Let's solve the art gallery (vertex cover) problem step by step: ### **Step 1: Define Variables** Let \( x_j \) be a binary variable indicating if an attendant is placed at door \( j \), for \( j = 1, 2, 3, 4, 5, 6 \). \[ x_j = \begin{cases} 1 & \text{if attendant is placed at door } j \\ 0 & \text{otherwise} \end{cases} \] ### **Step 2: Model Constraints** Each **room** is covered by the two adjacent doors. We need to ensure **every room** is supervised (i.e., at least one attendant at its doors). Label the **rooms** (for convenience): - Top left: Room A - Top right: Room B - Bottom left: Room C - Bottom right: Room D From the diagram, the **doors** are as follows: - Door 1: between Room A and Room C (left vertical) - Door 2: between Room A and Room B (top horizontal) - Door 3: between Room C and Room D (bottom horizontal) - Door 4: between Room B and Room D (right vertical) - Door 5: between Room A and Room D (middle vertical) - Door 6: between Room B and Room C (middle horizontal) But let's clarify which doors are adjacent to each room: #### **Room A (top left):** - Door 1 (left) - Door 2 (top) - Door 5 (middle vertical, between A and D) #### **Room B (top right):** - Door 2 (top) - Door 4 (right) - Door 5 (middle vertical) - Door 6 (middle horizontal, between B and C) #### **Room C (bottom left):** - Door 1 (left) - Door 3 (bottom) - Door 6 (middle horizontal) #### **Room D (bottom right):** - Door 3 (bottom) - Door 4 (right) - Door 5 (middle vertical) - Door 6 (middle horizontal) But from the image, we see that not all rooms are adjacent to all doors, so let's match them precisely by the boundaries: - **Room 1 (top left):** Door 1 (left), Door 2 (top) - **Room 2 (top right):** Door 2 (top), Door 5 (right) - **Room 3 (bottom left):** Door 1 (left), Door 3 (bottom) - **Room 4 (bottom right):** Door 3 (bottom), Door 5 (right) - The vertical doors in the middle (door 4, door 6) are between rooms (central position), so let's match them accordingly. Let's label the doors as per the diagram: - Door 1: left vertical (between rooms 1 and 3) - Door 2: top horizontal (between rooms 1 and 2) - Door 3: bottom horizontal (between rooms 3 and 4) - Door 4: bottom left vertical (between rooms 3 and 4) - Door 5: top right vertical (between rooms 2 and 4) - Door 6: middle horizontal (between rooms 2 and 4) Now, let's clarify which doors cover which rooms: | **Room** | **Doors** | |----------|---------------| | 1 | 1, 2 | | 2 | 2, 5, 6 | | 3 | 1, 3, 4 | | 4 | 3, 4, 5, 6 | ### **Step 3: Write Constraints** Let \( x_j \) be the variable for door \( j \), for \( j = 1, 2, 3, 4, 5, 6 \). - Room 1: \( x_1 + x_2 \geq 1 \) - Room 2: \( x_2 + x_5 + x_6 \geq 1 \) - Room 3: \( x_1 + x_3 + x_4 \geq 1 \) - Room 4: \( x_3 + x_4 + x_5 + x_6 \geq 1 \) ### **Step 4: Objective Function** Minimize total attendants: \[ \text{Minimize } Z = x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \] With \( x_j \in \{0, 1\} \). --- ### **Step 5: Solve the Integer Program** Let's try to minimize the number of attendants by hand: #### **Try 2 attendants:** Suppose we pick the minimum possible, 2. Is it possible? - If \( x_1 = 1 \), \( x_5 = 1 \): - Room 1: \( x_1 + x_2 = 1 + 0 = 1 \) (OK) - Room 2: \( x_2 + x_5 + x_6 = 0 + 1 + 0 = 1 \) (OK) - Room 3: \( x_1 + x_3 + x_4 = 1 + 0 + 0 = 1 \) (OK) - Room 4: \( x_3 + x_4 + x_5 + x_6 = 0 + 0 + 1 + 0 = 1 \) (OK) **So, YES! Two attendants suffice: at doors 1 and 5.** ### **Step 6: Final Answer** \[ \boxed{2} \] **The smallest number of attendants required is 2, placed at doors 1 and 5.** ### **Summary Table** | Door | Attendant? | |------|------------| | 1 | Yes | | 2 | No | | 3 | No | | 4 | No | | 5 | Yes | | 6 | No | --- **Final Answer:** > The minimum number of attendants required is **2**, placed at doors **1 and 5**.

