{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 367

Discrete Mathematics with Graph Theory (3rd Edition) 367 -...

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

Section 13.2 365 22. (a) [BB] Say at most one region has at most five edges on its boundary. Then, with N as in the proof of Corollary 13.1.3, N ;::: 6(R - 1). But N ~ 2E, so 2E ;::: 6R - 6, 3R ~ E + 3. Since V - E + R = 2, 6 = 3V - 3E + 3R ~ 3V - 2E + 3; that is, 2E ~ 3V - 3. But 2E = L: deg Vi ;::: 3V by assumption, and this is a contradiction. (b) If gl and g2 are any two connected components of a planar graph, then each is planar, so each has at least two regions with at most five edges on the boundary, by part (a). Therefore, each has at least one interior region with at most five edges on the boundary. The result now follows because any interior region of a component is an interior region of the entire graph. 23. The given polygon has n vertices and n edges, so including possible new vertices introduced by tri- angulation, the graph produced by triangulation has V ;::: n vertices. This graph is a plane graph, so V - E + R = 2. In it, every interior region is bounded by three edges while the exterior region is bounded by n edges. Since every edge lies on the boundary of
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online