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Talk:Commutator

2013-04-12T12:31:21Z

Circular17: /* Relationship of NA’=N and NB’=N */

== Addition of commutator GIF ==
I change PI to PN for consistency reasons.
I got rid of the table because it is now extraneous. However it is commented out in the raw text of this talk page in case you want to re-include it.


Circular17, I don't want to mess with your work and make large changes, but I think the part that is directly above the gif and reproduced below is not longer needed. I think in this case "a picture/animation says a thousand words" 
xxxxxxxxxxxxxxxxxxxx 
We notice that N A and N B are used as temporary storage :
* The content of the intersection, PN, is first hidden by A in its storage N A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of N A' is brought by A, and is then hidden by B in N B, so that the only transformation applied to these pieces is A.
* The content of N B' is brought by B, and then is placed immediately in N A' by A'. So the transformation applied to these pieces is B A'.

In other words, all the commutator does is :
* PA goes A
* PB goes B A'
* PN goes B'
xxxxxxxxxxxxxxxxx 

That being said, I think what might be really helpful is if below the animation, we talk about a "real life example". I'll see if I can put something together.[[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 15:18, 28 November 2012 (EST)

== Changes by Circular17 ==

[[User:Circular17|Circular17]] comment >> http://en.wikipedia.org/wiki/User_talk:Dave3457#Commutator_on_Speedsolving_wiki

:I would be happy to look over your work. Keep in mind that I not sure that I always receive an email notification when a page I'm watching on Speedsolving Wiki (SSW) is changed . I did receive one for your March 30 changes however. My watchlist only showed one of your March 30 edits. (Maybe that's intended)
: I am looking things over now, it might take a couple of days.
: Aren't commutators wonderful ! :) I just discovered (using a program) a wonderful 2-gen one that twists 2 corners. Its the only one of its kind that I know of, it may be my favorite comm being a 2-gen and all. Check it out below. (Note: I personally cut and paste new algorithms I'm learning into CubeTwister)
:Its not that hard to understand how this comm works, you don't have to memorize it, follow the two corner-edge pairs on each side of the twisted corner and how the twisted corner moves between them.
:R2DRD'RDR2 D2 R2D'R'DR'D'R2 D2 (16,22)
:[ R2DR D' RDR2, D2] OR [ R2DR D' RDR2, D] - [[User:Dave3457|Dave3457]] ([[User talk:Dave3457#top|talk]]) 04:05, 31 March 2013 (UTC)

::I don't know how these notifications works.
::That's an interesting sequence indeed.
::I would reflect it like this :
::R2U'R'UR'U'R2 U' R2URU'RUR2 U
::so that it can be used for orienting corners of the last layer on top.
::or for left hand :
::L2ULU'LUL2 U L2U'L'UL'U'L2 U'
::--[[User:Circular17|Circular17]] ([[User talk:Circular17|talk]]) 02:46, 31 March 2013 (EDT)
:::Ya, any commutator I post will certainly be designed for those who solve the top layer last. Myself, I don't solve for speed and solve the bottom layer last. In fact I often do not rotate the cube at all but keep each face pointing in the same direction all of the time. That way I don't use muscle memory to perform an algorithm and force myself to understand the algorithms more deeply.
:::- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 23:07, 8 April 2013 (EDT)
:I created two images to help with the defining of the variables and took what you wrote and expanded it. I've noticed that your writing is very "dense". Judging from your WP User page, I'm assuming that English is not your first language although you wouldn't guess judging from your grammar. (you do occasionally "miss" an odd word however). (For the record, I fully appreciate that my own grammar and spelling is very poor).
::You are right, English is not my first spoken language. But I started it very early with programming languages.
:With regards to your writing, may I suggest that you make it less "dense". I suspect that you are a mathematician. [I can't be sure because I can't read your user page :)] Whether or not you are a mathematician you tend to take the symbolic shortcuts that mathematicians tend to.
::You are right, I've studied mathematics and IT.
:For example I changed "When '''J''' and '''K''' have an empty intersection, A and B commute,..." to "When '''J''' and '''K''' have no intersection, A and B commute,..."
::Oh that's right. I'll try to think about it. It's not really a shortcut, but that there is a mathematical object describing the intersection which is called the empty set. But I completely understand that most people would be confused.
:That being said I appreciate that as things get more technical as you go into more and more detail, not using the concise language of mathematics becomes more and more difficult.
::Yeah, that's why I've done that, because I was lost myself.
:Because of my desire to share the wonders of the commutator to the general non-mathematical population I, personally, will be focusing on the basics here and will be trying to make those basics as clear as possible using images and "extended" language. (Also I'm not sure I'm going to understand, or be able to take the time to understand, the later stuff)
::Your explanations are very good. Thanks.
:I hope you don't mind but I'm planning on modifying and "extending" the first and second sections quite heavily.
::Have fun :)
:I've only glanced at the new stuff that you have written, but from what I've read it seems pretty "heady".
::Yes, it was hard for me to do this mathematical work, so I can understand that it's not very readable.
:My general thoughts at this point are that we can expect each reader to stop reading the page when things get to difficult for them. Or to put it another way, they will stop when the amount of effort required is not worth the reward. Toward this end, I personally, would strongly suggest beginning with the easiest stuff and working toward the more difficult, even if its not the "correct" order from a mathematician's point of view. For example, you presently begin with the example case [M', U2] = M' U2 M U2 which rotates edges, however the case you mention later, [R' D' R D R' D' R, U] which twists two corners would be much easier to understand. I could see me creating a series of images or an animation for that one.
::Absolutely. I'm ok with that.

:Also, I'm thinking that, at least when we are dealing with the basic stuff, that things would be easier to understand if we dealt with an example right from the start rather than talk in the abstract. For example you wrote...
:::If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.
: While I know I'm speaking from the point of view of a guy with only an average IQ, I think a lot of people besides me would not follow that easily either. If a diagram was associated with it I'm sure it would be a lot easier to follow.
::In fact, I'm thinking about specific moves while writing this. My goal was to make it general so that it can be applied to any situation. The quirk part and so on are the result of that exhaustive description.
:Personally, I thinking that since the B move is usually the rotation of the Up side, we should start with '''A inplace''' situations rather than the '''B inplace''' situations and that way we can talk about the easier to understand examples ...

