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 Post subject: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sun Jun 08, 2014 6:05 am 
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Hi,

- intro: I have some difficulties to understand 4D Rubik´s cube. On the other hand, I believe they can be all (or at least most of them) solved by only one picture.

- notation: to avoid confusion, lets call 3D "faces" of 4D tesseract the cells.

- presumption: if we make one face of 1x1x1 cube transparent (I mean only partially transparent, not totally transparent), we are able to see all of its 6 faces just in one look. Same situation for 3D Rubiks cube.

-query: on MC4D (see http://www.superliminal.com/cube/cube.htm) there is shown 3x3x3x3 tesseract. How would this tesseract look like if we make 8th cell (the one which is hidden at the moment) transparent? In other words, can you draw 4D Rubik´s cube so that all cells would be visible? I can only imagine it in case of NxNxNxN tesseract where N=1.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Mon Jun 09, 2014 3:42 pm 
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It's impossible to project all 8 cells of the tesseract into 3D space without distorting the relation of cells.

In my simulator, I 'm drawing 2 opposite cells each at the center of two horizontally aligned windows and 6 others around them. You can bring any cell to the center by arrow keys.
If this is not the best solution, I think it's kind of a good compromise. I think MC4D too has its own way to show the 8th hidden cell. But I don't remember now...

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 10, 2014 1:52 pm 
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gelatinbrain wrote:
It's impossible to project all 8 cells of the tesseract into 3D space without distorting the relation of cells.


I am not fully convinced about it. Could you please explain me why it is impossible in more details? Or, better yet, could you tell me what is wrong with my presumptiom mentioned above and its analogy related to the 3x3x3x3 tesseract?

gelatinbrain wrote:
In my simulator, I 'm drawing 2 opposite cells each at the center of two horizontally aligned windows and 6 others around them. You can bring any cell to the center by arrow keys.


I was aware of that feature. Thank you for mentioning it nevertheless.

gelatinbrain wrote:
If this is not the best solution, I think it's kind of a good compromise. I think MC4D too has its own way to show the 8th hidden cell. But I don't remember now...


On MC4D applet, you can press ctrl+mouse click (for example) to "switch" central cell (it will not work if you click on central cell, you have to click on other cell).

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 10, 2014 1:56 pm 
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Isn't the 8th cell everything outside the tesseract, i.e., infinite 3D space minus a cube?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 10, 2014 5:09 pm 
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bcube wrote:
gelatinbrain wrote:
I am not fully convinced about it. Could you please explain me why it is impossible in more details?

The 8th cell is, just like the central cell, surrounded by other 6 cells, but connected by the opposite edges. It is kind of omnipresent surrounding all other cells.
That's why it is impossible to project on 3D without distorting anything,
just like you can't project the globe onto a planar map without distorting anything. This is what I meant.

But you can draw each hyper edge of 8th cell or, 6 copis of this cell each with different orientation behind 6 surrounding cells. If this is what you meant, I think it's possible.

If you are familiar with my program, you can see, with arrow keys, the 8th cell fading in from behind one of surrounding cells. It's possible to show this always. And it's visible even without transparency.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 2:05 am 
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gelatinbrain wrote:
The 8th cell is, just like the central cell, surrounded by other 6 cells, but connected by the opposite edges. It is kind of omnipresent surrounding all other cells.
That's why it is impossible to project on 3D without distorting anything

I still do not know why it is impossible. As I was saying, I think I can perfectly imagine/draw it in case of 1x1x1x1 tesseract. If it is possible to draw it for cases N>1, feel free to use distortion - it is allowed.

gelatinbrain wrote:
But you can draw each hyper edge of 8th cell or, 6 copis of this cell each with different orientation behind 6 surrounding cells. If this is what you meant, I think it's possible.

I am sorry, you lost me. As far as cell does not consist of hyper edges (I think it consists of 3D stickers only which form (some of them) hyper edge subsequently), I do not know what you could draw. As a result of typing error, I assume you wanted to write "6 copies" (if not, what did you mean by "copis"?). I think it is not what I meant - I wanted to see 8th cell only once (somehow), not 6x. Trouble is, I can not imagine how could I see 8th cell only once (but I do not see reason why should not I either (based on analogy for 3D Rubik´s cube)). The goal is to draw as few cells as possible, i.e., ideally 8. In any case, feel free to draw anything. I am looking forward to seeing it.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 2:43 am 
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See my post above. The "distortion" you talk about would require inverting and wrapping a cube around the tesseract in infinite 3D space.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:21 am 
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KelvinS wrote:
See my post above. The "distortion" you talk about would require inverting and wrapping a cube around the tesseract in infinite 3D space.


I suppose by "cube" you mean the cell (8th, hidden cell to be specific). Net of 1x1x1x1 tesseract is composed of 8 identical 3D cubes (i.e. with same proportions). I admit it is impossible to physically make this body but we are able to imagine it approximately (by changing proportions and mutual angles among those 8 cubes). So I have no problem to visualize inverted and wrapped cube (called cell) around 6 other cells (central cell is excluded). So what 3D space is infinite? Finite proportion of each cell is given... I am really trying hard to understand your statement but I am incapable of to do it. Maybe in graphical form it would hit me...

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:39 am 
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Just imagine that all the infinite space outside the tesseract is one cell. It is mathematically impossible to "distort" a finite space into an infinite one by applying any combination of finite scale factors: Finite x Finite = Finite.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:58 am 
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KelvinS wrote:
Just imagine that all the infinite space outside the tesseract is one cell.

First please tell me what one cell is related to? Am I getting it right that you are talking about 8 cells (3D cube shaped, which form tesseract) + one independent, irrelevant cell "floating around"? If so, why is 9th cell involved?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 4:22 am 
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There are 8 cells in a 4D hypercube, but only 7 cells visible in a 3D tesseract. The 8th cell of a 4D hypercube occupies all of infinite 3D space outside of the tesseract. Basically the missing cell has to be turned inside out so that all the faces of that cell touch the right faces of the 6 outer cells of the tesseract.

