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358
Solutions to Review Exercises
11. First select edges
AD, BE, CF, GJ
and I
J
of
weight 1. Then select edges
AB, CG,
I
K
of
weight 2. Since
G1
would complete a circuit, we
cannot select this edge. Since
J
K
and
FG
would
also complete circuits, we cannot select these
either. We select next
CE
of weight 5. Since
BC
and
EF
would complete circuits, we cannot
select them, so we next select
HI
or
H K
of
weight 7. We now have 10 edges, the necessary number for a spanning tree, so we are done. The
weight of our tree is 5
+
6
+
5
+
7
=
23.
12. Let us start at vertex
A.
We first select
AD,
then
AB.
Next
BE
is selected and following that comes
EC.
Then come
CF
and
CG.
Next we select
GJ
and then
J1.
Then comes
1K
and finally
1H.
The
tree obtained by Prim is the same as that obtained by Kruskal in Exercise 11.
13. We use Kruskal's algorithm. The edge of least weight is
VI V2,
of weight 3 and the next edge chosen is
VI v3,
of weight 4. There are two edges of weight 5,
VI V4
and
V2V3,
but
V2V3
will complete a circuit. We
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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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