2
$\begingroup$

I was presented with the problem of breaking the unit cube $[0,1] \times [0,1] \times [0,1] $ into tetrahedron shapes. The first two pieces are easy, but it's not so easy to visualize after that. I found:

  • $\{ (0,0,0), (1,0,0), (0,1,0), (0,0,1) \}$

  • $\{ (1,1,1), (1,1,0), (1,0,1), (0,1,1) \}$

What remains is a triangular prism. Then maybe I think it readily splits into 2 or 3 pieces. In any case, they are difficult to draw and keep track of all the data.

Also I think I could have started differently with this other tetrahedron:

  • $\{ (0,0,0), (1,1,0), (0,1,1), (1,0,1) \}$

enter image description here

There should be lots of solutions, but I neither have the date to store into a computer nor do I have picture of even a single one.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ What's your question and how is it related to computer science? $\endgroup$ – David Richerby Mar 28 '18 at 14:35
  • $\begingroup$ @DavidRicherby it's Discrete Math. E.g. here's a CS course on polyhedral meshes. You see it sometimes with Computer Aided Design. $\endgroup$ – john mangual Mar 28 '18 at 14:36
  • 1
    $\begingroup$ Discrete math is Mathematics. The course you link to is called "Algorithms for polyhedra" (my emphasis). Your question doesn't seem to be about algorithms or any other computational concept. And what is your question, anyway? $\endgroup$ – David Richerby Mar 28 '18 at 14:40
  • $\begingroup$ @DavidRicherby here are 400 other questions with tag [Computational Geometry] $\endgroup$ – john mangual Mar 28 '18 at 14:41
  • 4
    $\begingroup$ I'm not asking about those questions. I'm asking about your question. What is computational about your question? $\endgroup$ – David Richerby Mar 28 '18 at 14:42
2
$\begingroup$

The simplest way is a partition into $3! = 6$ tetrahedra:

  1. $\{x,y,z : 0 \leq x \leq y \leq z \leq 1\}$.
  2. $\{x,y,z : 0 \leq x \leq z \leq y \leq 1\}$.
  3. $\{x,y,z : 0 \leq y \leq x \leq z \leq 1\}$.
  4. $\{x,y,z : 0 \leq y \leq z \leq x \leq 1\}$.
  5. $\{x,y,z : 0 \leq z \leq x \leq y \leq 1\}$.
  6. $\{x,y,z : 0 \leq z \leq y \leq x \leq 1\}$.

The corresponding tetrahedra have the following vertices:

  1. $(0,0,0),(0,0,1),(0,1,1),(1,1,1)$.
  2. $(0,0,0),(0,1,0),(0,1,1),(1,1,1)$.
  3. $(0,0,0),(0,0,1),(1,0,1),(1,1,1)$.
  4. $(0,0,0),(1,0,0),(1,0,1),(1,1,1)$.
  5. $(0,0,0),(0,1,0),(1,1,0),(1,1,1)$.
  6. $(0,0,0),(1,0,0),(1,1,0),(1,1,1)$.
Improve this answer
$\endgroup$
2
  • $\begingroup$ thanks. i seem to have trouble stating the problem in an algorithmic way. the point is that dividing the cube into simplexes is a fundamental problem. and there are many solutions. infinitely many?? $\endgroup$ – john mangual Mar 30 '18 at 14:43
  • 1
    $\begingroup$ It's not an algorithmic question. It's a mathematical question. $\endgroup$ – Yuval Filmus Mar 30 '18 at 14:45
0
$\begingroup$

John your second and third pictures are perfect, you have your 5 tetrahedrons there. You just can't see the fifth one behind the shaded area in the third picture but you can see it in the second picture following the dotted lines. So you've got it right.

Improve this answer
$\endgroup$
1
  • $\begingroup$ there are other decompositions i wonder if there is algorithm to find them all $\endgroup$ – john mangual Mar 28 '18 at 21:02

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.