BOUNDS FOR THE CRDT ALGORITHM€¦ · 1. Introduction In [11] Driscoll and Vavasis introduced the CRDT (Cross Ratios and Delaunay Triangulations) algorithm for computing conformal

BOUNDS FOR THE CRDT ALGORITHM

CHRISTOPHER J. BISHOP

Abstract. In [11] Driscoll and Vavasis described their CRDT algorithm for nu-merical computation of conformal maps from polygons to the unit disk. Althoughthe method works well in practice, it is not proven to converge in all cases and theyformulated a number of specific conjectures to quantify its behavior. In this notewe prove one of these conjectures: that the initial guess of the CRDT algorithmalways lies within a uniformly bounded distance of the correct answer.

Date: March 31, 2007.1991 Mathematics Subject Classification. Primary: 30C35, Secondary: 30C85, 30C62 .Key words and phrases. numerical conformal mappings, Schwarz-Christoffel formula, hyperbolic

3-manifolds, Sullivan’s theorem, convex hulls, quasiconformal mappings, quasisymmetric mappings,medial axis, CRDT algorithm.

The author is partially supported by NSF Grant DMS 04-05578.1

BOUNDS FOR THE CRDT ALGORITHM 1

1. Introduction

In [11] Driscoll and Vavasis introduced the CRDT (Cross Ratios and Delaunay

Triangulations) algorithm for computing conformal maps. Roughly speaking, it pro-

ceeds in three steps: (1) add extra vertices to the boundary of the polygon so that is

can be triangulated with pieces that are not too “thin”, (2) use the Delaunay triangu-

lation of the new polygon to construct an initial guess for the n images of the vertices

on the unit circle and (3) compute a conformal map using the Schwartz-Christoffel

formula with the guessed prevertices and modify the guesses based on the result. The

final step is iterated until the desired accuracy is achieved.

Driscoll and Vavasis formulated a number of specific questions regarding the be-

havior of the CRDT algorithm. In particular, they asked if steps (1) and (2) always

give an initial guess that is within a bounded distance of the true Schwartz-Christoffel

parameters when measured in terms of a metric derived from cross ratios. The pur-

pose of this note is to give an example that their specific formulation is false, but

prove that a slight modification is true; one merely has to replace the cross ratio of

four points by the conformal modulus of a related quadrilateral. This version follows

immediately from the following stronger result where the distance between n-tuples

on the unit circle, T, is measured in a quasiconformal sense, i.e., if w = w1, . . . , wnand z = w1, . . . , zn we define

dQC(w, z) = inflog K : ∃ K-quasiconformal h : D → D such that h(z) = w.

(The definitions and basic properties of quasiconformal maps are reviewed in Appen-

dix B.)

Theorem 1. There is a C < ∞ (independent of P ) so that the initial guess w of the

CRDT algorithm satisfies dQC(w, z) ≤ C.

The result is illustrated in Figure 1 which shows a target polygon (on the left) and

the Schwartz-Christoffel image using the CRDT initial guess as the parameters (in

the center). On the right we show the Schwartz-Christoffel image if we simply take

the parameters to be equidistributed on the circle (a common initial guess for some

iterative methods).

Given a generalized quadrilateral, i.e., a Jordan domain Ω and four distinct, ordered

points z = z1, z2, z3, z4 on the boundary, we can conformally map it to a rectangle

2 CHRISTOPHER J. BISHOP

Figure 1. On the left is the target polygon and in the center is theSchwartz-Christoffel image using the correct angles and the initial guessof the CRDT algorithm. By Theorem 1, this is the image of the tar-get polygon under a quasiconformal mapping sending vertices to ver-tices and with uniformly bounded QC constant. On the right is theSchwartz-Christoffel image using equidistributed parameters.

f : Ω → R with the four points mapping to the corners. We define ModΩ(z) =

|f(z2)−f(z1)|/|f(z2)−f(z3)| (the eccentricity of the rectangle). A K-quasiconformal

map from one generalized quadrilateral to another can change the modulus by at most

a multiplicative factor of K. It follows immediately from this fact and Theorem 1

that if P is an n-gon, z ⊂ T are conformal prevertices of P and w ∈ T is the initial

guess of the CRDT algorithm, then

1

K≤ ModD(z′)

ModD(w′)≤ K,

for any 4 vertices z′ = zj1, zj2, zj3 , zj4 ⊂ z, w′ = wj1 , wj2 , wj3 , wj4 ⊂ w. Equiva-

lently,

Corollary 2. − log K ≤ log ModD(z′) − log ModD(w′) ≤ log K.

On page 1792 of [11] Driscoll and Vavasis asked if this held with the conformal

modulus replaced by the cross ratio of the four points, in the case when the four points

correspond to the vertices of two adjacent triangles in the Delaunay triangulation of

P . We will see in Section 7 that the cross ratio version is false, so (2) is the appropriate

modification of their conjecture.

The proof of Theorem 1 reduces to showing that the CRDT initial guess (which is a

map from the vertices of P into the unit circle, T) can be extended to a quasiconformal

map from Ω, the interior of P , to the unit disk, D. To prove this we introduce a new

domain containing Ω. Suppose Ω has been triangulated. Each triangle defines a


unique circle passing through its three vertices. Each edge of the triangle is a chord

of the corresponding circle and divides the disk it bounds into two crescents, which we

denote the “inner” and “outer” crescent in the obvious way. (the inward normal to P

points into the inner crescent). Along each edge e of P we attach the corresponding

outer crescent, but crescents from different edges are considered disjoint (even if they

overlap in the plane). The resulting object R may be a plane domain, but more

generally is a Riemann surface if the outer crescents for different sides of P overlap.

See Figures 1 and 1.

Figure 2. To any polygon P we associated a Riemann surface R asfollows. Take a Delaunay triangulation of the polygon. For each tri-angle, form the circle containing the three vertices. For each boundarye of P is on the boundary of one triangle and divides the correspond-ing disk into two crescents. We attach the ‘outer’ crescent to P alongthe edge e, but make no other identifications. For the polygon shownabove, R is a planar domain, but it need not be in general.

If the surface R is actually a planar domain, then the CRDT map agrees with a map

ι : ∂Ω → T (called the “iota map”) which is defined for any simply connected plane

region and always has an 8-quasiconformal extension to the interiors, independent of

the geometry of the domain (we will explain in Section 4 why the two maps agree).

The ι map was originally invented in the context of hyperbolic 3-manifolds [24], [12],


Figure 3. Delaunay circumcircles can have large overlap. For this n-gon, each arm contributes a disk to R as indicated by the dashed circle.Clearly we can arrange for there to be points in the plane covered by≃ n points on the surface R.

but its connection to planar conformal maps has been explored in a series of papers

by Epstein, Marden, Markovic and the author [5], [6], [7], [8], [9], [13], [14], [15], [16].

If the surface R is not planar, it still comes with an obvious projection into the

plane. Although this map may be many-to-one, there is a certain sense in which the

surface is almost planar. We will prove the following easy result.

