# My ultimate commutator challenge



## mrCage (Jun 2, 2011)

Show that every even permutation on a nxnxn cube can be obtained by a series of commutators. If 2 consecutive comms is a single comm (somehow) then every even permutation is also a direct commutator. No restriction on number of layers used here ...

Per


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## Lucas Garron (Jun 2, 2011)

mrCage said:


> Show that every even permutation on a nxnxn cube can be obtained by a series of commutators. If 2 consecutive comms is a single comm (somehow) then every even permutation is also a direct commutator. No restriction on number of layers used here ...



Well, the first part is pretty trivial to see with some cubing intuition, especially since most of this relies on 3-cycles. The second one is gnarlier, and I never managed to get a satisfactory answer (despite Bruce's reference).


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## mrCage (Jun 2, 2011)

Lucas Garron said:


> Well, the first part is pretty trivial to see with some cubing intuition, especially since most of this relies on 3-cycles.


 
Elaborate. I fail to see the triviality...

Per


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## Lucas Garron (Jun 2, 2011)

mrCage said:


> Elaborate. I fail to see the triviality...
> 
> Per



EP, CP: Generated by 3-cycles. Easy with commutators.
EO, CO: Orient two at a time using commutators. Also easy.


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## TMOY (Jun 3, 2011)

Either I missed somethng or the 2nd part is trivial too, you just have to use the first part and a basic recursion.(if N-1 comms can be replaced by one single comm, than N comms are N-1 comms + 1 comm -> 1 comm + 1 comm = 2 commq -> 1 comm).

The really nontrivial part would be to prove that 2 comms can actually be merged into one.


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## mrCage (Jun 3, 2011)

Lucas Garron said:


> EP, CP: Generated by 3-cycles. Easy with commutators.
> EO, CO: Orient two at a time using commutators. Also easy.


 
True if we have even permutation in every orbital effected by the "total permutation". Hmmm .... And where are the centers in your sketchy proof?? I didn't mean solely 3x3x3 cube but any size nxnxn regular cube!!

Per


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## glazik (Jun 8, 2011)

Hi,

It has already been proved that the alternating group An, which is the group of even permutations on the set {1...n}, consists only of (direct) commutators for n >= 5.

More details, including links to published papers on group theory and examples applied to a 7x7x7 cube using an on-line app, are given in document CommutatorSubgroup.

I hope this may be of help.


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## Ravi (Aug 30, 2011)

mrCage said:


> True if we have even permutation in every orbital effected by the "total permutation".


 
If we don't have an even permutation in every orbit, then we aren't in the commutator subgroup. Commutators (and products of commutators) can only do even permutations, even if you restrict to a particular orbit (such as edges or corners on a 3x3). So the commutator subgroup G' is contained in the group H of operations that do even permutations on every orbit. But if we mod out orientations (which, as Lucas pointed out, are easy) then H/{orientations} = H_perm is just a direct product of alternating groups A_n (where n=8, 12, or 24, the sizes of the orbits). Then, as glazik pointed out, every element of A_n is a commutator, so everything in H_perm (and therefore H) should be a single commutator.

There's one catch, though: glazik's article only says that everything in A_n is a commutator of two elements of S_n, not necessarily of A_n. So it could be that some element of H_perm is a commutator of two permutations, but that one or both of these is an odd permutation on some orbit; this may give an impossible position of the NxNxN supercube. The question is: is every element of A_n (at least for n=8, 12, and 24) a commutator of two other elements of A_n?

Perhaps we can use E. Bertram's result from glazik's link (bottom of the first page) to prove this. If we set l = n-1 (which is odd for even n), then we get that every element of A_n can be written as a product of two l-cycles, and therefore as a commutator of one l-cycle (an even permutation) with some tau in S_n. But it may be that tau is an odd permutation--for example, (1234567) and (1234576) are conjugate in S_8, but only by the odd permutation (67). So we must instead use l = n-3. Then we can force tau to be an even permutation, so every element of A_n will be a commutator of two even permutations. The final problem is that in the case n=8, n-3 is less than the bound of 3n/4 required by Bertram. For this case, I went through all the possible cycle structures of permutations in A_8, and with a bit of permutation-matrix-multiplying dirty work from Mathematica, I believe I have found direct commutators for all of them. (Exercise for the reader?) Now let's go back up the chain of reasoning and see what we've proved:

- For n=8 or (from Bertram, using l = n-3) n>=12, every element of A_n is a commutator in A_n.
- Since H_perm is a direct product of groups A_n with n=8, 12, or 24, every element of H_perm is a commutator of elements in H_perm (and therefore in the NxNxN supercube group G).
- Since orientations are easy, every element of H is a commutator in G.
- Since all commutators in G belong to G' by definition of the commutator subgroup, and since G' is contained in H, it follows that {commutators} = G' = H.

Thus the commutator subgroup of the NxNxN supercube group is simply the set of operations that are even permutations on all orbits, and all such operations can be written as single commutators.

It's interesting to note, however, that a similar statement about commutator subgroups of more general groups is not true. Dummit and Foote's _Abstract Algebra_ (page 180 in the third edition) gives a rather convoluted-looking group of order 96 in which one element belongs to the commutator subgroup despite not being a commutator of any two elements in the group.


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## reThinking the Cube (Aug 30, 2011)

Ravi said:


> If we don't have an even permutation in every orbit, then we aren't in the commutator subgroup. Commutators (and products of commutators) can only do even permutations, even if you restrict to a particular orbit (such as edges or corners on a 3x3)...



Hmmm.



Lucas Garron said:


> I think he meant to write [R U R2, R U2 R2]. The Sune is one of the most important algs in cubing, and it is curious that the alg itself can be written as a commutator (because it's not obvious at a glance).
> See http://www.speedsolving.com/forum/showthread.php?5783-Math-Problem-12


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## Kirjava (Aug 31, 2011)

reThinking the Cube; are you implying that the Sune produces an odd number of swaps in each orbit?


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## reThinking the Cube (Sep 1, 2011)

Kirjava said:


> reThinking the Cube; are you implying that the Sune produces an odd number of swaps in each orbit?


 
NO, and that also implies that Sune+U cannot be written as a direct commutator, but Sune+U2 can.


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## mrCage (Sep 2, 2011)

There seems to be a confusion as to which sequence *exactly* is a sune: R' D' R D' R' D2 R or R' D' R D' R' D2 R D2.
Both permutations can be achieved be a singular commutator.

Per


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## Kirjava (Sep 2, 2011)

reThinking the Cube said:


> NO


 
Might be worth elaborating on what you mean in the future then. If your post just consists of two quotes by other people it's quite difficult to understand what you're going on about.


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## Lucas Garron (Sep 2, 2011)

mrCage said:


> There seems to be a confusion as to which sequence *exactly* is a sune: R' D' R D' R' D2 R or R' D' R D' R' D2 R D2.
> Both permutations can be achieved be a singular commutator.
> 
> Per


When a speedcuber is talking about a Sune, it's generally assumed that R U R' U R U2 R'. The inverse is an anti-Sune, and both algs have 48 symmetries, of which your former is one.


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## Ravi (Sep 14, 2011)

Ravi said:


> Then, as glazik pointed out, every element of A_n is a commutator, so everything in H_perm (and therefore H) should be a single commutator.


 
Wait, now I can't remember why I thought it follows that everything in H is a single commutator. Orientations are easy to write as products of commutators, and maybe even single commutators... but can they just be inserted into an existing permutation commutator? I'm too tired to make sense of this right now. It should at least still be true that H = G' and that everything in H_perm is a single commutator.


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## Christopher Mowla (Sep 15, 2011)

TMOY said:


> The really nontrivial part would be to prove that 2 comms can actually be merged into one.


Do you mean that a composite commutator composed of two commutators has the same result as if the two commutators are performed one after the other in either order?

For example,
[U' L' U, R'] [U' M2 U2 M2 U2 U, R'] can be merged to

U' L' U
U' M2 U2 M2 U2 U
(R')
U' U2 M2 U2 M2 U
U' L U
(R)
= [U' L' U U' M2 U2 M2 U2 U, R'] = [U' L' M2 U2 M2 U', R'].

The restriction for this is, given a commutator [A, B] and a commutator [C, D], the two can definitely be merged if B = D and A does not affect pieces in C and vice versa. In addition, both A and C affect different pieces touched by the move(s) B = D.

In my example, A = U' L' U, B = R' = R' = D, and C = U' M2 U2 M2 U2 U.

Of course, the two commutators in my example affect two different piece types. You could merge two commutators which affect just corners, but I think one of the two commutators would need to be much longer than its counterpart so that it restores the portion of the cube affected by the first part of the other.


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## macky (Sep 15, 2011)

cmowla, I didn't work out your example, but the question is whether or not it is possible to write the product of any two commutators as a single commutator. So we want to show either
(T) that the product of any two commutators can be written as a single commutator, or that
(F) some product of two commutators cannot be written out as a single commutator.
Your general approach merely shows one special case where merging is possible (by simple considerations, it seems); I'm sure you and others could come up with many more. But unless it suggests a general approach towards a constructive proof of (T), it does nothing to resolve our question.

And Per, I hope you realize that being good at FMC doesn't say much about being "brainy" on cube theory.


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## mrCage (Sep 15, 2011)

macky said:


> And Per, I hope you realize that being good at FMC doesn't say much about being "brainy" on cube theory.


 
Yes i know! I'm, still a bit uncertain how i would approach my own challenge.

Here are s few clues i would follow: 
if 2 comms can always be combined into one, then so can any any number of comms. Proof by induction.

If we limit to one orbital it's not all that hard so see that an even permutation can be achieved by comms. Can we somehow "parallellize" the comm's first parts to schieve a general proof?? Hence combinig all orbitals involved. 

I wish Jaap would come with some input 

Per


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## Christopher Mowla (Sep 15, 2011)

macky said:


> cmowla, I didn't work out your example, but the question is whether or not it is possible to write the product of *any* two commutators as a single commutator.


I don't see why this would be necessary to prove that any even permutation can be written as a single commutator because my approach doesn't limit the length of the commutators combined (and hence the two commutators could theoretically cover every permutation when combined). There many commutators which cause the same permutation of pieces as other commutators but are obviously composed of different moves.

If you look at my definition, isn't that what is meant by merging commutators? In what other combination can two commutators be merged into one, besides the trivial case when the same exact commutator is applied to more than one orbit of wings or big cube center pieces?

If my approach does not apply to your definition of merging two commutators into one commutator, please define what you think that means. If you cannot define it, then how can anyone prove or disprove it? 



macky said:


> I'm sure you and others could come up with many more.


Sure, anyone who can follow my definition can come up with more examples.


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## macky (Sep 15, 2011)

[edit] Ravi, I haven't ignored your post.



cmowla said:


> I don't see why this would be necessary to prove that any even permutation can be written as a single commutator because my approach doesn't limit the length of the commutators combined (and hence the two commutators could theoretically cover every permutation when combined). There many commutators which cause the same permutation of pieces as other commutators but are obviously composed of different moves.



You're correct. But I was answering your question


cmowla said:


> TMOY said:
> 
> 
> > The really nontrivial part would be to prove that 2 comms can actually be merged into one.
> ...


because it wasn't clear to me from this that you understood what TMOY said.

"Merging," as TMOY or whoever first used it used it, means writing some [A, B][C, D] in the form [E, F] for some E and F. You didn't define anything; you gave one set of conditions (B = D etc) under which merging is possible, as [A, B][C, B] = [AC, B].



cmowla said:


> If you look at my definition, isn't that what is meant by merging commutators? In what other combination can two commutators be merged into one, besides the trivial case when the same exact commutator is applied to more than one orbit of wings or big cube center pieces?


That's part of TMOY's question. In fact, if you can find even one product of two commutators that cannot be merged (provably, not just because the obvious methods don't work or "according my experience and theory"), then that's an even permutation that cannot be written as a single commutator and a counterexample to Per's question.

But ok, you seem to be trying to answer Per's question directly without inducting, by writing any even permutation first in the specific form you brought up, as [A, B][C, B] with A, B, C satisfying certain conditions. Your example does not seem to indicate a general approach for doing this. Maybe you can; you've certainly worked more closely with commutators than most cubers (including me). Which is why I hope that _you_ in particular understand the obvious reductions suggested in the first posts, so that you can put your intuition to best use.


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## Christopher Mowla (Sep 16, 2011)

macky said:


> you gave one set of conditions (B = D etc) under which merging is possible, as [A, B][C, B] = [AC, B].


Well, you forgot to mention that pieces affected by A != pieces affected by C. That's very important to remember. In fact, that very fact can be used as a generalized process of doing this. I made an effort to come up with a different example where B != D (since that appeared to be too trivial).

With [A, B] = [R U2 R',F] and [C, D] = [[B' D': M' D L' D' M D L D'], S' R2 S R2],
both C and D are "independent" from A, B and [A, B].

That is, [R U2 R',F] [[B' D': M' D L' D' M D L D'], S' R2 S R2] can be merged to
[F: [F' [B' D': M' D L' D' M D L D'] , S' R2 S R2 R U2 R']]

(Notice that I conjugated once to make this work, but this is still a commutator nonetheless.)

Does this example show things better for you? If you were to ask *me* when two commutators cannot be merged, if their pieces are not independent (whether both pieces from one commutator are from the other commutator--this example--or if one piece from each is independent--first example), one could merely attempt to combine any two commutators that do not follow the restrictions I gave. To me, that's proof that those particular two commutators cannot be merged into one (well, not to one that yields the same permutation as the product of the two commutators).

Why? If no pieces from the commutators are independent, then one will conflict with the other, at some level or another. You can still possibly merge them into a commutator, but that commutator will not have the exact same effect on the cube as their product due to the moves from one commutator affecting pieces the other affects.

I still don't see why when two commutators cannot merge doesn't mean one of them cannot merge with another that does the same permutation of the other, as long as the new commutator being used follows the restrictions I illustrated here.

If I am still not understanding what is being asked, I apologize. I mainly wanted to show another example which elaborated my method more.


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## Stefan (Sep 16, 2011)

cmowla said:


> those particular two commutators cannot be merged into one (well, not to one that yields the same permutation as the product of the two commutators).



*Any* two can be merged into one (if I understood Bruce/Joyner correctly). You seem to misunderstand what's meant with "merging" - you don't have to reuse "the moves from" the given commutators to build your output commutator. Think about the cube group elements, not about some move sequences representing them.


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## TMOY (Sep 18, 2011)

If I understand Bruce/Joyner correctly, their result is not sufficient to prove Per's challenge. It only implies that any non-disorienting even permutation of the cube (that is any even permutation A such that A^n has no pieces in place but disoriented for any n) can be written as a commutator, you still have to deal with the disorienting ones.


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## Stefan (Sep 19, 2011)

I just checked again (page 229 here) and I think it says that half of all cube group elements are commutators and that the requirement is exactly that corner permutation and edge permutation are even.


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## macky (Sep 19, 2011)

Stefan, that links gives me a 403. But the condition you mention is the "obvious" one Ravi gave, and Lucas and TMOY disagree in their interpretation of Bruce/Joyner. I think the result you're citing is the rather trivial one (as Lucas and Ravi showed) about the commutator subgroup (that is, the subgroup _generated by_ the commutators).

Can someone send me Joyner?


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## cuBerBruce (Sep 19, 2011)

macky said:


> Stefan, that links gives me a 403. But the condition you mention is the "obvious" one Ravi gave, and Lucas and TMOY disagree in their interpretation of Bruce/Joyner. I think the result you're citing is the rather trivial one (as Lucas and Ravi showed) about the commutator subgroup (that is, the subgroup _generated by_ the commutators).
> 
> Can someone send me Joyner?


I was able to access it using a Google "Quick View" link rather than the normal link.



Spoiler



http://docs.google.com/viewer?a=v&q...bO2OTH&sig=AHIEtbRPxlJIIErmVbTZdtRULoP_LKDdfQ


But yes, I think I have to agree with macky that Joyner is only proving that the elements can be written as a finite product of commutators, not necessarily a single commutator.


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## Stefan (Sep 19, 2011)

The link works fine for me. Try again? Alternatively, google _joyner adventures in group theory_ and you'll probably find it.

Oops, yes, I might have misunderstood Joyner's definition of the commutator subgroup. In my defense, his definition...



Joyner said:


> The group G′ generated by all the commutators {[g, h] | g, h belong to G}.



... doesn't even look like a complete sentence to me (and I find it unclear).


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## qqwref (Sep 19, 2011)

G' is generated by the set of all commutators in G - his set as written means {commutator of g and h, such that g and h belong to G}.

Can anyone find a cube position in the commutator subgroup which can provably NOT be created by a single commutator?


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## siva.shanmukh (Nov 25, 2011)

*Commutator for every even move sequence*

Since commutators produce only even permutations in any given orbit in an nxnxn super cube. I questioned myself if the following statement is true.

_All states that occur on an nxnxn super cube by applying a move sequence which has even number of quarter turns for each layer(assuming the number of layers = ceil(n/2)) can be represented as a commutator or a conjugate._

The answer is obviously yes, since the above case produces an even permutation in each orbit and we can always come up with a commutator/conjugate which does this.

_But my question is if any even move sequence (Even with respect to each layer) be expressed as a commutator/conjugate which is simplified by move cancellations?_

For example

R U' [U R U R U, z' y2] U R' is a commonly used U perm in the form of C A B A' B' C'

Can we write any even move sequence (Even with respect to each layer) in the same way as the above alg?


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## Christopher Mowla (Nov 25, 2011)

It's interesting that you mentioned the supercube and not just the regular 6-colored cubes. One exception that I believe is a possibility is just a half turn, say, U2.

We can express U2, for example, as a commutator if we consider the non-supercube by letting X be an N-Perm in the U layer and letting Y be U.
[R' U R2 B2 U R' B2' R U' B2 R2' U' R U', U].

However, representing U2 on the supercube as a commutator might not be possible (the commutator above twists the composite center 180 degrees too). I would be very interested to see someone do this (if it's possible).

Interesting enough though, if we consider the same case for the inner layer by letting X be an "interior N-Perm" in slice r and Y being r, then it does work on the supercube inner slice.
[D2 r' D2 F2 l' U2 l U2 r' U2 r U2 F2 r D2 r D2, r]

This problem is too complex for me to even guess at an answer for it in general, but I think the first example I gave above might be a trivial counterexample for the nxnxn supercube, which is what I am understanding to be your stated problem.


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## TMOY (Nov 25, 2011)

siva.shanmukh said:


> Since commutators produce only even permutations in any given orbit in an nxnxn super cube. I questioned myself if the following statement is true.
> 
> _All states that occur on an nxnxn super cube by applying a move sequence which has even number of quarter turns for each layer(assuming the number of layers = ceil(n/2)) can be represented as a commutator or a conjugate._
> 
> The answer is obviously yes, since the above case produces an even permutation in each orbit and we can always come up with a commutator/conjugate which does this.


Did you read the thread ? Your "obvious" assertion not only is far from obvious, but is not proved yet, not even for the 2^3 cube.

Actually, thanks to Bruce/Joyner proving it for any size of cube isn't more difficult than proving it for the 3^3 but we still need to complete the proof of Per's challenge.


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## qqwref (Nov 25, 2011)

cmowla said:


> However, representing U2 on the supercube as a commutator might not be possible (the commutator above twists the composite center 180 degrees too). I would be very interested to see someone do this (if it's possible).


On the 3x3x3, at least, no commutator can twist a single face by 180 degrees. A little proof:
- Let q be the sum of all centers' orientations, mod 4. Let q(A) be the effect on q when we perform sequence A. Observe that the order we do operations doesn't affect q (that is, q(AB) = q(BA)), and also cube rotations don't affect q (that is, q(A) = q([any rotation] A)).
- A commutator of form A B A' B' must thus have the following effect on q: q(ABA'B') = q(AA'BB') = q(identity) = 0.
- Since a 180-degree center twist affects q by 2, a commutator of form A B A' B' cannot accomplish this.

On even cubes we can, though: try [Lw Rw Dw2 Lw' Rw', U] on the 4x4x4.


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## siva.shanmukh (Nov 25, 2011)

TMOY said:


> Did you read the thread ? Your "obvious" assertion not only is far from obvious, but is not proved yet, not even for the 2^3 cube.
> 
> Actually, thanks to Bruce/Joyner proving it for any size of cube isn't more difficult than proving it for the 3^3 but we still need to complete the proof of Per's challenge.


 
My statement is now not just far from obvious but wrong like qqwref formalized. But it is not yet wrong for even super cubes like qqwref exemplified. And the spirit of making that statement is to signify the difference of the 4 centers of each orbit and not the non-moving center.

Can we try to prove/disprove my statement either by ignoring centers or by restricting the move sequence, that we start with, to have only total face quarter turns to be a multiples of four for each face.

i.e.,

_If any move sequence, even with respect to each slice and multiple of 4 with respect to each face be expressed as a commutator/conjugate which is simplified by move cancellations?_


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## Christopher Mowla (Nov 25, 2011)

@qqwref,

Thanks! Obviously, as siva.shanmukh noted, it follows that it is not possible on any odd cube size. Just adjusting your commuator a little, we can see that the closest we can get to doing U2 on the 5x5x5 and larger odd supercubes is moving all of the pieces except the fixed center piece: [Lw Rw (Uw2 y2) Lw' Rw', U'] (on the 5x5x5). 

We can treat this algorithm as a conjugate of U2 in order to have all pieces in the U-Layer rotated 180 degrees [[Lw Rw (Uw2 y2) Lw' Rw', U']:U2], however, you have already proved that U2 cannot be expressed as a commutator on the 3x3x3 odd supercube, and thus we cannot express the above as a conjugated commutator for larger odd cube sizes because the composite center on the nxnxn odd cube behaves exactly like the fixed center on the 3x3x3 supercube.