:* A = "rotating a corner" like [R' D' R D R' D' R, U]
:* A = "flipping an edge" like [R' E' R2 E2 R', U]
:* A = "exchanging two corners" like [R' L' D2 R L, U]
:* A = "exchanging two edges" like [M2 D2 M2, U]
::Ok.
:'''Odds and ends:'''
:...I noticed that you referred to a single corner being "rotated". Personally I prefer to reserve the word "rotate" to when I'm dealing with 3-cycles and such. I instead try to refer to them as being "twisted"
:- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 00:45, 9 April 2013 (EDT)
::Ok.
:I want to make a strong suggestion. Do you know how later on you use subscripts to identify pieces such as '''P'''A,'''P'''B,'''P'''N . May I suggest that we also use subscripts to identify the locations of '''J''' and '''K'''. May I suggest the capital letter '''L''' which would stand for '''L'''ocation. ie '''J'''=>'''L'''A, '''K'''=>'''L'''B. This would greatly relieve the "mental strain" one experiences trying to remember what they stand for. Since '''J''' and '''K''' are bolded one could simply do a mass "find and replace" operation. (although not in the images).
:While technically we should do it for '''N''', '''N'''A' and '''N'''B' as well, given how often those locations of referenced, I think that might be to much clutter and not really necessary. I would be happy to do the switch over.
:- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 15:23, 9 April 2013 (EDT)
::I was tempted to use the letter L at some point, but I noticed that it can be confused with the move L (left). With subscript however, this may be ok. About '''N''', are you thinking about '''L'''N ? Not sure it would be better, so I suppose we can keep it that way. And it sounds good, I think. N like intersection, like N-tersection. --[[User:Circular17|Circular17]] ([[User talk:Circular17|talk]]) 07:49, 12 April 2013 (EDT)

== Relationship of NA’=N and NB’=N ==

:Circular17, I suspect that there is something misleading about the way things are being presented right now. There is a bit of a suggestion that NA’=N and NB’=N are not perfectly symmetrical. (or maybe they aren’t)
:Specifically…the two NA’=N examples that I have discussed…
:[R' D' R D R' D' R, U] and [M2 D2 M2, U]
:have their NB’=N symmetric counterparts which accomplish the exact same thing…
:[U, R’ D R D’ R’ D R] and [U, M2 D2 M2]
:It seems that in these two cases [A,B] = [B,A’].
:However it isn’t true for…
:[U2,M] which is a NA’=N version
:and
:[M', U2] which is a NB’=N version
:ie [A,B] ≠ [B,A’] in this case.
:I guess it is because of the “overlap”. That Quirk thing that you talk about but that I don’t understand yet. Am I right on that?
:Whatever the case, we don’t want to give the impression that NA’=N and NB’=N are any different if they aren’t. I guess it comes down to whether or not you can do the same thing with both by just exchanging and reversing things. To what degree are they different?
:-- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 04:29, 12 April 2013 (EDT)
::I suppose you are talking about [A,B]' = (ABA'B')' = BAB'A' = [B,A]. It's true.
::NA'=N and NB'=N are symmetrical cases. But it is important to distinguish them, because A and B do not have the same roles, exchanging them is like a time reversal. --[[User:Circular17|Circular17]] ([[User talk:Circular17|talk]]) 08:31, 12 April 2013 (EDT)

Talk:Commutator

2013-04-12T11:51:24Z

Circular17: /* Changes by Circular17 */ same indent

Talk:Commutator

2013-04-12T11:49:41Z

Circular17: /* Changes by Circular17 */ response

== Addition of commutator GIF ==
I change PI to PN for consistency reasons.
I got rid of the table because it is now extraneous. However it is commented out in the raw text of this talk page in case you want to re-include it.


Circular17, I don't want to mess with your work and make large changes, but I think the part that is directly above the gif and reproduced below is not longer needed. I think in this case "a picture/animation says a thousand words" 
xxxxxxxxxxxxxxxxxxxx 
We notice that N A and N B are used as temporary storage :
* The content of the intersection, PN, is first hidden by A in its storage N A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of N A' is brought by A, and is then hidden by B in N B, so that the only transformation applied to these pieces is A.
* The content of N B' is brought by B, and then is placed immediately in N A' by A'. So the transformation applied to these pieces is B A'.

In other words, all the commutator does is :
* PA goes A
* PB goes B A'
* PN goes B'
xxxxxxxxxxxxxxxxx 

That being said, I think what might be really helpful is if below the animation, we talk about a "real life example". I'll see if I can put something together.[[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 15:18, 28 November 2012 (EST)

== Changes by Circular17 ==

[[User:Circular17|Circular17]] comment >> http://en.wikipedia.org/wiki/User_talk:Dave3457#Commutator_on_Speedsolving_wiki

:I would be happy to look over your work. Keep in mind that I not sure that I always receive an email notification when a page I'm watching on Speedsolving Wiki (SSW) is changed . I did receive one for your March 30 changes however. My watchlist only showed one of your March 30 edits. (Maybe that's intended)
: I am looking things over now, it might take a couple of days.
: Aren't commutators wonderful ! :) I just discovered (using a program) a wonderful 2-gen one that twists 2 corners. Its the only one of its kind that I know of, it may be my favorite comm being a 2-gen and all. Check it out below. (Note: I personally cut and paste new algorithms I'm learning into CubeTwister)
:Its not that hard to understand how this comm works, you don't have to memorize it, follow the two corner-edge pairs on each side of the twisted corner and how the twisted corner moves between them.
:R2DRD'RDR2 D2 R2D'R'DR'D'R2 D2 (16,22)
:[ R2DR D' RDR2, D2] OR [ R2DR D' RDR2, D] - [[User:Dave3457|Dave3457]] ([[User talk:Dave3457#top|talk]]) 04:05, 31 March 2013 (UTC)

