Discrete Mathematics with Graph Theory (3rd Edition) 379

Discrete Mathematics with Graph Theory (3rd Edition) 379 -...

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Chapter 13 (e) The graph is not planar, by Kuratowski's Theorem, since it contains K.3,3 as a subgraph, as shown. 3. (a) The graph is not planar. It has a subgraph homeomorphic to K. 3,3 as shown on the left. B G R 377 (b) Since the graph contains triangles, its chromatic number is at 3. A 3-coloring is shown on the right, so the chromatic number is 3. 4. (a) It cannot contain a subgraph homeomorphic to K.5 or K.3,3 because each of these graphs contain more than one circuit. (b) A near-tree is planar, so V - E + R = 2. There is just one circuit, so R = 2 and V = E. (c) We know that E deg v = 2/E/ = 2/V/. Suppose such a graph has a vertex of degree 3. If all others have degree greater than 2, then E deg v > 2/V/, a contradiction. 5. Exercise 6 of Section 13.1 tells us that 3E :::; 5V - 10. Substituting E = V + R - 2 gives the desired result. 6. (a) E :::;
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Unformatted text preview: 3V - 6 = 3(30) -6 = 84. (b) If the vertices of degree 1 are removed together with the edges with which they are incident, we are left with a subgraph 1i which has 15 vertices, and which is still connected and planar. The number of edges in 1i is at most 3(15) -6 = 39. Since 15 edges were removed from g, the number of edges in 9 is at most 39 + 15 = 54. 7. Here is K.2,3. This is homeomorphic to a graph with four vertices ("remove" a vertex of degree 2) and hence to a subgraph of K.4. 8. True. Since 9 and 1i are homeomorphic, each can be obtained from some graph K. by adding vertices of degree 2. If K. were not connected, then 9 would also not be connected. Also, if K. contained a circuit, so would g. Since 9 is a tree, it follows the K. must also be a tree. But introducing vertices of degree 2 to a tree still leaves a tree. So 1i is a tree....
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