Floyd-Warshall: all shortest paths

Question

I\'ve implemented Floyd-Warshall to return the distance of the shortest path between every pair of nodes/vertices and a single shortest path between each of the

Answer 1

If you just need the count of how many different shortest path exist, you can keep a count array in addition to the shortestPath array. Here's is a quick modification of the pseudocode from wiki.

procedure FloydWarshall ()
    for k := 1 to n
        for i := 1 to n
            for j := 1 to n
                if path[i][j] == path[i][k]+path[k][j] and k != j and k != i
                    count[i][j] += 1;
                else if path[i][j] > path[i][k] + path[k][j]
                    path[i][j] = path[i][k] + path[k][j]
                    count[i][j] = 1

If you need a way to find all the paths, you can store a vector/arraylist like structure for each pair to expand and collapse. Here is a modification of the pseudocode from the same wiki.

procedure FloydWarshallWithPathReconstruction ()
    for k := 1 to n
        for i := 1 to n
            for j := 1 to n
                if path[i][k] + path[k][j] < path[i][j]
                    path[i][j] := path[i][k]+path[k][j];
                    next[i][j].clear()
                    next[i][j].push_back(k) // assuming its a c++ vector
                else if path[i][k] + path[k][j] == path[i][j] and path[i][j] != MAX_VALUE and k != j and k != i
                    next[i][j].push_back(k)

Note: if k==j or k==i, that means, you're checking either path[i][i]+path[i][j] or path[i][j]+path[j][j], both should be equal to path[i][j] and that does not get pushed into next[i][j].

Path reconstruction should be modified to handle the vector. The count in this case would be each vector's size. Here is a modification of the pseudocode (python) from the same wiki.

procedure GetPath(i, j):
    allPaths = empty 2d array
    if next[i][j] is not empty:
        for every k in next[i][j]:
            if k == -1: // add the path = [i, j]
                allPaths.add( array[ i, j] ) 
            else: // add the path = [i .. k .. j]
                paths_I_K = GetPath(i,k) // get all paths from i to k
                paths_K_J = GetPath(k,j) // get all paths from k to j
                for every path between i and k, i_k in paths_I_K:
                    for every path between k and j, k_j in paths_K_J:
                        i_k = i_k.popk() // remove the last element since that repeats in k_j
                        allPaths.add( array( i_k + j_k) )

    return allPaths

Note: path[i][j] is an adjacency list. While initializing path[i][j], you can also initialize next[i][j] by adding a -1 to the array. For instance an initialization of next[i][j] would be

for every edge (i,j) in graph:
   next[i][j].push_back(-1)

This takes care of an edge being the shortest path itself. You'll have to handle this special case in the path reconstruction, which is what i'm doing in GetPath.

Edit: "MAX_VALUE" is the initialized value in the array of distances.

Answer 2

The 'counting' function in the current approved answer flails in some cases. A more complete solution would be:

procedure FloydWarshallWithCount ()
for k := 1 to n
    for i := 1 to n
        for j := 1 to n
            if path[i][j] == path[i][k]+path[k][j]
                count[i][j] += count[i][k] * count[k][j]
            else if path[i][j] > path[i][k] + path[k][j]
                path[i][j] = path[i][k] + path[k][j]
                count[i][j] = count[i][k] * count[k][j]

The reason for this is that for any three vertices i, j, and k, there may be multiple shortest paths that run from i through k to j. For instance in the graph:

       3             1
(i) -------> (k) ---------> (j)
 |            ^
 |            |
 | 1          | 1
 |     1      |
(a) -------> (b)

Where there are two paths from i to j through k. count[i][k] * count[k][j] finds the number of paths from i to k, and the number of paths from k to j, and multiplies them to find the number of paths i -> k -> j.