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

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 ($$\frac{4 \times 6 + 6 \times 8}{2}=36$$) These examples seem to be known as Permutohedron: https://en.wikipedia.org/wiki/Permutohedron

• Since you can "build" a unit cube using 12 tetrahedron congruent to one with vertices at $(0,0,0)$, $(\frac12,-\frac12,\frac12)$, $(\frac12,\frac12,\frac12)$ and $(-\frac12,\frac12,\frac12)$, you can tessellate $\mathbb{R}^3$ by tetrahedrons. Feb 14 '20 at 20:40
• @achille hui, I agree +1, thanks. how about those you cannot "build" before tessellation? Feb 14 '20 at 21:19
• maybe of interests too: math.stackexchange.com/questions/3548449/… Feb 16 '20 at 19:25

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

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).

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: • I am interested in the minimum, and the simplest case. But thanks +1 Feb 15 '20 at 3:18
• do you know other minimum and the simplest cases? Feb 16 '20 at 7:03
• maybe of interests too: math.stackexchange.com/questions/3548449/… Feb 16 '20 at 19:25

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