| Here's a sketch of a proof using tools from Markov chain monte carlo. Writing on mobile so forgive any typos. Suppose you start by drawing a point uniformly at random from the triangle. We'll call this step 0. With probability 1 this point is not in the sierpinski triangle, but if we consider finite approximations of the sierpinsky triangle it is in the level 0 sierpinski triangle (see this random illustration I found on Google [1]). In particular, since the Level 0 sierpinski triangle is just the whole triangle, the initial point is uniformly distributed in the level 0 sierpinski triangle. This is a base case. Then we can do an induction step: If we have a point which is uniformly distributed in the level k sierpinski triangle, applying the transformation (pick one vertex of the triangle at random and move halfway towards it) results in a point which is uniformly distributed in the level k+1 sierpinski triangle. This is because each corner of the level k+1 sierpinski triangle is a scaled copy of the level k sierpinski triangle. Putting the base case and induction step together, we deduce that when we repeatedly apply the transformation of picking a vertex and moving halfway towards it, at each step k we have a point which is uniformly distributed in the level k sierpinski triangle. This is enough to show that if we start with a sample of N points drawn uniformly from the triangle, after k steps we'll have N points drawn uniformly from the level k sierpinski triangle (i.e. an arbitrarily good approximation of the sierpinski triangle, if we make k large enough). It's not quite enough to show that if we start with a single point and record the location after each step we'll end up with something that looks like the sierpinski triangle. For that we need to establish ergodicity of the Markov chain. The details are technical but not difficult to show; basically the Markov chain needs to not get caught in fixed-length loops (aperiodicity) and never get stuck unable to reach part of the triangle (irreducibility). Both these conditions are satisfied, so the Markov chain is ergodic and starting from a single point will approximate a sierpinski triangle. I've glossed over a bunch of details but hopefully that shows the basic idea. [1] https://www.researchgate.net/profile/Ali-Bashir-4/publicatio... |
