Let's solve each question step by step: --- ### **Question 1** **Step 1: Construct the Graph from the Adjacency List** - **Belmont:** Marin City, Sonoma - **Hillsborough:** Livermore, Marin City, Sonoma - **Livermore:** Hillsborough, Sonoma - **Marin City:** Belmont, Hillsborough, Sonoma - **Sonoma:** Belmont, Hillsborough, Livermore, Marin City **Step 2: Find the Shortest Path from Belmont to Livermore** Let’s try to find the minimum transfers (edges): - **Belmont → Sonoma → Livermore** - Belmont → Sonoma (direct) - Sonoma → Livermore (direct) - Path length: 2 edges, so **1 transfer** (since each edge after the first is a transfer). **Final Answer:** **1** --- ### **Question 2** **Step 1: Write Down the Weights from the Graph for Path H → G → J → F** - **H → G:** 7 - **G → J:** 2 - **J → F:** 1 **Step 2: Add the Weights** - 7 + 2 + 1 = **10** But let's check the provided options. None match 10, so let's double-check the weights from the image: - **H → G:** 7 - **G → J:** 2 - **J → F:** 1 Wait, in the image, J → F is 1. So, total: **7 + 2 + 1 = 10**. But this is not in the options. However, let's check for possible negative weights or misreading: - H → G = 7 - G → J = 2 - J → F = 1 Again, total = **10** (But no 10 in options). Let’s double-check if any edge is negative: - No negative on these edges. The closest is 5 (could there be a mistake in the direction or in the image?). But the correct calculation is **7 + 2 + 1 = 10**. Since 7 is the closest among options, but the calculation is 10. However, based on the path and the weights, **10** is correct, but as per the options, **None match exactly**. If you must select from the provided options and trust the weights, **none exactly match**. --- ### **Question 3** **Step 1: Edge Listing** - A → C (3), C → E (8), B → A (5), B → D (1), C → D (4), D → E (9) **Step 2: Dijkstra’s from B** - **Initial:** - B: 0 - A: ∞ - C: ∞ - D: ∞ - E: ∞ - **Iteration 1:** - From B: - B → A: 5 - B → D: 1 - Update: - A: 5 - D: 1 - **Select min (D = 1), mark D as visited** - **Iteration 2:** - From D (current distance = 1): - D → E: 1 + 9 = 10 - Update: - E: 10 - Remaining: A (5), C (∞), E (10) After **second iteration**, **tentative distance to C** remains **∞** (no direct path from B or D to C). **But waiting for A to be selected, so after 2nd iteration:** - C: ∞ But as per options, the closest is **7**, but by Dijkstra, it's still **∞** after 2nd iteration. --- ### **Question 4** Does Dijkstra's algorithm initialize all vertices' distances to infinity and all predecessors to undefined/null? **Answer:** **True** --- ### **Question 5** Bellman-Ford time complexity is **O(VE)**. **Final Answers:** --- ## **Summary Table** | Q | Step-by-Step | Final Answer | |---|--------------|-------------| | 1 | Belmont → Sonoma → Livermore (1 transfer) | **1** | | 2 | H → G (7), G → J (2), J → F (1), total 10 | **None of the options match exactly, but calculation is 10** | | 3 | After 2nd iteration from B: C still ∞ | **∞ (not in options)** | | 4 | Initialization to ∞ and null | **True** | | 5 | Time complexity of Bellman-Ford | **O(VE)** | --- If you have to select from the given options for Q2 and Q3, please check the weights or provide clarifications. For the others, the step-by-step reasoning and answers are above.

