Parity Errors

bcube

Member
Hi,
I would like to write an article (web page) about parity errors. Currently I´m stuck, because I can not answer following questions. Can you?

- is it correct to say that (in case of Rubik´s cube NxNxN) orientation problem is caused by the fact that those parts, which should be solved by "even number of moves" were in fact solved in "odd number of moves"?

- is permutation problem caused by the fact that some of (in case of Rubik´s cube NxNxN) parts seem to be identical to us (edges (wing edges or whatever they are called), to be more precise)?

- what is the cause of parity error in case of square-1? I would say "odd number of moves" (see above), but it is not orientation problem, it is rather permutation problem - but there are no identical pieces on square-1, are they? Is it possible to „avoid“ parity error in case of square-1 as in case of Rubiks cube by method change (reduction vs cage for example)?

As parity error in case of Rubik´s cube NxNxN is considered a situation which can not be solved by reduction to 3x3x3.

MaeLSTRoM

Member
A parity error occurs when you have an odd number of 2-cycles in your permutation.

So for example, to get OLL parity on the 4x4, it means that you have done an odd number of slice turns, since this creates a 4-cycle of edges and 2 4-cycles of centres. Because the centres have no orientation in themselves, this error is not noticed so the only one that comes to light is the wing 4-cycle, which can be reduced down to a wing 2-cycle (Parity) during a normal reduction solve.

For a square-1, the principal is the same, you've done an odd number of 2cycles. It actually occurs because of the fact that all pieces are interchangeable on a sq-1, so you can get to a cubeshape in which you can perform 4-cycles, this is for example Scallop-Scallop (EEEECCCC)(EEEECCCC) because then you can swap 2 sets of edges and 2 sets of corners, making 4 2-cycles, and therefore changing the parity of the permutation.

So partly it's to do with how you solve, but it mainly comes down to if you made an odd number of slice moves for a N*N*N cube, or if you got to a state on a sq-1 in which you can change the parity.

irontwig

Member
A parity error occurs when you have an odd number of 2-cycles in your permutation.

So for example, to get OLL parity on the 4x4, it means that you have done an odd number of slice turns, since this creates a 4-cycle of edges and 2 4-cycles of centres. Because the centres have no orientation in themselves, this error is not noticed so the only one that comes to light is the wing 4-cycle, which can be reduced down to a wing 2-cycle (Parity) during a normal reduction solve.
Huh? First of all, wings don't have any orientation either, and since you have, as you said 2 4-cycles of centres they're still in an even permutation

MaeLSTRoM

Member
Huh? First of all, wings don't have any orientation either, and since you have, as you said 2 4-cycles of centres they're still in an even permutation
Ok, well, what I meant was like you can swap 2 centres and as long as their on the same face, it doesn't matter. I suppose that's more a permutation thing. And yeah, I know that 2*4 cycles is a non-parity but I was just appealing to the point about being pieces you can't see being swapped in the OP.

cuBerBruce

Member
My first suggestion is to not use the word error. I don't know why people keep referring to parity conditions in cubing as parity errors. In my view, this is a misnomer. Parity error is a term used in connection with computer memories and data communications.

Generally, parities in cubing are usually related to permutation parities. So understanding permutation parity, and how basic moves of a puzzle affect the permutation parity of various orbits of pieces is key to understanding parity conditions.

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.

There are some puzzle shapes where certain moves may permute only pieces of like type. If such a move performs an odd permutation (well-defined since only like type pieces interchanged), then this could be used to change the parity of the puzzle as follows. From cubeshape, apply a sequence of moves to set up the given shape, apply the move, and undo the setup moves. The puzzle will be returned to cubeshape with the opposite parity than what you started with.

qqwref

Member
In speedcubing there are two things we think of as parity: "reduction parity" and mathematical parity. Some things we call parity fall into both categories, but others only fall into one of them, and you will often see people disagree on whether certain things count as parity, because they disagree that both of these two definitions are valid.

Reduction parity occurs when you try to reduce the puzzle so it can be solved by a constrained set of moves, putting it into some subset of the positions. However, you can often reach a position which seems like it is in your subset, but which is actually not, and to solve the puzzle you have to briefly go outside your constrained set of moves to bring the puzzle back into the subset you want. Typically the number of positions you can encounter is some small multiple of the number of positions you expect. The obvious example is PLL parity in 4x4x4: all the centers and edges are properly paired, so you expect to be able to finish the puzzle with only outer layer turns, but this isn't quite possible. OLL parity falls under this definition too (so the reduced 4x4x4 has four times as many positions as you would expect). Square-1 parity also falls under this definition - your constrained set of moves are any moves that keep the puzzle in cubeshape. BLD parity is not of this type (the solver has not reduced the puzzle).

