I wrote:
The following observation (which may already have been obvious to you, in which case I apologize) may help to clarify arguments like Tom R's: The per-square winding numbers completely determine the lines and directions of the grid path: wherever you have two adjacent squares with equal PSWN there is no path-segment between them; wherever you have two with PSWN differing by 1 there is a path-segment with the higher-PSWN square on its left; there cannot be two with PSWN differing by more than 1.
Let me elaborate on this a little and see whether the result is a version of Tom's argument that Jim finds more acceptable. 1. Construct the optimal configuration when a disconnected path is permitted. This is the "obvious" thing with nested rectangles, because there just isn't space to wind more times around any individual square than you get in this configuration. 2. Obviously we cannot simply say "suppose the optimum actually-feasible path is obtained by small modifications to this infeasible optimum" without further justification. 3. But the justification we need is not hard to find. Consider those per-square winding numbers for the actual optimum. They are squarewise <= those for the infeasible optimum. And (see above) the PSWNs determine the path, as well as vice versa. So the question is: among all ways to reduce the PSWNs, what is the smallest reduction that yields a configuration corresponding to an actual path? 4. Whatever that reduction is, we can imagine doing it one square (and one unit of winding for that square) at a time. When we reduce one PSWN by 1, the number of path components cannot reduce by more than one -- for we cannot have components of more than two disconnected paths adjacent to a single square. 5. For a 2k-by-2k square, we begin with k path components. If the sum of PSWNs is j less than for the infeasible optimum, by #4 we have at least k-j components at the end. So if we end up with just one component, we must have j >= k-1. 6. We can in fact achieve j = k-1, by Tom's construction. This is therefore the actual optimum. (I have contributed no actual mathematics here, I think; just expressed Tom's idea a bit differently. Jim, does it help?) -- g