A FAST QC-MAPPING THEOREM FOR POLYGONS · 2016. 11. 28. · A FAST QC-MAPPING THEOREM FOR POLYGONS CHRISTOPHER J. BISHOP Abstract. Given a simple n-gon P in the plane, normalized

A FAST QC-MAPPING THEOREM FOR POLYGONS

CHRISTOPHER J. BISHOP

Abstract. Given a simple n-gon P in the plane, normalized to contain the unitdisk, we define a map from P to the unit circle so that (1) the map extends to be8-quasiconformal on the interior, (2) it contracts arclength on the boundary and(3) the images of all n vertices can be computed in time O(n). Thus we obtaina fast K-QC-approximation to the Riemann map with constant K independent ofthe domain.

Date: March 2, 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

A FAST QC-MAPPING THEOREM FOR POLYGONS 1

1. Introduction

In general, there is no simple formula for the Riemann mapping of a simply con-

nected domain to the unit disk, but we will show that there is a simple, fast, geometric

way to define a K-quasiconformal map to the disk, with K independent of the choice

of domain. In order to quantify what we mean by “fast” we will consider the case

when Ω is bounded by a simple polygon P with n ordered vertices v = vkn1 . If

f : D → Ω is conformal then we let z = f−1(v) denote the conformal preimages of

the vertices. We will define a map ι : P → ∂D and let w = ι(v) be our approximate

preimages. We will consider our approximation fast if we can compute the n points

w with work O(n). The closeness of w to z is measured by the metric

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

Theorem 1. There is a C < ∞ so that if Ω is bounded by a simply polygon P with

n vertices we can find points w = w1, . . . , wn = ι(v) ⊂ T so that

1. dQC(w, z) < log 8 where z denotes the conformal prevertices.

2. All n points in w can be computed in O(n) steps.

3. If D ⊂ Ω, then |wk − wk+1| ≤ |vk − vk+1| for k = 1, . . . n (indices interpreted

mod n), i.e., the map ι contracts arclength on the boundary.

A “step” is an infinite precision arithmetic operation or an evaluation of exp(x).

Since the time to apply the algorithm is dominated by the time to read the data

and write the answer, one can think of this as saying there is a formula, valid for all

n-gons, which gives a uniform approximation to the Riemann map.

The Schwarz-Christoffel formula gives an exact formula for the Riemann map of the

disk onto a polygonal region Ω: if the interior angles of P are απ = α1π, . . . , αnπ,

and z = z1, . . . , zn are the preimages of the vertices, then

f(z) = A+ C

∫ z n∏k=1

(1 −w

zk

)αk−1dw.

See e.g., [17], [28], [36]. However, the points z = z1, . . . zn are unknown until we

know f , so the formula seems circular. There are various iterative methods for finding

the points z starting from an initial guess (e.g., n uniformly distributed points on T,

[17], [27]), and Theorem 1 says that an initial guess can be computed in O(n) steps

2 CHRISTOPHER J. BISHOP

Figure 1. On the left is the target polygon. In the center is theSchwartz-Christoffel image if we use the prevertex guesses given byTheorem 1. On the right is the image if we use equidistributed prever-tices on the circle (a common default for some methods).

that is guaranteed to be in a dQC-ball of fixed size around the correct answer. See

Figure 1.

A more intricate method for computing the prevertices with arbitrarily small dQC-

error in linear time is described in [4]. Moreover, the map ι given in Theorem 1

was previously described in [3], so the novel aspects of this paper are the introduc-

tion of the medial axis from computational geometry in order to compute the map

explicitly, the linear time bound on this computation and the constant “1” in part

(3) of the theorem. The map ι in Theorem 1 has equivalent definitions coming from

3-dimensional hyperbolic geometry and computational geometry in the plane. The

first definition leads to the accuracy claim (1) in Theorem 1 and the second leads to

the speed of computation (2).

Briefly, given a simply connected planar domain Ω, one defines the “dome” SΩ over

Ω as the upper envelope of all open hemispheres in R3+ whose base disk is in Ω. This

is a surface in R3+ which meets R

2 along ∂Ω. Thurston [35] observed that there is an

isometry ι from SΩ with its hyperbolic path metric to the hyperbolic unit disk. The

restriction ι : ∂Ω → ∂D is the map we use as the fast approximation to the Riemann

mapping. It follows from a theorem of Sullivan [34] and Epstein and Marden [18]

that this map has a K-quasiconformal extension to the interior, which gives part (1)

of the theorem. We will discuss this further in Section 2.