Lemma 3. If γ is a connected curve on R which projects into a circle in the plane

then the projection is at most 3 to 1.

With this geometric property, we can use extremal length arguments to estimate

the hyperbolic metric on R and extend the quasiconformal bounds for the ι map to

an analogous map ϕ : R → D.

Theorem 4. If P is a planar polygon with vertices v = v1, . . . , vn and RP is

the Riemann surface constructed from a Delaunay triangulation of P as above, then

there is a K1-quasiconformal mapping of ϕ : RP → D so that ϕ(v) = w, the CRDT

approximation. K1 is independent of n and the polygon P .

We then have to show that there is quasiconformal map Ω → R with uniform

bounds that pointwise fixes the vertices. The composition of these two maps is

then a quasiconformal map Ω → R → D with uniform QC-constant, which proves

Theorem 1. Unfortunately, there is no such uniform bound for all polygons, but there

is a bound if the polygon has been suitable modified by adding extra vertices to some

of its edges. This is exactly what the first step of the CRDT algorithm does.


Theorem 5. If P is a planar polygon and we form a new polygon P ′ by adding ver-

tices of angle π according to the CRDT algorithm, then there is a K2-quasiconformal

map φ : P ′ → RP ′ which fixes each vertex of P ′ ( RP ′ is the surface constructed from

a Delaunay triangulation of P ′ as above). K2 is independent of n and the polygon P .

Although the CRDT algorithm gives an initial guess which is within a uniformly

bounded distance of the correct answer, there is no uniform bound on how long it

takes because the first step of CRDT may add an arbitrarily large number of extra

vertices (consider a 1×r rectangle for f ≫ 1). In [9] a method is given for computing

a initial guess which has the same uniform distance bounds to the correct answer,

but can be performed in time O(n), with constant independent of the geometry.

Moreover, in [8], an algorithm is given which computes the correct prevertices to

with dQC distance ǫ in time O(n) and constant depending only on ǫ (and no worse

than O(| log ǫ| · log | log ǫ|)).The rest of the paper is organized as follows:

Section 2: We define a tree-of-disks map and show how a Delaunay triangula-

tion of a polygon gives rise to such a map.

Section 4: We explain why the CRDT initial guess is a tree-of-disks map.

Section 5: We prove Theorem 4.

Section 6: We prove Theorem 5.

Section 7: We show the Driscoll-Vavasis conjecture is false for cross ratios.

Appendix A: Definitions and basic properties of modulus and cross ratios.

Appendix B: Definitions and basic properties of quasiconformal mappings.

2. Tree-of-disks maps

A linear fractional (or Mobius) transformation is a map of the form z → (az +

b)(/cz+d). This is a 1-1, onto, holomorphic map of the Riemann sphere S = C∪∞to itself. Such maps form a group under composition and are well known to map

circles to circles (if we count straight lines as circles which pass through ∞). Mobius

transforms are conformal, so they preserve angles.

The non-identity Mobius transformations are divided into three classes. Parabolic

transformations have a single fixed point on S and are conjugate to the translation

map z → z +1. Elliptic maps have two fixed points and are conjugate to the rotation


z → λz for some |λ| = 1. The loxodromic transformations also have two fixed points

and are conjugate to z → λz for some |λ| < 1. If, in addition, λ is real, then the map

is called hyperbolic.

The hyperbolic metric on D is given by dρD = 2|dz|/(1 − |z|2). Geodesics for this

metric are circles orthogonal to the boundary. The orientation preserving isometries

are exactly the Mobius transformations which preserve the disk, which all have the

form z → eiθ(z − a)/(1 − az), for some θ ∈ R and a ∈ D. The hyperbolic metric ρΩ

on a simply connected domain Ω (or Riemann surface) is defined by transferring the

metric on the disk to Ω by the Riemann map. We will sometimes write ρ for any

hyperbolic metric when the domain is clear from context.

A crescent is a simply connected planar domain bounded by two circular arcs

which meet at two distinct points (called the vertices). The angle θ of a crescent

is the interior angle at these vertices. Any two crescents with the same angle are

Mobius equivalent. To any crescent and a choice of an edge, we associate the elliptic

Mobius transformation that has the two vertices a, b as fix points and maps the given

edge to the other edge. This is given by

τa,b,θ(z) =(beiθ − a)z + ab(1 − eiθ)

(eiθ − 1)z + (b − aeiθ).(1)

The formula can be easily derived by sending a and b to 0 and ∞ by the map

w = τ(z) = (z−a)/(z−b), then multiplying by eiθ and then applying the inverse map

z = τ−1(w) = (bw − a)/(w − 1). Geometrically, a crescent is foliated by circular arcs

orthogonal to both boundaries and this elliptic transformations identifies endpoints

of leaves. See Figure 4.

Figure 4. The orthogonal foliation of a crescent

Suppose we have a finite collection D of disks in the plane and an adjacency

relation between them that makes the collection into the vertices of a tree. Suppose

the disk D0 has been designated the root of the tree. Then any other disk D has a


unique “parent” D∗ which is adjacent to D but closer to the root. Assume that for

every (non-root) disk we are given a map τD : D → D∗. The we can define a map

σD : D → D0 as follows. If D = D0, the map is the identity. Otherwise, there is a

unique shortest path of disks D0, . . . , Dk = D between D0 and D. Note that each

disk is preceded by its parent. Thus σ = τD1 · · · τDk

is a mapping from D to D0

as desired. Later we will refer to this as a “tree-of-disks” map.

Lemma 6. With notation as above, assume that every map τD is Mobius. Then

given n points v = v1, . . . , vn, with vk ∈ ∂Dk for k ∈ 1, . . . , n, we can compute

the n image points σ(v) in at most O(n) steps.

The simple proof of this (essentially just recording cross ratios at each step and

then reconstructing the points at the end) is given in [8] and [9]. The lemma does

not require a disk and its parent to actually touch (e.g., in [9] the lemma is applied

to the disks corresponding to the vertices of the medial axis of a polygon and these

need not overlap). However, in this note we are only interested in the case when

the bounding circles meet in at least two points. In this case there is an “obvious”

Mobius transformation from the child to the parent: either the disks agree and we

take the identity map, or they intersect in exactly two points and we take an elliptic

map that fixes the two intersection points and sends the child disk to the parent disk.

We shall call such a map an “elliptic tree-of-disks”.

One way that these arise is from a triangulation of a polygon P . This is a division

of Ω, the interior of P , into triangles using a maximal, disjoint set of diagonals (open

line segments in Ω whose endpoints are vertices of P ). The triangles of a triangulation

form the vertices of a tree with the obvious adjacency relation (iff they share an edge).

Each triangle defines a unique disk whose bounding circle passes through its three

vertices and circles corresponding to adjacent triangles intersect in at least two points

(and possibly be identical). Thus if we choose a root element of the triangulation we

get a corresponding tree-of-disks map from the vertices of P to the root circle.

