MATH 4171 Graph Theory
SPRING 2006
Homework Set IV Solutions
1.
Let
G
= (
X, Y, E
) be an
r
regular (
r >
0) bipartite graph. Prove that
G
has a perfect matching.
Proof.
Let
V
=
X
∪
Y
. We prove that
β
1
≥

V

2
, which means that
G
has a matching
M
with

M
 ≥

V

2
. Since
M
covers 2

M
 ≥ 
V

vertices, it means that every vertex of
G
is incident with
an edge in
M
, thus, by definition,
M
is a perfect matching.
By K¨onig theorem,
α
=
β
1
. So it is enough for us to show that
α
≥

V

2
. By definition, we need
to show that
α
= min
{
S

:
S
⊆
V
is a vertex cover of
G
} ≥

V

2
. For this, we only need to show
that

S
 ≥

V

2
, for every vertex cover
S
of
G
.
Let
S
⊆
V
be a vertex cover. Then
V
−
S
is an independent set. It follows that
E
(
S, V
−
S
),
the set of edges between
S
and
V
−
S
, consists of all edges that are incident with at least one
vertex in
V
−
S
. Since
G
is
r
regular, this observation implies that

E
(
S, V
−
S
)

=
r

V
−
S

. On
the other hand, the
r
regularity also implies that each vertex in
S
is incident with at most
r
edges
in
E
(
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 Spring '08
 Lax,R
 Graph Theory, Vertex, independent set, Bipartite graph, perfect matching

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