Heesch's Tiling Problem

1. INTRODUCTION. Let T be a tile in the plane. By calling T a tile, we mean that T is a topological disk whose boundary is a simple closed curve. But also implicit in the word " tile " is our intent to use congruent or reflected copies of T to cover the plane without gaps or overlapping; that is, we want to tessellate the plane with copies of T. In a minor abuse of language, one often speaks of T (as opposed to copies of it) as tiling or tessellating the plane, in the sense that T generates a tessellation. A tessellation by T may or may not be possible, so in order to learn something of T 's abilities with regard to tessellating the plane, we perform the following procedure: around a centrally placed copy of T , we attempt to form a full layer, or corona, of congruent copies of T. We require as part of the definition that no point of T should be visible from the exterior of a corona to a Flatland creature in this plane. Also, we should form the corona without allowing gaps or overlapping, just as if we were building a tessellation. If a corona can be formed, then we attempt to surround this corona with yet another corona, and then another, and so on; if we get stuck, we go back and change a previously placed tile and try again. If T tessellates the plane, then this procedure will never end. On the other hand, if T does not tessellate the plane, and if we check all of the possible ways of forming a first corona, a second corona, and so forth, we will find that there is a maximum number of coronas that can be formed. This maximum number of layers that can be formed around a single centrally placed copy of T is called the Heesch number of T and is denoted by H (T). We consider a few examples before proceeding. Consider first a regular hexagon. All bees know that a regular hexagon tessellates the plane, so H = ∞ for a regular hexagon. Next consider a regular pentagon. It is not hard to see that not even a single corona of copies of a regular pentagon can be formed around a central copy, whence H = 0 for a regular pentagon. These two examples exhibit the …