1. **Problem statement:** We have a connected graph with vertices A, B, C, D, E, F, G, H, I, J and edges forming triangles and diamond shapes. We need to remove the least number of edges so that the resulting graph has an Euler circuit, and then find such an Euler circuit. 2. **Euler circuit conditions:** A graph has an Euler circuit if and only if it is connected and every vertex has an even degree (an even number of edges incident to it). 3. **Step 1: Identify vertices with odd degree.** - Count the degree of each vertex. - Vertices with odd degree must be fixed by removing edges. 4. **Step 2: Remove edges to make all degrees even.** - Remove the minimum number of edges that change odd degrees to even. - Removing an edge reduces the degree of its two endpoints by 1. 5. **Step 3: Find an Euler circuit in the modified graph.** - Use Fleury's algorithm or Hierholzer's algorithm to find a circuit that uses every edge exactly once. --- **Since the exact edges are not listed, the general approach is:** - Identify odd degree vertices. - Remove edges connecting pairs of odd degree vertices to make their degrees even. - The minimal removal is the minimal matching of odd degree vertices. **Example:** Suppose vertices B, D, G, and I have odd degree. - Remove edges BD and GI (or other pairs) to make all degrees even. **Euler circuit example:** Start at vertex A and follow edges so that each edge is used exactly once, returning to A. --- **Final answers:** - a. Remove edges: BD, GI (example minimal set) - b. Euler circuit: A-B-C-D-E-H-G-F-J-I-H-D-C-B-A (example path) Note: The exact edges depend on the graph's edge list, which is not fully specified here.