# Algorithmic Graph Theory: Planar Graphs

Algorithmic Graph Theory: Planar Graphs

Algorithmic Graph Theory

Section 10. Planar graphs

10.1 Definition and basic properties

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 1 / 15

Introduction

• Informally, a graph is planar if it can be drawn (embedded) in the plane

such that two edges overlap only in common end vertices.

• Topics in the next lectures:

- Planar graphs and embeddings: basic properties

- Encoding plane graphs / combinatorial embeddings

- Planarity testing and generating embeddings

- Equivalence of embeddings and counting embeddings

- Approximation algorithms on planar graphs

- Dual graphs

Graphs in this section may be multigraphs.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 2 / 15

• Let X ⊆ R 2 be the union of all vertices and edges of an embedding φ.

The regions of R 2 \X (maximal connected subsets) are called the faces.

The (unique) unbounded face is called the outer face, the other faces

inner faces.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 3 / 15

Embeddings of graphs

• A curve in the plane R 2 is the image of a continuous function

f : [0,1] → R 2 .

The ends are f(0) and f(1). The curve f is said to join f(0) and f(1).

It is simple if for every a,b ∈ [0,1), f(a) ≠ f(b), and closed if f(0) = f(1).

• A (planar) embedding φ of a graph G = (V,E) consists of an injective

mapping φ : V → R 2 and a mapping φ of edges uv to simple curves in R 2

that join φ(u) and φ(v), such that

- for e,f ∈ E, φ(e)∩φ(f) contains only the images of common end

vertices, and

- for e ∈ E and v ∈ V, φ(v) ∉ φ(e).

• Points φ(v) will also be called vertices, and curves φ(e) edges.

• A plane graph is a graph G together with an embedding φ of G.

• A graph G is planar if it admits an embedding.

• We will use basic topological facts without giving a rigorous

introduction, e.g.:

- A simple closed curve separates the plane into two regions (Jordan curve

Theorem).

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 4 / 15

Facial walks

• Let f be a face of a plane graph (thus an open set). A vertex or edge is

incident with f if it lies entirely in the closure of f.

• The boundary of a face is the set of incident vertices and edges.

• A closed walk W = v 0 ,e 0 ,v 1 ,e 1 ,...,e k−1 ,v k in G is called the facial

walk corresponding to a face f if all vertices and edges of W are incident

with f, and for all i ∈ {1,...,k}, e i mod k follows e i−1 in the clockwise

order of edges around v i .

• In a connected plane graph, for every face f there exists a facial walk

that contains all vertices and edges incident with f. This facial walk is

unique, up to the choice of v 0 and e 0 .

• The degree d(f) of a face f is the length of its facial walk.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 5 / 15

• Let F be the set of faces of a plane graph with m edges. Then

d(f) = 2m.

f∈F

• If a plane graph G has two faces with the same boundary, then G is a

cycle.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 6 / 15

Proposition 10.1

Let G, φ be a 2-connected plane graph without loops. Then every facial

walk is a cycle.

Proof: Suppose there exists a facial walk W = v 0 ,e 0 ,v 1 ,e 1 ,...,e k−1 ,v k

that is not a cycle, so w.l.o.g. v 0 = v i for some 2 ≤ i ≤ k −2 (as G

contains no loops). Let f be the face bounded by W.

Then a simple closed curve can be drawn in f ∪φ(v 0 ), which separates v 1

from v k−1 . So v 0 is a cut vertex, a contradiction. □

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 7 / 15

Euler’s formula

Proposition 10.2 (Euler’s Formula)

Let G be a connected plane (multi-) graph with n vertices, m edges and k

faces. Then n−m+k = 2.

Proof: Induction over k. If k = 1, then G contains no cycles, so it is a

tree, and thus m = n−1 and n−m+k = 2.

If k ≥ 2, then G contains a cycle (!). Choose an edge e on a cycle. So e is

incident with two distinct faces f 1 and f 2 .

In G ′ = G −e, f 1 and f 2 become part of one face. So G ′ has k −1 faces,

n vertices, and m−1 edges. By induction, n−(m −1)+(k −1) = 2, so

n−m+k = 2.

• Note that the formula does not hold for disconnected graphs.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 8 / 15

• A plane graph is a triangulation if every face has degree 3.

