Find maximal subgraph containing only nodes of degree 2 and 3

Question

I'm trying to implement a (Unweighted) Feedback Vertex Set approximation algorithm from the following paper: FVS-Approximation-Paper. One of the steps of the algorithm (described on page 4) is to compute a maximal 2-3 subgraph of the input graph.

To be precise, a 2-3 graph is one that has only vertices of degree either 2 or 3.

By maximal we mean that that no set of edges or vertices of the original graph can be added to the maximal subgraph without violating the 2-3 condition.

The authors of the paper claim that the computation can be carried out by a "simple Depth First Search (DFS)" on the graph. However, this algorithm seems to elude me. How can the maximal subgraph be computed?

Answer 1

I think I managed to figure out something like what the authors intended. I wouldn't call it simple, though.

Let G be the graph and H be an initially empty 2-3 subgraph of G. The algorithm bears a family resemblance to a depth-first traversal, yet I wouldn't call it that. Starting from an arbitrary node, we walk around in the graph, pushing the steps onto a stack. Whenever we detect that the stack contains a path/cycle/sigma shape that would constitute a 2-3 super-graph of H, we move it from the stack to H and continue. When it's no longer possible to find such a shape, H is maximal, and we're done.

In more detail, the stack usually consists of a path having no nodes of degree 3 in H. The cursor is positioned at one end of the path. With each step, we examine the next edge incident to the end. If the only incident edge is the one by which we arrived, we delete it from both G and the stack and move the end back one. Otherwise we can possibly extend the path by some edge e. If e's other endpoint has degree 3 in H, we delete e from G and consider the next edge incident to the end. If e's other endpoint has degree 2 in H but is not currently on the stack, then we've anchored this end. If the other end is anchored too, then add the stack path to H and keep going. Otherwise, move the cursor to the other end of the stack, reversing the stack. The final case is if the stack loops back on itself. Then we can extract a path/cycle/sigma and keep going.

Typing this out on mobile, so sorry for the terse description. Maybe I'll find time to implement it.