The triangulation is called Delaunay if for every interior edge e of the triangulation,

the two triangles sharing edge e have opposite angles summing to at most π. An

equivalent formulation is as follows. If T1, T2 are closed triangles sharing the edge

e and D1, D2 are the open associated disks, then T1 6⊂ D2 and T2 6⊂ D1 (i.e., each


triangle has a vertex not on e; this vertex must be outside or on the circle for the

other triangle, but not strictly inside it). See Figure 5.

Figure 5. The triangles on the left satisfy the Delaunay condition,but the ones on the right do not.

Figure 6. A Delaunay and non-Delaunay triangulation. We havedraw a circumcircle on the right which fails the Delaunay condition.

Every polygon has a Delaunay triangulation which is unique except when adjacent

triangles define four vertices on a common circle: then we may “flip” the diagonal of

the resulting quadrilateral and get another Delaunay triangulation. Note however,

that the corresponding family of disks and the tree-of-disk map on the vertices does

not depend on the choice of Delaunay triangulation. Delaunay triangulations have

many nice properties and have been intensively studied (e.g., they are dual to Voronoi

diagrams; they maximize the minimum angle in the triangulation [21]; they minimize

the largest circumcircle [10], [23]). The basic facts can be found in various sources

such as [2], [4], [17], [18].

Suppose we have two adjacent triangles T1, T2 which satisfy the Delaunay condition

and have associated disks D1, D2. Assume these disks are distinct and the bound-

aries intersect at exactly two points a, b. Let γ1 and γ2 be the hyperbolic geodesics

connecting a and b in D1 and D2 respectively. Then γ1 ∪ γ2 forms the boundary of a


crescent C in D1 ∪ D2 with some angle θ. This is called the “normal crescent” since

its edges are perpendicular to the boundary of D1 ∪ D2 (and to distinguish it from

the “tangential crescent” D1 \ D2 which we shall mention later; note that both have

the same vertices and the same angle and can be mapped to each other by a π/2

rotation around the vertices). If D1 = D2 then the normal crescent has angle 0 and

consists of a single geodesic arc.

Also note that (D1∪D2)\C consists of two components, each a crescent. If T1 and

T2 satisfy the Delaunay condition, then one component is bounded by γ1 and an arc

of ∂D1 and the other is bounded by γ2 and an arc of ∂D2 (and otherwise the roles

of the γ’s are switched). The first component lies inside D1 and we call it the main

component of D1 with respect to D2. See Figure 2.

γ1 γ2

D1

D2

Figure 7. Given two adjacent triangles satisfying the Delaunay con-dition we define the normal crescent (shaded region) to be boundedby the hyperbolic geodesics in the corresponding circumcircles. TheDelaunay condition insures that γ1 to the left of γ2 in this figure.

A triangle T1 in our triangulation shares one, two or three edges with other trian-

gles and hence has the same number of main components with respect to neighbors.

Let G1 ⊂ D1 denote the intersection of these components (G for “gap”). If the trian-

gulation is Delaunay, then the gaps for each triangle are disjoint and gaps for adjacent

triangles are separated by the crescent for that pair. From now on, we assume this

is the case. See Figure 2 for an example of this “gap/crescent” decomposition.

Define a Riemann surface R by taking the disjoint union of gaps and crescents and

attaching them along the common edges. Alternatively we may leave the crescents

out and attach two gaps if they share a common crescent, by identifying points on the

two sides of the crescent which are identified by the elliptic transformation associated

to the crescent. In this case, it is easy to see that the resulting surface is the disk


Figure 8. A gap/crescent decomposition for a tree of disks. Whenwe collapse the crescents by elliptic maps the gaps come together toform a disk. This defines a continuous map ϕ from the tree of disksto the root disk. We have drawn arcs to indicate crescents with zeroangle; this occurs when adjacent triangles have all four vertices on asingle circle. We have futher subdivided the gaps by adding “boundarycrescents” whose edges complete the hyperbolic triangulation of thedisk.

and each gap is mapped into the disk by a Mobius transformation. This map can be

extended continuously to all of R by mapping each foliation leaf in a crescent to the

single point where its two endpoints have already been sent. See Figure 2.

If this map ϕ : R → D were quasiconformal with uniformly bounded constant,

we would have a proof of Theorem 1. This is not true because ϕ is not even a

homeomorphism in general (it can collapse two points of a crescent to the same

point). However, in Section 5 we shall show it is a quasi-isometry (with uniform

constants) of from the hyperbolic metric on R to the hyperbolic metric on D. By

results discussed in Appendix B, this implies there is a K-quasiconformal map from R

to D with the same boundary values with uniformly bounded K, proving the theorem.

3. Alternate descriptions of ϕ

This section is not needed for the proof of Theorem 1, but presents three alternate

ways of visualizing the map ϕ : R → D. The proof of the theorem will continue in

the next section.

3.1. Angle scaling. It is perhaps helpful to visualize the crescents as being collapsed

continuously, as illustrated in Figure 3.1. This is called an angle scaling family (these


play an important role in the linear time conformal mapping algorithm described in

[8]).

Figure 9. Here is an angle scaling family showing the the crescentsbeing collapsed continuously. In each picture the angles of the crescentsare multiplied by a factor t = 1, .8, .6, .4, .2, 0.

3.2. The medial axis flow: Another way to visualize the map on ∂R is by following

a foliation defined on R \ D0, where D0 is the root disk. Note that R \ D0 can be

written as a union of crescents of the form D∗ \D where D∗ is the parent of D. This

decomposes R as a union of the root disk and a finite number of tangential crescents.

As mentioned earlier, tangential and normal crescents are in a 1-to-1 correspondence

and corresponding crescents have and same vertices and angle; they just give alternate

ways of visualizing the decomposition of the surface. See Figure 3.2.

Each crescent has a foliation by circular arcs orthogonal to the boundaries. Op-

posite endpoints of each leaf are identified by the elliptic transformations in our

tree-of-disks map, so following a foliation leaf from ∂R through the crescents to ∂D0

gives the desired map. The process is illustrated in Figure 3.2

3.3. Hyperbolic domes: Given a planar domain Ω, the dome of Ω is a surface S in

the upper half-space, R3+, which is the upper boundary of the union of all hemispheres


Figure 10. The tangential and normal decompositions of a finiteunion of disks. In each case we have split the region into one disk(white) and several crescents (shaded) which use the same vertices andangles in both cases. In the tangential decomposition the disk is evi-dent. In the normal decompostion it must be assembled from Mobiusimages of the gaps between crescents, as in Figure 2.

centered on R2 whose bases on R2 are disks inside Ω. We are most interested in the

case when Ω is a simply connected plane domain which is a finite union of disks,

in which case the dome consists of finitely many faces lying on hemispheres, joined

along infinite hyperbolic geodesics. See Figure 3.3.

There is a natural map r : Ω → S defined as follows. Given a point z ∈ Ω consider

the family of balls in the upper half-space tangent to R2 at z. Take a small ball

(which must lie completely under S) and expand it until it makes first contact with

S at the point r(z). This turns out to be the extension to R2 of the nearest point

retraction from R3+ to the region above S (which is hyperbolically convex). The gaps

in Ω are the points which map to the hyperbolic faces of the dome. The normal

crescents are the points which map to a geodesic which joins two faces.