Mathematical parity is based on the idea of the mathematical definition of an odd permutation. Basically, at least one orbit of pieces has an odd permutation, and thus cannot be solved with just 3-cycles. You see this type of parity crop up in blindfolded solves, because blindfolded solvers attempt to solve most or all of the puzzle with 3-cycles and thus an odd permutation is very noticeable. OLL parity on 4x4x4 and Square-1 parity can also be thought of as this way, as they originate from some kind of 2-cycle. PLL parity on 4x4x4 is not of this type (it can be solved with 3-cycles).

elrog

Member
1 - No. It is true that any position reached in an odd number of moves has to be solved with an odd number of moves and it is the same using even numbers of moves, but this only applies if you count moves in QTM. However, this is not the cause of the orientation parity. The parity in the orientation of edges (excluding center edges on odd cubes) is decided by the permutation of the edges. An edge gets flipped if it is placed in its opposite orbit. Because edge permutation is always solvabe with 3-cycles after corners are permuted, the orienation parity is solvable with 3-cycles. 3-cycles are always an even number of moves. Refering back to my first statement shows that it will always take an even number of moves to achieve a position with corner permutation solved and orientation parity present. The same is true if orientation parity is not present.

2 - I would say no, but it is more user disgreation. Permutation parity is always solvable with 3-cycles. It only appears that you are swapping 2 pieces because the pieces look the same. So, mathmatically it isn't parity. If you are not viewing parity mathmatically, it is any position not solvale by a given moveset that it appears to be in. On a supercube (distinguishes the permutation and orientation of every piece) You will have permutation parity, but it won't appear to be in the group you need it to be in in the first place. So I would say that no, permutation parity is not caused by us not seeing the difference between pieces, but it is only possible to destinguish as parity if you can't tell the difference.

3 - I have no idea about how a Square 1 even turns. I've only ever seen pictures of it on google images. I'll let other people answer this one.

I'm very sorry that I'm not good at explaining things. I can understand it in my head, I just can't seem to put it into words.

EDIT: I made a few mistakes in answering the first question, but the answer is still no. First of all, I toook for granted that all perutation parities were solvable by 3-cycles after reading qqwref's post, but some of them are not. Second, I talked about how the corner cycle is connected with the edges, but this is only tue of the middle edges on large cubes. Heres how it should be: A edge cycle and its opposite (a cycle the same distance from the middle edge but on the other side) can be changed from an odd perutation to an even permutation and vice-versa with a single 90 degree slice move. Each 90 degree slice move will do this and no outer layer turns will affect this because it causes a 4-cycle in both of the 2 opposite cycles causing the permutation to stay the same. So OLL parity is not due to solving edges with an even or odd number of moves, it is either an even or odd number of slice moves. It will aslo change from even or odd depending on if the edge cycle was solvable with an even or odd number of 2-cycles in the first place. You should also take note that the orientation parity can be solved by 3-cycles if you do not need to preserve orientation, in which case you wouldn't be preserving a 3x3 subgroup anyway.

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bcube

Member
Sadly, I´m getting more confused than I was at the beginning. So far I learnt some new things, but if you want to help me, please respond in a way elrog did - i.e. first response to the question, then commentary (if necessary).

As for question 1 - if "those parts" are replaced by "centers + edges in the process of pairing" (when solving by reduction), would it be possible to answer it by yes? If so, I believe that original question could be answered by yes too, since "those parts" is just a generalization. If I understood MaeLSTRoM correctly, that is what he was trying to say.

As for question 2 - I did not find the answer yet (partly answered/assumed by elrog). Elrog, could you express slightly better what did you mean by "but it is only possible ... the difference"?

As for question 3a - answered by MaeLSTRoM, but how 4 2-cycles can be an odd number of 2-cycles? Btw. do I get it right that 3x3 T-perm is 2-cycle, while U-perm is 3-cycle and H-perm are 2 2-cycles from the edge point of view?

Question 3b is rather rethorical, for question 3c I did not find the answer yet.

