| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the shortest path in an undirected graph, we can use a breadth-first search (BFS) algorithm. However, I'll provide the shortest path here based on analyzing the connections: Starting from node 1, the shortest path to node 14 can be found by traversing the graph in the most direct way possible. Here is the shortest path: ``` [1, 2, 6, 10, 9, 14] ``` Explanation: - From node 1, you can go to nodes 0, 2, or 3. Choose node 2 because it leads closer to node 14. - From node 2, you can go to nodes 1, 5, 6, or 7. Choose node 6 because it leads closer to node 14. - From node 6, you can go to nodes 2, 7, 8, 9, or 10. Choose node 10 because it leads closer to node 14. - From node 10, you can go to nodes 6, 8, 9, or 14. Choose node 9 because it leads closer to node 14. - From node 9, you can go to nodes 6, 8, 10, 11, 12, 13, or 14. Choose node 14, which is the destination. Thus, the shortest path from node 1 to node 14 is [1, 2, 6, 10, 9, 14]. |
[1, 2, 6, 9, 11, 14] |
| deepseekr1 | The shortest path from node 1 to node 14 is determined using Breadth-First Search (BFS), which efficiently finds the shortest path in an unweighted graph. Here's the step-by-step breakdown: 1. **Start at node 1**. 2. **Adjacent nodes**: 0, 3, 2. Explore node 2 first (shorter path potential). 3. **From node 2**, adjacent nodes: 5, 7, 6. Proceed to node 6. 4. **From node 6**, adjacent nodes: 10, 9. Move to node 9. 5. **From node 9**, adjacent nodes: 11. Proceed to node 11. 6. **From node 11**, reach node 14. **Path**: [1, 2, 6, 9, 11, 14] **Verification**: Each consecutive pair in the path corresponds to an edge in the graph, confirming validity. No shorter path exists, as verified by BFS traversal. **Answer**: \boxed{[1, 2, 6, 9, 11, 14]} |
[1, 2, 6, 9, 11, 14] |
| gpto1 | To solve this problem, we'll use Breadth-First Search (BFS) since all edges in the graph are undirected and unweighted, making BFS ideal for finding the shortest path in terms of the number of edges. **Step 1: Represent the Graph** First, we'll represent the graph using an adjacency list. We'll go through each edge and populate the adjacency list accordingly. ```python # Initialize adjacency list adjacency_list = {i: [] for i in range(15)} # Nodes 0 through 14 # List of edges edges = [ (0,1),(1,3),(0,3),(3,4),(0,4),(1,2),(2,5),(2,7),(2,6),(6,7),(6,10), (5,8),(10,9),(5,10),(6,8),(7,8),(6,9),(7,10),(8,10),(9,11),(9,12), (9,13),(13,12),(13,11),(11,14) ] # Populate adjacency list for u, v in edges: adjacency_list[u].append(v) adjacency_list[v].append(u) ``` **Step 2: Perform BFS** We'll perform BFS starting from node `1` and keep track of the parents to reconstruct the path later. ```python from collections import deque # Initialize BFS variables queue = deque([1]) visited = set([1]) parent = {} # BFS loop while queue: current_node = queue.popleft() if current_node == 14: break for neighbor in adjacency_list[current_node]: if neighbor not in visited: visited.add(neighbor) parent[neighbor] = current_node queue.append(neighbor) ``` **Step 3: Reconstruct the Shortest Path** Using the `parent` dictionary, we'll reconstruct the path from node `14` back to node `1` and then reverse it. ```python # Reconstruct path path = [] current_node = 14 while current_node != 1: path.append(current_node) current_node = parent[current_node] path.append(1) path.reverse() ``` **Step 4: Present the Solution** The shortest path from node `1` to node `14` is: ```python print(path) ``` **Answer:** The shortest path from node 1 to node 14 is: **[1, 2, 6, 9, 11, 14]** |
[1, 2, 6, 9, 11, 14] |
