How
Bellman Ford's algorithm works
Bellman
Ford algorithm works by overestimating the length of the path from
the starting vertex to all other vertices. Then it iteratively
relaxes those estimates by finding new paths that are shorter than
the previously overestimated paths
Steps
1)
This step initializes distances from source to all vertices as
infinite and distance to source itself as 0. Create an array dist[]
of size |V| with all values as infinite except dist[src] where src is
source vertex.
2)
This step calculates shortest distances. Do following |V|-1 times
where |V| is the number of vertices in given graph.
…..a) Do following for each edge u-v
………………If dist[v] > dist[u] + weight of edge uv, then update dist[v]
………………….dist[v] = dist[u] + weight of edge uv
…..a) Do following for each edge u-v
………………If dist[v] > dist[u] + weight of edge uv, then update dist[v]
………………….dist[v] = dist[u] + weight of edge uv
3)
This step reports if there is a negative weight cycle in graph. Do
following for each edge u-v
……If dist[v] > dist[u] + weight of edge uv, then “Graph contains negative weight cycle”
The idea of step 3 is, step 2 guarantees shortest distances if graph doesn’t contain negative weight cycle. If we iterate through all edges one more time and get a shorter path for any vertex, then there is a negative weight cycle
……If dist[v] > dist[u] + weight of edge uv, then “Graph contains negative weight cycle”
The idea of step 3 is, step 2 guarantees shortest distances if graph doesn’t contain negative weight cycle. If we iterate through all edges one more time and get a shorter path for any vertex, then there is a negative weight cycle
How
does this work?
Like other Dynamic Programming Problems, the algorithm calculate
shortest paths in bottom-up manner. It first calculates the shortest
distances which have at-most one edge in the path. Then, it
calculates shortest paths with at-most 2 edges, and so on. After the
i-th iteration of outer loop, the shortest paths with at most i edges
are calculated. There can be maximum |V| – 1 edges in any simple
path, that is why the outer loop runs |v| – 1 times. The idea is,
assuming that there is no negative weight cycle, if we have
calculated shortest paths with at most i edges, then an iteration
over all edges guarantees to give shortest path with at-most (i+1)
edges (Proof is simple, you can refer this
or MIT
Video Lecture)
Example
Let us understand the algorithm with following example graph. The images are taken from this source.
Let us understand the algorithm with following example graph. The images are taken from this source.
Let
the given source vertex be 0. Initialize all distances as infinite,
except the distance to source itself. Total number of vertices in the
graph is 5, so all
edges must be processed 4 times.
Let
all edges are processed in following order: (B, E), (D, B), (B, D),
(A, B), (A, C), (D, C), (B, C), (E, D). We get following distances
when all edges are processed first time. The first row in shows
initial distances. The second row shows distances when edges (B, E),
(D, B), (B, D) and (A, B) are processed. The third row shows
distances when (A, C) is processed. The fourth row shows when (D, C),
(B, C) and (E, D) are processed.

The
first iteration guarantees to give all shortest paths which are at
most 1 edge long. We get following distances when all edges are
processed second time (The last row shows final values).
The
second iteration guarantees to give all shortest paths which are at
most 2 edges long. The algorithm processes all edges 2 more times.
The distances are minimized after the second iteration, so third and
fourth iterations don’t update the distances.