boundary of , the homothet with and on its boundary and its center on , that does not intersect . Hence, the total length of the path is upper bounded by the length of the union of the boundaries of these homothets above . By construction, none of the homothets corresponding to consecutive vertices along the path contain any of the other vertices along the path. Hence, we can apply Lemma 7 and it follows that the total length of the path is at most , completing the proof.

We are now ready to prove that the constrained generalized Delaunay graph is a spanner. Das and Joseph [12] showed that any plane graph that satisfies the diamond property and the good polygon property (similar to the visible-pair -spanner property) is a spanner. Subsequently, Bose et al. [6] improved slightly on the spanning ratio. They showed that a geometric (constrained) graph is a spanner of the visibility graph when it satisfies the following properties:

1. is plane.

2. satisfies the -diamond property.

3. The spanning ratio of any one-sided path in is at most .

4. satisfies the visible-pair -spanner property.

In particular, is a -spanner for

 t=2κκ′⋅max(3sin(α/2),κ).

It follows from Lemmas 5, 6, and 8 that the constrained generalized Delaunay graph satisfies these four properties. Moreover, even though in general the constrained generalized Delaunay graph is not a triangulation, if for a specific convex shape it is, it satisfies the visible-pair 1-spanner property: Since every face consists of three vertices that are pairwise connected by an edge, the shortest path between two vertices and on this face has length . Therefore, we obtain the following theorem:

###### Theorem 9

The constrained generalized Delaunay graph is a -spanner for

 t=⎧⎪⎨⎪⎩2κC⋅max(3sin(αC/2),κC), \emph{if G is a triangulation}2κ2C⋅max(3sin(αC/2),κC), \emph{otherwise.}

## 4 The Constrained Empty-Rectangle Delaunay Graph

In this section, we look at the case where the empty convex shape is an arbitrary rectangle. We assume without loss of generality that the rectangle is axis-aligned. We do not, however, assume anything about the ratio between the height and width of the rectangle. We first show that if two visible vertices cannot see any vertices in on one side of , then no vertex in on the opposite side of can see any vertices beyond either.

###### Lemma 10

Let and be two vertices that can see each other, such that is not vertical, and let be any convex polygon with and on its boundary. If the region of below does not contain any vertices visible to and , then no point in above can see any vertices in below .

Proof. We prove the lemma by contradiction, so assume that there exists a vertex in below that is visible to , but not to and . Since is a convex polygon and and lie on opposite sides of , the visibility edge intersects . Let be this intersection (see Figure 8).

Hence, and are visibility edges. Since is not a vertex, it is not the endpoint of any constraints intersecting the interior of triangle . It follows from Lemma 2 that there exists a convex chain of visibility edges between and and this chain is contained in . However, this implies that , the neighbor of along this chain, is visible to and lies in below . Next, we apply Lemma 2 on triangle and find that the neighbor of along the chain from to is visible to both and and lies in below , contradicting that this region does not contain any vertices visible to and .

Next, we introduce some notation for the following lemma. Let and be two vertices of the constrained generalized Delaunay graph that can see each other. Let be a rectangle with and on its West and East boundary and let , , and be the Northwest, Northeast, and Southwest corner of . Let be any points on in the order they are visited when walking from to (see Figure 9). Let and . Consider the homothets of with and on their respective boundaries, for , such that , where , , are the Northwest, Northeast, and Southwest corner of .

###### Lemma 11

We have

 k−1∑i=0(|miai|+|aibi|+|bimi+1|)=|pa|+|ab|+|bq|.

Proof. Let . We first show that for every we have that , for . Since is a homothet of and the slopes of and are equal, we have that . Furthermore, by construction , and since the slopes of and are equal, we also have that . Finally, since is a homothet of , we have that , which gives and .

Hence, since , for , we get

 k−1∑i=0(|miai|+|aibi|+|bimi+1|) =k−1∑i=0(c⋅|mimi+1|) =c⋅|pq| =|pa|+|ab|+|bq|,

proving the lemma.

Before we prove the bound on the spanning ratio of the constrained generalized Delaunay graph, we first bound the length of the spanning path between vertices and for the case where the rectangle is partially empty. We call a rectangle half-empty when contains no vertices in below that are visible to and no vertices in below that are visible to . We denote the - and -coordinate of a point by and .