| GPT-4o | To find the shortest path in an undirected graph, a Breadth-First Search (BFS) algorithm can be used. BFS is ideal for unweighted graphs as it explores all nodes at the present depth level before moving on to nodes at the next depth level. Let's represent the graph using an adjacency list and implement the BFS algorithm to find the shortest path from node 1 to node 14. ### Step-by-step Solution: 1. **Construct Adjacency List:** ```python from collections import deque, defaultdict edges = [ (0, 1), (1, 3), (0, 3), (3, 4), (0, 4), (1, 2), (2, 5), (2, 7), (2, 6), (6, 7), (6, 10), (5, 8), (10, 9), (5, 10), (6, 8), (7, 8), (6, 9), (7, 10), (8, 10), (9, 11), (9, 12), (9, 13), (13, 12), (13, 11), (11, 14) ] graph = defaultdict(list) for u, v in edges: graph[u].append(v) graph[v].append(u) ``` 2. **Implement BFS to find the shortest path:** ```python def bfs_shortest_path(graph, start, goal): queue = deque([(start, [start])]) visited = set() while queue: current_node, path = queue.popleft() if current_node == goal: return path visited.add(current_node) for neighbor in graph[current_node]: if neighbor not in visited: queue.append((neighbor, path + [neighbor])) visited.add(neighbor) return None start_node = 1 goal_node = 14 shortest_path = bfs_shortest_path(graph, start_node, goal_node) ``` 3. **Get the Result:** ```python print(shortest_path) ``` ### Execution of the Code: Running the above code will give us the shortest path from node 1 to node 14. Let's break down the BFS traversal for the given graph. Here's the shortest path found using BFS: ```python [1, 2, 6, 9, 11, 14] ``` ### Explanation: - Starting from node 1, the algorithm explores its neighbors: [0, 3, 2]. - From node 2, it explores: [1, 5, 7, 6]. - From node 6, it explores: [2, 7, 10, 8, 9]. - From node 9, it explores: [10, 11, 12, 13, 6]. - From node 11, it explores: [9, 13, 14]. - Node 14 is reached. Thus, the shortest path from node 1 to node 14 is: \[ \text{[1, 2, 6, 9, 11, 14]} \] So, the solution to the problem is: ```plaintext Solution: [1, 2, 6, 9, 11, 14] ``` |
[1, 2, 6, 9, 11, 14] |
| llama3.1-70B-Instruct | from collections import deque def shortest_path(graph, start, end): queue = deque([[start]]) visited = set([start]) while queue: path = queue.popleft() node = path[-1] if node == end: return path for neighbor in graph[node]: if neighbor not in visited: visited.add(neighbor) queue.append(list(path) + [neighbor]) return None # Define the graph as an adjacency list graph = { 0: [1, 3, 4], 1: [0, 3, 2], 2: [1, 5, 7, 6], 3: [0, 1, 4], 4: [0, 3], 5: [2, 8, 10], 6: [2, 7, 10, 9, 8], 7: [2, 6, 8, 10], 8: [5, 6, 7, 10], 9: [6, 10, 11, 12, 13], 10: [5, 6, 7, 8, 9], 11: [9, 13, 14], 12: [9, 13], 13: [9, 11, 12], 14: [11] } print(shortest_path(graph, 1, 14)) # Output: [1, 0, 4, 3, 1, 2, 6, 9, 11, 14] |
[1, 2, 6, 9, 11, 14] |
| llama3.1-405B-Instruct | To find the shortest path from node 1 to node 14 in the given undirected graph, we can use a breadth-first search (BFS) algorithm. BFS is particularly useful for finding the shortest path in an unweighted graph, as it explores all the nodes at a given depth level before moving on to the next level. Here's a step-by-step approach to solving this problem: 1. **Start with Node 1**: Begin the search from node 1. 2. **Explore Neighbors**: Explore all the neighbors of node 1. These are nodes 0, 2, and 3. 3. **Mark Visited Nodes**: Mark nodes 0, 2, and 3 as visited to avoid revisiting them. 4. **Next Level of Neighbors**: Move on to the next level of neighbors. For node 0, the unvisited neighbors are nodes 4. For node 2, the unvisited neighbors are nodes 5, 6, and 7. For node 3, the unvisited neighbor is node 4. 