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Chapter 10
287
7. (a) This graph is Hamiltonian:
AFGBCDHEA
is a Hamiltonian cycle.
(b) This graph is not Hamiltonian. The comer vertices,
A, G, E,
and
C
each have degree 2, so any
Hamiltonian cycle would have to include the eight edges adjacent with these vertices. Thus any
Hamiltonian cycle would have to include the proper cycle
ABC D EFG H A,
which cannot be.
8. The graph is not Hamiltonian as we explain. Since 1{
were a Hamiltonian circuit. Since vertices
B, E,
and
H
have degree 2, 1{ would have to contain edges
AB,
BC, GH, HI, FE,
and
ED.
Since 1{ cannot contain
a proper cycle, not both edges
AG
and
I C
could be in
1{. By symmetry, we may assume that
IC
is not in
H.
ABC
B
F
E
D
Then, since 1{ contains precisely two edges incident with
C, CD
is in 1{. Since
DC
and
DE
are now
in 1{,
DI
cannot be in
H.
But now only one edge incident with
I
is in 1{, a contradiction.
9. EACBDJGI H F E
is a Hamiltonian cycle.
10. (a) Any Hamiltonian cycle must contain precisely two edges incident with
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 Summer '10
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 Graph Theory

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