Discrete Mathematics with Graph Theory (3rd Edition) 385

Discrete Mathematics with Graph Theory (3rd Edition) 385 -...

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Chapter 13 383 Case 2: We draw a graph 91 whose vertices are those of the Hamiltonian cycle E, D, G, F, H, A, B, C, E arranged in a circle. There are five interior edges, which we label a, b, c, d, e as shown on the left be- low. Now we draw a graph 7t whose vertices are labeled a, b, c, d, e and two vertices are joined by an edge if and only if the corresponding edges cross in 91. E C D~B G~A F H c Since 7t contains a triangle, no 2-coloring exists, but there is a 2-coloring of 7t " {e} as shown on the left. By pulling the white edges outside the circle, 91 ,,{ e} can be redrawn as a planar graph, so removing the single adjacency constraint DH gives the hospital board a floor plan which comes as close as possible to the one desired. 24. (a) Kuratowski's Theorem says that a graph is planar if and only if it has no subgraph homeomorphic to K5 or to K 3,3.
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Unformatted text preview: The given graph is not planar because it has a sub graph isomorphic to K 3,3 with bipartition sets {A, C,H} and {B, G,E} as shown. 91 " {e} as a plane graph E C c A B (b) One Hamiltonian cycle is A, B, H, G, C, D, F, E. We redraw 9 as a graph 91 by putting these vertices around a circle and labeling interior edges a, b, c, d, e, f. We then draw a new graph, 7t, whose vertices are labeled a, b, c, d, e, f and an edge indicates that the corresponding edges of 91 cross. In order to 2-color 7t, we must remove two edges. On the right below, we show a 2-coloring of 7t " {a, c}. 91 A B E~H F~G D C c a~ f A 2-coloring of 7t, {a, c} b e ; f Then we redraw 91 " {a, c} as a planar graph by pulling white edges d and e outside the circle. (c) Finally, we draw 91 " {a, c} as a plane graph with straight edges (on the left) and then a corre-sponding floor plan....
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