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 Post subject: QJ Trajbers octahedron edge swapping
PostPosted: Mon Apr 25, 2011 2:43 am 
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Is it possible to get two edges needing swapping on the qj trajbers octahedron? I wouldn't have thought it was, but I seem to have it.

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Last edited by rline on Mon May 16, 2011 5:56 am, edited 1 time in total.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 2:45 am 
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Are the corners solved? If yes, then something's gone weird.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 3:13 am 
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Luke wrote:
Are the corners solved? If yes, then something's gone weird.

Well, I have 2 edges needing to be swapped, and 2 corners needing to be swapped. That's not possible is it? I'd claim that maybe one of my students switched stickers somehow, but then I solved it myself yesterday just fine. I do edges first, then corners. I solved all but 2 edges, then tried to solve corners, and now have two of them left as well.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 3:17 am 
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That's solvable. There are two two cycles, so they cancel each other out. Try and solve it like a super 3x3x3.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 3:22 am 
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Luke wrote:
That's solvable. There are two two cycles, so they cancel each other out. Try and solve it like a super 3x3x3.

I don't have a super 3x3x3. How should I think about it in that case? (I do have a crazy earth if that helps.)

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 3:26 am 
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Look up on YouTube how to solve a super cube. It's meerly a 3x3x3 where the center's orientation is visible. It's not hard to solve at all.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 4:19 am 
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Luke wrote:
Look up on YouTube how to solve a super cube. It's meerly a 3x3x3 where the center's orientation is visible. It's not hard to solve at all.

OK. Will do. Thanks for your help Luke.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 10:43 am 
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or don't. Just solve the edges first like you would with a normal solution. than solve the corners.
if you learned a solutin were it is other way round than solve corners first, than go for the edges. if the center in the upper row is twisted, meaning the tip of the octahedron, just swap two adjacent edges, than swap them again. that will have the rotated the center. Corner algs don't affect it, so you will after be able to just finish the solve.
that's all you need to know for the super 3x3x3.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon Apr 25, 2011 6:25 pm 
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First thanks for your reply.

Well all I know how to do is solve edges first and then corners (ult solution). So that's what I'd been doing. For the edges I just have a 3-cycle so I don't really know how to "swap two edges". So with my 3-cycle, the centers aren't rotated at all.

alaskajoe wrote:
or don't. Just solve the edges first like you would with a normal solution. than solve the corners.
if you learned a solutin were it is other way round than solve corners first, than go for the edges. if the center in the upper row is twisted, meaning the tip of the octahedron, just swap two adjacent edges, than swap them again. that will have the rotated the center. Corner algs don't affect it, so you will after be able to just finish the solve.
that's all you need to know for the super 3x3x3.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Wed Apr 27, 2011 6:02 pm 
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I think the most straightforward way of solving the Trajbers Octahedron using the Ultimate Solution is this:
1. Pick a corner to be the base. The base is the middle of the bottom layer.
2. Twist the bottom layer so that the 4 corners in the middle layer are in the correct places relative to the bottom corner.
3. Twist the middle layer corners so the colors line up correctly relative to the bottom corner and each other.
4. Use simple moves (similar to Burgo's method of getting the bottom layer inner edges of the Crazy 3x3x3 Earth) to insert the 4 bottom layer edges.
5. Use simple moves to insert 3 of the middle layer edges.
Note: in steps 4 and 5 make sure you end up with the corners still oriented correctly after each edge is inserted.
6. Twist the top so the top corner is solved.
7. Insert the 2 top layer edges that are farthest from the unsolved edge in the middle layer. (Make sure you keep the top corner solved while doing so.)
8. This leaves 3 edges to solve. Use the EPS of the Ultimate Solution.
9. Use the CPS of the Ultimate Solution to get all the centers into place.

Using this method no special super cube center twisting algorithms are needed. It makes me wonder if I can solve all the puzzles I have previously used such an algorithm for without it. hmmm...

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Thu Apr 28, 2011 5:56 pm 
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This method is nice and simple. However,

Quote:
7. Insert the 2 top layer edges that are farthest from the unsolved edge in the middle layer. (Make sure you keep the top corner solved while doing so.)

After doing this (and all other steps in order) successfully

Quote:
8. This leaves 3 edges to solve. Use the EPS of the Ultimate Solution.

