Chapter 12
357
(c) Let there be
x
vertices of degree 3 and
y
of degree 5. Altogether, then, there are
n
=
2
+
x
+
y
vertices, so
L:
deg
v
=
2(n
1) says 2
+
3x
+
5y
=
2(1
+
x
+
y).
This simplifies to
x
+
3y
=
0,
so
x
=
y
=
0, giving the tree precisely two vertices, a contradiction.
6. (a) Since the sum of the degrees of the vertices is 12, the remaining five vertices must all have degree
1. The vertex of degree 5 must be adjacent to the vertex of degree 2; otherwise, it would be
adjacent to all five vertices of degree
1,
leaving no vertex which could be adjacent to the other
vertex of degree 2. Then the vertex of degree 5 will be adjacent to four of the vertices of degree 1,
and the vertex of degree 2 will be adjacent to the other. The only possibility is shown on the left.
k
BG
kGB
A
DAD
F
E
F
E
(b) Two labeled nonisomorphic trees are shown above.
7. K8 has 8
6
spanning trees, by Theorem 12.2.3.
8. With vertices labeled as shown, the matrix defined in
Kirchhoff's Theorem is
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 Summer '10
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 Graph Theory, vertices, Spanning tree, KGB

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