Running Floyd-Warshall algorithm on graph with negative cost cycle
I think the answer is 4, and here is why.

Assume the graph is fully connected with all edges having weight of -1.

Now let's consider the three loops of Floyd-Warshall algorithm:

for k = 1 to n:
    for i = 1 to n:
        for j = 1 to n:


Since -2 is "shorter" than -1, after we finish k = 1, the weight for i -> k = 1 -> j is -2 for most i and j (exceptions would be i = k and j = k).

After we finish k = 2, the weight for i -> k = 2 -> j is -4 for most i and j. This is because i -> 1 -> 2 -> 1 -> j is the shortest, giving us -4.

And so on and so on for the exponential growth.

Floyd-Warshall algorithm does not guarantee that we will find a simple shortest path, that is, a path containing only one instance of each vertex.