Problem

Find minimum cost walk between all pairs of vertices. Negative cost edges are ok; negative cost cycles are not.

Matrix multiplication

Define w(k,x,y) = the cost of minimum cost walk of length at most k from x to y, or ∞ if no such walk exists.

We have a recurrence relation. The base case is:

• w(1,x,x) = 0;
• w(1,x,y) = cost(x,y), if (x,y) is an edge of the graph;
• w(1,x,y) = ∞, if x≠y and (x,y) is not an edge of the graph;

The actual recurrence is:

• w(k+l,x,y) = min(w(k,x,z)+w(l,z,y)).

The idea is to compute w(k,x,y) for a value of k based upon the already computed values for k/2. We need to get to a k greater than n.

The number of operations for computing all w(k,x,y) for a given k is O(n3). Doing this up to a k greater than n will take O(*n^3log n).

To retrieve the path, we can define a second array, f(k,x,y) = the next vertex after x on the walk of cost w(k,x,y).

When the minimum in the recurrence is reached for some intermediate vertex z, we set f(2k,x,y) = f(k,x,z).

Floyd-Warshall algorithm

It is also based on dynamic programming. We need to have a numbering of all vertices of the graph, X={z0,z1,...,zn-1}.

Then, we define w(k,x,y) = the cost of minimum cost walk from x to y, using, as intermediate vertices, only those in the set {z0,z1,...,zk-1}.

The recursion starts like for the matrix multiplication algorithm. Then, we have:

• w(k+1,x,y) = min(w(k,x,y), w(k,x,zk) + w(k,zk,y)).

Finally, wn,x,y is allowed to use all vertices in the graph.