And instead of asking what commutators are here, you should go search for what they are.
This is the notation Alejandro used:
[A,B] is a commutator. It is solved as A B A' B'.
C^D is a conjugate. It is solved as D C D'. Or, [A,B]^C is solved as C A B A' B' C'.
按大部分人的习惯，我把魔方改为了黄顶红前
[java3=300,300]
[param=scrptLanguage]SupersetENG[/param]
[param=scrpt]L'U'R2U'LUR2U'L'UUL \n CU U' L2U'R'UL2U'RU U CD \n U' RDR'URD'R' \n D ML' U L' U' ML U L U' D' \n TL U ML' U' R U ML U' R' TL' \n CL U ML' U' R' U ML U' R CR \n L' U L U' ML' U L' U' TL \n CR ML2 U' L2 U ML2 U' L2 U CL \n L2 CD TL' RUR'F'RUR'U'R'FR2U'R'U'TL CU L2[/param]
[param=initScrpt]U2R2BD2BL2B'U2FL2D'B'U'R'D'F2URD2BF' CU CB[/param]
[param=stickersRight]5,5,5,5,5,5,5,5,5[/param]
[param=stickersDown]1,1,1,1,1,1,1,1,1[/param]
[param=stickersLeft]2,2,2,2,2,2,2,2,2[/param]
[param=stickersUp]4,4,4,4,4,4,4,4,4[/param]
[/java3]
Alejandro Orozco 30.90 世界纪录的解法。怎么步骤这么短？
Hi, my name is Alejandro Orozco, for people who have been asking for the scramble, here it is, I also wrote the solution.
white up and green in front. 大家好，我叫Alejandro Orozco,很多人问我30.90秒世界记录的打乱法，下面就是了，我还写了还原方法，打乱的魔方放置是“白上绿前”：
[M',UL'U']^D 将[M',UL'U']写成普通步骤是M'(UL'U')M(ULU')，然后M'(UL'U')M(ULU') ^ D 写成普通步骤就是DM'(UL'U')M(ULU')D' [UM'U',R]^l 将[UM'U',R]写成普通步骤是(UM'U')R(UM'U')R'，然后(UM'U')R(UM'U')R' ^ X' 写成普通步骤就是x' (UM'U')R(UM'U')R' x [UM'U',R']^X' 将[UM'U',R']写成普通步骤是(UM'U')R'(UMU')R,然后(UM'U')R'(UMU')R ^ X' 写成普通公式就是
x' (UM'U')R'(UMU')R x L'ULU'M'UL'U'l ( [M,ULU']^X this one is more optimal but slower) 也可以用[M,ULU'] ^ X，将[M,ULU']展开为普通公式就是M(ULU')M'(UL'U')， 然后(UM'U')R'(UMU')R ^ X 写成普通公式就是 x M(ULU')M'(UL'U') x'，转法更优化一些，不过转起来会慢些) [M2,U'L2U]^X 将[M2,U'L2U]写成普通公式是M2(U'L2U)M2(U'L2U)，然后M2(U'L2U)M2(U'L2U) ^ X
写成普通公式就是x M2(U'L2U)M2(U'L2U) x' 也就是棱块还原步骤为：
DM'UL'U'MULU'D'
lUM'U'RUMU'R'l'
x'UM'U'R'UMU'Rx
L'ULU'M'UL'U'l
xM2U'L2UM2U'L2U x'