producing tetrahedra within a cube

Question

A common math problem involves dividing a cube into regular and irregular tetrahedra, where the points of the cube must also be the points of the tetrahedra. A problem I'm working on seems to be going the other way, i.e., "How many unique ways are there to arrange (regular and irregular) tetrahedra within a cube?"

EDIT: It is not necessary to fill the cube, each partially filled cube counts as an arrangement.

First, I would need to find how many ways there are to produce a unique tetrahedron within a cube such that the 4 points of the tetrahedron are 4 of the 8 points of the cube.
Next, for each case of number 1, I would need to find out how many ways there are to fit an additional tetrahedron without overlapping.
Next, for each case of number 2, I would need to fit one additional,
And another,
And one more time.
At this point I believe there is only on possible addition for each remaining case before the cube is full.

I've played with this problem for a while, but am having trouble coming up with a systematic approach to ensure my solution is complete. Any advice?

Are these regular tetrahedra? – J. M. ain't a mathematician May 31 '13 at 17:28 — J. M. ain't a mathematician, May 31 '13 at 17:28
No, any tetrahedra. I made the adjustment to the question. – ridthyself May 31 '13 at 17:35 — ridthyself, May 31 '13 at 17:35

Ross Millikan · Answer 1 · 2013-05-31T18:04:17.510

1

I would look at it as a tree search. There are ${8 \choose 4}=70$ sets of four corners. All but the $6$ faces and $6$ diagonal planes form tetrahedra. The hard part is writing a routine that takes two vertex lists and determines whether two tetrahedra overlap, and are therefore incompatible. Having done so, start with tetrahedron number 1. That is your first possibility. Then see if number 2 fits in without conflict. If so, 1,2 is another. Keep trying to add tetrahedra. Then backtrack and add more. Essentially it is a depth-first search of the tree. $2^{58}$ is a scary number, but you won't have to run through nearly that many.

edited May 31 '13 at 18:04

answered May 31 '13 at 17:50

Ross Millikan

374,822

All right, Mr. Millikan-- you've come to my aid once more. I'll post here my code here once I get it working, along with my solution to see if I missed anything. – ridthyself May 31 '13 at 17:59
Brute force tells me there are 40 ways to fill a cube with tetrahedra without adding any vertices. It's going to take considerable work arriving at some sort of proof, though. – ridthyself Jun 12 '13 at 18:23
40 is incorrect. – ridthyself Dec 24 '13 at 14:15

producing tetrahedra within a cube

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