MATH 239 Assignment 6
This assignment is due at noon on Friday, July 24, 2009,
in the drop boxes opposite the Tutorial Centre, MC 4067.
1. For each of the three connected graphs depicted below, determine if it is planar. If it is planar,
exhibit a planar embedding. If it is not planar, exhibit a subgraph that is an edge subdivision
of
K
5
or
K
3
,
3
.
2
1
3
4
5
6
A
B
C
D
E
F
G
H
t
u
v
w
x
y
z
Solution:
The ﬁrst graph is not planar. It contains the following subgraph which is isomorphic to
K
3
,
3
.
2
1
3
4
5
6
1
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View Full DocumentThe second graph is planar. A planar embedding is as follows.
B
D
H
F
E
G
C
A
The third graph is not planar. It contains the following subgraph which is an edge subdivision
of
K
3
,
3
.
t
u
v
w
x
y
z
2. Let
G
be a connected planar graph with
p
vertices, where
p
≥
3. Suppose that there exists a
planar embedding of
G
having
p
faces.
(a) Let
q
denote the number of edges in
G
. Show that
q
= 2
p

2.
(b) Prove that
G
is not 2colourable.
(c) Show that
G
is sometimes 3colourable, by ﬁnding an example of such a graph
G
which
is 3colourable. Justify your answer.
(d) Show that
G
is sometimes not 3colourable, by ﬁnding an example of such a graph
G
which is not 3colourable. Justify your answer.
Solution:
(a) From Euler’s formula, we have
p

q
+
s
=
c
+ 1. Since
G
is connected, we have
c
= 1,
and by hypothesis, we have
s
=
p
. Therefore
p

q
+
p
= 2, or
q
= 2
p

2.
(b) We ﬁrst show that
G
is not a tree. A tree has 1 face in any planar embedding, and
p
≥
3
by hypothesis, so a tree cannot have
p
faces. Hence
G
is not a tree. We conclude that
G
has at least one cycle.
Note that 2colourable is equivalent to bipartite. If
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 Spring '09
 M.PEI
 Math, Combinatorics, Graph Theory, Planar graph, Graph coloring, Petersen, Four color theorem

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