::I don't know how these notifications works.
::That's an interesting sequence indeed.
::I would reflect it like this :
::R2U'R'UR'U'R2 U' R2URU'RUR2 U
::so that it can be used for orienting corners of the last layer on top.
::or for left hand :
::L2ULU'LUL2 U L2U'L'UL'U'L2 U'
::--[[User:Circular17|Circular17]] ([[User talk:Circular17|talk]]) 02:46, 31 March 2013 (EDT)
:::Ya, any commutator I post will certainly be designed for those who solve the top layer last. Myself, I don't solve for speed and solve the bottom layer last. In fact I often do not rotate the cube at all but keep each face pointing in the same direction all of the time. That way I don't use muscle memory to perform an algorithm and force myself to understand the algorithms more deeply.
:::- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 23:07, 8 April 2013 (EDT)
:I created two images to help with the defining of the variables and took what you wrote and expanded it. I've noticed that your writing is very "dense". Judging from your WP User page, I'm assuming that English is not your first language although you wouldn't guess judging from your grammar. (you do occasionally "miss" an odd word however). (For the record, I fully appreciate that my own grammar and spelling is very poor).
::You are right, English is not my first spoken language. But I started it very early with programming languages.
:With regards to your writing, may I suggest that you make it less "dense". I suspect that you are a mathematician. [I can't be sure because I can't read your user page :)] Whether or not you are a mathematician you tend to take the symbolic shortcuts that mathematicians tend to.
::You are right, I've studied mathematics and IT.
:For example I changed "When '''J''' and '''K''' have an empty intersection, A and B commute,..." to "When '''J''' and '''K''' have no intersection, A and B commute,..."
::Oh that's right. I'll try to think about it. It's not really a shortcut, but that there is a mathematical object describing the intersection which is called the empty set. But I completely understand that most people would be confused.
:That being said I appreciate that as things get more technical as you go into more and more detail, not using the concise language of mathematics becomes more and more difficult.
::Yeah, that's why I've done that, because I was lost myself.
:Because of my desire to share the wonders of the commutator to the general non-mathematical population I, personally, will be focusing on the basics here and will be trying to make those basics as clear as possible using images and "extended" language. (Also I'm not sure I'm going to understand, or be able to take the time to understand, the later stuff)
::Your explanations are very good. Thanks.
:I hope you don't mind but I'm planning on modifying and "extending" the first and second sections quite heavily.
::Have fun :)
:I've only glanced at the new stuff that you have written, but from what I've read it seems pretty "heady".
::Yes, it was hard for me to do this mathematical work, so I can understand that it's not very readable.
:My general thoughts at this point are that we can expect each reader to stop reading the page when things get to difficult for them. Or to put it another way, they will stop when the amount of effort required is not worth the reward. Toward this end, I personally, would strongly suggest beginning with the easiest stuff and working toward the more difficult, even if its not the "correct" order from a mathematician's point of view. For example, you presently begin with the example case [M', U2] = M' U2 M U2 which rotates edges, however the case you mention later, [R' D' R D R' D' R, U] which twists two corners would be much easier to understand. I could see me creating a series of images or an animation for that one.
::Absolutely. I'm ok with that.

:Also, I'm thinking that, at least when we are dealing with the basic stuff, that things would be easier to understand if we dealt with an example right from the start rather than talk in the abstract. For example you wrote...
:::If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.
: While I know I'm speaking from the point of view of a guy with only an average IQ, I think a lot of people besides me would not follow that easily either. If a diagram was associated with it I'm sure it would be a lot easier to follow.
::In fact, I'm thinking about specific moves while writing this. My goal was to make it general so that it can be applied to any situation. The quirk part and so on are the result of that exhaustive description.
:Personally, I thinking that since the B move is usually the rotation of the Up side, we should start with '''A inplace''' situations rather than the '''B inplace''' situations and that way we can talk about the easier to understand examples ...

:* A = "rotating a corner" like [R' D' R D R' D' R, U]
:* A = "flipping an edge" like [R' E' R2 E2 R', U]
:* A = "exchanging two corners" like [R' L' D2 R L, U]
:* A = "exchanging two edges" like [M2 D2 M2, U]
::Ok.
:'''Odds and ends:'''
:...I noticed that you referred to a single corner being "rotated". Personally I prefer to reserve the word "rotate" to when I'm dealing with 3-cycles and such. I instead try to refer to them as being "twisted"
:- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 00:45, 9 April 2013 (EDT)
::Ok.
::I want to make a strong suggestion. Do you know how later on you use subscripts to identify pieces such as '''P'''A,'''P'''B,'''P'''N . May I suggest that we also use subscripts to identify the locations of '''J''' and '''K'''. May I suggest the capital letter '''L''' which would stand for '''L'''ocation. ie '''J'''=>'''L'''A, '''K'''=>'''L'''B. This would greatly relieve the "mental strain" one experiences trying to remember what they stand for. Since '''J''' and '''K''' are bolded one could simply do a mass "find and replace" operation. (although not in the images).
::While technically we should do it for '''N''', '''N'''A' and '''N'''B' as well, given how often those locations of referenced, I think that might be to much clutter and not really necessary. I would be happy to do the switch over.
::- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 15:23, 9 April 2013 (EDT)
:::I was tempted to use the letter L at some point, but I noticed that it can be confused with the move L (left). With subscript however, this may be ok. About '''N''', are you thinking about '''L'''N ? Not sure it would be better, so I suppose we can keep it that way. And it sounds good, I think. N like intersection, like N-tersection. --[[User:Circular17|Circular17]] ([[User talk:Circular17|talk]]) 07:49, 12 April 2013 (EDT)

== Relationship of NA’=N and NB’=N ==

:Circular17, I suspect that there is something misleading about the way things are being presented right now. There is a bit of a suggestion that NA’=N and NB’=N are not perfectly symmetrical. (or maybe they aren’t)
:Specifically…the two NA’=N examples that I have discussed…
:[R' D' R D R' D' R, U] and [M2 D2 M2, U]
:have their NB’=N symmetric counterparts which accomplish the exact same thing…
:[U, R’ D R D’ R’ D R] and [U, M2 D2 M2]
:It seems that in these two cases [A,B] = [B,A’].
:However it isn’t true for…
:[U2,M] which is a NA’=N version
:and
:[M', U2] which is a NB’=N version
:ie [A,B] ≠ [B,A’] in this case.
:I guess it is because of the “overlap”. That Quirk thing that you talk about but that I don’t understand yet. Am I right on that?
:Whatever the case, we don’t want to give the impression that NA’=N and NB’=N are any different if they aren’t. I guess it comes down to whether or not you can do the same thing with both by just exchanging and reversing things. To what degree are they different?
:-- [[User:Dave3457|Dave3457]] ([[User talk:Dave3457|talk]]) 04:29, 12 April 2013 (EDT)

Commutator

2013-04-12T11:22:16Z

Circular17: /* Effect */ in general but we can find some special cases for example if A = B'

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows one to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==
[[File:COMM Effects section Definitions.png|450px|thumb|right|This image defines the terms '''J''', '''K''', '''N''', '''N'''A’,'''N'''B’ used to describe how the commutator works. ]]
To help us discuss how a commutator works we will first define some terms. 
The image on the right gives a visual indication of which pieces of a cube are affected by the different sequences of a commutator. In the example commutator chosen, the sequence A is F and the sequence B is U, giving the simple commutator [F, U].
Referring to the image, the sequence A changes the pieces at a set of locations we will call '''J'''. The sequence B changes the pieces at a set of locations we will call '''K'''. We will call the intersection of '''J''' and '''K''', '''N'''. If '''J''' and '''K''' have no intersection, such as when A equals L and B equals R, then A and B are said to commute, and the commutator [L, R] does nothing. However if, as in the image, when '''J''' and '''K''' do have an intersection, A and B generally do not commute and the cube, as a result, is changed by the commutator. Note that the changes which this particular commutator make to the cube are rather difficult to understand, however more straight forward commutator examples will be discussed in a moment.

