Planar Graphs II
Margaret M. Fleck
30 April 2010
This lecture continues the discussion of planar graphs (section 9.7 of
Rosen).
1
Announcements
Makeup quiz last day of classes (at the start of class).
Your room for the final exam (Friday the 7th, 710pm) is based on the
first letter of your last name:
•
AH Roger Adams Lab 116
•
IW Business Instructional Facility (515 E. Gregory opposite Armory)
1001
•
XZ Roger Adams Lab 116
The conflict exam (Monday the 10th, 1:304:30pm) is in 269 Everett Lab.
2
A corollary of Euler’s formula
Suppose
G
is a connected simple planar graph, with
v
vertices,
e
edges, and
f
faces, where
v
≥
3. Then
e
≤
3
v

6.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Proof: The sum of the degrees of the regions is equal to twice the
number of edges. But each region must have degree
≥
3. So we
have 2
e
≥
3
f
. Then
2
3
e
≥
f
.
Euler’s formula says that
v

e
+
f
= 2, so
f
=
e

v
+2. Combining
this with
2
3
e
≥
f
, we get
e

v
+ 2
≤
2
3
e
So
e
3

v
+ 2
≤
0. So
e
3
≤
v

2. Therefore
e
≤
3
v

6.
We can also use this formula to show that the graph
K
5
isn’t planar.
K
5
has five vertices and 10 edges.
This isn’t consistent with the formula
e
≤
3
v

6.
Unfortunately, this trick doesn’t work for
K
3
,
3
, which isn’t
planar but satisfies the equation (with 6 vertices and 9 edges).
3
Another corollary
In a similar way, we can show that if
G
is a connected planar simple graph
with
e
edges and
v
vertices, with
v
≥
3, and if
G
has no circuits of length 3,
then
e
≤
2
v

4.
Proof: The sum of the degrees of the regions is equal to twice the
number of edges. But each region must have degree
≥
4 because
we have no circuits of length 3.
So we have 2
e
≥
4
f
.
Then
1
2
e
≥
f
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 fleck
 Graph Theory, Planar graph, Euler characteristic, Roger Adams Lab

Click to edit the document details