Christopher Mowla
Premium Member
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?
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?