# Learn Algorithm Design by Building a Shortest Path Algorithm - Step 45

### Tell us what’s happening:

Describe your issue in detail here.

I having trouble with this code. I’m not sure what I am missing.

``````my_graph = {
'A': [('B', 3), ('D', 1)],
'B': [('A', 3), ('C', 4)],
'C': [('B', 4), ('D', 7)],
'D': [('A', 1), ('C', 7)]
}

def shortest_path(graph, start):
unvisited = list(graph)
distances = {node: 0 if node == start else float('inf') for node in graph}
paths = {node: [] for node in graph}
paths[start].append(start)

while unvisited:
current = min(unvisited, key=distances.get)
for node, distance in graph[current]:
if distance + distances[current] < distances[node]:
distances[node] = distance + distances[current]

# User Editable Region

if paths[node] != [] and paths[node][-1] == node:
paths[node] = paths[current]

# User Editable Region

else:
paths[node].extend(paths[current])
paths[node].append(node)
unvisited.remove(current)

print(f'Unvisited: {unvisited}\nDistances: {distances}\nPaths: {paths}')

#shortest_path(my_graph, 'A')

``````

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### Challenge Information:

Learn Algorithm Design by Building a Shortest Path Algorithm - Step 45

You appear to have created this post without editing the template. Please edit your post to Tell us what’s happening in your own words.
Learning to describe problems is hard, but it is an important part of learning how to code.
Also, the more you say, the more we can help!

This works but there is a shorter way to do this. You can check if a variable is truthy or non-empy just by using it by itself as a condition.

Thank you! That helped me solve it!

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