# My ultimate commutator challenge

#### Christopher Mowla

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?

#### Christopher Mowla

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).

#### Renslay

##### Member
Wow. Interesting!

#### Christopher Mowla

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']
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.
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

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

##### blindmod

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'?

#### Christopher Mowla

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

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
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”.

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.
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.
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}
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}
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.

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]}
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
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}
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}
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.

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'
)'
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
)'
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'
/put 11 adj. to 1 in DR
F2 [R',D r' D'] F2
/put 3 adj. to 10 in DL
F2 R' [L,D r D'] R F2
/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
/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]
)'
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'
)'
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).
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'
/put 11 adj. to 1 in DR
F2 [R',D r' D'] F2
/put 3 adj. to 10 in DL
F2 R' [L,D r D'] R F2
/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
/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'
)'
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'].
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:
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.
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
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
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
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).

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.

Last edited:

#### siva.shanmukh

##### Member
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.

#### Christopher Mowla

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.

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.

Last edited:

#### Christopher Mowla

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)
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?
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:
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.
$$\left( \text{Number of pieces the piece type has} \right)!$$ 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
$$\left( \left( \text{cycle c} \right)^{\#\text{of occurrences of cycle c}}\times \left( \#\text{of occurrences of cycle c} \right)! \right)$$ times...
$$\left( \left( \text{1-cycles} \right)^{\#\text{untouched}}\times \left( \#\text{untouched} \right)! \right)$$
("# Untouched" means the amount of pieces which were not touched by any of the cycles multiplied in the denominator. They can be thought of as 1-cycles, where they are cycled with themselves and thus do not change position (possibly they could change orientation, but that still won't affect their 1-cycle status).
(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

##### Member
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)

#### Christopher Mowla

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)

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,
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}.
(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.

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#### Stefan

##### Member
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?

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

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

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.

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#### Stefan

##### Member
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

Just to give an update,
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.
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

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)
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).

#### qqwref

##### Member
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).