362
(0
This is not planar.
It
has a subgraph
homeomorphic to Ks as shown.
Solutions to Exercises
6. Let
N
be as described in the proof of Corollary 13.1.3. Then
5R
~
N
~
2E,
so
5R
~
2E.
Thus,
5(E

V
+
2)
~
2E,
giving
3E
~
5V

10.
7. [BB] We know from Theorem 13.1.4 that
E
~
3V

6. Substituting in
E
=
V
+
R

2, we obtain
V
+
R
 2
~
3V
 6, or
R
~
2V
 4, as required.
8. (a) [BB]
A
E
=
3
=
3(3)  6
=
3V
 6.
(b) Look at the proof of Theorem 13.1.4. We can only have
E
=
3V

6 if
3R
=
N
=
2E,
and
3R
=
N
means that every region is a triangle.
9. (a)
If
g
has three vertices, it must have at least two edges since it's connected.
It
can't have three
edges since it is not a triangle. Thus,
E
=
2
~
2
=
2V
 4 as required.
If
g
has four vertices,
it has at least three edges. Since it contains no triangles, it is either a tree or a 4cycle. In either
case,
E
~
4
=
2V
 4 as required.
Now assume
V
~
5, so
E
~
4. As in the proof of Theorem 13.1.4, count the edges on the
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details