Discrete Mathematics with Graph Theory (3rd Edition) 382

Discrete Mathematics with Graph Theory (3rd Edition) 382 -...

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380 Solutions to Review Exercises Below we draw 91 " {a} as a plane graph and then as a plane graph with straight edges. Finally, we produce the corresponding floor plan (which even satisfies an requirement not requested). 7 8 1 2 ::l c.... L 6 5 5 3 1 2 (b) We label the vertices of 9 7 4 as shown on the left and redraw it as 91. 4 8 5 8 9 91 6 3 5 a a The chromatic number of f R the new graph 1i is 3 because 1i contains e triangles. By removing vertex f, we obtain a R 0 graph whose chromatic e 1i" {f} number is 2. d d We draw 91 " {f} as a plane graph and then as a plane graph with straight edges. Finally, we
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Unformatted text preview: produce a floor plan. 1 2 1 2 7 <P---~4 7 8 8 21. (a) By Kuratowski's Theorem, the graph here is not planar: It contains K 3,3 as a subgraph, with bipartition sets {A, B, F} and {C, D, E}. (b) Call the graph 9. ACBEGFDA is a Hamiltonian cycle. On the left, we show these vertices in order on a circle and draw 9 again. Call the new graph 91. Next, we label the interior edges, a, b, e, d, e, and f, and draw a graph 1i (on the right) whose vertices correspond to a, ... , f....
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