Section
10.1
265
12. (a) [BB]
Kn
is Eulerian
++
n
is odd.
(b) [BB]
Kn
has an Eulerian trail
if
and only
if
n
=
2.
For
n
=
2,
certainly
00
has an Eulerian
trail. For
n
>
2,
if
two vertices have odd degree, then there are other vertices
of
odd degree so no
Eulerian trail can exist.
13. (a) Since
Km,n
is connected, we seek only necessary and sufficient conditions for every vertex to be
even.
In
Km,n,
every vertex in the bipartition set with m vertices has degree
n
and every vertex
in the other bipartition set has degree
m.
Hence,
Km,n
is Eulerian
if
and only
if
m and
n
are both
even.
(b)
Km,n,
m
:::;
n
has an Eulerian trail
if
and only
if
m
=
1,
n
=
2
or
m
=
2 and
n
is odd.
By Theorem to.1.5, complete bipartite graphs with the specified conditions on m and
n
have
Eulerian trails. On the other hand, since
Km,
n
has
n
vertices
of
degree m and m
of
degree
n,
if
m
>
2,
then each bipartition set contains more than two vertices (since
n
~
m)
so
it
is impossible
for there to exist precisely two vertices
of
odd degree.
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 Summer '10
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 Graph Theory, Existence, vertices, Graph theory objects, Eulerian path, Eulerian Trail

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