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Can you restate this? I don't quite understand what you just said. (This isn't a joke, as I have better things to do than to make fun of people, if that's what you are thinking).

When I edited my first post, I said I made some mistakes. You said I was wrong and that I didn't make a mistake, I just worded it wrong. My reply: wording things wrong is a mistake.

I don't understand how that didn't make sense unless it was my spelling.

Also, when I said that not all permutation parities can be solved with 3-cycles, I was talking about these cases:

I'm now aware after reading your post that this would be an OLL parity because they take and odd number of slice moves to solve. I still think of this as permutation because the edges are in fact flipped correctly.

When I read that all permutation parities are solvable with 3-cycles, I included these and made the statement that the single edge flip parity could be solved with 3-cycles because it was possible to convert it to one of these cases using a 3 cycle.

(**) If we like using the reduction method, when we get PLL parity

or some other version of it, we can think of the cause being from not being able to see proper way to pair the dedges so that there is an "even permutation of the composite edges". (So, if we were to look at the above image as a 2-cycle of composite edges, that's an odd permutation of composite edges on the 4x4x4). Just as it's impossible to prevent OLL parity unless you actually determine whether the permutation of the wing edges is odd or even at the start of a scramble (from which you make an effort to solve the centers using either an even or odd number of inner layer quarter slice turns to make the overall permutation of the wing edges even), it is much more difficult to prevent PLL parity because it's a question of HOW you pair the dedges after the centers are already solved, rather than the number of inner layer slice quarter turns you used to solve the centers with from the scramble you started with.

Just a quick note how to prevent PLL parity every solve (from a reasonably scrambled 4x4x4).

Spoiler

Always plan your entire edge pairing process so that you will always have just two composite edges unsolved as the end of the edge pairing process. From there, you can determine whether the permutation of the composite edges will be odd or even based on which way you solve the last two composite edges. You do the two edge pairing algorithm in the correct direction to complete the edge pairing process).

PLL parity can not be determined considering edge pieces alone. You need to use centers as a reference to determine the permutation parity of both the corners and the dedges. If those parities are different (one of them even, the other odd), then you have PLL parity. Similarly, cmowla's method to guarantee PLL parity being avoided will not work unless you are also careful to put corners into an even permutation, which is probably not worth it in a reduction-style speedsolve.

Of course, on odd size cubes, PLL parity will show up with ending up with two composite edges having a wrong central edge piece. You might say it's the wing edges that are actually wrong because the central edges can not be wrong (as far as parity is concerned). You might argue reduction isn't complete yet in this situation. Basically you use central edge pieces as reference to where the wings must go so PLL parity is basically a non-issue on odd size cubes.

Also, when I said that not all permutation parities can be solved with 3-cycles, I was talking about these cases:

I'm now aware after reading your post that this would be an OLL parity because they take and odd number of slice moves to solve. I still think of this as permutation because the edges are in fact flipped correctly.

When I read that all permutation parities are solvable with 3-cycles, I included these and made the statement that the single edge flip parity could be solved with 3-cycles because it was possible to convert it to one of these cases using a 3 cycle.

In these examples, edge pairing is incomplete so you do not have a reduction state. Therefore PLL parity is not even applicable. If you are viewing the edges as individual pieces, then indeed you can call this an odd permutation of edge pieces. Because of unclear wording in the original post, people tended to assume by permutation and orientation parities, bcube was probably talking about the two types of reduction parities commonly called PLL parity and OLL parity.

Since all cases of reduction states with OLL parity have an odd permutation of wing edges (per wing orbit, but let's talk 4x4x4 for simplicity where there is only one orbit of edges), many people talk about any position (not necessarily being a reduction state) having an odd permutation of wing edges as OLL parity.

Thank you very much for your post, cmowla, and for a huge effort you put in it. Although I appreciate help from the others (more now than before because of your explanations), your post is the most helpful for me. I believe it is an excellent source for both beginners (like me) and advanced cubers, even for experts!

I appreciate the fact that you added images and said the same thing in another way that might make sense to people who think more like you, but I could take your posts and pick apart every sentence if I wanted to take the time to make you look stupid.

Please do not look at ot that way. I truly believe his only intention was not only to help me, but also to help others in understanding concerning parity things. I also think his intention was fulfilled/reached. After all, in my opinion, it is totally fine to be less accurate or even wrong, if you can admit it after reasonable statements are made.

