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This is a follow-up to Puzzle about 6 infinite cylinders in space

Question:

Given six identical infinite (no caps) cylinders is there a beautiful arrangement in space such that each touches each other.

What you need to know is that beauty to a complete philistine such as myself means symmetry.

A symmetry in turn is for the purpose of this puzzle a nondistorting map of space onto itself that exchanges some cylinders but leaves the configuration as a whole in place.

Bonus: There are philistines and philistines. For a distinguished philistine such as myself mere beauty does not cut it. Only perfection will do.

A perfection within the confines of this puzzle is a symmetry or group of symmetries that acts transitively on the cylinders. Or in English: by possibly repeated possibly mixed application we can send any cylinder to any other.

Example: The finite cylinder solutions given here are symmetric but not perfect. The second, 7-pencil solution can be made into a perfect 6-pencil by leaving out the central, upright one.

As this is presumably rather hard partial answers (like considering only certain given symmetries) are welcome.

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I don't think a perfect arrangement is possible, but I have no proof. The best I've been able to find with computer assistance is one with 3-fold symmetry.

enter image description here

Here is a SageMath link where you can view it. The script is based on the one by PM 2Ring from their answer to the linked 6-cylinder question.

The locations for those 6 cylinders are given by the numbers below. The first triplet is a set of coordinates of a point on the central axis of a cylinder. The second triplet is a unit vector pointing along that central axis. The cylinders have unit radius.

(-0.858249176129892, 1.72213550263185, 0); ( 0.485021981582762, 0.756317033025778, 0.43901961566269)
(-1.06228850597332, -1.60433334062147, 0); (-0.897500754706579, 0.041882840931651, 0.43901961566269)
( 1.92053768210321, -0.117802162010372,0); ( 0.412478773123817,-0.798199873957429, 0.43901961566269)
( 3.8754403071009,   1.12206374493148, 0); ( 0.178347779811389,-0.972696164251417,-0.148506705191813)
(-2.90945586132661,  2.7951978843338,  0); ( 0.753205698499714, 0.640801790150925,-0.148506705191813)
(-0.965984445774283,-3.91726162926528, 0); (-0.931553478311103, 0.331894374100492,-0.148506705191813)
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  • $\begingroup$ Nice! Would it be possible to share the equations this is based on? (The linked sage script starts from precomputed specs, right?) $\endgroup$ – loopy walt Apr 7 at 10:18
  • $\begingroup$ @loopywalt I don't think the messy program I wrote is worth sharing, but what it did was pick two random lines which form the axes of two of the cylinders. The other 4 cylinders are 60 and 120 degree rotations of those. The score of this cylinder arrangement is the distance between the two cylinders that are furthest apart. Then it keeps randomly perturbing the two base cylinders to improve the score (rejecting the perturbation if the score gets too bad or the cylinders intersect). After many runs, it sometimes gets an almost zero score, i.e. a solution. $\endgroup$ – Jaap Scherphuis Apr 7 at 10:26
  • $\begingroup$ @loopywalt I have now added an explanation of what the computed numbers in the script mean. $\endgroup$ – Jaap Scherphuis Apr 7 at 11:18

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