5. **Continue BFS**: Continue this process, exploring each level of neighbors until we reach node 14. 6. **Path to Node 14**: Once node 14 is found, backtrack to find the shortest path from node 1 to node 14. Given the complexity of manually performing BFS on such a large graph, let's simplify the explanation by focusing on the key steps leading to node 14: - From node 1, we can reach node 2. - From node 2, we can reach nodes 5, 6, and 7. - Node 6 can reach node 10. - Node 10 can reach node 9. - Node 9 is directly connected to node 14. Therefore, one of the shortest paths from node 1 to node 14 is: **1 -> 2 -> 6 -> 10 -> 9 -> 14**. This path is derived by tracing back the steps from node 14 to node 1 after performing the BFS. Note that in an undirected graph, the direction of edges is not considered, so each step in the path represents a direct connection between nodes regardless of the direction of the edge. |
[1, 2, 6, 9, 11, 14] |
| DeepSeek-Coder-V2 | To find the shortest path from node 1 to node 14 in the given undirected graph G, we can use the Breadth-First Search (BFS) algorithm. BFS is well-suited for finding the shortest path in an unweighted graph. Here is the step-by-step solution: 1. **Initialize the BFS:** - Start from node 1. - Use a queue to keep track of nodes to visit, initialized with the starting node and its path. - Use a set to keep track of visited nodes to avoid revisiting them. 2. **Perform BFS:** - Dequeue the front node from the queue. - If the node is the target node (node 14), return the path. - Otherwise, enqueue all unvisited neighbors of the current node along with the path to reach them. 3. **Repeat until the target node is found or the queue is empty.** Here is the implementation of the BFS algorithm to find the shortest path from node 1 to node 14: ```python from collections import deque def bfs_shortest_path(graph, start, target): queue = deque([(start, [start])]) visited = set() while queue: node, path = queue.popleft() if node == target: return path if node not in visited: visited.add(node) for neighbor in graph[node]: if neighbor not in visited: queue.append((neighbor, path + [neighbor])) return None # Define the graph edges = { (0, 1), (1, 3), (0, 3), (3, 4), (0, 4), (1, 2), (2, 5), (2, 7), (2, 6), (6, 7), (6, 10), (5, 8), (10, 9), (5, 10), (6, 8), (7, 8), (6, 9), (7, 10), (8, 10), (9, 11), (9, 12), (9, 13), (13, 12), (13, 11), (11, 14) } # Create adjacency list representation of the graph graph = {} for edge in edges: if edge[0] not in graph: graph[edge[0]] = [] if edge[1] not in graph: graph[edge[1]] = [] graph[edge[0]].append(edge[1]) graph[edge[1]].append(edge[0]) # Find the shortest path shortest_path = bfs_shortest_path(graph, 1, 14) print(shortest_path) ``` Running this code will output the shortest path from node 1 to node 14: ``` [1, 2, 5, 10, 9, 11, 14] ``` Thus, the shortest path from node 1 to node 14 is represented as the list `[1, 2, 5, 10, 9, 11, 14]`. |
[1, 2, 6, 9, 11, 14] |
| Qwen2-72B-Instruct | To find the shortest path in an undirected graph, we can use a breadth-first search (BFS) algorithm. However, I'll provide the shortest path here based on analyzing the connections: Starting from node 1, the shortest path to node 14 can be found by traversing the graph in the most direct way possible. Here is the shortest path: ``` [1, 2, 6, 10, 9, 14] ``` Explanation: - From node 1, you can go to nodes 0, 2, or 3. Choose node 2 because it leads closer to node 14. - From node 2, you can go to nodes 1, 5, 6, or 7. Choose node 6 because it leads closer to node 14. - From node 6, you can go to nodes 2, 7, 8, 9, or 10. Choose node 10 because it leads closer to node 14. - From node 10, you can go to nodes 6, 8, 9, or 14. Choose node 9 because it leads closer to node 14. - From node 9, you can go to nodes 6, 8, 10, 11, 12, 13, or 14. Choose node 14, which is the destination. Thus, the shortest path from node 1 to node 14 is [1, 2, 6, 10, 9, 14]. |
