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
gf
has
n

2 vertices,
gf
has at most
(n22)
edges, so
(
n
1)
(n

2)
2
+
2 
degu

degv::;
1£1
degu

degv::;
2
and
(
n
1)
(n

2)
deg
u
+
deg
v
~
2

2
+
2
_
(n

1)(n

2)
(n

2)(n

3)
2

2

2
+
(n

2)[(n

1)

(n

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
).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Graph Theory, vertices, Graph theory objects, Michael Watson, Gabriel Andrew Dirac

Click to edit the document details