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Timeline for Optimization of points on a plane

Current License: CC BY-SA 3.0

18 events
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Dec 9, 2015 at 20:48 comment added Halbort I reached an equilibrium that involved a 7-gon. I think all regular polygons with some centroid points are equillibirum. The pentagon seems to be the most stable though.
Dec 9, 2015 at 3:13 vote accept Halbort
Dec 8, 2015 at 23:40 comment added Per Alexandersson @Halbort: Oh, there was a brace missing at some point - it should work now. I use Mathematica 9, by the way.
Dec 8, 2015 at 23:39 history edited Per Alexandersson CC BY-SA 3.0
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Dec 8, 2015 at 22:53 comment added Halbort I keep getting {{k,Ceiling[Length[newLists$109483]/2],Length[newLists$109483]};\ newLists$109483=SortBy[newLists$109483,Fitness];gg=ListPlot[newLists$\ 109483[[1]],Axes->False,PlotStyle->PointSize[0.02`]]} does not have \ appropriate bounds. >> as an error.
Dec 8, 2015 at 22:41 comment added Halbort I copy pasted it directly. I cannot see anything happening. Could you help me?
Dec 8, 2015 at 22:38 comment added Halbort Where does the image get stored?
Dec 8, 2015 at 21:29 comment added Per Alexandersson @GerhardPaseman: Yeah, you are right, seems like 6-gons, 5-gons and 4-gons are stable, but maybe not higher...
Dec 8, 2015 at 21:28 comment added Per Alexandersson @Halbort: Open a notebook, paste the code, and run. It dynamically updates a picture with the best configuration, and does 1800 generations.
Dec 8, 2015 at 20:05 comment added Halbort How do I run this mathematica code on my machine?
Dec 8, 2015 at 20:00 comment added Gerhard Paseman You may be right, but then I think you would have seen that in your simulation. I suspect regular n-gons are minima only for n up to at most 6, because of the combinatorial explosion of diagonals. Gerhard "Leaves The Numerics To Others" Paseman, 2015.12.08
Dec 8, 2015 at 19:48 comment added Per Alexandersson @GerhardPaseman: Yes, but I think you will find that a regular n-gon, with the remaining dots in the center is a local minima... Perhaps one can argue that the ONLY local minima are of this form, and then it is easy to find which of these is global.
Dec 8, 2015 at 19:46 comment added Gerhard Paseman It should be not too hard to show minimality by abstractly pushing dots around. Things inside the convex hull should gravitate toward a centroid, and points not too far outside the convex hull of the remaining points should be absorbable when there are at least 4 remaining points. Gerhard "Hand Waving Gives Some Elevation" Paseman, 2015.12.08
Dec 8, 2015 at 18:40 vote accept Halbort
Dec 9, 2015 at 0:44
Dec 8, 2015 at 18:40 comment added Halbort I just realized that I messed up.
Dec 8, 2015 at 18:32 comment added Per Alexandersson @Halbort: I get it to be $16/\sqrt{3}$...
Dec 8, 2015 at 18:20 comment added Halbort For $n = 4$, if you make an equilateral triangle with one point in the center I think you get $4\sqrt{3}$
Dec 8, 2015 at 18:05 history answered Per Alexandersson CC BY-SA 3.0