There is a set of locations that we will be calling '''N'''A'. This name is appropriate because the pieces can be identified by applying the inverse of A to the locations '''N'''. Referring to the lowest right hand cube, these particular locations are identified using two shades of grey. Pieces that are lighter grey are pieces that will be brought into the intersection region '''N''', pieces that are darker grey are pieces that are already within the intersection region and will remain in the region. Such pieces will however be moved within the region, as in this case, or will be twisted in place.

All of the above statements regarding the '''N'''A' region is also true for '''N'''B' region.

The only pieces that are affected by any commutator are the ones located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up after the sequences A and B are executed, will be put back into their original locations with sequences A' and B'. 
This restoration of all the pieces that are not in the union of '''N''', '''N'''A' and '''N'''B' is the reason the commutator is such a powerful tool.

Click the following thumbnail for an image similar to the one just discussed which instead uses the more common commutator [B' D B, U'].
[[File:COMM Effects section Example.jpg|70px ]]
=== Trivial case ===

When '''J''' and '''K''' have no intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

[[File:COMM_Effects_Twist_Corn.png|450px|thumb|right|This commutator twists two corners in place. It is an example of a commutator in which '''N'''A'='''N'''. [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy Applet of commutator] ]]
The image on the right breaks down a commutator that twists two corner pieces. In the type of commutator being discussed in this section, the sequence A does not move pieces into the intersection region but only moves or twists pieces that are already within the region. As a result, the '''N'''A' region and intersection region '''N''' are the same. In this particular commutator, the sequence A twists the single piece that is in the intersection region counter-clockwise. (Refer to the cube image which shows the effects of sequence A on the cube) Note that, as for all commutators, sequence A moves or twists all of the pieces that are in the '''J''' region, however any changes to that region which are outside of the intersection region are undone by sequence A'. If one performs this commutator on a cube or runs this [http://alg.garron.us/?alg=R-_D-_R_D_R-_D-_R__U_R-_D_R_D-_R-_D_R_U-&scheme=brogwy applet] they will see that the piece at '''N'''B' gets twisted clockwise when sequence A' is restoring the rest of the '''J''' region.

To appreciate sequence B's interaction with the intersection region one can observe the general nature of a commutator. Looking at the mathematical structure of a commutator, A B A' B' one can see that it has two overlapping [[conjugates]]. ie A B A' B' = (A B A') B' = A (B A' B'). (If one is unfamiliar with conjugates it is critical that one read the lead of the conjugate page now before continuing.) It is the action of the second conjugate, B A' B', that is apparent in this type of commutator. Because sequence A' only alters the contents of the intersection region, sequence B can be viewed as the initial setup move of a conjugate, A' can be viewed as the actual manipulation move and B' can be viewed as the restoring move. Regarding this specific commutator, because sequence A twists the corner piece at '''N'''A' counter-clockwise, the sequence A', which is acting as the middle manipulation sequence of a conjugate, twists the corner piece at '''N'''B' clockwise.
 
[[File:COMM_Eff_Exchange_Edges.jpg|450px|thumb|right|It is an example of a commutator in which '''N'''A'='''N'''. [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy Applet of commutator] ]]
In this next example, the sequence A again does not move pieces into the intersection region resulting in the '''N'''A' region and intersection region '''N''' being the same. However in this commutator, the sequence A moves pieces around within the intersection region. Specifically, it exchanges the only two pieces that are in the intersection region.

Everything else discussed regarding the previous example applies here as well. 
Here is an [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy applet of this commutator]

In summary: Sequence A initially alters the pieces that are in the intersection region, then sequence B, acting as the setup move of a conjugate, moves a different group of pieces into that intersection region. Sequence A' then, while restoring the rest of region '''J''', "reverse alters" those new pieces and sequence B' finally returns both altered groups of pieces back to their original locations.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')
* the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R] equivalent to R'U, thus going to the top-front-right position, replacing a part of PB

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are [[conjugate|conjugated]]. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-04-12T11:17:58Z

Circular17: /* Effect */ common rather than general

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows one to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==
[[File:COMM Effects section Definitions.png|450px|thumb|right|This image defines the terms '''J''', '''K''', '''N''', '''N'''A’,'''N'''B’ used to describe how the commutator works. ]]
To help us discuss how a commutator works we will first define some terms. 
The image on the right gives a visual indication of which pieces of a cube are affected by the different sequences of a commutator. In the example commutator chosen, the sequence A is F and the sequence B is U, giving the simple commutator [F, U].
Referring to the image, the sequence A changes the pieces at a set of locations we will call '''J'''. The sequence B changes the pieces at a set of locations we will call '''K'''. We will call the intersection of '''J''' and '''K''', '''N'''. If '''J''' and '''K''' have no intersection, such as when A equals L and B equals R, then A and B are said to commute, and the commutator [L, R] does nothing. However if, as in the image, when '''J''' and '''K''' do have an intersection, A and B do not commute and the cube, as a result, is changed by the commutator. Note that the changes which this particular commutator make to the cube are rather difficult to understand, however more straight forward commutator examples will be discussed in a moment.

There is a set of locations that we will be calling '''N'''A'. This name is appropriate because the pieces can be identified by applying the inverse of A to the locations '''N'''. Referring to the lowest right hand cube, these particular locations are identified using two shades of grey. Pieces that are lighter grey are pieces that will be brought into the intersection region '''N''', pieces that are darker grey are pieces that are already within the intersection region and will remain in the region. Such pieces will however be moved within the region, as in this case, or will be twisted in place.

All of the above statements regarding the '''N'''A' region is also true for '''N'''B' region.

The only pieces that are affected by any commutator are the ones located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up after the sequences A and B are executed, will be put back into their original locations with sequences A' and B'. 
This restoration of all the pieces that are not in the union of '''N''', '''N'''A' and '''N'''B' is the reason the commutator is such a powerful tool.

Click the following thumbnail for an image similar to the one just discussed which instead uses the more common commutator [B' D B, U'].
[[File:COMM Effects section Example.jpg|70px ]]
=== Trivial case ===

When '''J''' and '''K''' have no intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

[[File:COMM_Effects_Twist_Corn.png|450px|thumb|right|This commutator twists two corners in place. It is an example of a commutator in which '''N'''A'='''N'''. [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy Applet of commutator] ]]
The image on the right breaks down a commutator that twists two corner pieces. In the type of commutator being discussed in this section, the sequence A does not move pieces into the intersection region but only moves or twists pieces that are already within the region. As a result, the '''N'''A' region and intersection region '''N''' are the same. In this particular commutator, the sequence A twists the single piece that is in the intersection region counter-clockwise. (Refer to the cube image which shows the effects of sequence A on the cube) Note that, as for all commutators, sequence A moves or twists all of the pieces that are in the '''J''' region, however any changes to that region which are outside of the intersection region are undone by sequence A'. If one performs this commutator on a cube or runs this [http://alg.garron.us/?alg=R-_D-_R_D_R-_D-_R__U_R-_D_R_D-_R-_D_R_U-&scheme=brogwy applet] they will see that the piece at '''N'''B' gets twisted clockwise when sequence A' is restoring the rest of the '''J''' region.