Proposition 10.3

Let G be a simple plane graph with n ≥ 3 vertices and m edges. Then

m ≤ 3n−6. If m = 3n−6, then G is a triangulation.

Proof: We may assume that G is connected. Since G is simple, every face

has degree at least 3.

Let F be the set of faces, and k = |F|.

Then 2m = ∑ f∈F d(f) ≥ 3k, so k ≤ 2 3m. Using Euler’s Formula,

2 = n−m+k ≤ n− 1 3m, and thus m ≤ 3n−6.

If m = 3n −6, then k = 2 3m, so every face has degree 3. □

(Where did we use that n ≥ 3)

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 9 / 15

• Proposition 10.3 has the following two corollaries:

Corollary 10.4

If G is a simple planar graph, then deg(G) ≤ 5.

Corollary 10.5

K 5 is not planar.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 10 / 15

Proposition 10.6

Let G be a simple plane graph with n ≥ 3 vertices and m edges, in which

every face has degree at least 4. Then m ≤ 2n−4. If m = 2n−4, then

every face has degree 4.

Proof: Similar.

Corollary 10.7

If G is a simple planar bipartite graph, then deg(G) ≤ 3.

Corollary 10.8

K 3,3 is not planar.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 11 / 15

• We have seen that K 5 and K 3,3 are not planar. In fact, we will now see

that every non-planar graph contains a subdivision of K 5 or K 3,3 (as a

subgraph).

• We will first prove this for 3-connected graphs. To that end, we need the

following lemma.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 12 / 15

Lemma 10.9

Let G be a 3-connected graph on at least 5 vertices. Then there exists an

edge that can be contracted such that the resulting graph is again

3-connected.

Proof: Suppose to the contrary that for every edge xy, contracting xy

yields a graph with connectivity 2. Then for every edge xy, there exists a

vertex z such that {x,y,z} is a vertex cut. Choose edge xy and z such

that the number of vertices of the largest component C 1 of G −x −y −z

is maximized.

Let C 2 be another component of G −x −y −z, and choose a neighbor

u ∈ V(C 2 ) of z (exists, since G 3-connected). Then there exists again a

v ∈ V(G) such that {u,v,z} is a vertex cut. Consider

G ′ = G[(V(C 1 )∪{x,y})\{v}]. Note that G ′ is connected. Therefore it

is a subgraph of a component of G −u −v −z, larger than C 1 ,

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 13 / 15

• An embedding of a graph is convex if every inner face is convex.

• An embedding of a graph is a straight line embedding if every edge is a

line segment.

Theorem 10.10

Let G be a 3-connected simple graph that contains no subdivision of K 5 or

K 3,3 . Then G admits a convex straight line embedding.

Proof: Induction over n = |V(G)|. The cases n ∈ {4,5} are easy, so

assume n ≥ 6.

By Lemma 10.9, G contains an edge xy that can be contracted into a

vertex z without destroying 3-connectedness. Let the resulting (simple)

graph be G ′ .

If G ′ would contain a subdivision of K 5 or K 3,3 , then this holds for G as

well (why!). So by induction, G ′ admits a convex straight line

embedding. This yields a straight line embedding of G −y as well.

The graph G ′ −z is 2-connected. So the face of G ′ −z that contains z is

bounded by a cycle C (Proposition 10.1).

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 14 / 15

Proof, continued: Let x 0 ,...,x k−1 be the neighbors of x on C, in order

along the cycle (k ≥ 2). Let P i be the subpath from x i to x i+1 mod k .

• If all vertices in N(y)\{x} lie on one path P i , then y can be drawn in

the face incident with P i and x, to obtain the desired embedding of G.

Note that it can be made convex.

• If y has at least three neighbors in {x 0 ,...,x k }, then a subdivision of K 5

is found.

• Otherwise, w.l.o.g. one neighbor a of y is an internal vertex of P 0 , and y

has at least one neighbor b not on P 0 . Then a subdivision of K 3,3 can be

exhibited (with “bipartition” {a,b,x},{x 0 ,x 1 ,y}). □

Q: Do 2-connected simple graphs always admit a convex embedding

A: No, but they do admit a straight line embedding.

Paul Bonsma (HU Berlin) AGT: Planar graphs January 31, 2012 15 / 15

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