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 3-coloring is shown. For the graph on the right, X(Q) = 4. Since the graph contains a copy of lC4, X(Q) ~ 4, and a 4-coloring is shown. Each graph reminds us that the converse to the Four-Color 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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