How would you draw an inside-out person so that all the internal organs are on the outside and yet still touch each other in all the right places? You would need infinite distortion.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 5:44 am 
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bcube wrote:
Net of 1x1x1x1 tesseract is composed of 8 identical 3D cubes (i.e. with same proportions).

Maybe you could use this to imagine a hypercube from one view. Take the net, instead of an already 4-dimensionally folded tesseract, separate the cells, and now you can see every cell and everything on them. This would require some mental folding of the net into a tesseract, but you'd be able to see everything with just 3D rotations.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 10:54 am 
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KelvinS wrote:
Isn't the 8th cell everything outside the tesseract, i.e., infinite 3D space minus a cube?


KelvinS wrote:
The 8th cell of a 4D hypercube occupies all of infinite 3D space outside of the tesseract. Basically the missing cell has to be turned inside out so that all the faces of that cell touch the right faces of the 6 outer cells of the tesseract.

How would you draw an inside-out person so that all the internal organs are on the outside and yet still touch each other in all the right places? You would need infinite distortion.


Actually no! It sure seems like it would be though and this was what I initially thought too. If you watch an animation like:

Image

If you follow the central cell you can see that it looks like it turns inside out and our normal 3D intuition says that when something turns inside out it must "explode" and go from a finite volume to an infinite volume of all of the surroundings minus the cube in the center. Actually the cube isn't turning inside out, it's undergoing a rotation in the 4th dimension which causes it to appear to be the mirror of itself in 3D. That is, the right side becomes the left side and vice-versa.

We don't have any normal intuition with 3D objects turning into their own mirror but we do have a trivial intuition of 2D objects being flipped over and therefor being mirrored.

Imagine a 2D coin. In 3D it would be "infinitely thin", and "transparent" with only one side that you can see from both sides. Flipping it over would look something like:

Code:
O -> 0 -> () -> | -> () -> 0 -> O


Where there is a moment when rotating the 2D coin in 3D makes it look like a line edge-on and then when it starts going back to a circle it has been mirrored with right being left and left being right. At no point did the coin ever take up more volume.

That's exactly what's happening when you see the central cubic cell on a rotating tesseract look like it becomes zero volume for a moment and then turn inside out.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:00 pm 
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Thanks, So then where exactly *IS* the missing 8th cell in the animated image above??? Now I'm really confused myself!

Edit: The more I think about it, the more I think I must be right. Just imagine folding up the 8-cell net of a tesseract, and the 8th cell would be inside out and extending into infinite 3D space as I described above.

Image

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Last edited by KelvinS on Wed Jun 11, 2014 3:24 pm, edited 2 times in total.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:16 pm 
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bcube wrote:
gelatinbrain wrote:
But you can draw each hyper edge of 8th cell or, 6 copis of this cell each with different orientation behind 6 surrounding cells. If this is what you meant, I think it's possible.

I am sorry, you lost me. As far as cell does not consist of hyper edges (I think it consists of 3D stickers only which form (some of them) hyper edge subsequently), I do not know what you could draw.

I meant by "hyper-edge" an entity connecting two neighboring cells. just as an edge of a cube connecting two faces .
On the 4DMC projection model, they are representated by the space between cells.
I meant by this also a facet of a cell facing to this hyper-edge. This is what I meant when I wrote "drawing edge". I'm sorry if this has caused your confusion.

bcube wrote:

As a result of typing error, I assume you wanted to write "6 copies" (if not, what did you mean by "copis"?). .

Yes, this is just a type error. :oops:


With the projection model of MC4D and mine too, the 8th cell exists at the opposite of the central cell by any 6 surrounding cells.

In another word, it is located at the extension of any of 3 axes by both direction, but turning different facets to these vectors.

Imagine a unit tesseract with cell's radius 1. If the center coordinate of the central cell is (1,0,0,0), the center coordinate of the hidden cell is (-1,0,0,0). In the picture below, the center coordinate(w,x,y,z) of the orange cell(center at left) is (1,0,0,0). The center of the olive cell(center at right) is (-1,0,0,0).
Attachment:
temp.jpg
temp.jpg [ 43.56 KiB | Viewed 1531 times ]


So logically the 8th cell could be drawn at opposite sides of the central cell by any surrounding cells. This is what I meant. And for me it looks a natural extension of the MC4D model.

Maybe you have a different idea. Although I don't get it clearly, I agree that if the distorsion is allowed , other ways of projection is possible.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:29 pm 
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KelvinS wrote:
Thanks, So then where exactly *IS* the missing 8th cell in the animated image above??? Now I'm really confused myself!


I think it's easiest to understand by analogy. Take a cube that has been projected from 3D into 2D:
Attachment:
orthocube.png
orthocube.png [ 7.26 KiB | Viewed 1528 times ]


I have colored each face of the cube a different color. You can think of the blue face in the background and the red face in the foreground. It's easy to imagine the faces are transparent so that you can see through the red face that is blocking the view of all of the others.

So in my cube drawing there is the front face and the back face, and then the 4 faces between the two that join them together. 6 faces total.

In the tesseract animation there is the near cell (the big outer cube) and the far cell (the small central cube), and then then 6 surrounding cubes that join the two together. 8 cells total.

If you wanted to view all of the faces of a Rubik's cube at once you have a problem of the front face blocking the view of the back face like so:
Attachment:
rubikscubenottrans.jpg
rubikscubenottrans.jpg [ 10.1 KiB | Viewed 1528 times ]


The hidden 8th cell on MC4D is the solution to that problem. By removing the cell closest to the camera (the outer cell blocking the view of the rest of the cells) you can see the remaining 7 cells. This is like imagining my crappy 2D cube drawing as a box where the red face is the lid to the box and it's been removed so you can see the other 5 faces of the box at the same time.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 3:34 pm 
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So using your analogy of the 3D cube projected in 2D, where the back face coincides with the large square encompassing the entire figure, the 8th cell of a tesseract coincides with the large cube encompassing the 7 cells? In that case, the real difference is that you're saying the 8th cell shares the same 6 faces, but is not inverted. In other words it's internal not external to the tesseract as displayed, right? Or wrong? Please answer clearly: Where is it? In particular, where are the 8 corners of the 8th cell? And is the inside of the 8th cell between those 8 corners, or outside of them?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 4:04 pm 
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KelvinS wrote:
So using your analogy of the 3D cube projected in 2D, where the back face coincides with the large square encompassing the entire figure, the 8th cell of a tesseract coincides with the large cube encompassing the 7 cells? In that case, the real difference is that you're saying the 8th cell shares the same 6 faces, but is not inverted. In other words it's internal not external to the tesseract as displayed, right? Or wrong?
Sounds right to me.