For some reason I only looked at the 3x3x3 and not the 4x4x4 for the outer layer turns. Thanks again for pointing out that we can divide a face into 4 equal parts, which means we can do an X-perm 2 2-cycle of \( 1\times \frac{n}{2}\times \frac{n}{2} \) blocks!


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## Christopher Mowla (Apr 26, 2012)

*Possible Proof for Wing Edges*

I think I have an idea of how to prove this, at least for a single orbit of wings in the nxnxn.

Almost everyone who knows how to solve a 4x4x4 cube knows Stefan's PLL Parity algorithm, r2 F2 U2 r2 U2 F2 r2. Some know that this is just a conjugate of the move r2, as it can be expressed as [r2 F2 U2: r2].

Now, I can comfortably make the claim (without any possibility of doubt) that ANY 2 2-cycle of wing edges case in the "cage" of big cubes can also be made with a conjugate of the move r2. Like Stefan's alg, they of course will not be supercube safe, but I will later show that this will not be an issue (if you are concerned about having a supercube safe commutator).

For those who have never seen the two 2 2-cycle of wings algorithms I made in the past, see Algs_Sorted.pdf, 2 2-cycles.pdf, and the F3L documents in this post just to get an idea of what I mean by 2 2-cycle conjugates, in general.

In addition, we can create every 2 2-cycle of wing cases in the "cage" of big cubes by just conjugating the move r2.

Note that we can generate any even permutation with products of 2 2-cycles because every 2 2-cycle can be expressed as a product of two overlapping 3-cycles.

So suppose we have a conjugate A for one 2 2-cycle of wings A X A', where X = r2. In addition, we have a different conjugate B, for another 2 2-cycle of wings: B X B'. (Note that in both algorithms, X is present because I previously claimed that every 2 2-cycle case can be expressed as a conjugate of the same move, r2).

Now, a few posts ago in this thread, I showed that the move r2 can be expressed as the commutator [D2 r' D2 F2 l' U2 l U2 r' U2 r U2 F2 r D2 r D2, r] for the nxnxn.

So both A X A' and B X B' are commutators since the equality [O:[P,Q]] = [[O],[O:Q]] holds.

If we multiply these two commutators, we have
A X A' B X B'

Since r2 = (r2)'
= A X A' B X' B'

Now, the commutator B' A X A' B X' can be conjugated to obtain the above:
B
B' A X A' B X'
B'
= A X A' B X' B'
, which shows that the product (A X A') (B X B') merges to the commutator [B: [B' A, X] ].

Now, if we were to multiply three 2 2-cycles together, *where each 2 2-cycle is composed of only one conjugation of the move r2*, then we would have a problem because we would have an odd number of Xs (X = r2 = commutator). Since a commutator must have an even number of every move it contains, then these three 2 2-cycles could not possibly be merged to a single commutator.

However, note that we can freely create each and every 2 2-cycle of wings with two 2 2-cycle of wings. Here are two examples of this:
([l2 U' B2 U' x U' R' U: l2] ) (y' [R B: [l2 U' F2 U': l2]] y)
(y2 [r2 U F2 U: r2] y2) ([y x U2 l2 U F' L U R2 F': r2])

Therefore, we can always have an even number of Xs in our product.

Now let's confirm that we can merge three 2 2-cycle of wings (which are commutators) as one commutator.

Let one 2 2-cycle of wings algorithm be C X C' and another be D X D', and let's suppose that combining them yields a 2 2-cycle of wings.

Before we found that A X A' B X B' merged to the commutator [B: [B' A, X] ] and therefore A X A' B X B' is a commutator. Similarly, C X C' D X D' is also a commutator.

The equality [O:[P,Q]] = [[O],[O:Q]] tells us that a conjugated commutator is a commutator (which shows why A X A' and B X B' are commutators), which also means that a shifted commutator is a commutator (which will justify what follows).

Let's combine these four 2 2-cycle algorithms (which is combining three 2 2-cycles, by definition of (C X C')(D X D') ).

(A X A' B X B') (C X C' D X D')
Since X = X',
(A X A' B X' B') (C X C' D X' D')
Since [O:[P,Q]] = [[O],[O:Q]], we rotate the moves in parenthesis and still have a product of two commutators.

Moving A to the end and D' to the beginning,
(X A' B X' B' A) (D' C X C' D X')

Since [O:[P,Q]] = [[O],[O:Q]], we can also rotate pieces at the end positions of the parenthesis to the beginning or end of the entire product to still have a product of two commutators.
Rotating X to X', they cancel to leave us with:
(A' B X' B' A) (D' C X C' D)

Moving C' D from the end to the beginning of the entire product,
C' D A' B X' B' A D' C X
= [C' D A' B, X'], which is a single commutator, which means that (A' B X' B' A) (D' C X C' D) is also a single commutator.

To handle merging four 2 2-cycles in which their effect is literally four 2 2-cycles (no consecutive pair of 2 2-cycles merges to a single 2 2-cycle), we still can use the above argument.

Now this can be extrapolated to any number number of 2 2-cycles of wings.

Since every even permutation of wings can be reached with a product of 2 2-cycle wing algorithms, every 2 2-cycle of wings can be expressed as a commutator, and these commutators can be merged into one commutator, 
then every even permutation of wings can be expressed as a single commutator.

Lastly, as I mentioned at the beginning about the issue that a single 2 2-cycle conjugate is not supercube safe, note that we can construct all of these algorithms to affect the same centers. Since there must be an even number of Xs in order for all 2 2-cycle algorithms to merge as a single commutator, then doing a 2 2-cycle of four 1x(n-2) center blocks an even number of times yields the identity (the centers on supercubes would be untouched once the entire commutator is fully executed). But this is not at all required, as we can have consecutive 2 2-cycle algorithms involve different 1x(n-2) center blocks.

EDIT:

Maybe I should make it clear how we can extrapolate it to any number of 2 2-cycles of wings.

A X A B X B can be interpreted as either one 2 2-cycle (if the 2 2-cycles of A X A and B X B overlap two pieces) or two 2 2-cycles. So, to represent any number greater than or equal to one 2 2-cycle, we have

(A X A' B X B')(C X C' D X D')(E X E' F X F')....(Y X Y' Z X Z')...

I believe if we just look at the following example, we can see enough evidence. (Note that X = X').
(A X A' B X B')(C X C' D X D')(E X E' F X F')
We just rotate the outer most parenthesis so that X will be on the ends:
(X A' B X B' A)(C X C' D X D')(F' E X E' F X)
Then we cancel the Xs
(A' B X B' A)(C X C' D X D')(F' E X E' F)
We can continue rotating the outer most
(X B' A A' B)(C X C' D X D')(E' F F' E X) = (X)(C X C' D X D')(X)
Until we can get rid of the outer-most parenthesized groups completely
(C X C' D X D')
So that we will eventually end up with the single product of (C X C' D X D'), which merges to a single commutator.

Similar to this entire argument with conjugates, every ODD permutation of wing edges on even cubes at least, can be represented as a product of conjugates of the moves r and r', based on the method I showed in my 4-cycle conjugates in this post.


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## Christopher Mowla (Apr 27, 2012)

*Possible Proof for all pieces of the nxnxn cube*

I think I have now found a way to prove Per's challenge in its entirety.

For those who read my previous post before I edited it (several hours after I first posted it), I changed the first sentence. That post was merely supporting the idea that all even permutations of one orbit of wing edges in the nxnxn cube could be expressed as a single commutator. However, I believe I have now generalized it to the entire cube.

My previous post focused on taking just one orbit of wings at a time. But obviously this is not at all near to what Per's challenge was.

To start off, I want to give a few examples of how to express moving different piece types separately.

The idea is to have Rw2 instead of just r2, as in my previous post. That is, Rw2 represents all \( \left\lfloor \frac{n}{2} \right\rfloor \) slices on the right half of the nxnxn cube (for \( n\ge 1 \)).

Recall that in my previous post, I just called r2 X. Here, we just call Rw2 X. And just as r2 could be expressed as a commutator, so can Rw2.
Rw2 = [D2 Rw' D2 F2 Lw' U2 Lw U2 Rw' U2 Rw U2 F2 Rw D2 Rw D2, Rw].

And of course since Rw2 = Rw2', then X = X'.

So if we can see that all piece types (corners, middle edges, wing edges, and non-fixed centers) can be handled separately with conjugates of the same move, Rw2, then just as I was able to merge the commutators in my previous post as one (because they all consisted of the same move, X) we can merge all of these commutators into one commutator because we claimed that the conjugates A, B, C, D, etc. did not have to be the same.

*The Method*
All we need to do to affect each and every piece type with a product of conjugates ([A:Rw2][B:Rw2])([C:Rw2][D:Rw2]),etc., is to 

1) Let A = Moves which ultimately just affect the piece type of our choice in the layers Rw2.
2) Form the conjugate [A:Rw2]
3) Do another conjugate [B:Rw2] in which B are moves which also only affect the piece type of our choice in the layers Rw2.

(A must not equal B, otherwise we will have the identity).

So we will be able to affect each and every piece type with a PAIR of conjugates of Rw2.
That is,
[A: Rw2]
[B: Rw2]

This is actually no different than my previous post because we needed to have an EVEN number of Xs (here, Rw2s) in order to be able to merge all of them into a single commutator. Wing edges are the only piece type which do not require a pair of conjugates of r2 to be able to visually just swap wings (because swapping same color centers with each other cannot be seen on a regular cube), but we still needed to have an even number of Xs to be able to merge all conjugations of X into a single commutator.

* Examples*

Corners
http://alg.garron.us/?alg=[[F-,_RB_R-]:Rw2]
[[F-,_RB_R-]-:Rw2]&cube=7x7x7&notation=WCA[[F', RB R']:3r2]
[[F', RB R']':3r2]

Wing Edges(Note that we can take all orbits at once or only a subset)
[[2F', U F' U']: 3r2]
[[2F', U F' U']': 3r2]

Middle Edges
[S' U F' U' S U F U': 3r2]
[M' U' R U M U' R' U: 3r2]

X-Center Pieces
[e' m' e 2R U' 2L' U 2R' U' 2L U e' m e: 3r2]
[m e' m' e 2R U' 2L' U 2R' U' 2L U e' m e: 3r2]

...etc.

As you might be thinking, yes these pairs of conjugations of Rw2 affect more than 4 pieces (thus they are not 2 2-cycles). However, since any even permutation can be decomposed to a product of 3-cycles, every cycle type can generate all even permutations. In addition, there is no limitation to the number of moves it takes to show that Per's challenge is true or false, and there is no limitation to what cycle type the commutators being merged ought to be (each can be a different cycle type if necessary). So actually making a commutator that generates a random even permutation scramble of the nxnxn is very difficult, but it is definitely possible. However, being able to construct an actual algorithm isn't necessary to see that they exist. Besides, we could just construct simple ones if we need to see concrete examples.

Therefore, even though I thought that strictly having just 2 2-cycles in my previous post was necessary, the only necessity for proving Per's challenge was being able to construct commutators which all have a common element (in my previous post, it was the move r2, but here, it is Rw2 for the general case).

Hence, I think this post and my last post might be all we need to confirm that Per's challenge is indeed true for the nxnxn (not for the nxnxn supercube, of course).


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## macky (May 9, 2012)

cmowla said:


> I think I have an idea of how to prove this, at least for a single orbit of wings in the nxnxn.



Hi cmowla,
Your proof is wrong.



cmowla said:


> ...which shows that the product (A X A') (B X B') merges to the commutator [B: [B' A, X] ].


Up to here is fine (though this is a direct calculation, and you don't need the fact that r2 is a commutator). The problem is the following.



cmowla said:


> (A X A' B X' B') (C X C' D X' D')
> Since [O:[P,Q]] = [[O],[O:Q]], we rotate the moves in parenthesis and still have a product of two commutators.
> 
> Moving A to the end and D' to the beginning,
> (X A' B X' B' A) (D' C X C' D X')


The identity [O:[P,Q]] = [[O],[O:Q]] follows from the fact that conjugation is a homomorphism (in fact an isomorphism). Here, you conjugated the part within the first set of parentheses by A, and the second part by D. There's no reason to expect that this takes a direct commutator to a direct commutator. If you're not convinced, try following your argument in reverse to write down (A X A' B X' B') (C X C' D X' D') explicitly as a direct commutator.


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## oll+phase+sync (May 15, 2012)

Stefan said:


> I just checked again (page 229 here) and I think it says that half of all cube group elements are commutators and that the requirement is exactly that corner permutation and edge permutation are even.


 
I always wondered, how big the distance is form any non-commutator case to the nearest commutator Sub Group element. 

Is it just a quater turn, half turn, more?


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## aronpm (May 15, 2012)

oll+phase+sync said:


> I always wondered, how big the distance is form any non-commutator case to the nearest commutator Sub Group element.
> 
> Is it just a quater turn, half turn, more?


 
1 QTM


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## oll+phase+sync (May 16, 2012)

aronpm said:


> 1 QTM


 
How to proof?


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## Christopher Mowla (Sep 4, 2012)

Hey guys,

I don't know whether or not it is possible to solve all even permutations on the regular nxnxn or on the nxnxn supercube with only one commutator (well, I already pointed out that one center cannot be rotated 180 degrees with a commutator for the supercube non-fixed center pieces), but:

(*Disclaimer: *Although I am not finished verifying these results yet, this is what I have so far.)

Based on a systematic method I created (I started with pure theory and proof and then tested it today on the 3x3x3, as you will see the results I found below) I'm guessing that it takes no more than two commutators to solve every even permutation of the 3x3x3 regular cube (and the 2x2x2 cube) (including every possible orientation of edges and corners (I'm going to try to make the superflip with a product of two commutators tomorrow and show you guys...if that hasn't been done already, that is), and it takes no more than 3 commutators to solve every even permutation (of every orbit) of the nxnxn supercube. The only piece type which my results do not make any promise for is the 6 fixed center pieces on the odd nxnxn supercube. That is, with my current method for solving every possible even permutation with 3 (at most 4, if so) or less commutators cannot "reach" the supercube fixed centers, (at least not when there is an even number of them rotated 90 or -90 degrees instead of an even number of them rotated 180 degrees).

Just for now, here is my first official product of two commutators to solve a random scramble of the 3x3x3. It's 132 htm, but I have used CubeExplorer to get it that low (it might be 10 or so moves less than this...you can take each X and Y and try to reduce the amount of htm that way).

Notice that it's a 9-cycle of edges and a 7-cycle of corners. One of the 3 correctly positioned edges is unoriented.

*Scramble*
L2 F' L2 B U L R' U2 R2 B' L F' D2 U2 L' R' F U B U D2 B D' B' R' 

*Solution*
[R U2 R' F2 R B' R' D' F R' D R B' D L' B2 R2 U', D2 U2 L R2 F L' D' L' R2 U2 B2 D' U R D' B2 F2 U']
[L2 U2 R U2 B2 U L2 U' B2 U2 R' U2 L2, B2 U2 R F' L' B2 D F' D2 B2 F2 R' B' L' B2 U B']

View at alg.garron.us.

I was mainly posting this to ask you guys if any of these "claims" (which again, I have not finished validating yet with real examples) are something new, or has it been known (or proven), for example, that every even permutation (and orientation) of the 3x3x3 (and 2x2x2) can be solved with a product of two commutators?


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## Christopher Mowla (Sep 4, 2012)

Apparently the Superflip must be a trivial case, because it was very simple to write it as one commutator.
[D R' D' B' R2 B U' B' F2 R B D R2 D' F U2, D2 L2 F2 L2 R2 F2 R2 U2]

And here's one supertwist as well.
[U B2 D' B2 R2 F2 L U' L' F2 L' D' L D R2 U', D F2 R2 F2 D' U R2 F2 R2 U']

So we shouldn't let orientations scare us, because they are trivial to handle indeed.

Using my method (which I will explain soon, but I'm going to solve a 5x5x5 with 3 commutators first to verify my theory with an extreme example), not every "ugly" scramble requires two commutators for the 3x3x3 (or 3 commutators for the nxnxn): they can be handled in one commutator. For example, if you look at what cycle type the first and second commutators for the solution to that random 3x3x3 scramble in my last post separately, I can reach those cycle types in one commutator. So, it's possible that only a subset of all even permutations on the regular nxnxn require more than one commutator. The amount of pieces affected by a scramble does not correspond to how many commutators it takes to solve that even permutation: it depends on the cycle types (orientations of corners and middle edges do not have any affect on the amount of commutators required to solve any given case).

If someone wants to prove the efficiency of my method isn't efficient, just find one commutator that solves that random 3x3x3 scramble in did in my last post. My method isn't efficient for the worst cycle cases _if only one piece type is being cycled_, but if more than one piece type is affected by a scramble, it's pretty efficient (I cannot claim it to be optimal, because that would require that we prove that MrCage's theory is false).


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## Renslay (Sep 4, 2012)

Wow. Interesting!


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## Christopher Mowla (Sep 28, 2012)

cmowla said:


> Scramble
> L2 F' L2 B U L R' U2 R2 B' L F' D2 U2 L' R' F U B U D2 B D' B' R'
> 
> Solution
> ...





cmowla said:


> If someone wants to prove the efficiency of my method isn't efficient, just find one commutator that solves that random 3x3x3 scramble in did in my last post.


[F' L2 B D2 U2 R U2 F D' R B R B2 D' R D' F D', B' R2 F2 R2 B' U2 L U F R2 D2 F2 U' B R2 B2 U L2]

It took me a while to verify my original results and find and verify new results, but I'm finally finished. I have created a PDF document which contains proofs for each one of the following 3 theorems and a corollary.

*Theorem 1*: All even permutations of one orbit of n objects, where n > 2, can be solved with one commutator.

*Corollary (1b)*: The nxnxn supercube (and the regular supercube) can be solved with one commutator (not including orientations of corners and middle edges and not including the fixed center pieces on the odd supercube) if at least two pieces are solved in every piece orbit (an even permutation must be present in every orbit as well, of course).

*Theorem 2*: Only a subset of even permutations and orientations of the regular nxnxn cube and the nxnxn supercube can be solved with one commutator. 

*Theorem 3*: Every possible even permutation of the regular nxnxn cube and the nxnxn supercube, including orientations of the middle edges and corners, can be solved with a product of 2 commutators (except for a subset of fixed center positions on the odd supercube).

_Note that the proofs of all four of these statements require that statements in the *Prerequisite Information* and *The Method* sections are true. *Theorems 2* and *3* are dependent on *Theorem 1’s* corollaries to be true._


cmowla said:


> My method isn't efficient for the worst cycle cases if only one piece type is being cycled...


Now it is, as I made improvements to my method, and I proved Theorem 3 using the method.

(Document has been moved to post #44).​
*Remark*: Some may still be skeptical of my proof of Theorem 2 (which in short says that no, not every even permutation and orientation of the nxnxn can be solved in one commutator AND Mr. Cage's guess for a way to go about proving that is true IS wrong) being correct, since I basically used one method for all of them. They are very convincing to me because I can easily write any permutation of one orbit of pieces as one commutator without even thinking about cube moves or cubes for that matter (it's all based on the structure of cycles). In addition, there does not exist a single scramble yet that I have seen which was previously known to be able to be written as a commutator that I myself cannot write as one commutator using my method.

Best of all, from all of my theory one can create real examples without brute force. The most you'll have to do (besides basic "busy work") will feel like balancing a chemical equation in chemistry.

Note that I DID NOT provide any real examples (with worked out and detailed solutions) in the paper. I didn't create any big cube examples because there are billions of billions...of positions I could solve, both for the nxnxn super and the regular nxnxn cube.

Lastly, I create the solutions to these in cubetwister.

Here are sticker images I created (you just drag them into CubeTwister under Cubes->Stickers->Image). Except for the v-cube 7, I just used images made by Randelshofer and modified them a bit. (So I left the copyright statement in the images.) For the 6x6x6 and 7x7x7, I colored the orbits of oblique centers differently to hopefully make these images more user-friendly.

3x3x3, 4x4x4, 5x5x5, 6x6x6, 7x7x7


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## Christopher Mowla (Oct 3, 2012)

No comments? No scramble requests?

Anyway, here is some theory for what we could call "The Ultimate Conjugate Challenge." The following was much easier to prove than the theory about commutators. Note that the following is an independent source from the PDF document I made on commutator theory. I renumbered lemmas and theorems.

*Theorem*: The entire nxnxn supercube (and the nxnxn regular cube) can be solved with one conjugate.
(The proof has been put in the document in post #44).


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## Noahaha (Oct 3, 2012)

cmowla said:


> No comments? No scramble requests?



I'm also surprised about this. What you have proved seems utterly amazing to me.

Just one question. For the ultimate conjugate challenge, what's stopping me from having my conjugate be: U <solve cube to one move away> U'?


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## Christopher Mowla (Oct 3, 2012)

Noahaha said:


> I'm also surprised about this. What you have proved seems utterly amazing to me.
> 
> Just one question. For the ultimate conjugate challenge, what's stopping me from having my conjugate be: U <solve cube to one move away> U'?


Absolutely nothing! You've got the concept. I first started with "Theorem 1" which seemed to be pretty neat because you don't have to destroy the centers to complete all orbits of wing edges, but then it developed into this simple idea. I just kept my "Y" all the way to the end for mere consistency.

However, even the example you mentioned isn't as trivial as you might think. The trivial case is when moves completely cancel. Suppose you solve some supercube with a regular speedsolving solution. Let the first m moves be A and let the rest of the moves be B. That is, your regular solution is simply AB. We can write this as a conjugate by simply rewriting as: ABAA' = [A:BA].

"Theorem 1" and its corollary is probably interesting to K4 users, where as if we look at the cube as a whole, it seems pretty "pointless", despite that we do not have the trivial structure above.