Thurston had observed that the dome of Ω with the hyperbolic path metric it

inherits from H is isometric to the hyperbolic disk. We denote this isometry by

ι : S → D. If R is a planar domain, then the composition ι r : Ω → S → D is

exactly the same as ϕ.


Figure 11. A geometric interpretation of the initial step of CRDT.Given a polygon, we find the Delaunay triangulation and correspondingdisks, choose a root disk and find the tangential crescents, and foliateeach crescent with orthogonal arcs. Finally, follow the foliation fromthe vertices to points on the boundary of the root disk. Note thatthis example is not quite planar beause the two disks corresponding toopposite sides of the slit overlap.


Figure 12. On the left is a finite union of disks with its gap/crescentdecomposition. On the right is the dome of this region. Under theretraction map r : Ω → S, gaps map to faces and crescents are collapsedto geodesics between faces. Faces on the dome are shaded in alternatingcolors to make them easier to see, but they all correspond to whitegaps on the left. The shaded crescents on the left are collapsed togeodesic edges separating faces on the right. Geodesics on the domecorresponding to crescents of zero angle have not been drawn, althoughthey are drawn on the left.

It is interesting to compare Figures 3.2 and 12. Both involve finite unions of disks

which approximate the same polygonal region (shown in the upper left of Figure 3.2).

The first figure uses the circumcircles of the Delaunay triangulation and produces a

collection of disks that covers the polygon. The second figure uses a collection of

disks that are all inside the region (and are chosen using a discrete sampling of the

medial axis if the region, i.e., all the disks meet the boundary in at least two points).

Both collections of disks define a tree-of-disks map which approxmiates the Riemann

mapping onto the disk. However, it remains to be investiaged which gives a better

approximation, or if some other scheme for approximating by disks would do even

better.

4. The initial guess of CRDT is a tree-of-disks map

We now return to the proof of Theorem 1. In this section, we show that the initial

guess of CRDT as defined by Driscoll and Vavasis in terms of cross ratios, agrees

with the elliptic tree-of-disks map ϕ defined at the end of Section 2. We begin with

a brief review of CRDT.


The CRDT algorithm is a method of finding the conformal prevertices of a given

polygon P . The algorithm starts by adding new vertices of angle π to the edges of

P to get a new polygon P ′ which has a Delaunay triangulation with certain nice

properties. We will discuss how this is done in Section 6, and here we simply assume

we are given P ′. The second part of the algorithm uses a Delaunay triangulation of

P ′ to produce guesses for the conformal prevertices of P ′ (and hence for P whose

vertices are included among those of P ′). The third step of the CRDT algorithm is

to solve for the actual prevertices using an iterative procedure that starts with the

guess from the second step. We will not consider this part in detail.

How is the second step done? Suppose P is a simple n-gon and that T =

T1, . . . Tn−2 is a Delaunay triangulation of P . Let Dk be the circumcircle asso-

ciated to each triangle Tk, k = 1, . . . , n − 2. The CRDT algorithm chooses points

w = w1, . . . wn ⊂ T corresponding to the vertices v = v1, . . . , vn as follows.

Choose some root triangle from the triangulation and map its vertices to any three

points in T with the correct orientation. In general, suppose Q is the quadrilat-

eral formed by two adjacent triangles T1 and T2 (which have vertices vj1 , vj2 , vj3 and

vj1 , vj3 , vj4 in counterclockwise order respectively). Also suppose that we have already

defined wj1 , wj2 , wj3 . Then wj4 is uniquely determined by the condition

cr(wj1 , wj2 , wj3 , wj4) = −|cr(vj1 , vj2 , vj3 , vj4)|.(2)

It is easy to see by induction that this uniquely determines the points w up to a

Mobius transformation of the circle.

Next we will define a map ϕ : v → T which gives the same points w up to a Mobius

transformation. Given two adjacent triangles, the corresponding disks intersect at

two points and we take the elliptic Mobius transformation as in (1) between them.

This defines a “tree of disks” as in Lemma 6 and so we can define a corresponding

map by composing these maps along paths to the root.

We claim ϕ(v) is the same as the initial guess of the CRDT algorithm. Suppose

Q is the quadrilateral formed by Tk and its parent Tj (which have vertices v1, v3, v4

and v1, v2, v3 in counterclockwise order respectively). After conjugating by a Mobius

transformation which sends v1 → 0, v2 → 1 and v3 → ∞, the elliptic map τk is

conjugated to the Euclidean rotation around 0 which sends the image to v4 onto the


negative real axis. Thus

cr(v1, v2, v3, τk(v4)) = −|cr(v1, v2, v3, v4)|.

The Mobius maps τ1, . . . , τk−1 are then applied to all four points v1, v2, v3, τk(v4) and

hence do not change the cross ratio. Thus

cr(ϕ(v1), ϕ(v2), ϕ(v3), ϕ(v4)) = −|cr(v1, v2, v3, v4)|.

Therefore w = ϕ(v) is exactly the same as the CRDT guess for the prevertices.

5. Proof of Theorem 4

In this section we will show that given a planar polygon P , the map ϕ : R → D in

Section 4 is a quasi-isometry with constants independent of P . If R is planar then

this is a known result (e.g., [12], [7], [5]). Here we give a proof for the general case

using the planar case as our guide.

To prove such a result, we need to use some geometric property of R that comes

from the way it was constructed. The following turns out to be convenient.

Lemma 7. Suppose R is obtained from the Delaunay triangulation of a simple poly-

gon as above. If γ is a simple curve on R that projects into a circle z : |z − x| = rthen the projection is at most 3 to 1 (and hence γ has length ≤ 6πr).

Proof. R consists of P and a finite number crescents, one attached to each edge of P .

If γ is a circle in R then its projection to the plane is 1-to-1. If γ is an arc in R which

projects into a circle in the plane, then it leaves every crescent it enters, with at most

two exceptions (the crescents containing it endpoints). If we remove the part of γ

in these two exceptional crescents, the rest projects 1-to-1 to the plane because the

projection of P ⊂ R to P ⊂ R2 is 1-to-1. Thus the projection is at most 3-to-1.

Note that this is Mobius invariant; if we apply a Mobius transformation to R,

image satisfies the same condition. The 3-to-1 can be attained in some cases; see

Figure 13).

The next step is to show that the hyperbolic metric on R satisfies an estimate

similar to that for planar domains. If Ω is a simply connected planar domain, then

(e.g., [19], [22])|dz|

2dist(z, ∂Ω)≤ dρΩ ≤ 2|dz|

dist(z, ∂Ω).


Figure 13. This is an example where the projection onto a circlecan be 3-to-1. The solid circular arcs show outer crescents in ∂R thatare attached to the shaded polygon (they are attached to the two ‘top”edges of the trianular region in the center). The dashed circle can belifted to a curve in R which is covered three times near the bottom.

