Bellman-Ford Algorithm

Bellman-Ford Algorithm
Runtime	$O(V^{2}+E)$
Class	Dynamic Programming Graph Algorithms Shortest Path Problem

Approach: dynamic programming

All shortest path must have $\leq |V|-1$ edges. If this condition is not satisfied, there is a cycle in in the path, and therefore it is not the shortest.

The idea is to add one edge at a time, seeing if the edge should be included in the shortest path.

Recurrence

Let OPT(n-1, a) be the length of the shortest path from source node $s$ to node $a$ with at most $n-1$ edges.

The main loop iterates $n-1$ times. Inside each iteration, each edge is relaxed once.

By relaxing every edge, we add a possible edge to the shortest path. Since there is at most n - 1 edges in the shortest path, we iterate n - 1 times.

If we don't add the edge, the length is OPT(n - 2, a)

If we add the edge, the length would be OPT(n - 2, b) + w(b, a)

$OPT(n-1,a)=min(w(b,a)+OPT(n-2,b)\forall (b,a)\in E)$

Implementation

Time complexity: $O(V^{2}+VE)$

• Sparse: $O(V^{2})$
• Dense: $O(V^{3})$

// G: Graph with vertices (V) and edges (E)
// w[e]: weight of edge e
// S: starting node
Bellman-Ford(G, w, S)
    V, E = G
    pi[v] = null for all v  // traceback
    
    // initialize all shortest path algo
    d[v] = infty for all v
    d[s] = 0
    for i from 1 to |V| - 1:
        for all (u,v) in E
            if d[v] > d[u] + w(u, v):
                d[v] = d[u] + w(u,v)
                pi[v] = u;
    
    for all (u, v) in E:
        if d[v] > d[u] + w(u, v):
            // has negative cycle
            return false

This accounts for negative edges but not negative cycles. Bellman Ford can detect it by running an extra time, since if there is a negative cycle, the run time will improve.