The only wording I'd change is that in my 2D cube analogy I'd say the red face is the front face encompassing the other 5 faces. In reality the other 5 faces aren't inside of the red face, they're behind it which means they get projected into 2D in such a way that they get drawn inside of it.

KelvinS wrote:
Please answer clearly: Where is it? In particular, where are the 8 corners of the 8th cell? And is the inside of the 8th cell between those 8 corners, or outside of them?


Take this image. I think you can count 7 cells. The center one and the 6 distorted cubes that are packed around the center one:
Attachment:
Schlegel_wireframe_8-cell_all.png
Schlegel_wireframe_8-cell_all.png [ 307.15 KiB | Viewed 1513 times ]


Here is the 8th cell with the other 7 faded out:
Attachment:
Schlegel_wireframe_8-cell_highlight.png
Schlegel_wireframe_8-cell_highlight.png [ 229.84 KiB | Viewed 1513 times ]


The analogy is the same with the cube projected into 2D. The 5 faces of that cube were behind the red face so they got drawn inside of the red face. They're behind it in 3D which is why they're inside of it in 2D. In the tesseract case, the 7 cells are all behind the outer cell shown above so they get drawn inside of it. They're behind in the 4th dimension though which is why they're fully inside of it in 3D.


For what it's worth, you can look at a cube corner-on like so and see three faces. The remaining 3 back faces can be exploded out like so:
Attachment:
rubiks_corner_explode.gif
rubiks_corner_explode.gif [ 5.78 KiB | Viewed 1513 times ]

Viewing the cube like this it's easier to see that the faces are spacially separated from each other and no face is actually inside of another face.

Here is the analogy to that for the tesseract:
Attachment:
mc4d_corner_view_explode.png
mc4d_corner_view_explode.png [ 48.68 KiB | Viewed 1513 times ]

Viewing the tesseract like this makes it easier to see each cell is spacially separated from each other and none is actually inside of another cell.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 4:15 pm 
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Thanks for confirming where the 8 corners of the 8th cell are, that really helps, and I think we both agree here. :)

But where I think we (potentially still) disagree, is whether the internal "body" of the 8th cell lies between those 8 corners and within the tesseract (as you say), or outside the tesseract (as I still tend to think). I think this purely by looking at what faces of the 8 cells touch each other when you fold up the net of a tesseract, which implies an stereoscopic inversion.

See the image here under "Unfolding of a tesseract" and consider what would happen to the top cube when you fold up the structure so that the right faces touch each other:

http://www.herenow4u.net/index.php?id=cd6129

Edit: Or are you saying that the 8th cell *should* be stereoscopically inverted because we are effectively looking at it through/from the "inside" of the tesseract, just like we see the reverse side of the back face of a 3D cube through/from the inside of the cube? That would make sense! :D

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 5:57 pm 
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KelvinS wrote:
But where I think we (potentially still) disagree, is whether the internal "body" of the 8th cell lies between those 8 corners and within the tesseract (as you say), or outside the tesseract (as I still tend to think).
[...]
Or are you saying that the 8th cell *should* be stereoscopically inverted because we are effectively looking at it through/from the "inside" of the tesseract, just like we see the reverse side of the back face of a 3D cube through/from the inside of the cube? That would make sense! :D


Yeah I think you get it now but allow me to give my perspective on it :D .

First check out this cube where the 3 on the back face is written backwards:
Attachment:
cube_inverted_face.png
cube_inverted_face.png [ 9.4 KiB | Viewed 1474 times ]

You can imagine that as this cube is rotating, when the 3 is in the front it looks like a "3" and when it goes behind we look through the cube and see it written backwards.

If you thought purely in 2D instead of 3D this behavior would be very strange. As the cube is rotating you'd see a square shrink until it was an infinitely thin line and then it would start expanding again but it would expand as the mirror of itself. It would then shrink back to a line and then expand back to the original version of itself.

Check out this animation and when you look at it, try very hard not to think of it as a cube spinning but just look at one square as it shrinks and expands (you may need to click on the image to view the animation):
Attachment:
rotating_cube.gif
rotating_cube.gif [ 745.95 KiB | Viewed 1474 times ]


If your brain only had a 2D understanding of things, seeing the square shrink to a line and then expand as the mirror of itself would be very strange. Because we understand 3D though we understand that really what is happening is the 2D face is rotating in a 3rd dimension and it doesn't actually change size and seeing the mirror version of it can easily be understood as us seeing the square from the other side as it flips around. The area doesn't change and the area between the 4 corners stays between them. The odd thing is that the corners on the right swap places with the corners on the left.

That's exactly what is happening when you see a tesseract rotating in 3D:
Image

The outer cube shrinks into a flat plane and then start expanding again as the mirror of itself. Its volume doesn't change though and the volume stays between its 8 corners. It's just that the corners on the right swap with the corners on the left. The volume stays between them though.

Also, the only reason the cell (cube) in the center of the tesseract looks smaller than the outer one is that it's farther away. Just like in my colorful 2D cube from a few posts back the blue square in the middle is smaller than the outer red square because the blue square is farther away.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 6:31 pm 
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Wow, what a fantastic explanation, thanks Brandon! :D

So now I guess that answers why it is not easy to show the contents of the 8th cell on a tesseract, because it would obfuscate the contents of the other 7 cells, just like the front side of a cube gets in the way and stops us from seeing the back side - unless of course the front side is partially transparent.