To appreciate sequence B's interaction with the intersection region one can observe the general nature of a commutator. Looking at the mathematical structure of a commutator, A B A' B' one can see that it has two overlapping [[conjugates]]. ie A B A' B' = (A B A') B' = A (B A' B'). (If one is unfamiliar with conjugates it is critical that one read the lead of the conjugate page now before continuing.) It is the action of the second conjugate, B A' B', that is apparent in this type of commutator. Because sequence A' only alters the contents of the intersection region, sequence B can be viewed as the initial setup move of a conjugate, A' can be viewed as the actual manipulation move and B' can be viewed as the restoring move. Regarding this specific commutator, because sequence A twists the corner piece at '''N'''A' counter-clockwise, the sequence A', which is acting as the middle manipulation sequence of a conjugate, twists the corner piece at '''N'''B' clockwise.
 
[[File:COMM_Eff_Exchange_Edges.jpg|450px|thumb|right|It is an example of a commutator in which '''N'''A'='''N'''. [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy Applet of commutator] ]]
In this next example, the sequence A again does not move pieces into the intersection region resulting in the '''N'''A' region and intersection region '''N''' being the same. However in this commutator, the sequence A moves pieces around within the intersection region. Specifically, it exchanges the only two pieces that are in the intersection region.

Everything else discussed regarding the previous example applies here as well. 
Here is an [http://alg.garron.us/?alg=M2_D2_M2_U_M2_D2_M2_U-__&scheme=brogwy applet of this commutator]

In summary: Sequence A initially alters the pieces that are in the intersection region, then sequence B, acting as the setup move of a conjugate, moves a different group of pieces into that intersection region. Sequence A' then, while restoring the rest of region '''J''', "reverse alters" those new pieces and sequence B' finally returns both altered groups of pieces back to their original locations.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')
* the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R] equivalent to R'U, thus going to the top-front-right position, replacing a part of PB

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are [[conjugate|conjugated]]. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:42:46Z

Circular17: /* Conjugated part */ space

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

The conjugation is often used to provide a setup for some transformation. The conjugation ABA' is very similar to B, except that B is applied on other pieces as it would be without conjugation. For example, if a sequence exchange some pieces, or rotate some pieces, you can choose to exchange other pieces or rotate other pieces :

[F': [M2D2M2, U]]

where [M2D2M2, U] exchange the four top edges, so applying F' exchange the front-right edge instead of the front-top edge.

The intersection is '''N''' = front-top edge

The slot is '''N'''A' = '''N'''F = front-right edge

The conjugated part is:

'''C''' = ('''N'''B inter '''N''')B'A' = empty

Surprisingly, in this example, the conjugated part is empty. It means that no individual piece is actually following the whole conjugation. F' and [M2D2M2, U] are only connected so that some pieces are transfered through the intersection '''N'''. The transfer occur between both semi-conjugated parts.

In this simple case, we do not really distinguish the semi-conjugated parts. But it is possible to take this difference into account, and also to use a conjugated part.

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually.

'''H'''1 is extracted from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

==== Conjugated part ====

The conjugated part is used to move pieces inside the slot (the pieces do not enter nor leave the slot):

RUR'

where the intersection is '''N''' = top-right edge, top-right-front corner and top-right-back corner

and the slot is '''N'''A' = '''N'''R' = front-right edge, front-right-top corner and front-right-bottom corner

The conjugated part is :

'''C''' = ('''N'''B inter '''N''')B'A' = top-right-front corner

This piece is moved to the front-right-bottom corner and rotated so that the right facelet is moved to the bottom facelet.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:42:24Z

Circular17: /* Semi-conjugated parts */ typo

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

The conjugation is often used to provide a setup for some transformation. The conjugation ABA' is very similar to B, except that B is applied on other pieces as it would be without conjugation. For example, if a sequence exchange some pieces, or rotate some pieces, you can choose to exchange other pieces or rotate other pieces :

[F': [M2D2M2, U]]

where [M2D2M2, U] exchange the four top edges, so applying F' exchange the front-right edge instead of the front-top edge.

The intersection is '''N''' = front-top edge

The slot is '''N'''A' = '''N'''F = front-right edge

The conjugated part is:

'''C''' = ('''N'''B inter '''N''')B'A' = empty

Surprisingly, in this example, the conjugated part is empty. It means that no individual piece is actually following the whole conjugation. F' and [M2D2M2, U] are only connected so that some pieces are transfered through the intersection '''N'''. The transfer occur between both semi-conjugated parts.

In this simple case, we do not really distinguish the semi-conjugated parts. But it is possible to take this difference into account, and also to use a conjugated part.

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually.

'''H'''1 is extracted from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

==== Conjugated part ====

The conjugated part is used to move pieces inside the slot (the pieces do not enter nor leave the slot) :

RUR'

where the intersection is '''N''' = top-right edge, top-right-front corner and top-right-back corner

and the slot is '''N'''A' = '''N'''R' = front-right edge, front-right-top corner and front-right-bottom corner

The conjugated part is :

'''C''' = ('''N'''B inter '''N''')B'A' = top-right-front corner

This piece is moved to the front-right-bottom corner and rotated so that the right facelet is moved to the bottom facelet.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:38:35Z

Circular17: /* Conjugated part */ not enter nor leave

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

The conjugation is often used to provide a setup for some transformation. The conjugation ABA' is very similar to B, except that B is applied on other pieces as it would be without conjugation. For example, if a sequence exchange some pieces, or rotate some pieces, you can choose to exchange other pieces or rotate other pieces :

[F': [M2D2M2, U]]

where [M2D2M2, U] exchange the four top edges, so applying F' exchange the front-right edge instead of the front-top edge.

The intersection is '''N''' = front-top edge

The slot is '''N'''A' = '''N'''F = front-right edge

The conjugated part is:

'''C''' = ('''N'''B inter '''N''')B'A' = empty

Surprisingly, in this example, the conjugated part is empty. It means that no individual piece is actually following the whole conjugation. F' and [M2D2M2, U] are only connected so that some pieces are transfered through the intersection '''N'''. The transfer occur between both semi-conjugated parts.

In this simple case, we do not really distinguish the semi-conjugated parts. But it is possible to take this difference into account, and also to use a conjugated part.

