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arXiv:1205.2548v1 [cs.CG] 11 May 2012

Outerplanar graph drawings with few slopes

Kolja Knauer1,⋆, Piotr Micek2,⋆⋆, and Bartosz Walczak2,⋆⋆

1Institut f¨ ur Mathematik, Technische Universit¨ at Berlin,

knauer@math.tu-berlin.de

2Theoretical Computer Science Department,

Faculty of Mathematics and Computer Science, Jagiellonian University,

{micek,walczak}@tcs.uj.edu.pl

Abstract. We consider straight-line outerplanar drawings of outerpla-

nar graphs in which the segments representing edges are parallel to a

small number of directions. We prove that ∆ − 1 directions suffice for

every outerplanar graph with maximum degree ∆ ? 4. This improves the

previous bound of O(∆5), which was shown for planar partial 3-trees, a

superclass of outerplanar graphs. The bound is tight: for every ∆ ? 4

there is an outerplanar graph of maximum degree ∆ which requires at

least ∆−1 distinct edge slopes for an outerplanar straight-line drawing.

1 Introduction

A straight-line drawing of a graph G is a mapping of the vertices of G into distinct

points in the plane and of the edges of G into straight-line segments connecting

the points representing their end-vertices and passing through no other points

representing vertices. If it leads to no confusion, in notation and terminology, we

make no distinction between a vertex and the corresponding point, and between

an edge and the corresponding segment. The slope of an edge in a straight-line

drawing is the family of all straight lines parallel to the segment representing

this edge. The slope number of a graph G, introduced by Wade and Chu [1], is

the smallest number s such that there is a straight-line drawing of G using s

slopes.

Since at most two edges at each vertex can use the same slope, ⌈∆

bound for the slope number of a graph with maximum degree ∆. In general,

graphs with maximum degree ∆ ? 5 may have arbitrarily large slope number,

see [2,3]. If the maximum degree of a graph is at most 3 then the slope number

is at most 4 as shown by Mukkamala and Szegedy [4], improving a result of

Keszegh, Pach, P´ alv¨ olgyi, and T´ oth [5]. The question whether the slope number

of graphs with maximum degree 4 is bounded by a constant remains open.

The situation is different for planar straight-line drawings, that is, straight-

line drawings in which no two distinct edges intersect in a point other than a

2⌉ is a lower

⋆Supported by DFG grant FE-340/8-1 as part of ESF EuroGIGA project GraDR.

⋆⋆Supported by MNiSW grant 884/N-ESF-EuroGIGA/10/2011/0 as part of ESF Eu-

roGIGA project GraDR.

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common endpoint. It is well known that every planar graph admits a planar

straight-line drawing [6,7,8]. The planar slope number of a planar graph G is the

smallest number s such that there is a planar straight-line drawing of G using

s slopes. Keszegh, Pach, and P´ alv¨ olgyi [9] show that the planar slope number

is bounded by a function of maximum degree. Their bound is exponential and

their proof is non-constructive. Jel´ ınek, Jel´ ınkov´ a, Kratochv´ ıl, Lidick´ y, Tesaˇ r,

and Vyskoˇ cil [10] give an upper bound for the planar slope number of planar

graphs of treewidth at most 3, which is O(∆5).

In the present paper we consider drawings of outerplanar graphs. As outer-

planar graphs have treewidth at most 2, they admit planar drawings with O(∆5)

slopes. A straight-line drawing of a graph G is outerplanar if it is planar and

all vertices of G lie on the outer face. The outerplanar slope number of an out-

erplanar graph G is the smallest number s such that there is an outerplanar

straight-line drawing of G using s slopes. Dujmovi´ c, Eppstein, Suderman, and

Wood [11] consider the outerplanar slope number as a function of the number

of vertices. We provide a tight bound for the outerplanar slope number in terms

of the maximum degree.

Main Theorem. The outerplanar slope number of every outerplanar graph with

maximum degree ∆ ? 4 is at most ∆ − 1.

It is easy to see, that the tightness of this bound is witnessed by a cycle with

at least 2∆ − 3 vertices, where to each vertex we attach ∆ − 2 leaves.

Note that the tight bounds for the outerplanar slope number with respect to

the maximum degree ∆ are: 1 for ∆ = 1 and 3 for ∆ ∈ {2,3}. The latter bound

from above is implied by our theorem.

The proof of our theorem is constructive and yields a linear-time algorithm

to produce drawings of the claimed kind.

2 Basic definitions

Suppose we are given an outerplanar drawing of a connected graph G with

maximum degree ∆ ? 4. This drawing determines the cyclic ordering of edges

around every vertex. We produce an outerplanar straight-line drawing of G with

few edge slopes which preserves this ordering at every vertex. Our construction

is inductive: it composes the entire drawing of G from drawings of subgraphs of

G that we call bubbles.

We distinguish the outer face of G (the one that is unbounded in the given

drawing of G and contains all vertices on the boundary) from the inner faces.

The edges on the boundary of the former are outer edges, while all remaining

ones are inner edges. A snip is a simple closed counterclockwise-oriented curve

γ which

– passes through some pair of vertices u and v of G (possibly being the same

vertex) and through no other vertex of G,

– on the way from v to u goes entirely through the outer face of G crossing

no edges on the way,

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– on the way from u to v (considered only if u ?= v) goes through inner faces

of G possibly crossing some inner edges of G, each at most once.

