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 2color 7t, we must remove two edges. On the right below, we show a 2coloring of 7t " {a, c}. 91 A B E~H F~G D C c a~ f A 2coloring 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 corresponding floor plan....
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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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