14085_planar

Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1.4

1

Planar Graphs

Tucker, Applied Combinatorics, Section 1.4, prepared by Patti Bodkin

Tucker, Applied Combinatorics, Sect. 1.4

2Jan. 27, 2003

A graph is called planar if it can be drawn on a plane without edges crossing.

A plane graph is a drawing of planar graph in the plane.

a

e

d

b

f

c

Figure 1

a

e

f

bd

c

This is a plane graph of Figure 1.


3Jan. 27, 2003

Here are two ways to determine if a graph is planar:

The first is a practical heuristic, which the author refers to as the circle-chord method. It consists of a step-by-step method of drawing the graph, edge-by-edge without crossing any edges.

The second consists of theoretical results, such as Kuratowski’s Theorem, or Euler’s Formula.


4Jan. 27, 2003

Step One: Find a circuit that contains all the vertices of the graph.(Recall: a circuit is a path that ends where it starts)

**Note: step one is not always possible, and is often difficult**

Circle-Chord Method--Planar

aa

a b c d

e hgf

The circuit for this figure is highlighted in red


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Step Two: Draw this circuit as a large circle.

a b c d

e hgf gh

f

bc

e

d

a


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Step Three: Choose one chord, and decide to draw it either inside or outside the circle. If chosen correctly, it will force certain other chords to be drawn opposite to the circle. (Inside if the first chord was drawn outside, and vice versa.)

gh

f

bc

e

d

a

Since the chords could be drawn without crossing, this graph is planar.

a b c d

e hgf


7Jan. 27, 2003

Circle-Chord Method--Nonplanar

Step One: Find a circuit that contains all the vertices of the graph.(Recall: a circuit is a path that ends where it starts)

**Note: step one is not always possible, and is often difficult**

a

f

e

d

c

b

h

g

The circuit for Figure 2 is highlighted in red

Figure 2


8Jan. 27, 2003

Step Two: Draw this circuit as a large circle.a

f

e

d

c

bh

g

a

f

e

d

c

bh

g

The remaining edges (in black) must be drawn either inside or outside the circle. These edges are called the chords.



9Jan. 27, 2003

Step Three: Choose one chord, and decide to draw it either inside or outside the circle. If chosen correctly, it will force certain other chords to be drawn opposite to the circle. (Inside if the first chord was drawn outside, and vice versa.)

a

f

e

d

c

bh

g

Because these lines cross, this is not the plane graph for Figure 2. However, this does not mean that the graph is not planar. Often, it is very difficult to find the correct planar graph.

a

f

e

d

c

bh

g



10Jan. 27, 2003

Recall: and configurations.5K3,3K

We say that a subgraph is a configuration if it can be obtained from a by adding vertices in the middle of some edges. A configuration is defined similarly.

3,3K

3,3K

5K

Here a has been subdivided by adding vertices in the middle of some of the edges.

5K


11Jan. 27, 2003

Kuratowski’s Theorem

A graph is planar if and only if it does not contain a subgraph that is a or configuration.

If the graph is nonplanar than it contains one of the configurations with added vertices.

The configuration is more common in nonplanar graphs than .

f

e

d

c

bh

g

e

d

c

b

g

aa

H and F were removed to show the configuration

3,3K5K

3,3K 5K

3,3K

Left Side

Right Side


12Jan. 27, 2003

A graph is defined to be a connected planar graph if it’s in “one-piece”, i.e. there are paths between every pair of vertices.

Euler’s Formula

Assume that is a connected planar graph, with:V = # of vertices E = # of edges R = # of regions.

Euler’s Formula states that if is a connected planar graph, then any plane graph depiction of has regions.

Figure 3:a

e

cd

b

G

GG

2 VER

G


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Euler’s Formula

a

c2211

2 VER

d

2541

a

c

b

e

a

c

b

e

2552

a

e

cd

b

2585


14Jan. 27, 2003

Proof of Euler’s Formula

Draw a plane graph depiction of edge-by-edge.

Let denote the (connected) plane graph obtained after n edges have been added, and let , , , denote the number of vertices, edges and regions in respectively.

Initially there is , which consists of one edge, it’s two end vertices and the one (unbounded) region.

