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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).

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

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

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

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

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

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

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

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

All configurations of the 2x2x2 Rubik's cube can be solved with (1 turn) + (1 commutator)

Well of course I did not solve tens of millions of cases by hand, but I did some serious case reduction to solve about 1000 cases which represented all even permutations of the corners (and thus all even permutations of the 2x2x2).

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

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

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

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

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

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

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

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

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

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

That is, if there are some inner layer slices in an odd permutation, and/or if the outer layer slices are in an odd permutation, then we just have to do at most \( \frac{n}{2} \) quarter turn slices (to one half of an even cube) to make the permutation of all of the slices even. From there, we can without a doubt use one commutator to solve the rest.

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

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

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

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

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

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

If the outer layer slices are in an even permutation and the sum of the center rotations is a multiple of 360 (an even multiple of 180, if you will), then no outer layer slice turn needed.

If the outer layer slices are in an even permutation and the sum of the center rotations is an odd multiple of 180, then a half turn outer layer slice is needed.

If the outer layer slices are in an odd permutation, chose to do an outer layer quarter turn in the direction to cause the sum of the center rotations to become a multiple of 360.

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

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

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

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

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

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

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

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

Therefore, the answer to the question we have all been waiting for for 2.5 years:
, mrCage (if you're still reading the forums even though you haven't been actively lately),

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

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

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

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

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

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

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

Therefore, the answer to the question we have all been waiting for for 2.5 years:
, mrCage (if you're still reading the forums even though you haven't been actively lately),

[...]

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

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

Therefore, the answer to the question we have all been waiting for for 2.5 years:
, mrCage (if you're still reading the forums even though you haven't been actively lately),

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

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

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

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

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

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