Algorithms are Obsolete. about coursera-stanford-graph-search-shortest-paths-and-data-structures HOT 3 OPEN

Yes they are. I tried doing the questions on my own but was struggling a lot. Decided to look up for solutions to see how they work and found this. I have been trying to make this work but these are obsolete and do not run or even if they do, the solutions are wrong.

from coursera-stanford-graph-search-shortest-paths-and-data-structures.

heres the code that works if anyone is still looking for the answers

import time

file_path = "clustering1.txt"

with open(file_path, "r") as file:
num_nodes = int(file.readline())
clusters = {x: [x] for x in range(1, num_nodes + 1)}
edges = []
for line in file:
node1, node2, cost = map(int, line.strip().split())
edges.append((node1, node2, cost))
sorted_edges = sorted(edges, key=lambda x: x[2])

k = 4

def get_key_by_value(d, k):
for leader, nodes in d.items():
if k in nodes:
return leader

def update_clusters(c, a, d):
c[a].extend(c[d])
del c[d]

iteration = 0

start_time = time.time()

for edge in sorted_edges:
node1, node2, cost = edge
key1 = get_key_by_value(clusters, node1)
key2 = get_key_by_value(clusters, node2)
if key1 == key2:
continue
elif len(clusters[key1]) >= len(clusters[key2]):
update_clusters(clusters, key1, key2)
else:
update_clusters(clusters, key2, key1)
if len(clusters) == k:
break

max_spacing = None

for edge in sorted_edges:
node1, node2, cost = edge
key1 = get_key_by_value(clusters, node1)
key2 = get_key_by_value(clusters, node2)
if key1 != key2:
max_spacing = cost
break

end_time = time.time()
execution_time = end_time - start_time

print("Execution time: {:.6f} seconds".format(execution_time))
print("Execution time: {:.6f} minutes".format(execution_time / 60))
print("Execution time: {:.6f} hours".format(execution_time / 3600))
print("Maximum spacing of a 4-clustering: ", max_spacing)

from coursera-stanford-graph-search-shortest-paths-and-data-structures.

i was able to do it

from coursera-stanford-graph-search-shortest-paths-and-data-structures.