G

nG

nvne nr nG

1G

a

c

1G

We obtain from by adding an edge at one of the vertices in .2G1G 1G

Then , , , and so Euler’s formula is valid for , 11 e 11 r 21 v 1G2211:2111 versince


15Jan. 27, 2003

Proof of Euler’s Formula (cont.)

In general, is obtained from , by adding an edge at one of the vertices of .

The new edge might link two vertices already in .

If it does not, the other end vertex of the edge is a new vertex that must be added to .

nG1nG thn

1nG

1nGthn

nG

Now we will use the method of induction to complete the proof.

We have shown that the theorem is true for .

Next we assume that it is true for for any , and prove it is true for .

Let be the edge that is added to to get .

There are two cases to consider…

1G

1nG 1n nG

),( yx thn 1nGnG


16Jan. 27, 2003


In the first case, and are both in . Then they are on the boundary of a common region of , possibly the unbounded region.

See Figure 4Edge splits into two regions.

x y1nG

K 1nG

),( yx K

x

y

K

Figure 4Then,

So each side of Euler’s formula grows by 1. Hence if the formula was true for , it will also be true for . 1nG nG

111 ,, 11 nnnnnn vveerr


17Jan. 27, 2003


In the second case, one of the vertices is not in say it is . Then adding implies that is also added, but no new regions are formed (no existing regions are split). See Figure 5

1nG xyx,),( yx x

11 111 ,, nnnnnn vveerr

1nG

Thus and the value on each side of Euler’s formula is unchanged.

The validity of Euler’s formula for implies its validity for . By

induction, the formula is true for all and for the full graph .

nG

sGn ' G

x

y

K

Figure 5


18Jan. 27, 2003

Worked Example

Book Example # 5

How many regions would there be in a plane graph with 10 vertices each of degree 3? (Recall: the degree of a vertex is the number of edges incident to the vertex)

By the theorem in Section 1.3, the sum of the degrees, 10 x 3, equals , and so = 15. By Euler’s formula, the number of regions is:

e2e r

2 ver 21015 7


19Jan. 27, 2003

Corollary to Euler’s Formula (Thm. 2)

If is a connected planar graph with , then .1e 63 ve

Proof:

The degree of a region is defined to be the number of edges on its boundary.

To be precise, we say that if an edge occurs twice along a boundary, as does in region in Figure 6, the edge is counted twice in region degree; for example, region has degree 10 and region has degree 3 in the figure.

G

L

K

x

y

Figure 6

),( yxK sK '

K L


20Jan. 27, 2003

Each region in a plane graph must have degree , for a region of degree 2 would be bounded by 2 edges joining the same pair of vertices and a region of degree 1 would be bounded by a loop edge, but parallel edges and loops are not allowed in graphs (in this book.)

Since each region has degree , the sum of degrees of all regions will be at least . But this sum of degrees of all regions must equal , since this sum counts each edge twice. That is each of an edge’s two sides is part of some boundary.

Thus,

= (sum of regions’ degrees)

3

3r3 e2

e2 .3

2,3 rer


21Jan. 27, 2003

Corollary to Euler’s Formula (Thm. 2)

Proof: (cont.)

Combining this inequality with Euler’s formula we have:

Solving for e, we obtain:L

K

x

y

Figure 4:

In this case, edge is counted twice in the deg( ). The deg( ) is 3.

),( yxK

L

23

2 vere

63 ve


22Jan. 27, 2003

CAREFUL! Do not interpret the corollary as meaning: If , then a connected graph is planar, because there are many nonplanar graphs which also satisfy this equation.

63 ve

For example, has 6 vertices and 9 edges. So when you substitute into the equation, you get: , which holds. However, is not planar.

3,3K6)6(39 3,3K


23Jan. 27, 2003

FOR THE CLASS TO TRY:

Use the Corollary to Euler’s Formula to prove whether the following configuration is planar.

Substituting into the Corollary:

5K

5K

Configuration

156)73( 63v

1115

7v

5K

11e

However, the corollary states that must be true in a connected planar graph, and so the configuration cannot be planar.5K

63 ve

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14085_planar