5 ms·
Tell us some more serious math! You’re doing great so far and that’s where the “huh?” ends up answered.
by floatingatoll 5y ago
Tell us some more serious math! You’re doing great so far and that’s where the “huh?” ends up answered.
- scotty79 5y agoI'd be happy to tell you if I actually knew it. I just know it exists. Sorry if I gave the wrong impression. If I were to look for it I think I'd search for proof that Iterated Function System has a fixed point or attractor https://en.m.wikipedia.org/wiki/Iterated_function_system https://en.m.wikipedia.org/wiki/Iterated_function_system
- klyrs 5y agoThe point here is that you'll almost certainly end up inside the Sierpinski triangle within a logarithmic number of steps. Let's do the 1-dimensional version, the Cantor set. init_x = (anything) while 1: draw(x) if flip_coin() == HEADS: x = (x+1)/2 else: x = x/2 Supposing we start with a large negative number N, and our coin is secretly TAILS on both sides, our sequence will be N, N/2, N/4, N/8, N/16... So we'll rapidly converge on the point 0, but never quite make it into the interval [0, 1] because N is negative. But, if we draw our point with a diameter of say, 1/128, it will rapidly become visually indistinguishable from zero. If the coin is fair, then something really cool happens -- if we hit a HEADS and x was already in the interval [-1, 0] then our result will be in the interval [0, 1/2] -- which is inside the interval [0, 1] where the problem is easier to think about. Approaching the problem in this way requires a bunch of case analysis, but the gist is that an attractor will quickly suck any point into the fractal -- if you get really unlucky, it will only get very close to the fractal in question.