The medial axis of Ω consists of the centers of all disks in Ω whose boundary hits

∂Ω in two or more points; it has other names, such as the ridge set in [7], [23].


Figure 2. Examples of medial axes of polygons

The medial axis is a one dimensional object that is used to encode the shape

of 2-dimensional objects and was introduced by Blum [6].In the computer science

literature, a few papers consider its mathematical properties (e.g., [12], [13], [14], [15],

[16], [33], [38]), but most deal with algorithms for computing it and with applications

to areas like pattern recognition, robotic motion, classification of biological shapes,

control of cutting tools, sphere packing and mesh generation. A sample of such papers

includes: [9], [24], [25], [26], [29], [30], [32], [37], [39], [40].

Fix some medial axis disk D0 in Ω, and suppose D is a medial axis disk other than

D0. Then Ω \ D has at least two components and exactly one of these intersects

D0. Take the boundary arcs of D corresponding to the other components (the ones

that don’t hit D0). Doing this for every medial axis disk foliates Ω \D0 and the field

of orthogonal directions to this foliation is Lipschitz with respect to the hyperbolic

metric on Ω. Thus orthogonal trajectories exist and one can show they connect a

point x ∈ ∂Ω with ι(x) ∈ ∂D0 (we will call this the medial axis flow). See Figure

3. Although the definitions make sense for any simply connected domains, in this

paper we will only be interested in polygons and finite unions of disks, when they are

easiest to understand.

If Ω is bounded by an n-gon then the medial axis is a tree with O(n) vertices

and edges. The edges correspond to 1 parameter families of disks which meet ∂Ω at

exactly two points and come in three types (see Figure 4):

1. a line segment that is equidistant from two vertices

2. a line segment that is equidistant from two lines

3. a parabolic arc that is equidistant from a line and a vertex.


Figure 3. The top figure shows the medial axis of the domain, themiddle shows a choice of a root disk, D, and the foliation of Ω \ D,and the bottom shows the orthogonal trajectories that define the ι mapfrom ∂Ω to ∂D.

2a

2b

1 3

Figure 4. The different types of medial axis edges

It is a theorem of Chin, Snoeyink and Wang [10], [11] that the medial axis of

polygon with n vertices can be computed in O(n) steps. This depends on a deep

and difficult result of Chazelle [8] that a polygon can be triangulated in linear time.

However, special case, such as convex polygons are easier to deal with [1].

The rest of the paper is structured as follows:


Section 2: We give the definition of the ι map for general simply connected do-

mains and its connection to hyperbolic 3-dimensional geometry. We deduce part

(1) of Theorem 1

Section 3: We show how to compute the ι map for a finite unions of disks and

deduce some properties for general domains, including part (3) of Theorem 1.

Section 4: We compute ι for polygons, giving part (2) of Theorem 1.

2. The ι map

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 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. The hyperbolic metric on the upper half space R3+ is

dρR3+

= |dz|/dist(z,R2). As before, geodesics are circles orthogonal to the boundary,

and orientation preserving isometries are exactly the Mobius transformations (every

such transformation acting on R2 extends to an orientation preserving isometry of

R3+, and conversely). We will sometimes write ρ for any hyperbolic metric when the

domain is clear from context.

Given a closed set E ⊂ R2 we let C(E) ⊂ H

3 be the hyperbolic convex hull of E.

This is the smallest hyperbolically convex set in R3+ which contains all the infinite

hyperbolic geodesics (i.e., circles orthogonal to R2) with both endpoints in E. The

complement of C(E) is the union of all hyperbolic half-spaces in R3+ which do not

intersect the set of geodesics; i.e., it is the union of all hemispheres centered on R2

whose bases on R2 miss E. We are most interested in the case when Ω is a simply

connected plane domain and E = Ωc is its complement. Let S = SΩ be the boundary

component of C(Ωc) which separates Ω from C(Ωc). This is called the “dome” of Ω.

S is the boundary of the union of all hemispheres centered on R2 whose bases are

disks contained in Ω.

In fact, we do not have to consider all subdisks of Ω. It is not hard to show that

every point on the dome of Ω is on a hemisphere which intersects the dome along

an infinite hyperbolic geodesic (e.g., see Lemma 3 of [4]). Thus the base is a disk D

whose boundary hits ∂Ω in at least two points (the ends of the geodesic). In other

words, the dome of Ω is the upper envelope of hemispheres corresponding to medial

axis disks and each point on the dome corresponds to some point of the medial axis.


