A cube can be cut and flattened out onto a table in a way that the faces stay connected and none of them overlap. There are $384$ ways to make the cuts and $11$ distinct meshes emerge (see here). And similarly, it is possible to flatten a Tesseract (4-d cube) into 3-d space. This time, there are 261 distinct 3-d meshes. This is covered here: https://unfolding.apperceptual.com where Turney uses a new concept (I guess not new anymore) of "paired trees". And in this post: 3D models of the unfoldings of the hypercube?, the 3-d models of the meshes were visualized.

But a 4-d cube can also be flattened into 2-d space in a way that none of the faces overlap with each other (just as a 3-d cube can; by cutting some edges and rotating the faces down while keeping them connected). Below is an example of such a flattening. In the picture, the faces are named according to the centers of their faces in the original Tesseract. For example, if a face center is $(-1,0,0,1)$ then its name becomes -00+ and so on. The grey lines indicate that the faces aren't connected and just happen to lie adjacent. My question is - how many such distinct meshes exist (similar to how there were $11$ for the cube)? In the absence of an exact answer, I'd be interested in tight upper and lower bounds or a computational technique that could solve this in a reasonable amount of time or a computational technique for the bounds.

enter image description here

My attempt so far:

The trick involving paired trees Turney used for counting the 3-d flattening of a Tesseract doesn't seem to apply here easily. The reason is that if you're going one dimension down, any two "faces" are either directly connected, or have just one "hop" between them (connected through another face). For the case of 2-d faces of a Tesseract, the faces could have $0,1,2$ or $3$ hops between them and extending the concept of paired trees, leave alone counting such trees becomes way too complicated.

I came up with a way to start with one mesh and generate others from it in a kind of random manner (currently writing a paper about the approach, will post to ArXiv and add here once done). This provides a computational lower bound. I'm currently at about $514$ (but grows every day). However, I wouldn't know if and when I'm done with all the meshes with this approach. For an upper bound, one way would be to count the spanning trees of the graph formed by the faces. This is what gave us the $384$ for the cube. Kirchoff's theorem can be used to count these and it turns out to be $~2 \times 10^{20}$ for a Tesseract.

We could loop through the spanning trees, generate the meshes (if possible, which most of the time won't be) and see how many distinct ones emerged. But even if we spend 1 milli-second on each of them, it would take more than the current age of the universe to get through.

We can probably tighten this upper bound by leveraging the symmetries of the Tesseract and reduce by a factor of $~100$. Still, saying the number of meshes is somewhere between $514$ and $2 \times 10^{18}$ isn't very impressive.

  • $\begingroup$ This will not help in the bounds, but one route is to first unfold to $\mathbb{R}^3$ and then unfold to $\mathbb{R}^2$ (rather than directly from 4D to 2D). This was used in Diaz, Giovanna, and Joseph O'Rourke. "Hypercube Unfoldings that Tile R^3 and R^2." arXiv:1512.02086 (2015). $\endgroup$ Aug 15, 2022 at 11:44
  • $\begingroup$ Reading through the paper, but I found it easier to go from 4-d to 2-d than going from 4-d to 3-d to 2-d. This is because in the 3-d version, you have to keep track of which faces are repeated and this becomes an additional headache. $\endgroup$
    – ryu576
    Aug 16, 2022 at 6:36
  • 1
    $\begingroup$ While 4-d to 2-d has been what most have attempted so far, I think going to 3-d first will make the attempt easier, & more human readable. I believe there is a 3-d polyhedron who's net will relate near directly to the tesseract's net. Currently studying the deltoidal icositetrahedron; I think taking it's net and treating 16 edges as higher dimensional will reveal some interesting things -- or at least I hope. Really excited for any answer to your question! $\endgroup$
    – Ash
    Aug 27, 2022 at 23:49


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.