1. There are 4 roads from other cities coming in to vertex A. 2. Graph I is connected because all vertices have at least one path connecting them. 3. This graph is not an Euler circuit because not all edges have been covered. 4. In order for a graph to have an Euler circuit, all vertices must have even valence. Graph II is an Euler circuit because all vertices have even valence. 5. In order for a circuit to be an Euler circuit, each path must be covered once and only once. Since some edges in this graph have been covered twice, this graph cannot be an Euler circuit. 6. This graph is an Euler circuit. Each edge has been covered once and only once and all edges have been covered. 7. Two edges need to be added to this graph to make it an Eulerized graph. Adding the two edges to the vertices with odd valences will make all valences even. 8. Graph D is the best Eulerization of the graph. B is not Eulerized, and A and C have more paths being traveled more than once than graph D does. Since the purpose of Euler circuits is to find the optimal solution, the ideal solution is to have less edges being retraced. 9. Three real world applications in which a worker would want to find an Euler circuit on a street network would be a garbage collector, a street sweeper, and mail delivery. 10.A) An Euler circuit would be ideal. There are a small number of vertices (houses) and finding a path without having to retrace would most likely be simple.
B) An Eulerized graph would be the solution in this case. It would most likely be impossible to inspect every traffic light in a city without retracing at least a few streets. An Eulerized graph would help plan out a path with the least amount of retracing.
C) Again, an Euler circuit would be possible. A small number of vertices to visit would make finding a path without retracing possible.
D) An Eulerized graph would be the best solution in this situation. With all city streets needing to be inspected, it would...