EDIT:
I have worked out an example, and what Noahaha is true for some cases, but in general, the structure of my proof is necessary.

That is, 
1) Choose any algorithm "Y" (except for the scramble itself because that's trivial).
2) Execute Y on a solved supercube. That is the "solved state" in which you need to get your cube in.
3) Now execute scramble + Y to a solved supercube. Do moves to (scramble + Y) to get the "solved state."
4) Add Y' to get the real solved state.
DONE.


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## Christopher Mowla (Oct 12, 2012)

First of all, I realized that the proof for middle edge orientation (for now what I now call *Theorem 3*) was incomplete. If 2, 6, or 10 edges are flipped, the previous procedure does not apply directly because it violates the cube law of allowed edge orientations. I have, however, now proven edge orientation for 2, 6 and 10 flips.

I have also added in more prerequisite information, reformatted and renumbered all theorems and lemmas, added in 2 important corollaries (and refer to one of them throughout the proofs of what I now call *Theorem 2* and *Theorem 3*), swapped the appendixes (so that you do not need to print out the lists of cycle classes if you don't want to), and I have simplified the proof for the 3 corner twist. (I probably made other changes, but these seem to be the most major changes. Everything else is pretty much the same, if you've already read the document).

I have updated the download link in my post before my last. I bet you guys will like the format much better now.

Now enough of theory. Let's do an example.

*The 5x5x5 Supercube Solved with 2 Commutators*


Spoiler



First of all, if you don’t have cubetwister installed on your computer, you need to install it. If you don’t, it will almost be impossible to follow, much less replicate, this process.

Once you have it installed, download this cubetwister file and put it anywhere on your computer that you wish. Open the cubetwister program and find and open this file with it. Left click the script “Algorithm Factory” in the left margin and select the “5x5x5 Numbered” and “WCA 5x5x5 Notation” (by left clicking the * looking icon).

The following is a random 60 move scramble from qqtimer (WCA Notation).

_Also note that the solution to this scramble will be in WCA Notation as well (the same version of WCA notation at alg.garron.us)._

L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

Note that there is an even number of outer layer turns and an even number of inner layer turns. Thus the above scramble generates a position in the commutator subgroup (not necessarily true for fixed centers on the odd supercube, however, as *Lemma 4* in my document states).

Here is a 1642 btm 2 commutator solution to this random 5x5x5 supercube scramble (except for fixed centers).

[
f2D'fLfrB'D2l'L'Dbr'b'M'UF'r'FrLUfF2lF2l'DbL'S'F2l'F2lbfL'f'DR'fU2b2D'b2R'bS'RSR'f'D'f'D'fDf2R'fR2bR'b'fD2f'b'RbR'U'bU'b2D2bD2b'R2bF2Db'R2bB2D'b'R2bRFR'fR2f'D'fR2f'URFR'b'R2bB'D'b'R2bR'b'R2bU'BD'fR2f'B2Rb'R2bU'R2UfR2f'RSR2S'B'FR'SR2S'B'R'SR2S'UDSR2S'F'R'SR2S'F'B'R'SR2S'B2FR'S'U2SB'D'SR2S'R'F'RSR2S'LFU'f2U2R2f2R2U2f2R2bUf'U'b'UfU'RbU'S'Ub'U'SU2bUf'U'b'UfRbU'f'S'Ub'U'fSBwRulu'R'ul'u'B'U'S'U'f'USU'fU2b'FL2B'R'x'y'DRU'B'RFU2LD'B'LFR2UD2L2F2UD2B2U
,
R2F2L'F2U2L2U'B2DFL2D'R2U2LB2R2UR2F'y2DFR2L2U2Bd'B2dB'EBd'B'E'B'R'uRdR'u'RB2d'RuE'R'dRu'ER'uRd'R'u'RdBdRd'R'uRdR'u'd'Bu2B2R2u2R2B2u2B2R'E'R2ER'E'R2ER'D2R'EB2E'BDUB'EB2E'BUDBLEB2E'R'UB2REB2E'RD'B2R'EB2E'B'E'R2EFuR2u'R2d'R2dF'uR2u'F2uR2u'L'F2dB2d'LU2FuR2u'R'uR2u'R'D'Rd'R2dR2L2u'B2uBD'B'u'B2uR2uF2u'd'R'dBR'uRu'BR'd'F2dF2uRu'Bu'Ru'BuR'u2RuF'u2BlD2l'Fd'F'dR'FrD2r'D'fDf'D'RlBl'B'lB'l2D2lE'dBd'R'B'u2LR'S'R'SRE'B'b2U'SD'S'D'l'F'D'rB'MUM'
]


[
FL'D'LF'U2FL'DLF'U2R'B'EBUB'E'BU'RRFE'F'U'FEF'UR'FlrUl'U'r'UlU'l'F2BU2r'U'l'UrU'lU'Dbf'D'bDfD'b'Db'D2bDf'D'b'DfUb'f'L'bLfL'b'Lbu2d2R'u2Rd2R'u2Ru2R'uFd2R'u'Rd2R'uRF'u'RU'B'F'l'UrU'lUr'U2l'UrU'lUr'F2
,
D'B'DS2D'BDS2LU'BE'B'UBEB'L2UF'EFU'F'E'FL2RBd'B'u'BdB'uR'L'B'd'F'd'FuF'dFu'dBUL2Ub'U'f'UbU'fL2U2FlF'r2Fl'F'r2U2l2Ur2U'l2Ur2U2F'rFl2F'r'Fl2d'FdF'u'Fd'F'udL2u2Ld2L'u2Ld2LB2L'd'LuL'dLu'Ld'LuL'dLu'L2rBl'B'r'BlB2rBl'B'r'BlB2D2F2DB2D'B'F2DF'D'BDB2FD
]

The scramble does not generate a fixed center position which can be solved in the commutator subgroup, period, as it is the “cycle class” {180,-90,-90,-90,-90} which sums to an odd multiple of 180. Therefore, the 2 commutator solution doesn’t affect the fixed centers at all.

Note that solving the 5x5x5 supercube wasn’t more difficult than solving the regular 5x5x5 cube because numbering the non-fixed center pieces helps to find and setup pieces more efficiently.

Before I show the actual cube moves I used to create the 2 commutator solution above with, I must show the “plans” I made _first_. If you don’t make a plan, you’re setting yourself up for misery (and you most likely will never be able to solve the scramble you are working on because it becomes trial and error rather than having a systematic process to follow).

Note that the term “iteration” is proper to use with written number solutions, but they are “conjugates” when actually make moves which follow these written number solutions (the written number solutons are like “plans” for “how to build”—how to solve—an nxnxn cube scramble).

I still call the conjugates “iterations” in the actual move solutions to be consistent with their “plans”.



Spoiler: The Written Plan (with comments)






Spoiler: Corners



Let {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP]} = the solved state.
By the scramble, we have {6[SUP]+[/SUP],1[SUP]-[/SUP],8[SUP]0[/SUP],3[SUP]+[/SUP],4[SUP]+[/SUP],2[SUP]-[/SUP],5[SUP]0[/SUP],7[SUP]-[/SUP]} (A 5-cycle and a 3-cycle)

That is, (1→2→6)(3→4→5→7→8)

Iteration I (orientation fix):
(1[SUP]-(+)[/SUP]↔6[SUP]+(0)[/SUP]) (3[SUP]+(-)[/SUP]↔2[SUP]-(0)[/SUP]) (4[SUP]+(+)[/SUP]↔7[SUP]-(-)[/SUP]) (5[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) *4 2-cycle*
→ {1[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],8[SUP]0[/SUP],4[SUP]0[/SUP]}
Iteration II (solve at least 2 pieces):
(2[SUP]0(0)[/SUP]↔6[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (7[SUP]0(0)[/SUP]↔5[SUP]0(0)[/SUP]) (3[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *4 2-cycle*
→ {3[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],1[SUP]0[/SUP],4[SUP]0[/SUP],8[SUP]0[/SUP]}
Iteration III: (We now have a 5-cycle)
(7[SUP]0(0)[/SUP]↔3[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *2 2-cycle*
→ {7[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],4[SUP]0[/SUP],1[SUP]0[/SUP],8[SUP]0[/SUP]}
Iteration IV:
(4[SUP]0(0)[/SUP]↔6[SUP]0(0)[/SUP]) (7[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *2 2-cycle*
→ {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP]}

Notes:
By default, since we have a 5-cycle and a 3-cycle, we cannot solve the permutation in one commutator because we have other piece orbits involved (if at least two middle edges and at least two X-center pieces were untouched by the scramble—as the two of the + center pieces are—then we would be able to solve this corner cycle class in one commutator, but we don’t have that). That is, the orientations were not the reason why the corner solution had to be solved with 2 commutators instead of one.

The superscripts, just in case you haven’t noticed, are of the form:
[SUP]current orientation (the change in orientation when put into swap slot)[/SUP].
When we actually solve using cube moves, 6[SUP]+(0)[/SUP], for example, means to put 6 rotated clockwise in the swap slot (you add both signs in the superscript to get the required orientation in which the corner piece needs to be in when it is in a swap slot).

If you are swapping all 8 corners and the sum of all signs in all swaps in your written solution does not add to ±3, ±6, or 0 (or if you are swapping 12 edges and the sum is an odd integer), then you must revise your solution. In *Table 2* in the document, there is more than one option to choose from. So make it work. For example, here is an “alternate solution” for the first iteration to the corner scramble.
(1[SUP]-(+)[/SUP]↔6[SUP]+(0)[/SUP]) (3[SUP]+(+)[/SUP]↔2[SUP]-(-)[/SUP]) (4[SUP]+(+)[/SUP]↔7[SUP]-(-)[/SUP]) (5[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) → {1[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],8[SUP]0[/SUP],4[SUP]0[/SUP]}
The sum of the signs is 0 (the net sign for 1) + 1 (the net sign for 6) + -1 (the net sign for 3) + 1 (the net sign for 2) + -1 (the net sign for 4) + 1 (the net sign for 7) + 0 (the net sign for 5) + 0 (the net sign for 8
= 0 + 1 – 1 + 1 – 1 + 1 + 0 + 0 = 1, which is not ±3, ±6, or 0. Thus we would have to violate the cube law for orientation of corners in order to pull this stunt off! LOL. But the sum of the first iteration for the solution which was shown first is 0 which is acceptable.





Spoiler: Middle Edges



Let {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}= the solved state.
By the scramble, we have {8[SUP]+[/SUP],12[SUP]+[/SUP],9[SUP]+[/SUP],5[SUP]0[/SUP],10[SUP]0[/SUP],6[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]+[/SUP],3[SUP]0[/SUP],11[SUP]+[/SUP],4[SUP]+[/SUP],1[SUP]0[/SUP]} (5-cycle, 4-cycle, and a 2-cycle)

That is, (1→12→2→7→8)(3↔9)(4→11→10→5)

Iteration I (Orientation Fix):
(8[SUP]+(+)[/SUP]↔12[SUP]+(0)[/SUP]) (9[SUP]+(+)[/SUP]↔7[SUP]+(0)[/SUP]) (11[SUP]+(+)[/SUP]↔4[SUP]+(0)[/SUP]) (5[SUP]0(0)[/SUP]↔10[SUP]0(0)[/SUP])* 4 2-cycle*
→ {12[SUP]0[/SUP],8[SUP]0[/SUP],7[SUP]0[/SUP],10[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],2[SUP]0[/SUP],9[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],11[SUP]0[/SUP],1[SUP]0[/SUP]}
Iteration II (solve at least 2 pieces):
(1[SUP]0(0)[/SUP]↔12[SUP]0(0)[/SUP]) (2[SUP]0(0)[/SUP]↔7[SUP]0(0)[/SUP]) (3[SUP]0(0)[/SUP]↔9[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔10[SUP]0(0)[/SUP])* 4 2-cycle*
→ {1[SUP]0[/SUP],8[SUP]0[/SUP],2[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}
Iteration III (we know have a 3-cycle):
(2[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (1[SUP]0(0)[/SUP]↔4[SUP]0(0)[/SUP])
→ {4[SUP]0[/SUP],2[SUP]0[/SUP],8[SUP]0[/SUP],1[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}* 2 2-cycle*
Iteration IV:
(3[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (1[SUP]0(0)[/SUP]↔4[SUP]0(0)[/SUP])
→ {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}* 2 2-cycle*

Notes: The cycle class {5,4,2} can be solved in one commutator regardless if the corners, X-center pieces, or the + center pieces could be solved in one commutator. That is, the orientation case forced us to need to use 2 commutators instead of one.





Spoiler: Wing Edges



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {20,18,5,12,2,1,19,6,7,15,4,21,3,*14*,13,11,24,8,23,16,17,*22*,9,10} (an 18-cycle and a 4-cycle).

That is, (1→6→8→18→2→5→3→13→15→10→24→17→21→12→4→1→16→20)(7→9→23→19)(14)(22)

The two bold numbers above were untouched by the scramble. If at least two numbers (pieces) are untouched by a scramble, it is a given that we can solve that orbit in one commutator (see *Corollary (1)* in the document).

To solve this massive scramble with ease, we need to make a “sketch” first, and then use *Lemma 1* to transform that into the exact scramble we have.

Using the technique (m > n) in the proof of *Lemma 6* (on pages 16-17 in the document), we can obtain a solution for the cycle class {18,4} pretty easily.

_Sketch_
*By Lemma 1*, we can represent _any_ {18,4} as the following:
{18,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,22,19,20,21,23,24}

By the technique (m > n), the solution to this cycle class is
Iteration I: (18↔17)(1↔16)(2↔15)(3↔14)(4↔13)(5↔12)(6↔11)(7↔10)(8↔9) and (22↔20) *10 2-cycle*
→ {17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,17,20,19,22,21,23,24}
Iteration II: (1↔17)(2↔16)(3↔15)(4↔14)(5↔13)(6↔12)(7↔11)(8↔10) and (22↔21)(19↔20) *10 2-cycle*
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

_Back to our specific version of the cycle class {18,4}_
We create a “translation key” by numbering each number in “arrowed cycle representation” according to our sketch.






By “translating” from the solution to our sketch to our actual scramble (by using the key above), we get the one commutator solution:

Iteration I: (20↔16)(1↔11)(6↔4)(8↔12)(18↔21)(2↔17)(5↔24)(3↔10)(13↔15) and (19↔9) *10 2-cycle*
→ {16,21,24,8,17,11,9,4,7,13,6,18,10,14,15,1,5,12,23,20,2,22,19,3}
Iteration II: (1↔16)(6↔11)(8↔4)(18↔12)(2↔21)(5↔17)(3↔24)(13↔10) and (19↔23)(7↔9) *10 2-cycle*
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





Spoiler: X-Centers



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {15,21,13,4,22,18,12,2,1,14,24,9,19,16,3,6,11,23,20,8,5,10,17,7} (23-cycle)

That is,





(As you can see, I made a similar “translation key” as was done for wing edges).

_Sketch_
By *Lemma 1*, the following representation represents _all_ 23-cycles.
{23,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24}

Unfortunately, the 23-cycle is one of the very worst cases for big cube parts. Only one piece is solved (4 in our actual scramble) (or 24 in the representation above) and therefore we cannot have the luxury promised by *Corollary (1)* to be able to solve it in one commutator. We could solve it in one commutator using an odd number 2-cycle solution if the corners, middle edges, and + center pieces all had at least 2 pieces solved. The + center pieces do (as we will see next), but the middle edges and corners do not.

In the document, I actually solved the _sketch_ above to prove that all even permutation cycle classes for 24 objects could be solved in 2 commutators or less (considering permutations only—not taking orientations into account).

Just copying and pasting the solution directly from page 22 of the document,

Let's break it up into 5 sections by marking 8 numbers/pieces.
{*23*,1,2,3,*4*,*5*,6,7,8,*9*,*10*,11,12,13,*14*,*15*,16,17,18,*19*,20,21,22,24}

Iteration I: (4↔23)(5↔9)(10↔14)(15↔19)* 4 2-cycle*
{4,1,2,3,*23*,9,6,7,8,*5*,14,11,12,13,*10*,19,16,17,18,*15*,20,21,22,24}

Iteration II: (20↔23)(5↔15)(22↔24)(7↔9)* 4 2-cycle*
{4,1,2,3,*20*,7,6,9,8,*15*,14,11,12,13,*10*,19,16,17,18,*5*,23,21,24,22}

Iteration III: (1↔4)(2↔3)(11↔14)(12↔13)(16↔19)(17↔18)(21↔23)(22↔24)* 8 2-cycle*
{1,4,3,2,*20*,7,6,9,8,*15*,11,14,13,12,*10*,16,19,18,17,*5*,21,23,22,24}

Iteration IV: (2↔4)(12↔14)(17↔19)(22↔23)(5↔20)(10↔15)(6↔7)(8↔9)* 8 2-cycle*
= solved state.

*Note*: If an even permutation cycle class does not have at least 2 pieces solved and the number of required 2-cycles for one commutator is odd, the solution to the 23-cycle above is yet another solving technique which you can use (splitting up the large cycles into increments of 4…).

_Back to our scramble_.
If we translate the solution to the _sketch_ to our actual scramble, we find that one solution to our 23-cycle is the following:

Iteration I: (7↔15)(24↔18)(6↔22)(5↔20)* 4 2-cycle*
→ {7,21,13,4,6,24,12,2,1,14,18,9,19,16,3,22,11,23,5,8,20,10,17,15}
Iteration II: (19↔15)(24↔5)(3↔4)(17↔18)* 4 2-cycle*
→ {7,21,13,3,6,5,12,2,1,14,17,9,15,16,4,22,11,23,24,8,20,10,18,19}
Iteration III: (1↔7)(9↔12)(16↔22)(14↔10)(21↔20)(2↔8)(13↔15)(3↔4)* 8 2-cycle*
→ {1,20,15,4,6,5,9,8,7,10,17,12,13,22,3,16,11,23,24,2,21,14,18,19}
Iteration IV: (9↔7)(14↔22)(2↔20)(3↔15)(24↔19)(6↔5)(11↔17)(23↔18)* 8 2-cycle*
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





Spoiler: + Centers



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {20,21,3,14,17,1,18,11,2,6,12,5,22,4,24,16,23,10,13,19,7,15,9,8} (20-cycle and a 2-cycle).

That is,





The corresponding _sketch_ based on the translation key above is from the m > n technique on pages 16-17. Note that since (20+2)/2 = 11 = an odd integer, we do not need to use the 2 unsolved pieces as a “dummy” pair to be able to solve this cycle class in one commutator (which doesn’t conflict with any other piece orbits).

_Sketch_
{20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,22,21,23,24}

Iteration I: (20↔19)(1↔18)(2↔17)(3↔16)(4↔15)(5↔14)(6↔13)(7↔12)(8↔11)(9↔10) *10 2-cycle*
→ {19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,20,22,21,23,24}
Iteration II: (1↔19)(2↔18)(3↔17)(4↔16)(5↔15)(6↔14)(7↔13)(8↔12)(9↔11) and (22↔21) *10 2-cycle*
→{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

Using the translation key to translate the solution to the _sketch_ into a solution for our specific version of the cycle class {20,2}, we get:
Iteration I: (20↔19)(1↔13)(6↔22)(10↔15)(18↔24)(7↔8)(21↔11)(2↔12)(9↔5)(23↔17) *10 2-cycle*
→{19,11,3,14,23,13,24,21,12,22,2,9,6,4,18,16,17,15,1,20,8,10,5,7}
Iteration II: (1↔19)(6↔13)(10↔22)(18↔15)(7↔24)(21↔8)(2↔11)(9↔12)(23↔5)(4↔14) *10 2-cycle*
→{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





Now the actual cube move solution begins. For convenience, this is the official plan we created (without all of the explanation and notes about techniques, etc.) so that we can refer to it often as we make moves on the cube.