Let's carefully analyze and solve this problem step by step, ensuring clarity at each stage. --- ### **Problem Restatement:** We have a layout of rooms and doors (see the provided diagram). The goal is to place the **minimum number of attendants** at certain doors so that **each room is supervised**. Each attendant can supervise the room through the door they are stationed at. **Given:** - Variables \( x_j \) = 1 if an attendant is placed at door \( j \), 0 otherwise. - Each room must be supervised by **at least one** door (i.e., at least one attendant at an adjacent door). --- ### **Step 1: Identify the Doors and Rooms** From the diagram, the doors are numbered \(1\) through \(6\). The rooms are: - **Room 1 (top-left):** adjacent to doors 1 and 2 - **Room 2 (top-right):** adjacent to doors 2, 5, and 6 - **Room 3 (bottom-left):** adjacent to doors 1, 3, and 4 - **Room 4 (bottom-right):** adjacent to doors 3, 4, 5, and 6 --- ### **Step 2: Define Variables** Let: \[ x_1, x_2, x_3, x_4, x_5, x_6 \in \{0, 1\} \] Indicate whether an attendant is placed at the corresponding door. --- ### **Step 3: Write Constraints for Each Room** To ensure each room is supervised, for each room, the sum of \( x_j \) over the doors adjacent to it must be **at least 1**: - **Room 1:** doors 1 and 2 \[ x_1 + x_2 \geq 1 \] - **Room 2:** doors 2, 5, 6 \[ x_2 + x_5 + x_6 \geq 1 \] - **Room 3:** doors 1, 3, 4 \[ x_1 + x_3 + x_4 \geq 1 \] - **Room 4:** doors 3, 4, 5, 6 \[ x_3 + x_4 + x_5 + x_6 \geq 1 \] --- ### **Step 4: Objective Function** Minimize the total number of attendants: \[ Z = x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \] where each \( x_j \in \{0, 1\} \). --- ### **Step 5: Find the Minimum Number of Attendants** Let's try to find the smallest number of attendants by exploring possible combinations. --- ### **Attempt 1: Try with 1 attendant** Is it possible with only **one** attendant? - Suppose \( x_1 = 1 \), others 0: - Room 1: \( x_1 + x_2 \geq 1 \) → OK (since \( x_1=1 \)) - Room 3: \( x_1 + x_3 + x_4 \geq 1 \) → OK - But check room 2: \( x_2 + x_5 + x_6 \geq 1 \) → all zeros, **not satisfied** - Similarly, if \( x_2=1 \), others zero: - Room 2: OK - Room 1: needs \( x_1 + x_2 \geq 1 \) → OK - Room 3: needs \( x_1 + x_3 + x_4 \geq 1 \) → all zeros for \( x_1, x_3, x_4 \), **fails** - Similar reasoning applies for any single door: it cannot cover all rooms. **Conclusion:** **At least 2 attendants are needed.** --- ### **Attempt 2: Try with 2 attendants** Check possible pairs of doors: - **Pair (door 1 and door 5):** - \( x_1=1, x_5=1 \), others 0. Verify constraints: - Room 1: \( x_1 + x_2 \geq 1 \) → \( 1 + 0 \geq 1 \) OK - Room 2: \( x_2 + x_5 + x_6 \geq 1 \) → \( 0 + 1 + 0 \geq 1 \) OK - Room 3: \( x_1 + x_3 + x_4 \geq 1 \) → \( 1 + 0 + 0 \geq 1 \) OK - Room 4: \( x_3 + x_4 + x_5 + x_6 \geq 1 \) → \( 0 + 0 + 1 + 0 \geq 1 \) OK All rooms are covered with just **2 attendants** at doors 1 and 5. --- ### **Final Step: Confirm minimality** Since we've shown: - 1 attendant is insufficient. - 2 attendants can cover all rooms (e.g., at doors 1 and 5). **Therefore, the minimum number of attendants required is 2.** --- ### **Final Answer:** \[ \boxed{\text{Minimum attendants} = \mathbf{2}} \] **Placement:** Attendants are placed at **doors 1 and 5**. --- **Summary:** - **Variables:** \( x_1, x_2, x_3, x_4, x_5, x_6 \) - **Constraints:** as above - **Solution:** \( x_1 = 1, x_5 = 1 \), others 0 - **Minimum number of attendants:** **2** --- If you need further details, feel free to ask!