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 combinatorial 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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Re: Optimal Tesseract Slice

Postby gonegahgah » Sat Sep 02, 2017 7:56 am

I've been working on algorithms to produce a uniquely rotated cube. As I mentioned this needs 3 angles to generate all unique varieties.
I'm wondering what the minimum number of angles are required to have all the unique positions of a tesseract?

I realise that for a square in 2D space we have only:
- 4 faces with 4 major aspects that can only be in 360° (circle) of unique directions requiring just 1 angle.
and that for a cube in 3D space we have:
- 6 faces with 6 major aspects that can be a sphere of unique directions requiring 2 angles, and
- a square face that can rotate around itself requiring 1 angle, so 3 in total.
so for a tesseract in 4D space do we have:
- 8 faces with 8 major aspects that can be a hypersphere of unique directions requiring 3 angles, and
- a cube face that can be in a sphere of unique directions requiring 2 angles, and
- a square on the cube that can rotate around itself requiring 1 angle. so 6 in total...

So is that accurate?
Will the minimum number of angles I'll need to use, to represent all unique orientations of a tesseract, be 6 angles?
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Re: Optimal Tesseract Slice

Postby Catisfluffy » Mon Sep 04, 2017 2:15 am

Well, there are 3 angles needed to get 1 vertex to the right position, and another 3 to get the rest in position. It's like 2 angles to get a cube vertex in the right position, and 1 angle to orient the rest.
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Re: Optimal Tesseract Slice

Postby ICN5D » Mon Sep 04, 2017 2:33 pm

If it's making slices of n-cubes that you want, then, you only need two angles for the cube. Yes, there are 3 unique ways to rotate in 3D, but you only need 2, to make all unique slices. Using just two angles, you can take slices square, line, and vertex first.

If you slice a cube with plane xy , then you only need to rotate on the xz and yz planes. The xy rotation will just spin the slice in the xy plane, and provide nothing unique.


As for a tesseract, there are 6 ways to rotate in 4D. But, in practice you only need 3 angles, to take slices of it at cube, square, line, and vertex first.

If you 3D slice a tesseract with the xyz plane, then you only need to rotate on the xw, yw, and zw planes. The other 3 rotations available are xy, xz, and yz, which will only spin the slice in our slicing 3-plane, and show us nothing unique.


Note however, this applies to projections as well!


The general pattern:

1. Take the total number of ways to rotate in nD space, which is the number of 2D planes available, 2 choose n.

2. Subtract the 2D planes of rotation out of the slicing space, to get unique angles for slice/projection.



So, if we extended this to 4D slices of a 5D cube:

* 5D space has 10 unique ways to rotate, since there are 10 distinct 2D planes.

* A 4D slice/projection has 6 unique rotation planes in it, which we don't need.

* (10 ways to rotate in 5D) - (6 ways to rotate in 4D slice) = 4 unique angles to slice/project
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Re: Optimal Tesseract Slice

Postby Hugh » Tue Sep 05, 2017 3:01 pm

ICN5D wrote:If you 3D slice a tesseract with the xyz plane, then you only need to rotate on the xw, yw, and zw planes. The other 3 rotations available are xy, xz, and yz, which will only spin the slice in our slicing 3-plane, and show us nothing unique.


Very interesting discussion!

I've quoted this in the VRI thread, ICN5D.

A re-orientation of our viewpoint slicing 3-plane can be looking at the same 4D area, from different perceived directions, yet showing nothing unique within each view.

I don't mean to highjack this thread, so any thoughts about this idea would be much appreciated in the VRI one. Thanks. :)

http://hi.gher.space/forum/viewtopic.php?p=25620#p25620
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Re: Optimal Tesseract Slice

Postby gonegahgah » Sun Sep 10, 2017 4:40 am

Hi ICN5D.
I'm looking to present the whole 4D shape on a 2D screen in a new fashion - by rotated projection.
So I will need to track all the orientations. I'm finding I can use 3 rotations to do that for a cube and I'm guessing I'll find that 6 will do for a tesseract when I look to extend up.
I'm working on the algorithms at the moment to produce the rotations.
Using normal axial rotation approaches creates overlapping results which is probably workable.
However, I'm working more with rotations that will provide unique orientations for every combination of three angles.

