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import heapq
import math
def length_haversine(p1, p2):
lat1 = p1.lat
lng1 = p1.lng
lat2 = p2.lat
lng2 = p2.lng
lat1, lng1, lat2, lng2 = map(math.radians, [lat1, lng1, lat2, lng2])
dlat = lat2 - lat1
dlng = lng2 - lng1
a = math.sin(dlat / 2) ** 2 + math.cos(lat1) * math.cos(lat2) * math.sin(dlng / 2) ** 2
c = 2 * math.asin(math.sqrt(a))
return 6372797.560856 * c # return the distance in meters
def get_closest_node_id(nodes, source_node):
""" Search through all nodes and return the id of the node
that is closest to 'source_node'. """
min_node = None
min_value = None
for node_id, node in nodes.items():
length = length_haversine(source_node, node)
if min_node is None or length < min_value:
min_node = node_id
min_value = length
return min_node
def find_shortest_path(nodes, source_id, target_id):
""" Return the shortest path using Dijkstra's algortihm. """
# queue contains multiple (walk_dist, (node_0, node_1, ... node_n))-tuples
# where (node_0, node_1, ... node_n) is a walk to node_n
# and walk_dist is the total length of the walk in meters
queue = [(0, (source_id,))]
visited = set()
while queue:
# consider an unchecked walk
walk_dist, walk = heapq.heappop(queue)
walk_end = walk[-1]
if walk_end == target_id:
# you have reached your destination
return walk
if walk_end in visited:
# there exists a shorter walk to walk_end
continue
# otherwise this is the shortest walk to walk_end
visited.add(walk_end)
# consider all our neighbours
for neighbour in nodes[walk_end].neighbours:
if neighbour in visited:
# there exists a shorter walk to neighbour
continue
# otherwise this MIGHT be the shortest walk to neighbour
# so put it in the queue
new_dist = walk_dist + length_haversine(nodes[walk_end], neighbour)
new_walk = walk + (neighbour.id,)
heapq.heappush(queue, (new_dist, new_walk))
# no path found
return None
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