Optimal Tesseract Slice

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Optimal Tesseract Slice

Postby gonegahgah » Fri Jul 28, 2017 2:59 am

Yesterday I was thinking about what a 3D slice of a Rubik's® tesseract would look like...

I realised that we tend to look at a cube so that we can see three faces and that this changes the dead centre view for a 2Der...
When a 2Der looks at a square they would probably hold it so that they can see at least 2 faces (that we call edges).

So it occured to me to think about what is the optimal viewing angle for a 4Der as they would hold the tesseract so that they can see 4 faces (that we call cubes) at once.

I drew up the following as a result:
Image

This shows 2Der's optimal view of a square and a 3Der's optimal view of a cube.
It also shows a 2Der's slice of the 3Der's optimal view straight on.
It also hopefully conveys properly a 3Der's slice of a 4Der's optimal view of a tesseract straight on?

Initialy for the later I drew only the first image but I realised that the 4Der would have 360° of optimal view compared to our single primary optimal view.
So I'm guessing that our 3D slice of the 4Der's optimal view from straight on would be as shown; or anywhere between those as well for a full 360°?

The grey areas are the edges of each shape as experienced by the dimensional being to whom the object belongs.
Are my depictions correct? Or have I assumed something incorrectly?
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Re: Optimal Tesseract Slice

Postby gonegahgah » Fri Jul 28, 2017 7:44 am

Or is this more correct?
Image

I realised that there can only be one plane of any colour per any 3D slice...
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Re: Optimal Tesseract Slice

Postby gonegahgah » Fri Jul 28, 2017 9:01 am

I think I'm getting closer?
Image

I realised that the 2Der square has a green line at top, the 3Der cube has a green square at top, so the 4Der tesseract must have a green cube at top.
The 2Der then sees a green line at top when looking directly at the optimal 3D view of a cube.
The 3Der should then see a green square (squashed to diamond) at top when looking directly at the optimal 4D view of a tesseract.
The 4D optimal view can still be any of 360° of orientations where we are concerned...
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Re: Optimal Tesseract Slice

Postby Mercurial, the Spectre » Fri Jul 28, 2017 1:54 pm

You don't have to look at one of the cells. If you want 4 cubes visible, look at the tetrahedral-symmetric projections of the tesseract. Evidently, the vertex-first section provides the best possible orientation.

To a 3D person, one would have a rhombic dodecahedron hull. Four of the cubes would be visible (by symmetry, representing the vertices of the tetrahedron). Cross sections would start from a tetrahedron, which morphs into a truncated tetrahedron. If you continue the truncation so that they reach the midpoints you get the octahedron, its maximal slice. Then it reverses from that figure, back into a tetrahedron, albeit inverted.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Mon Jul 31, 2017 5:18 am

Hi Mercurial, bit complex for my understanding just yet...

By "vertex first" I did notice that that was the case for 2D and 3D.
In 4D I'm wondering if there are 3 vertex that can always be forward-most to a 3D observer while keeping a visually centered apex?
And I wonder if all 3 vertex appear forward-most to the 4D observer when they have a visually centered apex?

If that is the case the perhaps the following is getting closer to correct?
Image
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Re: Optimal Tesseract Slice

Postby Klitzing » Mon Jul 31, 2017 1:36 pm

Let
Code: Select all
x4o = Dynkin symbol of square
x4o3o = Dynkin symbol of cube
x4o3o3o = Dynkin symbol of tesseract
x4o3o3o3o = Dynkin symbol of penteract
...

where x denotes an edge length of one unit.

Let q be an according edge length of sqrt(2), then the vertex first representation of those, in layerwise orthogonally sectioning subpolytopes would be
Code: Select all
x4o = o || q || o 
      (point, (diametral) line segment, (opposite) point)
x4o3o = o3o || q3o || o3q || o3o
        (point, triangle, triangle in dual orientation, point)
x4o3o3o = o3o3o || q3o3o || o3q3o || o3o3q || o3o3o
          (point, tetrahedron, octahedron, dual tetrahedron, point)
x4o3o3o3o = o3o3o3o || q3o3o3o || o3q3o3o || o3o3q3o || o3o3o3q || o3o3o3o
            (point, pentachoron, rectified pentachoron, rectified pentachoron in inverted orientation, dual pentachoron)
...


