I"m wondering about this picture of a tesseract

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I"m wondering about this picture of a tesseract

Postby tryagain » Wed Jun 07, 2006 3:13 am

Picture:
Image

I’ve been looking at this picture of a tesseract, and I wondered if there an error on it. Since everything from 2D is proportional to 3D just as 3 D is proportional to 4D,

If we open up a square, we get 4 lines:
Image

If we open up a cube, we get four squares in a line and two squares sticking out:
Image

So if we open up a 4-cube, shouldn’t we get one with 6 cubes in a line, and two cubes sticking out like this:?
Image

or this?
Image

I dont see why a tesseract opened up would look almostl ike a cube opened up, when a square opened up (a line) is completely different than a cube opened up

p.s. sry if the pics are small, just click on them if u cant see it


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Postby PWrong » Wed Jun 07, 2006 8:18 am

A tesseract has eight cubes in it. Your last two pictures have ten. :?
Nice pictures though
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Postby bo198214 » Wed Jun 07, 2006 8:29 am

PWrong wrote:Nice pictures though


yup :)
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Postby wendy » Wed Jun 07, 2006 8:35 am

The top piccie is correct. The other two have extra cubes that ought not be there.

Orthogonal columns of 4 cubes in a line also work.

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Postby Keiji » Wed Jun 07, 2006 9:29 am

erm, you are saying:

Square -> 4 lines
Cube -> 4 squares + 2 on the sides
Tesseract -> 6 cubes + 4 on the sides

how is that more logical than the tesseract having 4 cubes + 4 on the sides? It isn't. :P

Tesseract has 4 cubes + 4 on the sides like the picture shows.
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Postby tryagain » Wed Jun 07, 2006 6:55 pm

oh, i didn't see the extra two cubes on the front and back...
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Postby PWrong » Thu Jun 08, 2006 6:46 am

Oh, I see now. So you think there should be six cubes in a line, plus two on the sides?

There are 11 distinct nets for a cube, and I'm not sure how many for the tesseract. However, I don't think yours is one of them.
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Postby Keiji » Thu Jun 08, 2006 7:28 am

Two on the sides wouldn't work, as we have seen from various folding animations.
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Re

Postby Kevin_Ahab » Tue Jul 25, 2006 3:18 pm

The way I see it, a n-dimensional hypercubes net is simply an n-1 dimensional hypercube with an n-1 dimensional hypercube touching each of its n-2 dimensional cells, and on one of those, another n-1 dimensional hypercube going further out. i.e.

Point: -n/a

Line: -unrepresnetnable in Pointspace

Square: -a line (n-1-d-hypercube) with a line from each of its points (n-2-d-hypercubes), and from one of those, another line

Cube: -a square with a square from each of its lines, and from one of those, another square

Tesseract: -a cube with a cube from each of its squares, and from one of those, another cube

This can be continued, but this forum is for the Fourth Dimension.

This is true becuase:

To find the number of n-spaces in an x-dimensional hypercube, simply: 1) double the number of n-spaces in the x-1-dimensional hypercube and 2) add this to the number of n-1-spaces in the x-1-dimensional hypercube.
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Postby pat » Wed Jul 26, 2006 7:58 pm

It's important to remember that there's more than one way to unfold a shape. The key part is how the pieces line up together. I personally, find the first picture you showed to be the easiest to see how they fit together.

For example, each of these is an unfolded cube:
Image

But, the 4-cube can't be six 3-cubes in a row with two 3-cubes tacked on the sides. That would imply there is a run of six 3-cubes that you can go through where when you enter the 3-cube through face AA, you exit the cube through the face opposite AA. There are no such runs in the 4-cube.

Start in any 3-cube that makes up the surface of the 4-cube. Exit through a (2-cube) face and into another 3-cube. Exit that 3-cube through the (2-cube) face opposite the one you entered on. Lather. Rinse. Repeat. No matter which 3-cube you started in or which face of the 3-cube you exited through to start off, you will end up back in the original 3-cube on your fourth step.
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Postby bo198214 » Wed Jul 26, 2006 8:08 pm

hey why not add all also the tesseract unfoldments?
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Postby pat » Wed Jul 26, 2006 8:16 pm

Because, I don't have that much time at the moment.

I should also note that the above may not be an exhaustive list of cube unfoldings. They were just all of the distinct 2-d ones that I came up with that do not involve adjacent squares with a cut between them.
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Postby bo198214 » Wed Jul 26, 2006 8:18 pm

*nod*
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Postby pat » Wed Jul 26, 2006 10:36 pm

Here are some unfolded hypercubes. This is nowhere near exhaustive.

Image
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Postby pat » Thu Jul 27, 2006 4:08 am

wendy wrote:Orthogonal columns of 4 cubes in a line also work.


Really? I can't get there.

In this picture, the lower green cube is between the blue cubes and between the magenta cubes. So, I can't find a way to make two orthogonal columns of 4 cubes since both columns have to share two cubes.

Image

Ah, unless you meant this:

Image
oops, I got the overlapping wrong on the top cube of the vertical column in this diagram

Which, of course, has to be what you meant... otherwise there would only have been seven cubes if the two columns of four shared a cube.

I got it now.
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Postby bo198214 » Thu Jul 27, 2006 7:48 am

Thanks pat, I was anyway thinking about programming a folding mode into 4d building blocks. Though I think that really interesting objects like the 3-torus looks rather awkward when surface is built by cubes ...
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Just ran across this...

Postby pat » Fri May 11, 2007 11:13 pm

There was an article in Volume 17, Issue 1 of the Journal of Recrational Mathematics (1984/85) titled "Unfolding the tesseract". The article is by Peter Turney. Here's the abstract:
The author shows that there are 261 distinct unfoldings of a tesseract (hollow hypercube) into order-8 polycubes in 3-space. This answers a question of M. Gardner [Scientific American 215 (1966), no. 5, 138-143].
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