Let's solve the problem step by step: --- ### **Step 1: Draw the Graph** Let's label the vertices as the rooms: \(A, B, C, D, E\) and the outdoors as \(F\). Now, let's list the connections (edges) based on the floorplan: - \(F\) is connected to \(A\) and \(B\) - \(A\) is connected to \(B\) and \(C\) - \(B\) is connected to \(D\) and \(E\) - \(A\) is connected to \(D\) - \(C\) is connected to \(D\) - \(D\) is connected to \(E\) - \(C\) is connected to \(F\) - \(E\) is connected to \(F\) Let's write out all edges: - \(FA, FB, FC, FE\) - \(AB, AD, BC, BD, BE, CD, DE\) But from the diagram, let's clarify the actual connections: 1. \(F\)–\(A\) 2. \(F\)–\(B\) 3. \(F\)–\(E\) 4. \(F\)–\(C\) 5. \(A\)–\(B\) 6. \(A\)–\(D\) 7. \(A\)–\(C\) 8. \(B\)–\(E\) 9. \(B\)–\(D\) 10. \(C\)–\(D\) 11. \(D\)–\(E\) 12. \(C\)–\(E\) (This doesn't exist) From the image, there is **no direct edge** from \(C\) to \(E\). --- ### **Step 2: Check for an Eulerian Circuit** Recall the rule: - **Eulerian Circuit**: All vertices must have even degree. Let's count the degree of each vertex: - **A:** Connected to \(F, B, C, D\) ⇒ degree **4** (even) - **B:** Connected to \(F, A, D, E\) ⇒ degree **4** (even) - **C:** Connected to \(F, A, D\) ⇒ degree **3** (odd) - **D:** Connected to \(A, B, C, E\) ⇒ degree **4** (even) - **E:** Connected to \(F, B, D\) ⇒ degree **3** (odd) - **F:** Connected to \(A, B, C, E\) ⇒ degree **4** (even) So, **C** and **E** have odd degrees (3 each). --- ### **Step 3: Conclusion** Since there are more than 0 vertices with odd degree (in fact, there are 2), **it is NOT possible to find a circuit through the house that uses each doorway once** (i.e., an Eulerian circuit). If there are exactly 2 vertices of odd degree, an **Eulerian path** (not a circuit) exists. But the problem asks for a circuit. --- ### **Final Answer** \[ \boxed{\text{DNE}} \] **Explanation:** There are two vertices (C and E) with odd degree, so an Eulerian circuit does not exist.

Let's analyze the problem step-by-step to determine whether an Eulerian circuit exists in the graph representing the house's rooms and doorways. --- ### **Step 1: Understand the problem** - **Vertices:** Rooms \(A, B, C, D, E\) and outdoors \(F\). - **Edges:** Doorways connecting rooms. - **Goal:** Find a path that uses each doorway exactly once and starts and ends at the same vertex (Eulerian circuit). --- ### **Step 2: Draw and identify the graph** Based on the floorplan, the connections are: - \(F\) connected to \(A, B, C, E\) - \(A\) connected to \(F, B, C, D\) - \(B\) connected to \(F, A, D, E\) - \(C\) connected to \(F, A, D\) - \(D\) connected to \(A, B, C, E\) - \(E\) connected to \(F, B, D\) ### **Step 3: List all edges** - \(F - A\) - \(F - B\) - \(F - C\) - \(F - E\) - \(A - B\) - \(A - C\) - \(A - D\) - \(B - D\) - \(B - E\) - \(C - D\) - \(D - E\) --- ### **Step 4: Count degrees of each vertex** Number of edges incident to each vertex: - \(A\): connected to \(F, B, C, D\) → degree **4** (even) - \(B\): connected to \(F, A, D, E\) → degree **4** (even) - \(C\): connected to \(F, A, D\) → degree **3** (odd) - \(D\): connected to \(A, B, C, E\) → degree **4** (even) - \(E\): connected to \(F, B, D\) → degree **3** (odd) - \(F\): connected to \(A, B, C, E\) → degree **4** (even) ### **Step 5: Determine the existence of an Eulerian circuit** - **Eulerian circuit exists if and only if**: - All vertices have **even degree**. - The graph is connected. In our case: - Vertices \(C\) and \(E\) have **odd degrees** (3). - The other vertices have **even degrees**. Since **not all vertices have even degrees**, **an Eulerian circuit does not exist**. --- ### **Step 6: Check for Eulerian path** - **Eulerian path exists if**: - Exactly **two** vertices have **odd degree**. - The graph is connected. Our graph has **two vertices with odd degree (C and E)**, and the graph is connected. **Therefore**, an **Eulerian path** (a path that uses each doorway exactly once, but does not necessarily start and end at the same vertex) **exists**. --- ### **Step 7: Final answer** Since the question asks specifically for a **circuit** (which requires starting and ending at the same vertex), and our graph does **not** meet the criteria for an Eulerian circuit, the answer is: \[ \boxed{\text{DNE}} \] --- ### **Final conclusion:** **It is not possible to find a circuit through the house that uses each doorway once.** **Answer: DNE**