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually. '''H'''1 is extract from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

==== Conjugated part ====

The conjugated part is used to move pieces inside the slot (the pieces do not enter nor leave the slot) :

RUR'

where the intersection is '''N''' = top-right edge, top-right-front corner and top-right-back corner

and the slot is '''N'''A' = '''N'''R' = front-right edge, front-right-top corner and front-right-bottom corner

The conjugated part is :

'''C''' = ('''N'''B inter '''N''')B'A' = top-right-front corner

This piece is moved to the front-right-bottom corner and rotated so that the right facelet is moved to the bottom facelet.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:36:41Z

Circular17: /* Usage */ conjugated part

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

The conjugation is often used to provide a setup for some transformation. The conjugation ABA' is very similar to B, except that B is applied on other pieces as it would be without conjugation. For example, if a sequence exchange some pieces, or rotate some pieces, you can choose to exchange other pieces or rotate other pieces :

[F': [M2D2M2, U]]

where [M2D2M2, U] exchange the four top edges, so applying F' exchange the front-right edge instead of the front-top edge.

The intersection is '''N''' = front-top edge

The slot is '''N'''A' = '''N'''F = front-right edge

The conjugated part is:

'''C''' = ('''N'''B inter '''N''')B'A' = empty

Surprisingly, in this example, the conjugated part is empty. It means that no individual piece is actually following the whole conjugation. F' and [M2D2M2, U] are only connected so that some pieces are transfered through the intersection '''N'''. The transfer occur between both semi-conjugated parts.

In this simple case, we do not really distinguish the semi-conjugated parts. But it is possible to take this difference into account, and also to use a conjugated part.

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually. '''H'''1 is extract from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

==== Conjugated part ====

The conjugated part is used to move pieces inside the slot :

RUR'

where the intersection is '''N''' = top-right edge, top-right-front corner and top-right-back corner

and the slot is '''N'''A' = '''N'''R' = front-right edge, front-right-top corner and front-right-bottom corner

The conjugated part is :

'''C''' = ('''N'''B inter '''N''')B'A' = top-right-front corner

This piece is moved to the front-right-bottom corner and rotated so that the right facelet is moved to the bottom facelet.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:23:49Z

Circular17: /* Conjugated part */ correction

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

==== Conjugated part ====

The conjugated part is used to move pieces inside the slot :

RUR'

where the slot is '''N'''A' = '''N'''R' = front-right edge, front-right-top corner and front-right-bottom corner

The conjugated part is :

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually. '''H'''1 is extract from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-02T10:11:39Z

Circular17: usage and link with commutators

A '''conjugate''' is a sequence of moves that consists in doing a sequence A, then a sequence B, and finally the inverse of the sequence A. In other words, a conjugate is an algorithm like A B A', where A and B can be any sequence of moves. In intuitive terms, it means: do something to set up another task which does something useful, and undo the setup.

== Mathematical definition ==

Given a group, a '''conjugate''' is an element of the form ghg', where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A'

It is sometimes written as [A: B] = A B A'

== Effect ==

In A B A', the sequence A is the setup move or sequence and A' puts the cube in its original state. In between, there is an embedded sequence B. Alone, this sequence would affect some set of locations '''K''' on the cube. Because of the conjugation, the affected pieces will be '''K'''A', which means that the locations will be transformed by the inverse of A.

Indeed the sequence B will be applied on a cube which space has been modified by A. The sequence B is supposed to change some pieces in the locations '''K''' and A consists in bringing the piece you want to change to the location '''K'''. Once you've done the sequence B, you bring those pieces back with A'.

=== Trivial case ===

If the set of locations '''K''' is not modified by A, then the conjugation does not change anything, and A B A' = B.

=== General case ===

Pieces can be affected by the following subsequence of ABA':
* by B only: this part is outside of the conjugation
* by AB: this is the first semi-conjugated part
* by BA': this is the second semi-conjugated part
* by ABA': this is the conjugated part

Let's call '''J''' the set of pieces modified by A, and '''N''' the intersection of '''J''' and '''K'''.

The initial location of the outside part is:
'''O''' = (('''K''' \ '''N''')B \ '''N''')B'

The initial location of the first semi-conjugated part is:
'''H'''1 = ('''N'''B \ '''N''')B'A'

The initial location of the second semi-conjugated part is:
'''H'''2 = '''N'''B' \ '''N'''

The initial location of the conjugated part is:
'''C''' = ('''N'''B inter '''N''')B'A'

The slot location is:
'''N'''A'

=== Usage ===

==== Conjugated part ====

The conjugated part is used to provide a setup for some transformation. Regarding this part, the conjugation ABA' is very similar to B, except that B is applied on other pieces as it would be without conjugation. For example, if a sequence exchange some pieces, or rotate some pieces, you can choose to exchange other pieces or rotate other pieces :

[F': [M2D2M2, U]]

where [M2D2M2, U] exchange the four top edges, so applying F' exchange the front-right edge instead of the front-top edge.

==== Semi-conjugated parts ====

Semi-conjugated parts can be used to move pieces inside or outside a slot. This is widely used to insert or extract a pair of pieces in [[F2L]] or to insert or extract corners or edges individually. '''H'''1 is extract from the slot and '''H'''2 is inserted into the slot.

Example: RU'R'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U' \ '''N''')UR' = top-right-front corner and front-right edge

These pieces are extracted from the slot and replace the top-back edge and top-back-left corner.

'''H'''2 = '''N'''U \ '''N''' = top-front-right corner and top-front edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-bottom corner.

Example: RUR'

'''N''' = top-right edge, top-right-back and top-right-front corners

Slot location : '''N'''A' = '''N'''R' = top-right-front corner, front-right edge and front-right-bottom corner

'''H'''1 = ('''N'''U \ '''N''')U'R' = front-right edge and front-right-bottom corner

These pieces are extracted from the slot and replace the top-front edge and top-front-left corner.

'''H'''2 = '''N'''U' \ '''N''' = top-back-left corner and top-back edge

These pieces are inserted into the slot, replacing front-right edge, and front-right-top corner.

=== Link with commutators ===

The outside part '''O''' is only affected by B', so it can be put back into place by adding B' to the sequence, giving ABA'B', which is the [[commutator]] of A and B.

The first semi-conjugated part '''H'''1, which was affected by AB, is then only affected by A, becoming the container of pieces PA.

The second semi-conjugated part '''H'''2 is split into two parts:
* one part which is affected by B', so affected by BA'B', and is then a conjugated part of the commutator
* one part which is not affected by B', so affected only by BA', becoming the container of pieces PB.

The conjugated part is split into two parts:
* one part which is affected by B', becoming the quirk part
* one part which is not, becoming a conjugated part of the commutator

Pieces that were not affected by the conjugate, but that are affected by the last B' move, are the PN pieces of the corresponding commutator.

