Computations with reciprocal edges

[HarrisonSantiago]

Hello, I'm attempting to calculate the homology of a directed flag complex and am struggling to understand the results from some toy examples with reciprocal edges. As detailed in #42 , when looking at a directed graph with two nodes who's adjacency matrix is [0 1; 1 0] there is no explicit identification of those two edges, which leads to a 1D hole. This makes sense, and similarly, a graph with an adjacency matrix of [0 1 1; 1 0 0; 1 0 0] leads to 2 1D holes.

However, a graph with an adjacency matrix (Ex1) of [0 1 1; 1 0 0; 0 1 0] leads to a single 1D hole, and a 2-simplex. On the other hand, an adjacency matrix (Ex2) of [0 1 1; 1 0 1; 0 0 0] leads to 2 2-simplices and no 1D holes.

I'm assuming in the first case the 1D hole between 0 and 1 disappears, and the new hole is formed by the cycle (0 -> 2 -> 1 -> 0), and then in the second example this hole converts to a 2-simplex when we change direction of the edge between 1 and 2.

However, this falls apart with the introduction of a third example (Ex3) with an adjacency matrix of [0 1 1; 1 0 1; 1 0 0]. Here we get 3 2-simplices, and no holes. Using the logic from above, this should result in 2 2-simplices and 1 hole.

Could you explain how graphs with reciprocal edges are handled? I'm not sure why flagser ignores the hole determined by (0->1 and 1->0) in Ex1 and Ex2, and then further why the hole determined by (0 -> 1 -> 2 -> 0) is ignored in Ex3.