There were in fact only 2 edges left to solve, which brings me right back to the original problem :)

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Thu Apr 28, 2011 6:35 pm 
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Last edited by robertpauljr on Wed May 11, 2011 3:36 pm, edited 1 time in total.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sat Apr 30, 2011 2:09 am 
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Last edited by robertpauljr on Wed May 11, 2011 3:37 pm, edited 1 time in total.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 09, 2011 3:14 pm 
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I think that it is necessary to clarify the terminology.
There is NO PARITY on a Trajber's Octahedron (or a 3x3x3 Supercube).

The Trajber's Octahedron has visually 6 corners but if you want to compare it with a normal 3x3x3 they are the centres of a rotation and correspond to the 3x3 centres. The 8 triangles correspond to the corners of a 3x3.
Some posts in this thread used the word "corner" refering to the Octahedron triangles, others used corner for the "visual" corners which correspond to the cenres of the 3x3.
I would prefer to emphasize the similarity to the 3x3. From now on, I'm using 3x3 terminology.
What is new compared with the ordinary 3x3 is the orientation of centres (pieces with four stickers).
That's why the solution is identical to the the solution of a 3x3 Supercube (BTW, if you want to get one, Olivér Nagy offers very nice Supercube stickers: I have made my first Supercube just adding orientation to the centre stickers of a 3x3x3 back in 1980. :) You could use little pieces of adhesive tape, a pen or whatever.
It may be easier to practise with a 3x3x3 Supercube first.
You can easily solve a 3x3 Supercube layer by layer. On a Trajber's Octahedron you have to maintain the orientation of the centres of the first and second layer (the middle belt of the Octahedron consisting of 4 centres and four edges) When you are done with the first two layers, orient the last layer edges correctly (e.g. by L U F U' F' L' or similar sequences. This orientation is harder to see on the Octahedron than on a 3x3), rotate the last layer centre correctly, 3-cycle the edges by Sune (R U R' U R U2 R') which does not rotate the last layer "centre". Solve the "corners" (triangles on the Octahedron) with your usual sequence.

EDIT:
robertpauljr wrote:
If the EPS is R U' R' U, then could we name R U R' U' EPS2?

R' U R U' (EPS)
R2
F' U' F U (EPS2)
R F R' F' (EPS2)
R2
U'

swaps 2 edges. Try it on a normal solved cube, and on the solved octahedron.

Does this help?
It swaps two edges AND two corners (you cannot swap just two edges on a 3x3) and it turns the centre (the tip of the T O) by U', in addition.

rline, if you still suspect a "parity" on your Trajber's Octahedron, please, take photos of all the faces.

EDIT2: Alternatively to doing the centres first, you could look for algos that just turn the centres (the T O tips), only.
A single centre can be turned by 180 degrees only. You can turn two centres, one clockwise, one anticlockwise. I think it is easier to do the centres earlier.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Wed May 11, 2011 3:46 pm 
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Using Konrad's terminology of calling the octahedron corners, centers, since they correspond to 3x3x3 cube centers, and they are indeed the center of rotation for each twistable layer, here is how I would attack the puzzle in a way that would require no algorithms except the EPS and CPS of the Ultimate Solution.

I think the most straightforward way of solving the Trajbers Octahedron using the Ultimate Solution is this:
1. Pick a center to be the base. The base is the middle of the bottom layer.
2. Twist the bottom layer so that the 4 centers in the middle layer are in the correct places relative to the bottom center.
3. Twist the middle layer centers so the colors line up correctly relative to the bottom center and each other.
4. Use simple moves to insert the 4 bottom layer edges.
5. Use simple moves to insert 3 of the middle layer edges.
Note: in steps 4 and 5 make sure you end up with the centers still oriented correctly after each edge is inserted.
6. Twist the top so the top center is solved.
7. Insert the 2 top layer edges that are farthest from the unsolved edge in the middle layer. (Make sure you keep the top center solved while doing so.)
8. This leaves 3 edges to solve. Use the EPS of the Ultimate Solution.
9. Use the CPS of the Ultimate Solution to get all the triangle pieces into place.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 3:23 am 
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Can I have a ruling please? Is this legal?

