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
Complete graphs K n
Check
examples-See Fig01 for
K4, K5, K2,3,
K3,3
Cyc le graphs
Cn
W heels
W n
Complete bipartite graphs K n,m
Q3, Q4
Check examples—
Fig02 for Q3, Q4
n-c ubes
Sketch05
Find planar representations for the following:
A
B
D
C
K 4
I
E
H
G
F
K 2, 3
N
J
O
M
P
K
L
3-c ube
Sketch06
Is graph A BCD EF planar?
A
B
C
D
E
F
Sketch07
Is graph ABCD EFGH planar?
A
E
C
B
F
G
D
H
Proof that K3,3 is not planar
see Fig 01
and
Sketch 08
and
Math Teacher
article
T
U
X
V
Y
W
Proof
Consider a subgraph
T
U
1
X
2
V
…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
X
W
Case 2
Case 2: W is inside the graph (region 1). This
forms regions 1a and 1b
Again, Y is adjacent to U, V, and W….
T
U
W
X
V
Is K5 planar? (see Fig01)
Claim: K5 is nonplanar.
Proof: By contradiction…
Suppose there is a planar representation of K5.
So v1, v2, v3, v4, v5 form a pentagon.
1
2
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 __________
3
Find # of regions, edges, vertices to
discover Euler’s formula
r
W3
K2,3
Q3
e
v
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 outline
By 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)e
Using 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 ≤ 5
A 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 – 4
Corollary 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 = ________ ≥ 4r
Solving 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:
clear
beyond 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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E
A
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C
D
E
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G
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More – handout ex 1-2
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C
1
2
Planar?
B
G
H
Q
R
F
A
N
D
E
P
Handout 3-4
S
M
3
J
K
L
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4
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Handout 5-7
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5
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Z
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A
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6
A
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7
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Ex- book 7, 23
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A
G
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E