NOTES FOR MATH 481LECTURE 26
VIVEK DHAND
1.
Bipartite Graphs
A graph
G
= (
V,E
) is
bipartite
if we can split up the vertices into two sets
X
and
Y
sothat every edge has one endpoint in
X
and the other endpoint in
Y
. In other words,no two vertices in
X
are adjacent, and no two vertices in
Y
are adjacent. We will referto the sets
X
and
Y
as the
partite sets
of the bipartite graph.
Theorem.
Let
G
be a graph.
G
is bipartite if and only if it contains no odd cycles.
Proof.
Let
G
be a bipartite graph. Suppose (
v
0
,...,v
k
) is a cycle in
G
. Since
v
i
and
v
i
+1
are adjacent, they must lie in diﬀerent partite sets. So
v
1
,v
3
,v
5
,...
must lie inone partite set and
v
0
,v
2
,v
4
,...
must lie in the other. Since
v
k
−
1
and
v
0
are adjacent,we see that
k
−
1 must be odd, so
k
must be even. This proves that bipartite graphscannot contain odd cycles.Now suppose
G
is a graph which does not contain any odd cycles. Note that
G
=
K
1
is clearly bipartite and does not contain any odd cycles (a cycle must have at least3 vertices). Therefore, we may assume that
G
has at least two vertices. Also,
G
isbipartite if and only if each connected component of
G
is bipartite, so we may assumethat
G
is connected. Pick a vertex
v
and deﬁne
X
=
{
x
∈
V

the shortest path from
x
to
v
has even length
}
Y
=
{
y
∈
V

the shortest path from
y
to
v
has odd length
}
We claim that these sets
X
and
Y
make
G
into a bipartite graph. Let
x
1
and
x
2
be vertices of
X
and suppose they are adjacent. Note that
v
is not adjacent to
x
1
,(otherwise the shortest path between them would be length one), so
v
=
x
2
. Similarly,
v
is not adjacent to
x
2
, so
v
=
x
1
. Let
P
1
= (
v,v
1
,...,v
2
k
) be a shortest path from
v
to
v
2
k
=
x
1
and let
P
2
= (
v,w
1
,...,w
2
l
) be a shortest path from
v
to
w
2
l
=
x
2
. Notethat
P
1
and
P
2
have even length.If
P
1
and
P
2
have no vertices in common except
v
, then we have found an odd cycle:(
v,v
1
,...,x
1
,x
2
,...w
1
,v
).If
P
1
and
P
2
do have common vertices in addition to
v
, then let
v
be the last suchvertex. Note that taking
P
1
from
x
1
to
v
is the shortest path from
x
1
to
v
(if therewas a shorter path, we could just take that path and then follow the rest of
P
1
for a
2 VIVEK DHAND
shorter path to
v
). The same goes for
P
2
: it is the shortest path from
x
2
to
v
. Now,the length of the routes that
P
1
and
P
2
take from
v
to
v
must be the same. If one wasshorter than the other, we would be able to shorten either
P
1
or
P
2
. Therefore, we mayassume that
v
=
v
i
=
w
i
for some
i
. This implies that (
v
,v
i
+1
,...,x
1
,x
2
,...,w
i
+1
,v
)is an odd cycle.The same proof works for adjacent vertices
y
1
,y
2
∈
Y
, except that
P
1
and
P
2
will haveodd length.We have shown that each connected component of
G
is bipartite, so
G
is bipartite.
Example.
K
n,m
is the
complete bipartite graph
. It is the bipartite graph of maximalsize with

X

=
n
and

Y

=
m
.