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I'd like to make a maze or a game that is played in the Order-6 cubic honeycomb. I like the fact that there are infinitely many cells at each vertex. However, I need some kind of coordinates here.

The coordinate system must be capable of these actions:

  • If I venture into the same cell by two different ways, I should be able to recognise that both endpoints are the same.
  • My orientation should not change by travelling into another cell (i.e. there is no preferred orientation). For example: I go one cell up, one cell forwards, one cell down, one cell backwards, one cell up, one cell forwards. Now I should end in the same cell as I began but also upside down.
  • Travelling like 20 cells in one direction shouldn't have any distorsive effects.

My only idea is storing a way from origin to the given cell but:

  • How do I get if two ways lead to the same point?
  • What if the path is unneccessarily long?

EDIT: Here is an image showing what I mean by the "orientation should not change". The arrow shows my orientation which does not change (from my perspective, there are only right angles). The picture shows intended behaviour. enter image description here

Hume2
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  • "My orientation should not change by travelling into another cell" ... "Now I should end in the same cell as I began but also upside down." ??? If your orientation does not change, why would you be upside down? Hyperbolic space is still orientable. – Paul Sinclair May 30 '19 at 04:39
  • I draw an arrow pointing up into one cell. Then I perform these 6 steps. After that, I should see the arrow pointing down. – Hume2 May 30 '19 at 05:18
  • I added a picture which explains what I mean. – Hume2 May 30 '19 at 12:04
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    That is not keeping the same orientation as you move from one cell to the next. I don't have an answer for your coordinate question, but suggest that you look at the dual tesselation Order-4 hexagonal tiling honeycomb. There the cells become nodes and moving from cell to cell corresponds to travel along edges. This converts your hypergeometric tesselation problem into a graph, which likely will be easier to figure out. – Paul Sinclair May 31 '19 at 04:29

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