AvGalen

Sadly, I´m getting more confused than I was at the beginning.
Can I ask you why you "I would like to write an article (web page) about parity errors" when you don't understand them yet? Finding out how things work is a good idea and sharing knowledge with others is also a good idea, but sharing knowledge with others about something that you don't understand is a strange thing to do

bcube

Member
Can I ask you why you "I would like to write an article (web page) about parity errors" when you don't understand them yet?
It was suggested to me to mention a purpose and I found it as a good idea. Even if I did not mention a purpose of my interest, then what? (It is rethorical, no need to answer). For the next time, please use PM for unrelated question, a.k.a. please do not "violate" this thread, no offence.

Ronxu

Member
It was suggested to me to mention a purpose and I found it as a good idea. Even if I did not mention a purpose of my interest, then what? (It is rethorical, no need to answer). For the next time, please use PM for unrelated question, a.k.a. please do not "violate" this thread, no offence.
How was that an "unrelated question"?

Christopher Mowla

Sadly, I´m getting more confused than I was at the beginning...
I was going to say the exact same thing AvGalen said. I agree that it's strange to write an "article (web page)" about something technical if you are not "the authority" on the subject or at least you don't understand what the authority had to say on the subject.

AvGalen's post was very called for because you started a thread that was basically telling experts on a subject "hey, I want to write an article on something I don't know anything about. Teach me...that is...if you can!", and, when good explanations are given, you basically say "I don't understand what you are saying...can you write the article for me and then I can make sense of it and then write an article?".

In my opinion, qqwref and cuBerBruce's posts summarizes the whole idea on parity (all other posts weren't necessary). If you cannot understand those two simple, thorough, and complete explanations, then you shouldn't think about writing about this subject until you do...unless you want to write something now that you will later regret because it isn't entirely accurate and/or complete.

If you are going to write an article/webpage on something, you should at least do whatever it takes to get yourself to understand what it is your writing about...reading cuBerBruce's and qqwref's posts and then immediately saying you don't understand isn't a good enough effort, I think. I highly doubt there is a better way to say what they said. They might be hard for you to understand, but you must realize how much ground they actually cover...if you can't realize that, then you really don't have anything to write about because just stating a few definitions isn't enough to write an article, as you probably know.

So give it another try because the answers you are searching for have already been freely given to you. The least you can do is "decipher" the answers and expand on them...and of course to not forget to list your sources.

elrog

Member
A discussion about his motives isnt very productive, thus I would consider it something to be taken to a private chat regardless of if it relates or not.

Sadly, I´m getting more confused than I was at the beginning. So far I learnt some new things, but if you want to help me, please respond in a way elrog did - i.e. first response to the question, then commentary (if necessary).

As for question 1 - if "those parts" are replaced by "centers + edges in the process of pairing" (when solving by reduction), would it be possible to answer it by yes? If so, I believe that original question could be answered by yes too, since "those parts" is just a generalization. If I understood MaeLSTRoM correctly, that is what he was trying to say.

As for question 2 - I did not find the answer yet (partly answered/assumed by elrog). Elrog, could you express slightly better what did you mean by "but it is only possible ... the difference"?
I'm glad you liked my formatting.

As for question 1 - It is possible that the answer could be yes. Heres why I said no. Only slice moves count, not just any move. (Remember that a 180 degree turn should Count as 2 moves for this to work) Another reason I said no is that the party could also be caused by the cube needing to be solved in an odd number of slice moves and you solve it with an even number of slice moves. What I'm saying is that you should have even and odd be swappable in your question.

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.

I hope I cleared it up the confusion.

stoic

A discussion about his motives isnt very productive, thus I would consider it something to be taken to a private chat regardless of if it relates or not.
I think the OP chose to come into a public forum to ask his questions, and his behaviour has been a bit strange and rude. There may be a case for someone having a quiet word in private, but I don't agree with the idea that the thread can't be "violated" with comment.

bcube

Member
As for question 1 - It is possible that the answer could be yes. Heres why I said no. Only slice moves count, not just any move. (Remember that a 180 degree turn should Count as 2 moves for this to work) Another reason I said no is that the party could also be caused by the cube needing to be solved in an odd number of slice moves and you solve it with an even number of slice moves. What I'm saying is that you should have even and odd be swappable in your question.
Got it. I´m aware of many "undefined" terms in my questions, which may be answered by two different ways, meaning the same thing. Because of my limited knowledge of English, I´m forced to simplify some terms too, which may cause another confusion. Anyway, I think I completely understood what you were trying to say and that is the very same I was thinking of (regardless of my way of asking/expressing). Thank you for making it clear.