Every snip γ defines a bubble H in G as the subgraph of G induced on the vertices

lying on or inside γ. Note that H is a connected subgraph of G as γ crosses no

outer edges. The oriented simple path P from u to v in H going counterclockwise

along the boundary of the outer face of H is called the root-path of H. If u = v

then the root-path consists of that single vertex only. The roots of H are the

vertices u and v together with all vertices of H incident to the edges crossed by

γ. Note that vertices of H not being roots cannot have edges to G−H. Note also

that the root-path and the roots of H do not depend on the particular snip γ

used to define H. The order of roots along the root-path gives the root-sequence

of H. A bubble with k roots is called a k-bubble. A special role in our proof is

played by 1- and 2-bubbles.

Bubbles admit a natural decomposition, which is the base of our recursive

drawing.

Lemma 1. Let H be a bubble with root-path v1...vk. Every component of H−

{v1,...,vk} is adjacent to either one vertex among v1,...,vkor two consecutive

vertices from v1,...,vk. Moreover, there is at most one component adjacent to

viand vi+1for 1 ? i < k.

Proof. Let C be a connected component of H − {v1,...,vk}. As H itself is

connected, C must be adjacent to a vertex from v1,...,vk. In order to get a

contradiction suppose that C is connected to two non-consecutive vertices vi

and vj. Let P be a simple vi,vj-path having all internal vertices in C. Let

P′= vi...vj be the subpath of the root-path of H connecting viand vj. Since

v1...vk is the root-path of H, all edges connecting the internal vertices of P′

to G − H are inner edges. Hence, also the edges of P′lie on inner faces which

are not faces of H. The symmetric difference of all these inner faces considered

as sets of edges is a simple cycle containing P′as a subpath. Let P′′denote

the other vi,vj-subpath of that cycle. It is internally disjoint from P and P′.

Moreover, P′′and P together enclose P′and thus the internal vertices of P′do

not lie on the outer face—contradiction.

Now, to prove the second statement, suppose that for some i two components

C and C′of H − {v1,...,vk} are adjacent to both vi and vi+1. We find two

internally disjoint vi,vi+1-paths P and P′through C and C′, respectively. As in

the above paragraph we use that vivi+1is contained in an inner face, which is not

a face of H. The third path P′′is obtained from that face by deleting the edge

vi,vi+1. It follows that P,P′,P′′form a subdivision of K2,3, which contradicts

outerplanarity of G.

⊓ ⊔

Lemma 1 allows us to assign each component of H −{v1,...,vk} to a vertex

of P or an edge of P so that every edge is assigned at most one component.

For a component C assigned to a vertex vi, the graph induced on C ∪ {vi} is

called a v-bubble. If P consists of a single vertex with no component assigned to

it, we consider that vertex alone to be a v-bubble. For a component C assigned

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H1

H2

H3

H4

H5

H6

H7

H8

H9

u

v

w

P

Fig.1. A 3-bubble H with root-path P (drawn thick), root-sequence u,v,w (connected

to the remaining graph by dotted edges), and splitting sequence into v- and e-bubbles

(H1,...,H9). For example, (H2,H3,H4) is a 2-bubble.

to an edge vivi+1, the graph induced on C ∪ {vi,vi+1} is called an e-bubble.

If no component is assigned to an edge of P then we consider that edge alone

an e-bubble. All v-bubbles of vi in H are naturally ordered by their clockwise

arrangement around vi in the drawing. All this leads to a decomposition of

the bubble H into a sequence (H1,...,Hb) of v- and e-bubbles such that the

naturally ordered v-bubbles of v1precede the e-bubble of v1v2, which precedes

the naturally ordered v-bubbles of v2, and so on. We call this sequence the

splitting sequence of H and write H = (H1,...,Hb). Since no single-vertex v-

bubbles occur in the splitting sequence unless H is itself a single-vertex v-bubble

the splitting sequence of H is unique. Note that v- and e-bubbles are special kinds

of 1- and 2-bubbles, respectively. Every 1-bubble is a bouquet of v-bubbles. The

splitting sequence of a 2-bubble may consist of several v- and e-bubbles. For an

illustration see Fig. 1.

The general structure of the induction in our proof is covered by the following

lemma (see Fig. 2):

Lemma 2.

2.1. Let H be a v-bubble rooted at v. Let v1,...,vkbe the neighbors of v in H

in clockwise order. Then H −v is a k-bubble with root-sequence v1,...,vk.

2.2. Let H be a v-bubble rooted at v0. Let v0...vnbe an induced path going from

v0counterclockwise along the outer face of H and such that v1,...,vn−1are

not cut-vertices in H. Then H − {v0,...,vn} has a unique component H′

adjacent to both v0and vn. Moreover, let X be the subgraph of H induced

on v0,...,vnand the vertices of H′. Let v1

0,...,vk0

0,v1be the neighbors of

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Fig.2. Three ways of obtaining smaller bubbles from v- and e-bubbles. The new root-

path is drawn thick.

v0in X in clockwise order. For 1 ? i ? n − 1 let vi−1,v1

the neighbors of vi in X in clockwise order. Let vn−1,v1

neighbors of vn in X in clockwise order. Then H′is a bubble with root-

sequence v1

coincide for 0 ? i ? n − 1.

