Extend triangular grid to 3D space?

I am interested in properties of regular neighbourhood meshes, but I feel like I'm missing keywords to investigate further.

In the 2D world I, like both triangular neighbourhood and square neighbourhood, because:

• they are both regular, which means to me:
• there is always same distance between two close neighbours (say, 1)
• there is always the same angle between directions towards two closest neighbours (60° for the triangle, 90° for the square)

But I like triangular best, because:

• it offers more neighbours than the square (6 against 4 only)
• 2 closest neighbours are also neighbours themselves (whereas in the square world, if your he-friend goes one step north and your she-friend goes one step east, they are now more than 1 step away)

In the 3D world, I only know the square neighborhood well.

But I don't like how it only offers 6 neighbours and how this last "neighbours neighbouring" feature is not respected.

Is there another regular 3D neighbourhood with more than 6 neigbours per node and featuring "close-neighbours-neighbouring"?

My interpretation of the close-neighbours-neighbouring property is as follows:

For each node, the minimum separation among all distinct pairs of its neighbours is the same as the distance between the node and its neighbours.

Then the fcc lattice (right side of the diagram below) satisfies the two properties, having 12 neighbours per node. The spaces between the nodes then form the alternated cubic/tetrahedral–octahedral honeycomb.

• Well, I think you nailed it :) Wonderful lattice to me. Thanks a lot! Oct 26, 2017 at 19:48
• Of course, I forgot asking: do you think this lattice is the only solution? Oct 30, 2017 at 12:38
• @iago-lito You could also try taking the 2D triangular lattice and stacking it so that triangles align. The cells gotten are triangular prisms. But I find this rather inelegant. Nov 15, 2017 at 18:00
• It sure is not ;) In addition it does not fulfill my requirements about regular angles and close-neighbours-neighbouring. I'll forget about it ;) Nov 20, 2017 at 18:24
• However, since the sphere layers may be stacked in any order (ABCABABABCABABCAB.. etc), you may construct any lattice you want appart from regular ABCABCABC.. but none will exhibit nice straight infinite directions like the regular does ;) Nov 20, 2017 at 18:27

No; my understanding of the "close neighbors neighboring" property is that the cells would have to be tetrahedra, and there is no regular tetrahedral tiling of space.

You can get something close to a tesselation that you want by computing a (dual of a) cetroidal Voronoi tesselation of a region of space. It will be irregular (so will not have perfectly equal distances and angles) of course.

• I am sorry, but your link seems broken. Why would they need to be tetrahedra? Oct 26, 2017 at 19:38

Consider that a rhombic dodecahedron tiles Euclidean 3 space.