You even could provide the representations of the sectioning heights between those vertex layers by Interpolation as e.g. o3t3r3o3o, where t represents the current square sectioning length (i.e. positive but smaller than q) and r = q-t is the other square sectioning length wrt. the next layer.

--- rk
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Re: Optimal Tesseract Slice

Postby ICN5D » Sat Aug 05, 2017 6:23 pm

I've been thinking about this one a bit. I just noticed a general pattern with how many vertices an observer can see at once, when looking at n-cubes from vertex-first.

• We have an n-dimensional observer looking at n-cubes.

• The eyes of the observer collect n-1 dimensional scans of their n-D world.

• We are only looking at vertex-first angles of n-cubes


Considering these rules, an n-D observer can see all but one vertices at once. No matter how many vertices an n-cube has, there will always be just one, lonely vertex on the far side, obscured by the rest of the surface. There will always be one vertex in the center, surrounded by an array of vertices that share a common n-1D circumsphere.



The pattern goes like this:

• A 2D observer looking at a 1D scan of a square (has 4 vertices) corner first, will see a single vertex at the center, surrounded by a 0-sphere array of 2 more vertices, with the last one occluded.

• A 3D observer looking at a 2D scan of a cube (has 8 vertices) corner first, will see a single vertex at the center, surrounded by a 1-sphere array of 6 more vertices, with the last one occluded.

• A 4D observer looking at a 3D scan of a tesseract (has 16 vertices) corner first, will see a single vertex at the center, surrounded by a 2-sphere array of 14 more vertices, with the last one occluded.

• A 5D observer looking at a 4D scan of a penteract (has 32 vertices) corner first, will see a single vertex at the center, surrounded by a 3-sphere array of 30 more vertices, with the last one occluded.

• A 6D observer looking at a 5D scan of a hexeract (has 64 vertices) corner first, will see a single vertex at the center, surrounded by a 4-sphere array of 62 more vertices, with the last one occluded.

In general:

• A nD observer looking at a n-1D scan of an n-cube (has 2n vertices) corner first, will see a single vertex at the center, surrounded by an (n-2)-sphere array of (2n-2) more vertices, with the last one occluded.
in search of combinatoric objects of finite extent
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Re: Optimal Tesseract Slice

Postby gonegahgah » Thu Aug 10, 2017 5:39 am

Thanks guys for looking at this for me.
Unfortunately my simple idea of extending to each successive dimension from the last dimension to work out the optimal view for each being didn't work out; not the way I was trying to do it. I ended up with the following:
Image

I was hoping the 4D cube face would extend nicely from the 3D square face (which had extended nicely from the 2D line face).
Unfortunately it is not like that mathematically and I can't get the cube to be an extension of the square face; no matter how I rotate it.
A centered diagonal can be put in place for the cube face. However, the rotation of the cube face can not be made to match the square area.

I'll study what you guys have written to see if it helps me.
I''m working out the formulas so I can program this so that might also give me the answer...
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Re: Optimal Tesseract Slice

Postby gonegahgah » Thu Aug 10, 2017 11:43 am

ICN5D wrote:• A 2D observer looking at a 1D scan of a square (has 4 vertices) corner first, will see a single vertex at the center, surrounded by a 0-sphere array of 2 more vertices, with the last one occluded.
• A 3D observer looking at a 2D scan of a cube (has 8 vertices) corner first, will see a single vertex at the center, surrounded by a 1-sphere array of 6 more vertices, with the last one occluded.
• A 4D observer looking at a 3D scan of a tesseract (has 16 vertices) corner first, will see a single vertex at the center, surrounded by a 2-sphere array of 14 more vertices, with the last one occluded.

Ah interesting, thank you ICN5D :)
What do you mean by the last one being occluded?

Sorry I get it now! It means, "Plus a final one that is at the back". I read it as "the last of the shown number of array of vertices being occluded" which was wrongly read.
So originally I thought you meant the 2nd of the 2 for the 2D square, the 6th of the 6 for the 3D cube, and the 14th of the 14 for the 4D tesseract.
But you actually meant the 4th and final vertex of the 2D square, the 8th and final vertex of the 3D cube, etc.
Now I understand. Nicely done ICN5D, thanks :) That should make it easier to picture. Certainly makes it easier to work out how to line it up!
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