From: John Conway <email@example.com> Date: Mon, 13 Dec 93 09:42:47 EST To: firstname.lastname@example.org, email@example.com Subject: Re: Aperiodic Space-filling tile?
My tile is a very simple one. It is aperiodic (only) in the sense that no tiling with it admits a translational symmetry. This seems to be the standard sense now, and it is a good one because in particular it implies that the tiles can't be grouped into equal clumps that are all embedded similarly in the tiling. Take a triangular prism: /|\ / | \ /__| \ \ \ | \ \ | \__\| Make it "generic" - in other words, avoid certain equalities between the sides and angles. In particular, don't make the side faces be rectangles. [The apparent right angles in the triangles of my sketch are there only because of the limitations of this keyboard.] Now take another such prism, again as generic as you can get it, subject to the condition that one of its parallelogram faces has the same shape as one of the ones on the first prism, oriented in such a way that you can stick the two prisms together to make a "biprism" in which no two triangles share an edge. Then this biprism is my tile. Which biprisms are sufficiently "generic", you may ask. I don't know a very easy way to find out, though almost all of them are, of course. Here is what's needed: 1) It must be impossible to fit the pieces together in any way except those in which the constituent prisms line up in parallel, triangle-to-triangle: /|\ / | \ /__| \ \ \ | \ \ | \ \__\| \ \ \ | \ \ | \__\| This sticks the biprisms together in two-dimensional "waffles" which are then stacked to form the whole tiling. 2) The angle of rotation between two adjacent "waffles" must not be 60 or 90 degrees. I just worked out one case. I took a rhombus made of two triangles root2,root2,1 for the common face, and took points vertically above and below the midpoints of the edges to be the other four vertices: /\ / \ A B / \ / \ \ / \ / D C \ / \/ A and C are above the plane, B and D below it, so that the "roof edges" of the prisms are AC and BD. I made the angles of the triangular faces at these vertices be right angles, but that gives a very flat biprism. I suggest to make them something like 30 degrees for a better shape. [This will still be generic enough.] I imagine you will be a bit disappointed, because the tiling is fairly "regular". But in fact the tiles are necessarily embedded in infinitely many distinct ways, however you do it. JHC