Because of unclear wording in the original post, people tended to assume by permutation and orientation parities, bcube was probably talking about the two types of reduction parities commonly called PLL parity and OLL parity.

As for my answer to question 2 - Permutation parity is not caused by not seeing the difference in pieces. The same situation would happen regardless of if you can see the difference or not. I was just saying that if you could see the difference in pieces, the cube wouldn't look like it was in a 3x3 state in the first place and thus, it is not parity (I would just say your not done reducing the cube to a 3x3 state yet). If you can't tell the difference, this is considered parity.

What causes permutation parity? - The user not paying attention to edge permutation while solving the last 2 or 3 edges causes permutation parity. You can make a cube appear to be in the 3x3 state, but it isn't because it has 2 edges swapped by using 2 3-cycles which combine to make 2 2-swaps of edges. Counting 2 edge pieces as one whole edge makes it look as if only 2 edges were swapped. If the user payed attention to wether the edges should be solvable with an even or odd number of moves, (you can determine this from the corners permutation because they are linked on a 3x3) the user could adjust to do a 3-cycle that puts the edges together while not permutating the same ones.

Square-1 parity basically requires the puzzle to be put into "cubeshape" (i.e., top and bottom layers both square-shaped) first. Because corners and edges are different in size, you can't simply describe arbitrary states of the puzzle as a permutation of 16 pieces (ignoring the middle layer). Or you would at least have to make up some arbitrary rules on how to order the pieces in to describe an arbitrary state as a permutation from the solved state.

Are you, by that, implying that it is not possible to "avoid" parity (PLL parity if that makes sense) in case of Square-1 by method change?

Now, let me ask you guys maybe the last serie of questions concerning parities: can be mathematically expressed a probability that OLL parity is met, when speedsolving 4x4x4 using reduction (i.e. not paying attention to parity at all)? Same question for PLL parity and finally a probability for a case, when both parities are met? Is value of PLL parity probability same for 5x5x5 (as for 4x4x4) and are parity probabilities depending on size of the cube (in a sense of number of layers of the cubes)?

last serie of questions concerning parities: can be mathematically expressed a probability that OLL parity is met, when speedsolving 4x4x4 using reduction (i.e. not paying attention to parity at all)?

Once you get into cubeshape, Square-1 parity is well-defined and you can't change the parity without going out of cubeshape and getting back to cubeshape.

If you have Square-1 parity (i.e. odd parity) and have the bottom layer solved, the top layer will be a parity PLL case.

If you want to "avoid" parity on Square-1 you have to have a system to get into cubeshape with even parity. For example you could first get the puzzle to shield-shield shape. Then if you rotate layers so you have only corners on the right side of the divider, a "/" move will keep the puzzle in shield-shield shape, but will result in getting the opposite parity when you get into square-square shape (aka "cubeshape") by whatever alg you use to get from shield-shield to square-square (such as "(-2,4)/(0,1)/(3,3)/"). But this requires some sort of recognition system of distinguishing between the two parity cases when in shield-shield shape. The difficulty of doing this fast is why people generally do not use such a system for speedsolving.

Is value of PLL parity probability same for 5x5x5 (as for 4x4x4) and are parity probabilities depending on size of the cube (in a sense of number of layers of the cubes)?

On the 5x5x5, PLL parity shows up as wings not matching up with the correct central edge piece. It's perhaps a matter of debate I suppose whether or not you consider "reduction" to be complete or not if this is the case. In any case, on the 5x5x5, you can easily fix PLL parity long before you get to the PLL step of the 3x3x3 phase of the solve. So when you get to the PLL step, the probability of PLL parity should be 0 since you would have fixed it at the point where you were finishing the reduction phase.

EDIT: And this would be the case for any odd size cube, 5x5x5 or larger.

EDIT2: For OLL parity, the probability is 0.5 for each orbit of wing pieces. So on 6x6x6, for instance, the probability of having OLL parity in at least one orbit of wings is 3/4. That is, 1 - 0.5^2. I note that OLL parity is also easily detected easily at the end of the reduction phase on odd size cubes, because the central edge serves as a reference to determine whether wings are swapped or not.