# Learn Algorithm Design by Building a Shortest Path Algorithm - Step 49

### Tell us what’s happening:

I thought this was the right answer, any chance i could get a hint at what I did wrong?

``````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, target = ''):
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]
if paths[node] and paths[node][-1] == node:
paths[node] = paths[current][:]
else:
paths[node].extend(paths[current])
paths[node].append(node)
unvisited.remove(current)

/* User Editable Region */

targets_to_print = paths[target] if target != '' else graph

shortest_path(my_graph, 'A')

/* User Editable Region */

``````

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### Challenge Information:

Learn Algorithm Design by Building a Shortest Path Algorithm - Step 49

remove `paths` and `!=''` from the syntax

` val_1 if condition else val_2`
this is the correct syntax.

Explanation
suppose there’s a variable a. a gets assigned val_1 if condition is true. here lets replace condition with a variable b. if b means if b has a value not 0 value (0 means false) then it becomes true.

I got that solved. Anyone who might get confused by val_1, you can re-read the instruction and take it to be literal.

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