So I guess you would have to do the same with the tesseract and show the 8th cell as partially transparent in order to see the other 7 cells "behind" it in the 4th dimension?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 6:35 pm 
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KelvinS wrote:
So I guess you would have to do the same with the tesseract and show the 8th cell as partially transparent in order to see the other 7 cells "behind" it in the 4th dimension?

You could do that but the cell of a 3^4 is itself like a Rubik's cube with 27 cubbie "stickers" so that would make for a lot of noise to overlay on top of everything else. Also, even if you could show the 8th cell in a way that didn't get in the way of somebody viewing and understanding the other 7, it would either get in the way of the mouse trying to click the others or wouldn't be able to be manipulated by the mouse at all. Neither one of those options is very practical.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 6:37 pm 
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Ah well I have a simple solution for that: You just need a 4D mouse! :lol:

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 6:48 pm 
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KelvinS wrote:
Ah well I have a simple solution for that: You just need a 4D mouse! :lol:


From the future 4D user interface helpdesk;
Help, somehow I've accidentally inverted my mouse cursor while doing 4D rotations and now it clicks the backside of buttons. I want to be able to press the buttons into the screen but instead my cursor now presses them out of the screen and it even looks backwards!
Attachment:
mouse_cursor_mirror.png
mouse_cursor_mirror.png [ 6.78 KiB | Viewed 1449 times ]


No doubt the solution will involve the poor user looking at their screen in a mirror and retracing their 4D rotation steps :lol:

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 6:53 pm 
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I have a solution for that too: Have you tried turning your computer off and back on again? :lol:

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Wed Jun 11, 2014 9:57 pm 
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Brandon Enright wrote:
KelvinS wrote:
Ah well I have a simple solution for that: You just need a 4D mouse! :lol:


From the future 4D user interface helpdesk;
Help, somehow I've accidentally inverted my mouse cursor while doing 4D rotations and now it clicks the backside of buttons. I want to be able to press the buttons into the screen but instead my cursor now presses them out of the screen and it even looks backwards!
Attachment:
mouse_cursor_mirror.png


No doubt the solution will involve the poor user looking at their screen in a mirror and retracing their 4D rotation steps :lol:


This reminds me of the short story "Technical Error" by Arthur C. Clarke. It's about a man that is accidentally rotated through the 4th dimension while working on a superconducter. It's a great story, but I can't find any PDFs of it online.

http://en.wikipedia.org/wiki/Technical_Error

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Thu Jun 12, 2014 11:19 am 
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Brandon Enright wrote:
Here is the analogy to that for the tesseract:
Attachment:
mc4d_corner_view_explode.png
mc4d_corner_view_explode.png [ 48.68 KiB | Viewed 1197 times ]

Viewing the tesseract like this makes it easier to see each cell is spacially separated from each other and none is actually inside of another cell.
I've always struggled with higher dimentions and I'm finding this thread very educational. Thanks for the very clear explainations. Some questions about the above image:

(1) I can easily see all the cells are cubes except for the purple cell in the back. I assume three faces of it match up with the green, yellow, and medium blue cells which touch the cyan cell but my brain cant put a cube in that position. That corner that touches the green, yellow, and medium blue cells looks like it must be concave.

(2) The notion of in front of and behind in 4D, I also struggle with. If we view the big cubes as closer to us and the smaller ones as father away why are small cubes drawn in front of many of the larger ones? This is easily seen when looking at the green, yellow, and medium blue cells.

Carl

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Thu Jun 12, 2014 11:49 am 
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wwwmwww wrote:
(1) I can easily see all the cells are cubes except for the purple cell in the back. I assume three faces of it match up with the green, yellow, and medium blue cells which touch the cyan cell but my brain cant put a cube in that position. That corner that touches the green, yellow, and medium blue cells looks like it must be concave.

(2) The notion of in front of and behind in 4D, I also struggle with. If we view the big cubes as closer to us and the smaller ones as father away why are small cubes drawn in front of many of the larger ones? This is easily seen when looking at the green, yellow, and medium blue cells.


bcube contacted me by PM more than a week ago while I was very busy asking for an explanation. I threw together a video https://www.youtube.com/watch?v=ZF7nu2fN0f8 showing how you can view all 8 cells at once like my screenshot above. Unfortunately I did this right before bed and I didn't think to test that my microphone was setup properly so there is audio but it's extremely quiet. You have to turn your volume way up to hear anything. Even if you can't hear it, you can at least see some 4D rotations in action.

I'm finding it very hard to explain exactly how far away each cell is from the camera in each dimension. This page describes it well http://steve.hollasch.net/thesis/chapter4.html but having a math understanding of it doesn't exactly make for an intuitive understanding of it.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Thu Jun 12, 2014 2:54 pm 
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Warning: this is going to be very confusing

gelatinbrain wrote:
bcube wrote:
gelatinbrain wrote:
But you can draw each hyper edge of 8th cell or, 6 copis of this cell each with different orientation behind 6 surrounding cells. If this is what you meant, I think it's possible.

I am sorry, you lost me. As far as cell does not consist of hyper edges (I think it consists of 3D stickers only which form (some of them) hyper edge subsequently), I do not know what you could draw.

I meant by "hyper-edge" an entity connecting two neighboring cells. just as an edge of a cube connecting two faces .


Sadly, if anything, I am more confused than I was before. Two neighboring cells are connected by "hyper center" (see purple and green center from 3D point of view.)

Attachment:
The attachment hyper center.jpg is no longer available


"Hyper edge" connect three cells (see purple, yellow and green edge from 3D point of view).

Attachment:
The attachment hyper edge.jpg is no longer available



gelatinbrain wrote:
On the 4DMC projection model, they are representated by the space between cells.
I meant by this also a facet of a cell facing to this hyper-edge.


I am totally lost here.

gelatinbrain wrote:
This is what I meant when I wrote "drawing edge". I'm sorry if this has caused your confusion.


