I recently came back from a Dagstuhl I organized with Phil, Claire and Daniel on optimization algorithms for planar graphs. I think I finally caught up for the week away and sat down this afternoon to think about research. A problem suggested at the workshop was **TSP in 1-planar graphs**. 1-planar graphs are those graphs that can be drawn in the plane such that any edge has at most one other edge crossing it.

We have PTASes for TSP in bounded-genus graphs and in particular, these PTASes solve the *subset *TSP problem wherein you want to find the shortest tour that visits an input subset of vertices. In general graphs, we would just take the metric completion (create the complete graph on that input subset of vertices where the weight of each edge is the length of the shortest path in the original graph) and solve the problem in that completion. In planar graphs you could do this, but you would lose planarity. The subset version is inherently interesting: if you are designing a delivery route, you definitely don’t want to visit *every intersection* in your region.

One would suppose that for 1-planar graphs, one would also want to give a PTAS for the subset version of the problem. However, we can argue that isn’t the case. Take a general graph G and consider any drawing of that graph in the plane. An edge in this drawing will have any number of other edges crossing it, creating a sequence of crossing points along the edge. Subdivide this edge by introducing a vertex between every consecutive crossing point and distribute the weight of the original edge among these new edges. Now you have a 1-planar graph H. If you could find a (1+eps)-approximate tour that visits the subset of vertices that correspond to the vertices of G, this ordering of vertices would correspond to a (1+eps)-approximate tour in G. However, metric TSP is APX-hard. So, there can’t be a PTAS for subset TSP in 1-planar graphs.

There might still be a PTAS for TSP (visiting all the vertices) in 1-planar graphs. This doesn’t seem as interesting a problem though.