1 Graph Transitive Closure and BMM

Proof. Consider a graph like the one below, where there are three rows of n vertices. Given two boolean matrices A and B, we can create such a graph by adding an edge between the i vertex of the first row and the j of the second row iff Aij == 1. Construct edges between the second and third rows in a similar fashion for B. Thus, the product AB can be computed by taking the transitive closure of this graph. It is equivalent to BMM by simply taking the ∧ of ij → jk. Since the graph has 3n nodes, given a T (n) algorithm for TC, we have an O(T (3n)) algorithm for BMM. (Of course, it takes n time to create the graph, but this is subsumed by T (n) as T (n) ≥ n as one must at least print the output of AB.)