Bellman Ford Algorithm Negative Weights in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance array with infinity.

DSA C

for (int i = 0; i < V; i++) {
    dist[i] = [1];
}

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

A1

B0

CINT_MAX

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of INT_MAX causes incorrect shortest path calculations.

Using negative values for initial distances is incorrect.

We use INT_MAX to represent infinity for initial distances.

2fill in blank

medium

Complete the code to relax edges inside the main loop.

DSA C

if (dist[u] != [1] && dist[u] + weight < dist[v]) {
    dist[v] = dist[u] + weight;
}

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

A0

Bweight

C-1

DINT_MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against 0 instead of INT_MAX causes wrong relaxations.

Not checking distance before relaxation leads to overflow.

We check if dist[u] is not infinity (INT_MAX) before relaxing.

3fill in blank

hard

Fix the error in the negative cycle detection condition.

DSA C

if (dist[u] != [1] && dist[u] + weight < dist[v]) {
    printf("Graph contains negative weight cycle\n");
    return 0;
}

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

AINT_MAX

B-1

C0

Dweight

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against 0 or -1 causes incorrect negative cycle detection.

Not checking distance leads to runtime errors.

We must check dist[u] is not INT_MAX before detecting negative cycles.

4fill in blank

hard

Fill both blanks to complete the edge relaxation loop.

DSA C

for (int i = 1; i <= [1] - 1; i++) {
    for (int j = 0; j < [2]; j++) {
        int u = edges[j].src;
        int v = edges[j].dest;
        int weight = edges[j].weight;
        if (dist[u] != INT_MAX && dist[u] + weight < dist[v]) {
            dist[v] = dist[u] + weight;
        }
    }
}

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

AV

BE

CV-1

DE-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using V-1 instead of V for outer loop causes insufficient iterations.

Using E-1 for inner loop misses the last edge.

The outer loop runs V-1 times and the inner loop runs over all E edges.

5fill in blank

hard

Fill all three blanks to print the shortest distances after Bellman-Ford completes.

DSA C

printf("Vertex Distance from Source\n");
for (int i = 0; i < [1]; i++) {
    if (dist[i] == [2]) {
        printf("%d -> INF\n", [3]);
    } else {
        printf("%d -> %d\n", i, dist[i]);
    }
}

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

AV

BINT_MAX

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of INT_MAX causes wrong INF detection.

Printing wrong vertex index causes incorrect output.

Loop over all vertices V, check if distance is INT_MAX to print INF, and print vertex i.

Practice

1. What is the main advantage of the Bellman-Ford algorithm compared to Dijkstra's algorithm?

easy

A. It only works on directed acyclic graphs.

B. It runs faster on all graphs.

C. It uses less memory.

D. It can handle graphs with negative weight edges.

Solution

1. Step 1: Understand the limitation of Dijkstra's algorithm

   Dijkstra's algorithm cannot correctly handle graphs with negative weight edges because it assumes that once a node's shortest path is found, it cannot be improved.

2. Step 2: Recognize Bellman-Ford's capability

   Bellman-Ford algorithm can handle negative weight edges and detect negative weight cycles, making it more versatile.

3. Final Answer:

   It can handle graphs with negative weight edges. -> Option D

4. Quick Check:

   Negative weights handled = A [OK]

Hint: Bellman-Ford works with negative edges; Dijkstra does not [OK]

Common Mistakes:

• Thinking Bellman-Ford is always faster
• Assuming it uses less memory
• Believing it only works on acyclic graphs

2. Which of the following is the correct way to initialize the distance array in Bellman-Ford algorithm in C?

easy

A. int dist[V]; for(int i=0; i

B. int dist[V]; for(int i=0; i

C. int dist[V]; for(int i=0; i

D. int dist[V]; for(int i=0; i

Solution

1. Step 1: Understand distance initialization

   Distances to all vertices except the source should be set to a very large number (infinity), here represented by INT_MAX, to indicate they are initially unreachable.