Thus each point of the dome of a polygon is either on a geodesic face, or corresponds

to one of the three types of medial axis edges. See Figure 5.

Figure 5. The dome of a “corner”. On the left is a polygon andits medial axis. On the right is the dome of the polygon, with sectionsshaded according to the part of the medial axis they correspond to.The darkest shading are geodesic faces (these correspond to verticesof the medial axis of degree ≥ 3); the lightest are Euclidean cones orcylinders (edge-edge bisectors in the medial axis). The medium shadingcorresponds to the parabolic bisector of a point and a line.

Let ρS denote the intrinsic path metric on S (using hyperbolic arclength). The

most important facts about the dome of Ω are the following two theorems.

Theorem 2 (Thurston). There is an isometry ι from (S, ρS) to (D, ρ).

Theorem 3 (Sullivan-Epstein-Marden). There is a K0 < ∞ so that for any hyper-

bolic simply connected domain Ω ⊂ R2 (other than the complement of a circular

arc), there is a K0-biLipschitz map σ from (Ω, ρ) to (S, ρS) which extends continu-

ously to the identity on ∂Ω. Consequently, there is a universal K < ∞ so that σ is

K-quasiconformal.

The first result appears in Thurston’s notes [35] (with more detailed proofs in the

papers of Epstein-Marden [18] and Rourke [31]). The second was apparently known

to Thurston and appeared in Sullivan’s paper [34] in the case when Ω is invariant

under a convex co-compact group of Mobius transformations. Epstein and Marden

[18] proved the more general statement quoted above. Alternate proofs are given in

[2], [5], [19]. The complement of a circular arc is only exceptional because in that case


S needs to be interpreted as a two-sided “folded” surface. This is carefully explained

in [21].

Epstein and Marden’s proof of Theorem 3 in [18] gives biLipschitz constant K0 ≈

88.2 and quasiconformal constant K ≈ 82.6 and they conjectured K0 = K = 2 is

correct. In [5] it is proven that one can take K = 7.82 and K0 = 13.3, but the

construction there is not group invariant. In [20], Epstein, Marden and Markovic

showed that K > 2 for the group invariant case using a result of McMullen’s on the

Teichmuller space of a punctured torus. See [19] for a different proof of this. More

recently Epstein and Markovic [22] have shown that K > 2.1 in the non-invariant

case, by showing the K is at least this large when Ω is the complement of a certain

logarithmic spiral.

When Ω is a finite union of disks, the dome S = SΩ and the map ι : S → D have a

particularly simple form. In this case the surface S is a finite union of geodesic faces.

Two of these can meet along infinite geodesic at some angle (called the bending angle

for the geodesic). Such a surface is called finitely bent and we will also refer to Ω as

a finitely bent domain. See Figure 6.

Figure 6. A finitely bent dome.

It is easy to visualize the ι map for finitely bent domains. Start with the case of

two overlapping disks, as shown in Figure 2. The dome is the union of two geodesic

faces, each sitting on a hemisphere, and joined along an infinite hyperbolic geodesic.

There is a 1-parameter family of elliptic Mobius transformations which leave this

geodesic fixed and rotate around it. Using these maps, we can leave one face fixed

and rotate the other until it is “flush” with the first one, i.e, we map the dome to

a hemisphere by a map which is the identity on one face and a hyperbolic isometry


on the other. This map is therefore an isometry of the hyperbolic path metric on

the dome to the path metric on the hemisphere (which equals the usual hyperbolic

metric and is isometric to the hyperbolic disk).

Figure 7. A dome consisting of two geodesic faces joined along aninfinite geodesic. By bending the dome along the geodesic we get aone-parameter, isometric family of surfaces ending with a hemisphere,which is obviously isometric to the hyperbolic disk.

For a general finitely bent domain, the faces form the vertices of a tree (with the

obvious adjacency relation: two faces are adjacent in the tree if they are joined along

a bending geodesic). Choose a root face and define the “parent” of each non-root

face: the unique adjacent face that is closer to the root. We start at a leaf of this

tree (i.e., a face with no children), and apply the elliptic rotation which isometrically

maps the corresponding face into the dome of the base disk of the parent. Now repeat

until the whole dome has been mapped to the hemisphere of the base of the root face.

See Figure 8.

