import heapq as heap # function which takes graph, vertex set, start and goal, uses djikstra algorithm to establish shortest paths from start # every other vertex in graph and returns shortest path between start and goal def dijkstra_algorithm(graph, point_set, start, goal): # dictionary which will store for every vertex it's distance to start dist = dict() # dictionary which keeps if vertex was already popped from queue visited = dict() # dictionary which keeps for every vertex which other vertex is previous on the shortest path from start prev = dict() # list which is used to keep priority queue with vertexes to be explored by the algorithm queue = [] # creating dictionary which for every vertex keeps all it's neighbors for edge in graph.keys(): point_set[edge[0]].append(edge[1]) point_set[edge[1]].append(edge[0]) for vertex in point_set: dist[vertex] = float('inf') prev[vertex] = None visited[vertex] = False dist[start] = 0 heap.heappush(queue, (0, start)) visited[start] = True while len(queue) > 0: current = heap.heappop(queue) for neighbor in point_set[current[1]]: new_dist = 0 if neighbor > current[1]: new_dist = dist[current[1]] + graph[(neighbor, current[1])] else: new_dist = dist[current[1]] + graph[(current[1], neighbor)] if new_dist < dist[neighbor]: dist[neighbor] = new_dist prev[neighbor] = current[1] if not visited[neighbor]: visited[neighbor] = True heap.heappush(queue, (new_dist, neighbor)) temp = goal shortest_path = [goal] while prev[temp] is not None: shortest_path.append(prev[temp]) temp = prev[temp] return shortest_path, dist[goal] if __name__ == "__main__": g = { (2, 1): 3, (3, 2): 2, (5, 3): 1, (9, 5): 5, (10, 9): 4, (9, 4): 3, (4, 1): 4, (7, 1): 6, (3, 1): 4, (6, 2): 3, (8, 6): 8, (8, 3): 2 } v = dict() for i in g.keys(): v[i[0]] = [] v[i[1]] = [] print(dijkstra_algorithm(g, v, 7, 10))