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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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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