Hi Catisfluffy.
That's mostly it.
Though its worth noting that 4th and 5th rotation orientate the cube 'face' at their current outwards direction.
The 6th and final rotation orientates two of the cube's faces on that cube.
Essentially that is the same as what you said.
It just adds recognition to the lower dimensional faces and the rotations they add. ie square rotations, + cube rotations , + hypercube rotations.

Hi Hugh. Nice to hear from you again.
Hopefully one day when I have my projection method developed you can play with it and get a greater sense of full 4D space.
I think of VRI as being 3.ȯ4D (that's a dotted 0). That is it could be thought of as only 4 directions of an infinite hypersphere of directions provided by 4D over 3D.
So that's a very tiny subset of 4D, but it must be quite disorientating none-the-less...
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Re: Optimal Tesseract Slice

Postby ICN5D » Sun Sep 10, 2017 4:41 pm

Hey Hugh! I don't have any more insights quite yet. I've been prepping for this hurricane. We'll be getting the strongest winds at 2am EDT , since we'll be on the east side of the eye wall. Worst place to be, for a north-bound storm.



gonegahgah wrote:I'm looking to present the whole 4D shape on a 2D screen in a new fashion - by rotated projection.
So I will need to track all the orientations. I'm finding I can use 3 rotations to do that for a cube and I'm guessing I'll find that 6 will do for a tesseract when I look to extend up.
I'm working on the algorithms at the moment to produce the rotations.
Using normal axial rotation approaches creates overlapping results which is probably workable.
However, I'm working more with rotations that will provide unique orientations for every combination of three angles.


This is what I'm planning to do, as well. I don't know how you intend on defining the surfaces, but I'm using parametric equations for the 1D and 2D elements. Like I said, all you need is 3 rotate parameters : xw, yw, and zw. And, you want to combine them into a perspective projection:

3 Rotation Parameters:


• XW
--------
x = (X)*cos(a) - (W)*sin(a)
y = Y
z = Z
w = (X)*sin(a) + (W)*cos(a)

• YW
--------
x = X
y = (Y)*cos(b) - (W)*sin(b)
z = Z
w = (Y)*sin(b) + (W)*cos(b)

• ZW
--------
x = X
y = Y
z = (Z)*cos(c) - (W)*sin(c)
w = (Z)*sin(c) + (W)*cos(c)

use angles a,b,c to rotate



• Now, depending on what program you use, you may need to combine these as one single function:

x =(X)*cos(a) - ((X)*sin(a) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(a))*sin(a)
y = (Y)*cos(b) - ((X)*sin(a) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(a))*sin(b)
z = (Z)*cos(c) - ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*sin(c)
w = (Z)*sin(c) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(c)



• Then, do a perspective projection onto plane xyz (flattening the w-axis, which is important for this specific rotation) , with camera distance 'd' , which you may use d = 5 or 6 , if using edge length 2 for your tesseract :

x = (x)/(w+d)
y = (y)/(w+d)
z = (z)/(w+d)



• Which comes out in a combined equation (it's nasty) :

x = ((X)*cos(a) - ((X)*sin(a) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(a))*sin(a))/((Z)*sin(c) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(c)+d)

y = ((Y)*cos(b) - ((X)*sin(a) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(a))*sin(b))/((Z)*sin(c) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(c)+d)

z = ((Z)*cos(c) - ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*sin(c))/((Z)*sin(c) + ((Y)*sin(b) + ((X)*sin(a) + (W)*cos(a))*cos(b))*cos(c)+d)



You will have to implement the above function with every single one of the elements you intend on projecting. If you are making a tesseract with 1D and 2D elements, using edge length of 2, that's:

• 32 line segments ( 4 ways to align a group of 8 lines) :

r(x,y,z,w) = { t , ±2 , ±2 , ±2 }
r(x,y,z,w) = { ±2 , t , ±2 , ±2 }
r(x,y,z,w) = { ±2 , ±2 , t , ±2 }
r(x,y,z,w) = { ±2 , ±2 , ±2 , t }

