Sorry, but as I said, I find the "algorithm description" in the paper you cited totally incomprehensible. It could be an alchemist's recipe or Archimedes' exclamation when stubbed his toe getting out of the bath for all the sense I can make of it. Constructed to impress rather than convey information. Further, I personally think that (much of) the literature on the subject is wrong, but I haven't reached the point of proof yet.

I did come across this description of Aho's algorithm which at a cursory glance seems to be infinitely clearer:

The following procedure is an implementation of the algorithm presented by Aho et al. (1981) with modifications for dealing with incompatibilities.

1. A subset of the orthologous-repeats table is created, in which only “relevant” rows (organisms) are considered (initially all rows, since all organisms are being considered). Within this subset of rows, only those columns in which at least two rows have a 1 and one row has a 0 are considered.
2. Using this subset of the original repeat occurrence table, a graph is created by iterating through the columns. If two rows both have a 1 in a given column, an edge of weight 1 is created between the two corresponding organisms. If an edge already exists between those two organisms, its weight is incremented by 1.
3. Multiple connected components are sought within the graph. If the graph contains a single connected component, weak edges must be eliminated. This is accomplished by removing edges, beginning with those of weight 1 and incrementally removing edges of greater weight, until multiple connected components arise.
4. Steps 1-3 are recursively applied to each connected component containing more than two organisms. The “relevant” rows in each run are the organisms within the connected component.

Consider the above example illustrated in Table 3B. The phylogenetic inference of column i is supported by column h, and column k is supported by column l. Thus, in the shared-repeat graph, edges (A, B) and (C, D) have weight 2, while the edge (A, C) has weight 1. Removing the minimum weight edge is akin to removing column j, which has the least support.

I haven't pursued it because I'd rather finish or abandon my own experiments before I get familiar with someone else's method; but maybe it will help you to bring your code together. Good luck.