7x7 OLL Parity

[Skywalker2026]

Question about 7×7 OLL parity after edge pairing

On the yellow (Y) face of a 7×7 cube, after all edges have been paired, is it possible to encounter an OLL parity in which one edge has a pattern such as

YYYYBBY

or

YBBYYYY

where B is blue?

If either or both of these are possible OLL parity cases after edge pairing, which algorithm(s) should be used to transform the affected edge to

YYYYYYY?

In particular, I am trying to understand whether the location of the flipped wing pieces (left versus right of the midge) requires different algorithms, or whether the same parity algorithm can handle both cases.

Thanks!

 

[stoic]

No; try it on a solved cube to see. If you perform an OLL parity alg but include both slices you’d end up with
YBBYBBY.
The single slice equivalents are YBYYYBY and YYBYBYY.
The reason being, the wings don’t just flip singly in isolation; they exchange places with the equivalent wing on the opposite slice.

 

[Skywalker2026]

Thank you, stoic. That makes sense now. I had incorrectly assumed that a wing might flip independently.

So, after edge pairing, the possible patterns I was considering are constrained by the equivalent wings exchanging places on opposite slices. This also explains why the examples I saw were symmetric, such as YBYYYBY, YYBYBYY, and YBBYBBY.

Your suggestion to try the parity algorithm on a solved cube is very helpful. I will experiment with it to understand this better.

Many thanks!

 

[Skywalker2026]

One more question, stoic.

I have a related question. During edge pairing, when working on the last edge, could asymmetric patterns such as YYYYBBY or YBBYYYY occur?

More generally, if we consider the five edge pieces between the two corners as either Y or B, are all 2⁵ = 32 possible combinations possible during the last-edge pairing stage, or are there still constraints on which patterns can occur?

Many thanks!

 

[stoic]

No, for the same reason.
It might be more helpful/correct to think of this as “edge parity” rather than OLL parity as you don’t get OLL parity (or PLL parity!) on odd-numbered NxNxNs. (Happily, it’s solvable with the OLL parity alg you already know. Later, you’ll learn some of the last 2 edges cases and also develop strategies to avoid it.)
On 7x7x7, once you’ve paired all the edges reduction is complete and there’s no more parity to contend with.

 

[Skywalker2026]

Thank you very much, stoic. This clears up my misunderstanding.

I now understand that on a 7×7 this should be called edge parity, rather than OLL parity, and that once all the edges are paired, reduction is complete and there is no OLL or PLL parity to deal with.

I also understand your answer to my edge-pairing question: even while working on the last edge, asymmetric patterns such as YYYYBBY or YBBYYYY cannot occur, for the same reason you explained earlier—the wings do not flip singly in isolation, but exchange places with the equivalent wing on the opposite slice.

If I understand this correctly, during the pairing of the last edge, the five edge pieces therefore cannot be treated as five independent Y/B choices. Since the equivalent wings on opposite slices act as pairs, it seems that there are only four possible patterns:

YYYYYYY
YBYYYBY
YYBYBYY
YBBYBBY

Please correct me if I have misunderstood this.

This is very helpful. I learned something important today. Many thanks!

 

[bcube]

On the yellow (Y) face of a 7×7 cube, after all edges have been paired, is it possible to encounter an OLL parity in which one edge has a pattern such as

YYYYBBY

or

YBBYYYY

where B is blue?

Here is another way how you can think about it. As you can see, the B wing edges are not in the same wing orbit. So ask instead: can I swap 2 pieces from different orbits with each other? If that's difficult to answer, simplify the case: can I swap 2 pieces from different orbits with each other on a 3x3x3 cube?

By experimenting, you should come to a conclusion that each wing edge can occupy 24 places on cube size bigger than 3x3x3. Try it on the simplest case of 4x4x4 cube if that's not obvious. In each edge spot like UF, a wing edge can occupy 2 places in its orbit, and wing edge orientation is determined by its permutation. Therefore, you can't orient wing edge in its orbit without moving it to the other place in its specific edge spot. Thus, stop thinking about orientation and start thinking about permutation in what you call an OLL parity.

or whether the same parity algorithm can handle both cases.

I personally use 1 parity algorithm for each wing edge case. The best part? My parity algorithm is only 1 move long. Can you figure out my parity algorithm?

 

[Skywalker2026]

Thank you, bcube, for helping me again. Your explanation gives me another way to understand the problem.

I think I now understand your first point. In my examples YYYYBBY and YBBYYYY, the two B wing pieces are from different wing orbits. So instead of asking whether those wings can be independently “flipped,” I should first ask whether two wing pieces belonging to different orbits can be swapped. Since a wing piece remains within its own orbit, those patterns cannot arise in the way I originally imagined after all 12 edges have been paired.

Your comment that wing orientation is determined by permutation is particularly helpful. It also helps me understand why thinking of an individual wing as simply flipping in place can be misleading. What appears visually to be an orientation issue is really constrained by the permutation of the wing pieces within their respective orbits.

As for your “one-move parity algorithm,” I accepted your challenge and did some experimenting on my 7×7.

I started with the YBYBY case. I thought perhaps I could move the two B wings to an adjacent edge, use the ordinary edge-pairing/flip procedure there, and thereby avoid using the familiar 18-move parity algorithm.

Interestingly, the experiment fixed the current edge:

YBYBY → YYYYY

but the adjacent edge involved in the operation became:

BYBYB

When I then tried to pair that adjacent edge normally, its midge dictated the final orientation, so I obtained:

BYBYB → BBBBB

rather than YYYYY.

So my attempted shortcut did not eliminate the problem; it effectively transferred it from one edge to another.

I therefore have not yet figured out your one parity algorithm for each wing edge case. I will experiment some more before asking you for the answer.

Thank you again. Thinking in terms of wing orbits and permutation rather than independent wing orientation has been very helpful.

 

[Skywalker2026]

Dear stoic and bcube,

I wanted to thank you again for your excellent advice and suggestions. I have written a new Notebook #9 to help me remember and better understand how to solve edge parity when I encounter it again.

Notebook #9 – Understanding Edge Parity on Odd-Order Rubik's Cubes

I greatly appreciate your help and the knowledge you shared with me.

Thank you again!