Tweaking Floyd-Warshall Algorithm to detect cycles

Question

Cheers, I am trying to solve the problem of minimum length cycle in a directed graph, and I came across a solution that suggested that I should tweak the Floyd-Warshall algorithm to solve that. It stated that instead of setting path[i][i] = 0 I should instead set path[i][i] = INFINITY, but I don't exactly understand why that is the case! I find that the main diagonal of the array used by Floyd-Warshall does not change, so how can it help me to see the path of the cycle? I understand that the generated array of the algorithm helps me find the shortest path of a pair. e.g. path[i][j] gives me the shortest path from i to j but, although the intuition stays the same, I see that nothing changes, and I can't take the desired result.

I even tried visualizing the process, using this website here, I generated a graph which contained many cycles inside it, but although the diagonal is initialized with infinity, it does not get changed. Can anyone explain what am I missing or what I can do to solve my problem?

That trick will only work for directed graphs, because it finds paths like A->B->A, which are only cycles in directed graphs. Are you working with directed graphs? — Matt Timmermans, Apr 28 '22 at 13:42
Yes I am working with directed graphs. But again, it doesn't seem to do the trick as I explained above. Can you explain why or how that works? — krikowian, Apr 28 '22 at 14:40

score 2 · Answer 1 · answered Apr 28 '22 at 15:20

There is an implementation detail in the way the Floyd-Warshall algorithm is coded for the animation site; it prevents you from seeing the results that you expect.

Download the source code and look at Floyd.js to see the condition that is not supposed to be there:

for (var k = 0; k < this.size; k++) {
    for (var i = 0; i < this.size; i++) {
        for (var j = 0; j < this.size; j++) {
            if (i != j && j != k && i != k) // <<== This is the problem
                ...

The algorithm never computes a path from a node to itself (i.e. when i == j) through a third node, so it never detects cycles. Essentially, the condition makes an assumption that the pass to itself cannot be improved, which is not correct in case when the main diagonal is set to INFINITY.

score 2 · Accepted Answer · answered Apr 28 '22 at 15:22

For directed graphs, the idea is that you're just changing your path matrix so that, instead of storing the length of shortest path from i to j, path[i][j] stores the length of the shortest non-empty path, i.e., it only includes paths with at least one edge. Of course that only affects paths from a vertex to itself.

So now, we initialize path[i][i] with infinity instead of 0, because we haven't at that time found any non-empty paths from the vertex to itself.

We then do the normal Floyd-Warshall iterations after initializing the rest of the matrix according to the edges:

for k in |V|:
    for j in |V|:
        for i in |V|:
            path[i][j] = min(path[i][j], path[i][k] + path[k][j])

Lets say there is a simple cycle 1 -> 2 -> 1. Then, when (i,j,k) == (1,1,2), we do path[1][1] = min(path[1][1], path[1][2] + path[2][1])

This changes path[1][1] from infinity to the cycle length.

If you modified an implementation and it doesn't do this, then that implementation was probably optimized to ignore the diagonal altogether.

Tweaking Floyd-Warshall Algorithm to detect cycles

2 Answers2