Bellman Ford Algorithm Negative Weights in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What problem does the Bellman Ford algorithm solve?

It finds the shortest paths from a single source vertex to all other vertices in a graph, even if some edges have negative weights.

intermediate

How does Bellman Ford handle negative weight edges differently from Dijkstra's algorithm?

Bellman Ford can handle negative weight edges by relaxing all edges multiple times, while Dijkstra's algorithm assumes all edge weights are non-negative and may fail with negative weights.

beginner

What is the main step repeated in the Bellman Ford algorithm?

Relaxing all edges in the graph repeatedly (|V| - 1) times, where |V| is the number of vertices.

intermediate

How does Bellman Ford detect a negative weight cycle?

After |V| - 1 relaxations, it performs one more relaxation step. If any distance can still be reduced, a negative weight cycle exists.

beginner

What is the time complexity of the Bellman Ford algorithm?

O(|V| * |E|), where |V| is the number of vertices and |E| is the number of edges.

What does the Bellman Ford algorithm do if it finds a negative weight cycle?

A Reports that no shortest path exists due to the cycle

B Ignores the cycle and continues

C Stops and returns the current distances

D Converts negative weights to positive

Bellman Ford detects negative weight cycles and reports that shortest paths are undefined because the cycle can reduce path cost indefinitely.

How many times does Bellman Ford relax all edges in the worst case?

A |V| - 1 times

B |V| times

C |E| times

D Once

Bellman Ford relaxes all edges |V| - 1 times to ensure shortest paths are found.

Which of these algorithms can handle graphs with negative edge weights?

A Dijkstra's algorithm

B Prim's algorithm

C Bellman Ford algorithm

D Kruskal's algorithm

Only Bellman Ford can handle negative edge weights safely.

What is the main data structure used to store distances in Bellman Ford?

A Stack

B Array or map of distances

C Queue

D Linked list

Distances from the source to each vertex are stored in an array or map.

If no negative weight cycles exist, what does Bellman Ford guarantee?

A No paths found

B Longest paths from source to all vertices

C Only paths with positive weights

D Shortest paths from source to all vertices

Bellman Ford guarantees shortest paths if no negative weight cycles exist.

Explain how the Bellman Ford algorithm detects negative weight cycles in a graph.

Think about what happens if distances keep decreasing after expected relaxations.

Describe the main steps of the Bellman Ford algorithm for finding shortest paths.

Focus on initialization, relaxation, and cycle detection.

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