Testing planarity part 1

Testing planaritypart 1

Thomas van Dijk

Preface

Appendix of Planar Graph Drawing Quite hard to read So we’ll try to explain it, not just tell you

about it

Preface

“There are a number of efficient algorithms for planarity testing, which are unfortunately all difficult to implement.” (www.mathworld.com)

Preface

“There are a number of efficient algorithms for planarity testing, which are unfortunately all difficult to implement.” (www.mathworld.com)

Implementation in Mathematica was actually bugged until version 4.2.1 (!) (Also mathworld.com)

What is planarity testing?

Whether a graph can be drawn on the plane without edge crossings

Equivalently, whether a “planar embedding” of the graph exists

Embedding

Adjacency-list representation of the graph Order in the list defines clockwise order of

the edges

Embedding


the edges

1

4

2

3

2 4 3 4 3 1

3 1 21 2 4

Embedding


the edges

1

4

2

3

2 4 3 4 3 1

3 1 21 2 4

1

2

3

4

2 4 3

3 4 1

1 4 21 2 3

Overview

“Vertex addition algorithm” First presented by Lempel et al (’67) Improved to linear time by

Booth and Lüker (’76) We present this second algorithm

Main idea

Start with one vertex Add vertices and their edges one by one

make sure we don’t break planarity If you can’t add the next vertex, graph is

not planaradd the vertices in an order such that if we

fail, we that know that using any other order wouldn’t have worked either

We need to know about

Some general observations st-Numbering Bush forms / PQ-Trees

O(m) is O(n) for planar graphs

Remember from Euler’s Theorem we have thatm ≤ 3n-6for simple plane graphs

Algorithm can just reject graphs with too many edges.

So from now on, O(m) is O(n)

Biconnected components

Definition:A graph G is biconnected ifffor every two distinct vertices there exist two internally disjoint paths between them

Also iff G contains no cut-vertices i.e.: cannot be made unconnected by

removing a single vertex


Theorem:A graph is planar iffall its biconnected components are planar

Proof:Induction on the number of biconnected components


If the graph is planar, clearly its biconnected components are planar

A graph is planar iff its connected components are planar

A connected graph with no biconnected components is a tree; trees are planar


If a graph has one biconnected component, then all vertices not in it are in trees. So the graph is planar iff the biconnected component is planar

Biconnected components Induction step: The next biconnected component is

connected to the rest of the graph by a cut-vertex v.

Biconnected components Induction step: The next biconnected component is

connected to the rest of the graph by a cut-vertex v.

Since the biconnected component itself is planar, an embedding with v on the exterior exists. (As Hans showed us.)


The biconnected components of a graph can be found in linear time(If you are interested, an exercise in Cormen

et al explains it) So from now on, we can assume

graphs are biconnected.

st-Numbering

From the ‘main idea’ slide:“Add the vertices in an order such that if we fail, we that know using any other order wouldn’t have worked either”

st-Numbering

Special nodes: source s (“1”) and sink t (“n”)

“1” and “n” adjacent

1

8

st-Numbering



j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k) 1

8

st-Numbering



j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k)

So 2 must be next to 1and n-1 must be next to n

2 71

8

st-Numbering



j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k)

So 2 must be next to 1and n-1 must be next to n

4 6

2

5

71

3

8

Some st-numbering examples

12

1

11

2

10

3

9

4

8

5

7

6

31

5

2

4

st-Numbering

Theorem:Every biconnected graph has an st-numbering

Proof:We give an algorithm and show that it works if the graph is biconnected

Our graph is biconnected, so anst-numbering exists.

Finding an st-numbering

First step, standard DFS. Can visit children in any order Calculate and store per vertex:

Parent“Depth first search number:” DFNLow-point: the lowest DFN among nodes that

can be reached by a path using any amount of tree edges followed by 1 back edge

DFS

G

1

2

3

4

5

6 DFN

DFS

G

1

2

3

4

5

6 DFN

1

1

1

1

3

3LOW

DFS

G

Finding an st-numbering

Push the vertices onto a stack and pop them off

Four rules for what to push and pop Assign increasing numbers to the vertices

when they are popped off for the last time

Example follows

Initial setup

The node with DFN 2 is called the ‘source’ The node with DFN 1 is called the ‘sink’

Stack with sourceon top of sink

Source and sinkare “old”

c

f

d

b

e

a

ac

Rule “2”

There is a “new” tree edge: follow it and take the path that leads to its low point

Push the vertices on the path ontothe stack

Mark all as old.

c

f

d

b

e

a

abc

No matches for vertex a

There are now no rule matches for a. Pop it off the stack and give it the next

available number This node is ready

c

f

d

b

e

1

bc

Rule “3”

There is a “new” back edge into this node: from the other end, go back up tree edges to an “old” vertex

Push the vertices onthe path ontothe stack

Mark all as old c

f

d

b

e

1 bfedc

No matches for vertex b

So the vertex is ready and gets its number

c

f

d

2

e

1fedc

No matches for vertex f

So the vertex is ready and gets its number

c

3

d

2

e

1

edc

Et cetera…

6

3

5

2

4

1

Correctness

All vertices are numbered The numbering is indeed an st-numbering

Algorithm recap

For linear complexity, we need thatO(m) = O(n).

For correctness, we need that the graph is biconnected

After the break …

Ron will tell you about:bush forms and PQ-treeshow it all fits together into a planarity testing

algorithm

Any questions?

Any questions?

http://www.planarity.net/

Testing planarity part 1

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