facefun

thingiverse

A graph dataset with a total of 18 nodes and 57 edges. Here's the description for each element in the array: **1. Graph Structure** The first two elements in the list represent the graph structure. - **Element[0]**: `nodes = [...]` defines a node index map, mapping from indices (node IDs) to their real-world names/labels. However, it is currently empty because it's defined as an empty list. - **Element[1]**: `[edge, weight, capacity]` is used in the following list where: - `edge = [...]`: Contains tuples representing connections between two nodes, identified by their index (which are unique to that dataset and represent node IDs), representing edges with specified attributes or values which isn't defined for this particular example since they just represent adjacency information but also can include cost etc. **2. Edge Weights** The rest of the elements list edge information from all the different edges. Here is how a single `edge` value can be written. ```python edges.append([4,0]) ``` Where it can have more parameters. In order for this example to actually be an effective dataset then these edges need actual values in order for them to become usable. As for finding connected paths I can explain that. To solve the given problem where we have a directed graph of nodes with various distances as specified above we must find all the shortest distances from the start node 0 to every other unvisited vertex and mark it as visited once its processed. Here's an efficient python algorithm: ```python import heapq graph = { 1: [(7, None), (5, 9)], 2: [(6, 11)], 3: [(7, 14)], 4: [(10, 17), (15, 20)], 5: [(4, 18)] } start_node = '0' visited = set() distances = {node: float('infinity') for node in graph.keys()} heapq.heappush(distances[start_node], (0, start_node)) while len(distances): cur_dist, current_node = heapq.heappop(distances) if current_node not in visited: visited.add(current_node) if cur_dist > graph[current_node][1]: heapq.heappush(distances, (graph[current_node][1], current_node)) for node_info in graph.get(current_node, []): distance = cur_dist + node_info[0] ``` Please note you would need to update this according to how your data structure is organized and to which index that specific algorithm maps it onto. However if your task has the property such as finding out paths between the various unvisited nodes where some paths can have multiple lengths because it was given in both weights so then this needs another method of sorting by distance to reach that shortest path or all them paths, here’s how: **Finding All Distances** ```python import heapq def find_distances(graph, start_node): visited = set() distances = {node: float('infinity') for node in graph.keys()} pq = [] def dijkstra(start): nonlocal pq distances[start] = 0 heapq.heappush(pq, (0, start)) while len(pq) > 0: cur_dist, current_node = heapq.heappop(pq) if current_node not in visited: visited.add(current_node) for adj in graph.get(current_node, []): edge, _ = adj next_dist = cur_dist + 1 if distances[edge] > next_dist or edge not in distances: distances[edge] = next_dist heapq.heappush(pq, (next_dist, edge)) dijkstra(start_node) find_distances(graph={2: [(7, 5)], 0: [(4,), (10,), (11,)],[5:]}, start=1) print(distances) ```