4 ms·
Random Hexagon Fractal
- gus_massa 7mo agoUsing the baricenter, the position of the new point is New Point C = (Vertex An + Vertex A{n+1} + Point C) /3 where An and A{n+1} are consecutive vertices. It can be simplified to New Point C = (2 \* Midpoint Mn + Point C) /3 This method defines a linear IFS [1], so you get a nice regular fractal as expected. https://en.wikipedia.org/wiki/Iterated_function_system https://en.wikipedia.org/wiki/Iterated_function_system --- The incenter is more complicated to calculate using the coordinates, and I officially hate anyone that chose the incenter for this operation. Anyway, I'm very confused because I expect the atractor to be symmetric. I'm wondering if it needs more iteration to get close enough. It looks like an interesting problem, but as I just said, calculating the coordinates of the incenter is a complicated.