[1, 2, 6, 9, 11, 14] |
| baichuan4 | To find the shortest path from node 1 to node 14 in the given undirected graph, we can use Dijkstra's algorithm or any other suitable shortest path algorithm. Let's apply Dijkstra's algorithm to find the solution. First, we initialize the distance of each node from node 1. The initial distances are as follows: ``` Distance from Node 1: 0 - infinity 1 - 0 2 - infinity 3 - infinity 4 - infinity 5 - infinity 6 - infinity 7 - infinity 8 - infinity 9 - infinity 10 - infinity 11 - infinity 12 - infinity 13 - infinity 14 - infinity ``` Now, let's start the algorithm by relaxing the edges and updating the distances: 1. Start at Node 1. Since it is the starting node, its distance is 0. 2. Node 1 is connected to Nodes 0, 2, and 3. Update their distances: - Distance of Node 0 from Node 1: 0 + 1 (edge weight, assumed to be 1 for all edges) = 1 - Distance of Node 2 from Node 1: 0 + 1 = 1 - Distance of Node 3 from Node 1: 0 + 1 = 1 3. Node 0 is connected to Nodes 1, 3, and 4. We only need to update the distances of Nodes 3 and 4 because Node 1 is the starting node: - Distance of Node 3 from Node 0: 1 + 1 = 2 (but we already have a shorter path via Node 1, so no update needed) - Distance of Node 4 from Node 0: 1 + 1 = 2 4. Node 2 is connected to Nodes 1, 5, 6, and 7. Update their distances: - Distance of Node 5 from Node 2: 1 + 1 = 2 - Distance of Node 6 from Node 2: 1 + 1 = 2 - Distance of Node 7 from Node 2: 1 + 1 = 2 5. Node 3 is connected to Nodes 1, 0, 4, and has already been visited. Update the distance of Node 4: - Distance of Node 4 from Node 3: 1 + 1 = 2 (but we already have a shorter path via Node 0, so no update needed) 6. Continue this process until we reach Node 14 or there are no more nodes to visit. After running Dijkstra's algorithm, we get the following shortest distances from Node 1: ``` Distance from Node 1: 0 - 1 1 - 0 2 - 1 3 - 1 4 - 2 5 - 2 6 - 2 7 - 2 8 - 3 9 - 3 10 - 3 11 - 4 12 - 4 13 - 4 14 - 4 ``` Now, we can trace back the shortest path from Node 14 to Node 1 by following the predecessors or parents of each node in the shortest path tree. The shortest path from Node 1 to Node 14 is: ``` 1 -> 2 -> 6 -> 10 -> 9 -> 11 -> 14 ``` Therefore, the shortest path from Node 1 to Node 14 is represented as the list [1, 2, 6, 10, 9, 11, 14]. |
[1, 2, 6, 9, 11, 14] |
| doubao | One possible shortest path from node 1 to node 14 is [1, 2, 5, 8, 10, 9, 11, 14]. | [1, 2, 6, 9, 11, 14] |
| minimax | To find the shortest path from node 1 to node 14 in the given undirected graph, we can use Dijkstra's algorithm, which is a common algorithm for finding the shortest paths between nodes in a graph, which may represent, for example, road networks. However, since I am a text-based model and cannot execute code, I will describe the steps you would take to apply Dijkstra's algorithm to this problem: 1. Initialize the distance to all nodes as infinity, except for the starting node (node 1), which should be set to 0. 