2.3. Let H be an e-bubble with roots u,v. Let u1,...,uk,v be the neighbors of

u in H in clockwise order and u,v1,...,vℓbe the neighbors of v in H in

clockwise order. Then H −{u,v} is a bubble with root-sequence u1,...,uk,

v1,...,vℓ, where ukand v1may coincide.

i,...,vki

n,...,vkn

i,vi+1be

be the

n

0,...,vk0

0,v1

1,...,vk1

1,...,v1

n,...,vkn

n, where vki

i

and v1

i+1may

Proof. First we prove 2.1. Since H is a v-bubble, H − v is connected. The sym-

metric difference of the inner faces of H incident to v, considered as sets of

edges, gives a simple clockwise cycle in H passing through v and v1,...,vkin

this order. Let γ be a closed curve going counterclockwise from vkthrough the

outer face of G to v1and then through the inner faces of H at v, crossing the

edges vv2,...,vvk−1in this order, back to vk. Clearly, γ is a snip defining the

bubble H − v with root-sequence v1,...,vk.

Next we prove 2.2. Since none of v1,...,vn−1is a cut-vertex in H, the graph

H−{v0,...,vn} has a component adjacent to both v0and vn. Moreover, since the

path v0...vnconsists only of outer edges, such a component is unique. Thus H′is

well defined. Now, the symmetric difference of the inner faces of X incident to any

of v0,...,vn, considered as sets of edges, gives a simple clockwise cycle in H pass-

ing through vn,...,v0and then through v1

in this order. Let γ be a closed curve going counterclockwise from vkn

the outer face of G to v1

0and then through the inner faces of H at v0,...,vn,

crossing the edges v0v2

order, back to vkn

n. Clearly, γ is a snip defining the bubble H′with root-sequence

v1

0,v1

Finally we show 2.3. Since H is an e-bubble, H −{u,v} is connected. Again,

the symmetric difference of the inner faces of H incident to u or v, considered

0,...,vk0

0,v1

1,...,vk1

1,...,v1

n,...,vkn

n through

n

0,...,v0vk0

0,v1v1

1,...,v1vk1

1,...,vnv1

n,...,vnvkn−1

n

in this

0,...,vk0

1,...,vk1

1,...,v1

n,...,vkn

n.

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as sets of edges, gives a simple clockwise cycle in H passing through v, u and

v1,...,vkin this order. Let γ be a closed curve going counterclockwise from

vℓthrough the outer face of G to u1and then through the inner faces of H

at u and v, crossing the edges uu2,...,uuk,vv1,...,vvℓ−1in this order, back

to vℓ. Clearly, γ is a snip defining the bubble H − {u,v} with root-sequence

u1,...,uk,v1,...,vℓ.

⊓ ⊔

3 Bounding regions

Depending on the maximum degree ∆ of G define the set S of ∆ − 1 slopes to

consist of the horizontal slope and the slopes of vectors f1,...,f∆−2where

fi= (−1

2+

i−1

∆−3,1) for i = 1,...,∆ − 2.

An important property of S is that it cuts the horizontal segment L from (−1

to (1

2,1) into ∆−3 segments of equal length

straight-line drawing of G using only slopes from S and preserving the given

cyclic ordering of edges at each vertex of G.

The essential tool in proving that our construction does not make bubbles

overlap are bounding regions. Their role is to bound the space of the plane

occupied by bubbles. The bounding region for a bubble is parametrized by ℓ and

r which depend on the degrees of the roots in the bubble. Let v be a point in

the plane. For a vector x let R(v;x) = {v + αx : α ? 0}. We define LB(v;ℓ)

to be the set consisting of v and all points p with py ? vy. If ℓ > 0 then we

furthermore require that

2,1)

1

∆−3. We construct an outerplanar

– p lies on R(v;f1) or to the right of it

– px> vx

– p lies to the right of R(v;fℓ+

– p lies to the right of R(v;f∆−2)

if ℓ = 1,

if ∆ = 4 and ℓ = 2,

if ∆ ? 5 and 2 ? ℓ ? ∆ − 2,

if ℓ = ∆ − 1.

1

∆−4f1)

See Fig. 3 for an illustration. Similarly, RB(v;r) consists of v and all points p

with py? vy. If r < ∆ − 1 we furthermore require that

– p lies to the left of R(v;f1)

– px< vx

– p lies to the left of R(v;fr+

– p lies on R(v;f∆−2) or to the left of it

if r = 0,

if ∆ = 4 and r = 1,

if ∆ ? 5 and 1 ? r ? ∆ − 3,

if r = ∆ − 2.

1

∆−4f∆−2)

Now, for points u,v in the plane such that uy = vy and ux ? vx we define

bounding regions as follows:

B(uv;ℓ,r) = LB(u;ℓ) ∩ RB(v;r)

¯B(uv;ℓ,r;h) = B(uv;ℓ,r) ∩ {p : py< uy+ h}

for 0 ? ℓ,r ? ∆ − 1,

for 0 ? ℓ,r ? ∆ − 1 and h > 0.

We denote B(vv;ℓ,r) simply by B(v;ℓ,r) and¯B(vv;ℓ,r;h) simply by¯B(v;ℓ,r;h).

