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318
Chapter
11
Review
1. There are four odd vertices, labeled
A, B, C,
D
in the figures to the right. In the
unweighted case, at least two additional
edges are required;
in
fact, two suffice, as
shown. The table shows the calculations
required for the weighted graph.
It
is
necessary to duplicate four edges, as shown.
Solutions to Review Exercises
ABC
CffP
3
3
1
Weighted Case
Partition into pairs
Sum of lengths of shortest paths
{A,B},{C,D}
{A, C}, {B, D}
{A,D},{B,C}
3+4=7
4+3=7
4+1=5
2. This graph has exactly two vertices of odd degree,
C
and
G.
The shortest path from
C
to
G
is
CADG,
of length 8. (This follows from Exercise 27 of the Review Problems for Chapter 10; it also follows
quickly by inspection.) Hence, we solve the problem by duplicating edges
CA, AD
and
DG.
3. Let us label the bipartition sets
{UI, U2, U3}
and
{v!,
V2, V3, V4, V5, V6}.
Our graph has six vertices of
odd degree, namely,
VI, •.
.
,V6.
The shortest path between any two of these will always be of length
two, passing through one of
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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, Addition

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