###### Lemma 12

Let and be two vertices that can see each other. Let be a rectangle with and on its boundary, such that it is half-empty. Let and be the corners of on the non-half-empty side. The constrained generalized Delaunay graph contains a path between and of length at most .

Proof. We prove the lemma by induction on the rank of when ordered by size, for any two visible vertices and , such that is half-empty. We assume without loss of generality that lies on the West boundary, lies on the East boundary and that is half-empty below . This implies that and are the Northwest and Northeast corner of . We also assume without loss of generality that the slope of is non-negative, i.e. and (see Figure 10). Note that this can be achieved by swapping and , if needed.

We note that the case where lies on the West boundary, lies on the North boundary and is half-empty below can be viewed as a special case of the one above: We shrink until one of and lies in a corner. This point can now be viewed as being on both sides defining the corner and hence and are on opposite sides: If lies in the Southwest corner, we treat it as lying on the South boundary while lies on the North boundary. If lies in the Northeast corner, we treat it as lying on the East boundary while lies on the West boundary. An analogous statement holds for the case where lies on the West boundary, lies on the North boundary and is half-empty above .

Let be the Southwest corner of . Let be a homothet of that is contained in and whose West boundary is intersected by . Let , , be the Northwest, Northeast, and Southwest corner of and let be the intersection of and . We call homothet similar to if and only if .

Base case: If is a half-empty rectangle of smallest area, then does not contain any vertices visible to both and : Assume this is not the case and grow a rectangle similar to from to . Let be the first vertex hit by that is visible to and lies in . Note that this implies that is contained in . Therefore, is smaller than . Furthermore, is half-empty: By Lemma 10, the part below the line through and does not contain any vertices visible to or in , and the part between the line through and and the line through and does not contain any vertices visible to or since is the first visible vertex hit while growing . However, this contradicts that is the smallest half-empty rectangle.

Hence, does not contain any vertices visible to both and , which implies that is an edge of the constrained generalized Delaunay graph. Therefore the length of the shortest path from to is at most .

Induction step: We assume that for all half-empty rectangles smaller than the lemma holds. If is an edge of the constrained generalized Delaunay graph, the length of the shortest path from to is at most .

If is not an edge of the constrained generalized Delaunay graph, there exists a vertex in that is visible from both and . We grow a rectangle similar to from to . Let be the first vertex hit by that is visible to and lies in and let and be the Northwest and Northeast corner of (see Figure 10). Note that this implies that is contained in . We also note that is not necessarily an edge in the constrained generalized Delaunay graph, since if it is a constraint, there can be vertices visible to both and above . However, since is half-empty and smaller than , we can apply induction on it and we obtain that the path from to has length at most when lies on the East boundary of , and that the path from to has length at most when lies on the North boundary of .

Let be the projection of along the vertical axis onto . Since is contained in , can see . Since and are visibility edges and is not the endpoint of a constraint intersecting the interior of triangle , we can apply Lemma 2 and obtain a convex chain of visibility edges (see Figure 10). For each of these visibility edges , there is a homothet of that falls in one of the following three types (see Figure 11): (i) lies on the North boundary and lies in the Southeast corner, (ii) lies on the West boundary and lies on the East boundary and the slope of is negative, (iii) lies on the West boundary and lies on the East boundary and the slope of is not negative. Note that the case where lies on the South boundary and lies on the North boundary cannot occur, since the slope of any is at most that of . Also note that the case where lies on the South boundary and lies on the East boundary cannot occur, since we can shrink the rectangle until lies in the Southwest corner, resulting in a Type (iii) rectangle. Let and be the Northwest and Northeast corner of . We note that by convexity, these three types occur in the order Type (i), Type (ii), and Type (iii).

Let be the projection of along the vertical axis onto , let be the homothet of with and on its boundary that is similar to , and let and be the Northwest and Northeast corner of . Using these , we shift Type (ii) and Type (iii) rectangles down as far as possible: We shift down until either or lies in one of the North corners or the South boundary corresponds to the South boundary of . In the latter case, and are the same rectangle.

Since all rectangles are smaller than , we can apply induction, provided that we can show that is half-empty. For Type (i) visibility edges, the part of the rectangle that lies below the line through and is contained in , which does not contain any visible vertices, and the region of