Spoiler: The Written Plan






Spoiler: Corners



Let {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP]} = the solved state.
By the scramble, we have {6[SUP]+[/SUP],1[SUP]-[/SUP],8[SUP]0[/SUP],3[SUP]+[/SUP],4[SUP]+[/SUP],2[SUP]-[/SUP],5[SUP]0[/SUP],7[SUP]-[/SUP]} (A 5-cycle and a 3-cycle)

That is, (1→2→6)(3→4→5→7→8)


Iteration I (Orientation Fix, using *Table 2* in the document)
(1[SUP]-(+)[/SUP]↔6[SUP]+(0)[/SUP]) (3[SUP]+(-)[/SUP]↔2[SUP]-(0)[/SUP]) (4[SUP]+(+)[/SUP]↔7[SUP]-(-)[/SUP]) (5[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) *4 2-cycle*
→ {1[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],8[SUP]0[/SUP],4[SUP]0[/SUP]}
Iteration II (solve at least 2 pieces):
(2[SUP]0(0)[/SUP]↔6[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (7[SUP]0(0)[/SUP]↔5[SUP]0(0)[/SUP]) (3[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *4 2-cycle*
→ {3[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],1[SUP]0[/SUP],4[SUP]0[/SUP],8[SUP]0[/SUP]}
Iteration III:
(7[SUP]0(0)[/SUP]↔3[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *2 2-cycle*
→ {7[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],6[SUP]0[/SUP],5[SUP]0[/SUP],4[SUP]0[/SUP],1[SUP]0[/SUP],8[SUP]0[/SUP]}
Iteration IV:
(4[SUP]0(0)[/SUP]↔6[SUP]0(0)[/SUP]) (7[SUP]0(0)[/SUP]↔1[SUP]0(0)[/SUP]) *2 2-cycle*
→ {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP]}





Spoiler: Middle Edges



Let {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}= the solved state.
By the scramble, we have {8[SUP]+[/SUP],12[SUP]+[/SUP],9[SUP]+[/SUP],5[SUP]0[/SUP],10[SUP]0[/SUP],6[SUP]0[/SUP],2[SUP]0[/SUP],7[SUP]+[/SUP],3[SUP]0[/SUP],11[SUP]+[/SUP],4[SUP]+[/SUP],1[SUP]0[/SUP]} (5-cycle, 4-cycle, and a 2-cycle)

That is, (1→12→2→7→8)(3↔9)(4→11→10→5)


Iteration I (Orientation Fix, using *Table 1* in the document)
(8[SUP]+(+)[/SUP]↔12[SUP]+(0)[/SUP]) (9[SUP]+(+)[/SUP]↔7[SUP]+(0)[/SUP]) (11[SUP]+(+)[/SUP]↔4[SUP]+(0)[/SUP]) (5[SUP]0(0)[/SUP]↔10[SUP]0(0)[/SUP])* 4 2-cycle*
→ {12[SUP]0[/SUP],8[SUP]0[/SUP],7[SUP]0[/SUP],10[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],2[SUP]0[/SUP],9[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],11[SUP]0[/SUP],1[SUP]0[/SUP]}
Iteration II (solve at least 2 pieces):
(1[SUP]0(0)[/SUP]↔12[SUP]0(0)[/SUP]) (2[SUP]0(0)[/SUP]↔7[SUP]0(0)[/SUP]) (3[SUP]0(0)[/SUP]↔9[SUP]0(0)[/SUP]) (4[SUP]0(0)[/SUP]↔10[SUP]0(0)[/SUP])* 4 2-cycle*
→ {1[SUP]0[/SUP],8[SUP]0[/SUP],2[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}
Iteration III:
(2[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (1[SUP]0(0)[/SUP]↔4[SUP]0(0)[/SUP]) → {4[SUP]0[/SUP],2[SUP]0[/SUP],8[SUP]0[/SUP],1[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],3[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}* 2 2-cycle*
Iteration IV:
(3[SUP]0(0)[/SUP]↔8[SUP]0(0)[/SUP]) (1[SUP]0(0)[/SUP]↔4[SUP]0(0)[/SUP]) → {1[SUP]0[/SUP],2[SUP]0[/SUP],3[SUP]0[/SUP],4[SUP]0[/SUP],5[SUP]0[/SUP],6[SUP]0[/SUP],7[SUP]0[/SUP],8[SUP]0[/SUP],9[SUP]0[/SUP],10[SUP]0[/SUP],11[SUP]0[/SUP],12[SUP]0[/SUP]}* 2 2-cycle*





Spoiler: Wing Edges



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {20,18,5,12,2,1,19,6,7,15,4,21,3,14,13,11,24,8,23,16,17,22,9,10} (an 18-cycle and a 4-cycle).

That is, (1→6→8→18→2→5→3→13→15→10→24→17→21→12→4→1→16→20)(7→9→23→19)(14)(22)

Iteration I: (20↔16)(1↔11)(6↔4)(8↔12)(18↔21)(2↔17)(5↔24)(3↔10)(13↔15) and (19↔9) *10 2-cycle*
→ {16,21,24,8,17,11,9,4,7,13,6,18,10,14,15,1,5,12,23,20,2,22,19,3}
Iteration II: (1↔16)(6↔11)(8↔4)(18↔12)(2↔21)(5↔17)(3↔24)(13↔10) and (19↔23)(7↔9) *10 2-cycle*
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





Spoiler: X-Centers



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {15,21,13,4,22,18,12,2,1,14,24,9,19,16,3,6,11,23,20,8,5,10,17,7} (23-cycle)

That is, (1→9→12→7→24→11→17→23→18→6→16→14→10→22→5→21→2→8→20→19→13→3→15)(4)


Iteration I: (7↔15)(24↔18)(6↔22)(5↔20)* 4 2-cycle*
→ {7,21,13,4,6,24,12,2,1,14,18,9,19,16,3,22,11,23,5,8,20,10,17,15}
Iteration II: (19↔15)(24↔5)(3↔4)(17↔18)* 4 2-cycle*
→ {7,21,13,3,6,5,12,2,1,14,17,9,15,16,4,22,11,23,24,8,20,10,18,19}
Iteration III: (1↔7)(9↔12)(16↔22)(14↔10)(21↔20)(2↔8)(13↔15)(3↔4)* 8 2-cycle*
→ {1,20,15,4,6,5,9,8,7,10,17,12,13,22,3,16,11,23,24,2,21,14,18,19}
Iteration IV: (9↔7)(14↔22)(2↔20)(3↔15)(24↔19)(6↔5)(11↔17)(23↔18)* 8 2-cycle*
→ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





Spoiler: + Centers



Let {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} = the solved state.
By the scramble, we have {20,21,3,14,17,1,18,11,2,6,12,5,22,4,24,16,23,10,13,19,7,15,9,8} (20-cycle and a 2-cycle).

(1→6→10→18→7→21→2→9→23→17→5→12→11→8→24→15→22→13→19→20)(4↔14)(3)(16)

Iteration I: (20↔19)(1↔13)(6↔22)(10↔15)(18↔24)(7↔8)(21↔11)(2↔12)(9↔5)(23↔17) *10 2-cycle*
→{19,11,3,14,23,13,24,21,12,22,2,9,6,4,18,16,17,15,1,20,8,10,5,7}
Iteration II: (1↔19)(6↔13)(10↔22)(18↔15)(7↔24)(21↔8)(2↔11)(9↔12)(23↔5)(4↔14) *10 2-cycle*
→{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}





We will handle each piece type one at a time, combine all solutions into 4 (large) composite conjugates, rewrite as two large commutators, and finally shorten and simplify the 2 commutators into how it was first presented.

Copy and paste each of the following into the cubetwister script and clip the check mark to apply to the cube, one at a time, and follow along with the comments (I cannot promise perfect spelling or complete sentence structure, as I was commenting as I went) and, at the same time, look at the “plan” above to see why we chose to do what we did.

Also, for the z algorithms (z1, (z1)', z2, and (z2)'), choose the “5x5x5 super” cube to see how they affect the centers.

Lastly, note that I put the scramble at the beginning of each of them.



Spoiler: Cube Move Solution






Spoiler: Corner Solution



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/--------------------Iteration I--------------------
/put 1 0 across from 6+ in U layer.
[B, U'F' U]
/put 7+ in U layer
[F, U B' U']
/put 4- opp. of 7+ in U layer
U[F', U' B U] U'
/put 5 0 opp of 8 0 and 2- opp 3 0 in D layer
F'U F[F2, D' B' D] F' U' F

/diagonal 4 2-cycle in U and D Layers = z1
B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
(
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
)'

/--------------------Iteration II--------------------
/2 0 is opp of 6 0 already in U layer.
/put 4 0 opp. of 8 0 in D layer
F R [L,F' R' F] R' F'
/put 7 0 opp. of 5 0 in U layer and put 1 0 opp. of 3 0 in D layer
B U' R' B' R [L2, D R2 D'] R' B R U B'

/diagonal 4 2-cycle in U and D Layers = z1 = (z1)'
B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'

(
F R [L,F' R' F] R' F'
B U' R' B' R [L2, D R2 D'] R' B R U B'
)'

/--------------------Iteration III--------------------
/7 0 is already opp. to 3 0 in U layer.
/Put 4 0 in U Layer
[L2, U' R U]
/put 1 0 opp. of 4 0 in U Layer
D' [R2, U' L2 U] D

/Diagonal 2 2-cycle in U Layer = z2
L R U2 L' R' U F' B' U2 F B U'

(
[L2, U' R U]
D' [R2, U' L2 U] D
)'

/--------------------Iteration IV--------------------
/put 4 0 opp. to 6 0 in U layer
D' [R2, U' L2 U] D
/put 1 0 opp. to 7 0 in U Layer
D [L2, UR' U'] D'

/Diagonal 2 2-cycle in U Layer = z2 = (z2)'
L R U2 L' R' U F' B' U2 F B U'

(
D' [R2, U' L2 U] D
D [L2, UR' U'] D'
)'





Spoiler: Middle Edge Solution



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/--------------------Iteration I--------------------
/put 8 0 opp. of 12 + in U layer
F' R'[L', U M' U'] R F
/put 9 0 opp. of 7 + in D layer
D [M',D' R D] D'
/put 11 0 in U Layer (4+ is also moved opp. of 11 0 in U layer)
L[M', B L' B'] L'
/put 5 0 opp. of 10 0 in D layer
F' [B, D' S'D] F

/double H perm 4 2-cycle in U and D layers = z1
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]

(
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
)'

/--------------------Iteration II--------------------
/put 1 0 opp. of 12 0 in U layer
U' R [L', U M' U'] R' U
/put 4 0 opp. of 10 0 in U layer
L[R', U M U'] L'
/put 3 0 opp. of 9 0 in D layer
D' B[F', D S'D'] B' D
/put 7 0 opp. of 2 0 in D layer
B'[F, D' S D] B

/double H perm 4 2-cycle in U and D layers = z1 = (z1)'
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]

(
U' R [L', U M' U'] R' U
L[R', U M U'] L'
D' B[F', D S'D'] B' D
B'[F, D' S D] B
)'


/--------------------Iteration III--------------------
/put 8 0 opp. of 2 0 and 1 0 opp. of 4 0 in U layer
R F [E', F' U' F] F' R'

/H perm 2 2-cycle in U layer = z2
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'

(
R F [E', F' U' F] F' R'
)'

/--------------------Iteration IV--------------------
/put 4 0 opp. of 1 0 in U layer
F' R' [E, R U R'] R F
/put 3 0 opp. of 8 0 in U layer
U' R' [L2,U' M U] R U

/H perm 2 2-cycle in U layer = z2 = (z2)'
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'

(
F' R' [E, R U R'] R F
U' R' [L2,U' M U] R U
)'





Spoiler: Wing Edge Solution



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/--------------------Iteration I--------------------
/put 16 adj. to 20 in UF
[l',B L' B']
/put 8 adj. to 12 in UL
R2 [b2, U B' U'] R2
/put 21 adj. to 18 in UR
F2 L'[R, U l U'] L F2
/put 5 adj. to 24 in UB
r [u', L' DL] r'
/19 is already adj. to 9 in DB
/put 11 adj. to 1 in DR
F2 [R',D r' D'] F2
/2 is already adj. to 17 in DF
/put 3 adj. to 10 in DL
F2 R' [L,D r D'] R F2
/6 and 4 are already adjacent in FR
/put 13 in FL
R2 [L2, F r F'] R2
/put 15 adj. to 13 in FL
R2 [L2,F' r' F] R2

/10 2-cycle = z1
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'

(
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
)'

/--------------------Iteration II--------------------
/put 1 adj. to 16 in UF
[l, D' L D]
/put 4 adj. to 8 in UL
D2 [b2, UB' U'] D2
/put 3 adj. to 24 in UB
D' [r',UR' U'] D
/put 18 adj. to 12 in UR
F [U',R d R'] F'
/put 2 adj. to 21 in DR
F' [RF' R', f'] F
/5 is already adj. to 17 in DF
/put 10 adj. to 13 in DL
F2 R [L,D' l D] R' F2
/put 23 adj. to 19 in DB
[u,R D' R']
/6 and 11 are already in FL
/put 9 adj. to 7 in FR
[R f R', B2]

/10 2-cycle = z1 = (z1)'
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'

(
[l, D' L D]
D2 [b2, UB' U'] D2
D' [r',UR' U'] D
F [U',R d R'] F'
F' [RF' R', f'] F
F2 R [L,D' l D] R' F2
[u,R D' R']
[R f R', B2]
)'





Spoiler: X-Center Solution



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/--------------------Iteration I--------------------
/put 7 in Ulb (opp. to 15)
[b',U' f U]
/put 6 in Dlb (opp. to 22)
L[b,D f'D'] L'
/put 24 in Dlf (opp. to 18)
D' [l', D' r D] D
/put 5 in Ulf
[l',U' r U]
/put 20 in Urb (opp. to 5)
B2 [r', U' l U] B2

/4 2-cycle = z1
[[l' r' u2 l r, U], x2]

(
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
)'

/--------------------Iteration II--------------------
/put 5 in Dlf (opp to 24)
D2 [r, D l' D'] D2
/put 3 in Urb (opp. to 4)
U' [l,U r' U'] U
/put 17 in Urf
U2 [l,U r'U'] U2
/put 18 in Ulb (opp to 17)
B2[l, U r' U'] B2
/put 15 in Drf
L' [f', D b D'] L
/put 19 in Dlb (opp to 15)
F[l,D' r' D] F'

/4 2-cycle = z1= (z1)'
[[l' r' u2 l r, U], x2]

(
D2 [r, D l' D'] D2
U' [l,U r' U'] U
U2 [l,U r'U'] U2
B2[l, U r' U'] B2
L' [f', D b D'] L
F[l,D' r' D] F'
)'

/--------------------Iteration III--------------------
/put 1 in Ulb (opp to 7)
F'[l', U r U'] F
/put 20 in Ulf (opp. to 21)
[l', U' r U]
/put 8 in Drb (opp to 2)
D[b,D f' D'] D'
/put 9 in Drf (opp to 12)
L2 [f', D b D'] L2
/13 is already opp. 15 in the F layer
/put 4 in Flu (opp. to 3)
B2 [u2, F' d F] B2
/put 16 in Bld
[d,B' u' B]
/put 22 in Bru (opp. to 16)
B2 [d', B' u B] B2
/put 10 in Brd
[d, B u' B']
/put 14 in Blu (opp. to 10)
R' [u', B d B'] R

/8 2-cycle = z2
[[[l' r' u2 l r, U], x2], x]

(
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R
)'

/--------------------Iteration IV--------------------
/put 2 in Ulf (opp. to 20)
L2 [f, U b' U'] L2
/put 3 in Urf (opp. to 15)
U2 [l, U r' U'] U2
/put 6 in Dlb (opp. to 5)
r F' r [l, D r' D'] r' F r'
/put 7 in Drf (opp. to 9)
[r', D' l D]
/put 17 in Fld (opp. to 11)
F2[u2,Fd' F'] F2
/put 19 in Bld (opp. to 24)
L' [d', B' uB] L
/put 14 in Brd (opp. to 22)
[d, B u' B']
/put 18 in Frd
[d, F' u' F]
/put 23 in Flu (opp. to 18)
R [u, F' d' F] R'

/8 2-cycle = z2 = (z2)'
[[[l' r' u2 l r, U], x2], x]

(
L2 [f, U b' U'] L2
U2 [l, U r' U'] U2
r F' r [l, D r' D'] r' F r'
[r', D' l D]
F2[u2,Fd' F'] F2
L' [d', B' uB] L
[d, B u' B']
[d, F' u' F]
R [u, F' d' F] R'
)'





Spoiler: + Center Solution



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/--------------------Iteration I--------------------
/put 19 in UMb (opp. to 20)
L' [b', U S U'] L
/put 13 in DlS (opp. to 1)
B [l',D M' D'] B'
/put 24 in DMf (opp. to 18)
[M', D' l' D]
/put 11 in UlS (opp. to 21)
U [M, U r'U'] U'
/put 6 in FMu (opp. to 22)
F [E', F d' F'] F'
/put 8 in FrE (opp. to 7)
F2[E, F u' F'] F2
/put 12 in LdS (opp. of 2)
B2 [d,L' E L] B2
/put 9 in LEf (opp. of 5)
L[d, L' E L] L'
/put 23 in BMu (opp. of 17)
R2 [u', B E B'] R2
/put 10 in BEr
R' [E, B u' B'] R
/put 15 in BEl (opp. of 10)
R' [E, B' u' B] R

/10 2-cycle = z1
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R
)'


/--------------------Iteration II--------------------
/put 1 in UMb (opp. to 19)
B2 [M, U r' U'] B2
/put 6 in DSl (opp. to 13)
F [l,D' M D] F'
/put 2 in USl (opp. to 11)
B [l,U' M U] B'
/put 5 in DMb (opp. to 23)
[M',D l' D']
/21 is already opp. to 8 in F
/put 9 in FMu (opp. to 12)
L' [u', F E F'] L
/put 4 in BEl (opp. to 14)
B[d, B' E B] B'
/put 15 in BMu (opp. to 18)
[u', B E B']
/put 24 in LEf (opp. to 7)
R d [E2, L d' L'] d' R'
/put 10 in LSu
R E' [u2, L' E' L] E R'
/put 22 in LSd (opp. to 10)
R' E' [d2,L E' L'] E R

/10 2-cycle = z1 = (z1)'
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
B2 [M, U r' U'] B2
F [l,D' M D] F'
B [l,U' M U] B'
[M',D l' D']
L' [u', F E F'] L
B[d, B' E B] B'
[u', B E B']
R d [E2, L d' L'] d' R'
R E' [u2, L' E' L] E R'
R' E' [d2,L E' L'] E R
)'





Now we can combine them all into one macro 4 iteration (conjugate) solution as follows (because they are a direct product).


Spoiler



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/2328 btm

/--------------------Iteration I--------------------
/***Corner solution***
/put 1 0 across from 6+ in U layer.
[B, U'F' U]
/put 7+ in U layer
[F, U B' U']
/put 4- opp. of 7+ in U layer
U[F', U' B U] U'
/put 5 0 opp of 8 0 and 2- opp 3 0 in D layer
F'U F[F2, D' B' D] F' U' F

/***Middle Edge solution***
/put 8 0 opp. of 12 + in U layer
F' R'[L', U M' U'] R F
/put 9 0 opp. of 7 + in D layer
D [M',D' R D] D'
/put 11 0 in U Layer (4+ is also moved opp. of 11 0 in U layer)
L[M', B L' B'] L'
/put 5 0 opp. of 10 0 in D layer
F' [B, D' S'D] F

/***Wing Edge Solution***
/put 16 adj. to 20 in UF
[l',B L' B']
/put 8 adj. to 12 in UL
R2 [b2, U B' U'] R2
/put 21 adj. to 18 in UR
F2 L'[R, U l U'] L F2
/put 5 adj. to 24 in UB
r [u', L' DL] r'
/19 is already adj. to 9 in DB
/put 11 adj. to 1 in DR
F2 [R',D r' D'] F2
/2 is already adj. to 17 in DF
/put 3 adj. to 10 in DL
F2 R' [L,D r D'] R F2
/6 and 4 are already adjacent in FR
/put 13 in FL
R2 [L2, F r F'] R2
/put 15 adj. to 13 in FL
R2 [L2,F' r' F] R2

/***X-center solution
/put 7 in Ulb (opp. to 15)
[b',U' f U]
/put 6 in Dlb (opp. to 22)
L[b,D f'D'] L'
/put 24 in Dlf (opp. to 18)
D' [l', D' r D] D
/put 5 in Ulf
[l',U' r U]
/put 20 in Urb (opp. to 5)
B2 [r', U' l U] B2

/***+ Center Solution
/put 19 in UMb (opp. to 20)
L' [b', U S U'] L
/put 13 in DlS (opp. to 1)
B [l',D M' D'] B'
/put 24 in DMf (opp. to 18)
[M', D' l' D]
/put 11 in UlS (opp. to 21)
U [M, U r'U'] U'
/put 6 in FMu (opp. to 22)
F [E', F d' F'] F'
/put 8 in FrE (opp. to 7)
F2[E, F u' F'] F2
/put 12 in LdS (opp. of 2)
B2 [d,L' E L] B2
/put 9 in LEf (opp. of 5)
L[d, L' E L] L'
/put 23 in BMu (opp. of 17)
R2 [u', B E B'] R2
/put 10 in BEr
R' [E, B u' B'] R
/put 15 in BEl (opp. of 10)
R' [E, B' u' B] R

/***corner z1***
/diagonal 4 2-cycle in U and D Layers = z1
B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
/***Middle Edge z1***
/double H perm 4 2-cycle in U and D layers = z1
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]
/***Wind Edge z1***
/10 2-cycle = z1
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'
/***X-center z1***
/4 2-cycle = z1
[[l' r' u2 l r, U], x2]
/***+ Center z1
/10 2-cycle = z1
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'


(
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F

F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F

[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2

[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2

L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R
)'

/--------------------Iteration II--------------------
/***Corner solution***
/2 0 is opp of 6 0 already in U layer.
/put 4 0 opp. of 8 0 in D layer
F R [L,F' R' F] R' F'
/put 7 0 opp. of 5 0 in U layer and put 1 0 opp. of 3 0 in D layer
B U' R' B' R [L2, D R2 D'] R' B R U B'

/***Middle Edge solution***
/put 1 0 opp. of 12 0 in U layer
U' R [L', U M' U'] R' U
/put 4 0 opp. of 10 0 in U layer
L[R', U M U'] L'
/put 3 0 opp. of 9 0 in D layer
D' B[F', D S'D'] B' D
/put 7 0 opp. of 2 0 in D layer
B'[F, D' S D] B

/***Wing Edge Solution***
/put 1 adj. to 16 in UF
[l, D' L D]
/put 4 adj. to 8 in UL
D2 [b2, UB' U'] D2
/put 3 adj. to 24 in UB
D' [r',UR' U'] D
/put 18 adj. to 12 in UR
F [U',R d R'] F'
/put 2 adj. to 21 in DR
F' [RF' R', f'] F
/5 is already adj. to 17 in DF
/put 10 adj. to 13 in DL
F2 R [L,D' l D] R' F2
/put 23 adj. to 19 in DB
[u,R D' R']
/6 and 11 are already in FL
/put 9 adj. to 7 in FR
[R f R', B2]

/***X-center Solution***
/put 5 in Dlf (opp to 24)
D2 [r, D l' D'] D2
/put 3 in Urb (opp. to 4)
U' [l,U r' U'] U
/put 17 in Urf
U2 [l,U r'U'] U2
/put 18 in Ulb (opp to 17)
B2[l, U r' U'] B2
/put 15 in Drf
L' [f', D b D'] L
/put 19 in Dlb (opp to 15)
F[l,D' r' D] F'