2. Set the starting node as current and mark all other nodes as unvisited. Create a set of all the unvisited nodes called the unvisited set. 3. For the current node, ***pare the newly calculated tentative distance to the current assigned value and assign the smaller one. For example, if the current node A is marked with a distance of 6, and the edge connecting it with a neighbor B has length 2, then the distance to B through A will be 6 + 2 = 8. If B was previously marked with a distance greater than 8 then change it to 8. If not, leave it as is. 4. Once we have considered all of the unvisited neighbors of the current node, mark the current node as visited. A visited node will not be checked again. 5. If the destination node has been marked visited (when planning a route between two specific nodes) or if the smallest tentative distance among the nodes in the unvisited set is infinity (when planning a complete traversal; occurs when there is no connection between the initial node and remaining unvisited nodes), then stop. The algorithm has finished. 6. Otherwise, select the unvisited node that is marked with the smallest tentative distance, set it as the new "current node," and go back to step 3. Since I cannot execute the algorithm, I will provide a conceptual solution based on the graph's structure: - Start at node 1. - The immediate neighbors are nodes 0, 3, and 2. - From node 1, the shortest path to node 14 would involve going through the nodes with the least cumulative distance. - We would continue to expand from the closest nodes until we reach node 14. Given the graph's structure, the shortest path from node 1 to node 14 is likely to go through nodes 2, 6, 9, and then to node 14. However, without executing the algorithm, I cannot provide the exact shortest path. To get the exact shortest path, you would need to implement Dijkstra's algorithm in a programming language of your choice or use a graph analysis tool that can perform this computation for you. |
[1, 2, 6, 9, 11, 14] |
| kimi | To find the shortest path from node 1 to node 14 in an undirected graph, you can use algorithms like Dijkstra's or the Breadth-First Search (BFS) algorithm. Since the graph is undirected and we're looking for the shortest path, BFS is a suitable choice because it explores the nearest nodes first. Here's a step-by-step solution using BFS: 1. Start at node 1. 2. Explore all the neighbors of node 1 (nodes 0, 2, and 3) and mark them as visited. 3. From each of these neighbors, explore their neighbors that haven't been visited yet, and mark them as visited. 4. Continue this process until you reach node 14. Here's the path in a list format: 1. Start at node 1. 2. From node 1, go to node 2 (path: [1, 2]). 3. From node 2, go to node 5 (path: [1, 2, 5]). 4. From node 5, go to node 8 (path: [1, 2, 5, 8]). 5. From node 8, go to node 6 (path: [1, 2, 5, 8, 6]). 6. From node 6, go to node 9 (path: [1, 2, 5, 8, 6, 9]). 7. From node 9, go to node 11 (path: [1, 2, 5, 8, 6, 9, 11]). 8. From node 11, go to node 14 (path: [1, 2, 5, 8, 6, 9, 11, 14]). So the shortest path from node 1 to node 14 is [1, 2, 5, 8, 6, 9, 11, 14]. |
[1, 2, 6, 9, 11, 14] |
| wenxinyiyan | To find the shortest path from node 1 to node 14, we can use a graph traversal algorithm such as Breadth-First Search (BFS). Here's the step-by-step process to find the shortest path: 1. Start with node 1 as the source node. 2. Initialize an empty queue and enqueue node 1. 3. Mark node 1 as visited. 4. While the queue is not empty, dequeue a node from the queue. 5. Check if the dequeued node is the target node 14. If it is, trace back the path from the target node to the source node and return the path. 6. If the dequeued node is not the target, iterate over its neighbors. For each unvisited neighbor, mark it as visited, enqueue it into the queue, and record its predecessor as the dequeued node. 7. Repeat steps 4-6 until the target node is found or the queue becomes empty. Using this process, the shortest path from node 1 to node 14 in the given graph can be found as follows: 1 -> 2 -> 5 -> 10 -> 6 -> 9 -> 14 Therefore, the shortest path from node 1 to node 14 is represented as the list: [1, 2, 5, 10, 6, 9, 14]. |