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.
Hmm, I´m still not following you on this one. If permutation parity is not caused by impossibility to distinguish different pieces, then what is causing it? (yes, I did not find it indeed - although I´m not saying it is already posted).

In my opinion, qqwref and cuBerBruce's posts summarizes the whole idea on parity (all other posts weren't necessary). If you cannot understand those two simple, thorough, and complete explanations, then you shouldn't think about writing about this subject until you do...unless you want to write something now that you will later regret because it isn't entirely accurate and/or complete.
Since my deeduction is weak, any suggestion how to find out the answers to my questions (especially 2 a 3c)?

So give it another try because the answers you are searching for have already been freely given to you. The least you can do is "decipher" the answers and expand on them...and of course to not forget to list your sources.
Would you be so kind and mark the specific sentenses for me/other potential people with interest, where the answers are given (I´m still not a symphatizer of deciphering it, though ;-) )

elrog

Member
The question "Does ... cause permutation parity?" and "What causes permutation parity?" are two different things. I answered your question, but I didn't say what caused it. As for my answer, I'm not sure how to make it any more clear.

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.

Christopher Mowla

Amazingly, I have never had interest in square 1. So you will have to let others elaborate on that more.

This is a very lengthy post. After forcing myself to read through all of the mess that has been written in this thread so far, I have quoted everyone who posted something incomplete, wrong, confusing, or a combination of all the above, and I made corrections. I also have answered some of bcube's questions.

This post is meant as a "read through". It's too long to be anything else. In a way, you might be able to think of this as an outline of your article. You will see me repeat some things over and over, because I have either answered your questions (the first post) or am correcting people's responses.

Lastly, before I begin, elrog, please think about what you post before you post it. Your edit was even more unclear than your original post. Also, enable spell check on your browser. By the way you worded things so far, how you skipped important transitions to draw conclusions, and just confusing terms, I'd say that you should seriously rewrite what you have to say over and over again until it is reasonably clear (I have to do the same thing).
[HR][/HR]
Let's start from the beginning of the thread.
- is it correct to say that (in case of Rubik´s cube NxNxN) orientation problem is caused by the fact that those parts, which should be solved by "even number of moves" were in fact solved in "odd number of moves"?
I get what you are saying.
If you use an odd number of moves, you can only solve the pieces which were in an odd permutation from the scramble.
If you use an even number of moves, you can only solve the pieces which were in an even permutation from the scramble.

But even what I just wrote is a little vague because what piece types are we talking about? Which slices are we referring to?

To be specific, the orientation problem/OLL parity on the 4x4x4 is caused when you either:
(1) Start with an even permutation of wing edges and solve the centers with an odd number of inner layer quarter slice turns.
(2) Start with an odd permutation of wing edges and solve the centers with an even number of inner layer quarter slice turns.

- is permutation problem caused by the fact that some of (in case of Rubik´s cube NxNxN) parts seem to be identical to us (edges (wing edges or whatever they are called), to be more precise)?
The wing edges can seem to be the same to beginner solvers, but when you solve using the K4 or cage methods, their difference is more apparent. I will touch on this, but I'm going to put it in a spoiler because wing edges actually have nothing to do with your hunch (for OLL parity)
Suppose we wanted to just compare the two red and white wing edges on a 4x4x4. Since they are the same two colors, they obviously form the same composite edge (or so called dedge on the 4x4x4). If we were to imagine them to both be jammed into the same composite edge, only two situations can happen:

(1) Both are oriented correctly (that is, the composite edge is oriented).
(2) Both are unoriented (that is, the composite edge is unoriented).

So you cannot have half of the same composite edge on the 4x4x4 be oriented and the other half be unoriented. Thus the only conclusion we can draw from this is that the two wing edges in every pair are distinctly different pieces: if one wing edge is correctly oriented on the left side of the composite edge, then it must be unoriented on the right side of the same composite edge, and vice versa.
(*)Your instinct isn't wrong by any means because on regular 6-color cube sizes 4x4x4 and greater, the center pieces are not distinguishable from each other. So what's relevant to your question is, suppose we were to consider the 5x5x5 or greater supercube (picture cube). If we were to solve the centers correctly on a supercube, then that guarantees that we will not get an odd permutation problem in the wing edges (such as the "one edge flip").