/***+ Center Solution
/put 1 in UMb (opp. to 19)
B2 [M, U r' U'] B2
/put 6 in DSl (opp. to 13)
F [l,D' M D] F'
/put 2 in USl (opp. to 11)
B [l,U' M U] B'
/put 5 in DMb (opp. to 23)
[M',D l' D']
/21 is already opp. to 8 in F
/put 9 in FMu (opp. to 12)
L' [u', F E F'] L
/put 4 in BEl (opp. to 14)
B[d, B' E B] B'
/put 15 in BMu (opp. to 18)
[u', B E B']
/put 24 in LEf (opp. to 7)
R d [E2, L d' L'] d' R'
/put 10 in LSu
R E' [u2, L' E' L] E R'
/put 22 in LSd (opp. to 10)
R' E' [d2,L E' L'] E R

/***corner (z1)' ***
/diagonal 4 2-cycle in U and D Layers = z1 = (z1)'
B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
/***middle edge (z1)' ***
/double H perm 4 2-cycle in U and D layers = z1 = (z1)'
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]
/***Wing Edge (z1)'***
/10 2-cycle = z1 = (z1)'
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'
/***X-Center (z1)'***
/4 2-cycle = z1= (z1)'
[[l' r' u2 l r, U], x2]
/***+ Center (z1)'***
/10 2-cycle = z1 = (z1)'
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
F R [L,F' R' F] R' F'
B U' R' B' R [L2, D R2 D'] R' B R U B'

U' R [L', U M' U'] R' U
L[R', U M U'] L'
D' B[F', D S'D'] B' D
B'[F, D' S D] B

[l, D' L D]
D2 [b2, UB' U'] D2
D' [r',UR' U'] D
F [U',R d R'] F'
F' [RF' R', f'] F
F2 R [L,D' l D] R' F2
[u,R D' R']
[R f R', B2]

D2 [r, D l' D'] D2
U' [l,U r' U'] U
U2 [l,U r'U'] U2
B2[l, U r' U'] B2
L' [f', D b D'] L
F[l,D' r' D] F'

B2 [M, U r' U'] B2
F [l,D' M D] F'
B [l,U' M U] B'
[M',D l' D']
L' [u', F E F'] L
B[d, B' E B] B'
[u', B E B']
R d [E2, L d' L'] d' R'
R E' [u2, L' E' L] E R'
R' E' [d2,L E' L'] E R
)'


/--------------------Iteration III--------------------
/***corner solution***
/7 0 is already opp. to 3 0 in U layer.
/Put 4 0 in U Layer
[L2, U' R U]
/put 1 0 opp. of 4 0 in U Layer
D' [R2, U' L2 U] D

/***Middle Edge solution***
/put 8 0 opp. of 2 0 and 1 0 opp. of 4 0 in U layer
R F [E', F' U' F] F' R'

/***X-Center Solution***
/put 1 in Ulb (opp to 7)
F'[l', U r U'] F
/put 20 in Ulf (opp. to 21)
[l', U' r U]
/put 8 in Drb (opp to 2)
D[b,D f' D'] D'
/put 9 in Drf (opp to 12)
L2 [f', D b D'] L2
/13 is already opp. 15 in the F layer
/put 4 in Flu (opp. to 3)
B2 [u2, F' d F] B2
/put 16 in Bld
[d,B' u' B]
/put 22 in Bru (opp. to 16)
B2 [d', B' u B] B2
/put 10 in Brd
[d, B u' B']
/put 14 in Blu (opp. to 10)
R' [u', B d B'] R

/***corner z2***
/Diagonal 2 2-cycle in U Layer = z2
L R U2 L' R' U F' B' U2 F B U'
/***Middle edge z2***
/H perm 2 2-cycle in U layer = z2
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'
/***X-Center z2***
/8 2-cycle = z2
[[[l' r' u2 l r, U], x2], x]


(
[L2, U' R U]
D' [R2, U' L2 U] D

R F [E', F' U' F] F' R'

F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R
)'

/--------------------Iteration IV--------------------
/***corner solution***
/put 4 0 opp. to 6 0 in U layer
D' [R2, U' L2 U] D
/put 1 0 opp. to 7 0 in U Layer
D [L2, UR' U'] D'

/***middle ede solution***
/put 4 0 opp. of 1 0 in U layer
F' R' [E, R U R'] R F
/put 3 0 opp. of 8 0 in U layer
U' R' [L2,U' M U] R U

/***X-Center Solution***
/put 2 in Ulf (opp. to 20)
L2 [f, U b' U'] L2
/put 3 in Urf (opp. to 15)
U2 [l, U r' U'] U2
/put 6 in Dlb (opp. to 5)
r F' r [l, D r' D'] r' F r'
/put 7 in Drf (opp. to 9)
[r', D' l D]
/put 17 in Fld (opp. to 11)
F2[u2,Fd' F'] F2
/put 19 in Bld (opp. to 24)
L' [d', B' uB] L
/put 14 in Brd (opp. to 22)
[d, B u' B']
/put 18 in Frd
[d, F' u' F]
/put 23 in Flu (opp. to 18)
R [u, F' d' F] R'

/***corner (z2)' ***
/Diagonal 2 2-cycle in U Layer = z2 = (z2)'
L R U2 L' R' U F' B' U2 F B U'
/***middle edge (z2)' ***
/H perm 2 2-cycle in U layer = z2 = (z2)'
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'
/***X-Center (z2)'
/8 2-cycle = z2 = (z2)'
[[[l' r' u2 l r, U], x2], x]

(
D' [R2, U' L2 U] D
D [L2, UR' U'] D'

F' R' [E, R U R'] R F
U' R' [L2,U' M U] R U

L2 [f, U b' U'] L2
U2 [l, U r' U'] U2
r F' r [l, D r' D'] r' F r'
[r', D' l D]
F2[u2,Fd' F'] F2
L' [d', B' uB] L
[d, B u' B']
[d, F' u' F]
R [u, F' d' F] R'
)'


Now, removing comments,


Spoiler



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R

B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'
[[l' r' u2 l r, U], x2]
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R
)'



F R [L,F' R' F] R' F'
B U' R' B' R [L2, D R2 D'] R' B R U B'
U' R [L', U M' U'] R' U
L[R', U M U'] L'
D' B[F', D S'D'] B' D
B'[F, D' S D] B
[l, D' L D]
D2 [b2, UB' U'] D2
D' [r',UR' U'] D
F [U',R d R'] F'
F' [RF' R', f'] F
F2 R [L,D' l D] R' F2
[u,R D' R']
[R f R', B2]
D2 [r, D l' D'] D2
U' [l,U r' U'] U
U2 [l,U r'U'] U2
B2[l, U r' U'] B2
L' [f', D b D'] L
F[l,D' r' D] F'
B2 [M, U r' U'] B2
F [l,D' M D] F'
B [l,U' M U] B'
[M',D l' D']
L' [u', F E F'] L
B[d, B' E B] B'
[u', B E B']
R d [E2, L d' L'] d' R'
R E' [u2, L' E' L] E R'
R' E' [d2,L E' L'] E R

B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'
[[l' r' u2 l r, U], x2]
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
F R [L,F' R' F] R' F'
B U' R' B' R [L2, D R2 D'] R' B R U B'
U' R [L', U M' U'] R' U
L[R', U M U'] L'
D' B[F', D S'D'] B' D
B'[F, D' S D] B
[l, D' L D]
D2 [b2, UB' U'] D2
D' [r',UR' U'] D
F [U',R d R'] F'
F' [RF' R', f'] F
F2 R [L,D' l D] R' F2
[u,R D' R']
[R f R', B2]
D2 [r, D l' D'] D2
U' [l,U r' U'] U
U2 [l,U r'U'] U2
B2[l, U r' U'] B2
L' [f', D b D'] L
F[l,D' r' D] F'
B2 [M, U r' U'] B2
F [l,D' M D] F'
B [l,U' M U] B'
[M',D l' D']
L' [u', F E F'] L
B[d, B' E B] B'
[u', B E B']
R d [E2, L d' L'] d' R'
R E' [u2, L' E' L] E R'
R' E' [d2,L E' L'] E R
)'



[L2, U' R U]
D' [R2, U' L2 U] D
R F [E', F' U' F] F' R'
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R

L R U2 L' R' U F' B' U2 F B U'
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'
[[[l' r' u2 l r, U], x2], x]

(
[L2, U' R U]
D' [R2, U' L2 U] D
R F [E', F' U' F] F' R'
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R
)'



D' [R2, U' L2 U] D
D [L2, UR' U'] D'
F' R' [E, R U R'] R F
U' R' [L2,U' M U] R U
L2 [f, U b' U'] L2
U2 [l, U r' U'] U2
r F' r [l, D r' D'] r' F r'
[r', D' l D]
F2[u2,Fd' F'] F2
L' [d', B' uB] L
[d, B u' B']
[d, F' u' F]
R [u, F' d' F] R'

L R U2 L' R' U F' B' U2 F B U'
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'
[[[l' r' u2 l r, U], x2], x]

(
D' [R2, U' L2 U] D
D [L2, UR' U'] D'
F' R' [E, R U R'] R F
U' R' [L2,U' M U] R U
L2 [f, U b' U'] L2
U2 [l, U r' U'] U2
r F' r [l, D r' D'] r' F r'
[r', D' l D]
F2[u2,Fd' F'] F2
L' [d', B' uB] L
[d, B u' B']
[d, F' u' F]
R [u, F' d' F] R'
)'


We have 4 conjugates of the form (A z1 A') (B (z1)' B') (C z2 C') (D (z2)' D') (all z’s even permutations). On page 10 of the document, we can rewrite this as the two commutators [A z1 A', B A'] [C z2 C', D C'].


Spoiler



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'
/4116 btm

[
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R

B2 F2 L R' B2 U2 L2 R2 U2 F2 L R'
[M2 U2 M2 U M2 U2 M2 U', R L' f2 S2 b2 L R' x2]
[l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U', z2]
x U l' U2 l2 F2 r' D2 r' D2 r2 F2 l' U2 U' x'
[[l' r' u2 l r, U], x2]
[[[S' u2: [M u d M', U2]],x2],x] z [S' u2: [M u d M', U2]] z'

(
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R
)'

,

F R [L,F' R' F] R' F'
B U' R' B' R [L2, D R2 D'] R' B R U B'
U' R [L', U M' U'] R' U
L[R', U M U'] L'
D' B[F', D S'D'] B' D
B'[F, D' S D] B
[l, D' L D]
D2 [b2, UB' U'] D2
D' [r',UR' U'] D
F [U',R d R'] F'
F' [RF' R', f'] F
F2 R [L,D' l D] R' F2
[u,R D' R']
[R f R', B2]
D2 [r, D l' D'] D2
U' [l,U r' U'] U
U2 [l,U r'U'] U2
B2[l, U r' U'] B2
L' [f', D b D'] L
F[l,D' r' D] F'
B2 [M, U r' U'] B2
F [l,D' M D] F'
B [l,U' M U] B'
[M',D l' D']
L' [u', F E F'] L
B[d, B' E B] B'
[u', B E B']
R d [E2, L d' L'] d' R'
R E' [u2, L' E' L] E R'
R' E' [d2,L E' L'] E R

(
[B, U'F' U]
[F, U B' U']
U[F', U' B U] U'
F'U F[F2, D' B' D] F' U' F
F' R'[L', U M' U'] R F
D [M',D' R D] D'
L[M', B L' B'] L'
F' [B, D' S'D] F
[l',B L' B']
R2 [b2, U B' U'] R2
F2 L'[R, U l U'] L F2
r [u', L' DL] r'
F2 [R',D r' D'] F2
F2 R' [L,D r D'] R F2
R2 [L2, F r F'] R2
R2 [L2,F' r' F] R2
[b',U' f U]
L[b,D f'D'] L'
D' [l', D' r D] D
[l',U' r U]
B2 [r', U' l U] B2
L' [b', U S U'] L
B [l',D M' D'] B'
[M', D' l' D]
U [M, U r'U'] U'
F [E', F d' F'] F'
F2[E, F u' F'] F2
B2 [d,L' E L] B2
L[d, L' E L] L'
R2 [u', B E B'] R2
R' [E, B u' B'] R
R' [E, B' u' B] R
)'

]

[
[L2, U' R U]
D' [R2, U' L2 U] D
R F [E', F' U' F] F' R'
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R

L R U2 L' R' U F' B' U2 F B U'
[R' B': [E, B U B']] U [R' B': [E, B U B']] U'
[[[l' r' u2 l r, U], x2], x]

(
[L2, U' R U]
D' [R2, U' L2 U] D
R F [E', F' U' F] F' R'
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R
)'

,

D' [R2, U' L2 U] D
D [L2, UR' U'] D'
F' R' [E, R U R'] R F
U' R' [L2,U' M U] R U
L2 [f, U b' U'] L2
U2 [l, U r' U'] U2
r F' r [l, D r' D'] r' F r'
[r', D' l D]
F2[u2,Fd' F'] F2
L' [d', B' uB] L
[d, B u' B']
[d, F' u' F]
R [u, F' d' F] R'

(
[L2, U' R U]
D' [R2, U' L2 U] D
R F [E', F' U' F] F' R'
F'[l', U r U'] F
[l', U' r U]
D[b,D f' D'] D'
L2 [f', D b D'] L2
B2 [u2, F' d F] B2
[d,B' u' B]
B2 [d', B' u B] B2
[d, B u' B']
R' [u', B d B'] R
)'
]


Deleting all returns and some spaces, we have:


Spoiler



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'

/4056 btm 2 comm. solution.

[
[B,U'F'U][F,UB'U']U[F',U'BU]U'F'UF[F2,D'B'D]F'U'F F'R'[L',UM'U']RF D[M',D'RD]D' L[M',BL'B']L' F'[B,D'S'D]F [l',BL'B'] R2[b2,UB'U']R2 F2L'[R,UlU']LF2 r[u',L'DL]r' F2[R',Dr'D']F2 F2R'[L,DrD']RF2 R2[L2,FrF']R2 R2[L2,F'r'F]R2 [b',U'fU] L[b,Df'D']L' D'[l',D'rD]D [l',U'rU] B2[r',U'lU]B2 L'[b',USU']L B[l',DM'D']B' [M',D'l'D] U[M,Ur'U']U' F[E',Fd'F']F' F2[E,Fu'F']F2 B2[d,L'EL]B2 L[d,L'EL]L' R2[u',BEB']R2 R'[E,Bu'B']R R'[E,B'u'B]R B2F2LR'B2U2L2R2U2F2LR' [M2U2M2UM2U2M2U',RL'f2S2b2LR'x2] [l'U2l2F2r'D2r'D2r2F2l'U2Ul'U2l2F2r'D2r'D2r2F2l'U2U',z2] xUl'U2l2F2r'D2r'D2r2F2l'U2U'x' [[l'r'u2lr,U],x2] [[[S'u2:[MudM',U2]],x2],x]z[S'u2:[MudM',U2]]z' ( [B,U'F'U] [F,UB'U'] U[F',U'BU]U' F'UF[F2,D'B'D]F'U'F F'R'[L',UM'U']RF D[M',D'RD]D' L[M',BL'B']L' F'[B,D'S'D]F [l',BL'B'] R2[b2,UB'U']R2 F2L'[R,UlU']LF2 r[u',L'DL]r' F2[R',Dr'D']F2 F2R'[L,DrD']RF2 R2[L2,FrF']R2 R2[L2,F'r'F]R2 [b',U'fU] L[b,Df'D']L' D'[l',D'rD]D [l',U'rU] B2[r',U'lU]B2 L'[b',USU']L B[l',DM'D']B' [M',D'l'D] U[M,Ur'U']U' F[E',Fd'F']F' F2[E,Fu'F']F2 B2[d,L'EL]B2 L[d,L'EL]L' R2[u',BEB']R2 R'[E,Bu'B']R R'[E,B'u'B]R )'
,
F R [L,F' R' F] R' F' B U' R' B' R [L2, D R2 D'] R' B R U B' U' R [L', U M' U'] R' U L[R', U M U'] L' D' B[F', D S'D'] B' D B'[F, D' S D] B [l, D' L D] D2 [b2, UB' U'] D2 D' [r',UR' U'] D F [U',R d R'] F' F' [RF' R', f'] F F2 R [L,D' l D] R' F2 [u,R D' R'] [R f R', B2] D2 [r, D l' D'] D2 U' [l,U r' U'] U U2 [l,U r'U'] U2 B2[l, U r' U'] B2 L' [f', D b D'] L F[l,D' r' D] F' B2 [M, U r' U'] B2 F [l,D' M D] F' B [l,U' M U] B' [M',D l' D'] L' [u', F E F'] L B[d, B' E B] B' [u', B E B'] R d [E2, L d' L'] d' R' R E' [u2, L' E' L] E R' R' E' [d2,L E' L'] E R ( [B, U'F' U] [F, U B' U'] U[F', U' B U] U' F'U F[F2, D' B' D] F' U' F F' R'[L', U M' U'] R F D [M',D' R D] D' L[M', B L' B'] L' F' [B, D' S'D] F [l',B L' B'] R2 [b2, U B' U'] R2 F2 L'[R, U l U'] L F2 r [u', L' DL] r' F2 [R',D r' D'] F2 F2 R' [L,D r D'] R F2 R2 [L2, F r F'] R2 R2 [L2,F' r' F] R2 [b',U' f U] L[b,D f'D'] L' D' [l', D' r D] D [l',U' r U] B2 [r', U' l U] B2 L' [b', U S U'] L B [l',D M' D'] B' [M', D' l' D] U [M, U r'U'] U' F [E', F d' F'] F' F2[E, F u' F'] F2 B2 [d,L' E L] B2 L[d, L' E L] L' R2 [u', B E B'] R2 R' [E, B u' B'] R R' [E, B' u' B] R )'
]

[
[L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R LRU2L'R'UF'B'U2FBU' [R'B':[E,BUB']]U[R'B':[E,BUB']]U' [[[l'r'u2lr,U],x2],x] ( [L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R )'
,
D'[R2,U'L2U]D D[L2,UR'U']D' F'R'[E,RUR']RF U'R'[L2,U'MU]RU L2[f,Ub'U']L2 U2[l,Ur'U']U2 rF'r[l,Dr'D']r'Fr' [r',D'lD] F2[u2,Fd'F']F2 L'[d',B'uB]L [d,Bu'B'] [d,F'u'F] R[u,F'd'F]R' ( [L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R )'
]


We have a two commutator solution. Let’s call it [A,B][C,D]. We can shorten this solution by letting A, B, C, and D be scrambles and solving them separately. I have not had a lot of practice solving supercubes, so the following solutions are far from being optimal. I used CubeExplorer for the 3x3x3 part. (You must use center facelet twists for the 4x4x4 and greater cubes because the centers on those cubes contain non-fixed center pieces…and their exact positioning DOES MATTER).

*Important Note*. We do not have to worry about solving the fixed centers, even though I solved most of them in all 4 solutions.


