1. **Problem Statement:** Determine if the given graph with vertices A, B, C, D, E, F, G, H, I, J, K and specified edges has an Eulerian cycle or an Eulerian path. 2. **Definitions:** - An **Eulerian cycle** is a cycle that uses every edge exactly once and starts and ends at the same vertex. - An **Eulerian path** is a path that uses every edge exactly once but does not necessarily start and end at the same vertex. 3. **Key Theorems:** - A connected graph has an Eulerian cycle if and only if every vertex has an even degree. - A connected graph has an Eulerian path (but not a cycle) if and only if exactly two vertices have odd degree. 4. **Calculate the degree of each vertex:** - $\deg(A) = 3$ (connected to B, C, K) - $\deg(B) = 1$ (connected to A) - $\deg(C) = 2$ (connected to A, E) - $\deg(D) = 3$ (connected to I, K, E) - $\deg(E) = 6$ (connected to D, H, J, F, G, C) - $\deg(F) = 1$ (connected to E) - $\deg(G) = 1$ (connected to E) - $\deg(H) = 2$ (connected to E, J) - $\deg(I) = 1$ (connected to D) - $\deg(J) = 2$ (connected to H, E) - $\deg(K) = 2$ (connected to A, D) 5. **Count vertices with odd degree:** - Odd degree vertices: A(3), B(1), D(3), F(1), G(1), I(1) - Number of odd degree vertices = 6 6. **Check connectivity:** - The graph is connected as all vertices are reachable through edges. 7. **Conclusion:** - Since more than two vertices have odd degree, the graph does **not** have an Eulerian cycle. - Since the number of odd degree vertices is not exactly two, the graph does **not** have an Eulerian path either. **Final answer:** The graph has neither an Eulerian cycle nor an Eulerian path because it has 6 vertices with odd degree, which violates the necessary conditions for both.