# LL edges flipped by quarter turns



## Stefan (Apr 24, 2009)

I just made up a possibly reasonable usage of using an edge orientation scheme differing from the common one. It's just for the last layer (an outside layer).

Normally we say an edge is correctly oriented iff its last face sticker is on the last face. My proposal is to say an edge is correctly oriented iff it can be solved (relative to the centers) with an even number of quarter turns. So for example UR is incorrectly oriented at UF, but correctly oriented at FU (it can be solved for example by doing the two moves F R).

That scheme makes this short permutation algorithm orientation-preserving:
M' U M U2 M' U M

That's the U perm case for this scheme. The H perm is the same as in the common scheme. A possible Z perm alg is this: B2 M' B2 U M2 U' F2 M' F2. No idea about other "PLL"s for this scheme, I was mainly thinking about edge permutations and I just like the above alg.


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## Pedro (Apr 24, 2009)

orienting edges is just waste of time 

I use that U perm (and its variations, of course) quite a lot

just set up the *stickers* you want to cycle and apply the corresponding alg

just like your old method, but more efficient, in my view


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## Stefan (Apr 24, 2009)

You're thinking of blindsolving, I guess? I was more thinking about regular solving, as you rarely have a "last layer" and "PLL" in blindsolving. But yeah, why not use it for blindsolving as well and use the scheme for all twelve edges...


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## Waynilein (Apr 24, 2009)

I can see it being useful in BLD (if you do orientation and permutation separately), but what are the advantages of this system in normal solving? I think it would take longer to determine whether an edge is correctly oriented with this system than with the standard OLL.

I guess you could do something like a Petrus LL with this (except the edges are oriented according to quarter turn count), but I don't see it being useful in F2L in the sense that Petrus/ZZ edge orientation is useful.


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## cuBerBruce (Apr 24, 2009)

That EO scheme (applied to all edges, not just LL edges) can be useful in computer cube solver programs because of its 48x symmetry. The conventional EO scheme used in ZZ or BLD solving has only 16x symmetry. Programs often use lookup tables for pruning tables and calculating the effects of moves. These tables are often reduced in their size by using symmetry, and the higher the symmetry, the less memory that will be required for such tables.

I note that because of its symmetry, it doesn't matter which face of the cube is designated as the U face. The same edges are considered oriented or mis-oriented regardless of which way you decide to hold the cube.


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## SimonWestlund (Apr 25, 2009)

Pedro said:


> orienting edges is just waste of time
> 
> I use that U perm (and its variations, of course) quite a lot
> 
> ...



I do that, but if a piece is in the correct place but flipped, then I use flipping algs.


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## TheBB (Apr 25, 2009)

StefanPochmann said:


> My proposal is to say an edge is correctly oriented iff it can be solved (relative to the centers) with an even number of quarter turns.


This isn't well defined. On a solved cube, do U'. The UR edge can be solved now with U or R U' R' U, odd or even.
Edit: Actually it may be well defined, but those quarter turns must move the edge...
Edit: Oh sure, it is well defined, but only with that additional information .


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## Johannes91 (Apr 25, 2009)

TheBB said:


> Edit: Actually it may be well defined, but those quarter turns must move the edge...


Of course.

A common EO definition is that <U,D,F2,B2,R,L> preserve EO, and <F,B> flip the 4 edges on the face turned. This scheme can be defined similarly: <U2,D2,F2,B2,R2,L2> preserve EO, and <U,D,F,B,R,L> flip 4 edges.

Edit: I misused the notation, should've written {F,F',B,B'} and {U,U',D,D',F,F',B,B',R,R',L,L'} instead.


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## TheBB (Apr 25, 2009)

I found that out a minute before you posted . But anyway, you still need to point out (and prove) that there is no odd-numbered QTM algorithm that puts an edge back in its place in the same orientation. But that's not so hard.


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## cuBerBruce (Apr 25, 2009)

Johannes91 said:


> TheBB said:
> 
> 
> > Edit: Actually it may be well defined, but those quarter turns must move the edge...
> ...



Careful, Johannes, <U,D,F,B,R,L> is the entire (fixed-centers) cube group. What you mean is that each of the specified generators in <U,D,F,B,R,L> flip 4 edges, not the entire group!

Edit: And similarly there are elements of <F,B> that (for the common EO scheme) do not flip any edges (and some that flip 8).


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## Scigatt (May 1, 2009)

StefanPochmann said:


> I just made up a possibly reasonable usage of using an edge orientation scheme differing from the common one. It's just for the last layer (an outside layer).
> 
> Normally we say an edge is correctly oriented iff its last face sticker is on the last face. My proposal is to say an edge is correctly oriented iff it can be solved (relative to the centers) with an even number of quarter turns. So for example UR is incorrectly oriented at UF, but correctly oriented at FU (it can be solved for example by doing the two moves F R).
> 
> ...



First of all, I don't see why you would restrict that to the last layer, as it seem to work for all edges.

Secondly, I thought of this a while back, but I prefer to state it slightly differently.
'An edge flips whenever it moves from one middle slice to another(relative to centres)'

The thing I prefer about this way of saying it is that it justifies both the standard and 'isotropic' way of seeing edge flip.

The isotropic definition is a direct consequence of seeing it this way.

Explaining the standard method is a little more complicated. Without turning restrictions, edges can move freely between all 3 slices:

```
E <-> M
^     ^
 \   /
  v v
   S
```

However, if we forbid a certain type of turn, we prevent a direct movement of edges between two slices(For <F,B> the M<->E transition is stopped). This change makes the slice transition map 'linear'.

E<->S<->M

So, if we want to move an edge between S and E/M, we will need 1 slice transition. If we need to make a E<->M transition we will need 2. It is possible to utilize more edge transitions than necessary, but only by doubling back on the map, which always adds an even number of slice transitions. Therefore, in this scheme it is impossible to change the 'slice parity' of an edge without moving it to a different slice. The fact that this applies to the 'home' slice and that slice parity is a consistent extension to the concept of edge flip (That's what the isotropic definition says) implies that the standard edge flip scheme is valid.

Edit:



TheBB said:


> I found that out a minute before you posted . But anyway, you still need to point out (and prove) that there is no odd-numbered QTM algorithm that puts an edge back in its place in the same orientation. But that's not so hard.



Because I feel that I'm on a roll, I will try to prove this, despite it's obviousness.

First of all, you will need to consider opposite sides of the cube to be the same colour.

Looking at the stickers of an edge and the adjacent centre stickers, there are 4 possible states that an edge can be in:

Both edge stickers match adjacent centre stickers:R-state
Only one edge sticker matches with adjacent centre:S-state
One sticker matches with _far_ centre:H-state
Both edge stickers match far centre stickers:I-state

Define R-state edges as correctly flipped and I-state edges as incorrectly flipped.

Now we can begin to see how edges change states. Let's start with an R-state edge. There are two faces we can turn to move this edge. Ignoring half-turns, turning either side puts the edge in S-state.

Now let's look at an S-state edge. If we turn the face in which the stickers match, the edge is in R-state. If we turn the other face, the edge is in H-state.

For an H-state edge, turning the face with the centre that matches the far side leads to an I-state edge, turning the other face leads to an S-state edge.

Finally, for an I-state edge, turning either 'relevant' face leads to an H-state edge.

We can sum up this information in a edge state-transition map:

R<->S<->H<->I

What you're asking is essentially if there are any odd state-transition* algs that can preserve edge state. From the map, the answer is obviously no. Similarly, all algs that flip an edge in place must have an odd number of state-transitions.


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