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Section 10.1
267
24. (a)
If
0
has a vertex of degree 0,
0
would have the most edges in the case that each pair of the
remaining
n
 1 vertices were joined with an edge. The graph in all would have at most
(n~
1
) =
(nl)Jn2)
< m edges.
(b) Assume to the contrary that
0
is not connected. Then we can find vertices
u
and
v
not joined to
each other by a walk. In particular,
u
and
v
are not adjacent to any common vertex and they are not
adjacent to each other. We conclude that deg
u
+
deg
v
~
n
 2. [If deg
u
=
k,
then
u
is adjacent
to
k
vertices,
v
is not adjacent to these; neither is it adjacent to
u
or to itself, so deg
v
~
n

k
 2.]
The remaining
n
 2 vertices each have degree at most
n
 2 (none can be adjacent to both
u
and
v)
so the sum of the degrees of all vertices is at most
n

2
+
(n

2)(n

2)
=
n
2

3n
+
2
=
(n

l)(n

2).
Proposition 9.2.5 then says
1£1
~ (nl~n2),
a contradiction.
25. Assume that
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 Summer '10
 any
 Graph Theory

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