Note that the bottom border of a bounding region is always included, the left

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ℓ = 0

ℓ = 12345

v

Fig.3. Boundaries of LB(v;ℓ) for ∆ = 6. Vectors fi at v are indicated by thick arrows.

Vectors

2f1 at v + fi are indicated by thin arrows. Note that f3 lies on the boundary

of LB(v;4).

1

border (if exists) is included if ℓ = 1, the right border (if exists) is included

if r = ∆ − 2, and the top border (if exists) is never included. If ℓ > r then

B(v;ℓ,r) = {v}.

We use B(v;ℓ,r) and¯B(v;ℓ,r;h) to bound drawings of 1-bubbles H with

root v such that r − ℓ + 1 = dH(v). Note that every 1-bubble drawn inside

B(v;ℓ,r) can be scaled to fit inside¯B(v;ℓ,r;h) for any h > 0 without changing

slopes. We use B(uv;ℓ,r) and¯B(uv;ℓ,r;h) with u ?= v to bound drawings of

2-bubbles H whose root-path starts at u and ends at v, such that ℓ = ∆−dH(u)

and r = dH(v) − 1. Here H cannot be scaled if the positions of both u and

v are fixed, so the precise value of h matters. However, every 2-bubble drawn

inside B(uv;ℓ,r) can be scaled to fit inside¯B(uw;ℓ,r;h) for any h > 0 without

changing slopes, where w is some point of the horizontal line containing uv.

Lemma 3. Bounding regions have the following geometric properties (∗ denotes

any relevant value).

3.1. If u′v′⊆ uv, ℓ′? ℓ, and r′? r then B(u′v′;ℓ′,r′) ⊆ B(uv;ℓ,r).

3.2. If i < ℓ then a vector at u in direction fi points outside B(uv;ℓ,∗) to the

left of it. If i > r then a vector at v in direction fipoints outside B(uv;∗,r)

to the right of it.

3.3. If u,v,w are consecutive points on a horizontal line and ℓ−1 ? r+1 then

B(uv;∗,r) ∩ B(vw;ℓ,∗) = {v}.

Moreover, the following holds for ∆ ? 5.

3.4. For l < r, h > 0, u′= u + hfℓ, and v′= v + hfr we have¯B(u′v′;1,∆ −

2;

∆−4h).

3.5. If u,v,w are consecutive points on a horizontal line, ℓ′? ℓ, r′? ∆ − 3,

and r ? 1 then¯B(uv;ℓ′,r′;(∆−3

3.6. If u,v,w,x are consecutive points on a horizontal line, |uv| = |wx| ? |vw|,

ℓ ? 2, and r ? ∆ −3 then¯B(uv;∗,r;∆−3

h

∆−4) ⊆¯B(uv;ℓ,r;∆−3

∆−4)2|vw|) ⊆ B(uw;ℓ,r).

∆−4|uv|) ∩¯B(wx;ℓ,∗;∆−3

∆−4|wx|) = ∅.

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Proof. Statement 3.1 clearly follows from the definition. Statement 3.2 is implied

by the definition and for ∆ ? 5 by the fact that

fℓ+

1

∆−4f1

1

∆−4f∆−2= fr+1+

= fℓ−1+

1

∆−4f∆−3

1

∆−4f2

for ℓ = 2,...,∆ − 2,

fr+

for r = 1,...,∆ − 3.

Statement 3.2 directly yields 3.3: the vector at v in direction fr+1points outside

both B(uv;∗,r) and B(vw;ℓ,∗). To see 3.4 note that the point u′+

which is the top-left corner of¯B(u′v′;1,∆ − 2;

Hence it lies at the top-left corner of B(uv;ℓ,r). Similarly, the top-right corner

of¯B(u′v′;1,∆ − 2;

∆−4) lies at the top-right corner of B(uv;ℓ,r). To prove 3.5

it suffices to consider the case r = 1 and r′= ∆ − 3. The top-right corner of

¯B(uv;ℓ′,r′;(∆−3

α

∆−4f1,

1

∆−4f1).

h

∆−4), equals u + h(fℓ+

h

∆−4)2|vw|) is

v +∆−3

∆−4|vw|(f∆−3+

1

∆−4f∆−2) = v + |vw|(f∆−2+

= w +∆−3

1

∆−4f1+∆−3

1

∆−4f∆−2).

∆−4·

1

∆−4f∆−2)

∆−4|vw|(f1+

Therefore, it lies on the right side of B(uw;ℓ,1) and the conclusion of 3.5 follows.

Finally, for the proof of 3.6 it suffices to consider the case ℓ = 2, r = ∆−3, and

|uv| = |vw| = |wx| = λ. The top-right corner of¯B(uv;∗,∆ − 3;∆−3

top-left corner of¯B(wx;2,∗;∆−3

∆−4λ) are respectively

∆−4λ) and the

v + λ(f∆−3+

1

∆−4f∆−2) = v + λ(f∆−2+

1

∆−4f1)= w + λ(f1

1

∆−4f2),

1

∆−4f∆−3).

w + λ(f2

+

+

They coincide if ∆ = 5, otherwise the former lies to the left of the latter.

⊓ ⊔

4The drawing

We present the construction of a drawing first for ∆ ? 5 and then for ∆ = 4.

