328
Solutions to Exercises
Solving for
kl'
we get
kl
=
(3 
2)k3
+
(4 
2)k4
+
(5 
2)k5
+ .
.. +
2. This is exactly what we
were asked to establish since (i 
2)k
i
is just 2]deg
Vj

2], tlte sum over all vertices of degree i.
21. [BB] There are no cycles in tlte subgraph since there are no cycles in
CkH2k+2.
Also, given any two
C
vertices, tltere is a patlt between tltem in
CkH2k+2
(because
CkH2k+2
is connected). Any
H
vertex
on tltis patlt would have degree at least 2. Thus, tltere is none; tlte patlt consists entirely of
C
vertices
and hence lies witltin tlte subgraph. The subgraph is connected, hence a tree.
22. (a) Let
x
be tlte number of
H
vertices adjoined. Since Thad
k
 1 edges, and one new edge is added
for each
H,
g
has
(k1)
+x
edges. Therefore, E degvi
=
2(k1+x).
But Edegvi
=
4k+x
since each
C
has degree 4 and each
H
has degree
1.
Therefore,
4k
+
x
=
2k

2
+
2x
and
x
=
2k+2.
23.
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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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