It is also easy to interpret ι as a map in the plane. For each parent-child pair

of adjacent faces, these two faces lie on overlapping hemispheres, and the bases are

overlapping disks in the plane. If we remove the parent disk from its child, we are

left with a crescent. The elliptic transformation which rotates the child face also acts

on the plane and rotates the outer edge of the crescent to the inner edges. Every

crescent has a foliation by circular arcs perpendicular to both boundary arcs, this

rotation identifies the endpoints of leaves of this foliation. See Figure 9.

Any finitely bent domain Ω can be written as a root disk D and a finite union of

crescents (which form the vertices of a tree, just as their corresponding faces on the

dome did) and each crescent is foliated by orthogonal circular arcs. The ι map simply

moves a point from ∂Ω to ∂D by following the foliation paths through each crescent

until it reaches ∂D. See Figure 10. Several more examples are shown in Figure 11. It


Figure 8. We start with the dome of a finitely bent domain (herean approximation to a square). At each stage we located the leaves ofthe tree of faces and rotate them to become flush with the parent. Atthe end we have a single hemisphere.

Figure 9. The orthogonal foliation of a crescent

is surprising (at least to the author) that such a simple geometric construction yields

a map that is uniformly close to the Riemann map ∂Ω → ∂D.

Figure 11 shows some polygons which have been approximated by a finite union of

media axis disks, along with the flows for this approximation. For these domains, the

finite disk approximation gives a reasonable approximation to the ι map of the poly-

gon. However, in general, it requires an unbounded number of disks to adequately

approximate a polygon. For example, a 1 × r rectangle requires a finitely bent ap-

proximation with O(r) disks give a good approximation of ι. Thus, while finite disk

approximations may give a good way to approximate ι for many examples, to prove


Figure 10. On the top is the dome of a finitely bent domain (anapproximaton of the polygon in Figure 1) and its image after we havemade all the faces flush with the root face. On the bottom is thecorresponding picture in the plane. On the left is the decompositionof the domain into a root disk and crescents and the foliation of thecrescents by orthogonal arcs. On the right is are a few sample pathsgiven by following the foliations from ∂Ω to ∂D.

Theorem 1 we will have to compute ι directly for polygons. We introduce the general

idea using finite unions of disks in the next section and then deal with the case of

polygons after that.

3. The tree-of-disks construction

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. We do not require

that adjacent disks actually intersect. We also allow disks of zero radius, i.e., points,

but require that these be leaves of the tree. Suppose the disk D0 has been designated


Figure 11. Some more examples of the flow from ∂Ω to ∂D whichgives the ιmap. In each case we have approximated a polygonal domainby a finite union of disks, foliated the resulting crescents and drawn asample of the resulting paths.

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. We refer to this as a

“tree-of-disks” map.

Lemma 4. 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.

Proof. If D ∈ D has positive radius, choose three distinct reference points zD1 , z

D2 , z

D3

on ∂D; otherwise let this collection be empty. Every other point z on ∂D is uniquely

determined by cr(zD1 , z

D2 , z

D3 , z). Label each point v in v with the minimal k so

that v is on the boundary of a kth generation disk D. For k = 0 we do noth-

ing to the vertex. For k > 0, compute τD(v) ∈ ∂D∗ and record the cross ratio

cr(zD∗

1 , zD∗

2 , zD∗

3 , τD(v)). Also compute and record the images of the three reference

points for D, i.e., cr(zD∗

1 , zD∗

2 , zD∗

3 , τD(zDk )) for k = 1, 2, 3.


If a vertex is on ∂D0 then it maps to itself. If D is a first generation disk, then

we just compute τD(v) ∈ ∂D0 and compute the τD images of the three reference

points for D. For each child of D′ of D, we can now compute σD′ for any associated

vertices using the previously recorded cross ratios with respect to the reference points

for D and we can also compute the images on ∂D0 for the references points for D′.

In general, if D is a disk and we have already computed where the reference points

for its parent are mapped on ∂D0, we can use the recorded cross ratio information

to compute where the associated vertices and reference points for D map to. This

allows us to map every point of v to ∂D0 in O(n) steps.

If we restrict the map to ∂Ω = ∂SΩ, we have defined ι as a composition of elliptic

Mobius transformations on each circular arc in ∂Ω. Note that the crescents that we

use are always of the form W = D2 \D1 and that we are mapping the edge ∂W ∩∂D2

to the edge ∂W ∩ ∂D1. Thus we are in the case of the following lemma.