-1 < t < 1


• 24 squares ( 6 ways to align a group of 4 squares ) :

r(x,y,z,w) = { u , v , ±2 , ±2 }
r(x,y,z,w) = { u , ±2 , v , ±2 }
r(x,y,z,w) = { u , ±2 , ±2 , v }
r(x,y,z,w) = { ±2 , u , v , ±2 }
r(x,y,z,w) = { ±2 , u , ±2 , v }
r(x,y,z,w) = { ±2 , ±2 , u , v }

-1 < u,v < 1


These are the parametrized pieces I'm going to use, but not with a tripe rotate function. I'm doing a single rotation, but with adjustable edge lengths, so I can make rotating rectangular tesseracts. Hyperrectangles!
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Re: Optimal Tesseract Slice

Postby gonegahgah » Mon Sep 11, 2017 12:26 pm

Hi ICN5D, keep well with that weather!

It turns out after all of my mental anguishing that the process for a cube (still in 3D at the moment) is much shorter than I realised.
Considering an environment from front to back we need to:
1. Rotate the cube centrally about the z-axis (the one pointing towards us) - spin angle.
2. Rotate this centrally around the x-axis (the one pointing rightwards) towards the back - back angle.
3. Rotate this centrally around the z-axis (again the one pointing towards us) - side angle.
Surprisingly this generates all the unique orientations except for the 0 and 180 back angles. Those two might need extra handling.

This does parallel the dimensions somewhat. Spin angle is closely resembles the rotation of a square in 2D.
The back angle and side angle then uniquely locate these in 3D.
That does give me hope that these will likewise be the starting process for locating things in 4D and then adding 3 other actions on top...

I'm programming from scratch at the moment. I was using vectors but creating rotation functions is looking better.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Tue Sep 12, 2017 11:57 pm

I've been thinking about how 3D rotation can be stepped up to 4D rotation. I've come up with the following but I'm not sure it is complete?
Basically you start with the 3D moves and then add 3 extra rotations after that...
1. Rotate the cube centrally about the z-axis (the one pointing towards us) - spin angle.
2. Rotate this centrally around the x-axis (the one pointing rightwards) towards the back - back angle.
3. Rotate this centrally around the z-axis (again the one pointing towards us) - side angle.
4. rotate this directly between the x-w - spin4 angle.
5. rotate the cube back between the z-w - back4 angle.
4. rotate this between the x-w again - side4 angle.

I've realised that what I'm looking to create is a co-ordinate system that represents all the possible rotations.

The pattern from the above for any number of dimensions seems to be:
1. orientate the 'face' relative to the forward direction (this is done by using the full sequence for the dimension-1 one rotation). (2D and above)
2. spin the orientation into the extra dimension again with respect of the forward direction. (4D and above)
3. move it back along to a latitude. (3D and above)
4. move it to the correct sideways by rotation(s). (3D and above)

This seems to follow (I looked up the name) the triangular number system which goes: 1, 3, 6, 10, 15, etc.
ie. 2D = 1 rotation, 3D = 3 rotations, 4D = 6 rotations, 5D = 15 rotations...

Or more thoroughly:
2D does step 1 in one rotation (though backwards).
3D does step 1 in one rotation, step 3 in one rotation and step 4 in 1 rotation.
4D does step 1 in 3 rotations, step 2 in 1 rotation, step 3 in 1 rotation and step 4 in 2 rotations.
5D does step 1 in 6 rotations, step 2 in 1 rotation, step 3 in 1 rotation and step 4 in 3 rotations.

Now that last one actually has 11 (not 10) which makes me suspect there must be more?
If it does follow a pattern maybe it is the next one up which are the mersenne numbers.
These step up as: 1, 3, 7, 15, 31 and are the result of 2^n-1. So n would number of dimensions less one.
So maybe 4D actually requires 7 rotations and I just haven't identified one yet?
It always feels the extra dimensions always add some new trick so I'm inclined to think that this later may be closer?

