Bellman Ford Algorithm Negative Weights in DSA C++ - Build from Scratch

Bellman Ford Algorithm Negative Weights

📖 Scenario: You are working on a navigation system that finds the shortest path between cities. Some roads have toll discounts, which means the cost can be negative. You need to find the shortest path from a starting city to all other cities, even if some roads reduce the cost.

🎯 Goal: Build a program using the Bellman Ford algorithm to find the shortest distances from a starting city to all other cities in a graph that may have negative edge weights.

📋 What You'll Learn

Create a list of edges representing the graph with exact values

Add a variable for the number of vertices called V

Implement the Bellman Ford algorithm using the given edge list and V

Print the shortest distances from the source city after running the algorithm

💡 Why This Matters

🌍 Real World

Navigation systems and routing algorithms often need to handle roads with discounts or penalties, which can be modeled as negative weights.

💼 Career

Understanding Bellman Ford is important for software engineers working on graph algorithms, network routing, and optimization problems.

1

Create the graph edges

Create a vector of edges called edges with these exact entries: {0, 1, 5}, {1, 2, -2}, {0, 2, 4}, {2, 3, 3}, {3, 1, 1}. Each edge is a struct with src, dest, and weight.

DSA C++

# Your code here

Hint

Use a vector<Edge> and initialize it with the exact edges given.

2

Add the number of vertices

Add an integer variable called V and set it to 4 to represent the number of vertices in the graph.

DSA C++

    vector edges = {
        {0, 1, 5},
        {1, 2, -2},
        {0, 2, 4},
        {2, 3, 3},
        {3, 1, 1}
    };
    // Your code here

Hint

Set V to the total number of vertices, which is 4.

3

Implement the Bellman Ford algorithm

Create a vector dist of size V initialized with INT_MAX. Set dist[0] to 0 as the source vertex. Use a for loop to relax all edges V - 1 times. Inside the loop, use a for loop to iterate over edges and update dist if a shorter path is found.

DSA C++

    vector dist(V, INT_MAX);
    dist[0] = 0;
    // Your code here

Hint

Initialize distances, then relax edges repeatedly to update shortest paths.

4

Print the shortest distances

Use a for loop with variable i from 0 to V - 1 to print the shortest distance to each vertex in the format: "Distance to vertex i is dist[i]".

DSA C++

    for (int i = 0; i < V; i++) {
        // Your code here
    }
    return 0;
}

Hint

Print each vertex number and its distance on a new line.

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