To solve the problem:

1. Eulerize the Graph:

   • Eulerize the graph by ensuring all vertices have an even degree. You can add duplicate edges (multi-edges) between vertices to achieve this. The goal is to do this efficiently, adding the fewest number of edges possible.

2. Current Degrees:

   • Check the degrees of each vertex in the graph:
     • A: degree = 4 (even)
     • B: degree = 3 (odd)
     • C: degree = 3 (odd)
     • D: degree = 3 (odd)
     • E: degree = 3 (odd)

3. Fix Odd Degrees:

   • Pair up the odd-degree vertices. Adding edges between them makes their degree even:
     • Connect B to C.
     • Connect D to E.

   This results in all vertices having an even degree.

4. Find the Euler Circuit:

   • Now that the graph is Eulerized, find an Euler circuit (a path that visits every edge exactly once and starts and ends at the same vertex). One possible Euler circuit is:
     • A → B → C → E → D → A → E → B → D → C → A

Final Answer:

The Euler circuit is ABCEDAEBDCA.

Let me know if you'd like detailed steps or have further questions.

Related Questions:

1. What is the degree of each vertex after Eulerizing the graph?
2. Can you explain why adding the edges B–C and D–E was efficient?
3. How would you handle Eulerization if the graph had disconnected components?
4. What algorithm is commonly used to find an Euler circuit?
5. Can you generalize the conditions required for a graph to have an Euler circuit?

Tip:

When Eulerizing, aim to add edges between vertices that are close together to minimize additional path length.