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For a personal project involving a coordinate system for navigating the galaxy using 36/38 equidistant points around a circle, I need to find all intersections between 3+ lines using 3 pairs of points (6 unique points) so that it makes a valid "Address", I have looked up many articles on this website and all that I could find involved intersections between two lines only.

I started off doing this to visualize the lines themselves:

lines = CirclePoints[36];
Graphics[{Black, Line@Subsets[lines, {2}]}]

Then I managed to adapt the code to show two line intersections:

lines=Line/@Subsets[CirclePoints[36],2];
pts=Graphics`Mesh`FindIntersections[lines];
Graphics[{{Thickness[0.0005],lines},{Directive[Red,PointSize[0.005]],Point[pts]}}]

The code above works for less than 21 points but any higher and it displays nothing, I suspect its because of an old function, aside from that issue I am lost on how to check specifically for 3+ line intersections while ignoring just 2 line ones, is such a thing possible?

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Here we fixed one point say pts[[1]] and let the other 5 points are the subset of remain points,but it still take a long times. We only test n=18 or n=20.

k = 9;
n = 2 k;
pts = CirclePoints[n];
lines = Line /@ Subsets[pts, {2}];
sixsubsets = Prepend[1] /@ Subsets[Range[2, n], {5}];
intersectionpts = 
  RegionIntersection[Line[{pts[[#1]], pts[[#4]]}], 
     Line[{pts[[#2]], pts[[#5]]}], Line[{pts[[#3]], pts[[#6]]}]] & @@@
    sixsubsets;
Cases[intersectionpts, Except[EmptyRegion[2]]];
DeleteDuplicates[
  Table[TransformedRegion[#, 
      RotationTransform[j*2 π/n]] & /@ %, {j, 1, n}]];
Graphics[{lines, {Red, %}}]

enter image description here

enter image description here

Not so effective,only provide a possible way.

pts = CirclePoints[12];
lines = Line /@ Subsets[pts, {2}];
sixsubsets = {Line[{#1, #4}], Line[{#2, #5}], Line[{#3, #6}]} & @@@ 
   Subsets[pts, {6}];
intersections = RegionIntersection @@@ sixsubsets;
Graphics[{lines, {Red, intersections}}]

enter image description here

enter image description here

enter image description here

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  • $\begingroup$ Is it possible to reduce the # of comparisons by checking intersections on 1/nth of the circle? intersections are symmetrical on 2pi/n, so perhaps checking all intersections between one set of adjacent points aka(point 1 to n other) and (point 2 to n other) with the third line being created between any two points where atleast one is on the same hemisphere as the initial 2 adjacent points. e.g. 12 points, pick point 1&2, make subsets where lines are (1, [3 -11]) and (2, [4-12]) with the last list ([3,4,11,12], [3-12] ). Compare for intersects, then duplicate and rotate by pi/6. $\endgroup$
    – Auri Crystal
    Jun 30, 2021 at 23:51

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