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from __future__ import annotations


def print_distance(distance: list[float], src):
    print(f"Vertex\tShortest Distance from vertex {src}")
    for i, d in enumerate(distance):
        print(f"{i}\t\t{d}")


def check_negative_cycle(
    graph: list[dict[str, int]], distance: list[float], edge_count: int
):
    for j in range(edge_count):
        u, v, w = [graph[j][k] for k in ["src", "dst", "weight"]]
        if distance[u] != float("inf") and distance[u] + w < distance[v]:
            return True
    return False


def bellman_ford(
    graph: list[dict[str, int]], vertex_count: int, edge_count: int, src: int
) -> list[float]:
    """
    Returns shortest paths from a vertex src to all
    other vertices.
    >>> edges = [(2, 1, -10), (3, 2, 3), (0, 3, 5), (0, 1, 4)]
    >>> g = [{"src": s, "dst": d, "weight": w} for s, d, w in edges]
    >>> bellman_ford(g, 4, 4, 0)
    [0.0, -2.0, 8.0, 5.0]
    >>> g = [{"src": s, "dst": d, "weight": w} for s, d, w in edges + [(1, 3, 5)]]
    >>> bellman_ford(g, 4, 5, 0)
    Traceback (most recent call last):
     ...
    Exception: Negative cycle found
    """
    distance = [float("inf")] * vertex_count
    distance[src] = 0.0

    for i in range(vertex_count - 1):
        for j in range(edge_count):
            u, v, w = [graph[j][k] for k in ["src", "dst", "weight"]]

            if distance[u] != float("inf") and distance[u] + w < distance[v]:
                distance[v] = distance[u] + w

    negative_cycle_exists = check_negative_cycle(graph, distance, edge_count)
    if negative_cycle_exists:
        raise Exception("Negative cycle found")

    return distance


if __name__ == "__main__":
    import doctest

    doctest.testmod()

    V = int(input("Enter number of vertices: ").strip())
    E = int(input("Enter number of edges: ").strip())

    graph: list[dict[str, int]] = [dict() for j in range(E)]

    for i in range(E):
        print("Edge ", i + 1)
        src, dest, weight = [
            int(x)
            for x in input("Enter source, destination, weight: ").strip().split(" ")
        ]
        graph[i] = {"src": src, "dst": dest, "weight": weight}

    source = int(input("\nEnter shortest path source:").strip())
    shortest_distance = bellman_ford(graph, V, E, source)
    print_distance(shortest_distance, 0)

Bellman Ford

K
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and 3 more contributors

Problem Statement

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.

Approach

  • 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:
    1. 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.

Time Complexity

O(VE)

Space Complexity

O(V^2)

Founder's Name

  • Richard Bellman & Lester Ford, Jr.

Example

    # 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

Code Implementation Links

Video Explanation

A video explaining the Bellman-Ford Algorithm

Others

Sources Used: