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> For each Julia set, one of these 5 machines will correctly draw the set. That's really interesting. Does this essentially correspond to a proof of being able
by Xcelerate 2y ago
> For each Julia set, one of these 5 machines will correctly draw the set.
That's really interesting. Does this essentially correspond to a proof of being able to compute the correct set with probability no less than 1/5?
For the question "which of the 5 is correct?", is it presumed that there exists a proof that hasn't been found yet or that this is undecidable (e.g., within ZFC)?
- rssoconnor 2y ago> For the question "which of the 5 is correct?" There is a discontinuity as the parameter 'c' crosses a location that lies on the boundary of the Mandelbrot set, where the corresponding Juila set goes from a thick ring of disconnected points (Cantor-set like) to suddenly connected and the ring is completely filled in. One candidate machine will draw a filled in Julia set, and another candidate machine will draw the ring. For each Turing Machine one can construct a complex value c, that is just barely outside the Mandlebrot set if the machine halts, but is on the inside (specifically on the boundary) of the Mandelbrot set if the Machine does not halt. Thus being able to correctly draw the Julia set all of these points amounts to solving the halting problem. Though any individual point may or may not be solvable. Of course there is at least one Turing Machine that searches for an inconsistency in ZFC. If ZFC is consistent then this machine never halts, but ZFC cannot prove this fact.