# Locating Points “orderly” within a cube [closed]

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.

• Look for Tuples – OkkesDulgerci Jan 8 '19 at 19:50
• Why not a simple array of $8 \times 8 \times 8$ particles? – David G. Stork Jan 8 '19 at 20:33
• Graphics3D[Point[SpherePoints]] ? – OkkesDulgerci Jan 8 '19 at 20:54
• 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.. – OkkesDulgerci Jan 8 '19 at 21:37
• 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. – Michael E2 Jan 8 '19 at 22:04

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, 265]],
BoxStyle -> Directive[Thick, Red]
] • What is the difference between your solution and Graphics3D[Point[RandomSample[Tuples[Range[0, 1, 1/7], 3], 265]], BoxStyle -> Directive[Thick, Red]] – OkkesDulgerci Jan 8 '19 at 21:44
• 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 – PaladinDanse Jan 9 '19 at 10:07

What about $$265$$ points sampled uniformly from the cube? Somewhat orderly!

Graphics3D[Point /@ (Flatten@Table[{RandomReal[],RandomReal[],RandomReal[]},{265}])] • Yes. This gives me a nice grid but with 512 instead of 265 particles. – PaladinDanse Jan 8 '19 at 20:47