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Discrete Mathematics with Graph Theory (3rd Edition) 377

Discrete Mathematics with Graph Theory (3rd Edition) 377 -...

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Section 13.3 (c) 91 1 ~2 3~7 6 4 a Ji ert~J' d c Ji is not 2-colorable since it contains a triangle. If e is removed, we get the graph at the right. Shown are the graph 9' and a possible floor plan. ~, 3~7 13. A graph depicting this situation is shown. Ji a B~H LR BA DR 9 c * b Ii B1 'f K f d e FR 375 Since Ji has a triangle, X(Ji) = 3 and it is impossible to construct the desired house. We could make X(Ji) = 2 by removing any of d, g, h. The most practical of these to remove might be h in which case 91 becomes as shown on the left. One possible corresponding house plan is at the right.
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Unformatted text preview: FR r61 H LR BA ~ r--K DR 14. (a) [BB] We can immediately draw 91 for this problem (using the obvious Hamiltonian cycle that a golfer must follow). Both 91 and Ji are shown. 91 ~~ 2 Ji~ ga b 8 3 c f 7 4 ad e 6 5 Since Ji contains a triangle, X(Ji) = 3, so such a course is impossible. (b) X(Ji) would equal 2 if a, cor e were removed from Ji, that is, if we removed one of the restrictions that holes 1 and 5 share a common border, or that holes 2 and 6, or holes 4 and 8 share a common border....
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