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Section 10.2
269
(b) The graph in (b) is not Hamiltonian. Both edges incident with a vertex of degree two must be
on any Hamiltonian cycle. Thus, the inner cycle would have to be part of any Hamiltonian cycle.
Since this inner cycle excludes some vertices, it would be a proper cycle, a contradiction. The
graph does have a Hamiltonian path, however. Start at any outer vertex, go around the outside
stopping one short of the vertex where you started, then enter and go round the inner cycle.
(c) The graph in (c) is not Hamiltonian.
If
H were a Hamiltonian cycle,
then it would contain vertices
A
and
J,
these having degree 2, and
hence edges
AB, AG,
JI
and
JK.
Precisely two of the edges incident
with
K
would be in H, hence, precisely one of
DK, EK.
Suppose
DK
were in and
EK
out. Then edges
DE
and
EF
would be part
of1t.
At
this point, two edges incident at
D
are in H, so
DC
could not be in H.
As the only other edges incident with C, both C
B
and C
I
would have
to be H.
D
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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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