Math 103, Spring 2011, Solutions to Chapter 7 Homework Assignment
Problem 12
:
(a)
There are many spanning trees in the given network, hence many possible correct answers to the question.
Here is one spanning tree, though there are many more.
The key point is that we must draw a graph which
contains all the vertices of the original graph, and just enough of its edges to be a tree, i.e. connected and without
circuits.
Note that if we drew in one more edge we would create a circuit, whereas if we removed one edge we
would no longer have a network (i.e. our graph would no longer be connected); in either case we would no
longer have a tree.
(b)
To find the redundancy of the given network, compute
M  (N 
1
)
, where
M
equals the number of edges in
the graph, and
N
equals the number of vertices.
In this case
M
= 24 and
N
= 16, so
the redundancy equals
24  (16  1)
=
9
.
Conceptually, this means that 9 is just the right number of edges to remove from the original network to produce
a spanning tree.
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 Spring '07
 Berkowitz
 Math, Graph Theory, Skip, Kruskal's algorithm, Spanning tree, Kruskal

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