# Learn Algorithm Design by Building a Shortest Path Algorithm - Step 47

### Tell us what’s happening:

I have read already a post about this but I am unable to make it work using the slice syntax I have tried almost everything possible (at least in my head) it has to be something simple but I cant find the answer to progress. Please help!

### Your code so far

``````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 - Paso 47

Welcome to the forum @Peibowl

You should use the slice syntax to assign a copy of paths[current] to the neighbor node path.

Here is an article on Python Slicing you may find helpful.

Happy coding

1 Like

Problem here is I dont know what to slice. The exercise states " Fix that bug by assigning a copy of `paths[current]` to the neighbor node path."
In case this is not done by paths[node] = paths[current] I don’t really know how to create that copy of paths[current] as intended in the exercise. I am 100% lost here and for as much as I read about slicing I dont reach the solution.

I got it. I really don`'t find the sense of doing what the exercise admitted as valid to keep progressing. I dont really understand why it works that way and not the previous way implemented.

It would be nice if someone with a higher level of expertise in coding could throw some explanation about the difference.

Thank you in advance!

Hi @Peibowl

Assigning is a reference to something. If you change the something, then you change what the variable is referencing.

my_variable = another_variable

If later you make a change to another_variable, my_variable will reference that.

If you reference a copy of the another_variable, then the original value is still assigned, even if another_variable is later modified.

Happy coding