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There are (apparently) 261 distinct unfoldings of the 4D hypercube, a.k.a., the tesseract, into 3D.1 These unfoldings (or "nets") are analogous to the 11 unfoldings of the 3D cube into the plane.2 Usually only one hypercube unfolding is illustrated,


          TesseractUnf
          (Image from this link.)
the one made famous in Salvador Dali's painting Corpus Hypercubus. My question is:

Q. Has anyone made models/images of the 261 unfoldings as solid objects in $\mathbb{R}^3$?

(If not, I might do so myself.)


1Peter Terney, "Unfolding the Tesseract." Journal of Recreational Mathematics, Vol. 17(1), 1984-85.


2
CubeNets


Update. See also the followup question, "Which unfoldings of the hypercube tile 3-space: How to check for isometric space-fillers?."

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    $\begingroup$ An unusual question in that I am kinda hoping no one answers. $\endgroup$ Commented Mar 1, 2015 at 0:54
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    $\begingroup$ You might like a movie on the 2D case: etudes.ru/ru/etudes/cubisme $\endgroup$
    – Igor Pak
    Commented Mar 1, 2015 at 3:28
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    $\begingroup$ "Domino's Sugar? I'd like to place an order for $2088$ sugar cubes..." $\endgroup$ Commented Mar 1, 2015 at 5:34
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    $\begingroup$ @ManfredWeis: That is a legitimate question, but I would prefer not to answer it. $\endgroup$ Commented Mar 1, 2015 at 14:36
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    $\begingroup$ I notice that of the 11 unfoldings of the cube, 2 have mirror symmetries and the other 9 do not. If we count the "chiral" pairs separately, then we have a total of 20 unfoldings. From the beautiful figures in two of the answers, it is difficult for me to tell how many of the 261 unfoldings of the 4D hypercube have mirror symmetries. Does someone know? $\endgroup$
    – Menachem
    Commented May 23, 2018 at 18:44

2 Answers 2

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I implemented the ideas in the paper using Mathematica. I pushed it a bit further to actually generate the images below. You can download this Mathematica notebook to see the code and detailed explanation.

You might notice Dali's original in the middle of the third row from the bottom.

enter image description here

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  • $\begingroup$ Please share the code, Yes! :-) Do you verify the count 261? $\endgroup$ Commented Mar 4, 2015 at 1:50
  • $\begingroup$ Yes, it's 261. I did the cube with similar techniques and got 11 as well. I need to clean up the notebook just a bit and I'm not sure what the correct way to share it is. It's quite a lot of code that's probably not completely appropriate for this site. I could upload the notebook to my webspace and just post a link, if nothing else. $\endgroup$ Commented Mar 4, 2015 at 1:52
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    $\begingroup$ Thanks, Yes, a link is best. And, sure, clean up as you wish. This is great! Were Dali still with us, he'd have 260 new paintings to contemplate. :-) $\endgroup$ Commented Mar 4, 2015 at 1:53
  • $\begingroup$ @JosephO'Rourke The linked notebook is there now - hope you enjoy it! $\endgroup$ Commented Mar 4, 2015 at 2:33
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I used sage to make a 3d animation of all 261 unfoldings.

Here is a screenshot of the first few:

enter image description here

The file cube-unfoldings.txt contains all the unfoldings, each line contains a list of 8 points.

Edit: By popular, I add the (poorly commented) code:

unfolding the hypercube.ipynb: a jupyter notebook with sage code to generate the pairings for the unfoldings together and find the embeddings. To view the code easily, checkout the file at github

The animation on the website is made with threejs, and all the code is contained in the unfoldings.html, which you can also view on github.

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    $\begingroup$ Beautiful! ${}$ $\endgroup$ Commented May 21, 2018 at 11:06
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    $\begingroup$ Very nice! I had the idea of displaying these on the web using X3Ddom on my extended to do list - I guess I can cross that off my list. :) Can you make the code available? $\endgroup$ Commented May 23, 2018 at 15:07
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    $\begingroup$ @MoritzFirsching why not both? :) $\endgroup$
    – j.c.
    Commented May 23, 2018 at 16:59
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    $\begingroup$ @j.c. sure, I didn't mean 'or' to be exclusive.. $\endgroup$ Commented May 23, 2018 at 17:47
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    $\begingroup$ @MoritzFirsching Thanks! I think the code improves the answer a lot. $\endgroup$ Commented May 23, 2018 at 18:07

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