Question

for an assignment I am asked by to find find an algorithm that calculates the transitive closure of a directed graph using O(n 4 ) time. We already learned about the floyd warshall algorithm, which is a lot better, so can someone help me create one that runs in O(n4) time? is there such an algorithm?

I know it seems dumb of a question. I dont really understand why we are being asked to find the slower way to do it.

Answer 1

Floyd Warshall is O(n^3), and since O(n^3) is a subset of O(n^4), it is also O(n^4).

Thus, by setting a new graph G'=(V,E',w') where E' = V x V (clique, complete graph) and w'(u,v) = 1 if (u,v) is in E, otherwise INFINITY – by using Floyd-Warshall algorithm, each pair (u,v) that ends up with a value less then infinity is in the closure.

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A Theta(n^4) solution:

Q <- E (init) #(Q is a set)
for i from 1 to n:
   for each v in V:
     for each w in V:
        for each u in V:
            if (v,w) is in Q and (w,u) is in E:
               Q <- Q U {(v,u)}  #add (v,u) to Q

Complexity is trivially Theta(n^4), we only need to show it indeed finds the transitive closure.

By induction:

1. For shortest path of length 1 from u to v it is the base clause,
   since (u,v) is in E.
2. For each k:
   
   Each pair (u,v) with shortest path of length k>1 – there is w such that there is a path u -> ... -> w, and an edge (w,v). From induction hypothesis, in previous iterations we added (u,w) to Q, and thus the condition will yield true, and we will add (u,v) to the resulting set Q.

Similary show that if some pair (u,v) was added to Q, then there is a path u->..->w->v, and thus it was rightfully added.

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A second Theta(n^4) solution:

Set G' as described above, and for each vertex v in V run Bellman Ford from v.

Each run of BF is Theta(n^3)¹, running it n times is Theta(n^4)

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(1) Technically it is O(VE), but for none sparse graphs E is in Theta(V^2)