Both constructions follow the same idea but differ in technical details.

Lemma 4. Suppose ∆ ? 5.

4.1. Let H be a 1-bubble with root v. Suppose that the position of v is fixed. Let

ℓ and r be such that 0 ? ℓ,r ? ∆−1 and r−ℓ+1 = dH(v) ? ∆−1. Then

there is a straight-line drawing of H inside B(v;ℓ,r).

4.2. Let H be a 2-bubble with first root u and last root v. Suppose that the

positions of u and v are fixed on a horizontal line so that u lies to the left

of v. Let ℓ = ∆ − dH(u) and r = dH(v) − 1. Then there is a straight-line

drawing of H inside¯B(uv;ℓ,r;∆−3

∆−4|uv|) such that the root-path of H is

drawn as the segment uv.

4.3. Let H be a k-bubble with roots v1,...,vkin this order along the root-path.

If k = 1 then suppose dH(v1) ? ∆−2, otherwise suppose dH(v1),dH(vk) ?

∆ − 1. Suppose that for some λ > 0 the positions of v1,...,vkare fixed in

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this order on a horizontal line so that |v1v2| = ... = |vk−1vk| = λ. Then

there is a straight-line drawing of H inside¯B(v1vk;1,∆ − 2;∆−3

that the root-path of H is drawn as the segment v1vk.

∆−4λ) such

The drawings claimed above use only slopes from S and preserve the order of

edges around each vertex w of H under the assumption that if there are edges

connecting w to G − H then they are drawn in the correct order outside the

considered bounding region.

Proof. The proof constructs the required drawing by induction on the size of H.

The construction we are going to describe clearly preserves the order of edges

at every vertex of H and uses only slopes from S, and we do not explicitly state

this observation anywhere further in the proof.

Let H be a k-bubble with splitting sequence (H1,...,Hb). Assuming that

the lemma holds for any bubble with fewer vertices than H has, we prove that

H satisfies 4.1 if k = 1, 4.2 if k = 2, and 4.3 for any k.

First we prove 4.1 and 4.2 for the case that b ? 2 and one of H1,Hbis a v-

bubble. By symmetry we consider only the case of H1being a v-bubble. Let u and

v denote respectively the first and the last root of H. Define Y = (H2,...,Hb),

r1= ℓ + dH1(u) − 1, and ℓY = ∆ − dY(u). By the induction hypothesis we can

draw H1inside B(v1;ℓ,r1) and Y inside¯B(uv;ℓY,r;∆−3

drop the restriction on the height of the latter bounding region). We scale the

drawing of H1to make it so small that it lies entirely below all vertices of Y not

contained in the root-path and (for k = 2) below the horizontal line bounding

from above the requested bounding region of H. Since r1+ 1 = ℓY and by 3.2,

our scaled drawing of H1 lies to the left of the leftmost edge at the root u of

Y . Thus the drawings of H1and Y do not overlap. By 3.1 they both fit within

¯B(uv;ℓ,r;∆−3

∆−4|uv|).

Now we consider the case of H being a v-bubble rooted at v and such that

dH(v) ? ∆ − 1. We are to prove that 4.1 holds for H. If v is the only vertex of

H, the conclusion is trivial. Thus we assume that H has at least two vertices,

that is, ℓ ? r.

Suppose 1 ? ℓ,r ? ∆ − 2. Define H′= H − v. By 2.1 the graph H′is an

(r −ℓ+1)-bubble with root-sequence formed by the neighbors vℓ,...,vrof v in

H listed in the clockwise order around v. Put the vertices vℓ,...,vrat points

v + fℓ,...,v + fr, respectively. This way vℓ,...,vrlie on a common horizontal

line L and partition L into segments of length

induction hypothesis the 1-bubble H′can be drawn inside B(vr;0,dH′(vr) − 1)

as well as inside B(vr;∆ − dH′(vr),∆ − 1). Choose the former drawing if ℓ =

r = ∆−2, the latter if ℓ = r = 1, or any of the two otherwise. After appropriate

scaling the chosen drawing fits within B(v;r,r). If ℓ < r then apply the induction

hypothesis to draw H′inside¯B(vℓvr;1,∆−2;

bounding region is included in B(v;ℓ,r).

It remains to consider the following cases for v-bubble H: ℓ = 0 or r = ∆−1.

They cannot hold simultaneously as dH(v) ? ∆ − 1. By symmetry it is enough

to consider only one of them. Thus we assume 1 ? ℓ ? r = ∆ − 1.

∆−4|uv|) (in case k = 1 we

1

∆−3. Now, if ℓ = r then by the

1

∆−4). It follows from 3.4 that this

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If v has only one neighbor in H, say w, then ℓ = r = ∆ − 1 and clearly

H′= H − v is a 1-bubble rooted at w. Put w horizontally to the right of v.

Draw H′inside B(w;∆ − dH′(w),∆ − 1) by the induction hypothesis, scaling

the drawing appropriately to fit it within B(v;∆ − 1,∆ − 1).

Now suppose that v has at least two neighbors in H and therefore ℓ < r =

∆ − 1. Let P = w0...wnbe the simple path of length n ? 1 starting at w0= v

and going counterclockwise along the outer face of H so that

– the edge w0w1preceeds the root-angle in the clockwise order around w0= v,

– the vertices w1,...,wn−1have degree ∆ and are not cut-vertices in H,

– the vertex wnhas degree at most ∆ − 1 or is a cut-vertex in H.

