Discrete Mathematics with Graph Theory (3rd Edition) 384

# Discrete Mathematics with Graph Theory (3rd Edition) 384 -...

This preview shows page 1. Sign up to view the full content.

382 A 2-coloring of 1i . . . . {b} f R a c d e R W W W Solutions to Review Exercises 91 . . . . {b} as a plane graph 9 Bl FR Finally, we draw 91 . . . . {b} as a plane graph with straight edges (on the left) and then a corresponding floor plan. FR.~ ____ ------~LR DR A 23. A graph is planar if and only if it has no subgraph homeomorphic to JCs or JC3 ,3' The given graph has JC 3,3 as a subgraph, with bipar- tition sets {A, D, F} and {E, G, H}, as shown. E The paths through A,E,C,B,F,G,D,H,A and C E, D, G, F, H, A, B, C, E are Hamiltonian. D Case 1: We draw a graph 91 whose vertices are those of the Hamiltonian cycle A, E, C, B, F, G, D, H, A arranged in a circle. There are five interior edges, which we label
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a, b, c, d, e as shown on the left be-low. Now we draw a graph 1i 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. 91 1i [email protected] n C C G b e ]1>" c B F Since 1i contains a triangle, no 2-coloring exists, but there is a 2-coloring of 1i . . . . {d} as shown on the left. By pulling the white edges outside the circle, 91 . . . . { d} can be redrawn as a planar graph, so removing the single adjacency constraint AG gives the hospital board a floor plan which comes as close as possible to the one desired. ----------------- -----------n w R b W R c 91 . . . . {d} as a plane graph A H B F...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online