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336
(c) The edge ofleast weight incident with
E
is
CEo
The edge of
least weight among those adjacent to
CE
is
SC.
The edge of
least weight among those which, together with
CE
and
SC,
fonn a tree is
BC.
Then we must add, in order,
DE, AD,
and
DT.
Finally, so as not to duplicate the tree we obtained in
Exercise 1, we select C
F.
(d) Starting at vertex
E,
we select first edge
G E
of weight 1 and
then
GF
(to change the tree we had before). Of the edges
incident with
G, E
and
F,
those of least weight are
FE
and
GC.
We cannot select
FE
because it is incident with
two
of
the vertices already chosen, so we choose
GC,
then
BC, CD
and
AF.
We obtain another spanning tree of weight 10.
Solutions to Exercises
A
2
D
~T
C
6
F
B
C
D
F
E
3. [BB] We need five edges because there are six vertices in the
graph. We use Kruskal's algorithm. First select
BC
of weight
5, then
DF
of weight 6. Next we select
BF
and
AB,
both of
weight 7. We would like to select
AF
next, but cannot since it
completes the circuit
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 Summer '10
 any
 Graph Theory

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