9.7 Planar Graphs

Intro problem- 3 houses and 3 utilities

K3,3 problem: Can 3 houses be connected to 3 utilities so that no 2 lines cross?

Similarly, can an isomorphic version of K3,3 be drawn in the plane so that no two edges cross?

T X Y houses U V W utilities

Planar Def

Def:A graph is called planar if it can be drawn in the

plane without any edges drawing.

Sketchpad examples

Check examples--

See Fig01 for K4, K5, K2,3, K3,3

Complete graphs K n

Cycle graphsC n

WheelsW n

Complete bipartite graphs K n,m

Q3, Q4

Check examples—Fig02 for Q3, Q4

n-cubes

Sketch05Find planar representations for the following:

K 4

K 2, 3

3-cube

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ONM

LK

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Sketch06

Is graph ABCDEF planar?

FED

CBA

Sketch07

Is graph ABCDEFGH planar?

HGFE

DCBA

Proof that K3,3 is not planar

see Fig 01 and

Sketch 08 and Math Teacher article

T

U V W

X Y

Proof

Consider a subgraph T 2

U 1 V

X

…proof

Pf. 1: Case 1: W is outside the graph (region 2). This forms region 2a and 2b.Y must be adjacent to U, V, W….T

U V W X

Case 2

Case 2: W is inside the graph (region 1). This forms regions 1a and 1bAgain, Y is adjacent to U, V, and W….

T

U W V

X

Is K5 planar? (see Fig01)

Claim: K5 is nonplanar.

Proof: By contradiction…

Suppose there is a planar representation of K5. 2So v1, v2, v3, v4, v5 form a pentagon.

1 3

5 4{v1,v3} must be present. WLOG, let it be on the inside.Then construct {v2,v4} and {v2,v5} on the outside.So __________ are on the __________

Find # of regions, edges, vertices to discover Euler’s formula

r e v

W3

K2,3

Q3

Theorem 1: Euler’s Formula

Thm: Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar representation of G. Then r = _______

Proof:First, specify a planar representation of G.We will prove by specifying a sequence of subgroups G1, G2, … Ge =G,

adding an edge at each step. This is possible because G is connected.

Arbitrarily pick an edge of G to obtain G1.Obtain Gn from Gn-1 by arbitrarily adding an edge that is incident with

a vertex in Gn-1, adding the other vertex if necessary.

…proof outlineBy induction: Basis: e=1 G1

r 1 = ___ e 1 = ___ v 1 = ___ So_________ Inductive step: Assume n and show n+1. This means:

Assume r n = e n – v n + 2 and add {a n+1, b n+1 } to G n to obtain G n+1 and show ___________…

Case 1: an+1, bn+1 Gn

R is split into 2 regoins. r n+1 = ___ e n+1 = ___ v n+1 = ___ So ___________

Case 2: an+1 Gn but bn+1 Gn

r n+1 = ___ e n+1 = ___ v n+1 = ___

Question: How do you prove a graph is either planar or not planar?

• To prove it is…

• To prove it isn’t…

Corollary 1: e≤3v-6

Corollary 1:If G is a connected planar simple graph with e edges and v vertices where v≥3, then

e≤3v-6. Def: deg(R )= number of edges on the boundary of region R

Proof:Assume G is simple. Therefore it has no loops or multiple edges. So it has no regions of

degree 1 or 2. A planar representation of G has r regions, each of degree at least 3.

Note: 2e = ≥ 3r

So r ≤ (2/3)eUsing Euler’s Theorerm, r = ______ ≤ _____

…

Ex: Use the contrapositive of Corollary 1 to prove that K5 is nonplanar.

Q: Can you use Corollary 1 to show K3,3 is nonplanar?

Cor. 2– region degree ≤ 5A Corollary of Cor. 1 is the following: Cor. 2: If G is a connected planar simple graph, then G has a

vertex of degree not exceeding 5.Proof:• Case 1: G has 1 or 2 vertices: result _______• Case 2: G has at least 3 vertices. By ___, we know e ≤

____ so 2e ≤ ____ • To show result, assume degree of every vertex is ______.

Then because 2e = _____ by _______, we have 2e ≥ 6v (why?).

• But this contradicts ___. So there must be a vertex with degree ≤ 5.

Corollary 3: e ≤ 2v – 4Corollary 3: If G is a connected planar simple graph with e

edges, v vertices, v≥3, and no circuits of length 3, then e≤2v – 4.

Proof .Assume G is simple. Consider a planar representative of G.Therefore it has no loops or multiple edges, which would

create regions of degree 1 or 2.With no circuits of length 3, there are no regions of degree 3.Therefore, all regions are at least degree 4.So 2e = ________ ≥ 4rSolving for r…By Euler’s Formula…

K3,3

Ex: Use Corollary 2 to prove that K3,3 is nonplanar.

Is Q4 planar or not? Prove.

Kuratowski’s Theory

Def: Replacing {u,v} with {u,w} and {w,v} is an elementary subdivision. Def: G1=(V1,E1) and G2=(V2,E2) are homeomorphic if they can be

obtained from the same graph by a sequence of elementary subdivisions.

Kuratowski’s Theorem:A graph is nonplanar iff it contains a subgraph homeomorphic to K3,3

or K5. Proof:clearbeyond scope of class

examples

• Are the following planar or not? Why?

Peterson ex Ex: Use the two Euler Corollaries on the Peterson example (See examples in notes)

Cor. 1: e ≤ 3v - 6

Cor. 3: e ≤ 2v - 4

Kuratowski’s Theory

Ex: Use Kuratowski’s Theory on the Peterson example.

see written handout

More ex

• See handout for more examples using Euler and Kuratowski– See sketch06 and

sketch07 on sketchpad to the right

– See attached Sketchpad handout with 7 more ex

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HGFE

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More – handout ex 1-2

21 Planar?C

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Handout 3-4

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Handout 5-7

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Ex- book 7, 23A

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