Is there a tiling of the Euclidean space such that cells that share a vertex or an edge also share a face? My arts-student brother asked me a question that led to a morning spent constructiong duals of graphs with Geomag, with no success.
The premise is the observation that in the plane, the square tiling leads to ambiguity over the (colloquial) term "adjacency", i.e. whether two tiles sharing a vertex but not an edge count as adjacent.
This problem is avoided with the hexagonal tiling: all hexes that share a vertex actually share a whole edge.
The question is: can this be extended to 3D space?
Note that graph theory is not my specialty, so feel free to correct any misuse of terminology.
My investigation started with the observation that in the plane, the "disambiguity property" (i.e. that tiles sharing a vertex share a whole edge) is equivalent to requiring that each vertex has exactly three edges. If it had more, it would locally cut the plane in 4 parts such that two opposite parts do not share an edge (and if it were less, it wouldn't really be a vertex).
Therefore, the dual of a 2D disambiguous tiling is a tiling of triangles - and indeed, the dual of the regular-triangle tiling is the hexagonal tiling.
Unfortunately, this reasoning does not extend easily to 3D. In three dimensions, it stands to reason that in a disambiguous tiling must have vertices with exactly 4 edges. Therefore, its dual must be a tiling of triangular pyramids. We constructed such a tiling, and its dual, but tiling the dual appeared to produce something that is not a tiling of space - we couldn't even recognize clearly defined volumes.
Is this a known problem in literature?
 A: As you note, the hexagon tiling meets your criteria. This is because just three meet at any vertex. The first hexagon has two adjacent sides and each of these aligns with one of the remaining two hexagons.
In three dimensions, the criterion is met when just four polyhedra meet at a vertex. Less than four will not be sensible polyhedra; at least three faces must surround each vertex, so you must have your original polyhedron plus three others. More than four allows two to be only vertex-connected.
A convenient way to find such tilings is to take tilings of four-faced polyhedra (tetrahedra) and dualise the tiling. (In the plane you need to find tilings of triangles - that of equilateral triangles is dual to that of your hexagons).
The truncated octahedron is the only uniform polyhedron to meet this condition. There are three such tilings of two or more uniform polyhedra, see The Archimedean Honeycomb Duals which describes the dual tilings and notes the uniform tiling dual to each.
Similarly in the plane there is a tiling of right isosceles triangles which is dual to that of the octahedron and square, and so on.
Quite a few more spacefilling tetrahedra are known, though there is no proof the enumeration is complete. I believe that a list of the known examples exists, but I cannot recall it offhand. Dualising their tilings by hand can be rather laborious, as they are generally less regular than the uniform ones.
