One feature of the algorithm that's been missing so far is a proper rotational symmetry. I've already implemented a point symmetry to get the same result as a 2-fold rotation; it's fast and the code is simple but not generalizable to n-fold rotations. What I find especially cool is 3-fold symmetry. Here I'll explain how I coded the general symmetry and present some results.

According to the original algorithm, the rotational symmetry is applied to the activator and inhibitor arrays of each scale, before looking for the best variations. Each pixel/array element is averaged with its symmetrical points around the origin of the image. The number of averaged pixels depends on the symmetry order. First we'll see how the symmetry works in practice, then to which part of the image it should be applied, and finally an idea to organize all this data.

**Rotational symmetry**

For the moment, let's forget the Turing patterns, forget that we will work with data arrays and just focus on the point symmetry in a 2D coordinate system. We have a picture, assumed square, with a width 2r, and the origin at coordinates (0, 0).

Let's also assume that we want the picture to be point-symmetrical around the origin, with an order of 4. That means that the image will be four times the same sub-image, rotated around the origin.

There's a point at coordinates (x0, y0) which symmetrical points we want to find. If we draw a circle centered on the origin and passing by (x0, y0), the first symmetrical point is also on this circle but rotated by one fourth of a full circle, that is 2π/4. This gives the coordinates (x1, y1). The second symmetrical point (x2, y2) is found by the same method and a rotation of 2*2π/4. And the last point (x3, y3), same thing with a rotation of 3*2π/4. In the end, we have four points that are symmetrical around the origin.

In general, for a symmetry order *s*, the rotations of the (x0, y0) point around the origin will be the angles

where

Obviously, s = 1 is a special case with no symmetry at all.

Concretely, how to compute the coordinates of the points symmetrical to (x0, y0)? Knowing the angles
θ, the
rotation is done with

Looping through the values of *i*, it's easy to make a list of all symmetrical points to (x0, y0). And if these two formulas are applied to all points in the appropriate portion of the image (for a symmetry order s = 4, that's the top-right quarter), we can list all the symmetries in the image.

**Application**

For the project, we have to apply these rotations to a 2D array (activator or inhibitor) and average the values of the symmetrical points. For example, let's work on the activator array act[x][y], again with s = 4:

- Find the coordinates (x0, y0), (x1, y1), (x2, y2), (x3, y3) thanks to the formulas above.
- Compute the average avg = (act[x0][y0] + act[x1][y1] + act[x2][y2] + act[x3][y3]) / 4.
- Replace the values in the array, i.e. act[x0][y0] = avg, act[x1][y1] = avg, act[x2][y2] = avg and act[x3][y3] = avg.
- Loop steps 1 to 3 over all act[x][y] with (x, y) belonging to the top-right quarter.

Here, step 1 is done on-the-fly. This requires computing rotations each time we want to symmetrize an array. But if the symmetry order is fixed, it's more elegant to make a list of symmetrical coordinates at the beginning and just go through the list when repeating steps 2 and 3. Besides, it avoids computing cos and sin all the time, which could be computationally expensive (thanks to Jason Rampe for the idea).

First caveats about the rotation applied to a data array: the (x, y) values are integers, but we compute them with cos and sin functions which give floating-point values. So there are lossy conversions happening, and the rotation/symmetries are flawed. This isn't too worrisome if the resolution is large enough, because the couple of pixels badly symmetrized will be lost in the sea of correct ones. But it's still important to remember that the integer operations are not perfect. In addition, we're...

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