Note that if n = 1 then the second condition is satisfied vacuously. Since the de-

grees of w1,...,wn−1are at least 3 and by outerplanarity, P is an induced path.

Thus by 2.2 the graph H−P has exactly one component H′adjacent to both w0

and wn. All other components of H − P are adjacent to wn. Together with wn

they form a (possibly trivial) 1-bubble Y rooted at wn. Let X denote the sub-

graph of H induced on w0,...,wnand the vertices of H′. Define rX= dX(wn)−1

and ℓY = ∆ − dY(wn). Let wℓ

0

clockwise. Let w1

i

be the neighbors of wiin X ordered clockwise, for

1 ? i ? n − 1. Let w1

n

be the neighbors of wnin X ordered clockwise.

Note that w∆−2

i

and w1

H′is a bubble with root-sequence wℓ

For i = 0,...,n − 1 define

0,...,w∆−2

be the neighbors of w0in X ordered

i,...,w∆−2

n,...,wrX

i+1may coincide, for 0 ? i ? n−1. It follows from 2.2 that

0,...,w∆−2

0

,w1

1,...,w∆−2

1

,...,w1

n,...,wrX

n.

λi=

?

1

∆−2

∆−3

if w∆−2

i

if w∆−2

i

= w1

?= w1

i+1,

i+1.

Put the vertices w1,...,wnin this order from left to right on the horizontal line

going through w0in such a way that |wiwi+1| = λi. Put each vertex wj

wi+ fj. Note that if w∆−2

i

and w1

the same point. All wj

ilie on a common horizontal line L at distance 1 to the

segment w0wnand partition L into segments of length

iat point

i+1are the same vertex then they end up at

1

∆−3. Define

BX=¯B(w0wn;ℓ,rX;∆−3

BY =¯B(wn;ℓY,∆ − 1;1).

∆−4),

Draw H′inside¯B(wℓ

that if H′is a 1-bubble then n = 1 and the root of H′has at least two edges

outside H′(namely, the ones going to w0and w1), and therefore the first case of

the induction hypothesis yields its drawing inside the claimed bounding region.

By 3.4 this bounding region is contained in BX. Draw Y inside BY using the

induction hypothesis and scaling. This way it lies entirely below the line L. By

3.2 the drawing of Y lies to the right of the edge wnwrX

with the drawing of X. Clearly, BX and BY are contained in B(w0;ℓ,∆ − 1).

This completes the proof for the case of H being a 1-bubble.

0wrX

n;1,∆ − 2;

1

∆−4) using the induction hypothesis. Note

n and thus does not overlap

Page 11

Now, suppose k = 2. We are to show that 4.2 holds for H. Suppose first that

H is a single e-bubble with roots u and v. If H consists of the edge uv only

then the conclusion follows trivially. Thus assume that H is not the single edge

uv. This implies that dH(u),dH(v) ? 2 and therefore ℓ ? ∆ − 2 and r ? 1.

Let uℓ,...,u∆−2,v denote the neighbors of u in H ordered clockwise, and let

u,v1,...,vrbe the neighbors of v in H ordered clockwise. Let H′= H −{u,v}.

By 2.3 the graph H′is a bubble with all wi

0and wi

1being the roots. Define

h =

?

|uv|

∆−3

∆−2|uv|

if u∆−2= v1,

if u∆−2?= v1.

Put each vertex uiat point u+hfiand each vertex viat point v+hfi. Note that

if u∆−2and v1are the same vertex then they end up at the same point. All ui

and vilie on a common horizontal line and partition it into segments of length

h

∆−3. Draw H′inside¯B(uℓvr;1,∆−2;

bounding region is contained in¯B(uv;ℓ,r;∆−3

¯B(uv;ℓ,r;∆−3

∆−4|uv|).

Now, consider the case that H is a 2-bubble but not an e-bubble. Let w0...wn

be the root-path of H. Thus w0= u, wn= v, and n ? 2. Suppose that none of

the edges w0w1,...,wn−1wnis a bridge in H. This implies that dH(u),dH(v) ? 2

and therefore ℓ ? ∆ − 2 and r ? 1. We split H into the e-bubble H1 and the

rest Y = (H2,...,Hb) being a 2-bubble with roots w1 and wn. Define r1 =

dH1(w1) − 1 and ℓY = ∆ − dY(w1). The assumption that w0w1,...,wn−1wn

are not bridges yields r1? 1 and ℓY ? ∆ − 2. Let wℓ

neighbors of w0 in H1 ordered clockwise. Similarly, let w0,w1

neighbors of w1in H1ordered clockwise. Define

h

∆−4) using the induction hypothesis. This

∆−4h) by 3.4. Since h ? |uv|, we have

∆−4h) ⊆¯B(uv;ℓ,r;∆−3

0,...,w∆−2

0

,w1denote the

1,...,wr1

1

be the

α =

?

1

∆−3

∆−2

if w∆−2

0

if w∆−2

0

= w1

?= w1

1,

1.