Lemma 8. There is an M < ∞ so that the following holds. If R = RP is the surface

obtained from a Delaunay triangulation of a simple, planar polygon P , then

|dz|Mdist(z, ∂R)

≤ dρR ≤ 2|dz|dist(z, ∂R)

.

Proof. The proof of the right hand inequality is exactly the same as the planar case:

since R contains a disk D of radius r = dist(z, ∂R) around z, the monotonicity

property of hyperbolic distance [22] (also known as the Schwarz inequality) implies

dρR(z) ≤ dρD(z) =2

r|dz|.

To prove the lower bound, it suffices to consider the disk of radius r/2 around z and

show its hyperbolic diameter is bounded uniformly away from zero. This is equivalent

to showing that the modulus of the path family Γ connecting D = D(z, 12r) ⊂ R to

∂R is bounded away from zero.

Let v ∈ ∂R minimize dist(z, ∂R) and let v also denote the projection of v to the

plane. For t ∈ [12r, 3

2r], take a component γt of the lift of the circle w : |w − v| = t

which hits D. By Lemma 7 this arc has length ≤ 6πt and it must connect D to ∂R.

Thus if ρ is an admissible metric for the path family Γ,∫

γρ|dz| ≥ 1 for all γ ∈ Γ and


hence∫∫

R

ρ2dxdy ≥∫ 3r/2

r/2

∫

γt

ρ2tdtdθ

≥ [

∫ 3r/2

r/2

∫

γt

tdθdt]−1

∫ 3r/2

r/2

∫

γt

ρ2tdθdt

≥ [6π3

2r]−2

∫ 3r/2

r/2

1tdt

≥ [6π3

2]−2r−2 1

2[9

4r2 − 1

4r2]

= (6π3

2)−2.

Thus the modulus of Γ is bounded away from zero uniformly, as desired. This com-

pletes the proof of Lemma 8.

Lemma 9. Suppose G is a gap in R and z ∈ G. Let D be the disk associated to G.

Then1√2dist(z, ∂Rp) ≤ dist(z, ∂D) ≤ dist(z, ∂Rp).

Proof. The right hand inequality is obvious since D ⊂ R. To prove the left hand

inequality, note that if z = 0 or if the radial projection of z is in ∂R then we have

equality. Otherwise, the radial segment through z hits a geodesic γ in D bounding a

crescent. Replacing z with this point of intersection decreases dist(z, ∂D) more than

dist(z, ∂Rp) and for a point on γ, the inequality holds by a simple computation.

On each gap G, ϕ is an isometry from the hyperbolic metric of the corresponding

disk D to the unit disk. We have just seen that this metric (restricted to G) is

comparable to the hyperbolic on R and hence ϕ is Lipschitz on G. On the crescents

ϕ is clearly Lipschitz, and thus it Lipschitz on all of R.

To prove ϕ is a quasi-isometry, we need to prove the opposite direction, ρD(ϕ(z), ϕ(w)) ≥CρR(z, w) − C. It suffices to show that for some ǫ > 0 the preimage under ϕ of a

ball of hyperbolic radius ǫ in D has uniformly bounded hyperbolic diameter in R. If

this is true, then any two points distance d > ǫ apart in D can be connected by a

geodesic, and this geodesic can be cut into ≃ d/ǫ parts, each of which has preimage

of diameter C. Thus the two original points have all preimages within (C/ǫ)d of each

other, which is the desired lower bound.


The key step to proving this is the following.

Lemma 10. There is a ǫ0 > 0 and C < ∞ so that if ǫ < ǫ0 then the following

holds. Suppose τjM1 is a finite collection of elliptic transformations such that each

τj rotates by angle θj, that∑

j |θj| ≤ L and that each τj has one fixed point in D(0, ǫ)

and one fixed point outside D(0, 1/ǫ). Then |τ1 · · · τM(w)−w exp(i∑

j θj)| ≤ CǫL

for any w with 12≤ |w| ≤ 2.

Proof. Each elliptic transformation preserves a family of circles and by the hypothesis

on the fixed points, the circles in these families of diameter between 1/4 and 4 differ

by at most angle Cǫ from circles concentric with the origin. Thus if τj rotates by

angle θj it will differ from a rotation of angle θj around the origin by at most Cǫθj.

See Figure 14. Summing gives the desired result.

θ1

θ2

0 1

Figure 14. If the fixed points are ǫ-near 0 and ∞ the elliptic trans-formations look like Euclidean rotations up to error Cǫ.

Suppose ǫ < ǫ0/2 and consider a ball B of hyperbolic radius ǫ in D. Renormalizing

by a Mobius transformation we may assume it is centered at the origin. If is it

contained inside the image of one gap G ⊂ D ⊂ R, there is nothing to do, since the

preimage has smaller size in R. Thus B must hit one or more bending geodesics (i.e.,

images of crescents under ϕ). Let γj be an enumeration of these and let θj be the

angle measure of the corresponding crescents in R. Since each of these hits B, the

endpoints of all the γ’s must be clustered in two balls of (Euclidean) radius δ (which

tends to zero with ǫ) on opposite side of the circle. Without loss of generality we

may assume this balls are centered at −i and +i.


By applying a Mobius transformation taking the unit disk to the upper half-plane

we can instead assume these geodesics are in H and each has one endpoint in [0, δ]

and the other outside [1/δ,∞]. These geodesics also come with a natural left to right

ordering and each has an associated elliptic Mobius transformation which rotates

by angle θj around the endpoints of γj. If we apply these maps from rightmost to

leftmost, letting the map act on everything to the right of the corresponding geodesic,

we get the map ϕ on the gaps.

Let B2 be the ball of radius 1/2 centered at i. If∑

j θj = L ≥ 6π then Lemma 10

implies the preimage of B2 in R covers the unit circle more than 3 to 1 at some point.

This is a contradiction, so we must have L < 6π. But this means that the preimage

of B is a connected set in R which can be covered by a uniformly bounded number

of balls of radius 1/4 which are more than distance 1/4 from ∂R. Each element of

this cover therefore has bounded hyperbolic radius in R and hence the preimage of

B had uniformly bounded radius.

Thus ϕ is a quasi-isometry with uniformly bounded constants and thus there is

a K-quasiconformal map R → D (with uniformly bounded K) that has the same

boundary values. This completes the proof of Theorem 4

Note that the only place we used the fact that we have a Delaunay triangulation

is to say θj ≥ 0. For a general triangulation the θ’s might also be negative, and the

proof would break down at the point where we show∑

j |θj| is uniformly bounded.

6. Proof of Theorem 5

It would be convenient if there was a uniformly K-quasiconformal map Ω → R

fixing the vertices, but this is not true with K independent of P = ∂Ω. Instead, we

will see how to add new vertices of angle π to P to obtain a polygon P ′ for which it

is true. This is basically the first step of the CRDT algorithm as described in [11].

