# How to decompose a unit cube into tetrahedra?

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:

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

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.

• What's your question and how is it related to computer science? Commented Mar 28, 2018 at 14:35
• @DavidRicherby it's Discrete Math. E.g. here's a CS course on polyhedral meshes. You see it sometimes with Computer Aided Design. Commented Mar 28, 2018 at 14:36
• 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? Commented Mar 28, 2018 at 14:40
• @DavidRicherby here are 400 other questions with tag [Computational Geometry] Commented Mar 28, 2018 at 14:41

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. $(0,0,0),(1,0,0),(1,1,0),(1,1,1)$.
• 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?? Commented Mar 30, 2018 at 14:43
• It's not an algorithmic question. It's a mathematical question. Commented Mar 30, 2018 at 14:45

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.

• there are other decompositions i wonder if there is algorithm to find them all Commented Mar 28, 2018 at 21:02