First, here's a picture of one half of the octahedron, completely solved.
Attachment:
cube 001.jpg
cube 001.jpg [ 58 KiB | Viewed 2634 times ]


And now I turn it over, to reveal everything solved except for one center piece twisted 90 °.
Attachment:
cube 002.jpg
cube 002.jpg [ 57.16 KiB | Viewed 2634 times ]

Is this solvable or has someone switched my stickers?

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 3:24 am 
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That's impossible. Pull the cap off and put it on the right way.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 3:27 am 
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Luke,

Thankyou. I can die a happy man now that I know some student has done the dirty. All I have to do now is figure out which one...

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 3:48 am 
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Luke wrote:
That's impossible. Pull the cap off and put it on the right way.


How's that impossible Luke? It's just a centre rotation isn't it? It's just because he hasn't treated the top like a supercube? Am I wrong?

The solution is just to apply sune a few times to match the edges to that centre and recyce the corners?
Is it the 90 deg?

EDIT: Yes, I see what is happening, I didn't know that it couldn't be 90 deg on the last one, cool, learned something.
..You know everything..
Burgo.


Attachments:
centre-rotation.gif
centre-rotation.gif [ 26.57 KiB | Viewed 2620 times ]

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 4:26 am 
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His centre is rotated 90 degrees, which is impossible. What you are showing is a 180 degree rotation.

Let q be the sum of the rotation of the centres modulo 2 (each centre can be rotated 0, +1, +2 or -1). Any 90 degree rotation increases q by one (eq, if it is 0 it becomes 1 and vice-versa). Any 90 degree rotation also changes the permutation parity of the edges from odd to even and vice-versa.
In a solved cube, q=0 and the permutation parity of the edges is even. After a turn, this is changed to q=1 and an odd parity. After another turn, we get q=0 again and an even parity. Clearly, q=0 <=> even parity and q=1 <=> odd parity.
However, in his cube, q=1 but the permutation of the edges is even so something isn't right.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 4:37 am 
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Yeah, I just figured it out by watching it cycle.. You have a way with words Tom. Wow.

Burgo.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 5:08 am 
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Does anybody know where to get a nice one? The QJ one is ordinary (I really like the trajbers, I'd use it more if I had a good one).

EDIT: I see where you want to go and I just did a solve using Ultimate Solution, but what a pain. Some methods suit certain cubes, and Ultimate Solution is very clunky with this cube (In the last 5 edges I mean). With Ultimate you need to visualize the orientation of the edges, but the Trajbers makes this difficult. Add to this the need to orientate the last centre, and it's very visually difficult: requires a lot of double checking. If I were you here (wanted to use Ultimate for it), I would use Ultimate Solution for everything except the last 5 edges. Orientating the last 4 with FURU'R'F' and rotating with Sune RUR'UR U2 R'. Sune will also place an edge on the top centre for you, if one is not placed. Then use the correctly placed edge in the front position. It's a small deviation from your method, but it will save you a headache.

I have never used a centre orientating alg, don't even know one. Just place your edges carefully in the lower half, with Ultimate solution it is easy not to move centres, it's supercube friendly, and so are those 2 other algs.

Cheers,
Burgo.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Sun May 15, 2011 10:27 am 
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@Burgo: see my strategy above. It may be a slightly (very slightly) modified Ultimate Solution, but it works great and does not require learning the two algorithms, or any others for that matter. I have used it many many times on the Trajbers Octahedron and found it very simple and straightforward from the start.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 16, 2011 3:07 am 
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Thanks for talking me into persisting with that Robertpauljr,
I just sat down and had a little go with it again and I have come to terms with the visual differences now and the method is working well, I like it.
Cheers,
Burgo.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 16, 2011 4:21 am 
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Burgo wrote:
Does anybody know where to get a nice one? The QJ one is ordinary (I really like the trajbers, I'd use it more if I had a good one).....
I have a custom made Trajber's 3x3x3. To be completely honest, the QJ turns as well as the custom made (I had puchased it for 20 times the price of the QJ's) I think that the mass-produced Trajber's 4x4 are even better. So, I miss to understand why you call them ordinary? :o

@rline: I think that posting your picture earlier, would have saved you many headaches, right? :)

