face turning oactahedron : how to define edge orientation ?

[deadalnix]

Everything is in the title. I'm trying to figure a way to know if an edge is correctly oriented or not, and also to know which move change edge orientation.

 

[flee135]

I haven't touched mine in a while, but I'm pretty sure it's not possible.

 

[Lucas Garron]

I'm pretty sure it's not possible.

But of course it's possible.

Trivial, too, in a mathematical sense. Getting a nice EO scheme is a different issue, though.

deadalnix: Define a principal sticker for each edge, like 3x3x3 EO, and see what the moves do to those.

 

[qqwref]

Uh, there's no edge orientation on any face-turning octahedron. It's simply impossible for an edge to be flipped in place. In fact, if you group the stickers into two groups (so that in the solved state two colors being adjacent means they are in different groups), each face can only contain stickers from one of the groups.

 

[Lucas Garron]

qq: Really? I don't remember that from playing with FTO's. But I suppose so.


Anyhow, my point was: defining an EO scheme in general is normally straightforward, although (as in this case) possibly degenerate,

 

[TMOY]

Yes qqwref's right. Just look at which stickers are affected by a face turn: one group contains the stickers of the face itself and the opposite stickers of the three corners, the other group contains all other affected stickers. It's easy to see that the face turn willl preserve both groups globally.
The other consequences are the following ones:
- not all even center permutations are possible, there are two orbits of 12 centers each;
- a corner can have only two possible orientations, not four.

 

[deadalnix]

Yes, qqref is actualy right (be I didn't notice that before). Edge ARE oriented on FTO.

TMOY, I don't get your proof. I assume this is highly or I'm god damn stupid today.

EDIT: anyway, in a general sense, how to define the edge orientation on a n directional puzzle.

Cube is 3 directionnal. We can define axis A and axis B and apply some simple rules : A colors are on A face and not on A orbit. B colors are on B face and not B orbit.

This is working because each edge have an A and/or B sticker and edges are always on an A or B face.

But with 4 directions, I don't see how to do.

Lets define A, B and C with the same rules. If a AB edge is between a C face and a D face (last axis) ?

 

[jaap]

TMOY, I don't get your proof.

Imagine the FTO with only two colours. Faces alternate in colour, i.e. any two adjacent faces have different colours. The edge pieces all have two colours. If you do any move on the FTO with this colour scheme then nothing seems to happen. Therefore the edges cannot be flipped. By the same reasoning, the corner pieces cannot be turned 90 degrees.

EDIT: anyway, in a general sense, how to define the edge orientation on a n directional puzzle.

Start with the solved position. On every edge piece, draw cross on one of its two stickers - it doesn't matter which sticker, as long as every edge piece has exactly one cross. This pattern of crosses defines the edge orientation. On a piece of paper, draw a picture of this pattern of crosses.
Mix up the puzzle. Any edge piece, wherever it is, either has its cross matching the picture, in which case it is not flipped, or its cross does not match with the picture, in which case it is flipped.

From this it is also fairly easy to see why most puzzles have an edge flip parity constraint. Look at any edge piece. Turn a face containing that edge. The edge either flips or doesn't flip according to your pattern of crosses. Turn the face again and again until the edge piece is back where it started. Clearly the edge piece is still in the same orientation it started with, so on its travel around the face it must have gotten flipped an even number of times (incl. zero), not an odd number. If you now look at all the edge pieces in that face and do the move, then all their movements together are what that single edge just went through. Therefore it must be that an even number of those edges get flipped.

The same reasoning can be done on any face. As all face moves flip an even number of edges, the total number of flipped edges always remains even. In the same way you can prove the corner twist constraint.

 

[deadalnix]

You way to handle edge orientation is nice and really fit n directionnal.

Thx.