Lemma 5. Suppose Ω is a crescent which lies on one side of the line L passing

through the two vertices. Let γ1, γ2 be the circular arcs in ∂Ω with γ1 between γ2 and

L. If τ is an elliptic Mobius transformation fixing the two vertices and mapping γ2

to γ1 then |τ ′(z)| ≤ 1 on γ2.

Proof. To see this suppose τ(z) = (az + b)/(cz + d) where ad − bc = 1 (which we

can always assume by normalizing). Then a simple calculation shows |τ ′(z)| < 1

iff |1/c| < |z + d/c|. Note that −d/c = τ−1(∞). By normalizing by a Euclidean

similarity, we may assume the vertices are 1 and −1 and the crescent lies in the

upper half-plane. See Figure 12. Then −d/c is on the negative imaginary axis and

|τ ′(z)| < 1 outside a circle C centered at −d/c passing through −1 and 1 (since the

derivative of an elliptic transformation has modulus one at the two fixed points). Let

γ be the arc of this circle between 1 and −1 which lies in the upper half-plane. We

claim that γ2, the upper edge of our crescent, lies above γ.

Suppose the elliptic transformation is a rotation by θ around the points −1, 1.

Since γ2 and its image are both in the upper half-plane, θ < π. Therefore −d/c

lies on a circle which makes angle π − θ with the segment [−1, 1]. See Figure 12.

Hence the isosceles triangle with base [−1, 1] and vertex −d/c has two base angles of

ψ = (π− θ)/2 and the circle C makes angle π/2−ψ = θ/2 with [−1, 1]. Since γ1 lies

in H, γ2 makes angle of at least θ with [−1, 1] and hence lies above C.


θ ψψ

ψ ψ

π−2ψ

π−θ

θ

>θ

−d/c

π/2

γ2γ1

1−1γ

Figure 12. The setup in Lemma 5. We prove that γ2 lies above γby showing that γ makes angle θ/2 with [−1, 1], but γ2 makes angle> θ with the same segment.

This implies that the map ι : ∂Ω → ∂D0 can be chosen to have derivative< 1 on ∂Ω

(except possibly at the vertices). It is proven in [3] that the quasiconformal extension

of ι to the interior of Ω given by the Sullivan-Epstein-Marden theorem can be chosen

with derivative bounded by M independent of Ω.We can now record several useful

facts, by passing from finitely bent domains to general simply connected domains.

Lemma 6. If Ω is simply connected and contains D, then we can choose ι so that

ι : ∂Ω → T does not increase the length of any boundary arc.

Proof. Choose approximating finitely bent domains and pass to the limit.

Lemma 7. If Ω is simply connected and contains a disk D, then ι : ∂Ω → T is a

Mobius transformation when restricted to ∂D ∩ ∂Ω.

Proof. We can choose a medial axis disk D′ so that ∂D ∩ ∂Ω ⊂ ∂D′ ∩ ∂Ω, and

approximate Ω by a finite collection of medial axis disks which includesD′. Then ∂D∩

∂Ω will be on the boundary of each approximating domain and so the approximating

map will be Mobius. Since a (non-degenerate) limit of Mobius transformations is

Mobius, we are done.


Given a polygon Ω and a medial axis disks D0, D ⊂ Ω, assume D0 is the root of

the medial axis and define ΩD to be the union of all medial axis disk whose centers lie

on the path from D to D0. Let ιD be the the ι map for ΩD with root D0. Then the ι

map for Ω can be factored through ιD, at least for the boundary arc γthat separated

from D0 by D, e.g., see Figure 13. Thus the map on γ does not depend at all on

∂Ω \ (∂ΩD ∪ γ).

γ

0D D

Figure 13. The ι map of γ to ∂D0 can be written as a compositionof the imap onto ∂D followed by a Mobius transformation from D toD0 which is determined by the medial axis path from D to D0 andwhich only depends on the union of disks corresponding to this path(shaded region in the figure).

4. Constructing ι from the medial axis

By choosing several medial axis disks on the path from D to D0 we can also

write ιD as a composition of Mobius transformations corresponding to each subpath.

For a polygon, the medial axis is a tree and we can simply choose the intermedite

vertices. Then ιD is written as a composition of Mobius transformations each of

which corresponds to an edge of the medial axis.

Given a non-root vertex of the medal axis corresponding to the disk D, let ΩD be

the union of medial axis disks with centers on the edge between D and its parent.

