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274
Solutions to Exercises
with
u
or
v),
we are left with a graph
gf
with
1£1
degu
 degv edges. (By
£,
we mean the set of
edges of
g.)
Since
has
n
 2 vertices,
has at most
(n22)
edges, so
(
n
1)
(n

2)
2
+
2 
 degv::;
2
and
(
n

deg
u
+
v
~
2

2
+
2
_ (n

1)(n
 2)
(n

2)(n
 3)
2

2

2
+
(n

2)[(n

1) 

3)]
=
+2=n2+2=n
2
and the result follows.
18. [BB] Add a new vertex adjacent to all existing vertices and apply the result of Exercise 16.
19. (thanks to Michael Watson) Let
v
and
w
be two adjacent vertices. Each of these vertices is adjacent to
at least
~
 1
=
n
2
2
other vertices.
If
n
is odd, this means that both
v
and
w
are adjacent to at least
n
2
1
other vertices. Since there are only
n
 2 other vertices in all, it follows that they must both be
adjacent to a common vertex, and we have a triangle.
So assume
n
is even and that
9
does not
contain a triangle. Then
v
must be adja
cent to exactly half of the remaining
n
 2
vertices (let 8
1
be the set of such vertices)
and
w
must be adjacent to the other half
(denote this set 8
2).
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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

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