288
Solutions to Review Exercises
Case 3a.
AD
is in
1i
while
AE
is not. Since
1i
contains precisely two edges incident with
E,
both
BE
and
DE
are in
1i
and we have a proper cycle in
1i.
Case 3b.
AE
is in
1i
while
AD
is not. This time
BE
is not in
1i
(else,
1i
would contain a proper
cycle). We need two edges incident with
E
and with
B,
so both
DE
and
BD
are in
1i,
so
1i
does not contain a path from
D
to any ofthe vertices
F,
G,
H, I,
J.
This concludes the proof.
11. (a)
2n
edges.
(b)
If
a graph is Hamiltonian, any vertex is the starting point of a Hamiltonian cycle. So let
v
be a
fixed vertex. There are
n
choices for the second vertex in the cycle, then
n

1 for the third,
n

1
for the fourth (which is in the bipartition set not containing
v),
and so on. Altogether, there are
n[(n

1)!]2 different Hamiltonian cycles in
Kn,n.
2
(c)
Kn n
has
n
2
edges and each cycle requires
2n
edges, so the maximum is
~
= !!:
edge disjoint
,
2n
2
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 Summer '10
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 Graph Theory, Glossary of graph theory, hamiltonian cycle, Graph theory objects, hamiltonian cycles, Eulerian

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