Extend triangular grid to 3D space?

Question

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

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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)

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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"?

Answer 1

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.

Answer 2

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.

Answer 3

Consider that a rhombic dodecahedron tiles Euclidean 3 space.