There is no reason to be sorry. Am I missing something or you did not write "drawing edge" before?


gelatinbrain wrote:
In another word, it is located at the extension of any of 3 axes by both direction, but turning different facets to these vectors.


I believe I know what are you trying to say, I just do not get it by your words.

gelatinbrain wrote:
Imagine a unit tesseract with cell's radius 1. If the center coordinate of the central cell is (1,0,0,0), the center coordinate of the hidden cell is (-1,0,0,0).


Agreed.

gelatinbrain wrote:
In the picture below, the center coordinate(w,x,y,z) of the orange cell(center at left) is (1,0,0,0). The center of the olive cell(center at right) is (-1,0,0,0).
Attachment:
The attachment temp.jpg is no longer available


On that picture I see red cell with coordinates 1,0,0,0 and I do not see colour of opposite cell with coordinates -1,0,0,0.

gelatinbrain wrote:
So logically the 8th cell could be drawn at opposite sides of the central cell by any surrounding cells.


Again, I think I know what you are trying to say but again, I do not understand it by your words.

gelatinbrain wrote:
This is what I meant. And for me it looks a natural extension of the MC4D model.

Maybe you have a different idea.


The best I can come up with at this moment is probably to draw 7 cells and no 8th cell. Tricky part is to draw a colour of that no shown 8th cell into those 7 cells. We are able to draw more colours on one sticker in case of 3D Rubik´s cube. When we do so, supercube will appear.

Attachment:
The attachment supercube.jpg is no longer available


Do not ask me how 3x3x3x3 supercube would look like. I am not even sure if that supercube idea is feasible... If it is, I still would not be satisfied, because I wanted to draw it otherwise in my original post.


rayray_2561 wrote:
bcube wrote:
Net of 1x1x1x1 tesseract is composed of 8 identical 3D cubes (i.e. with same proportions).

Maybe you could use this to imagine a hypercube from one view.


I have thought of it before. I suppose it is technically possible. Maybe it is a big step to visualize 8th cell from MC4D/gelatinbrain´s applet point of view for me. Could any of you guys draw rotating 3x3x3x3 tesseract first? Just similarly to the one Brandon posted (but instead of 1x1x1x1 3x3x3x3 this time).

Brandon Enright wrote:
In the tesseract case, the 7 cells are all behind the outer cell shown above so they get drawn inside of it. They're behind in the 4th dimension though which is why they're fully inside of it in 3D.


Say what? In the past you said about 4D hypercube: "Our brains can almost but not quite understand it[...]". I wish you were wrong but now I see you were totally right!

Brandon Enright wrote:
Attachment:
The attachment supercube.jpg is no longer available

Viewing the tesseract like this makes it easier to see each cell is spacially separated from each other and none is actually inside of another cell.


Thank you for this projection. As I was saying via PM - if only it would be possible to draw it in "classical" point of view, i.e. "cube-like" shape instead of "tetrahedron-like" shape. One question: How each cell can be spatially separated from each other, if they are connected by stickers (for example hyper edge connects three cells - they are not spatially separeted, are they?)?

KelvinS wrote:
So I guess you would have to do the same with the tesseract and show the 8th cell as partially transparent in order to see the other 7 cells "behind" it in the 4th dimension?


My original point/query exactly.

Brandon Enright wrote:
You could do that but the cell of a 3^4 is itself like a Rubik's cube with 27 cubbie "stickers" so that would make for a lot of noise to overlay on top of everything else. Also, even if you could show the 8th cell in a way that didn't get in the way of somebody viewing and understanding the other 7, it would either get in the way of the mouse trying to click the others or wouldn't be able to be manipulated by the mouse at all. Neither one of those options is very practical.


Please be aware I do not target on usability/playable applet, i.e. practical things.


Brandon Enright wrote:
bcube contacted me by PM more than a week ago while I was very busy asking for an explanation.


I contacted you more than 3 weeks ago, so... technically, you are still right ;-)


Attachments:
supercube.jpg
supercube.jpg [ 15.38 KiB | Viewed 1159 times ]
hyper center.jpg
hyper center.jpg [ 81.31 KiB | Viewed 1159 times ]
hyper edge.jpg
hyper edge.jpg [ 81.73 KiB | Viewed 1159 times ]

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Thu Jun 12, 2014 3:50 pm 
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bcube wrote:
Warning: this is going to be very confusing
Sadly, if anything, I am more confused than I was before. Two neighboring cells are connected by "hyper center" (see purple and green center from 3D point of view.)
.

edge -> face :oops:

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 8:10 am 
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bcube wrote:
rayray_2561 wrote:
bcube wrote:
Net of 1x1x1x1 tesseract is composed of 8 identical 3D cubes (i.e. with same proportions).

Maybe you could use this to imagine a hypercube from one view.


I have thought of it before. I suppose it is technically possible. Maybe it is a big step to visualize 8th cell from MC4D/gelatinbrain´s applet point of view for me. Could any of you guys draw rotating 3x3x3x3 tesseract first? Just similarly to the one Brandon posted (but instead of 1x1x1x1 3x3x3x3 this time).

I definitely could not draw that, but look at some of the animations in this video, because those might help. The stuff about the rotating through planes is kind of confusing, but the unwrapping was really helpful for me to understand the tesseract. Also, I was able to model what one view in 3d-space might look like with the net view.
Gelatinbrain's "hyper-edge" means that dimension of said thing is 2 less than dimension of overall thing, i.e. line=1 dimensional, cube=3 dimensional, line is cube's edge. For tesseract=4 dimensional, hyper-edge=4-2=2 dimensional, which is a square. Hyper-face is then 1 dimension less, and hyper-vertex is 3 less.
Attachment:
File comment: No hyper-edge selected
Screen Shot 2014-06-14 at 8.48.27 AM.png
Screen Shot 2014-06-14 at 8.48.27 AM.png [ 32.46 KiB | Viewed 1046 times ]

Attachment:
File comment: One kind selected
Screen Shot 2014-06-14 at 8.48.20 AM.png
Screen Shot 2014-06-14 at 8.48.20 AM.png [ 37.86 KiB | Viewed 1046 times ]

