I've been playing with the unfoldings of a cube and a dicube. And it got me thinking of generalizing the unfolding process to different dimensions.

Let me explain. A 1x1 square which is a 2D object can be unfolded into a line of length 4 which is a 1D object. I believe this is the only unfolding possible in this case where the 2D object becomes a 1D object.

Moving up to the next dimension lets looks at a 1x1x1 cube. Its surface is made up of 6 1x1 squares and these can be unfolded into 11 different 2D objects (a subset of the hexominos) as seen here.



Now let's take this up 1 more dimension. We have the 1x1x1x1 hypercube. My 3D brain always struggles with 4D objects, but I believe it is fair to say the surface of a 1x1x1x1 hypercube is made up of 8 1x1x1 cubes.

I'm clearly not the first person to think of this as I found these nice animations on the web.

First what the 2D shadow of an unfolding cube looks like.


We can compare this to the 3D shadow of an unfolding hypercube.


So I bet you can guess what the next question is...

How many different order-8 polycubes can be produced by unfolding a hypercube into 3 dimensional space?

And there too it looks like I've been beaten to the punch. It turns out that question was aparently first asked 3 years before I was born. Here is a very nice article I found AFTER I started typing this post.

http://unfolding.apperceptual.com/

It turns out there are 261.

So... needing a new question how about this. 261 times 8 equals 2088. 2088 can be factored as 29*3*3*2*2*2. So it looks like the most compact cuboid that these MIGHT fit into would be one with these dimensions 29*9*8.

So anyone have a clue if the 261 unfoldings of a hypercube would fit into such a box? Or a proof that they wouldn't?

I'm very very tempted to give it a try if I can manage to get the 261 unfoldings into a format I can play with them. If they can then I may even try to see how cheap I could make the set of 261 pieces available on Shapeways.

Carl

The first way that comes to mind to prove that they can't fit would be to copy the proof for trying to pack tetrominos into a rectangle.
First you assume that the rectangle is like a chess board that has black and white tiles. Then you color your pieces to show that they must cover a different number of black/white tiles. If you look at the T piece, you'll see it will cover 3 white and 1 black (or visa-versa). Any rectangle you plan on covering will have the same number of black and white tiles. Because the T shape doesn't maintain this balance (like all the other pieces), you know you can't fit them in any rectangle.

You should be able to try a similar thing with the tesseract folding pieces. For each piece, count how many black/white pieces it has and just keep track of the imbalance between them. So if it has 2 white and 6 black, record that it has an imbalance of 4. After you process everything, you should have a list of numbers. If you cannot separate the list into 2 list of equal sums, then you've proven they won't fit.
Now I'm worried this won't work because there's lots of pieces with imbalances so if one piece has 4 extra white, another one can counter balance it with 4 black. But I'm hoping there will be an odd number of unbalanced pieces to force it to never fit.


Another hand-wavy proof is that it doesn't look like many of the pieces fit well against an edge or corner.