Disclaimer: i am bioinformatician and programmer, please excuse if my wording and definitions are far from elegant and occasionally imprecise.

Intro: I am interested in space tessellations of n dimensional spaces with following properties:

1 a tessellation must be made with only one type of element, with identical angles and edge lengths
e.g. cubic honeycomb is okay, pythagorean tiling is not (one type but two sub-types of different size) and https://en.wikipedia.org/wiki/Truncated_trihexagonal_tiling is also not (three types of elements)

2 for n dimensional space, each n dimensional element is built out of one type of n-1 dimensional elements,
e.g. cubic honeycomb is okay, hexagonal prismatic honeycomb is not

3 elements must connect only through n-1 dimensional elements
e.g. hexagons on a 2D plane are okay, squares are not okay for 2D plane (they connect to 8 others, 4 through edges, 4 through vertices)

Question: For 2D, hexagons are an example which satisfy these conditions. Are there any examples for N > 2?
Sub-question: Does https://en.wikipedia.org/wiki/Rhombic_dodecahedral_honeycomb satisfy these conditions? (i cannot come up with an idea on how to verify this)
Note 1: I am mainly interested in N = 3 and N = 4, but general solution would be best.
Note 2: If they exist, pointers what to look for (keywords, books, articles, etc) would be greatly appreciated

EDIT: A related question: Are there higher-dimensional tessellations touching only nearest neighbours?

  • $\begingroup$ Rhombic dodecahedral honeycomb fails the 3rd condition. Maybe you are asking for too much. $\endgroup$ – Ivan Neretin Sep 3 '18 at 16:30
  • $\begingroup$ @IvanNeretin Could you demonstrate how does it fail the 3rd condition? It's not obvious to me from the pictures I've found, and in general a method to check this properly would be nice. Also: i do not think that 'too much' is a thing in math. Either something exists, or not. If it only happens in 2D and not higher dimensions, it would be interesting to see why. $\endgroup$ – Empischon Sep 4 '18 at 7:50
  • 1
    $\begingroup$ Read the Wikipedia link till the words: "...The honeycomb is thus cell-transitive, face-transitive and edge-transitive; but it is not vertex-transitive, as it has two kinds of vertex. The vertices with the obtuse rhombic face angles have 4 cells. The vertices with the acute rhombic face angles have 6 cells." Now if there are 6 cells meeting at a vertex, then each cell must have 5 neighbors, but it only has 4 faces meeting at that vertex, hence there are four face-neighbors and one single-point-neighbor. $\endgroup$ – Ivan Neretin Sep 4 '18 at 8:12
  • $\begingroup$ @IvanNeretin Thanks! $\endgroup$ – Empischon Sep 4 '18 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.