To see that quasiconformal extension need not hold uniformly for all P , simply

consider a 1 × n rectangle. The four corners lie on the boundary of a disk D and

the Delaunay triangulation consists of adding a diagonal. The modulus of the path

family in the rectangle connecting the two sides of length 1 is 1/n. The modulus

of the generalized quadrilateral with domain D and the four corners as vertices is


approximately the logarithm of their cross ratio, i.e., is approximately 1/ log n. How-

ever, if ι had a K-quasiconformal extension to P with K independent of P , these two

moduli would be comparable within a factor of K of each other, so we deduce there

is no such extension for large n.

The problem with the previous example is that the quadrilateral formed by taking

adjacent triangles in the Delaunay triangulation is long and skinny. However, if all

the triangles in the triangulation are “round” (e.g., no small angles) then there is no

problem. The rest of this section is devoted to making this more precise. We first

introduce a geometric restriction on the Delaunay triangulation of P which insures

there is a uniform K-quasiconformal map Ω → R. After that, we will verify that the

first step of the CRDT algorithm produces a polygon whose Delaunay triangulation

has this property.

Lemma 11. Suppose P has a rooted triangulation such that each triangle either (1)

is isosceles with the base adjacent to its parent and base angle ≤ π/4, (2) is isosceles

with any angle and is a leaf of the triangulation tree with the base adjacent to its parent

or (3) has all angles in [θ, π − θ] for some θ > 0. Then there is a K-quasiconformal

map f : P → RP which fixes each of the vertices of P , and K depends only on θ.

Proof. First consider the case when T is isosceles. We will let the two equal base

angles be denoted α and the opposite angle it then π − 2α. The circumcircle of the

triangle defines two crescents attached to T along the non-base sides of T . We claim

that the interior angles of these crescents are also α. This is simple plane geometry

as illustrated in Figure 15. By definition ∠cad = α and ∠adc = π2− α. Since abd

is isosceles with base ad we have ∠bad = π2− α and ∠abd = 2α. Since triangles eab

and eca are similar, this means ∠eac = ∠eba = 2α and hence ∠ead = 2α − α = α,

as claimed.

If α ≤ π4, then the reflection of each crescent attached to T (across the edge shared

with T ) is a crescent inside T . Moreover the two crescents we get are disjoint. The

union of a crescent C and its reflection is a crescent C ′ of twice the angle and so we

can map C to C ′ be a 2-quasiconformal map which extends to the identity on the

rest of T . See Figure 16.

If T is isosceles but α > π4, the picture is illustrated in Figure 17. By assumption

the triangle is attached to its parent along the base, but not adjacent to any other


αα

π/2−α2α

π/2−α

b c d e

a

Figure 15. The isosceles case: small α

Figure 16. A crescent of angle α can be quasiconformally mappedto a crescent of angle β > α with K = β/α. Just use a Mobiustransformation to first map to a wedge and then use a map of the formreiθ → reiθβ/α.

triangle. Now we map the big crescent with angle 2α (this is the angle by the previous

paragraph) and base ab to the smaller crescent with angle α and the same base by

a 2 quasiconformal map. Then map the small crescent to the triangle abc by a K-

quasiconformal map which is the identity on ab and send d to c. This can be done

with a uniform K, independent of α. It easy to build such a map explicitly in this

case.

Finally consider a triangle T with all its angles bounded below by θ > 0. Consider

a crescent C of angle α attached to T along edge e. Then the angle bisectors of T

split T into three subtriangles, each of which contains a crescent with angle ≥ θ/2

and shares an edge with T . See Figure 18. Let C ′ be the crescent of angle θ/2


a

b

c

2α

αd

Figure 17. The isosceles case: large α. The crescent with base aband angle 2α can be 2-QC mapped to the crescent with the same baseand half the angle, and then mapped in a uniformly QC way to thelong narrow triangle abc. Both maps are the identity on ab.

adjacent to C. We can map C ∪ C ′ → C ′ by a quasiconformal map with constant

K = (α + θ/2)/(θ/2) which extends to the identity of the rest of T . Since α ≤ π, we

have K ≤ 1 + 2π/θ, as desired.

γβ

α

Figure 18. Mapping RP to R quasiconformally. Because every angleof the triangle is bounded away from zero each of the external crescentscan be mapped to one of the three disjoint internal crescents withuniform QC bounds.


Lemma 12 (Driscoll-Vavasis, [11]). Given any polygon P we can add vertices of

angle π to form a new polygon which satisfies the hypothesis of Lemma 11.

We will not repeat the proof from [11] here, but we will sketch their construction,

which has two steps. In the first step, every vertex with interior angle ≤ π/4 is

“chopped off” by taking the largest isosceles triangle T in P formed with v and

subsets of its two adjacent edges and adding two new vertices to P at the midpoints

of the two sides of T . A new polygon P ′ is formed by replacing v by these two new

vertices and the edge between them. This edge is protected: no new vertices may

be added to it later. The second step is iterative. For each edge e in the current

polygon compute its length L and the minimal distance D in P from e to any vertex

which is not one of its endpoints (distance is the path distance within the polygon).

If D < L/(3√

2), the edge e is split into three equal edges and the process continues.

See Figure 19. The filled dots indicate the original vertices and the open dots the

vertices added by the algorithm.

Figure 19. Adding extra points to a polygon in the CRDT algorithmto remove small angles from the triangulation

Driscoll and Vavasis give a lower bound r0 on the shortest edge that can be pro-

duced: r0 is the minimum over all unprotected edges of the path distance in P from

that edge to any non-adjacent edge. Since every step reduces the length of some edge

by a third, this proves the process terminates in a finite number of steps (and gives

an estimate for the number of vertices that have been added). Thus the resulting

polygon has a triangulation in which every triangle T is either (1) isosceles with angle

≤ π/4 or (2) D ≥ L/(3√

2) for any side of T . The later condition easily implies the

angles of T are bounded away from 0, so Lemma 12 holds.


As Driscoll and Vavasis note in their paper [11], the number of new points needed

can be arbitrarily large, depending on the geometry of P (if P has a long, narrow

corridor, then many new vertices have to be added).

7. A counterexample

Here we shall show that Corollary 2 does not hold with modulus replaced by cross

ratios. Consider the domain Ω bounded by a polygon P in Figure 20. It is the

square [−1, 1]× [−1, 1]) with thickened slit removed, [0, 1]× [−ǫ, ǫ]. The dotted lines

in the figure are the boundaries of a Delaunay triangulation of P and the shaded

quadrilateral Q is formed by the union of two adjacent triangles. The absolute value

of the cross ratio of the four vertices is

cr1 =|D − A||B − C||C − D||A − B| =

√

1 + (1 − ǫ)2 ·√

1 + (1 − ǫ)2

2ǫ · 2 =1

2ǫ− 1

2+

ǫ

4.