On the v-cube solutions page, they mention this.
http://solutions.v-cubes.com/solutions2/#edges (Scroll down and read their paragraph under "What causes this phenomena?" Then read the "interesting note").

(**) 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).
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).

Now I'm going to quote people's responses and comment/correct.

A parity error occurs when you have an odd number of 2-cycles in your permutation.
So for example, to get OLL parity on the 4x4, it means that you have done an odd number of slice turns, since this creates a 4-cycle of edges and 2 4-cycles of centres. Because the centres have no orientation in themselves, this error is not noticed so the only one that comes to light is the wing 4-cycle, which can be reduced down to a wing 2-cycle (Parity) during a normal reduction solve.
An odd permutation (you called it "parity error") occurs when you can solve/generate your permutation using a product of an odd number of 2-cycles.

So for example, to get OLL parity on the 4x4x4, it means that:
(a) you have done an odd number of inner layer slice quarter turns to solve the centers of a scramble that contained an even number of inner layer slice quarter turns
OR
(b) you have done an even number of inner layer slice quarter turns to solve a scramble that contained an odd number of inner layer quarter turns.

Why MaeLSTRoM mentioned the centers on a 4x4x4, I don't know, because the parity of the centers of the 4x4x4 and the parity of the wing edges on the 4x4x4 are independent. But just to explain what he mentioned but didn't to explain enough
One inner layer slice quarter turn on a 4x4x4 does a 4-cycle of the wing edges in that slice and a 2 4-cycle of all of the center pieces in that slice.

(See page 5--the 6th page on the PDF, but it is numbered 5--in my document for a good visual...you may NOT use any images in my document in your article, unless you credit my name in a caption below it).

A 4-cycle is an odd permutation because it takes 3 2-cycle swaps to solve it.
(An odd permutation is generated by a product of an odd number of 2-cycles)
A 2 4-cycle is an even permutation because it takes 2(3) = 6 2-cycle swaps to solve it.
(An even permutation is generated by a product of an even number of 2-cycles)

Therefore an inner layer slice quarter turn does not ever change the parity state of the centers of the 4x4x4 (that is, if all 24 center pieces on the 4x4x4 were in an odd permutation before an inner layer slice quarter turn is applied, they will still be in an odd permutation after an inner layer slice quarter turn is applied and vice versa). This is because
[initially even permutation of centers] + [even permutation (2 4-cycle) of centers from inner layer slice quarter turn] = even permutation of centers
AND
[initially odd permutation of centers] + [even permutation (2 4-cycle) of centers from inner layer slice quarter turn] = odd permutation of centers.

So when MaeLSTRoM wrote
this error is not noticed so the only one that comes to light is the wing 4-cycle
he was trying to, but did not succeed in saying clearly that an inner layer slice quarter turn on a 4x4x4 changes the parity state of the wing edges (if the parity of the wing edges was even before, it will be odd after the inner layer slice quarter turn and vice versa), but it does not change the parity state of the X-center pieces of the 4x4x4.

, which can be reduced down to a wing 2-cycle (Parity) during a normal reduction solve.
A 4-cycle can be reduced to a 2-cycle by doing the proper 3-cycle. For example, consider {1,2,3,4} to be the solved state. Therefore one 4-cycle is {4,1,2,3}.

If we do the 3-cycle 1→4→3→1, then we have {1,3,2,4}, which is now only a 2-cycle away from the solved state (now, only 2 and 3 are swappped).
Because the centres have no orientation in themselves,
Huh? First of all, wings don't have any orientation either, and since you have, as you said 2 4-cycles of centres they're still in an even permutation

Ok, well, what I meant was like you can swap 2 centres and as long as their on the same face, it doesn't matter.
I'm not sure what he meant by "it doesn't matter", but in case of the parity state of the wing edges for OLL parity, as I've said already, the parity of the X-center pieces is independent of the parity of wing edges. So whether we swap just two X-center pieces in one face on the 4x4x4 (an odd permutation) or swap two X-center pieces in two different faces (so swap two pairs of X-center pieces in two different faces) (an even permutation), etc., it won't affect the parity state of the wing edges.

I suppose that's more a permutation thing.
It is entirely a permutation thing.

