298
Solutions to Exercises
12. (a) Let VI be any odd vertex. Since VI acquires even degree in
gil,
it is incident with a new edge VI
W.
If
W
is odd, then VI
W
is the desired path. Otherwise,
W
is even in
g
and, still even in
gil,
and
hence incident with another new edge
wu.
If
u
is odd, then VI
wu
is the desired path. Otherwise,
continue until the first odd vertex WI is reached.
(b) Continuing as in (a), let PI be a path of new edges from VI to WI and suppose there is another odd
vertex
v2
(rt
{VI, WI}). Apply the procedure outlined in (a) to this vertex, but with two changes.
If
the path of new edges starting at
V2
reaches a vertex
W
on PI which was even in
g,
then there
must be another new edge incident with
W
(since
W
still has even degree in
gil).
Follow from
W
that edge which was not in Pl. Also, if the first odd vertex encountered on the path of new edges
from
V2
should happen to be VI (or WI or
V2),
then observe that there must be a third new edge
incident with VI (WI,
V2
respectively) because VI has even degree in
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 Summer '10
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 Graph Theory, Vertex, Shortest path problem, new edges, odd vertex

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