Attachment:
File comment: Another kind of turn
Screen Shot 2014-06-14 at 8.53.42 AM.png
Screen Shot 2014-06-14 at 8.53.42 AM.png [ 38.98 KiB | Viewed 1046 times ]

Attachment:
File comment: The kind you're probably most used to turning
Screen Shot 2014-06-14 at 8.55.23 AM.png
Screen Shot 2014-06-14 at 8.55.23 AM.png [ 39.69 KiB | Viewed 1046 times ]

Attachment:
File comment: The last kind
Screen Shot 2014-06-14 at 8.56.45 AM.png
Screen Shot 2014-06-14 at 8.56.45 AM.png [ 33.91 KiB | Viewed 1046 times ]

Attachment:
File comment: A kind I forgot originally
Screen Shot 2014-06-14 at 12.43.45 PM.png
Screen Shot 2014-06-14 at 12.43.45 PM.png [ 36.42 KiB | Viewed 999 times ]

Attachment:
File comment: Another one I forgot
Screen Shot 2014-06-15 at 8.05.44 AM.png
Screen Shot 2014-06-15 at 8.05.44 AM.png [ 27.47 KiB | Viewed 901 times ]

I think I showed the different types of hyper-edges you could've selected. It wasn't all of them, but you could rotate through 3D space and select them. I think this showed what would be selected for all types of turns. I hope that helped, even though I couldn't show you the rotating 3x3x3x3.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 10:01 am 
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rayray_2561 wrote:
Maybe you could use this to imagine a hypercube from one view.


bcube wrote:
I have thought of it before. I suppose it is technically possible. Maybe it is a big step to visualize 8th cell from MC4D/gelatinbrain´s applet point of view for me. Could any of you guys draw rotating 3x3x3x3 tesseract first? Just similarly to the one Brandon posted (but instead of 1x1x1x1 3x3x3x3 this time).

rayray_2561 wrote:
I definitely could not draw that, but look at some of the animations in this video, because those might help. The stuff about the rotating through planes is kind of confusing, but the unwrapping was really helpful for me to understand the tesseract.


Suprisingly, it is more clear to imagine rotating through planes than unwrapping as seen in that video for me. Maybe the reason is that I saw following pictures first, thus I was influenced by them (I think I understood them unlike the unwrapping in the video).

Attachment:
cube net.png
cube net.png [ 5.24 KiB | Viewed 1019 times ]

Attachment:
tesseract net.png
tesseract net.png [ 152.23 KiB | Viewed 1018 times ]



rayray_2561 wrote:
Also, I was able to model what one view in 3d-space might look like with the net view.
Gelatinbrain's "hyper-edge" means that dimension of said thing is 2 less than dimension of overall thing, i.e. line=1 dimensional, cube=3 dimensional, line is cube's edge. For tesseract=4 dimensional, hyper-edge=4-2=2 dimensional, which is a square. Hyper-face is then 1 dimension less, and hyper-vertex is 3 less.


Hmmm, it´s too complicated to think about it like that for me. I think I get your point (as well as gelatinbrain´s) now, but for me it is easier to think about separated parts (i.e. 2-colored center, 3-colored edge and 4-colored corner) forming a layer (from 3D point of view). I guess that it can be called "hyper-edge" from 3D point of view but I still find that more confusing than calling only one piece an edge (just like in 3D) insted of several pieces... Anyway, those two previous posts from gelatinbrain and you helped me to understand what gelatinbrain meant as "hyper-edge" (but I will still understand as "hyper-edge" something else). I guess it is a matter of non-existing sufficient notation for 3x3x3x3 hypercube, agreed?

gelatinbrain wrote:
In the picture below, the center coordinate(w,x,y,z) of the orange cell(center at left) is (1,0,0,0). The center of the olive cell(center at right) is (-1,0,0,0).


I agree now. For some reason I was beeing stupid and overlooked right part of the picture (probably because when I play with your applet, I look only at the left side of it).

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 11:41 am 
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bcube wrote:
I guess it is a matter of non-existing sufficient notation for 3x3x3x3 hypercube, agreed?

Yeah, I guess there isn't really any good way of describing it. I think that we should stick to hyper- things when talking about the tesseract, and use cell, center, edge, and corner when talking about the puzzle. My last question is: Did my pictures help anyone look at the 3x3x3x3 in a new way? Would it be easy to program a 3x3x3x3 with this kind of view?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 2:33 pm 
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rayray_2561 wrote:
My last question is: Did my pictures help anyone look at the 3x3x3x3 in a new way? Would it be easy to program a 3x3x3x3 with this kind of view?


YES! Your pictures make a lot of sense to me. Thank you for making them. If there was a simulator with this sort of view, I feel I would have a better shot a solving it.

-d


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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 8:44 pm 
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darryl wrote:
If there was a simulator with this sort of view, I feel I would have a better shot a solving it.

I would definitely too, so I'm wondering how quick it would be to make it.

P.S. I'm looking at you guys, gelatinbrain and Boris...

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 8:50 pm 
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rayray_2561 wrote:
darryl wrote:
If there was a simulator with this sort of view, I feel I would have a better shot a solving it.

I would definitely too, so I'm wondering how quick it would be to make it.

P.S. I'm looking at you guys, gelatinbrain and Boris...

Personally I think the Gelatinbrain view of the tesseract is exceptionally good. The two views let you see all 8 cells and it's obvious how they relate to each other.

Also, did you know you can click the grey petals (or white arrows on older versions) to do 4D rotations to change which cells are at the center of the two views?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 9:04 pm 
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I did know that, but 3D rotations and having all cells viewable is much more comprehendable for me.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sat Jun 14, 2014 9:16 pm 
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rayray_2561 wrote:
I did know that, but 3D rotations and having all cells viewable is much more comprehendable for me.

If you want an unfolded view like your images above, Gelatinbrain's MagicPolyhedra is very close to that already. You can see 7 cells in one view and the only missing cell is the center of the other view. Even better, the bottommost cell on the left is the topmost cell on the right and vice-versa. That means that you can see the center cell like how you've drawn it above as on the bottom of the cross.

