Tiling - Grading Non-Tiles

Basic intent: In addition to classifying polygons as those that can tile the plane and those that cannot, further probe the non-tiles and grade them in their ability to progress towards a full tiling.

I recently came to know about the Heesch problem. Basic question: which are polygons that can neatly surround themselves with copies but fail to tile the entire plane? Generalization: Which polygons can surround themselves 2,3, n times but fail to tile the plane? The number of 'coronas' a polygon can form around itself with copies of itself is its Heesch number.

Eg: A regular pentagon has Heesch number 0, it cant surround itself neatly even once.

The following polygon has Heesch number 1: Take a regular hexagon with side a and stretch 3 alternate sides to a larger length b (so the edge lengths are a,b,a,b,a,b) keeping all angles at 120 degrees. It can surround itself once with copies of itself - to see this just reflect the polygon about all its edges. See figure below. But it cannot surround the resulting layout neatly with copies of itself. (Thanks to Profs. Roman Karasev and Arseniy Akokpyan for telling me about this polygon).

Heesch himself had found a convex pentagon which has Heesch number 1. Details are here: https://www.uwgb.edu/dutchs/symmetry/heesch.htm

I dont yet know about convex polygons with higher but finite Heesch numbers.

One could try another tack to classify convex polygons which fail to tile the plane but are better than obvious non-tiles. The Heesch problem is about repeatedly surrounding a central tile with copies. So we tried another topology - tiling the plane by patching together 'walls' of polygons.

Let us define a WALL as a *simply connected* set of points formed with infinitely many polygonal regions and dividing the plane into two disjoint regions A and B such that the distance of any point P_A in region A to any point P_B in region B has a finite lower bound ( intuitively, A and B should be clearly separated by the wall and the wall should have no cavities inside it).

Given a polygon P, we could form a wall W1 out of copies of P, then form another wall W2 with copies of P, then see if the two walls can be 'patched' to form a thicker wall W (W1 and W2 could be identical). And if this process can be continued indefinitely, P can tile the plane. We could call W1 and W2 to be 'patchable' walls. We could say a wall has thickness t if it can be 'sliced' into t walls.

It is easy to see that copies of any convex polygon P can form walls in infinitely many ways. But for most polygons, even forming 2 patchable walls is impossible. For example, the hexagon with Heesch number 1 above can form 2 patchable walls .



The Heesch polygon (another convex polygon with Heesch number 1) too can form 2 patchable walls - resulting in a wall of thickness 2. In figure below, it is shown how copies of this tile can form a wall that is bounded below by an infinite straight line so an identical wall can be patched to this wall. But it appears that no further patching can be done on to the resulting 2- layer wall to form walls of higher thickness.



Question: Are there non-tiles which can form walls of thickness 3 (or in general n)?