[1, 2, 6, 9, 11, 14] |
| zhipu | To find the shortest path from node 1 to node 14 in the given undirected graph, we can use a graph traversal algorithm such as Breadth-First Search (BFS), which is well-suited for finding the shortest path in an unweighted graph like the one provided. Here's how we can apply BFS to find the shortest path: 1. Start from the source node (node 1) and enqueue it. 2. Mark the source node as visited. 3. Dequeue a node from the queue and examine it. 4. For each unvisited neighbor of the current node, mark it as visited and enqueue it, also keeping track of the path taken to reach it. 5. Repeat steps 3 and 4 until the destination node (node 14) is reached or the queue is empty. Let's perform the BFS step by step: - Start at node 1. Neighbors are 0, 3, and 2. Enqueue them and mark them as visited. - Dequeue node 0. Neighbors are 1 (already visited), 3 (already visited), and 4. Enqueue 4 and mark it as visited. - Dequeue node 3. Neighbors are 1 (already visited), 0 (already visited), and 4 (already visited). No new nodes to enqueue. - Dequeue node 2. Neighbors are 1 (already visited), 5, 7, and 6. Enqueue 5, 7, and 6 and mark them as visited. - Dequeue node 4. Neighbors are 0 (already visited), 3 (already visited). No new nodes to enqueue. - Dequeue node 5. Neighbors are 2 (already visited), 8, and 10. Enqueue 8 and 10 and mark them as visited. - Dequeue node 7. Neighbors are 2 (already visited), 6 (already visited), and 8 (already visited). No new nodes to enqueue. - Dequeue node 6. Neighbors are 2 (already visited), 7 (already visited), 10 (already visited), 8 (already visited), and 9. Enqueue 9 and mark it as visited. - Dequeue node 8. Neighbors are 5 (already visited), 6 (already visited), 7 (already visited), and 10 (already visited). No new nodes to enqueue. - Dequeue node 10. Neighbors are 5 (already visited), 6 (already visited), 7 (already visited), 8 (already visited), and 9 (already visited). No new nodes to enqueue. - Dequeue node 9. Neighbors are 6 (already visited), 10 (already visited), 11, 12, and 13. Enqueue 11, 12, and 13 and mark them as visited. - Dequeue node 11. Neighbors are 9 (already visited), 13 (already visited), and 14. Enqueue 14 and mark it as visited. We have reached node 14. Now, we need to backtrack to find the path: - Node 14 was reached from node 11. - Node 11 was reached from node 9. - Node 9 was reached from node 6. - Node 6 was reached from node 2. - Node 2 was reached from node 1. Thus, the shortest path from node 1 to node 14 is: [1, 2, 6, 9, 11, 14]. Here is the path represented as a list: ```python [1, 2, 6, 9, 11, 14] ``` This path ensures the minimum number of edges is traversed to get from node 1 to node 14 in the given graph. |
[1, 2, 6, 9, 11, 14] |
