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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.
1
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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