Can anyone help to resolve this?
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Re: Optimal Tesseract Slice

Postby gonegahgah » Wed Sep 13, 2017 3:12 am

I suspect that should really be:

1. orientate the 'face' relative to the forward direction (this is done by using the full sequence for the next lower dimension's orientation process).
2. move it back along to a latitude (1 rotation).
3. move to the correct sideways by rotation (full sequence for next lower dimension's orientation process).

Which makes 1 for 2D square, 3 for 3D cube and 7 for 4D tesseract.
Hopefully that is closer?

More precisely this consists for 4D tesseract as:
1. Rotate the cube centrally about the z-axis (the one pointing towards us; in 4D is around z-w plane) - spin angle.
2. Rotate this centrally around the x-axis (the one pointing rightwards; in 4D is around x-w plane) towards the back - back angle.
3. Rotate this centrally around the z-axis (again the one pointing towards us; in 4D is around z-w plane) - side angle.
4. Rotate this back around the x-w plane - back4 angle.
5. Rotate this around the x-z plane - side4 spin angle.
6. Rotate this around the z-w plane - side4 back angle.
7. Rotate this around the x-z plane - side4 side angle.

Or something like that? I'll have to test it out mathematically (good idea!)
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Re: Optimal Tesseract Slice

Postby gonegahgah » Wed Sep 13, 2017 4:03 am

Now I'm back to thinking it is only 6 steps.
The intention of rotations 4 to 6 is to keep the cube shape, which was orientated in steps 1 to 3, always facing outwards on the hypersphere...

1. Rotate the cube centrally about the z-axis (the one pointing towards us; in 4D is around z-w plane) - spin3 angle.
2. Rotate this back along the sphere around the x-axis (the one pointing rightwards; in 4D is around x-w plane) - back3 angle.
3. Rotate this centrally around the z-axis (again the one pointing towards us; in 4D is around z-w plane) - side3 angle.
4. Rotate the w-x plane about the y-z plane - to spin to required w-side angle - spin4 angle.
5. Rotate this back along the hypersphere along that new w-angle around the w-x plane - back4 angle (same rotation as step 2).
6. Rotate the x-y plane around the w-z plane - side4 angle (same rotation as step 3).

The top three imply automatic dual rotation around the w-axis in 4D and are shown like this to show that they are normal cube rotation co-ordinates.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Wed Sep 13, 2017 9:47 am

Maybe I can just rotate around the 360° sideways first?

1. Rotate the w-x plane about the y-z plane - to warp to required w-side angle - warp angle.
2. Rotate the cube centrally about the z-axis (the one pointing towards us; in 4D is around z-w plane) - spin angle.
3. Rotate this back along the sphere around the x-axis (the one pointing rightwards; in 4D is around x-w plane) - back angle.
4. Rotate this centrally around the z-axis (again the one pointing towards us; in 4D is around z-w plane) - side angle.

Maybe that will work just as well to provide a workable co-ordinate system? Does that look okay ICN5D?

Edit 2017-09-15:
Step 1 above is the only move from 4D into our space and steps 2-3 basically move the result of that around in our space.
Step 1 is insufficient to produce all the odd shapes that 4D can throw at us.
So I have to rule out these 4 steps... I'm not sure if any 4 step process would work?

ICN5D, you're saying you can get any cube position in just 2 rotations.
Assuming we have to do the same rotations (for different distances) in the same order, I can't seem to get that to work?
If for example we rotate the front face sideways first and then rotate - what now faces us - up or down then there seem to be some unreachable positions?
We can't for example get the front face rotated on its side?
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Re: Optimal Tesseract Slice

Postby ICN5D » Fri Sep 15, 2017 1:00 am

Oh sure, you can add in the third rotation. It won't show you any unique slices, other than the same slices being in different orientations. Here's what I mean. This is a cube sliced in 2D, with only 2 rotations plugged in:

https://www.desmos.com/calculator/fgqzkxuntu

The script starts off with a corner-first scan of a cube.

use 'a' to translate up/down in 3D

use 'b' and 'c' to rotate on two different planes.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Fri Sep 15, 2017 2:13 am

Hi ICN5D, now I understand.
Without coloured faces the cubes are basically symmetrical in many views so less angles are probably required.
I'm looking to have the faces have different colours and to produce unique orientations of those colours.
So that's why I need all three angles for a cube and possibly six angles for a tesseract.