Fix the position of w1on the segment w0wnso that α|w0w1| =∆−3

Put each vertex wi

that if w∆−2

0

and w1

All wi

1lie on a common horizontal line L at distance h to the segment

w0wnand partition L into segments of length

∆−4|w1wn| = h.

0at point w0+hfiand each vertex wi

1are the same vertex then they end up at the same point.

1at point w1+hfi. Note

0and wi

h

∆−3. Define

B1=¯B(w0w1;ℓ,r1;∆−3

BY =¯B(w1wn;ℓY,r;h).

∆−4h),

Let H′

wi

1being the roots. Draw H′

hypothesis. By 3.4 this bounding region is contained in B1. Moreover, r1? ∆−3

as w1w2is not a bridge. This together with 3.5, the choice of w1, and the fact that

h ? |w0wn| imply that B1⊆¯B(w0wn;ℓ,r;∆−3

of H, apply the induction hypothesis to draw Y inside BY. This way it lies

1= H1− {w0,w1}. By 2.3 the graph H′

1inside¯B(wℓ

1is a bubble with all wi

1;1,∆−2;

0and

0wr1

h

∆−4) using the induction

∆−4|w0wn|). To complete the drawing

Page 12

entirely below L and therefore does not overlap with the drawing of H′

also lies to the right of the edge w1wr1

If H is a 2-bubble with root-path w0...wnand some edge wiwi+1is a bridge

in H, then H splits into three bubbles: X with root-path w0...wi, Y = {wiwi+1}

being a trivial e-bubble, and Z with root-path wi+1...wn. Define

1. By 3.2 it

∆−4|w0wn|).

1. Clearly, BY ⊆¯B(w0wn;ℓ,r;∆−3

rX=

?

?

ℓ + dX(w0) − 1

dX(wi) − 1

if i = 0,

if i ? 1,

ℓZ=

r − dZ(wn) + 1

∆ − dZ(wi+1)

if i + 1 = n,

if i + 1 ? n − 1.

We are free to choose the positions of wi (unless i = 0) and wi+1 (unless

i + 1 = n) on the segment w0wn. So we take care that X and Z are small

enough, that is, that after applying the induction hypothesis to draw X inside

¯B(w0wi;ℓ,rX;|wiwi+1|) and Z inside¯B(wi+1wn;ℓZ,r;|wiwi+1|), these bounding

regions do not intersect and are both contained in¯B(w0wn;ℓ,r;∆−3

It remains to prove 4.3 for H. If k = 1 then the claim follows directly from 4.1

and 3.1 by scaling. Thus assume k ? 2. There is a splitting of H into 2-bubbles

X1,...,Xk−1so that the splitting sequences of X1,...,Xk−1together form the

splitting sequence (H1,...,Hb) of H. Therefore,

∆−4|w0wn|).

– the roots of Xiare vi,vi+1

– Xi−1∩ Xi= {vi}

for i = 1,...,k − 1,

for i = 2,...,k − 1.

Apply 4.2 to draw each Xiinside¯B(vivi+1;∆ − dXi(vi),dXi(vi+1) − 1;∆−3

Consecutive bounding regions do not overlap by 3.3, while non-consecutive ones

are disjoint by 3.6. By 3.1 they are all contained in¯B(v1vk;1,∆−2;∆−3

∆−4λ).

∆−4λ).

⊓ ⊔

Lemma 5. Suppose ∆ = 4.

5.1. Let H be a 1-bubble with root v. Suppose that the position of v is fixed. Let

ℓ and r be such that 0 ? ℓ,r ? 3 and r − ℓ + 1 = dH(v) ? 3. Then there is

a straight-line drawing of H inside B(v;ℓ,r).

5.2. Let H be a 2-bubble with first root u and last root v. Suppose that the

positions of u and v are fixed on a horizontal line so that u lies to the left

of v. Let ℓ = 4 − dH(u) and r = dH(v) − 1. Then there is a straight-line

drawing of H inside B(uv;ℓ,r) such that the root-path of H is drawn as

the segment uv.

The drawings claimed above use only slopes from S and preserve the order of

edges around each vertex w of H under the assumption that if there are edges

connecting w to G − H then they are drawn in the correct order outside the

considered bounding region.

Proof. The proof, like for Lemma 4, constructs the required drawing by induction

on the size of H. We proceed along the same lines as in the proof of Lemma 4,

focusing only on those details in which the two proofs differ.

Page 13

Let H be a 1- or 2-bubble with splitting sequence (H1,...,Hb). Assuming

that the lemma holds for any bubble with fewer vertices than H has, we prove

that H satisfies 5.1 or 5.2 if it is a 1- or 2-bubble, respectively.

If b ? 2 and one of H1,Hbis a v-bubble, then 5.1 or 5.2 follows for H by the

same argument as in the proof of Lemma 4.

Now we consider the case of H being a v-bubble rooted at v and such that

dH(v) ? 3. We are to prove that 5.1 holds for H. If v is the only vertex of H, the

conclusion is trivial. Thus we assume that H has at least two vertices, that is,

ℓ ? r. If 1 ? ℓ = r ? 2 then the same argument as in the proof of Lemma 4 gives

the required drawing of H. If ℓ = 1 and r = 2 then, like in the proof of Lemma

4, define v1and v2to be the two neighbors of v in H and draw the 2-bubble

H −v inside B(v1v2;1,2) using the induction hypothesis—this bounding region

is clearly contained in B(v;1,2).