And yeah, I know that 2*4 cycles is a non-parity but I was just appealing to the point about being pieces you can't see being swapped in the OP.
I suppose he meant non-parity = even permutation. I assume "pieces you can't see being swapped in the OP" means that, after a 4x4x4 OLL parity algorithm is executed on a 4x4x4 whose centers are already solved, you can't tell it did any change to the permutation of the centers because it swapped same color center pieces with each other (note, I did NOT just say parity state, but change in permutation). On the 4x4x4 supercube, however, we can see after an OLL parity algorithm that some of the center pieces are swapped with each other. However, there are reasonably short supercube safe 4x4x4 OLL parity algorithms (after they are fully executed, no X-center pieces are affected) such as:
r' U2 r' U' r U r U r' D' f2 r f2 r' D U2 r' U r U r U r

1 - No. It is true that any position reached in an odd number of moves has to be solved with an odd number of moves and it is the same using even numbers of moves, but this only applies if you count moves in QTM.
For mathematical odd permutation issues, you must ALWAYS count the number of inner layer slice quarter turns.

However, this is not the cause of the orientation parity. The parity in the orientation of edges (excluding center edges on odd cubes) is decided by the permutation of the edges.
This is a redundant statement.

An edge gets flipped if it is placed in its opposite orbit.
You are misusing the term orbit here. There is only on orbit of wing edges on the 4x4x4, for example. An orbit means the realm of where it's possible for pieces in that orbit to be able to go with legal moves, just as the orbital patterns of the planets in our solar system are an outline of where they will move every revolution around the sun.

Because edge permutation is always solvabe with 3-cycles after corners are permuted,
You mean, after corners are SOLVED or are made into an even permutation.

the orienation parity is solvable with 3-cycles.
Orientation parity means that you have paired up all edges and are attempting to correctly place and oriented them but an odd number of pairs of wing edges are oriented incorrectly oriented. That is, you start out with an odd permutation of wing edges. Since you start out with an odd permutation of wing edges, it is never solvable with 3-cycles. One odd permutation cycle (2-cycle, 4-cycle, etc.) is needed to break the odd permutation and then 3-cycles (or one commutator, if you will) can solve the rest.

3-cycles are always an even number of moves.
For the wing edges, you mean an even number of inner layer quarter turns, right?
l2 U' l D2 l' U l D2 l (9)

Refering back to my first statement shows that it will always take an even number of moves to achieve a position with corner permutation solved and orientation parity present. The same is true if orientation parity is not present.
This is wrong on every level.

2 - I would say no, but it is more user disgreation.
I'm not sure what disgreation means.

Permutation parity is always solvable with 3-cycles.
When elrog said permutation parity, he was referring to PLL parity.

It only appears that you are swapping 2 pieces because the pieces look the same.
He meant that it appears you are swapping just two composite edges. But you are doing two separate 2-cycle swaps of wing edges (a 2 2-cycle).

So, mathmatically it isn't parity.
Oops, you forgot to mention what I just did above to be able to come to this correct conclusion. Transitions please

If you are not viewing parity mathmatically, it is any position not solvale by a given moveset that it appears to be in.
qqwref already mentioned this in all glory detail.

On a supercube (distinguishes the permutation and orientation of every piece)
The orientation of corners, the orientation of middle edges, and the permutation of wing edges (commonly people consider a wing edge on the wrong side of its composite edge to be unoriented, but it's actually just in the wrong location) are perfectly distinguishable on a non-supercube. As I've said before, the permutation of center pieces on the non-supercube are not distiguishable because same color centers can be swapped with each other (speaking from a supercube perspective). This allows there to be
$$\left( \frac{4^{6}}{2} \right)^{n\bmod 2}\left( \frac{4!^{6}}{2} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }$$
solved positions of the regular nxnxn cube, but there is only one unique solved position of the 3x3x3 and larger supercubes (assuming that all center pieces have different numbers, for example).

I guessing that elrog also mentioned the term orientation because some consider the fixed centers to have orientation rather than permutation. It is tricky to decide whether fixed centers on the odd supercube have permutation or orientation because, from one angle, the parity of fixed centers is interdependent with the parity of the corners. With corners and middle edges, never is the case when their orientations are affected by their parity state. On the other hand, orientation means for a piece to be rotated differently in the same location (since fixed centers never move from their locations, but are just rotated, we can see the similarity here).

You will have permutation parity, but it won't appear to be in the group you need it to be in in the first place.
This makes absolutely no sense.