@robertpauljr: I'll try your (slightly modified) Ultimate solution, when I have got time for this.
When I teach somebody the 3x3, I always feel a tradeoff between the numbers of algorithms used and the easiness of strategy. A low number of algorithms, requires more thinking about the strategy. The classical Ultimate solution, as I recollect it, uses two algorithms, but getting the last edges right, is not really for beginners. (See Burgo's edges first method for teaching kids. He describes a new algorithm instead of using the Ultimate solution.) My own first 3x3 solution used two algorithms as well, but I have to admit that it had been very complicated and quite lengthy. :)

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 16, 2011 4:30 am 
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Hi Konrad,

By `ordinary` I mean `not excellent`. I would like a physically bigger puzzle. That does not jag or catch either: I mean the mechanism doesn't need to be a GuHong, but at the moment it is, well, not excellent. I was just wondering if I had overlooked a better quality one, because I would buy it.

I have the 4x4x4 Trajber's and I'm very happy with it too, I find it very visually difficult to picture it as a 4x4x4 though :roll: . Nice puzzle.

Cheers,
Burgo.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 16, 2011 4:54 am 
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Burgo wrote:
Hi Konrad,

By `ordinary` I mean `not excellent`. I would like a physically bigger puzzle. That does not jag or catch either: I mean the mechanism doesn't need to be a GuHong, but at the moment it is, well, not excellent. I was just wondering if I had overlooked a better quality one, because I would buy it.

I have the 4x4x4 Trajber's and I'm very happy with it too, I find it very visually difficult to picture it as a 4x4x4 though :roll: . Nice puzzle.

Cheers,
Burgo.
Actually, my custom made T O (my first custom puzzle ever!) is a bit bigger and heavier. Visually it is really nice and I like the weight and the over all feeling. Catching is the same! I think it is caused by the different forms of the "cubies".
I do not know, if others mass-produced exist, but I doubt it.

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 Post subject: Re: QJ Trajbers octahedron parity
PostPosted: Mon May 16, 2011 5:43 am 
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Konrad
Quote:
@rline: I think that posting your picture earlier, would have saved you many headaches, right?

Yes. I live and learn... :oops:

Quote:
The classical Ultimate solution......getting the last edges right, is not really for beginners.

I think I'd have to respectfully disagree with that. But I may be biased. :wink:

I can also say that the ult solution does work really well with the QJ trajbers octahedron. At least it did before someone switched my corner piece.

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 Post subject: QJ Trajbers octahedron edge swapping
PostPosted: Mon May 16, 2011 1:42 pm 
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rline wrote:
Is this solvable or has someone switched my stickers?

This pattern already was published on a forum by 5 months back:
http://twistypuzzles.com/forum/viewtopic.php?p=240052#p240052

The pattern is very easily solved:
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=17715&p=241755#p241755
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=19318&p=235091#p235091


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 Post subject: Re: QJ Trajbers octahedron edge swapping
PostPosted: Mon May 16, 2011 2:03 pm 
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pytlivyj_1 wrote:
rline wrote:
Is this solvable or has someone switched my stickers?

This pattern already was published on a forum by 5 months back:
http://twistypuzzles.com/forum/viewtopic.php?p=240052#p240052

The pattern is very easily solved:
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=17715&p=241755#p241755
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=19318&p=235091#p235091

In this thread we are not talking about the 4x4x4 Octahedron. We are talking about the 3x3x3 Octahedron. Can you rotate one center by 90˚ while the rest of the puzzle is solved with legal moves?

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 Post subject: Re: QJ Trajbers octahedron edge swapping
PostPosted: Mon May 16, 2011 3:19 pm 
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robertpauljr wrote:
pytlivyj_1 wrote:
rline wrote:
Is this solvable or has someone switched my stickers?

This pattern already was published on a forum by 5 months back:
http://twistypuzzles.com/forum/viewtopic.php?p=240052#p240052

The pattern is very easily solved:
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=17715&p=241755#p241755
http://twistypuzzles.com/forum/viewtopic.php?f=8&t=19318&p=235091#p235091

In this thread we are not talking about the 4x4x4 Octahedron. We are talking about the 3x3x3 Octahedron. Can you rotate one center by 90˚ while the rest of the puzzle is solved with legal moves?

I guess, here is a misunderstanding. A 90 degree turn of one centre of the 3x3x3 is not possible (see TomZ's precise explanation above) and a 3x3x3 Trajber's Octahedron is just a shape variant of a 3x3x3.

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