Question

part 2 of 3
score: 79.41%, 13.5 of 17 points
points: 0 of 1
the layout of a city with land masses and bridges is shown. use this map to answer the questions below.
...
a. it is possible to draw a graph that models the layout of the city, where vertices represent land masses and edges represent bridges. choose the graph below that models the city
(options: four graphs with vertices a, b, c, d)
b. use the graph from part (a) to answer the following question
is it possible for city residents to walk across all of the bridges without crossing the same bridge twice?
options: yes, no

Explanation:

Step1: Analyze the graph structure for part (a)

In graph - theory, when vertices represent land - masses and edges represent bridges. We need to check the number of edges (bridges) between vertices (land - masses).
For the first graph: Vertex \(A\) has degree \(2\), vertex \(B\) has degree \(3\), vertex \(C\) has degree \(2\), vertex \(D\) has degree \(1\).
For the second graph: Each vertex (\(A\), \(B\), \(C\), \(D\)) has degree \(2\).
For the third graph: Vertex \(A\) has degree \(2\), vertex \(B\) has degree \(2\), vertex \(C\) has degree \(2\), vertex \(D\) has degree \(2\).
For the fourth graph: Vertex \(A\) has degree \(2\), vertex \(B\) has degree \(3\), vertex \(C\) has degree \(3\), vertex \(D\) has degree \(2\).

Assuming the correct graph is the first one (since it has the appropriate number of connections based on the bridge - landmass relationship in the city layout).

Step2: Use Euler's theorem for part (b)

Euler's theorem states that a graph has an Eulerian path (a path that traverses every edge exactly once) if and only if it has at most two vertices of odd degree.
In the graph (assuming the first graph from part (a) where vertex \(B\) has degree \(3\) (odd) and vertex \(D\) has degree \(1\) (odd)).

Answer:

a. The first graph (assuming based on bridge - landmass connection analysis).
b. Yes (because the graph has exactly two vertices of odd degree (\(B\) and \(D\)), so there is an Eulerian path which means it is possible to walk across all bridges without crossing the same bridge twice).