Spoiler: Optimization






Spoiler: A



*A for commutator 1 (838 btm originally)*
[B,U'F'U][F,UB'U']U[F',U'BU]U'F'UF[F2,D'B'D]F'U'F F'R'[L',UM'U']RF D[M',D'RD]D' L[M',BL'B']L' F'[B,D'S'D]F [l',BL'B'] R2[b2,UB'U']R2 F2L'[R,UlU']LF2 r[u',L'DL]r' F2[R',Dr'D']F2 F2R'[L,DrD']RF2 R2[L2,FrF']R2 R2[L2,F'r'F]R2 [b',U'fU] L[b,Df'D']L' D'[l',D'rD]D [l',U'rU] B2[r',U'lU]B2 L'[b',USU']L B[l',DM'D']B' [M',D'l'D] U[M,Ur'U']U' F[E',Fd'F']F' F2[E,Fu'F']F2 B2[d,L'EL]B2 L[d,L'EL]L' R2[u',BEB']R2 R'[E,Bu'B']R R'[E,B'u'B]R B2F2LR'B2U2L2R2U2F2LR' [M2U2M2UM2U2M2U',RL'f2S2b2LR'x2] [l'U2l2F2r'D2r'D2r2F2l'U2Ul'U2l2F2r'D2r'D2r2F2l'U2U',z2] xUl'U2l2F2r'D2r'D2r2F2l'U2U'x' [[l'r'u2lr,U],x2] [[[S'u2:[MudM',U2]],x2],x]z[S'u2:[MudM',U2]]z' ( [B,U'F'U] [F,UB'U'] U[F',U'BU]U' F'UF[F2,D'B'D]F'U'F F'R'[L',UM'U']RF D[M',D'RD]D' L[M',BL'B']L' F'[B,D'S'D]F [l',BL'B'] R2[b2,UB'U']R2 F2L'[R,UlU']LF2 r[u',L'DL]r' F2[R',Dr'D']F2 F2R'[L,DrD']RF2 R2[L2,FrF']R2 R2[L2,F'r'F]R2 [b',U'fU] L[b,Df'D']L' D'[l',D'rD]D [l',U'rU] B2[r',U'lU]B2 L'[b',USU']L B[l',DM'D']B' [M',D'l'D] U[M,Ur'U']U' F[E',Fd'F']F' F2[E,Fu'F']F2 B2[d,L'EL]B2 L[d,L'EL]L' R2[u',BEB']R2 R'[E,Bu'B']R R'[E,B'u'B]R )'

f2 D' f L f r B' D2 l' L' D b r' b' M'
U F' r' F r LU f F2lF2 l' D b L' S' F2 l' F2 l
b f L' f' DR' f U2 b2 D' b' b' R' bS' R S R' f2
f D' f' D' f D f' f' R' fR2 bR' b' f D2 f'
b' R b R' U' b U' b'b' D2 b

D2b' R2 b
F2 D b' R2 b
B2 D'b' R2 b
R F R' f R2 f'
D' f R2 f'
U R F R' b' R2 b
B' D' b' R2 b
R' b' R2 b
U' B D' f R2 f'
B2 R b' R2 b
U' R2 U f R2 f'

R SR2 S'
B' F R' S R2 S'
B' R' S R2 S'
U D S R2 S'
F' R' S R2 S'
F' B' R' S R2 S'
B2 F R' S' U2 S
B' D' S R2 S'
R' F' R S R2 S'
L F U' f2 U2 R2 f2 R2 U2 f2

R2 b U f' U' b' U f U' R b U' S' U b' U' S
U2 b U f' U' b' U f R b U' f' S' U b' U' f S
Bw R u lu' R' u l' u' Bw'
b U' S' U'f' U S U' f U2 b'

F L2 B' R' x' y' D R U' B' R F U2 L D' B' L F R2 U D2 L2 F2 U D2 B2 U

= 273 btm
f2D'fLfrB'D2l'L'Dbr'b'M'UF'r'FrLUfF2lF2l'DbL'S'F2l'F2lbfL'f'DR'fU2b2D'b2R'bS'RSR'f'D'f'D'fDf2R'fR2bR'b'fD2f'b'RbR'U'bU'b2D2bD2b'R2bF2Db'R2bB2D'b'R2bRFR'fR2f'D'fR2f'URFR'b'R2bB'D'b'R2bR'b'R2bU'BD'fR2f'B2Rb'R2bU'R2UfR2f'RSR2S'B'FR'SR2S'B'R'SR2S'UDSR2S'F'R'SR2S'F'B'R'SR2S'B2FR'S'U2SB'D'SR2S'R'F'RSR2S'LFU'f2U2R2f2R2U2f2R2bUf'U'b'UfU'RbU'S'Ub'U'SU2bUf'U'b'UfRbU'f'S'Ub'U'fSBwRulu'R'ul'u'B'U'S'U'f'USU'fU2b'FL2B'R'x'y'DRU'B'RFU2LD'B'LFR2UD2L2F2UD2B2U





Spoiler: B



*B for commutator 1 (630 btm originally)*
F R [L,F' R' F] R' F' B U' R' B' R [L2, D R2 D'] R' B R U B' U' R [L', U M' U'] R' U L[R', U M U'] L' D' B[F', D S'D'] B' D B'[F, D' S D] B [l, D' L D] D2 [b2, UB' U'] D2 D' [r',UR' U'] D F [U',R d R'] F' F' [RF' R', f'] F F2 R [L,D' l D] R' F2 [u,R D' R'] [R f R', B2] D2 [r, D l' D'] D2 U' [l,U r' U'] U U2 [l,U r'U'] U2 B2[l, U r' U'] B2 L' [f', D b D'] L F[l,D' r' D] F' B2 [M, U r' U'] B2 F [l,D' M D] F' B [l,U' M U] B' [M',D l' D'] L' [u', F E F'] L B[d, B' E B] B' [u', B E B'] R d [E2, L d' L'] d' R' R E' [u2, L' E' L] E R' R' E' [d2,L E' L'] E R ( [B, U'F' U] [F, U B' U'] U[F', U' B U] U' F'U F[F2, D' B' D] F' U' F F' R'[L', U M' U'] R F D [M',D' R D] D' L[M', B L' B'] L' F' [B, D' S'D] F [l',B L' B'] R2 [b2, U B' U'] R2 F2 L'[R, U l U'] L F2 r [u', L' DL] r' F2 [R',D r' D'] F2 F2 R' [L,D r D'] R F2 R2 [L2, F r F'] R2 R2 [L2,F' r' F] R2 [b',U' f U] L[b,D f'D'] L' D' [l', D' r D] D [l',U' r U] B2 [r', U' l U] B2 L' [b', U S U'] L B [l',D M' D'] B' [M', D' l' D] U [M, U r'U'] U' F [E', F d' F'] F' F2[E, F u' F'] F2 B2 [d,L' E L] B2 L[d, L' E L] L' R2 [u', B E B'] R2 R' [E, B u' B'] R R' [E, B' u' B] R )'


M U' M' B r' D F l D S D S' U b2
B E R' S' R S R L' u2B R dB' d' E
l' D2 ll B l' B l B' l' R' D f D' f' D r D2 r' F' R d' F d F' l D2 l'
B' u2 F u2 uR' u' u' R u' B' u R' u B' u R' u' F2 d' F2 d R B' u R' u' R B' d' R d u F2u'

R2 u' B2 u
B DB' u' B2 u
L2 R2 d' R2 d
R' D R u R2 u'
R uR2 u'
F' U2 L' d B2 d'
F2 Lu R2 u'
F2 u R2 u'
F d' R2 d
R2 u R2 u'

F' E' R2E
B E B2 E'
R B2 D R' EB2 E'
R' B2 U' R E B2 E'
L' B' D' U' B' E B2 E'
BU' D' B' E B2E'
R D2 R E' R2E
R E' R2 E
RB2 u2 B2 R2 u2 R2 B2 u2

B' du Rd' R' u' R d R' d' B' d' R' u R dR' u' RE' uR' d' R Eu' R' d 
B2 R' u R d' R' u' R d
d'B E B d B'E' B d' B2d
B' U2 L2R2 F' D' y2 F R2 U' R2 B2 L' U2 R2 D L2 F' D' B2 U L2 U2 F2 L F2 R2 

= 266 btm
MU'M'Br'DFlDSDS'Ub2BER'S'RSRL'u2BRdB'd'El'D2l2Bl'BlB'l'R'DfD'f'DrD2r'F'Rd'FdF'lD2l'B'u2Fu'R'u2Ru'B'uR'uB'uR'u'F2d'F2dRB'uR'u'RB'd'RduF2u'R2u'B2uBDB'u'B2uL2R2d'R2dR'DRuR2u'RuR2u'F'U2L'dB2d'F2LuR2u'F2uR2u'Fd'R2dR2uR2u'F'E'R2EBEB2E'RB2DR'EB2E'R'B2U'REB2E'L'B'D'U'B'EB2E'BU'D'B'EB2E'RD2RE'R2ERE'R2ERB2u2B2R2u2R2B2u2B'duRd'R'u'RdR'd'B'd'R'uRdR'u'RE'uR'd'REu'R'dB2R'uRd'R'u'RBEBdB'E'Bd'B2dB'U2L2R2F'D'y2FR2U'R2B2L'U2R2DL2F'D'B2UL2U2F2LF2R2





Spoiler: C



*C of 2nd commutator (314 btm originally)*
[L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R LRU2L'R'UF'B'U2FBU' [R'B':[E,BUB']]U[R'B':[E,BUB']]U' [[[l'r'u2lr,U],x2],x] ( [L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R )'

F L' D' L F' U2 F L' D L F' U2
R' B' E B UB' E' B U'RR F E' F' U' F E F'UR'
F l rU l' U' r' Ul U' l' F
B F U2 r' U' l' U r U'l U'
D b f' D' b D f D' b'Db' D2 b D f' D'b' D f U b' f' L' b L f L'b' Lb
u2d2 R' u2 R d2 R' u2Ru2 R' u Fd2 R' u' R d2 R'uR F' u' RU' B' F' l' U r U' l U r' U'
U' l' U r U' l U r' F2
= (127 btm)
FL'D'LF'U2FL'DLF'U2R'B'EBUB'E'BU'RRFE'F'U'FEF'UR'FlrUl'U'r'UlU'l'F2BU2r'U'l'UrU'lU'Dbf'D'bDfD'b'Db'D2bDf'D'b'DfUb'f'L'bLfL'b'Lbu2d2R'u2Rd2R'u2Ru2R'uFd2R'u'Rd2R'uRF'u'RU'B'F'l'UrU'lUr'U2l'UrU'lUr'F2





Spoiler: D



*D of 2nd commutator (246 btm originally)*
D'[R2,U'L2U]D D[L2,UR'U']D' F'R'[E,RUR']RF U'R'[L2,U'MU]RU L2[f,Ub'U']L2 U2[l,Ur'U']U2 rF'r[l,Dr'D']r'Fr' [r',D'lD] F2[u2,Fd'F']F2 L'[d',B'uB]L [d,Bu'B'] [d,F'u'F] R[u,F'd'F]R' ( [L2,U'RU] D'[R2,U'L2U]D RF[E',F'U'F]F'R' F'[l',UrU']F [l',U'rU] D[b,Df'D']D' L2[f',DbD']L2 B2[u2,F'dF]B2 [d,B'u'B] B2[d',B'uB]B2 [d,Bu'B'] R'[u',BdB']R )'

D' F' B2 D' B' D F D' F2 B D B2 D' F2 D2
B2 l' B' r B lB' r' B2 l' B' r B lB' r' L2 u L'd' L u'L' d L'uL' d' L u'L' d L B2 L' d2L' u2 L d2L' u2 L2 d' u' F d F' u F d'F' d l2 F' r F l2F' r' F U2r2 U' l2 U r2 U' l2U'
U' r2Fl F' r2 F l' F' U2 L2 f' U b' U' f U bU' L2 U' B' d' uF' d' F u' F' dF d B L R u' B d' B' uB d B' R' L'
L' F' E F U F' E'F U' L L B E'B' U' B E B'U L' S2 D' B' D S2 D' B D

= (155 btm)
D'F'B2D'B'DFD'F2BDB2D'F2D2B2l'B'rBlB'r'B2l'B'rBlB'r'L2uL'd'Lu'L'dL'uL'd'Lu'L'dLB2L'd2L'u2Ld2L'u2L2d'u'FdF'uFd'F'dl2F'rFl2F'r'FU2r2U'l2Ur2U'l2U2r2FlF'r2Fl'F'U2L2f'Ub'U'fUbU'L2U'B'd'uF'd'Fu'F'dFdBLRu'Bd'B'uBdB'R'L2F'EFUF'E'FU'L2BE'B'U'BEB'UL'S2D'B'DS2D'BD





Now we substitute our new solutions for A, B, C, and D into our two commutators [A,B][C,D]. Note that we must take the inverse of each before doing so.

*However*, we do not need to take the inverse of A and C, since they are equal to their inverses (recall the structures [A z1 A', B A'][C z2 C', D C'], where z1 and z2 are a product of disjoint 2-cycles).



Spoiler



L Bw B' D2 U' Bw Lw Fw' U2 F2 Lw Bw' L B Lw Uw D Bw2 D B Dw' Fw F L' R2 U' Uw' Lw B Fw' D2 Uw' Dw2 Lw' Dw' U2 Bw2 B L2 F U2 R2 F' Rw' R2 B Lw U' Dw' B L' Lw Bw U' Bw' D' Dw' R2 B Uw'
/1642 btm 2 commutator solution

[
f2D'fLfrB'D2l'L'Dbr'b'M'UF'r'FrLUfF2lF2l'DbL'S'F2l'F2lbfL'f'DR'fU2b2D'b2R'bS'RSR'f'D'f'D'fDf2R'fR2bR'b'fD2f'b'RbR'U'bU'b2D2bD2b'R2bF2Db'R2bB2D'b'R2bRFR'fR2f'D'fR2f'URFR'b'R2bB'D'b'R2bR'b'R2bU'BD'fR2f'B2Rb'R2bU'R2UfR2f'RSR2S'B'FR'SR2S'B'R'SR2S'UDSR2S'F'R'SR2S'F'B'R'SR2S'B2FR'S'U2SB'D'SR2S'R'F'RSR2S'LFU'f2U2R2f2R2U2f2R2bUf'U'b'UfU'RbU'S'Ub'U'SU2bUf'U'b'UfRbU'f'S'Ub'U'fSBwRulu'R'ul'u'B'U'S'U'f'USU'fU2b'FL2B'R'x'y'DRU'B'RFU2LD'B'LFR2UD2L2F2UD2B2U
,
(MU'M'Br'DFlDSDS'Ub2BER'S'RSRL'u2BRdB'd'El'D2l2Bl'BlB'l'R'DfD'f'DrD2r'F'Rd'FdF'lD2l'B'u2Fu'R'u2Ru'B'uR'uB'uR'u'F2d'F2dRB'uR'u'RB'd'RduF2u'R2u'B2uBDB'u'B2uL2R2d'R2dR'DRuR2u'RuR2u'F'U2L'dB2d'F2LuR2u'F2uR2u'Fd'R2dR2uR2u'F'E'R2EBEB2E'RB2DR'EB2E'R'B2U'REB2E'L'B'D'U'B'EB2E'BU'D'B'EB2E'RD2RE'R2ERE'R2ERB2u2B2R2u2R2B2u2B'duRd'R'u'RdR'd'B'd'R'uRdR'u'RE'uR'd'REu'R'dB2R'uRd'R'u'RBEBdB'E'Bd'B2dB'U2L2R2F'D'y2FR2U'R2B2L'U2R2DL2F'D'B2UL2U2F2LF2R2)'
]

[
FL'D'LF'U2FL'DLF'U2R'B'EBUB'E'BU'RRFE'F'U'FEF'UR'FlrUl'U'r'UlU'l'F2BU2r'U'l'UrU'lU'Dbf'D'bDfD'b'Db'D2bDf'D'b'DfUb'f'L'bLfL'b'Lbu2d2R'u2Rd2R'u2Ru2R'uFd2R'u'Rd2R'uRF'u'RU'B'F'l'UrU'lUr'U2l'UrU'lUr'F2
,
(D'F'B2D'B'DFD'F2BDB2D'F2D2B2l'B'rBlB'r'B2l'B'rBlB'r'L2uL'd'Lu'L'dL'uL'd'Lu'L'dLB2L'd2L'u2Ld2L'u2L2d'u'FdF'uFd'F'dl2F'rFl2F'r'FU2r2U'l2Ur2U'l2U2r2FlF'r2Fl'F'U2L2f'Ub'U'fUbU'L2U'B'd'uF'd'Fu'F'dFdBLRu'Bd'B'uBdB'R'L2F'EFUF'E'FU'L2BE'B'U'BEB'UL'S2D'B'DS2D'BD)'
]


Lastly, we actually take the inverses of B and D to arrive at the form of the 2 commutator solution originally shown.


----------



## siva.shanmukh (Oct 17, 2012)

There have been so many posts since I have last seen this thread! And looks like cmowla's proof is 14 pages! This is going to take a while for me to go through it completely! A lot of work. Hats off just for the amount of it and the detail in which it is written.


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## Christopher Mowla (Oct 17, 2012)

siva.shanmukh said:


> There have been so many posts since I have last seen this thread! And looks like cmowla's proof is 14 pages! This is going to take a while for me to go through it completely! A lot of work. Hats off just for the amount of it and the detail in which it is written.


Thanks for taking interest, and welcome back!

I actually just did some major format editing for the "ultimate conjugate challenge" (I neglected it for a long time because I was trying to perfect the commutator document), and I merged it with the commutator document. I don't know if it was the best idea to merge them, but the document looks pretty good (it's so long, it's like a book now, even without the appendices). You should just look at the document below instead of the commutator one because it is up to date.

Commutator and Conjugate Theory​
You don't need to read past page 43 if you're only interested in the commutator proofs (which is the material which really matters...i.e., see the remark for the conjugate proof at the bottom of the page before Appendix A. That could have been enough proof for the non-rigorous people).

To everyone, I have added more explanation about the operation tables (in addition to the detailed example I just gave) in the commutator section "The Method" (to hopefully make things clearer if you all didn't see the example). I have added some images throughout the document as well.

Unfortunately when I first posted the 5x5x5 example, I didn't realize that the spoilers were going to squish the "permutation keys" for the scrambles of the wing edges, X-centers, and +-centers. I edited that post earlier today and just made the keys images so that the formatting I intended to display is kept. So if you were confused when you saw the "plan (with comments)", go look at it again.


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## Christopher Mowla (Jan 17, 2013)

If I did the following calculations correctly,

Ignoring the orientations of corners and middle edges completely (just calculating the number of permutations of the corners, middle edges, and big cube parts), the percentage of even permutations for the nxnxn supercube which *cannot* be solved with only one commutator (remember, I proved that 100% of permutations--including orientations of the corners and middle edges--of the regular nxnxn Rubik's cube can be solved with 2 commutators) is:

2x2x2: 0% (there are no conflicting orbits, as there is just the corner orbit of the 2x2x2)
3x3x3: 13.9264%
4x4x4: 1.3479%
5x5x5: 0.0486179%
6x6x6: 0.00071384%
7x7x7: 3.90594*10^-6%
8x8x8: 8.69994*10^-9%
9x9x9: 7.22148*10^-12%
10x10x10: 2.44007*10^-15%
11x11x11: 3.07255*10^-19%

Here's the Mathematica code/formula for these calculations (the 2x2x2 is omitted from the formula, of course)


Spoiler: Mathematica Code



Table[((11808))*((94122082279003380940800)/(4!^6))^Floor[(n-2)^2/4]*((94122082279003380940800)^Floor[(n-2)/2]*((113891040))^Mod[n,2])/((24-23Mod[n,2])*2^Mod[n,2])/(1/2 (((8!)*(24!/(4!^6))^Floor[(n-2)^2/4]*(24!^Floor[(n-2)/2]*(12!)^Mod[n,2])/((24-23Mod[n,2])*2^Mod[n,2])))),{n,3,11}]*100//N


How I did this calculation?


Spoiler



Simply put, a small subset of cycle classes (cycle types) for 8, 12, and 24 objects (8 corners, 12 middle edges, and 24 big cube parts) cannot be solved with an equal even number of 2-cycles in 2 iterations. *Theorem 1* in the document shows that if we only consider one orbit, no matter if it even has a million objects, every even permutation can be solved in 2 iterations with an equal number of either an odd or even number of 2-cycle swaps. The problem is, this small subset of permutations cannot be solved with an equal number of even number of 2-cycle swaps in just 2 iterations...they can be solved with an equal odd number of 2-cycle swaps in 2 iterations, however.

(Remember 2 "iterations" means 2 "conjugates", and from the description above, 2 conjugates which can be merged into one commutator).

The subset of cycle classes which cannot be solved with an equal even number of 2-cycles in 2 iterations are the following:


Spoiler



By *Corollary (1)* in the document, we only need to count the cycle classes in 7 and 8 objects for corners, 11 and 12 objects for middle edges, and 23 and 24 objects for big cube parts. The following is a list of only the subset of cycle classes mentioned.

*For Corners*
7 Pieces (1)
7

8 Pieces (2)
6,2
5,3

*For Middle Edges*
11 Pieces (1)
11

12 Pieces (4)
10,2
9,3
8,4
7,5

*For Big Cube Parts*
23 Pieces (15)
23
11,4,3,3,2
10,5,3,3,2
9,6,3,3,2
9,5,4,3,2
8,7,3,3,2
8,6,4,3,2
8,5,5,3,2
8,5,4,3,3
7,7,4,3,2
7,6,5,3,2
7,6,4,3,3
7,5,5,4,2
7,5,5,3,3
6,5,5,4,3

24 Pieces (11)
22,2
21,3
20,4
19,5
18,6
17,7
15,9
14,10
13,11
7,5,4,3,3,2
6,5,5,3,3,2


From here, we must count how many of each of these cycle classes there are. We can do this simply by following the guide I made in another thread.


cmowla said:


> \( \left( \text{Number of pieces the piece type has} \right)! \) divided by:
> 
> \( \left( \left( \text{cycle a} \right)^{\#\text{of occurrences of cycle a}}\times \left( \#\text{of occurrences of cycle a} \right)! \right) \) times
> \( \left( \left( \text{cycle b} \right)^{\#\text{of occurrences of cycle b}}\times \left( \#\text{of occurrences of cycle b} \right)! \right) \) times
> ...



(Note that "# of untouched" is 1! and 0! = 1 for these calculations because (8-7)!=(8-8)!=(12-11)!=(12-12)!=(24-23)!=(24-24)! = 1.

*Corners*
\( \left( \frac{8!}{7} \right)+\left( \frac{8!}{6\left( 2 \right)}+\frac{8!}{5\left( 3 \right)} \right)=\text{11808} \)

*Middle Edges*
\( \left( \frac{12!}{11} \right)+\left( \frac{12!}{10\left( 2 \right)}+\frac{12!}{9\left( 3 \right)}+\frac{12!}{8\left( 4 \right)}+\frac{12!}{7\left( 5 \right)} \right) \)

*Big Cube Parts*
(\( \frac{24!}{23}+\frac{24!}{11(4)(3^{2})(2!)(2)}+\frac{24!}{(10)(5)(3^{2})(2!)(2)}+ \)\( \frac{24!}{9(6)(3^{2})(2!)(2)}+\frac{24!}{9(5)(4)(3)(2)}+\frac{24!}{(8)(7)(3^{2})(2!)(2)}+ \)\( \frac{24!}{(8)(6)(4)(3)(2)}+\frac{24!}{8(5^{2})(2!)(3)(2)}+\frac{24!}{8(5)(4)(3^{2})(2!)}+ \)\( \frac{24!}{(7^{2})(2!)(4)(3)(2)}+\frac{24!}{7(6)(5)(3)(2)}+\frac{24!}{7(6)(4)(3^{2})(2!)}+ \)\( \frac{24!}{7(5^{2})(2!)(4)(2)}+\frac{24!}{7(5^{2})(2!)(3^{2})(2!)}+\frac{24!}{6(5^{2})(2!)(4)(3)} \))
+
( \( \frac{24!}{22(2)}+\frac{24!}{21(3)}+\frac{24!}{(20)(4)}+\frac{24!}{(19)(5)}+\frac{24!}{(18)(6)}+ \)\( \frac{24!}{(17)(7)}+\frac{24!}{(15)(9)}+\frac{24!}{(14)(10)}+\frac{24!}{(13)(11)}+ \)\( \frac{24!}{(7)(5)(4)(3^{2})(2!)(2)}+\frac{24!}{(6)(5^{2})(2!)(3^{2})(2!)(2)} \))= 94122082279003380940800.

