Answer
ME,NH,VT,MA,CT,RI,MA,NH is an Euler path.
Work Step by Step
Vertex ME and vertex NH are odd vertices. The other vertices are even vertices. This graph has exactly two odd vertices.
When a connected graph has exactly two odd vertices, then it has at least one Euler path. Any Euler path must begin at one odd vertex and end at the other odd vertex.
Let's start at vertex ME. The only possible move from here is to travel to vertex NH.
According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges NH-VT and NH-MA are not bridges, we can choose either of these edges as the next step in the path.
From vertex NH, the path can travel to vertex VT, and then to vertex MA.
Now, we can see that the edge MA-NH is a bridge, so according to Fluery's Algorithm, we must choose a different edge.
The path can travel to vertex CT. After this step, the path must then travel to vertex RI, then to vertex MA, and then finally to vertex NH, because these are the only available edges.
This path is ME,NH,VT,MA,CT,RI,MA,NH. This path travels through every edge of the graph exactly once, so it is an Euler path.
This is one Euler path but there are other Euler paths in this graph also.