We will call this the generalized crescent associated that edge of the medial axis.

Let ιD : ∂ΩD → ∂D∗ be the ι map for ΩD with root D∗. If x ∈ ∂Ω ∩ ∂CD, then

ι(x) can be computed by composing the maps ιD′ for the generalized crescents which

correspond to the path from D to D0.

Thus to compute ι for a polygon, we only have to compute it for each type of

generalized crescent. There are several different cases, depending on the type of edge


between D and its parent. The generalized crescent corresponding to each type of

medial axis edge has an associated flow as shown in Figure 14. Each of these defines

a Mobius transformation from its outer boundary to it inner boundary, so we can

compute ι at every vertex using Lemma 4. An example where we have joined the

flows from different generalized crescents is shown in Figure 15.

Figure 14. The polygon minus the root disk can be broken intopieces with these basic shapes corresponding to the cases consideredabove. The medial axis flow from the polygon to the disk is given byfollowing the flow on each piece.

Let D denote the disks that correspond to vertices of the medial axis and assume

we have chosen a root disk. Since the medial axis has edges that terminate at the

the convex vertices of the polygon, we will also add disks or radius 0 centered at each

vertex of angle < π. All that remains is to describe how to map a disk D1 ∈ D to its

parent D2 ∈ D.

Recall that the edges of the medial axis come in three types as illustrated in Figure

4. In that figure we have split case two into the subcases of parallel and non-parallel

lines since the formulas for ι will look slightly different in these cases. Thus the medial

axis is a finite tree whose vertices are the centers of disks in D and whose edges are

as above.

Case 0 (degenerate disks at convex vertices): Recall that this collection

includes the convex angles (interior angle < π) and that the corresponding disks

have zero radius. Map each such point to the closest point of its parent disk.


Figure 15. The decomposition of a polygon into generalized cres-cents and the medial axis flow onto the root disk.

Case 1 (equidistant from two points): If the two disks are D1 and D2 with

D2 being the parent, then the desired map is just the unique elliptic Mobius trans-

formation which fixes the two points, a, b of ∂D1 ∩ ∂D2 and maps ∂D1 \ D2 onto

∂D2 ∩D1. If the two boundary circles meet at angle α then this map is given by the

formula τ(z) = σ−1(eiασ), where σ(z) = (z − b)/(z − a) sends a to ∞ and b to zero.

Simplifying gives

τ(z) =(aeiα − b)z − ab(1 − eiα)

(eiα − 1)z + (a− eiαb).

Case 2a (equidistant from two parallel lines): Consider Figure 16. We

normalize so that D1 = D(0, 1) and D2 = D(A, 1). Note that on the dome of Ω,

the points above the centers of the disks D1, D2 are exactly distance A apart (the

straight line between them is a geodesic on the dome with this length). Since ι is

an isometry from the dome to the disk, it will map these points to points which are

exactly hyperbolic distance A apart in the disk.

With the points labeled as shown, i and −i will map to a and b respectively and 0

will map to a point c which is hyperbolic distance A from the center of D2. Thus the

desired map is τ(z) = σ(z) + A where σ is the unique Mobius transformation which

fixes 1 and −1 and maps 0 to the point −x on the negative real axis at hyperbolic


i

−i

2

A

A0a

b

c

Figure 16. Case 2a: between parallel lines

distance A from 0. Since the hyperbolic metric on the disk is given by

dρ =2|dz|

1 − |z|2,

the point w is the solution of

A =

∫ w

0

2

1 − r2dr = log

1 + w

1 − w,

which is

w =1 − e−A

1 + e−A.

Thus

σ(z) =z − w

1 − wz,

and this determines τ .

Case 2b (equidistant from two non-parallel lines): The situation is shown

in Figure 17. We normalize so that D1 = D(x, x sinα) and D2 = D(y, y sinα). We

will assume D1 is smaller than D2; the opposite case is handled similarly. The points

a, b are mapped to c, d respectively. The points on the dome above x and y are joined

by a geodesic which projects to the straight line between x and y. Above a point

t ∈ R+ the dome has height t sinα so the length of this geodesic on the dome is

A =

∫ y

x

dt

t sinα=

1

sinαlog

y

x.