== See also ==
* [[Commutator]]
* [[Commutators and Conjugates]]

[[Category:Puzzle theory]]

Conjugate

2013-04-01T09:57:24Z

Circular17: /* Effect */ general case

Commutator

2013-04-01T09:37:52Z

Circular17: /* Quirk and conjugated part */ line ending

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')
* the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R] equivalent to R'U, thus going to the top-front-right position, replacing a part of PB

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are [[conjugate|conjugated]]. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-04-01T09:37:32Z

Circular17: /* Quirk and conjugated part */ wiki conjugate

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')
* the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R] equivalent to R'U, thus going to the top-front-right position, replacing a part of PB

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are [[conjugate|conjugated]]. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-04-01T09:29:48Z

Circular17: /* Example */ explanations of the remaining corner

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')
* the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R] equivalent to R'U, thus going to the top-front-right position, replacing a part of PB

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-04-01T09:22:09Z

Circular17: /* Example */ summary

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :
* PN, i.e. front-right edge and front-right-bottom corner, goes F
* PA, i.e. top-right edge and top-right-back corner, goes R'
* PB, i.e. front-bottom edge, goes F'R
* the front-right-top corner goes U'
* the front-bottom-left goes D

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:37:34Z

Circular17: /* Summary of sets of pieces */ usage

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:36:11Z

Circular17: /* Summary of sets of pieces */ not necessarily complex in fact

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:34:13Z

Circular17: /* Example */ remove Q reference, because it could be C

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:32:59Z

Circular17: /* Overlap */ interference instead of quirk in order to reduce confusion

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:31:53Z

Circular17: /* Quirk part */ quirk and conjugate

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk and conjugated part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:26:26Z

Circular17: /* Quirk part */ both

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with both PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:25:49Z

Circular17: /* Quirk part */ line ending

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :

'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:24:21Z

Circular17: /* Quirk part */ typo

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.
'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :
'''C''' = ('''J''' union '''K''') \ ('''H'''A union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:23:51Z

Circular17: /* Quirk part */ conjugated part

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them PC and their starting location '''C'''.
'''C''' = ('''J''' union '''K''') \ ('''H'''A/ union '''H'''BA' union '''H'''A'B union '''Q''')

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

The conjugated part is :
'''C''' = ('''J''' union '''K''') \ ('''H'''A/ union '''H'''BA' union '''H'''A'B union '''Q''') = front-right-top corner and front-bottom-left corner

The front-right-top corner goes U' and the front-bottom-left corner goes D.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:07:38Z

Circular17: /* Classification */ correction

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call their starting location '''C'''.

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap with quirk part and without conjugated part: [R', F]
* overlap with conjugated part and without quirk part: [R', F']
* overlap with both quirk and conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:06:34Z

Circular17: /* Quirk part */ quirk correction

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of some pieces that go outside of the pseudo 3-cycle

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call their starting location '''C'''.

In the overlap case, either '''Q''' or '''C''' is non empty, so there are at least 4 sets of pieces: PA, PB, PN and PQ or PC.

In complex cases, there may be five sets with PQ and PC.

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T08:00:42Z

Circular17: /* Quirk part */ empty

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

The quirk part is :
'''Q''' = ('''H'''A' \ '''H'''B) A' = empty

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:53:07Z

Circular17: /* Example */ pieces

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner. So PN is the set of pieces at these locations.

'''H'''B is the front-right edge and the front-right-top corner. So PA is the top-right edge and the top-right-back corner.

'''H'''A' is the front-right edge and the front-right-top corner. So PB is the front-bottom edge only.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:49:49Z

Circular17: /* Example */ HA'

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner.

'''H'''B is the front-right edge and the front-right-top corner.

'''H'''A' is the front-right edge and the front-right-top corner.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:49:07Z

Circular17: /* Quirk part */ example

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

====== Example ======

Let's consider [R', F']

'''H'''A is the front-right edge and the front-right-bottom corner.
'''H'''B is the front-right edge and the front-right-top corner.

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:46:17Z

Circular17: /* Quirk part */ any part can go outside

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:35:25Z

Circular17: /* No overlap */ storage comment

A '''commutator''' is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

== Mathematical definition ==

Given a group, a '''commutator''' is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

== Fundamental principle ==

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing:
L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

== Effect ==

The sequence A changes a set of pieces located at some locations '''J''' on the cube. The sequence B affects some locations '''K'''. If '''J''' and '''K''' have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, '''J''' and '''K''' have an intersection '''N'''.

Pieces that are brought into the intersection by A are '''N'''A' (apply the inverse of A to the locations '''N'''). Pieces that are brought into the intersection by B are '''N'''B'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of '''N''', '''N'''A' and '''N'''B'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

=== Trivial case ===

When '''J''' and '''K''' have an empty intersection, A and B commute, so the commutator does nothing.

=== B inplace ===

If '''N'''B' = '''N''', the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be '''N''' and '''N'''A'. So those pieces will be in '''J''', i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the [[conjugate]] of B by A, and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part PQ (see general case).

==== Examples ====
Let's consider [M', U2] = M' U2 M U2

In this example, '''N''' are the top-front and top-back edges. '''N'''B' is '''N'''U2, which is exactly '''N''', and '''N'''A' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, '''N''' are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

=== A inplace ===

This is a symmetrical case. Affected pieces will only be '''N''' and '''N'''B'. So those pieces will be in '''K''', i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the [[conjugate]] of A' by B.

==== Examples ====

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples:
* A = "rotating a corner" like [R' D' R D R' D' R, U]
* A = "flipping an edge" like [R' E' R2 E2 R', U]
* A = "exchanging two corners" like [R' L' D2 R L, U]
* A = "exchanging two edges" like [M2 D2 M2, U]

=== Three different sets ===

When '''N'''A' and '''N'''B' are not included in '''N''', there are three sets involved in the commutator:
* '''N'''A' : pieces that are brought into the intersection by A
* '''N'''B' : pieces that are brought into the intersection by B
* '''N''' : the intersection of '''J''' and '''K''', i.e. where A and B interfere.

Note that '''N'''A' can have locations in common with '''N''' and '''N'''B' can have locations in common with '''N'''. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:
# moving pieces inside the intersection (from '''N''' to '''N''')
# bringing pieces into the intersection (from '''N'''A' \ '''N''' to '''N''' or from '''N'''B' \ '''N''' to '''N''')

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:
* Case 2 only: the three sets '''N'''A', '''N'''B' and '''N''' are completely separated (no overlap)
* Case 1 and 2 at the same time: the three sets overlap

==== No overlap ====

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
* A stores the content of the intersection PN in '''N'''A, and replace it with some pieces PA from '''N'''A'
* B stores PA in '''N'''B, and bring some pieces PB from '''N'''B' into the intersection
* A' retrieves PB from the intersection and places them in '''N'''A', and brings to the intersection previously stored pieces PN in '''N'''A
* B' retrieves PN from the intersection and places them in '''N'''B', and finally places PA in the intersection.

