My ultimate commutator challenge

[macky]

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

Hi cmowla,
Your proof is wrong.

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

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

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

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

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

[oll+phase+sync]

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

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

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

[aronpm]

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

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

1 QTM

[oll+phase+sync]

1 QTM

How to proof?

[cmowla]

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?

[cmowla]

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]

Wow. Interesting!

[cmowla]

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

[cmowla]

No comments? No scramble requests?

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

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

[Noahaha]

No comments? No scramble requests?

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

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