Why shortest path algorithms power navigation in Data Structures Theory - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the shortest path distance from the start node.

Data Structures Theory

distance = graph.get([1], float('inf'))

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Aneighbor

Bend_node

Cstart_node

Dpath

Attempts:

3 left

💡 Hint

Common Mistakes

Using the end node instead of the start node for initial distance.

Using a neighbor node which is not the starting point.

The shortest path algorithm starts by setting the distance to the start_node as known, usually zero or a base value.

2fill in blank

medium

Complete the code to update the shortest distance if a better path is found.

Data Structures Theory

if new_distance [1] distances[neighbor]:

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A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which would update for longer paths.

Using '==' which does not help find shorter paths.

We update the distance only if the new distance is less than the current known distance.

3fill in blank

hard

Fix the error in the code that selects the next node with the smallest distance.

Data Structures Theory

current_node = min(unvisited_nodes, key=lambda node: [1][node])

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Adistances

Bgraph

Cneighbors

Dpaths

Attempts:

3 left

💡 Hint

Common Mistakes

Using the graph structure instead of distances.

Using neighbors or paths which do not hold distance values.

The distances dictionary holds the current shortest distances to nodes, so we select the node with the smallest distance.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps nodes to their distances if the distance is less than 10.

Data Structures Theory

{node: [1] for node, dist in distances.items() if dist [2] 10}

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Adist

B>

C<

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' to filter distances.

Mapping nodes to themselves instead of their distances.

This comprehension creates a dictionary of nodes with distances less than 10, mapping each node to its distance.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase node names to their distances if the distance is greater than 5.

Data Structures Theory

{ [1]: [2] for [3], dist in distances.items() if dist > 5 }

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Anode.upper()

Bdist

Cnode

Ddistance

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original node name instead of uppercase for keys.

Using incorrect variable names in the loop.

This comprehension maps each node's uppercase name to its distance, but only includes nodes with distances greater than 5.