Hence the image of x in D2 is hyperbolic distance A from y and must be on the

segment [x, y]. Thus if w = (1 − e−A)/(1 + e−A) and σ(z) = z−w1−wz

, the desired map

τ : D1 → D2 is

τ(z) = y(sinα)σ(1

x sinα(z − x)) + y.


ysin( )αa

bd

0 x y

c2α

Figure 17. Case 2b: between non-parallel lines

Note that since sinα = |a−x|/|x|, there is no need to actually evaluate a trigonometric

function, i.e., we could have written

τ(z) =y|a− x|

xσ(

z − x

|a− x|) + y.

Case 3 (equidistant from a point and a line): This case is pictured in Figure

18. Suppose a is the vertex in question and L is the line. Let a∗ be the reflection of a

across L. Then D1 can be mapped to D2 by an elliptic Mobius transformation which

fixes a and a∗ and sends L to itself. This elliptic element rotates around a by some

angle θ. We will think of this as a composition of n separate rotations, each by angle

θ/n and consider what happens. Applying each of these rotations in turn produces

a sequence of disks Bk intermediate between D1 and D2. Let τk : Bk → Bk+1 be

the elliptic map that fixes the two points of intersection a, bk = ∂Bk ∩ ∂Bk+1 and

σk : Bk → Bk+1 be the elliptic rotation fixing a and a∗. Then τn · · · τ1 converges

to the desired τ : D1 → D2 as n→ ∞.

The map ι is invariant under compositions with linear fractional transformations,

so we can conjugate by one so that a → ∞, a∗ → i, and R maps to the unit disk

centered at i. This means the elliptics we are considering become Euclidean rotations.

See right side of Figure 18.

This map conjugates each Bk to a half-plane tangent to the unit disk around i,

τk to the rotation by angle θ/n around intersection of the corresponding lines (i.e.,

the point labeled f in Figure 18) and σk to Rθ/n, the rotation by θ/n around i. It

is easy to see that τ1(z) = Rθ/n(z − 2 tan(θ/2n)). Thus limn→∞ τn · · · τ1(z) =

Rθ(z− θ). Conjugating Rθ(z− θ) back to the original disks gives the desired Mobius

transformation from D1 to D2. We have now completed the proof of Theorem 1.


a

L

θ

d

a*

θ

c

b

L2

L1θ/n θ/n

i

e f

g

Figure 18. Case 3: equidistant from a point and a line. On the leftis the family of circles passing through a and tangent to L. On theright we have conjugated a to ∞ and L to an arc of the circle centeredat i.

References

[1] A. Aggarwal, L. J. Guibas, J. Saxe, and P. W. Shor. A linear-time algorithm for computing theVoronoı diagram of a convex polygon. Discrete Comput. Geom., 4(6):591–604, 1989.

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

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

[4] C.J. Bishop. Conformal mapping in linear time. preprint, 2006.[5] C.J. Bishop. An explicit constant for Sullivan’s convex hull theorem. In Proceedings of the

Ahlfors-Bers Colloqium, 2001. Amer. Math. Soc., Providence, RI. to appear.[6] H. Blum. A transfomation for extracting new descriptors of shape. In W.W. Dunn, editor, Proc.

Symp. Models for the perception of speech and visual form, pages 362–380, Cambridge, 1967.MIT Press.

[7] L.A. Caffarelli and A. Friedman. The free boundary for elastic-plastic torsion problems. Trans.Amer. Math. Soc., 252:65–97, 1979.

[8] B. Chazelle. Triangulating a simple polygon in linear time. Discrete Comput. Geom., 6(5):485–524, 1991.

[9] C.-S. Chiang and C.M. Hoffmann. The medial axis transform for 2d regions. ACM Transactionson graphics, 1982.

[10] F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in lineartime. In Algorithms and computations (Cairns, 1995), volume 1004 of Lecture Notes in Comput.Sci., pages 382–391. Springer, Berlin, 1995.

[11] F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in lineartime. Discrete Comput. Geom., 21(3):405–420, 1999.

[12] H. I. Choi, S. W. Choi, and H. P. Moon. Mathematical theory of medial axis transform. PacificJ. Math., 181(1):57–88, 1997.


[13] H.I. Choi, C.Y. Han, and J.-H. Yoon. Medial axis tranform distance and its applications 2000,April 2000 Kyung Moon, Seoul,Korea. In Geometric modeling and computer graphics, pages65–69.

[14] S. W. Choi and H.-P. Seidel. Hyperbolic Hausdorff distance for medial axis transformation.Graphical Models, 63:369–384, 2001.