It remains to consider the following cases for v-bubble H: ℓ = 0 or r = 3.

They cannot hold simultaneously as dH(v) ? 3. By symmetry it is enough to

consider only one of them. Thus we assume 1 ? ℓ ? r = 3.

If v has only one neighbor in H then draw H the same way as in the proof

of Lemma 4. Now suppose that v has at least two neighbors in H and therefore

ℓ < r = 3. Define path w0...wn, bubble H′, index rX, and vertices wj

the proof of Lemma 4. For i = 0,...,n − 1 define

ilike in

λi=

?

1

1 + ε

if w2

if w2

i= w1

i?= w1

i+1,

i+1,

where ε > 0 is small enough for the following argument to work. Draw vertices

wiand wj

ilike in the proof of Lemma 4 but with the new definition of λi. All

wj

ilie on a common horizontal line, and moreover the segments w2

w2

i?= w1

drawing of H, it suffices to draw H′inside BX, and then the remaining part of

H can be drawn like in the proof of Lemma 4. But here drawing H′inside BX

is more tricky.

If H′is a 1-bubble then n = 1, w2

B(w2

0;1,2) by induction hypothesis, scaling the drawing appropriately to fit it

within BX. Thus we assume that H′is a k-bubble with k ? 2. We split H′into

2-bubbles X1,...,Xk−1so that the splitting sequences of X1,...,Xk−1together

form the splitting sequence of H′. The roots of X1,...,Xk−1are pairs of vertices

consecutive in the sequence wj

i. Define ℓi= 4−dXi(ui) and ri= dXi(vi), where

ui and vi denote respectively the first and the last root of Xi. We draw each

Xi inside B(uivi;ℓi,ri) using the induction hypothesis. Since each root of H′

has degree at most 3 in H′, we have ℓi? 2 for 2 ? i ? k − 1 and ri ? 1 for

1 ? i ? k − 2. Thus the bounding regions for Xido not overlap. Moreover, for

2 ? i ? k−2 we clearly have B(uivi;ℓi,ri) ⊆ BX. Thus to complete the proof for

the case of H being a v-bubble, it remains to show that the drawings of X1and

Xk−1 are contained in BX. We do not necessarily have B(u1v1;ℓ1,r1) ⊆ BX.

However, this inclusion may not hold only if ℓ = 2 and w2

we have |u1v1| = ε and thus the drawing of X1indeed lies within BX provided

iw1

i+1(for

i+1) have length ε. Define BX= B(w0wn;ℓ,rX). To obtain the required

0= w1

1, and thus we can draw H′inside

0?= w1

1. In this case

Page 14

that ε is small enough. Similarly, the drawing of Xk−1in contained in BXfor ε

small enough.

Now, suppose that H is a 2-bubble. We are to show that 5.2 holds for H.

Suppose first that H is a single e-bubble with roots u and v. If H consists of the

edge uv only then the conclusion is trivial. Thus assume that H is not the single

edge uv. This implies that dH(u),dH(v) ? 2 and therefore ℓ ? 2 and r ? 1.

Define vertices uj,vjand bubble H′like in the proof of Lemma 4. Define

h =

?

|uv|

|uv| − ε

if u2= v1,

if u2?= v1,

where ε > 0 is small enough for the following to work. Put each vertex uiat

point u + hfiand each vertex viat point v + hfi, so that if u2and v1are the

same vertex then they end up at the same point. All uiand vilie on a common

horizontal line, and moreover the segment u2v1(if exists) has length ε. The

same argument as in the previous paragraph shows that H′can be drawn inside

B(uℓvr;1,2).

Finally, consider the case that H is a 2-bubble but not an e-bubble. Let

w0...wn be the root-path of H. Thus w0 = u, wn = v, and n ? 2. Suppose

that none of the edges w0w1,...,wn−1wn is a bridge in H. This implies that

dH(u),dH(v) ? 2 and therefore ℓ ? 2 and r ? 1. We split H into 2-bubbles

X1,...,Xn so that the roots of Xi are wi−1,wi and the splitting sequences

of X1,...,Xn together form the splitting sequence (H1,...,Hb) of H. Define

ℓi= 4 − dXi(wi−1) and ri= dXi(wi). Thus ℓ1= ℓ and rn= r. We draw each

Xi inside B(wi−1wi;ℓi,ri) using the induction hypothesis. Since there are no

bridges among w0w1,...,wn−1wn, we have ℓi = 2 for 2 ? i ? n and ri = 1

for 1 ? i ? n − 1. Thus the bounding regions for Xi do not overlap and are

contained in B(w0wn;ℓ,r). The case that some edge wiwi+1is a bridge in H is

handled the same way as in the proof of Lemma 4.

⊓ ⊔

Now, to prove the Main Theorem, pick any vertex v of G of degree less than

∆ (such a vertex always exists in an outerplanar graph), fix its position in the

plane, and apply 4.1 or 5.1 to the graph G considered as a 1-bubble with root v.

Acknowledgments

We thank V´ ıt Jel´ ınek and D¨ om¨ ot¨ or P´ alv¨ olgyi for introducing us to the problem

at the meeting in Prague in summer 2011.

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