2. Step 2: Set source distance to zero

   The distance to the source vertex itself must be zero because the shortest path from source to source is zero.

3. Final Answer:

   int dist[V]; for(int i=0; i -> Option B

4. Quick Check:

   Initialize all to max, source to 0 = C [OK]

Hint: Set all distances to INT_MAX except source = 0 [OK]

Common Mistakes:

• Initializing all distances to zero
• Using -1 for unreachable distances
• Setting all distances to source index

3. What will be the output distances array after running Bellman-Ford on this graph with edges:
edges = [(0,1,4), (0,2,5), (1,2,-3), (2,3,4)] starting from source 0?

medium

A. [0, 4, 1, 5]

B. [0, 4, 1, 9]

C. [0, 4, 5, 9]

D. [0, 4, -3, 4]

Solution

1. Step 1: Initialize distances

   dist = [0, INF, INF, INF]

2. Step 2: Relax edges repeatedly

   First iteration:
   dist[1] = min(INF, 0+4) = 4
   dist[2] = min(INF, 0+5) = 5
   dist[2] = min(5, 4 + (-3)) = 1
   dist[3] = min(INF, 1 + 4) = 5
   Second iteration:
   dist[1] stays 4
   dist[2] stays 1
   dist[3] stays 5
   No further updates.

3. Final distances:

   [0, 4, 1, 5]

4. Final Answer:

   [0, 4, 1, 5] -> Option A

5. Quick Check:

   Relax edges to update distances = A [OK]

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

Common Mistakes:

• Not relaxing edges enough times
• Ignoring negative edge effect
• Wrong initial distances

4. Identify the error in this Bellman-Ford code snippet in C:

for (int i = 0; i < V; i++) {
  for (int j = 0; j < E; j++) {
    int u = edges[j].src;
    int v = edges[j].dest;
    int w = edges[j].weight;
    if (dist[u] + w < dist[v]) {
      dist[v] = dist[u] + w;
    }
  }
}

medium

A. The outer loop should run V-1 times, not V times.

B. The inner loop should run V times, not E times.

C. The distance array is not initialized.

D. The weight comparison should be dist[v] + w < dist[u].

Solution

1. Step 1: Check loop iteration count

   Bellman-Ford requires relaxing edges exactly V-1 times to find shortest paths correctly.

2. Step 2: Identify the incorrect loop count

   The outer loop runs V times here, which is one extra iteration and unnecessary.

3. Final Answer:

   The outer loop should run V-1 times, not V times. -> Option A

4. Quick Check:

   Outer loop runs V-1 times = D [OK]

Hint: Relax edges exactly V-1 times for correct shortest paths [OK]

Common Mistakes:

• Running outer loop V times instead of V-1
• Confusing inner loop count with vertex count
• Incorrect weight comparison direction

5. Given a graph with edges
[(0,1,1), (1,2,-1), (2,3,-1), (3,1,-1)], what will Bellman-Ford detect when run from source 0?

hard

A. It will run infinitely without stopping.

B. It will find shortest paths without any issues.

C. It will detect a negative weight cycle.

D. It will give incorrect distances but no error.

Solution

1. Step 1: Analyze the cycle in the graph

   The edges (1->2), (2->3), and (3->1) form a cycle with total weight -1 + -1 + -1 = -3, which is negative.

2. Step 2: Understand Bellman-Ford's negative cycle detection

   Bellman-Ford detects negative weight cycles by checking for further relaxation after V-1 iterations. Here, it will detect the negative cycle.

3. Final Answer:

   It will detect a negative weight cycle. -> Option C

4. Quick Check:

   Negative cycle detected = A [OK]

Hint: Negative cycle if distances keep decreasing after V-1 relaxations [OK]

Common Mistakes:

• Assuming algorithm runs infinitely
• Believing it finds shortest paths despite negative cycle
• Ignoring cycle detection step