Your unfolded cross somewhat obscures the relationship between cells and it takes some thinking to see how the bottom cell relates to the other 6 cells it touches. With Gelatinbrain's view though that relationship is presented naturally.

My biggest complaint about how Gelatinbrain presents the 3^4 is that only 90 degree cell rotations are allowed and simulating twisting a cell about an edge or corner costs more moves. MC4D has its own problem in that there is no way to perform a 180 degree cell twist about a face without costing two moves.

I'd really like to have the movecount based on cell twists where any of the 24 possible moves (actually 23) only count as one move.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sun Jun 15, 2014 5:30 am 
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Brandon Enright wrote:

Your unfolded cross somewhat obscures the relationship between cells and it takes some thinking to see how the bottom cell relates to the other 6 cells it touches.


That´s why I suggested supercube idea. In my opinion it is not necessary to draw all 8 - you can draw only 7 cubes (without green one as seen on pictures of rayray). However, it would require to draw 2 colours (one of them is green in solved state of puzzle) for some cubies on the rest of the cubes (excluding blue one). Not to mention programming of such concept would be probably difficult, because the colors of 2-colored cubies are changed when making a move.

Personally, I am more excited about whether or not could someone draw rotating 3x3x3x3 hypercube. It doesn´t necessarily have to be animated/in .gif format. It would be totally ok if it was expressed such as following picture (ignore the letters).

Attachment:
rotating tesseract.jpg
rotating tesseract.jpg [ 142.59 KiB | Viewed 922 times ]


Because if someone is able to draw rotating 3x3x3x3 tesseract, it will be piece of cake to make transparent one of its cell. Therefore, it would be possible to draw it as initially intended.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Sun Jun 15, 2014 5:39 am 
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bcube wrote:
Because if someone is able to draw rotating 3x3x3x3 tesseract, it will be piece of cake to make transparent one of its cell. Therefore, it would be possible to draw it as initially intended.

I think you could do this very easily just by superimposing a semi-transparent 3x3x3 cube on top of the larger cell (outer 8 corners) of a tesseract projected in 3D as you have shown. BUT the brain would find it incredibly difficult to interpret, let alone solve.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 24, 2014 3:52 am 
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Alright. Just because nobody posted rotating 3x3x3x3 hypercube yet doesn´t automatically mean it it impossible to draw it. Allow me to ask you this: can you prove that it is possible (or impossible) to draw either 3x3x3x3 tesseract with one transparent cell or 3x3x3x3 rotating tesseract with regard to the conditions mentioned above?

If you are going to prove it is impossible, please implement the facts that it is possible in case of 1x1x1x1 tesseract (for both one transparent cell and rotating hypercube) into your proof.

KelvinS wrote:
I think you could do this very easily just by superimposing a semi-transparent 3x3x3 cube on top of the larger cell (outer 8 corners) of a tesseract projected in 3D as you have shown. BUT the brain would find it incredibly difficult to interpret, let alone solve.


I don´t think it matches required criteria. If you still think it does, could you please express it in graphical way?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 24, 2014 10:13 am 
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I am glad my hypercube sequence image has been of some help after so long. ;)

A real hypercube of the 3x3x3, i.e. a 3x3x3x3 can be possible, as long as:

1. All pieces should be flexible enough (be it telescopic rods, or elastic fabric, so that the issues
of the unfixed 3D length in the four dimensions is resolved).
2. During each "Inside Out" move, all faces turn inside out too, becoming mirror images of their previous selves.

Thus, the letters in the image above play a vital role.

Now, where is that algorithm in the case of super-stickering where everything is solved, except one or two
centers which are 90 degrees off?

:)


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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 24, 2014 11:15 am 
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kastellorizo wrote:
Now, where is that algorithm in the case of super-stickering where everything is solved, except one or two
centers which are 90 degrees off?

Are you talking about the traditional MC4D-style 3x3x3x3? By Center do you mean the cube in the center of a cell? If so there is a simple [4,3] commutator (actually [[1,1],[1:1]] if you break it down) that will twist two centers pure. In MC4D syntax:
Code:
MagicCube4D 2
@twist two centers 90 deg@@{4,3,3} 3.0@(g 78 25 132) 132,1,2 187,1,2 132,-1,2 187,-1,2 23,1,2 47,-1,1 23,-1,2 187,1,2 132,1,2 187,-1,2
132,-1,2 23,1,2 47,1,1 23,-1,2.

This macro syntax is just about impossible for a human to read so if you do mean those centers I could make a video demo showing how it works.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 24, 2014 7:57 pm 
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Brandon Enright wrote:
kastellorizo wrote:
Now, where is that algorithm in the case of super-stickering where everything is solved, except one or two
centers which are 90 degrees off?

Are you talking about the traditional MC4D-style 3x3x3x3? By Center do you mean the cube in the center of a cell? If so there is a simple [4,3] commutator (actually [[1,1],[1:1]] if you break it down) that will twist two centers pure. In MC4D syntax:
Code:
MagicCube4D 2
@twist two centers 90 deg@@{4,3,3} 3.0@(g 78 25 132) 132,1,2 187,1,2 132,-1,2 187,-1,2 23,1,2 47,-1,1 23,-1,2 187,1,2 132,1,2 187,-1,2
132,-1,2 23,1,2 47,1,1 23,-1,2.

This macro syntax is just about impossible for a human to read so if you do mean those centers I could make a video demo showing how it works.


Yeap, I was trying to point out that the commutator should not be hard to provide, if and only if
the 4D action is understood. And yes, by definition, we should be talking about the cell's central cuboid.

:)



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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jun 24, 2014 10:14 pm 
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kastellorizo wrote:
Yeap, I was trying to point out that the commutator should not be hard to provide, if and only if
the 4D action is understood. And yes, by definition, we should be talking about the cell's central cuboid.

