We start by constructing an undirected graph with five vertices labeled A, B, C, D, and E. The edges are AB, BC, CD, DE, EA, AC, BA, and EE. Note that BA is the same as AB in an undirected graph, and EE represents a self-loop at vertex E.
Now let's analyze the degree of each vertex. Vertex A has degree 3 connected to B, C, and E. Vertex B has degree 2 connected to A and C. Vertex C has degree 3 connected to B, D, and A. Vertex D has degree 2 connected to C and E. Vertex E has degree 3 with connections to D, A, and a self-loop.
According to Euler's theorem, a connected graph has an Euler circuit if and only if every vertex has even degree. In our graph, vertices A, C, and E have odd degrees of 3, while vertices B and D have even degrees of 2. Since we have three vertices with odd degree, this graph does not have an Euler circuit.
For a semi-Eulerian path to exist, a graph must have exactly two vertices with odd degree. However, our graph has three vertices with odd degree: A, C, and E. Since we have more than two odd-degree vertices, no semi-Eulerian path exists either. Therefore, this graph has neither an Euler circuit nor a semi-Eulerian path.
In conclusion, our undirected graph with vertices A, B, C, D, E has three vertices with odd degree. According to Euler's theorems, a graph needs all even-degree vertices for an Euler circuit, or exactly two odd-degree vertices for a semi-Eulerian path. Since we have three odd-degree vertices, this graph has neither an Euler circuit nor a semi-Eulerian path.