a a a b b b b c c A B'1 C'1 C1 B1 D C'2 C2 B2
Proto-Tile:

Display:

USAGE: The red points of the proto-tile on the right can be moved. Changing some settings require pressing Update. Setting can be reproduced with the URL after Update.

This is from a picture of the Voderberg Tiling on Wikipedia.

The idea is to tile the plane with tiles of a single shape and size in a way where there is no global symmetry.

The parameters of the tile are:

To see what n represents, look at these settings for n = 8.

With the vertices as labeled in the diagram, the tile is constrained by:

The uppercase letters also indicate equal angle measurements A, B, C, D, B′ = 360° - B, C′ = 360° - C of the interior angles of the polygon. The sum of internal angles of a polygon gives A + B + D = 180°. The lowercase letters a, b, c indicate equal lengths.

The congruent parts AB1C1C1B1 and AB2C2C2D are just rotations of each other by angle A. In fact, AB1D and DB1A are isosceles triangles with vertex angle A and base angles A′.

At each center of the spiral, the angles add up (n+1)A + B + D = 360° which gives A = 360°/n. The distance between the two centers d can be taken as the scaling parameter for the overall size, so the two big isosceles triangles are determined, and their base a. The point B2 is determined by A and a. Any one of the remaining four points determines b and B (or D = 180° - A - B), so determines the other three points and C and c.

The constraints allows 2(n+1) pieces to fit together at each center. After that, use recursion at the next vertex around the spiral that have not filled up to 360°. At first, the inside of the spiral have angles B + D and the filled part of the outside of the spiral have angle B, so the remainder is filled up with A + A + D. Later on, the inside of the spiral can also have have angles B + D + A but the filled part of the outside of the spiral still have angle B, so the remainder is filled up with A + D. These two patterns repeat forever and so fill up the plane.

Or, consider that the inside of each spiral is made up of either a dark colored box of two tiles in opposite directions, or a light colored tile. Each of these units give rise to a dark colored box on the outside of the spiral, and the light colored units give rise to an additional light colored unit diagonally in front of it. The dark colored boxes form a straight line, and the light colored units are the turns of the spiral.

Looking at the Full picture, the spirals are basically made up of n straight segments per half revolution. Every half revolution, the segments increase by a length of a or by two tiles.

The total number of tiles is 2n(half_revolutions)2 + 2. That's 2 initial tiles, plus n(2r - 1) tiles for the r-th half revolution for each of the 2 spirals.

The only problem is if the polygon intersects with itself, in which case the tiles will overlap.

There can be no self-intersections if B2B1C1 are colinear and C2C1C1 are colinear. The first colinear condition gives B = A′ + A, and the second colinear condition gives a quadratic equation in b. These are set by the buttons b+ and b-. But the quadratic equation only have real solutions when n ≥ 14.

Smaller n can still give a tile with no self intersections. Experimentally, n ≥ 8 works. Perhaps one can be calculated by maximizing the distance of C1 from C2C2.


IMPLEMENTATION NOTES:


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