254
Solutions
to
Exercises
puts 1 and 5 in the same set, a contradiction.
R
I§}
3
w~:
R
W
(e) This is bipartite, with bipartition sets indicated
R
and
W
on the graph at the right above.
(f)
This is not bipartite because it contains a triangle.
22. (a) [BB]
At
least two
of
the three vertices must lie in one
of
the bipartition sets. Since these two are
joined by an edge, the graph cannot be bipartite.
(b)
A graph which contains an ncycle with
n
odd cannot
be
partite. To see why, let
VI
V2V3
...
V2k+IVI
be
an odd cycle. Then
VI
and
V2
must lie in different bipartition sets, so
VI
and
V3
lie in the same
bipartition set. Continuing this way,
VI,
V3, V5,
...
,V2k+1
all lie in the same bipartition set. Since
VI
and
V2k+1
are adjacent, we have a contradiction.
23. (a)
Yes.
Assign to each vertex
of
the subgraph the bipartition set
it
had in the larger graph. Since each
edge in the subgraph is an edge in the graph, the sub graph can only contain edges joining vertices
in different bipartition sets.
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 Summer '10
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 Graph Theory, Sets, Vertex, vertices, Bipartite graph, Complete bipartite graph

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