To calculate the percentages, I just took the formula for the number of positions of the nxnxn Rubik's cube and omitted 3^7 (corner orientations) and 2^11 (middle edge orientations). I multiplied that by 1/2 (because only half of the permutations are even permutations). I made that the denominator of a fraction.

I then took the same adjusted formula (except that I didn't multiply by 1/2) and made that the numerator.

I substituted:
11808 = number of corner permutations which cannot be solved with an even # of 2-cycles in 2 iterations for 8!
113891040 = number of middle edge permutations " " for 12!
94122082279003380940800 for 24!

And of course, I multiplied by 100 to get a percentage


These percentages are for the nxnxn supercube (except for fixed centers on the odd supercube, as usual) because the cycle class {20,4} can look the same as the cycle class {20,(2,2)} on some non-fixed center orbit on a regular cube for a subset of the {20,4}'s permutation cases, for example. Therefore for the regular 6-color nxnxn Rubik's cube, the percentages are even less than the low percentages listed (but for the 6-colored 2x2x2 and 3x3x3, the percentages are the same, of course, because they do not have non-fixed centers).

I didn't attempt to calculate the exact number of even permutations including orientations as well because there is no set rule on all of the orientation possibilities...one would have to brute force this with a computer program (not just the brute force technique I used in the proof which minimized the number of tests required).

It's definitely possible, but probably not very easy to calculate, the actual percentages for the 6-color nxnxn cube (it's certainly an easier calculation to do than if we were to consider orientations as well).

So in short, MrCage's idea question that maybe one commutator is sufficient to solve all even permutations of the nxnxn wasn't too far fetched, as we can see that as n gets large, the percentage of permutations which cannot be solved with one commutator decreases exponentially. We've got to thank odd parity for this, uh?


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## Stefan (Jan 18, 2013)

*Re: My ultimate commutator challenge.*

Can you show a (preferably simple) 3x3x3 case that can't be solved with a single commutator? (Sorry if one has been posted already and I missed it, don't have time to read it all)


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## Christopher Mowla (Jan 19, 2013)

I just made some changes to my previous post. 35 cycle classes previously listed in that post which contained at least two equal even cycles , three or more 3s, three or more 5s have been eliminated because I found that they actually can be solved with an even number of 2-cycles in 2 iterations (I didn't need to know that they could be handled in one commutator when I wrote the theory). Therefore the percentages have gone down, but not substantially (less than 1% on all cube sizes > 2x2x2)



Stefan said:


> Can you show a (preferably simple) 3x3x3 case that can't be solved with a single commutator?


Since in the last post I mentioned percentages dealing with permutations only (and not orientations), when you say "simple", do you mean a 3x3x3 state in which all corner and middle edges are oriented correctly? If so,


Spoiler



Here is the complete set of 3x3x3 permutations (ignoring orientations) which cannot be solved with one commutator:

Whenever the corners have one of the cycle classes {{7},{6,2},{5,3}} and the middle edges have one of the cycle classes {{11},{10,2},{9,3},{8,4},{7,5}} simultaneously.

So you might consider the simplest subset of cases to be a corner cycle class {5,3} with a middle edge cycle class {7,5}.





Stefan said:


> (Sorry if one has been posted already and I missed it, don't have time to read it all)


In post #38, there was one example that couldn't be solved with one commutator because of middle edge orientations, but I don't believe I've posted any other 3x3x3 example which couldn't be solved with one commutator which were due to:

Corner orientations only,
Both middle edge and corner orientations,
Just the permutations,
The permutations and orientation of the middle edges, or
The permutations and orientation of the corners.

(and when I say "just corner orientations only", for example, this could mean either the pure twist or a complete 3x3x3 scramble. When I say "the permutations", I mean the interaction of the cycle types of the corners and middle edges.)

The simplest case is perhaps R U ?

If what I have written so far isn't enough to give you an idea, please let me know what you mean by "simple", and I'll be sure to make an example for you. If there are any specific 3x3x3 states which you would like to know if my method can solve with one commutator, let me know.


----------



## Stefan (Jan 19, 2013)

Sorry I was unclear. With "simple case" I meant one with many pieces still solved or with a short alg, so your R U is very nice in both regards, thanks. However, that one can easily be done with a single commutator: [R, L R U2 F B' U' D'] (the second part moves the pieces from U to R so that the third part (R' turn) effectively does a U turn). Am I misunderstanding something?



cmowla said:


> Here is the complete set of 3x3x3 permutations (ignoring orientations) which cannot be solved with one commutator:
> 
> Whenever the corners have one of the cycle classes {{7},{6,2},{5,3}} and the middle edges have one of the cycle classes {{11},{10,2},{9,3},{8,4},{7,5}} simultaneously.



Unless I misunderstand, that is also wrong. Here's a single commutator doing a 7-corners-cycle and an 11-edges-cycle:
[U D2 L' R F2 L R', B D' R B' D2 F' B']
It does the 7-cycle by overlapping two 4-cycles at one piece and the 11-cycle by overlapping two 6-cycles. First and third part of the commutator do the cycles, second and fourth part switch affected and non-affected pieces.

Also, judging by the title page of your document, a single in-place edge-flip seems to be your poster boy category of what's impossible, but here's a commutator that does that:
[U2 R L' B R2 D2 R2 B' R' L U' L2 F2 L2, U2 L' R F2 L R']


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## Christopher Mowla (Jan 19, 2013)

Thanks Stefan for the counterexamples. I'm speechless, especially for the 7-cycle and 11-cycle. I guess if anything, I developed a method for solving cubes in 2 commutators. I knew that my method couldn't do a pure 3 corner twist in one commutator without a modification (see page 12 of the PDF), but I didn't realize that it cannot handle certain permutations which can be solved with one commutator. Interesting.


----------



## Christopher Mowla (Jan 20, 2013)

Again I appreciate the feedback, Stefan. Finally someone gave me some (very useful too).

Now that I know it is possible for single commutators solve positions that I couldn't solve before, I just expanded my method to a higher level. Using my "expanded method",

I have found a single commutator for R U (this was very hard to make):
[D' R2 U' R2 B2 R2 B2 R2 D L' D2 L' D2 L2 B2 U R2, D2 F2 U' L2 U2 L2 R2 U L R2 U' B2 U' L2 F2 R B2 D]

and the exact same {11,7} you posted:
[L2 D2 F U2 B U2 B2 D F2 U2 R U2 F R' B2 L' B' U L, U2 B2 U2 F' U2 L2 U' B2 R' U' L2 D2 B' L D B2 D2 U2]

Now that I look at it again, I believe I was confusing material from my proofs for orientations to be handled in 2 commutators with "permutations of conflicting orbits" (which is false, as you pointed out). Still no excuse for not realizing that huge misconception.

It appears that all permutations can be solved with one commutator, because even the worst cycle classes do not need to use all slots in their orbits to be solved in their two iterations. We can always have 2 slots free, and so even if one or more orbits of pieces need to be solved in an odd number of 2-cycles, they do not have to touch any other orbits.

For orientations, I'm not sure yet. I'll have to look at what all of this means now that I did successfully compute single commutators for what I previously thought to be the worst cases (well, I thought corner twists which summed to +- 6 to be pretty bad, but I'll have to see how true that is now). It might be possible for me now to construct a proof for all orientation cases to be solved in one commutator.

If I stumble upon a case which my "expanded" method cannot handle during the process, I'll be sure to post it and ask you if you can find a single comm. for it.


----------



## Stefan (Jan 20, 2013)

Glad I could help (and it was fun, too). I guess the reason you're not getting much feedback is that it's so much and that it appears to be rather complicated, so I doubt many people read it (I know I don't). So it was very helpful to get nice examples to just try without having to learn all the theory 

And yeah, if you have another impossibility candidate, let us know.


----------



## Christopher Mowla (Jan 21, 2013)

Just to give an update,


cmowla said:


> Just for now, here is my first official product of two commutators to solve a random scramble of the 3x3x3. It's 132 htm, but I have used CubeExplorer to get it that low (it might be 10 or so moves less than this...you can take each X and Y and try to reduce the amount of htm that way).
> 
> Notice that it's a 9-cycle of edges and a 7-cycle of corners. One of the 3 correctly positioned edges is unoriented.
> 
> ...


Yesterday I made a one commutator solution that solves this state completely.
*Scramble*
L2 F' L2 B U L R' U2 R2 B' L F' D2 U2 L' R' F U B U D2 B D' B' R'
*Solution*
[B2 R U2 L2 B2 R2 B' L D' R U' B' L2 D B D' R2 B', B2 L' U2 R2 U L2 B' R B2 L D' B2 F' D2 R' B' F']

I created a "very bad" 3x3x3 scramble which I currently think is a very good position to test. This scramble:
Does a 7-cycle of corners and an 11-cycle of edges
Has a corner twist which sums to -6
Does an 8 edge flip
Has one edge flipped in its own location and has one corner twisted in its location
L2 B D2 U2 F R F R2 B2 U' F R2 B2 U' L D2 R' D2 B
and, even though the process to obtaining the solution was more involved, I found a one commutator solution in about 30 minutes 
[D U L2 U R2 D' L2 F' U' R2 U R' F2 L' R2 F R2 U2 R, B2 L R' B2 R U2 L B' U2 B' U L R2 D U' R2 F2 U L']

On page 12 of the PDF, I mention that my method cannot generate a 3 twist with one commutator, but I gave a modification of the method to create a 3 twist by using a 2 twist. [[L', U' R'U][U, F' D' F], F]. I also mentioned that this commutator cannot be merged with another to solve a composite case which includes a pure 3 twist. If we make Y (where we can denote the commutator above to be [X,Y]) an A perm in the face F instead of the move F, ([[L', U' R'U][U, F' D' F], R' D R' U2 R D' R' U2 R2]), then we do have a 3 twist commutator which can be merged with the middle edge orbit and all other orbits as well. I wasn't proud of this solution because it was too "creative" to be useful for commutator theory because, for more difficult cases, we won't be able to find a solution so easily without using a systematic process which allows us to view any permutation and orientation as the same problem. Using my updated method, I successfully create a one commutator solution for the pure 3 twist (which obviously can be merged with the permutation of any orbit, including the corner orbit, because I created it using my method): [B2 R2 B' L' B R2 B2 R B L B' R', R2 F' U2 B2 F R F R' B2 F' U2 F R F' R]
(For those interested how to make a 3 twist commutator in a different manner than just using a pure 2 twist, see what X and Y in [X,Y] do separately).

The examples in this post pretty much have solved all of the states which I previously thought were impossible. So it is possible that my "updated method" is "The Method" to create single commutators for _any_ even permutation of the nxnxn cube. I actually had this method all along, but I tried to fix some problems without using all of the available tools that I actually had available to me (I wasn't aware of them). So it was impossible for me to solve some states with one commutator using only a portion of my "tools". However, thanks to Stefan, he made me look deeper into my toolbox to find the required tools that were there from the beginning to fix previously impossible problems.

If anyone wants me to find a solution to any particular 2x2x2 or 3x3x3 scramble (or a specific partially solved case) (I can do bigger cubes too, but no larger than the 5x5, please), I can do that. I can even make a video explanation of the systematic process I use, if you couldn't follow the 5x5x5 supercube example I have already given in unabridged detail in writing (I will only do this for the 2x2x2 and the 3x3x3, and possibly for the 4x4x4, but no larger because it becomes impractical). *But be sure that your requested scramble/state is an even permutation* (that is, you can solve it using an even number of quarter turn slices for every set of slices). Make no mistake, the commutator solutions formed with this method are LONG. You can optimize their length after you have already found them. In addition, this method makes no claim that the commutator you find (even after finding the optimal length of X and Y) is the shortest commutator that can solve a given case, as there is more than one way to solve the same problem, and shorter commutators might exist which only can be used on the 2x2x2 or 3x3x3, but are not able to merge with other orbits for the nxnxn (the solutions created with my method are definitely able to be merged with all orbits of pieces in the nxnxn). If no one asks me for examples, I might eventually make tutorial videos and put them on YouTube.

Just one last quick note, if you're interested in trying this on your own, all you'll need is a pencil and paper, the program CubeTwister, these sticker images for Cubetwister 3x3x3, 4x4x4, 5x5x5, 6x6x6, 7x7x7, and, if you are solving the 2x2x2 or 3x3x3, some 3x3x3 solver (I use CubeExplorer) (for optimization only, of course). It might also help to have a notepad document open as well.


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## Christopher Mowla (Feb 11, 2013)

Stefan said:


> And yeah, if you have another impossibility candidate, let us know.


I was hoping to continue proving that all orientations of corners and middle edges (mixed with each other) could be solved with one commutator using my method, but I have come across a case which has stumped my "current" method.

The case is: a pure 12 flip (superflip) of middle edges + _any_ 7-cycle of corners (the specific orientation case of the corners doesn't matter).

If someone can find a single commutator that can generate any case like this, then my method cannot generate a single commutator solution for every case that _can_ be generated by a single commutator. If the parity of the corners wasn't dependent on the parity of the middle edges, if we could flip an odd number of middle edges on the cube, or if we somehow had 9 corners on a cube instead of 8, then this set of cases would be reachable with my method.:fp

Here is a detailed explanation why my method cannot do this category of cases (and, if my method isn't proven incomplete with a single commutator solution for this category of cases, this could be an excellent counterexample/proof that not every orientation case can be solved with one commutator)


Spoiler



Basically this is what my method is saying. A 7-cycle of corners needs a 3 2-cycle for iteration 1 and a 3 2-cycle for iteration 2. I cannot swap an additional dummy pair (to have a 4 2-cycle for iterations 1 and 2) because then I wouldn't be able to solve the 7-cycle in just 2 iterations. Thus with my method, the number of 2-cycle swaps of corners must be _odd_ in both iterations (for those who haven't read any of my examples or theory, both iterations must always have the same number of 2-cycles anyway...I am just mentioning "both" iterations for the sake of completeness).

To generate a 12 flip with one commutator I:
1) Do a pure 6 flip
2) Do a 6 2-cycle to put the 6 middle edges which were not flipped in step 1 in the slots of those which were flipped in step 1.
3) I undo the 6 pure flip in slots flipped in step 1, which means I now flip the remaining 6 oriented middle edges (undoing/inverse of a flip is a flip)
4) I undo the 6 2-cycle to solve back all middle edges (which are all now flipped).

Thus I need a 6 2-cycle to create a 12 flip, but the 7-cycle of corners requires an odd permutation which only allows me to have a 5 2-cycle at most to be able to manipulate middle edges.

So there is a clear contradiction.


Again, I believe I can solve all permutation cases and many orientation cases of the nxnxn cube with one commutator, so if someone wants an example solve, I will do what I can. But as far as trying to prove that every possible permutation of the nxnxn cube can be solved with one commutator, I might be able to prove that all even permutation configurations of the nxnxn even cube can be solved with one commutator, but not the nxnxn odd cube (with this method).


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## qqwref (Feb 11, 2013)

Could you generate a 12-flip with a 6-flip plus a 2-2-2-2-4 cycle pattern? E.g. [(R' F R U)5, D2 F2 R D R' U' B2 R2 F' U' F R2 F2 D' U F2 U2] (note that that last sequence performs a corner 2-cycle, so we can easily verify that the edge permutation is odd).


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## Christopher Mowla (Feb 11, 2013)

qqwref said:


> Could you generate a 12-flip with a 6-flip plus a 2-2-2-2-4 cycle pattern? E.g. [(R' F R U)5, D2 F2 R D R' U' B2 R2 F' U' F R2 F2 D' U F2 U2] (note that that last sequence performs a corner 2-cycle, so we can easily verify that the edge permutation is odd).


Thanks for the useful feedback, qqwref. With your help I did just successfully create a commutator which does a superflip + 7-cycle of corners:
[U2 L F2 D2 F2 U2 L2 B R B' L' U' R' D2 U' L' F2 L2 U2, U2 R2 U2 R2 B D L2 D2 B2 U' B R' U' L' U F' L2 R]

It seems like having a 4-cycle as a part of my z is okay if I am just doing pure orientation with the piece type which has that 4-cycle. The pure orientation _might_ be the only exception which allows z to be a cycle class other than a product of disjoint 2-cycles, but then again, this is the only case so far that appeared to not being able to be reached with the restriction of just having a product of disjoint 2-cycles. I'll have to look into this more to see if the parity of two different piece types can have any other potential conflicts. If not, then it will be a question of whether or not all corner orientations and edge orientations can be solved with one commutator (using my method) separately. If so, then they can unquestionably can be merged.

I have almost finished proving that 3-cycles of corners + any corner orientation case can be solved with one commutator, but that's as far as I've gotten because it's very difficult work. This is going to be a brute force search, and I don't have a program to do this for me, so it might take a while to go through all of the cases by hand. The amount of tests for cycle classes which involve more pieces will increase as well, which doesn't help. (I might need to do around 4000 tests just for corner orientations).

Thanks a bunch! I'm really hoping that the nxnxn cube can be solved with one commutator. I guess we will see.


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## Christopher Mowla (Jun 3, 2013)

It has been several months since I posted in this thread, but I have been working on this project off and on, and now I believe I have some results to state without proof.

_I will later provide proof in one form or another. For now, you'll have to take my word for it and try to come up with counter examples._

[1]


Stefan said:


> [...]if you have another impossibility candidate, let us know.





Spoiler: All of these



For the 3x3x3,
Either of the two middle edge cycle classes {6,3,2},{5,4,2} mixed with any of the corner cycle classes {7},{6,2},{5,3}.

Or, in general, any cycle class of a piece type that is marked as *odd* mixed with a cycle class of another piece type that is marked as *even* (and vice versa) from the following list. (Every cycle class not in this list are both odd and even, so they can be merged with each other and with all cycle classes in this list.)

Cycle classes in 7-8 Objects (for corners)
*Odd* (3)
{7}
{6,2}
{5,3}

Cycle classes in 11-12 Objects (for middle edges)
*Odd* (5)
{11}
{10,2}
{9,3}
{8,4}
{7,5}
*Even* (2)
{6,3,2}
{5,4,2}


Cycle classes in 23-24 Objects (for all non-fixed centers and wing edges)
*Odd* (14)
{23}
{9,5,4,3,2}
{8,6,4,3,2}
{7,6,5,3,2}
{22,2}
{21,3}
{20,4}
{19,5}
{18,6}
{17,7}
{16,8}
{15,9}
{14,10}
{13,11}
*Even* (47)
{18,3,2}
{17,4,2}
{16,5,2}
{16,4,3}
{15,6,2}
{15,5,3}
{14,7,2}
{14,6,3}
{14,5,4}
{13,8,2}
{13,7,3}
{13,6,4}
{12,9,2}
{12,8,3}
{12,7,4}
{12,6,5}
{11,10,2}
{11,9,3}
{11,8,4}
{11,7,5}
{10,9,4}
{10,8,5}
{10,7,6}
{9,8,6}
{15,4,3,2}
{14,5,3,2}
{13,6,3,2}
{13,5,4,2}
{12,7,3,2}
{12,6,4,2}
{12,5,4,3}
{11,8,3,2}
{11,7,4,2}
{11,6,5,2}
{11,6,4,3}
{10,9,3,2}
{10,8,4,2}
{10,7,5,2}
{10,7,4,3}
{10,6,5,3}
{9,8,5,2}
{9,8,4,3}
{9,7,6,2}
{9,7,5,3}
{9,6,5,4}
{8,7,6,3}
{8,7,5,4}


[2]
To believe it or not, I have successfully made a single commutator solution for all (8!/2)(3^7) corner cases by hand. Since there are no orbit parity conflicts between corners on the 2x2x2 (because corners are the only orbit of pieces on the 2x2x2), I claim that:

*All configurations of the 2x2x2 Rubik's cube can be solved with (1 turn) + (1 commutator)*​ 
_Well of course I did not solve tens of millions of cases by hand, but I did some serious case reduction to solve about 1000 cases which represented all even permutations of the corners (and thus all even permutations of the 2x2x2)._

One of the hardest corner configurations cases (by far) I had to solve using my method was surprisingly a 2 2-cycle. (The 7-cycle and all the cycle classes involving all 8 corners were "trivial" in comparison.)
Maybe cuBerBruce or someone can explain why?
D L2 U B2 R2 D L' F2 D' L D2 L2 U R' D R2 F2 R

I found 2 completely different solutions for this case, but this was my first.
[R F2 L F2 U2 B2 U R F2 D' L2 D' B2 D L' D' F2, L2 B2 D B2 D' L D' R' D2 B2 D2 B2 R B2 D B2 L]

Since the only cycle classes in corners which needed a specific number of 2-cycles in z1 (and thus z2) required an odd number of 2-cycles, I made an effort to solve all corner cases using an odd number of 2-cycles (in both iterations, of course). Thus, in theory, all of my corner solutions can be successfully merged with all single commutator solutions (generated with my method) of cycle classes of middle edges, wing edges and non-fixed centers not in the list as well as all of the cycle classes under the odd sections in that list.

So, as far as my method is concerned, 38/40 cycle classes for middle edges could be merged with my complete set of corner solutions for the 3x3x3, but I have not bothered to solve the edge orientations because I still could not claim that all configurations of the 3x3x3 could be solved with (1 turn) + (1 commutator) because of the two middle edge cycle classes {6,3,2} and {5,4,2}, not to mention that there are dozens of conflicts in every orbit of 24 pieces in the 4x4x4 and larger cubes. A major disappointment, but I can't help that.


