Bellman Ford Algorithm Negative Weights in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance array with infinity values.

DSA Go

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

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

A0

Bmath.MaxInt32

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing distances with 0 instead of infinity.

Using negative numbers for initial distances.

We use math.MaxInt32 to represent infinity for initial distances.

2fill in blank

medium

Complete the code to relax edges in the Bellman-Ford algorithm.

DSA Go

for i := 0; i < E; i++ {
    u := edges[i].src
    v := edges[i].dest
    w := edges[i].weight
    if dist[u] != math.MaxInt32 && dist[u] + w [1] dist[v] {
        dist[v] = dist[u] + w
    }
}

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

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causing incorrect relaxation.

Not checking if dist[u] is infinity before relaxing.

We relax edges only if the current distance to u plus weight w is less than the current distance to v.

3fill in blank

hard

Fix the error in the negative cycle detection condition.

DSA Go

for i := 0; i < E; i++ {
    u := edges[i].src
    v := edges[i].dest
    w := edges[i].weight
    if dist[u] != math.MaxInt32 && dist[u] + w [1] dist[v] {
        return true // Negative cycle detected
    }
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' in the condition.

Not checking if dist[u] is infinity.

Negative cycle exists if we can still relax an edge, meaning dist[u] + w is less than dist[v].

4fill in blank

hard

Fill both blanks to complete the main loop and relaxation step.

DSA Go

for i := 1; i < [1]; i++ {
    for j := 0; j < E; j++ {
        u := edges[j].src
        v := edges[j].dest
        w := edges[j].weight
        if dist[u] != math.MaxInt32 && dist[u] + w [2] dist[v] {
            dist[v] = dist[u] + w
        }
    }
}

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

AV

B<

C>

DE

Attempts:

3 left

💡 Hint

Common Mistakes

Using E instead of V for the outer loop.

Using '>' instead of '<' in the relaxation condition.

The outer loop runs V-1 times, and relaxation happens if dist[u] + w is less than dist[v].

5fill in blank

hard

Fill all three blanks to create a map of vertex to distance for vertices with distance less than infinity.

DSA Go

result := map[int]int{}
for i := 0; i < V; i++ {
    if dist[i] [1] math.MaxInt32 {
        result[[2]] = [3]
    }
}

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

A<

Bi

Cdist[i]

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' in the condition.

Using dist instead of dist[i] or wrong key in the map.

We add vertices with finite distances to the result map using their index and distance.

Practice

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

easy

A. It runs faster on all graphs.

B. It uses less memory.

C. It can handle graphs with negative weight edges.

D. It only works on directed acyclic graphs.

Solution

1. Step 1: Understand Dijkstra's limitation

   Dijkstra's algorithm cannot handle negative weight edges because it assumes once a node is visited with the shortest path, it won't be updated again.

2. Step 2: Bellman-Ford's capability

   Bellman-Ford algorithm relaxes edges multiple times and can detect negative weight edges, allowing it to work correctly even with negative weights.

3. Final Answer:

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

4. Quick Check:

   Negative weights handled = B [OK]

Hint: Bellman-Ford works with negative edges, Dijkstra doesn't [OK]

Common Mistakes:

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

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

easy

A. dist := make([]int, V); for i := range dist { dist[i] = math.MaxInt32 }

B. dist := make([]int, V); for i := range dist { dist[i] = 0 }

C. dist := make([]int, V); for i := range dist { dist[i] = -1 }

D. dist := make([]int, V); for i := range dist { dist[i] = 1 }

Solution

1. Step 1: Understand distance initialization

   Distances to all vertices except the source should be set to a very large number (infinity) to represent unknown distance.

2. Step 2: Use math.MaxInt32 for infinity

   In Go, math.MaxInt32 is used to represent a large integer value for initialization.

3. Final Answer:

   dist := make([]int, V); for i := range dist { dist[i] = math.MaxInt32 } -> Option A

4. Quick Check:

   Initialize distances to max int = A [OK]

Hint: Initialize distances to max int except source [OK]

Common Mistakes:

• Initializing distances to zero
• Using negative values for distances
• Not initializing distances at all

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

medium

A. [0, 4, 5, 9]

B. [0, 4, 1, 5]

C. [0, 4, 1, 9]

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

Solution

1. Step 1: Initialize distances

   dist = [0, ∞, ∞, ∞]

2. Step 2: Relax edges repeatedly

   After first relaxation: dist[1]=4, dist[2]=5
   After second relaxation: dist[2]=min(5,4 + (-3))=1
   After third relaxation: dist[3]=dist[2] + 4 = 1 + 4 = 5
   After further relaxations no changes occur.

3. Final Answer:

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

4. Quick Check:

   Relax edges stepwise = A [OK]

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

Common Mistakes:

• Not updating dist[2] after negative edge relaxation
• Adding weights incorrectly
• Confusing edge directions

4. Identify the bug in this Go code snippet for Bellman-Ford algorithm:

for i := 0; i < V; i++ {
  for _, e := range edges {
    if dist[e.src] + e.weight < dist[e.dest] {
      dist[e.dest] = dist[e.src] + e.weight
    }
  }
}

medium

A. Loop should run V-1 times, not V times.

B. The condition should use <= instead of <.

C. dist array is not initialized.

D. Edges should be relaxed only once.

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 off-by-one error

   The loop runs V times, which is one extra iteration and may cause incorrect behavior or unnecessary work.

3. Final Answer:

   Loop should run V-1 times, not V times. -> Option A

4. Quick Check:

   Relax edges V-1 times = A [OK]

Hint: Relax edges exactly V-1 times [OK]

Common Mistakes:

• Running loop V times instead of V-1
• Using wrong comparison operator
• Not initializing dist array

5. Given a graph with edges:
edges = []Edge{{0,1,1}, {1,2,-1}, {2,3,-1}, {3,1,-1}}
and source vertex 0, what does Bellman-Ford detect after running?

hard

A. No negative cycle detected; shortest paths found.

B. Distances remain infinite for all vertices except source.

C. Algorithm fails due to zero-weight edges.

D. Negative weight cycle detected involving vertices 1, 2, 3.

Solution

1. Step 1: Analyze edges for cycles

   The edges form a cycle 1 -> 2 -> 3 -> 1 with total weight -3 (negative cycle).

2. Step 2: Bellman-Ford cycle detection

   After V-1 relaxations, one more relaxation detects if any distance can still be reduced, indicating a negative cycle.

3. Final Answer:

   Negative weight cycle detected involving vertices 1, 2, 3. -> Option D

4. Quick Check:

   Negative cycle detection = D [OK]

Hint: Check for distance updates after V-1 relaxations [OK]

Common Mistakes:

• Ignoring negative cycles
• Assuming zero-weight edges cause failure
• Thinking no cycle exists