Please allow me to quibble a bit :P

The centers are cubes with 24 orientations. They can only be twisted in the same natural way we're used to. Twisting them only requires a few cells and I don't think any real understanding of 4D or even the 4D relationship between the cells is needed.

I think the only pieces that really require any sort of 4D understanding are the pieces with 4 cube "stickers", one on each of 4 adjacent cells. Here is a 4D twist of such a piece. Note that this is the only piece twisted or moved on the whole puzzle:
Attachment:
corner_inverted.png
corner_inverted.png [ 22.77 KiB | Viewed 630 times ]


See that I can twist this piece back to its original orientation with a 4D rotation of the whole tesseract:
Attachment:
mc4d_4d_rotation_inversion.png
mc4d_4d_rotation_inversion.png [ 29.67 KiB | Viewed 630 times ]


If there is a common theme in all of my posts in this thread, it's that you can best understand 4D by looking at how 2D transitions to 3D and then apply that same mechanism by analogy from 3D to 4D.

To understand this 4D corner twist, first look at a Rubik's cube. On a Rubik's cube the only moves available to you are 2D rotations on different 3D axes. So if you want to twist the UFR corner, you can do an F followed by a R move. This is two 2D twists that work together to create a 3D twist.

On the 3x3x3x3 the only moves available to you are 3D rotations on different 4D axes. If you want to perform a 4D twist of a corner you need to perform two 3D that work together to achieve a 4D twist.

So I'd argue that for the 3^N Rubik's cube, you really need to understand N - 1 dimensional twists and also understand a bit about how a bunch of N - 1 dimensional objects (cubes in the case of a 3x3x3x3) are connected together. There is very little N dimensional understanding needed for a 3^N puzzle.

And before anyone says "so if you solve a 3^5 that proves you understand 4D" I'd argue that you can actually use recursion and first reduce 3^5 to 3 * 3^4 and then reduce that to 3 * 3 * 3^3. You can just keep breaking down the complex higher dimensional problems into simpler problems, solve them there, and then translate those back up into the dimensions you need to.

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Mon Jul 07, 2014 11:33 am 
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kastellorizo wrote:
Thus, the letters in the image above play a vital role.


Relations among the cells are definitely nicely displayed on your picture. However, my intention was different. I was rather aiming on "not to necessarilly use animatated / .gif picture". From this point of view, I found the letters on your picture not so helpful.


bcube wrote:
can you prove that it is possible (or impossible) to draw either 3x3x3x3 tesseract with one transparent cell or 3x3x3x3 rotating tesseract with regard to the conditions mentioned above?

If you are going to prove it is impossible, please implement the facts that it is possible in case of 1x1x1x1 tesseract (for both one transparent cell and rotating hypercube) into your proof.


If there is no-one here to answer that question, would you recommend me some places/people where should I ask?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Mon Jul 07, 2014 12:23 pm 
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bcube wrote:
can you prove that it is possible (or impossible) to draw either 3x3x3x3 tesseract with one transparent cell or 3x3x3x3 rotating tesseract with regard to the conditions mentioned above?

If you are going to prove it is impossible, please implement the facts that it is possible in case of 1x1x1x1 tesseract (for both one transparent cell and rotating hypercube) into your proof.

You can definitely draw the 8th cell transparently so that all are visible. It would look something like:
Attachment:
mc4d_fake_8th_cell.png
mc4d_fake_8th_cell.png [ 78.87 KiB | Viewed 314 times ]

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jul 15, 2014 12:40 pm 
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I am sorry Brandon, I don´t see relations among the pieces on your picture. On the other hand, I don´t see a way how to visualize it better.


All in all, I appreciate your intention to help (especially Brandon, rayray_2561 but also others). It´s probably not your fault (it´s my unreasonably high demands) that I did not find what I was looking for yet. I will think your thoughts as well as mine over for a several weeks/months and if it is still unclear after such a time, I will be begging for an explanation.


One last question I bear in mind for now: why exactly do you think no one posted rotating 3x3x3x3 tesseract (considering rotating 1x1x1x1 hypercube is possible to visualize)?

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jul 15, 2014 2:49 pm 
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I've really enjoyed this thread. I have a question but to ask it properly I need to verify a few things first.

Lets start with a 2D Rubik's Cube.

Shape: Square
Faces: Lines (4 of them)
Stickers: Lines (3 on each face)
Turns available: 4 A rotation about the axis through the center of each face.

The 3D Rubik's Cube.

Shape: Cube
Faces: Squares (6 of them)
Stickers: Squares (9 on each face)
Turns available: 6 A rotation about the axis through the center of each face.

The 4D Rubik's Cube

Shape: Hypercube
Faces: Cubes (8 of them)
Stickers: Cubes (27 on each face)
Turns available: 24? It appears each face can be rotated on 3 axes and there are 8 faces so that is where I get this number.

So I'm confused now by the 2D Rubik's cube. Do I understand the turns available correctly? It appears that a rotation requires a face to go out of the plane so maybe it is more a mirroring operation than a rotation. And I'm not sure I'm correct about the 24 turns available on the 4D Rubik's cube. Having multiple axes of rotation through a single face doesn't seem like an expected generalization from 3D to 4D? There wasn't an increase from 2D to 3D. How many turns are available on a 5D Rubik's Cube. If each face becomes a Hypercube and the trend appears to say there would be 10 faces how many turns are possible? Guessing it may be 240. 24 axes for each Hypercube and 10 faces total. Is that correct?

Carl

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 Post subject: Re: visualization of 8th cell for 4D Rubik's cube
PostPosted: Tue Jul 15, 2014 3:00 pm 
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Carl, I think the difference is that from 2D to 3D, you moved the turning axis from the center of a line(2 vertices) to a face(>2 vertices), whereas from 3D to 4D you kept the turning axis on the same thing(face). For a true continuation of this pattern you would have a puzzle with 8 axes of rotation, one appearing the center of each cube. The actual axis would be the line passing through the center of the hypercube and the center of each cubic cell. This, however, is impossible to project to 3D (I think...) because we can't tell where the center of the hypercube is.

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