378
9. Here
X(Q)
=
4.
To see this, start with vertex
Vl
and proceed
clockwise. Since
Vl,
V2,
V3
fonn
a triangle, they must
be
given
different colors, say red, white and green.
To
avoid the use
of
a
fourth colour, we would have to color
V4
red (since
it
has
already been joined to green and white vertices). But then
V5
must
be
white and
V6
green. Now
V7
is adjacent to red, white
and green vertices, so we need a fourth colour, blue, for this.
10. For the graph on the left,
X(Q)
=
3. Since the graph
contains triangles,
X(Q)
~
3, and a 3coloring is
shown. For the graph on the right,
X(Q)
=
4.
Since
the graph contains a copy
of
lC4,
X(Q)
~
4, and a
4coloring is shown. Each graph reminds us that the
converse to the FourColor Theorem is false since in
Exercise 2 we noted that neither is planar, yet
X(9)
:::;
4 in each case.
11. (a) False. A counterexample is shown.
~
Solutions to Review Exercises
(b) True. Since
9
contains
lC4
as a subgraph, we know that
X(g)
~
4.
But
since
9
is planar,
X(g)
:::;
4,
so
X(Q)
=
4.
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 Summer '10
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 Graph Theory, vertices

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