So it is a 3-cycle of (PN, PA, PB) into (PA, PB, PN)

We notice that '''N'''A and '''N'''B are used as temporary storage:
* The content of the intersection, PN, is first hidden by A in its storage '''N'''A, and comes back with A', so that the only transformation applied to these pieces is B'.
* The content of '''N'''A' is brought by A, and is then hidden by B in '''N'''B, so that the only transformation applied to these pieces is A.
* The content of '''N'''B' is brought by B, and then is placed immediately in '''N'''A' by A'. So the transformation applied to these pieces is B A'.

Note that '''N'''A may or may not overlap with '''N'''A', and '''N'''B may or may not overlap with '''N'''B', but that it does not have any implication, except that the location where the pieces come from can also be the location where they are temporarily stored. The most obvious case is when '''N'''A equals to '''N'''A', or '''N'''B equals to '''N'''B', which means that pieces of the intersection '''N''' are swapped with pieces of '''N'''A' or '''N'''B', as in [R2, F2]

In other words, all the commutator does is:
* PA goes A
* PB goes B A'
* PN goes B'
[[File:Break down of Commutator A B A B If there is no overlap.gif]]

===== Examples =====

* [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front edge and PN is the top-right edge.
* [R' D' R, U'] is a 3-cycle of corners

==== Overlap ====

This is the most complex case, because it is like the "no overlap" case, but with an additional quirk.

===== Safe part =====

Some pieces in the intersection '''N''' are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call '''H'''B the location of pieces hidden by B:

'''H'''B = (('''N'''B) \ '''N''') B'

Similarly, pieces are hidden by A:

'''H'''A = (('''N'''A) \ '''N''') A'

We will also use pieces that are hidden by A' so that they are not affected by B':

'''H'''A' = (('''N'''A') \ '''N''') A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
* PA will be pieces that are brought into '''H'''B by A, i.e. pieces that are at the beginning in '''H'''BA'. This set may overlap with the intersection '''N'''.
* PB will be pieces that are brought into '''H'''A' by B, i.e. pieces that are at the beginning in '''H'''A'B' \ '''N'''
* PN will be pieces that are hidden by A, which are at the beggining in '''H'''A.

We get the same result as before:
* PA goes A
* PB goes B A'
* PN goes B'

====== Example ======

Let's consider [R', F]. The intersection '''N''' is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

'''H'''B = (('''N'''B) \ '''N''') B' = (('''N'''F) \ '''N''') F'
which is the set of locations containing the front-right edge and the front-right-bottom corner.

So PA are front-right-top corner and top-right edge.

'''H'''A = (('''N'''A) \ '''N''') A' = (('''N'''R') \ '''N''') R

So PN are front-right edge and front-right-bottom corner.

'''H'''A' = (('''N'''A') \ '''N''') A = (('''N'''R) \ '''N''') R'
which is the set of locations containing the front-right-top corner and the front-right edge.

PB are pieces at '''H'''A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.
* PA goes R', thus replacing PN
* PN goes F, thus replaces a part of PA and a part of PB
* PB goes FR, thus replacing a part of PA and the top-right-back corner (the quirk part '''Q''')

The quirkiness is that the locations of those pairs overlap, and that an extra location '''Q''' is used, the top-right-back corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges around it.

===== Quirk part =====

Some pieces are outside of the pseudo 3-cycle. Let's call their starting location the quirk part '''Q'''.

'''Q''' = ('''H'''A' \ '''H'''B) A'

Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
* PQ goes A B A' B'

The quirk part '''Q''' is used as an extra location for the pseudo 3-cycle, to store some pieces of PB. So '''Q''' has two meanings :
* the initial location of pieces affected by all moves PQ
* the final location of pieces of PB that go outside of the pseudo 3-cycle

So in the overlap case, there are at least 4 sets of pieces: PA, PB, PN and PQ

In complex cases, there may be a fifth set PC affected by a conjugate (see general case).

== General case ==

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 24 = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

=== Pieces affected by one element only ===

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

=== Pieces affected by two elements ===

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted PB.

=== Pieces affected by three elements ===

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted PN because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted PA.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them PC to express that they are affected by conjugates.

=== Pieces affected by all four elements ===

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted PQ.

=== Summary of sets of pieces ===

In all cases, pieces that are moved by the commutator are in one of these sets:
* PA : pieces moved into the intersection by A and then hidden by B
* PB : pieces moved into the intersection by B and then hidden by A
* PN : pieces hidden by A and then moved out of the intersection of B'
* PC : pieces affected by a conjugate, in the inplace cases or in complex overlap cases
* PQ : pieces moved by all elements of the sequence of the commutator, in the overlap case or in complex inplace cases

== Classification ==

If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.

If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.

So either the three sets PA, PB and PN are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection '''N''' only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus PC is not empty.

If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:
* trivial commutator which is equivalent to the identity: [L, R]
* double inplace, A and B move pieces inside the intersection
* single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
* single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
* no overlap, 3-cycle: [R' D' R, U']
* overlap without conjugated part: [R', F]
* overlap with conjugated part

== See also ==
* [[Commutators and Conjugates]]
* [[Conjugate]]

[[Category:Puzzle theory]]

Commutator

2013-03-31T07:25:44Z

Circular17: /* Quirk part */ double usage of Q

Commutator

2013-03-31T07:20:07Z

Circular17: /* Quirk part */ removed confusing comma

Commutator

2013-03-31T07:19:15Z

Circular17: /* Quirk part */ double usage of Q

Commutator

2013-03-31T07:16:07Z

Circular17: /* Example */ quirk explanation

Commutator

2013-03-31T07:10:46Z

Circular17: /* Quirk part */ rephrase

Commutator

2013-03-31T07:05:30Z

Circular17: removed spaces before :

Commutator

2013-03-31T07:03:04Z

Circular17: /* Safe part */ explanation of HA'

Commutator

2013-03-31T07:00:02Z

Circular17: /* Safe part */ typo

Talk:Commutator

2013-03-31T06:46:23Z

Circular17: /* Changes by Circular17 */

Commutator

2013-03-30T19:21:50Z

Circular17: /* Classification */ no need to mention quirk here

Commutator

2013-03-30T19:20:48Z

Circular17: /* Classification */ overlap necessarily with quirk part

Commutator

2013-03-30T19:12:51Z

Circular17: /* Classification */ 3-cycle example

Commutator

2013-03-30T19:12:00Z

Circular17: /* Classification */ possible

Commutator

2013-03-30T19:00:22Z

Circular17: /* Classification */ draft

Commutator

2013-03-30T18:52:43Z

Circular17: /* Classification */ start

Commutator

2013-03-30T18:22:48Z

Circular17: /* General case */ classification section

Commutator

2013-03-30T18:17:32Z

Circular17: /* B inplace */ quirk part