On the other hand, the conformal map f of Ω to the disk which sends −1/2 to 0 and

has positive derivative at −1/2 will map the edge [A,B] to an interval on T centered

at −1 and with length ≃ 1, but will map the edge [C,D] to an arc centered at 1 of

length ≃√

|C − D|. The cross ratios of these points will therefore be

cr2 =|f(D) − f(A)||f(B) − f(C)||f(C) − (D)||f(A) − f(B)| ≤ M√

ǫ,

for some constant M independent of ǫ, if ǫ is small enough. Hence

log cr1 − log cr2 >1

2log

1

ǫ− O(1),

which is as large as we wish if ǫ is small enough. This polygon does not satisfy the

C

D

2 ε

A

B

Figure 20. Cross ratios of f(Q) and ϕ(Q) scale at different powers of ǫ

edge separation condition of the CRDT algorithm, but applying that method to this


polygon will only add vertices to the horizontal edges adjacent to C and D and the

same quadrilateral will be present. Thus the conjecture fails even if we add the extra

vertices.

Appendix A. Conformal modulus and cross ratios

Suppose Γ is a family of locally rectifiable paths in a planar domain Ω and ρ is a

non-negative Borel function on Ω. We say ρ is admissible for Γ if

ℓ(Γ) = infγ∈Γ

∫

γ

ρds ≥ 1,

and define the modulus of Γ as

Mod(Γ) = infΩ

∫

M

ρ2dxdy,

where the infimum is over all admissible ρ for Γ. This is a well known conformal

invariant whose basic properties are discussed in many sources such as Ahlfors’ book

[1]. A generalized quadrilateral Q is a Jordan domain in the plane with four specified

boundary points x1, x2, x3, x4 (in counterclockwise order). We define the modulus of

Q, MQ(x1, x2, x3, x4) (or just MQ or M(Q) if the points are clear from context), as the

modulus of the path family in Q which connects the arc (x1, x2) to the arc (x3, x4).

This is also the unique positive real number M such that Q can be conformally

mapped to a 1×M rectangle with the arcs (x1, x2), (x3, x4) mapping to the opposite

sides of length 1. In this paper, we will be particularly concerned with the case when

Q = D and we are given four points in counterclockwise order on the unit circle.

Given a generalized quadrilateral Q with four boundary points x1, x2, x3, x4, the

quadrilateral Q′ with vertices x2, x3, x4, x1 is called the reciprocal of Q and it is easy

to see that Mod(Q′) = 1/Mod(Q).

Given four distinct points a, b, c, d in the plane we define their cross ratio as

cr(a, b, c, d) =(d − a)(b − c)

(c − d)(a − b).

Note that cr(a, b, c, z) is the unique Mobius transformation which sends a to 0, b

to 1 and c to ∞. This makes it clear that cross ratios are invariant under Mobius

transformations; that cr(a, b, c, d) is real valued iff the four points lie on a circle; and

is negative iff in addition the points are labeled in counterclockwise order on the

circle.


If the four points lie on T, then since cr and MD are both invariant under Mobius

transformations of the disk to itself, each must be a function of the other in this case.

The usual way to represent this function (e.g., as in Ahlfors’ book [1]) is to map the

disk to the upper half plane, H, sending the points a, b, c to 0, 1,∞ respectively and

d to −P = cr(a, b, c, d) ∈ (−∞, 0). Then MD(a, b, c, d) is the same as the modulus of

the path family in H connecting (−∞,−P ) to (0, 1). By symmetry, this is twice the

modulus of the path family of closed curves in the plane which separate [−P, 0] from

[1,∞]. We will denote this modulus by M(P ). The transformation z → (z−1)/(z+P )

sends 0, 1,∞,−P to − 1P, 0, 1,∞, so by Mobius invariance of modulus and the fact

that conjugate quadrilaterals have reciprocal moduli, we see that

MH(1

P) =

1

MH(P ),(3)

and hence MH(1) = 1 and M(1) = 1/2.

In [1] Ahlfors gives a formula (page 46, equation (16)) relating P and M

P + 1 = exp(2πM)1

16

∞∏

n=1

(1 + exp((1 − 2n)2πM)

1 + exp((−2n)2πM))8.

For M > 0 the infinite product converges and for M large (say M ≥ 1) we have∞∏

n=1

(1 + exp((1 − 2n)2πM)

1 + exp((−2n)2πM))8 = 1 + 8e−2πM + O(e−4πM).

Thus for P ≥ 1, (equivalently M ≥ 1), we have

log(P + 1) = 2πM − log 16 + 8e−2πM + O(e−4πM),

which implies

P ≃ exp(2πM).

For 0 < P ≤ 1, (equivalently 0 < M ≤ 1), we can use (3) to deduce

log(1

P+ 1) =

π

2M− log 16 + 8e−π/(2M) + O(e−π/M),

which implies

P ≃ exp(− π

2M).

In other words,

M ≃ 1

2πlog P, P ≫ 1,

M ≃ π

2| log P | , P ≪ 1,


Thus for x = x1, x2, x3, x4 ⊂ T, since ModD = 2M ,

MD(x) ≃ 1

πlog |cr(x)|, |cr(x)| ≫ 1,

MD(x) ≃ π

| log |cr(x)|| , |cr(x)| ≪ 1,

Another elegant connection between modulus and cross ratios is given in [3] (Bagby

shows that conformal modulus for a ring domain is given by minimizing an integral

involving logarithms of cross ratios).

Appendix B. Quasiconformal mappings

Quasiconformal mappings are a generalization of conformal mappings which play

an important role in modern analysis and a central role in the current paper. There

are (at least) three equivalent definitions of a K-quasiconformal mapping between

planar domains. Suppose f : Ω → Ω′ is a homeomorphism.

Geometric definition: for any generalized quadrilateral Q ⊂ Ω, Mod(Q)/K ≤Mod(f(Q)) ≤ KMod(Q).

Analytic definition: f is absolutely continuous on almost every vertical and

horizontal line and the partial derivatives of f satisfy |fz| ≤ k|fz| where

k = (K − 1)/(K + 1).

Metric definition: For every x ∈ Ω

lim supr→0

maxy:|x−y|=r |f(x) − f(y)|miny:|x−y|=r |f(x) − f(y)| ≤ K.

For a proof of the equivalence of the first two, see [1] and for a discussion of the

third and a generalization to metric spaces see [20] and its references. A mapping

is conformal iff it is 1-quasiconformal and the composition of a K1-quasiconformal

map with a K2-quasiconformal map is (K1K2)-quasiconformal. Thus the distance

dQC satisfies the triangle inequality.

Although quasiconformal maps must have derivatives almost everywhere, they can

be non-differentiable on smaller sets. For example, the famous von Koch snowflake is

a non-differentiable curve which is the image of the unit circle under a quasiconformal

map of the plane (such an image is called a quasicircle; they have a simple geometric

characterization, e.g., see [1]). Even though they don’t have to be differentiable

everywhere, by Mori’s theorem every K-quasiconformal map is Holder continuous of


order 1/K. Moreover, any quasiconformal map of D to itself extends continuously

to the boundary. The extension is also a homeomorphism and its restriction to the

boundary is a quasisymmetric homeomorphism, i.e., there is an M < ∞ (depending

only on K) so that 1/M ≤ |f(I)|/|f(J)| ≤ M , whenever I, J ⊂ T are adjacent

intervals of equal length. Conversely, any quasisymmetric homeomorphism of T can

be extended to a K-quasiconformal selfmap of the disk, where K depends only on

M .

