Learn Algorithm Design by Building a Shortest Path Algorithm - Step 28

Python
1

Tell us what’s happening:

Inside your new if statement you should assign 0 to the node in the distances dictionary.

I’ve tried various methods of inputting the assigned node into the “distances” dictionary and still can’t seem to figure it out.

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)]
}


# User Editable Region

def shortest_path(graph, start):
    unvisited = []
    distances = {}
    for node in graph:
        unvisited.append(node)
        if node == start:
            distances = {[node]: 0}

# User Editable Region

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/125.0.0.0 Safari/537.36

Challenge Information:

Learn Algorithm Design by Building a Shortest Path Algorithm - Step 28

2

Step 28

The distance from the starting node is zero, because the algorithm begins its assessment right from there.

After appending node to unvisited in your loop, create an if statement that triggers if the node is equal to the starting node. Then assign 0 to that node inside the distances dictionary.

I forgot to add the steps

3

Hi there and welcome to our community!

You can use bracket notation to access key/value pairs within a dictionary and to reassign values, or to add new key/value pairs.

EXAMPLE:

nato_alphabet = {
  'A': 'Alpha',
  'B': 'Beta',
  'C': 'Charlie'
}

print(nato_alphabet['B']) # Beta

letter = 'B'
nato_alphabet[letter] = 'Banana'
print(nato_alphabet[letter]) # Banana

nato_alphabet['B'] = 'Bravo'
print(nato_alphabet['B']) #Bravo

nato_alphabet['D'] = 'Delta'
print(nato_alphabet) # { 'A': 'Alpha', 'B': 'Bravo', 'C':'Charlie', 'D': 'Delta' }

You can access/modify/add a key/value pair within a dictionary explicitly or via a variable. Hope that makes sense!

1 Like
4

Thank you very much, i figured since node was not a key in the dictionary that i had to define it.

2 Likes