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

Recall & Review

beginner

What is the main purpose of the Bellman Ford algorithm?

The Bellman Ford algorithm finds the shortest paths from a single source vertex to all other vertices in a weighted 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 correctly process graphs with negative weight edges by relaxing all edges multiple times, while Dijkstra's algorithm assumes all weights are non-negative and can give wrong results if negative weights exist.

intermediate

What does it mean when Bellman Ford detects a negative weight cycle?

It means there is a cycle in the graph where the total sum of edge weights is negative, causing the shortest path to be undefined because you can keep going around the cycle to reduce the path cost infinitely.

beginner

In Bellman Ford, how many times do we relax all edges in the graph?

We relax all edges exactly (V - 1) times, where V is the number of vertices in the graph. This ensures the shortest paths are found if no negative cycles exist.

intermediate

What is the time complexity of the Bellman Ford algorithm?

The time complexity is O(V * E), where V is the number of vertices and E is the number of edges, because it relaxes all edges (V - 1) times.

What does the Bellman Ford algorithm detect besides shortest paths?

A Negative weight cycles

B Maximum flow

C Minimum spanning tree

D Strongly connected components

Bellman Ford can detect negative weight cycles by checking if any edge can still be relaxed after (V - 1) iterations.

How many times are edges relaxed in Bellman Ford for a graph with V vertices?

A V times

B V - 1 times

C E times

D E - 1 times

Edges are relaxed (V - 1) times to ensure shortest paths are found.

Which type of graph edge weights can Bellman Ford handle that Dijkstra cannot?

A Negative weights

B Positive weights only

C Zero weights only

D Non-integer weights

Bellman Ford works correctly with negative weight edges, unlike Dijkstra.

What happens if Bellman Ford finds an edge that can still be relaxed after (V - 1) iterations?

A The graph has no edges

B The graph is disconnected

C The algorithm finished successfully

D The graph has a negative weight cycle

Relaxing an edge after (V - 1) iterations means a negative weight cycle exists.

What is the worst-case time complexity of Bellman Ford?

A O(V + E)

B O(V^2)

C O(V * E)

D O(E^2)

Bellman Ford runs in O(V * E) time because it relaxes all edges (V - 1) times.

Explain how the Bellman Ford algorithm finds shortest paths in graphs with negative weights.

Think about how edges are relaxed and how cycles affect shortest paths.

Describe the significance of detecting a negative weight cycle in Bellman Ford.

Why does a negative cycle break shortest path calculations?

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: Identify Bellman Ford's capability

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

3. Final Answer:

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

4. Quick Check:

   Negative weights handled = D [OK]

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

Common Mistakes:

• Thinking Bellman Ford is always faster than Dijkstra
• Believing Bellman Ford cannot detect negative cycles
• Assuming Bellman Ford 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. vector<int> dist(V, 0); dist[src] = INT_MAX;

B. vector<int> dist(V, INT_MAX); dist[src] = 0;

C. vector<int> dist(V, -1); dist[src] = 0;

D. vector<int> dist(V, INT_MIN); dist[src] = 0;

Solution

1. Step 1: Understand distance initialization

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

2. Step 2: Set source distance to zero

   The source vertex distance is zero because the distance from source to itself is zero.

3. Final Answer:

   vector<int> dist(V, INT_MAX); dist[src] = 0; -> Option B

4. Quick Check:

   Initialize with INT_MAX and source zero = B [OK]

Hint: Initialize all distances to INT_MAX except source zero [OK]

Common Mistakes:

• Setting all distances to zero
• Using INT_MIN which is incorrect for distances
• Not setting source distance to zero

3. Consider the following graph edges and weights:

Edges: (0->1, 4), (0->2, 5), (1->2, -3), (2->3, 4), (3->1, 6)

What is the shortest distance from vertex 0 to vertex 3 after running Bellman Ford?

medium

A. 5

B. 9

C. 6

D. 7

Solution

1. Step 1: Initialize distances

   dist[0]=0, dist[1]=∞, dist[2]=∞, dist[3]=∞.

2. Step 2: Relax edges repeatedly

   After first iteration: dist[1]=4, dist[2]=5, dist[3]=∞.
   Relax (1->2): dist[2]=4 + (-3) = 1.
   Relax (2->3): dist[3]=1 + 4 = 5.
   Relax (3->1): dist[1]=min(4, 5 + 6=11) = 4.
   Second iteration: No further improvements.
   Final dist[3] = 5.

3. Step 3: Check for negative cycles

   No negative cycle detected.

4. Final Answer:

   5 -> Option A

5. Quick Check:

   Shortest path dist to 3 = 5 [OK]

Hint: Relax edges V-1 times, track minimum distances [OK]

Common Mistakes:

• Not updating distances after negative edge relaxation
• Confusing edge directions
• Stopping relaxation too early

4. The following C++ code snippet attempts to implement Bellman Ford but has a bug:

for (int i = 0; i < V; i++) {
  for (auto edge : edges) {
    if (dist[edge.u] + edge.w < dist[edge.v]) {
      dist[edge.v] = dist[edge.u] + edge.w;
    }
  }
}

What is the bug?

medium

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

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

C. The distance array is not initialized.

D. Edges should be relaxed in reverse order.

Solution

1. Step 1: Understand Bellman Ford iteration count

   Bellman Ford requires relaxing edges exactly V-1 times to find shortest paths without negative cycles.

2. Step 2: Identify the loop error

   The code runs the outer loop V times, which is one extra iteration and may cause incorrect detection of negative cycles or unnecessary work.

3. Final Answer:

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

4. Quick Check:

   Outer loop count = V-1 [OK]

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

Common Mistakes:

• Running outer loop V times instead of V-1
• Changing comparison operator incorrectly
• Ignoring distance initialization

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 vertex 0?

hard

A. Graph is disconnected.

B. Shortest distances without any issues.

C. A negative weight cycle reachable from source.

D. Algorithm will run infinitely.

Solution

1. Step 1: Identify 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 reachable from the source by checking if any edge can still be relaxed after V-1 iterations.

3. Step 3: Conclusion

   Since the negative cycle is reachable from vertex 0, Bellman Ford will detect it and report its presence.

4. Final Answer:

   A negative weight cycle reachable from source. -> Option C

5. Quick Check:

   Negative cycle detected = A [OK]

Hint: Negative cycle if edges relax after V-1 iterations [OK]

Common Mistakes:

• Ignoring negative cycles reachable from source
• Assuming algorithm runs infinitely
• Thinking graph is disconnected