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I have a small question about creating a 3D array of particles. It's necessary that 265 particles are located orderly within a unit cube. One hint is to use a simple cubic structure but I don't see how to realize that. So I'm looking for some commands to distribute the particles with the same distance to all it's neighbors but I have no ideas how to do that.

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closed as unclear what you're asking by Henrik Schumacher, Michael E2, m_goldberg, kirma, bbgodfrey Jan 13 at 22:45

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Look for Tuples $\endgroup$ – OkkesDulgerci Jan 8 at 19:50
  • $\begingroup$ Why not a simple array of $8 \times 8 \times 8$ particles? $\endgroup$ – David G. Stork Jan 8 at 20:33
  • $\begingroup$ Graphics3D[Point[SpherePoints[265]]] ? $\endgroup$ – OkkesDulgerci Jan 8 at 20:54
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    $\begingroup$ I am not sure this is possible since` 265^(1/3)=6.42316` particles per side of the cube which is not a whole number.. $\endgroup$ – OkkesDulgerci Jan 8 at 21:37
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    $\begingroup$ It's not clear to me that everyone know what the technical definition "located orderly" is. I can see from google that it's a term in science journal articles, but I don't know what it means. $\endgroup$ – Michael E2 Jan 8 at 22:04
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You could use the code HilbertCurve3D[n] by Michael Trott (page 93 of The Mathematica Guidebook for Programming) from this question. Given input n, the function returns $2^{3n}$ orderly points within the cube from {0,0,0} to {1,1,1}. Use RandomSample to select 265, or any number, of these points.

Graphics3D[
   Point[RandomSample[HilbertCurve3D[3], 265]],
   BoxStyle -> Directive[Thick, Red]
]

random sample of unit cube

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  • $\begingroup$ What is the difference between your solution and Graphics3D[Point[RandomSample[Tuples[Range[0, 1, 1/7], 3], 265]], BoxStyle -> Directive[Thick, Red]] $\endgroup$ – OkkesDulgerci Jan 8 at 21:44
  • $\begingroup$ Well I think these two solution are as good as it gets to solve further problems. Simple Cubic with some vacancy defect should do it too. Thanks alot $\endgroup$ – PaladinDanse Jan 9 at 10:07
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What about $265$ points sampled uniformly from the cube? Somewhat orderly!

Graphics3D[Point /@ (Flatten@Table[{RandomReal[],RandomReal[],RandomReal[]},{265}])]

265 points in cube

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  • $\begingroup$ Yes. This gives me a nice grid but with 512 instead of 265 particles. $\endgroup$ – PaladinDanse Jan 8 at 20:47

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