What if some roads actually pay you to travel them? How do you find the cheapest path then?
Why Bellman Ford Algorithm Negative Weights in DSA Javascript?
Imagine you have a map with cities connected by roads, some roads have tolls that can actually give you money back (negative cost). You want to find the cheapest way to travel from one city to all others.
Trying to find the cheapest path by checking all routes manually is slow and confusing, especially when some roads reduce your cost. You might miss cheaper paths or get stuck in loops.
The Bellman Ford algorithm smartly checks all roads multiple times to find the cheapest paths, even when some roads have negative costs. It also detects if there is a loop that keeps reducing cost endlessly.
function findCheapestPath(graph, start) {
// Manually check all paths, very complex and slow
// No clear way to handle negative costs
}function bellmanFord(graph, start) {
// Initialize distances
// Relax edges repeatedly
// Detect negative cycles
}You can reliably find shortest paths in graphs with negative weights and know if a negative cycle exists.
Calculating the cheapest cost to send packages between cities where some routes offer discounts or rebates.
Manual path checking is slow and error-prone with negative weights.
Bellman Ford handles negative weights and detects negative cycles.
It ensures correct shortest paths even with tricky costs.
Practice
Solution
Step 1: Understand Bellman-Ford's goal
The algorithm finds shortest paths from one source vertex to all others, handling negative weights.Step 2: Compare with other algorithms
Dijkstra's algorithm cannot handle negative weights, but Bellman-Ford can and also detects negative cycles.Final Answer:
To find the shortest path from a single source to all vertices, even with negative weights -> Option DQuick Check:
Bellman-Ford finds shortest paths with negative weights [OK]
- Confusing Bellman-Ford with Dijkstra's algorithm
- Thinking it finds longest paths
- Assuming it only detects cycles
Solution
Step 1: Initialize distances to Infinity
All vertices except the source start with distance Infinity to represent unknown paths.Step 2: Set source distance to zero
The source vertex distance is zero because the path from source to itself is zero.Final Answer:
const dist = new Array(V).fill(Infinity); dist[source] = 0; -> Option CQuick Check:
Distance array initialized with Infinity except source = 0 [OK]
- Filling array with zeros instead of Infinity
- Setting source distance to Infinity
- Using map without fill causing undefined values
edges = [[0,1,4],[0,2,5],[1,2,-3],[2,3,4]]Using Bellman-Ford from source 0, what is the distance array after all relaxations?
Solution
Step 1: Initialize distances
dist = [0, Infinity, Infinity, Infinity]Step 2: Relax edges |V|-1 = 3 times
First iteration:
- (0,1,4): dist[1] = 0 + 4 = 4
- (0,2,5): dist[2] = 0 + 5 = 5
- (1,2,-3): dist[2] = min(5, 4 + (-3)) = 1
- (2,3,4): dist[3] = 1 + 4 = 5
dist = [0,4,1,5]
Second and third iterations: no updates.Final Answer:
[0, 4, 1, 5] -> Option AQuick Check:
Distances after relaxations = [0,4,1,5] [OK]
- Not relaxing edges enough times
- Ignoring negative weight edge effect
- Incorrectly updating distances
for (let i = 0; i < V; i++) {
for (const [u, v, w] of edges) {
if (dist[u] + w < dist[v]) {
dist[v] = dist[u] + w;
}
}
}Solution
Step 1: Check loop iteration count
Bellman-Ford requires relaxing edges exactly V-1 times to find shortest paths.Step 2: Identify off-by-one error
The code runs the outer loop V times, which is one extra iteration and may cause incorrect behavior or unnecessary work.Final Answer:
The outer loop should run V-1 times, not V times -> Option AQuick Check:
Outer loop runs V-1 times in Bellman-Ford [OK]
- Running outer loop V times instead of V-1
- Mixing up edge and vertex iteration
- Wrong relaxation condition
edges = [[0,1,1],[1,2,-1],[2,3,-1],[3,1,-1]]and source 0, what does Bellman-Ford detect after running all relaxations?
Solution
Step 1: Analyze the cycle in the graph
The edges (1->2), (2->3), (3->1) form a cycle with total weight -3 (negative cycle).Step 2: Bellman-Ford detects negative cycles
After V-1 relaxations, one more iteration finds further distance reductions, indicating a negative cycle reachable from source.Final Answer:
Negative cycle detected reachable from source -> Option BQuick Check:
Negative cycle detection triggers on further relaxation [OK]
- Ignoring negative cycle presence
- Assuming zero-weight cycles cause failure
- Thinking distances remain Infinity
Ready to prove it?
Pass the test and this pattern moves unknown