[15] S. W. Choi and H.-P. Seidel. Linear one-sided stability of MAT for weakly injective domain. J.Math. Imaging Vision, 17(3):237–247, 2002.

[16] S.W. Choi and S.-W. Lee. Stability analysis of medial axis transform. In Proc. 15th ICPRBrcelona, Spain, volume 3, pages 139–142, 2000.

[17] T. A. Driscoll and L. N. Trefethen. Schwarz-Christoffel mapping, volume 8 of Cambridge Mono-graphs on Applied and Computational Mathematics. Cambridge University Press, Cambridge,2002.

[18] 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.

[19] 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.

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

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

[22] 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.

[23] W.D. Evans and D.J. Harris. Sobolev embeddings for generalized ridged domains. Proc. LondonMath. Soc. (3), 54(1):141–175, 1987.

[24] H. N. Gursoy and N. M. Patrikalakis. Automated interrogation and adaptive subdivisionof shape using medial axis transform. Advances in Engineering Software and Workstations,13(5/6):287–302, 1991.

[25] C.M. Hoffmann. Computer vision, descriptive geometry and classical mechanics. In ComputerGraphics and Mathematics, pages 229–244. Springer Verlag, Eurographics Series, 1992.

[26] R. A. Jinkerson, S. L. Abrams, L. Bardis, C. Chryssostomidis, A. Clement, N. M. Patrikalakis,and F. E. Wolter. Inspection and feature extraction of marine propellers. Journal of ShipProduction,, 9(2):88–106, 1993.

[27] P. K. Kythe. Computational conformal mapping. Birkhauser Boston Inc., Boston, MA, 1998.[28] Z. Nehari. Conformal mapping. Dover Publications Inc., New York, 1975. Reprinting of the

1952 edition.[29] N. M. Patrikalakis and T. Maekawa. Shape Interrogation for Computer Aided Design and Man-

ufacturing. Springer Verlag, 2002.[30] F. P. Preparata. The medial axis of a simple polygon. In Mathematical foundations of computer

science (Proc. Sixth Sympos., Tatranska Lomnica, 1977), pages 443–450. Lecture Notes inComput. Sci., Vol. 53. Springer, Berlin, 1977.

[31] C. Rourke. Convex ruled surfaces. In Analytical and geometric aspects of hyperbolic space(Coventry/Durham, 1984), pages 255–272. Cambridge Univ. Press, Cambridge, 1987.

[32] E.C. Sherbrooke, N. M. Patrikalakis, and E. Brisson. An algorithm for the medial axis transformof 3d polyhedral solids. IEEE Transactions on Visualization and Computer Graphics, 2(1):44–61, 1996.


[33] E.C. Sherbrooke, N. M. Patrikalakis, and F.-E. Wolter. Differential and topological propertiesof medial axis transforms. CVGIP: Graphical Model and Image Processing, 58(6):574–592, 1996.

[34] D. Sullivan. Travaux de Thurston sur les groupes quasi-fuchsiens et les varietes hyperboliquesde dimension 3 fibrees sur S1. In Bourbaki Seminar, Vol. 1979/80, pages 196–214. Springer,Berlin, 1981.

[35] W.P. Thurston. The geometry and topology of 3-manifolds. The Geometry Center, Universityof Minnesota, 1979.

[36] L. N. Trefethen and T. A. Driscoll. Schwarz-Christoffel mapping in the computer era. In Pro-ceedings of the International Congress of Mathematicians, Vol. III (Berlin, 1998), numberExtra Vol. III, pages 533–542 (electronic), 1998.

[37] J. Wang. Medial axis and optimal locations for min-max sphere packing. J. Comb. Optim.,4(4):487–503, 2000.

[38] E.-F. Wolter. Cut locus and the medial axis in global shape interrogation and representation.1993. MIT, Dept. of Ocean Engineering, Design Laboratoy Memorandum 92-2.

[39] Q. J. Wu. Sphere packing using morphological analysis. In Discrete mathematical problems withmedical applications (New Brunswick, NJ, 1999), volume 55 of DIMACS Ser. Discrete Math.Theoret. Comput. Sci., pages 45–54. Amer. Math. Soc., Providence, RI, 2000.

[40] C.-K. Yap. An O(n log n) algorithm for the Voronoı diagram of a set of simple curve segments.Discrete Comput. Geom., 2(4):365–393, 1987.

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

11794-3651

E-mail address : [email protected]

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