This preview shows page 1. Sign up to view the full content.
Section 13.2
365
22. (a) [BB] Say at most one region has at most five edges on its boundary. Then, with
N
as in the
proof of Corollary 13.1.3,
N
;:::
6(R

1). But
N
~
2E,
so
2E
;:::
6R

6,
3R
~
E
+
3.
Since
V

E
+
R
=
2, 6
=
3V

3E
+
3R
~
3V

2E
+
3; that is,
2E
~
3V

3. But
2E
=
L:
deg
Vi
;:::
3V
by assumption, and this is a contradiction.
(b)
If
gl
and
g2
are any two connected components of a planar graph, then each is planar, so each has
at least two regions with at most five edges on the boundary, by part (a). Therefore, each has at
least one
interior
region with at most five edges on the boundary. The result now follows because
any interior region of a component is an interior region of the entire graph.
23. The given polygon has
n
vertices and
n
edges, so including possible new vertices introduced by tri
angulation, the graph produced by triangulation has
V
;:::
n
vertices. This graph is a plane graph,
so
V

E
+
R
=
2. In it, every interior region is bounded by three edges while the exterior region
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