Tuesday, November 15, 2016

Fun with puzzle pieces

If all edges placed on a square template are identical with mirror symmetry over the midpoint, there are two distinct shapes that will tessellate, one with two “outs” that are adjacent and one with two “outs” that are opposite. Note that the shape must have two “out” edges and two “in” edges; if the number of “outs” is not equal to the number of “ins”, the pieces will not tessellate. Below is an illustration of the two possibilities with an edge that forms puzzle pieces.

Suppose that the edge does not have the bump centered in the middle but rather offset to one side. How many distinctly different shapes with these identical asymmetric edges will tile the plane?
I answer this question in a note published in the November 2016 issue (Vol 100 Issue 549, pp 511-516) of the Mathematical Gazette: there are 15 distinctly different shapes and all will tile the plane. Thirteen will tile the plane as Heesch types and two will tile it in a non-isohedral pattern.

The note in the Mathematical Gazette is limited to the square template. Exploring Tessellations: A Journey Through Heesch Types and Beyond extends the analysis. If the template is a rhombus or diamond (a square squashed), there are 30 distinct shapes of which 20 will tile the plane. If the template is a regular hexagon, there are 108 distinct shapes that have three “ins” and three “outs” when identical, asymmetric edges are fitted to the template. Thirty-four tile isohedrally as Heesch types and another nine tile anisohedrally. Sixty-five will not tile.

Exploring Tessellations: A Journey Through Heesch Types and Beyond also considers equilateral templates fitted with edges that have central rotation and some of these results have been mentioned in past posts on this blog. There are two distinct shapes if the template is an equilateral triangle, four for a square, seven for a rhombus, and nine for a regular hexagon. One of the nine for the regular hexagon will not tessellate. Also, all Heesch types except three that will not fit as equilateral polygons (CC3C3, CC4C4, and C3C3C6C6) can be formed with identical edges of central rotation.

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