1
$\begingroup$

This is a follow-up question to my previous question, which I felt was too unrelated to what I wanted: Expansion from 3D tetrahedron to 4D 5-cell via combining smaller parts

Anyway, if you don't know already, I'll say it again: I'm making n-dimensional maze puzzles. These are basically mazes where you move around in n-dimensional space (n is specified in the puzzle) and you have to get from point A to point B. The n-dimensional solid is split into its Platonic components (e.g. a 5D hypercube that's 3 cubes wide would become nine 3D cubes) and then those are cut into smaller cubes of the same width (e.g. width 3 would be 27 smaller cubes). After this, I make each mini-cube a wall or a space, and it is a maze then.

I need to have some more solids instead of just n-dimensional cubes. For example, the 5-cell. Problem is, I don't know how to represent a 5-cell.

The 4D hypercube can be represented as a point that extends by translating itself left/right in the first dimension, that extends by translating itself up/down to a square, that extends by translating itself forward/backward to a cube, that extends by translating itself in 4D directions to a hypercube. However, I don't know how this works for a 5-cell, or a 120-cell, or a whatever-cell. I need a way to take a 5-cell and turn it into tetrahedra, then make it into smaller tetrahedra, so I can have a base for my maze. If the cube can be turned into a hypercube by translating it in a 4D direction, how does the tetrahedron translate into a 5-cell?

If the tetrahedron wouldn't make a great maze using this method, let me know of a better method for a tetrahedron in any dimension. Thanks so much in advance. :)

(Sorry for reposting, but I think that previous question would not help, and I don't think editing my first question into a whole new question would be a good idea!)

$\endgroup$
1
$\begingroup$

Start with a regular tetrahedron represented in three dimensions with the vertices

$(1,1,1)$

$(1,-1,-1)$

$(-1,1,-1)$

$(-1,-1,1)$

in which each edge has length $\sqrt{8}$.

Add a zero for the fourth coordinate and then specify the fifth vertex as indicated below:

$(1,1,1,0)$

$(1,-1,-1,0)$

$(-1,1,-1,0)$

$(-1,-1,1,0)$

$(0,0,0,a)$

where $a$ is chosen so that the distance from the fifth vertex to the other four matches the $\sqrt{8}$ value in the base tetrahedron. Thereby, $a=\pm\sqrt{5}$.

If you want rational coordinates only, you need five dimensions (compare with an equilateral triangle needing three dimensions, see here). Here is one possible five-dimensional set of vertices (edge length =$\sqrt{2}$):

$(1,0,0,0,0)$

$(0,1,0,0,0)$

$(0,0,1,0,0)$

$(0,0,0,1,0)$

$(0,0,0,0,1)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What I need is how to represent a 5-cell as a bunch of tetrahedra. Perhaps I said my question wrong. I will edit it. $\endgroup$ – Jonah Ryman Feb 17 '19 at 19:33
  • $\begingroup$ Are you looking for a projection into three dimensions? $\endgroup$ – Oscar Lanzi Feb 17 '19 at 19:52
  • $\begingroup$ No. A 4D cube can be split into the important "parts:" the position at every unit of length. A 4D cube that is 5 inches long in 4D space can be turned into 5 5x5x5 3D cubes, omitting the inbetweens. This is how I make my mazes. Otherwise there would be infinite blocks to work with, and that would be a daunting task. I'm asking, how would you do this same thing with a 5-cell? $\endgroup$ – Jonah Ryman Feb 17 '19 at 20:02

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.