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