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Discrete Mathematics with Graph Theory (3rd Edition) 356

Discrete Mathematics with Graph Theory (3rd Edition) 356 -...

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354 Solutions to Exercises 7. (a) [BB] Choose a vertex v. For each vertex u adjacent to v, orient the edge uv in the direction u ---+ v. (b) The answer is yes. Since 9 is an Eulerian graph, there exists an Eulerian circuit. Now just orient the edges of this circuit in the direction of a walk along it. 8. [BB] Yes. To get from a to b in the new orientation, just find the path from b to a in the old orientation and follow it in reverse. 9. (a) [BB] False. The graph shown at the right cannot be given a strongly connected orientation, yet every edge is part of a circuit. (b) True. If u has degree 1 and u is incident with the edge e, then orienting e away from u makes it impossible to get to u from another vertex and orienting e towards u makes it impossible to leave u. Either of these situations contradicts the fact that the graph is strongly connected. (c) True. By Theorem 10.1.4, a connected graph with no odd vertices is Eulerian. Orienting the edges along an Eulerian circuit in the direction of the circuit produces a strongly connected orientation. 10. (a) [BB]
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