mth 231 - lecture 24

MTH 231 0

MTH 231 - Lecture 24

Yevgeniy Kovchegov

Oregon State University

MTH 231 1

Topics:

• Homeomorphic graphs.

• Planar graphs.

• Examples.

• Euler’s formula for planar graphs.

• Is K5 “planar” in the toral world?

MTH 231 2

Planar Graphs

K4

The above complete graph K4 is planar as illustrated on thefollowing picture:

a b

c d

MTH 231 3

Planar Graphs

However the complete graph K5 of five vertices is known to be

not planar.

MTH 231 4

Planar Graphs

There is another famous graph known to be not planar. It is

the complete bipartite graph K3,3 on 3 and 3 vertices. See

the figure above.

MTH 231 5

Homeomorphic graphs

Graphs G1 and G2 are homeomorphic if G1 and G2 can bereduced to isomorphic graphs by performing a sequence of seriesreductions:

v

v

v

1

2

⇒

v

v

1

2

A series reduction consists of deleting a vertex v of degree 2

(if it has no incident loops or parallel edges), and replacing its

incident edges (v, v1) and (v, v2) by one edge (v1, v2).

MTH 231 6

Homeomorphic graphs

Graphs G1 and G2 are homeomorphic if G1 and G2 can bereduced to isomorphic graphs by performing a sequence of seriesreductions.

Example.

Graphs and

are homoemorphic.

MTH 231 7

Planar Graphs

The Kuratowski’s Theorem (on p.724) states that a graph isnot planar if and only if it contains a subgraph homeomorphicto K5 or K3,3.

We will illustrate it with the following example:

g

h

a b

c d

e f

MTH 231 8

Planar Graphs

g

h

a b

c d

e f

Example. Consider the above graph. In order for us to figureout if it is planar or not, we try to draw it so that no two edgesintersect. In this case, we try and fail. Now, we try to proveit is not planar. By Kuratowski’s Theorem, we need to find asubgraph homeomorphic to K5 or K3,3.

Here is how we find such subgraph.

MTH 231 9

Planar Graphs

g

h

a b

c d

e f

⇒

g

h

a b

c d

e f

Example (continued). We notice that if we remove the edge

(g, h), the above subgraph will be homeomorphic to K3,3 as can

be seen after redrawing it as follows.

MTH 231 10

Planar Graphs

g

h

a b

c d

e f

⇔

g

h

a b

c d

e f

Example (continued). The above subgraph is indeed homeo-

morphic to K3,3 as both vertices, g and h are of degree two, and

can be removed.

MTH 231 11

Planar Graphs

g

h

a b

c d

e f

⇔

a b

c d

e f

Example (continued). Once g and h are removed, we get

K3,3.

MTH 231 12

Planar Graphs

g

h

a b

c d

e f

Example (continued). Thus the above graph is not planar.

MTH 231 13

Planar Graphs

Example.

g

a b

c d

e f

While another, slightly different graph above is planar as can be

seen by redrawing it in two steps.

MTH 231 14

Planar Graphs

g

a b

c d

e f

⇔

a b

cd e

f

g

Example (continued). First, reposition vertex e.

MTH 231 15

Planar Graphs

a b

cd e

f

g⇔

a

b

cd

e

f

g

Example (continued). Next, reposition vertex a, and no two

edges will intersect.

MTH 231 16

Euler’s Formula

a b

c d

As we know K4 is planar, and can therefore be drawn as above.

Now, let us count the number of vertices, edges and faces we

get.

MTH 231 17

Euler’s Formula

a b

c d

Here, the number of vertices is

v = 4,

the number of edges is

e = 6,and the number of faces is

f = 4

Thus the Euler’s formula for planar graphs holds for K4:

v − e + f = 2

MTH 231 18

Euler’s Formula

As it is shown in Theorem 1 on p.720, the Euler’s formula

v − e + f = 2

holds for all planar graphs.

For example, consider another planar graph:

a

b

cd

e

f

g

Here v = 7, e = 11 and f = 6. So Euler’s Formula holds hereas well:

v − e + f = 7− 11 + 6 = 2

MTH 231 19

Euler’s Formula

Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then

v − e + f = 2

holds for all planar graphs.

Proof: We will use induction on the number of vertices v.

Basis step: v = 1. If there are k loops, then e = k and f = k+1.Therefore

v − e + f = 1− k + (k + 1) = 2confirming the Euler’s formula for the basis step v = 1.

In general, by adding a loop, we always increase e and f by 1,thus keeping v − e + f constant.

MTH 231 20

Euler’s Formula

Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then

v − e + f = 2

holds for all planar graphs.

Proof (continued): We will use induction on the number ofvertices v.

Induction step: Suppose the formula is known for v = n. Weconstruct a connected planar graph G of n + 1 vertices froma connected planar subgraph Gn of n vertices, by adding the(n + 1)-st vertex vn+1, and connecting it to Gn by edges, andadding loops.

n

n+1

G

v

First we add vertex vn+1 andan edge connecting it to Gn.

Note that adding one vertexand one edge and no faces, willnot change v − e + f .

MTH 231 21

Euler’s Formula

Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then

v − e + f = 2

holds for all planar graphs.

Proof (continued): We will use induction on the number ofvertices v.

Induction step: Suppose the formula is known for v = n. Weconstruct a connected planar graph G of n + 1 vertices froma connected planar subgraph Gn of n vertices, by adding the(n + 1)-st vertex vn+1, and connecting it to Gn by edges, andadding loops.

n

n+1

G

v

Next we add more edges con-necting vn+1 to Gn, and loops.

Note that adding one edgeand one face will not changev − e + f .

MTH 231 22

Is K5 “planar” in the toral world?

MTH 231 23

Is K5 “planar” in the toral world?

MTH 231 24

Is K5 “planar” in the toral world?

MTH 231 25

Is K5 “planar” in the toral world?

Yes, K5 is planar in the toral world. That is, it can be drawn ona torus, with no two edges intersecting each other.

However, if we count the vertices, edges, and faces, we see thatthere v = 5, e = 10, and f = 5. Obtaining for K5 on a torus,

v − e + f = 0