Bellman-Ford algorithm in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance to the source vertex as zero.

Data Structures Theory

distance[source] = [1]

Drag options to blanks, or click blank then click option'

Ainfinity

B-1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the source distance to infinity or a negative number.

Forgetting to initialize the source distance.

The distance from the source vertex to itself is always zero in the Bellman-Ford algorithm.

2fill in blank

medium

Complete the code to relax an edge from u to v with weight w.

Data Structures Theory

if distance[u] != [1] and distance[u] + w < distance[v]:

Drag options to blanks, or click blank then click option'

A0

Binfinity

C-1

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if distance[u] is zero instead of infinity.

Not checking if u is reachable before relaxing.

We check if the distance to u is not infinity before relaxing the edge, to ensure u is reachable.

3fill in blank

hard

Fix the error in the loop that runs the relaxation steps for all vertices.

Data Structures Theory

for i in range([1] - 1):

Drag options to blanks, or click blank then click option'

Alen(graph)

Blen(graph) + 1

Clen(graph) * 2

Dlen(graph) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number of edges instead of vertices.

Using the full number of vertices instead of one less.

The Bellman-Ford algorithm requires |V| - 1 iterations to relax all edges, where |V| is the number of vertices.

4fill in blank

hard

Fill both blanks to check for negative weight cycles after relaxation.

Data Structures Theory

for u, v, w in edges:
    if distance[u] != [1] and distance[u] + w < [2]:

Drag options to blanks, or click blank then click option'

Ainfinity

Bdistance[v]

C0

Ddistance[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with distance[u] instead of distance[v].

Not checking if distance[u] is reachable.

After all relaxations, if we can still relax an edge, it means a negative weight cycle exists.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension for distances after relaxation.

Data Structures Theory

distances = {vertex: [1] for vertex in graph if graph[vertex] [2] [3]

Drag options to blanks, or click blank then click option'

A0

B!=

CNone

Dinfinity

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' in the condition.

Setting distance to infinity instead of zero.

This comprehension sets distance 0 for vertices that are reachable (not None).

Practice

1. What is the primary purpose of the Bellman-Ford algorithm in graph theory?

easy

A. To sort vertices in topological order

B. To find the minimum spanning tree of a graph

C. To detect cycles in an undirected graph

D. To find the shortest path from a single source to all vertices, even with negative edge weights

Solution

1. Step 1: Understand Bellman-Ford's goal

   The Bellman-Ford algorithm is designed to find shortest paths from one source vertex to all other vertices in a graph.

2. Step 2: Recognize its unique feature

   Unlike some algorithms, Bellman-Ford can handle graphs with negative edge weights and detect negative weight cycles.

3. Final Answer:

   To find the shortest path from a single source to all vertices, even with negative edge weights -> Option D

4. Quick Check:

   Bellman-Ford finds shortest paths with negative edges = A [OK]

Hint: Bellman-Ford handles negative edges and finds shortest paths [OK]

Common Mistakes:

• Confusing Bellman-Ford with algorithms for minimum spanning trees
• Thinking it only works with positive edge weights
• Assuming it sorts vertices or detects cycles in undirected graphs

2. Which of the following is the correct step repeated in the Bellman-Ford algorithm?

easy

A. Relax all edges |V|-1 times

B. Relax all edges once

C. Relax edges only if weights are positive

D. Relax edges in topological order

Solution

1. Step 1: Recall the relaxation process

   Bellman-Ford repeatedly relaxes all edges to update shortest path estimates.

2. Step 2: Identify the number of repetitions

   It performs this relaxation exactly |V|-1 times, where |V| is the number of vertices.

3. Final Answer:

   Relax all edges |V|-1 times -> Option A

4. Quick Check:

   Relax edges |V|-1 times = D [OK]

Hint: Relax edges |V|-1 times to find shortest paths [OK]

Common Mistakes:

• Relaxing edges only once
• Relaxing edges only if weights are positive
• Assuming topological order is required

3. Consider the graph with edges and weights:
A -> B (4), A -> C (5), B -> C (-3), C -> D (2)
Using Bellman-Ford from vertex A, what is the shortest distance to vertex C?

medium

A. 5

B. 1

C. -3

D. 4

Solution

1. Step 1: Initialize distances from A

   Distance to A=0, B=∞, C=∞, D=∞ initially.

2. Step 2: Relax edges repeatedly

   First relaxation: A->B=4, A->C=5. Then B->C updates C to 4 + (-3) = 1. C->D updates D to 1 + 2 = 3.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Shortest path A->B->C = 4 + (-3) = 1 [OK]

Hint: Check paths through neighbors to update shortest distances [OK]

Common Mistakes:

• Ignoring negative edge from B to C
• Taking direct edge A->C without relaxation
• Confusing edge weights with distances

4. You run Bellman-Ford on a graph and after |V|-1 iterations, you find an edge u -> v where distance[u] + weight(u,v) < distance[v]. What does this indicate?

medium

A. The graph contains a negative weight cycle reachable from the source

B. The algorithm has completed successfully with correct distances

C. There is a disconnected vertex in the graph

D. The graph is undirected

Solution

1. Step 1: Understand the relaxation check after iterations

   After |V|-1 relaxations, no edge should improve distances if no negative cycles exist.

2. Step 2: Interpret the condition

   If an edge can still relax a distance, it means a negative weight cycle exists reachable from the source.

3. Final Answer:

   The graph contains a negative weight cycle reachable from the source -> Option A

4. Quick Check:

   Extra relaxation means negative cycle = C [OK]

Hint: Extra relaxation after |V|-1 means negative cycle [OK]

Common Mistakes:

• Thinking algorithm finished correctly despite extra relaxation
• Confusing negative cycle with disconnected vertices
• Assuming graph type (directed/undirected) affects this check

5. Given a graph with vertices V = {S, A, B, C} and edges with weights:
S->A (1), A->B (-1), B->C (-1), C->A (-1)
What will Bellman-Ford detect when run from source S?

hard

A. It will find shortest paths without issues

B. It will report no path to vertex C

C. It will detect a negative weight cycle involving vertices A, B, and C

D. It will fail due to disconnected graph

Solution

1. Step 1: Analyze the cycle formed by A, B, and C

   The edges A->B (-1), B->C (-1), and C->A (-1) form a cycle with total weight -3, which is negative.

2. Step 2: Understand Bellman-Ford behavior on negative cycles

   Bellman-Ford detects negative weight cycles reachable from the source, which prevents shortest path calculation.

3. Final Answer:

   It will detect a negative weight cycle involving vertices A, B, and C -> Option C

4. Quick Check:

   Negative cycle A-B-C with sum -3 detected = B [OK]

Hint: Check cycles' total weight; negative sum means cycle detected [OK]

Common Mistakes:

• Assuming shortest paths exist despite negative cycle
• Ignoring the cycle formed by A, B, and C
• Thinking graph is disconnected when it is not