Bellman Ford Algorithm Negative Weights in DSA Rust - 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 edge weights are non-negative and may fail with negative weights.

intermediate

What does it mean if 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 shortest paths are found if no negative cycles exist.

beginner

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 that Dijkstra's algorithm cannot?

A Shortest path in unweighted graphs

B Maximum flow in networks

C Negative weight cycles

D Minimum spanning tree

Bellman Ford can detect negative weight cycles by checking for further relaxation after |V| - 1 iterations.

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

A |E| times

B |V| - 1 times

C |V| times

D Once

Bellman Ford relaxes all edges |V| - 1 times to guarantee shortest paths if no negative cycles exist.

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

A The graph has only positive weights

B The graph is disconnected

C The shortest path is found

D The graph has a negative weight cycle

Further relaxation after |V| - 1 iterations means a negative weight cycle exists.

Which of these is a correct time complexity for Bellman Ford?

A O(V * E)

B O(E^2)

C O(V + E)

D O(V^2)

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

Why can't Dijkstra's algorithm handle negative weight edges properly?

A It uses a greedy approach that assumes non-negative weights

B It only works on trees

C It requires sorted edges

D It uses dynamic programming

Dijkstra's greedy approach assumes edge weights are non-negative to guarantee correctness.

Explain how the Bellman Ford algorithm finds shortest paths in graphs with negative weights and how it detects negative weight cycles.

Think about repeated edge relaxation and what happens if distances keep decreasing.

Describe the difference between Bellman Ford and Dijkstra's algorithm in handling graphs with negative edge weights.

Focus on how each algorithm treats edge weights and cycles.

Practice

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

easy

A. It can handle graphs with negative weight edges.

B. It runs faster on all graphs.

C. It uses less memory.

D. It only works on directed acyclic graphs.

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 A

4. Quick Check:

   Negative weights handled = A [OK]

Hint: Bellman-Ford works with negative edges, Dijkstra doesn't [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 Rust code snippets correctly initializes the distance vector for Bellman-Ford with 5 vertices, starting from vertex 0?

easy

A. let mut dist = vec![i32::MIN; 5]; dist[0] = 0;

B. let mut dist = vec![0; 5]; dist[0] = i32::MAX;

C. let mut dist = vec![0; 5];

D. let mut dist = vec![i32::MAX; 5]; dist[0] = 0;

Solution

1. Step 1: Understand initialization for Bellman-Ford

   Distances to all vertices except the source should be set to a very large number (infinity), here represented by i32::MAX. The source vertex distance is set to 0.

2. Step 2: Check each option

   let mut dist = vec![i32::MAX; 5]; dist[0] = 0; correctly sets all distances to i32::MAX and source (index 0) to 0. Others either invert this or use wrong values.

3. Final Answer:

   let mut dist = vec![i32::MAX; 5]; dist[0] = 0; -> Option D

4. Quick Check:

   Initialize distances with max except source 0 = D [OK]

Hint: Set all distances max except source zero [OK]

Common Mistakes:

• Setting source distance to max instead of zero
• Initializing all distances to zero
• Using i32::MIN which is negative infinity

3. What will be the output distances vector after running one iteration of Bellman-Ford on this graph with edges:
edges = [(0, 1, 4), (0, 2, 5), (1, 2, -3), (2, 3, 4)] starting from vertex 0?
Initial distances: [0, i32::MAX, i32::MAX, i32::MAX]

medium

A. [0, 4, 5, i32::MAX]

B. [0, 4, 1, 5]

C. [0, 4, 5, 9]

D. [0, 4, 1, i32::MAX]

Solution

1. Step 1: Relax edges once from initial distances

   Start distances: [0, MAX, MAX, MAX]. Relax edges:
   - Edge (0->1,4): dist[1] = min(MAX, 0+4) = 4
   - Edge (0->2,5): dist[2] = min(MAX, 0+5) = 5
   - Edge (1->2,-3): dist[2] = min(5, 4-3) = 1
   - Edge (2->3,4): dist[3] = min(MAX, 1+4) = 5

2. Step 2: Determine distances after one iteration

   After relaxing all edges once, distances are [0, 4, 1, 5]. But since the question asks after one iteration (one full pass), all edges are relaxed once, so the final distances are [0, 4, 1, 5].

3. Final Answer:

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

4. Quick Check:

   Relax edges once = [0,4,1,5] [OK]

Hint: Relax all edges once to update distances [OK]

Common Mistakes:

• Not relaxing edges in order
• Ignoring negative edge effect
• Stopping after partial relaxation

4. Consider this Rust code snippet for Bellman-Ford. What is the bug causing incorrect detection of negative cycles?

for _ in 0..graph.len() {
    for &(u, v, w) in &edges {
        if dist[u] != i32::MAX && dist[u] + w < dist[v] {
            dist[v] = dist[u] + w;
        }
    }
}
// Check for negative cycle
for &(u, v, w) in &edges {
    if dist[u] != i32::MAX && dist[u] + w < dist[v] {
        return true; // Negative cycle detected
    }
}
return false;

medium

A. The outer loop should run graph.len() - 1 times, not graph.len() times.

B. The distance vector is not initialized properly.

C. The negative cycle check should be inside the outer loop.

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 (where |V| is number of vertices) to find shortest paths. Running it |V| times can cause incorrect behavior.

2. Step 2: Identify the bug in loop count

   The code runs the outer loop graph.len() times (equal to |V|), which is one too many. This can cause false detection of negative cycles.

3. Final Answer:

   The outer loop should run graph.len() - 1 times, not graph.len() times. -> Option A

4. Quick Check:

   Relax edges |V|-1 times = A [OK]

Hint: Relax edges exactly vertices minus one times [OK]

Common Mistakes:

• Running relaxation too many times
• Checking negative cycle inside main loop
• Incorrect distance initialization

5. You have a graph with 4 vertices and edges:
[(0, 1, 1), (1, 2, -1), (2, 3, -1), (3, 1, -1)]. Using Bellman-Ford starting from vertex 0, what will the algorithm detect?

hard

A. No negative cycle, shortest paths found.

B. Algorithm will run infinitely without detection.

C. Negative cycle detected involving vertices 1, 2, and 3.

D. Only positive weight cycles detected.

Solution

1. Step 1: Analyze the cycle formed by edges

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

2. Step 2: Bellman-Ford detects negative cycles reachable from source

   Since this negative cycle is reachable from vertex 0 via edge (0->1), Bellman-Ford will detect it during the negative cycle check.

3. Final Answer:

   Negative cycle detected involving vertices 1, 2, and 3. -> Option C

4. Quick Check:

   Negative cycle in reachable cycle = C [OK]

Hint: Check cycles reachable from source for negative sum [OK]

Common Mistakes:

• Ignoring negative cycle presence
• Assuming algorithm runs infinitely
• Thinking only positive cycles exist