[Math] L-systems and Sierpinski Triangle

Question

I was just shocked when I saw these consecutive outcomes of an L-system converging to the Sierpinski triangle (shown in the picture below).

I'm interested to know how could one arrange the rules of an L-system so that it would converge to a to the Sierpinski triangle. What did he/she have in mind?
Thanks in advance.

Answer 1

The $L$-system described in the Wikipedia page to which you link is:

variables : A B

constants : + −

start : A

rules : (A → B−A−B), (B → A+B+A)

angle : 60°

Here, A and B both mean "draw forward", + means "turn left by angle", and − means "turn right by angle" (see turtle graphics). The angle changes sign at each iteration so that the base of the triangular shapes are always in the bottom (otherwise the bases would alternate between top and bottom).

The way I think about this is that these rules express the fundamental fractal symmetry of the Sierpinski triangle, namely, the symmetry of the whole triangle with each of the three next-smaller triangles. If one images a "basic traverse" of the full main triangle to be traveling on a line segment from the bottom left to the bottom right, let us call this an A-move, for the fractal sits to our left as we make this move.

Thus, I would put two additional pictures before your four pictures, one consisting of a line segment at the bottom, and the next consisting of three movements, like half a hexagon, traversing one side each of the three next-smaller triangles.

If we consider replacing the basic A move with the moves arising from one level deeper in the fractal symmetry, then what we would do is follow that half hexagon, one edge each of the three next-smaller triangles: first going up on the left, then across to the right, then down at the right. Note that when we go up the (bottom half) of the left side of the big triangle, the corresponding triangle is on our right now, so this is a B move. But when we go across to the right, at the bottom of the top subtriangle, it is on our left, and then coming back down to the bottom right point, the third (lower-right) triangle is on our right. So we have replaced the basic A move (moving across the bottom of the full big triangle), with the moves B-A-B. Similarly, a B move would be replaced with A+B+A, with the corresponding interpretation of angle rotation (I believe it also makes sense to add an extra reorientation rotation at the start of this, but evidently this just gets canceled out.)

One could easily find other ways to make the traverse that would also express these fundamental symmetries, and arrive at different systems generating the Sierpinski triangle.