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
UI,
U2,
U3.
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 Summer '10
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 Graph Theory, Addition, shortest path, vertices, odd degree

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