MATH 222 FALL 2011, Assignment Six
(Answer Key)
1. Prove that if a graph
G
has exactly two vertices of odd degree (and an arbitrary
number of vertices of even degree), then there is a path in
G
joining these two
vertices.
It follows from the Handshaking Lemma that no graph can have exactly one vertex
of odd degree. Thus the two vertices of odd degree in
G
must be in the same
(connected) component, which implies that they are joined by a path in
G
.
2. Find a maximum matching in the graph below. Prove that it is a maximum matching
by exhibiting a minimum vertex cover.
1
2
7
3
5
9
10
8
6
4
Let
M
=
{
14
,
36
,
58
,
710
}
and
C
=
{
3
,
7
,
4
,
8
}
. Then
M
is a matching and
C
is a
vertex cover. Since

M

= 4 =

C

,
M
is a maximum matching and
C
is a minimum
vertex cover.
3. Let
G
= (
V,E
) be a simple graph with

V
 ≥
11. Prove either
G
or the complement
G
must be nonplanar. (Hint: Consider the number of edges in
G
and in
G
.)
Denote
n
=

V

. Suppose that both
G
and
G
are planar. Then
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 Spring '11
 HuangJing
 Math, Graph Theory, Vertex, independent set

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