Given a weighted directed graph G(V,E) and a source vertex s ∈ V, determine for each vertex v ∈ V the shortest path between s and v.
- Initialize the distance from the source to all vertices as infinite.
- Initialize the distance to itself as 0.
- Create an array dist[] of size |V| with all values as infinite except dist[s].
- Repeat the following |V| - 1 times. Where |V| is number of vertices.
- Create another loop to go through each edge (u, v) in E and do the following:
- dist[v] = minimum(dist[v], dist[u] + weight of edge).
- Lastly iterate through all edges on last time to make sure there are no negatively weighted cycles.
O(VE)
O(V^2)
- Richard Bellman & Lester Ford, Jr.
# of vertices in graph = 5 [A, B, C, D, E]
# of edges in graph = 8
edges [A->B, A->C, B->C, B->D, B->E, D->C, D->B, E->D]
weight [ -1, 4, 3, 2, 2, 5, 1, -4 ]
source [ A, A, B, B, B, D, D, E ]
// edge A->B
graph->edge[0].src = A
graph->edge[0].dest = B
graph->edge[0].weight = -1
// edge A->C
graph->edge[1].src = A
graph->edge[1].dest = C
graph->edge[1].weight = 4
// edge B->C
graph->edge[2].src = B
graph->edge[2].dest = C
graph->edge[2].weight = 3
// edge B->D
graph->edge[3].src = B
graph->edge[3].dest = D
graph->edge[3].weight = 2
// edge B->E
graph->edge[4].src = B
graph->edge[4].dest = E
graph->edge[4].weight = 2
// edge D->C
graph->edge[5].src = D
graph->edge[5].dest = C
graph->edge[5].weight = 5
// edge D->B
graph->edge[6].src = D
graph->edge[6].dest = B
graph->edge[6].weight = 1
// edge E->D
graph->edge[7].src = E
graph->edge[7].dest = D
graph->edge[7].weight = -3
for source = A
Vertex Distance from Source
A 0 A->A
B -1 A->B
C 2 A->B->C = -1 + 3
D -2 A->B->E->D = -1 + 2 + -3
E 1 A->B->E = -1 + 2
A video explaining the Bellman-Ford Algorithm
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