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
is bounded by
n
edges. Since every edge lies on the boundary
of
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 Summer '10
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 Graph Theory, Planar graph, edges, Andrew Wiles, interior region

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