However even then, if we gave the angles full range, there would be a duplicate series of unique orientations.
So I have to limit one of the angles (I limit the back one) to 0 to π and the other two angles go the full -π to <π.
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Re: Optimal Tesseract Slice

Postby ICN5D » Fri Sep 15, 2017 2:32 am

Ah, now I see what you're doing. Yep, you'll need all rotations available. I didn't think about the different coloring you are using. The equation for the 6 rotations in 4-space are going to be ridiculous complex, if you try to combine them the way I do. But, some other method can make it easier.
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Re: Optimal Tesseract Slice

Postby quickfur » Mon Oct 30, 2017 6:04 pm

The "optimal" view of a tesseract, at least in terms of maximizing the projection volume and seeing all 4 cells at a time, would be the vertex-first projection:

Image

The yellow dot in the center is the nearest corner (vertex) to the 4D viewpoint. You can see 4 cells surrounding it, each looks like a deformed cube (an artifact of perspective projection). If you compare this with the 3D view you had, you'll see that it is a direct analogy: the nearest corner (vertex) lies in the center of the projected image, surrounded by the facets that share that vertex. Each facet is somewhat "squished" due to the view angle: in the 3D case, the square faces appear as rhombuses; in the 4D case, the cube cells appear as squished cubes. The vertex-first view is also the maximal one: in 3D, it lets you see 3 of the 6 faces of the cube. In 4D, it lets you see 4 of the 8 cubes of the tesseract.

This is why I have been promoting the projection approach to 4D visualization all these years. It is the most direct and intuitive way to generalize the way our own eyes see 3D.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Thu Nov 02, 2017 11:33 am

Which is cool. The only thing I'm hoping to improve upon is the sense of visual orientation.
With straight projection the rotations with 4-space look magic even though we may understand what is happening.
I'm hoping that rotated projection will remove a bit of that magical appearance making it easier to work with.
At least I hope so.
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Re: Optimal Tesseract Slice

Postby quickfur » Thu Nov 02, 2017 4:54 pm

It is no more magic than the appearance of 3D rotations when projected on a 2D screen. We only don't notice it because our brain has grown accustomed to seeing it and interpreting it in a 3D way. A 2D being looking at a projected 3D projection would perceive it as something completely impossible. But if you try change the 3D representation so that it looks "normal" to a 2Der, then from our point of view as 3Ders the representation wouldn't make any sense, because that's not how 3D rotations work.

There's simply not enough degrees of freedom in a 2D rotation for it to completely represent a 3D rotation, the same way there's just not enough degrees of freedom in a 3D rotation to completely represent a 4D rotation. Somewhere along the line, you have to resort to "magic"-looking things (i.e., things that don't look like a rotation in the lower dimension) to represent the higher-dimensional rotation. I don't see any way around this -- you can't shoehorn a higher-dimensional construct into a lower-dimensional one without losing something.
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Re: Optimal Tesseract Slice

Postby gonegahgah » Fri Nov 03, 2017 12:12 am

That is true that our brains are used to interpreting 3D movement. You are right.
And no matter what we do there will always be some distortion when projecting higher dimensions to lower dimensions.
So even rotated projection adds distortion too. It just isn't possible not to.

Hopefully though it will add a greater sense of what being 4D is.
Maybe it is possible with practice to see your projection with as much automatic intuitiveness as seeing a 3D object.
I know I still get caught up thinking 3D at times...
I'm hoping that rotated projection will offer a more intuitive representation. As I say... hopefully.
I do know that it certainly presents more of the object (with distortion still) so that has to hopefully help the viewer.

I've worked out the 3D rotation math so far.
I am looking to work with geographic co-ordinates instead of cartesian co-ordinates for directional positioning.
I'm hoping this makes it easier to step from 3D to 4D.
But time is scarce at the moment. I'll hopefully get back to it in the new year...
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