Let's go through each question carefully, step-by-step, with detailed explanations and final answers. --- ### **Question 1: Minimum bus transfers from Belmont to Livermore** **Step 1: Understand the Network** - **Vertices (Cities):** Belmont, Marin City, Sonoma, Hillsborough, Livermore - **Connections (Edges):** Based on adjacency list: - Belmont ↔ Marin City, Sonoma - Marin City ↔ Sonoma, Hillsborough - Sonoma ↔ Hillsborough, Livermore - Hillsborough ↔ Livermore **Step 2: Find the shortest path in terms of number of transfers** - Starting point: **Belmont** - Goal: **Livermore** **Possible routes:** 1. **Belmont → Sonoma → Livermore** - Transfers: 2 edges, but transfers are counted as the number of times a commuter switches buses, i.e., number of edges traversed minus 1. - Since the journey involves 2 edges, number of transfers = 1 (transfers happen when moving from one route to another). 2. Other routes (e.g., Belmont → Marin City → Hillsborough → Livermore) involve more edges and thus more transfers. **Step 3: Conclusion** - The shortest route involves **1 transfer**. **Final Answer:** \[ \boxed{1} \] --- ### **Question 2: Path weight H → G → J → F** **Step 1: Identify the weights for each segment** From the diagram: - **H → G:** 7 - **G → J:** 2 - **J → F:** 1 **Step 2: Sum the weights** \[ 7 + 2 + 1 = 10 \] **Step 3: Check options** Options are: - 7 - 12 - -3 - 5 **Observation:** Our calculation yields **10**, but it's not in the options. Let's verify if the weights are correct: - The diagram indicates **H → G = 7** - **G → J = 2** - **J → F = 1** Total: **10** **Conclusion:** Since 10 is not listed, either the options are incorrect or there's a typo. But based on the diagram, **the total weight is 10**. **Final Answer:** \[ \boxed{10} \] *(Note: If your test requires choosing from options, and options don't include 10, choose the closest or clarify with your instructor. But mathematically, the total is 10.)* --- ### **Question 3: Dijkstra's Algorithm - Distance to C after 2 iterations** **Given graph edges and weights:** - A → C: 3 - C → E: 8 - B → A: 5 - B → D: 1 - C → D: 4 - D → E: 9 **Step 1: Initialization** - Start from **vertex B** - Tentative distances: | Vertex | Distance | Predecessor | |---------|------------|--------------| | B | 0 | - | | A | ∞ | null | | C | ∞ | null | | D | ∞ | null | | E | ∞ | null | **Step 2: First iteration** - From B: - Update A: 0 + 5 = 5 - Update D: 0 + 1 = 1 - Selected vertices to process next: D (smallest tentative distance = 1), then A (distance 5) **Updated tentative distances:** | Vertex | Distance | Predecessor | |---------|------------|--------------| | B | 0 | - | | D | 1 | B | | A | 5 | B | | C | ∞ | null | | E | ∞ | null | **Step 3: Second iteration** - Process **D**: - D → E: 1 + 9 = 10 → update E to 10 - Process **A**: - A → C: 5 + 3 = 8 → update C to 8 - A → D: already known, no change **Updated tentative distances after second iteration:** | Vertex | Distance | Predecessor | |---------|------------|--------------| | B | 0 | - | | D | 1 | B | | A | 5 | B | | C | 8 | A | | E | 10 | D | **Answer:** **Tentative distance to C after second iteration is 8.** **Final Answer:** \[ \boxed{8} \] *(Note: The options provided are 7, 14, 3, 5. Since 8 is not listed, double-check options. If options are only 7, 14, 3, 5, then the closest or the correct based on the calculation is **8**, which is not listed. Possibly a typo or misreading in options. The correct tentative distance after second iteration is **8**.)* --- ### **Question 4: Dijkstra's Algorithm Initialization** **Statement:** > Dijkstra's algorithm initializes all vertices' distances to infinity (∞) and all vertices’ predecessors to undefined (null). **Evaluation:** - **True**, because: - Distances are set to ∞ to indicate they are initially unreachable. - Predecessors are set to null or undefined, as no paths have been established yet. **Final Answer:** \[ \boxed{\text{True}} \] --- ### **Question 5: Complexity of Bellman-Ford Algorithm** **Recall:** - Bellman-Ford relaxes all edges V-1 times. - Each relaxation involves checking all E edges. **Time complexity:** \[ O(V \times E) \] **Final Answer:** \[ \boxed{O(VE)} \] --- ### **Summary of Final Answers:** | Question | Final Answer | |------------|--------------| | 1 | **1** | | 2 | **10** | | 3 | **8** | | 4 | **True** | | 5 | **O(VE)** | --- **Note:** For questions where options do not exactly match calculations, rely on the step-by-step reasoning. If there are discrepancies, double-check the provided diagrams and data.