| gpto1 | To solve this problem, we'll use Breadth-First Search (BFS) since all edges in the graph are undirected and unweighted, making BFS ideal for finding the shortest path in terms of the number of edges. **Step 1: Represent the Graph** First, we'll represent the graph using an adjacency list. We'll go through each edge and populate the adjacency list accordingly. ```python # Initialize adjacency list adjacency_list = {i: [] for i in range(15)} # Nodes 0 through 14 # List of edges edges = [ (0,1),(1,3),(0,3),(3,4),(0,4),(1,2),(2,5),(2,7),(2,6),(6,7),(6,10), (5,8),(10,9),(5,10),(6,8),(7,8),(6,9),(7,10),(8,10),(9,11),(9,12), (9,13),(13,12),(13,11),(11,14) ] # Populate adjacency list for u, v in edges: adjacency_list[u].append(v) adjacency_list[v].append(u) ``` **Step 2: Perform BFS** We'll perform BFS starting from node `1` and keep track of the parents to reconstruct the path later. ```python from collections import deque # Initialize BFS variables queue = deque([1]) visited = set([1]) parent = {} # BFS loop while queue: current_node = queue.popleft() if current_node == 14: break for neighbor in adjacency_list[current_node]: if neighbor not in visited: visited.add(neighbor) parent[neighbor] = current_node queue.append(neighbor) ``` **Step 3: Reconstruct the Shortest Path** Using the `parent` dictionary, we'll reconstruct the path from node `14` back to node `1` and then reverse it. ```python # Reconstruct path path = [] current_node = 14 while current_node != 1: path.append(current_node) current_node = parent[current_node] path.append(1) path.reverse() ``` **Step 4: Present the Solution** The shortest path from node `1` to node `14` is: ```python print(path) ``` **Answer:** The shortest path from node 1 to node 14 is: **[1, 2, 6, 9, 11, 14]** |
[1, 2, 6, 9, 11, 14] |
| deepseekv3 | To find the shortest path from node 1 to node 14 in the given undirected graph, we can use the **Breadth-First Search (BFS)** algorithm. BFS is suitable for finding the shortest path in an unweighted graph because it explores all nodes level by level, ensuring that the first time we reach the destination, we have found the shortest path. ### Step-by-Step Solution: 1. **Graph Representation**: The graph can be represented as an adjacency list: ``` 0: [1, 3, 4] 1: [0, 3, 2] 2: [1, 5, 7, 6] 3: [0, 1, 4] 4: [0, 3] 5: [2, 8, 10] 6: [2, 7, 10, 8, 9] 7: [2, 6, 8, 10] 8: [5, 6, 7, 10] 9: [6, 10, 11, 12, 13] 10: [5, 6, 7, 8, 9] 11: [9, 13, 14] 12: [9, 13] 13: [9, 11, 12] 14: [11] ``` 2. **BFS Implementation**: We start BFS from node 1 and keep track of the path using a parent map. When we reach node 14, we backtrack to reconstruct the path. 3. **BFS Execution**: - **Queue**: [1] - **Visited**: {1} - **Parent Map**: {1: None} - **Steps**: - Dequeue 1, enqueue its neighbors 0, 3, 2. - Dequeue 0, enqueue its neighbors 1 (already visited), 3 (already visited), 4. - Dequeue 3, enqueue its neighbors 0 (already visited), 1 (already visited), 4 (already visited). - Dequeue 2, enqueue its neighbors 1 (already visited), 5, 7, 6. - Dequeue 4, enqueue its neighbors 0 (already visited), 3 (already visited). - Dequeue 5, enqueue its neighbors 2 (already visited), 8, 10. - Dequeue 7, enqueue its neighbors 2 (already visited), 6, 8, 10. - Dequeue 6, enqueue its neighbors 2 (already visited), 7 (already visited), 10, 8, 9. - Dequeue 8, enqueue its neighbors 5 (already visited), 6 (already visited), 7 (already visited), 10. - Dequeue 10, enqueue its neighbors 5 (already visited), 6 (already visited), 7 (already visited), 8 (already visited), 9. - Dequeue 9, enqueue its neighbors 6 (already visited), 10 (already visited), 11, 12, 13. - Dequeue 11, enqueue its neighbors 9 (already visited), 13 (already visited), 14. - Dequeue 14: We have reached the destination. 4. **Reconstructing the Path**: From the parent map, we backtrack from node 14 to node 1: 14 ← 11 ← 9 ← 6 ← 2 ← 1 ### Final Path: The shortest path from node 1 to node 14 is: `[1, 2, 6, 9, 11, 14]` |
[1, 2, 6, 9, 11, 14] |