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## Christopher Mowla (Jun 4, 2013)

Hmm

Just putting a [,] in CubeTwister and inserting some moves to the left and right of the comma delimiter, it didn't take too long to generate a {7} + {5,4,2}.
[ R F, B' F D' U R' D' B' L' F' U L' R' B2 L' R2 D R']

The conditions in my last post were true for my restriction of just using disjoint 2-cycles in my conjugates, but since I saw that it was possible, I went back to the fundamentals of my method, and I successfully integrated a 3-cycle into the conjugates in addition to the disjoint 2-cycles. (This allowed me to solve the {5,4,2} using an odd number of 2-cycles in both conjugates for the middle edge commutator).

It didn't take long to translate my abstract solution into cube moves, so I solved the oriented case {7} + {5,4,2}.

*Case*
L' U2 R' U L2 U B2 D L B2 D' U' L2 F2 D2

*Solution*
[D2 F2 L B2 D2 B2 L' U' L2 F2 D' B2 L2 D' R2, L U2 L R U2 B2 L U' L' U2 L2 B2 D L R' B2 D' R' U]

I have to probably study how a 3-cycle can be used to solve middle edge orientations, should I need them, but, as far as corners are concerned, since I believe I have successfully solved all even permutations and orientations using only disjoint 2-cycles, I'll stick with disjoint 2-cycles for the time being.

I will attempt to solve all cycle classes of 23-24 objects using this new trick (and the {6,3,2} for the middle edges, of course) under the *even* lists to create odd solutions for them so that they can be merged with my complete set of corner solutions (which are all odd).

Sorry guys for rambling on...this method has been evolving since I started, and I guess it will continue to evolve should it need to evolve any further to reach all even permutations of the nxnxn cube with a single commutator...or so we all hope.


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## Christopher Mowla (Jun 7, 2013)

I have gone through all even and odd cases in my list in my last post, and I can be confident now to say that I have legitimate proof that:

*All configurations of the even nxnxn supercube can be solved with *\( \frac{n}{2} \) *slice turns + 1 commutator!*​ 
That is, if there are some inner layer slices in an odd permutation, and/or if the outer layer slices are in an odd permutation, then we just have to do at most \( \frac{n}{2} \) quarter turn slices (to one half of an even cube) to make the permutation of all of the slices even. From there, we can without a doubt use one commutator to solve the rest.

I am now going to begin racing to the finish line by working on the very last part of this project...middle edge orientations. I have already made a concrete and systematic plan of how I am going to tackle all (12!/2)(2^11) cases, so the busy work will begin shortly.


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## Stefan (Jun 7, 2013)

cmowla said:


> *All configurations of the even nxnxn supercube can be solved with *\( \frac{n}{2} \) *slice turns + 1 commutator!*​



This allows those n/2 turns to include half turns and to include the same slice several times, so it's a weaker statement than what I think you meant.


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## Christopher Mowla (Jun 7, 2013)

Yeah, I wasn't quite sure how to state it in one sentence (I should have given myself more time), but I see exactly what you mean. It's vague.

I guess for the even cube, I should have just said "slice quarter turns", but I am in the habit of just saying "slice turns" implying that they might need to be half turns in one case of the odd supercube, because, as qqwref proved, the subset of fixed center permutations which sum to an odd multiple of 180 are not in the commutator subgroup.

For those who have not read my document, they are (I called them "cycle classes" for fixed centers, although all 6 fixed centers are in a 1-cycle with themselves, but...)


Spoiler: Subset of supercube fixed centers which sum to an odd multiple of 180



0 Centers (0) (the unique solved case)

1 Center (1)
180

2 Centers (2)
-90,-90
90,90

3 Centers (2)
180,180,180
90,-90,180

4 Centers (4)
-90,-90,180,180
90,90,180,180
90,-90,-90,-90
90,90,90,-90

5 Centers (5)
180,180,180,180,180
-90,-90,-90,-90,180
90,90,-90,-90,180
90,90,90,90,180
90,-90,180,180,180

6 Centers (6)
-90,-90,180,180,180,180
90,90,180,180,180,180
-90,-90,-90,-90,-90,-90
90,90,-90,-90,-90,-90
90,90,90,-90,180,180
90,-90,-90,-90,180,180


So to make all configurations of the nxnxn the odd supercube into the commutator subgroup (assuming that I will successfully reach all middle edge orientations with a single commutator and assuming I make the time to verify that all fixed center positions which sum to a multiple of 360 can be reached with one commutator), besides applying an inner slice quarter turn to all slices (on half/one side of a big cube) which have odd permutations in order to make them have even permutations,



If the outer layer slices are in an even permutation and the sum of the center rotations is a multiple of 360 (an even multiple of 180, if you will), then no outer layer slice turn needed. 
If the outer layer slices are in an even permutation and the sum of the center rotations is an odd multiple of 180, then a half turn outer layer slice is needed. 
If the outer layer slices are in an odd permutation, chose to do an outer layer quarter turn in the direction to cause the sum of the center rotations to become a multiple of 360.


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## Christopher Mowla (Apr 29, 2014)

cmowla said:


> I am now going to begin racing to the finish line by working on the very last part of this project...middle edge orientations. I have already made a concrete and systematic plan of how I am going to tackle all (12!/2)(2^11) cases, so the busy work will begin shortly.


I now have effectively exhaustively solved every middle edge configuration in the commutator subgroup with a single commutator _with my method_.

Therefore, the answer to the question we have all been waiting for for 2.5 years:


mrCage said:


> Show that every even permutation on a nxnxn cube can be obtained by a series of commutators. If 2 consecutive comms is a single comm (somehow) then every even permutation is also a direct commutator. No restriction on number of layers used here ...
> 
> Per


, mrCage (if you're still reading the forums even though you haven't been actively lately),

*There does not exist any position in the commutator subgroup of the nxnxn cube which cannot be expressed as a single commutator!*

I would like to thank Stefan for helping me see that some positions which I originally didn't think were possible to solve with a single commutator _were_ possible. I would also like to thank qqwref for helping me handle the superflip case with my method and verifying that not all fixed center positions on the odd supercube which are in the commutator subgroup can be solved with a single commutator. Without you guys, it's _very_ possible that I could have still accomplished what I have, but it would have taken longer without the useful feedback!

I intend to publish my proof (which still needs to be made more concise and organized), and possibly attempt to earn a PHD by publication from it. So the official proof (which is quite beautiful, despite that it is lengthy--hundreds, perhaps a couple thousand pages) of this will be eventually shown, but until then, I will gladly accept scramble requests for any position of the nxnxn cube, and I will post a single commutator solution to it.

Please don't give me a scramble of a cube larger than the 7x7x7 because it's impractical. If the nxnxn cube configuration is not entirely in the commutator subgroup (for those who don't know what this means, it means that you cannot solve the scramble using an even number of slice quarter turns for every pair of slices, when looking at a single face as a frame of reference), I will simply do a 90 degree quarter turn of all r slices I need to in order to make it in the commutator subgroup first, and then I will find a single commutator solution to that case. Therefore, if you want a specific scramble solved with a single commutator alone (without first seeing a series of quarter turns followed by a single commutator), please be sure that the entire nxnxn cube state is indeed in the commutator subgroup.


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## Carrot (Apr 29, 2014)

cmowla said:


> I now have effectively exhaustively solved every middle edge configuration in the commutator subgroup with a single commutator _with my method_.
> 
> Therefore, the answer to the question we have all been waiting for for 2.5 years:
> , mrCage (if you're still reading the forums even though you haven't been actively lately),
> ...



Solve U2  


Spoiler



my shortest:


Spoiler



[L R U2 R' L', U]


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## Christopher Mowla (Apr 29, 2014)

Carrot said:


> Solve U2  (I didn't read up on what exactly the commutator subgroup consists of, but if U2 is in that, then I would be interested in seeing a commutator solution for it)


Yay, my first volunteer! You might find this post interesting.

Yes, that position is in the commutator subgroup because it's an even permutation of corners (and therefore middle edges, and vice versa).

EDIT:
You didn't notice that I answered your question in that thread?! Man, I bet you felt ignored! LOL.


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## Carrot (Apr 29, 2014)

cmowla said:


> Yay, my first volunteer! You might find this post interesting.
> 
> Yes, that position is in the commutator subgroup because it's an even permutation of corners (and therefore middle edges, and vice versa).
> 
> ...



Don't worry, I forgot all about that post, but thanks for taking your time twice  and I actually found a solution to all of them (except U D, and yes, in my first post I meant L R' and not M) while trying to solve U2 myself (I was stupid and missed the obvious solution)


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## qqwref (Apr 29, 2014)

cmowla said:


> I now have effectively exhaustively solved every middle edge configuration in the commutator subgroup with a single commutator _with my method_.
> 
> Therefore, the answer to the question we have all been waiting for for 2.5 years:
> , mrCage (if you're still reading the forums even though you haven't been actively lately),
> ...


Wow, congratulations! Good luck with the PhD!


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## 10461394944000 (Apr 29, 2014)

cmowla said:


> *There does not exist any position in the commutator subgroup of the nxnxn cube which cannot be expressed as a single commutator!*



o cool

L' Uw Rw' B' R2 Rw' Fw R Uw D2 Fw' U' Fw U2 Rw Uw U R B2 R Rw2 D2 Fw' F2 Rw' Fw2 D Uw' Fw R Uw2 R L' B Rw L F2 R2 Uw2 L gogogo


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## cmhardw (Apr 29, 2014)

cmowla said:


> I now have effectively exhaustively solved every middle edge configuration in the commutator subgroup with a single commutator _with my method_.
> 
> Therefore, the answer to the question we have all been waiting for for 2.5 years:
> , mrCage (if you're still reading the forums even though you haven't been actively lately),
> ...



Congratulations on your result! I am very excited for you, and I hope this proof will help you earn your PhD!


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## Christopher Mowla (Apr 30, 2014)

10461394944000 said:


> o cool
> 
> L' Uw Rw' B' R2 Rw' Fw R Uw D2 Fw' U' Fw U2 Rw Uw U R B2 R Rw2 D2 Fw' F2 Rw' Fw2 D Uw' Fw R Uw2 R L' B Rw L F2 R2 Uw2 L gogogo


Here you go, Ben! (I assumed that this was for the 4x4x4, so everyone please put the cube size with your scrambles!)

604 btm (Keep in mind that for cube sizes larger than the 3x3x3, my method will at best yield 4 times God's number for the supercube nxnxn of the nxnxn cube being solved. That's assuming we have a supercube optimal solver, which we don't.)
[
F L2 F D' R2 D F' L2 F D' R2 D F2 U' R2 U2 R2 U' L2 U R2 U' L U' R2 U L u B' F l F' L2 F l' B F' u' B' u L2 u' F D2 F' d2 F D2 B F' D B' d2 B D' U B' U' b2 r2 U B U' B' r2 b' B D l' B' L' R' B l B' R L B D' R U' b' D Bw U b' U' B' U D' b R' b r' b' L b r b2 r' b L' b' r f' L' f r f' L f r' u' r2 f D' b2 D f' r' D' b2 D l r' u R' Uw' r2 U l U' r2 U l2 u R B l2 B r2 u B' l2 B u' r2 l u' b' u l' B2 l u' b u l' U
,
B2 d' b d B2 d' b' d b l2 f' L f l2 r' f' L' f r f' b2 B' D2 b r b' r' D2 B r b r2 R' f R f' r d2 f R' f' R d2 f B' d2 f2 R' d R f2 d2 L d' L' Bw D2 L f' L' R B' L R' f L' R B R' D2 b' R' U' B F' u' B' F U2 B F' u B' F U' R d' b2 D B' D' B b2 d2 B' D B d' D2 B' d' B U D2 B' d B D' U' r Uw' B' u F u' F' B U' F u F' U2 r' D2 B U' B' D2 B U B' F D U2 L2 D' F D F' L2 U2 F D' F2
]

EDIT:
BTW, this was the first 4x4x4 scramble I solved with a single commutator (I never did anything except the 3x3x3 before this solve).


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## guysensei1 (Apr 30, 2014)

Umm... does this mean every normally solvable position on a 3x3 can be solved with a single commutator?
If not, what does 'even permutation' mean?


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## Christopher Mowla (Apr 30, 2014)

guysensei1 said:


> Umm... does this mean every normally solvable position on a 3x3 can be solved with a single commutator?
> If not, what does 'even permutation' mean?


For the 3x3x3, half of all of its positions can be solved with a single commutator. The other half first require that you do a single quarter turn to any face (not a slice like M, E, or S), and then they become solvable with a single commutator.

EDIT:
See pages 29-31 of my "# of Permutations on the nxnxn Cube" PDF which is linked in my signature, if you wish to know about even and odd permutations, or you always can check out Ryan Heise's pages.

EDIT2:
Of course, an easy way to tell if a certain 3x3x3 state is solvable with a single commutator (or a product of multiple commutators) is to scramble the cube with an even number of quarter turns.


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## 10461394944000 (Apr 30, 2014)

cmowla said:


> Here you go, Ben! (I assumed that this was for the 4x4x4, so everyone please put the cube size with your scrambles!)
> 
> 604 btm (Keep in mind that for cube sizes larger than the 3x3x3, my method will at best yield 4 times God's number for the supercube nxnxn of the nxnxn cube being solved. That's assuming we have a supercube optimal solver, which we don't.)
> [
> ...



doesn't seem to work


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## Christopher Mowla (Apr 30, 2014)

10461394944000 said:


> doesn't seem to work


I'm not quite sure what you mean. When I clicked the link in your quote, alg.garron.us didn't work, but the link in my post works.


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## 10461394944000 (Apr 30, 2014)

cmowla said:


> I'm not quite sure what you mean. When I clicked the link in your quote, alg.garron.us didn't work, but the link in my post works.



there we go

you should probably start using alg.cubing.net instead, pretty sure alg.garron.us isn't maintained anymore (and java is silly).


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## Jakube (Apr 30, 2014)

cmowla said:


> _R2_Uw2_L&cube=4x4x4¬ation=WCA



This appeared, when I clicked the quote-button. Speedsolving converts the not in notation into ¬. So maybe there happend some error with copy-paste, ben. 

Why does this happen? If I click the link, it direkts me to the correct link.


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## guysensei1 (Apr 30, 2014)

Does this theory apply to 2x2?
If yes,
F2 R2 F R2 U' F U2 F R2 U'


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## Christopher Mowla (Apr 30, 2014)

guysensei1 said:


> Does this theory apply to 2x2?
> If yes,
> F2 R2 F R2 U' F U2 F R2 U'


Yes, of course. nxnxn means nxnxn.

*Scramble* (you gave me an odd permutation...count your quarter turns! So I solved your scramble + the move R)
F2 R2 F R2 U' F U2 F R2 U' (R)

*Solution*
[U' R2 U' R2 U' R' U2 R', R F2 R2 U2 R' D B' U]


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## Carrot (Apr 30, 2014)

Can [a,b] [c,d] = [e,f] be written in a way such e and f are both constructed only using a,b,c,d?


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## whauk (Apr 30, 2014)

Carrot said:


> Can [a,b] [c,d] = [e,f] be written in a way such e and f are both constructed only using a,b,c,d?


not in general:



> However, the product of two or more commutators need not be a commutator. A generic example is [a,b][c,d] in the free group on a,b,c,d. It is known that the least order of a finite group for which there exists two commutators whose product is not a commutator is 96; in fact there are two nonisomorphic groups of order 96 with this property.


from: http://en.wikipedia.org/wiki/Commutator_subgroup#Commutators


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## bobthegiraffemonkey (Apr 30, 2014)

whauk said:


> not in general:



True, I believe that was already mentioned. However, since we have in this case that the product is always a commutator, does that make it possible to construct the product? I suspect not, but it would be interesting to know for sure.


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## sneze2r (Apr 30, 2014)

cmowla said:


> ...
> 
> 604 btm (Keep in mind that for cube sizes larger than the 3x3x3, my method will at best yield 4 times God's number for the supercube nxnxn of the nxnxn cube being solved. That's assuming we have a supercube optimal solver, which we don't.)
> [
> ...



...How did You find this out!?


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## Christopher Mowla (Apr 30, 2014)

Carrot said:


> Can [a,b] [c,d] = [e,f] be written in a way such e and f are both constructed only using a,b,c,d?


If [a,b] and [c,d] are able to be merged together with my method, then for permutations at least, [a,b][c,d] = [c,bd] != [c,db].

Note that I would have to solve all middle edge and corner cases in an alternate manner (a manner in which I thought was impractical when I started, as it is even more involved than my current one) in order to prove that it holds true for orientations of corners and middle edges mixed with their permutations.



sneze2r said:


> ...How did You find this out!?


By intensive research (brain storming, as I didn't use any existing sources) for 1.5 years. Well, my method itself took about half that time, but it did take that long to both create/fully develop the method and effectively solve every possible case that can arise on the nxnxn cube and to be certain that my proof is correct.


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## Coolster01 (May 1, 2014)

TOTAL INSANITY


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## Christopher Mowla (May 1, 2014)

Coolster01 said:


> TOTAL INSANITY


I actually thought that myself, especially when I was solving all corner and middle edge orientation cases by hand (and verifying that they are all correct with software that I wrote in Mathematica which is over 200 pages of code and documentation if printed. I needed to write software because there are more than *10* things that can be wrong with a solution)! I never thought I would finish...it's hard to believe it's only been a year and a half since I seriously started figuring all of this out from scratch.

If I try to picture all of the details of my proof at once, I feel _very_ intimidated and overwhelmed, because this project is bigger than I am, and its applications to cubing (as well as possible derivative works from it) are huge! My proof is only an "introduction" to commutator theory, because it introduces an entirely new universe, and we just travel from the earth to the sun. The way I chose to solve the corner and edge orientation cases, for example, was the easiest (and therefore I could solve the cases as fast as possible) because I had to solve _every_ case there is. When I release my method and theory, we can all work together to find a variety of solutions for any given case, form new hypotheses, and find ways to prove/disprove them.

EDIT:
Congratulations on your 2x2x2 avg WR!


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## sneze2r (May 2, 2014)

What i will write now might be obvious to You cmowla, but...

I was thinking about Your theorem when i was asleep and i think that it is possible to prove it using lemma:

Every even permutation P can be written as P=(A*B), that B has the same length as A.

So, on cube with applied algorithm P, when i do algorithm A, then on the cube i would have remaining situation B-1 which could be solved with C A-1 C, where alg C is setting up the remaining elements of the cube.


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## Christopher Mowla (May 2, 2014)

sneze2r said:


> Every even permutation P can be written as P=(A*B), that B has the same length as A.
> 
> So, on cube with applied algorithm P, when i do algorithm A, then on the cube i would have remaining situation B-1 which could be solved with C A-1 C, where alg C is setting up the remaining elements of the cube.


This general idea is correct for permutations (and is where my proof begins), and I believe at the beginning of this thread, Ravi implied something like this, but it's much more involved than this for permutations of all orbits on the nxnxn cube, and even more so for orientations of corners and middle edges.

In fact, I prove this lemma in my proof independently of any old proofs of it (I couldn't find any on the web, to be honest).


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## coldsun0630 (Feb 16, 2016)

*One Commutator Solve - My first attempt*

Scramble: U2 F' B' D2 L F2 L2 F U' B2 R2 L2 U2 F' U2 F' U2 B' L2 D2
Solution: [F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2, F2 L2 U D R D B2 D' L U' B' D R'] (view at alg.cubing.net)

Whoa, it actually works!
It was my first attempt, and also pretty successful. Well, I think it was not too hard. Below is how it works:


Spoiler



Scramble: U2 F' B' D2 L F2 L2 F U' B2 R2 L2 U2 F' U2 F' U2 B' L2 D2

1st cycle: UFR > DBR > FDR > LBD > UFR // UF > DF > RB > RU > UB > FU + DR > DL > RD
solve alg: F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2 (17f*)

2nd cycle: UFR > BLU > FUL > FLD > UFR // UB > DB > LF > UL > LB > BU + UF > FR > FU
setup alg: R B U' D R2 B' U' D2 R D2 F U' (12f*)

Final Output: [F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2, F2 L2 U D R D B2 D' L U' B' D R']


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## G2013 (Feb 16, 2016)

How do you (all) figure out which moves to do, to form the one commutator that solves a scramble? I can't find it possible.


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## coldsun0630 (Feb 17, 2016)

G2013 said:


> How do you (all) figure out which moves to do, to form the one commutator that solves a scramble? I can't find it possible.



First, I searched for the whole cycle on the scramble, each corners and edges. It's similar with the memorization on 3x3 BLD. (not 3OP, but BH)

Second, I cut the cycle into half. But, the two cycles(which was cut) should be same one, because the first part of the commutator will excute the same cycle. (the normal one, and the inverse one.)

Third, I chose any between the two cycles, then I solved the cycle. And this is the first part of the commutator. - That's what the '1st cycle' does.
(I didn't want to do FMC with this, so I just ran 'Cube Explorer' here. Of coursely, doing without computer program is posslble.)

Forth, I put setup moves at the second part of the commutator. Each pieces of '2nd cycle' should correspond to other pieces of '1st cycle', but in inverse way. Then it will solve the '2nd cycle' - That's what the '2st cycle' does.
For example:


Spoiler



- 1st cycle: UFR > DBR > FDR > LBD > UFR
- inverse of 1st cycle: UFR > LBD > FDR > DBR > UFR
- 2nd cycle: UFR > BLU > FUL > FLD > UFR

1. UFR should go to UFR (Just stay there!)
2. BLU should go to LBD
3. FUL should go to FDR
4. FLD should go to DBR


And Last, I merged them into one commutator. And this is just what you saw!


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## G2013 (Feb 17, 2016)

With CubeExplorer or not, it's still awesome. Thanks for replying


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