Quasiconformal maps are also a generalization of biLipschitz maps, i.e., maps which

satisfy

1

K≤ |f(x) − f(y)|

|x − y| ≤ K.

For K-quasiconformal selfmaps of the disk, there is almost a converse. Although a

quasiconformal map f : D → D need not be biLipschitz, it is a quasi-isometry of the

disk with its hyperbolic metric ρ, i.e., there are constants A,B such that

1

Aρ(x, y) − B ≤ ρ(f(x), f(y)) ≤ Aρ(x, y) + B.

This says f is biLipschitz for the hyperbolic metric at large scales. A quasi-isometry

is also called a rough isometry in some sources. In [14] Epstein, Marden and Markovic

show that any K-quasiconformal selfmap of the disk is a quasi-isometry respect to the

hyperbolic metric with A = K and B = K log 2 if 1 ≤ K ≤ 2 and B = 2.37(K − 1)

if K > 2.

A quasi-isometry of the disk to itself need not be quasiconformal (indeed, it need

not even be continuous), but there is a close connection in terms of boundary values.

If h : D → D is a quasi-isometry then a result of Vaisala [25] implies that h has

a continuous extension to the boundary and that this extension is quasisymmetric,

with a constant that depends only on rough isometry constants of h. Hence there

is a K-quasiconformal self-map H of the disk with these same boundary values, and

with K depending only on rough isometry constants of h. Thus to prove a circle

homeomorphism has a quasiconformal extension to the disk, it suffices to prove it

has a quasi-isometric extension.


References

[1] L. V. Ahlfors. Lectures on quasiconformal mappings. The Wadsworth & Brooks/Cole Mathe-matics Series. Wadsworth & Brooks/Cole Advanced Books & Software, Monterey, CA, 1987.With the assistance of Clifford J. Earle, Jr., Reprint of the 1966 original.

[2] F. Aurenhammer. Voronoi diagrams - a survey of a fundamental geometric data structure. ACMComp. Surveys, 23:345–405, 1991.

[3] T. Bagby. The modulus of a plane condenser. J. Math. Mech., 17:315–329, 1967.[4] M. Bern and D. Eppstein. Mesh generation and optimal triangulation. In Computing in Eu-

clidean geometry, volume 1 of Lecture Notes Ser. Comput., pages 23–90. World Sci. Publishing,River Edge, NJ, 1992.

[5] C. J. Bishop. Divergence groups have the Bowen property. Ann. of Math. (2), 154(1):205–217,2001.

[6] C. J. Bishop. Quasiconformal Lipschitz maps, Sullivan’s convex hull theorem and Brennan’sconjecture. Ark. Mat., 40(1):1–26, 2002.

[7] C. J. Bishop. An explicit constant for Sullivan’s convex hull theorem. In In the tradition ofAhlfors and Bers, III, volume 355 of Contemp. Math., pages 41–69. Amer. Math. Soc., Provi-dence, RI, 2004.

[8] C.J. Bishop. Conformal mapping in linear time. preprint, 2006.[9] C.J. Bishop. A fast QC-mapping theorem for polygons. preprint, 2007.

[10] E. F. D’Azevedo and R. B. Simpson. On optimal interpolation triangle incidences. SIAM J.Sci. Statist. Comput., 10(6):1063–1075, 1989.

[11] T. A. Driscoll and S. A. Vavasis. Numerical conformal mapping using cross-ratios and Delaunaytriangulation. SIAM J. Sci. Comput., 19(6):1783–1803 (electronic), 1998.

[12] D. B. A. Epstein and A. Marden. Convex hulls in hyperbolic space, a theorem of Sullivan,and measured pleated surfaces. In Analytical and geometric aspects of hyperbolic space (Coven-try/Durham, 1984), volume 111 of London Math. Soc. Lecture Note Ser., pages 113–253. Cam-bridge Univ. Press, Cambridge, 1987.

[13] D. B. A. Epstein, A. Marden, and V. Markovic. Complex angle scaling. In Kleinian groups andhyperbolic 3-manifolds (Warwick, 2001), volume 299 of London Math. Soc. Lecture Note Ser.,pages 343–362. Cambridge Univ. Press, Cambridge, 2003.

[14] D. B. A. Epstein, A. Marden, and V. Markovic. Quasiconformal homeomorphisms and theconvex hull boundary. Ann. of Math. (2), 159(1):305–336, 2004.

[15] D. B. A. Epstein, A. Marden, and V. Markovic. Complex earthquakes and deformations of theunit disk. J. Differential Geom., 73(1):119–166, 2006.

[16] D. B. A. Epstein and V. Markovic. The logarithmic spiral: a counterexample to the K = 2conjecture. Ann. of Math. (2), 161(2):925–957, 2005.

[17] S. Fortune. Voronoı diagrams and Delaunay triangulations. In Computing in Euclidean geom-etry, volume 1 of Lecture Notes Ser. Comput., pages 193–233. World Sci. Publishing, RiverEdge, NJ, 1992.

[18] S. Fortune. Voronoi diagrams and Delaunay triangulations. In Handbook of discrete and com-putational geometry, CRC Press Ser. Discrete Math. Appl., pages 377–388. CRC, Boca Raton,FL, 1997.

[19] J. B. Garnett and D. E. Marshall. Harmonic measure, volume 2 of New Mathematical Mono-graphs. Cambridge University Press, Cambridge, 2005.

[20] J. Heinonen and P. Koskela. Quasiconformal maps in metric spaces with controlled geometry.Acta Math., 181(1):1–61, 1998.


[21] C.L. Lawson. Software for C1 surface interpolation, pages ix+388. Academic Press [HarcourtBrace Jovanovich Publishers], New York, 1977. Publication of the Mathematics Research Cen-ter, No. 39.

[22] Ch. Pommerenke. Boundary behavior of conformal maps. Grundlehren der Math. Wis-senschaften, 299. Springer-Verlag, 1992.

[23] V. T. Rajan. Optimality of the Delaunay triangulation in Rd. Discrete Comput. Geom.,

12(2):189–202, 1994.[24] D. Sullivan. Travaux de Thurston sur les groupes quasi-fuchsiens et les varietes hyperboliques

de dimension 3 fibrees sur S1. In Bourbaki Seminar, Vol. 1979/80, pages 196–214. Springer,Berlin, 1981.

[25] J. Vaisala. Free quasiconformality in Banach spaces. II. Ann. Acad. Sci. Fenn. Ser. A I Math.,16(2):255–310, 1991.

C.J. Bishop, Mathematics Department, SUNY at Stony Brook, Stony Brook, NY

11794-3651

E-mail address : [email protected]

Documents

BOUNDS FOR THE CRDT ALGORITHM€¦ · 1. Introduction In [11] Driscoll and Vavasis introduced the CRDT (Cross Ratios and Delaunay Triangulations) algorithm for computing conformal