Packing/tessellating 3 dimensional space fully by polytopes? Give examples.

Question

What is a shortlist of first few simplest (say 5~10 simplest) possible shapes of polyhedra/polytopes (with a minimum number of edges shared) to pack the 3-dimensional flat space (say $\mathbb{R}^3$) fully?

By the simplest, I require it to be "with a minimum number of edges shared."

As far as I know,

(1) The cubic works:

(2) This polyhedron with

• 24 vertices ($4 \times 6$)
• 14 faces contain 6 squares and 8 hexagons
• 36 edges ($\frac{4 \times 6 + 6 \times 8}{2}=36$)

seem also work:

These examples seem to be known as Permutohedron: https://en.wikipedia.org/wiki/Permutohedron

Answer 1

Perhaps you would be interested in this MO question: How many vertices/edges/faces at most for a convex polyhedron that tiles space?, and this $38$-face tiler:

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Answer 2

It is not hard to see that a regular octahedra cannot tile space. But there is a subtle way to identify an octahedron that can.

Think of a body-centered cubic lattice. Select any face of a lattice cube; its corners and the body centers of the two cubes sharing that face define the six vertices of an octahedron. Congruent octahedra, in three different orientations, may be defined by choosing different lattice-cube faces, and they fill all of space (in fact, each point not along a boundary belongs to two of these octahedra, because there are two interlocking simple cubic lattices in the bcc lattice).

Answer 3

Triangular prisms or hexagonal prisms would do so as well: they simply extend the well-known regular 2D tilings into 3D via periodic stacking within the 3rd direction.

--- rk

Answer 4

If you wouldn't be affected when considering different types of polyhedra within your honeycomb (aka 3D tesselation), then you might look here for lots more.

--- rk