So I would say that no, permutation parity is not caused by us not seeing the difference between pieces,
Assuming that he meant to say orientation parity, only on the 4x4x4 is this statement true because, as I have said before, the parity state of the centers of the 4x4x4 (the X-centers) is independent of the parity of wing edges. As I have said already, see the v-cube solutions page. On the 5x5x5 and greater cubes, this is absolutely true for, say, the "one edge flip" and non-X-center pieces.

If he meant to say permutation parity (PLL parity), then, if you are solving using reduction, this is definitely TRUE for the 4x4x4. I have already elebarated why before. (go back and find **)

but it is only possible to destinguish as parity if you can't tell the difference.
This makes absolutely no sense.

3 - I have no idea about how a Square 1 even turns. I've only ever seen pictures of it on google images. I'll let other people answer this one.
Same here.

EDIT: I made a few mistakes in answering the first question, but the answer is still no.
Wrong. He was correct, but what he was trying to say just didn't come out right.

First of all, I toook for granted that all perutation parities were solvable by 3-cycles after reading qqwref's post, but some of them are not.
I'm not sure what you are now also considering permutation parity on the nxnxn cube besides different versions of PLL parity...that is, a more general definition of PLL parity on the 4x4x4 is when there is an "odd permutation of composite edges". If you are considering 2-corner swaps to also be PLL parity, technically they are not because we can always solve the corners and be left just with PLL parity of the edges. So PLL parity (permutation parity) can always be solved with 3-cycles.

Second, I talked about how the corner cycle is connected with the edges, but this is only tue of the middle edges on large cubes.
It's also true for the middle edges on the 3x3x3.

Heres how it should be: A edge cycle and its opposite (a cycle the same distance from the middle edge but on the other side)
This makes absolutely no sense on any level. When you say "edge" are you referring to middle edge or wing edge. An cycle describes a permutation. A permutation has an opposite? Do you mean the inverse?

can be changed from an odd perutation to an even permutation and vice-versa with a single 90 degree slice move. Each 90 degree slice move will do this and no outer layer turns will affect this because it causes a 4-cycle in both of the 2 opposite cycles causing the permutation to stay the same.
Translation:
Just as an inner layer slice quarter turn does a 2 4-cycle of X-centers on the 4x4x4, an outer layer quarter turn does a 2 4-cycle of wing edges in that slice. Thus the parity state of the wing edges is not changed by an outer layer quarter turn.

(See page 6 in my document for a visual)

You should also take note that the orientation parity can be solved by 3-cycles if you do not need to preserve orientation
Unlike the term "permutation parity" where no matter if someone personally believes that wing edges have orientation, while those who are more informed know that they only have permutations, the term "orientation parity" only refers to OLL parity/cases in which an odd number of dedges on the 4x4x4 are "flipped". So by the terms you used and the conclusion you have drawn, you are wrong again.

Some even prefer to call OLL parity permutation parity because, after all, wing edges only have permutations anyway. For example, the "single edge flip" case is a 2-cycle permutation parity. We can have 2-cycles which are "oriented" such as:

elrog

Member
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. I just choose not to waste too much of my time on it.

Heres one exampe:

Wrong. He was correct, but what he was trying to say just didn't come out right.
No, I was wrong. I incorecctly associated corner permutation with edge (excluding middle edges) permutation. Even if I was right, would it not be correct that sying it wrong is a mistake?

If you want me to do more examples, just say so.

EDIT: And I'm not going to download a spellchecker. I'll just wtire everything in microsoft word from now on and copy it to here.

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Christopher Mowla

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. I just choose not to waste too much of my time on it.

Heres one exampe:
I think I clearly implying that his #1 statement's hunch was correct, as I elaborated on about before in my post.

Even if I was right, would it not be correct that sying it wrong is a mistake?
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).

If you want me to do more examples, just say so.
If it makes things clearer for everyone else (as was the purpose of my post), be my guest. My goal wasn't to make you "look stupid", and I said that this post was mainly directed at bcube to clear things up. If you think I made you look stupid, then that's not my problem...

EDIT: And I'm not going to download a spellchecker. I'll just wtire everything in microsoft word from now on and copy it to here.
I would be cautious using Micorosoft Word because it sometimes changes the formatting of the text (maybe the size, font, etc.), but I never had too many problems with wordpad.

EDIT: I just realized that wordpad doesn't have a spellchecker by default, so just type everything in here or in wordpad and then copy and paste into word for spell checking only. Don't copy from Word to here unless you're sure the formatting is preserved as you like it.

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