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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)
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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

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