266
Solutions to Exercises
synunetric. Finally, if
U
'"
v
and
v'"
w,
then there is a walk
U
=
UO, Ub
...
,Uk
=
v
from
U
to
v
and
a walk
v
=
vO,vb .
.. ,Ve
=
w
from
v
to
w.
Butthenu
=
UO,Ub .
.. ,Uk
=
V
=
VO,V1, .
.. ,Ve
=
w
is
a walk from
U
to
w,
proving
U
'"
w
and establishing transitivity.
19. No. Let
9
be as shown and let circuit
C
be
ABC D A.
After removal of the edges of
C,
the
remaining graph
91
consists of the 3cycles
AF EA
and
DHGD.
This is not connected.
F
G
20. Let
A
be a vertex in
91
and
B
be a vertex in the circuit
C
1.
Since
9
is connected, there is a walk
in
9
from
A
to
B.
If
the edge
AU1
is in
C
1,
then
A
is a vertex in both
91
and
C
1
and we are done.
Otherwise, the edge
AU1
is not in
C1,
so
AU1
is in
91
and hence
U1
is in
91.
Repeat this argument.
Eventually, either we find a vertex
Ui
in both
91
and
C
1
or else none of the edges along the above path
is in
C
1.
In this case,
B
is in
91
as well as
C
1
and again, we are done.
21. We are given that there is an isomorphism
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 Summer '10
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 Logic, Graph Theory, Vertex, Planar graph, vertices, Vertextransitive graph

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