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MATH 4171 Graph Theory
SPRING 2006
Homework Set VI Solutions
1.
Let
G
= (
V, E
) be a plane graph of minimum degree at least three. Prove that
G
has a
region of size at most Fve.
Proof. Let
d
1
, d
2
, ..., d

V

be the degrees of
G
. Then 2

E

=
d
1
+
d
2
+
...
+
d

V

≥
3

V

, so

V
 ≤
2

E

/
3. Let
s
1
, s
2
, ..., s
r
be the sizes of regions of
G
. Then 2

E

=
s
1
+
s
2
+
...
+
s
r
. If all
regions have size at least six, then 2

E
 ≥
6
r
, so
r
≤ 
E

/
3. However, we deduce from Euler
formula that 2 =

V
  
E

+
r
≤
2
3

E
  
E

+
1
3

E

= 0, a contradiction, which proves that
s
i
≤
5 for at least one
i
.
2.
Decide if the two given
graphs are planar. You need to
justify your conclusions.
Solution.
The Frst graph is
planar.
A redrawing of this
graph is shown on the right.
The second graph is nonplanar,
because deleting the two center
vertices results in a subdivision
of
K
3
,
3
.
x
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This note was uploaded on 02/17/2009 for the course MATH 4171 taught by Professor Lax,r during the Spring '08 term at LSU.
 Spring '08
 Lax,R
 Graph Theory

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