Commutators and Conjugates - Invent Your Own Algorithms! [Rubik's Cube]

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  • เผยแพร่เมื่อ 21 ต.ค. 2024

ความคิดเห็น • 20

  • @mateszabo5535
    @mateszabo5535 5 หลายเดือนก่อน

    OLL: F [ R, U ] F' = F, ( R, U, R', U' ) F'

    • @EidenCubing
      @EidenCubing  5 หลายเดือนก่อน +1

      Correct! Well found!

  • @Shadowex2000
    @Shadowex2000 2 ปีที่แล้ว +6

    Wow. This is stunningly clear and concise. The best commutator explanation I have found. Thanks a lot!

  • @bananasauce1000
    @bananasauce1000 4 หลายเดือนก่อน +1

    this is way better and more clearly explained than j perm’s tutorial. thanks so much!

  • @sarahwatts8184
    @sarahwatts8184 ปีที่แล้ว +1

    Best tutorial ever I don’t know why you aren’t more popular

  • @yishnir
    @yishnir ปีที่แล้ว

    I really appreciate the work you put into this. This has been, by far, the clearest explanation I've seen so far. It was very useful giving the algorithm notation which let's me repeat the moves until I'm comfortable enough to 'watch' what the pieces are doing.

  • @tharunsrinivasan7060
    @tharunsrinivasan7060 2 ปีที่แล้ว +1

    Great video and it really helped me out!

  • @sakicuber
    @sakicuber 3 ปีที่แล้ว +1

    Great video!

  • @ytpah9823
    @ytpah9823 ปีที่แล้ว +1

    🎯 Key Takeaways for quick navigation:
    00:01 🧩 A commutator is a technique used to move three pieces on a Rubik's Cube in a cycle.
    00:31 🧩 Commutators can be used to cycle around three edge pieces or three corner pieces.
    01:00 🧩 To perform an edge commutator, start with an interchange move that affects only two of the pieces, followed by a sequence of moves to solve the desired sticker, then undo the interchange move and the second set of moves.
    01:59 🧩 Corner commutators are similar to edge commutators, but the interchange move is a normal move instead of a slice move.
    02:58 🧩 Some cases, like the U-perm, cannot be solved using a commutator because the interchange move must only affect two of the three pieces.
    03:26 🧩 Cases like the U-perm can be solved with a commutator by using a conjugate, which involves a setup move before the interchange move to move one of the three pieces out of the interchange layer.
    04:29 🧩 Conjugates can also be used with corner commutators, where a setup move is performed before the interchange move to solve the desired corner.
    04:58 🧩 In some cases, the commutator algorithm can be simplified by canceling out certain moves, resulting in a standard algorithm for that particular case.
    05:30 🧩 There is another algorithm from 2-look OLL that is also a conjugate, allowing for the creation of unique algorithms for solving the Rubik's Cube.

  • @MatthiasTroffaes
    @MatthiasTroffaes 3 ปีที่แล้ว

    Awesome stuff, and great explanation!

  • @SushantTari-vq8mc
    @SushantTari-vq8mc 5 หลายเดือนก่อน

    Nice video

  • @ASH135
    @ASH135 2 ปีที่แล้ว

    That was helpful

  • @alexgoes8973
    @alexgoes8973 ปีที่แล้ว

    Is the commutator the T oll

    • @tylerduncan5908
      @tylerduncan5908 ปีที่แล้ว +1

      I believe the case he's referring to is U oll (solved top cross with headlights on one side)
      It goes
      R2 D (R' U2 R) D' (R' U2 R')
      R2 is the conjugate or setup move.
      R2 -> C
      Then you have
      D -> A
      R' U2 R -> B
      D'-> A'
      R U2 R'-> B'
      Then undo setup
      R2 -> C'
      So when u you put it all together, you get C (ABA-¹B-¹) C-¹
      R2 D (R' U2 R) D' (R' U2 R) RR'
      The final R2 cancels with the R to get R'
      There are other corner oll's that you can solve with commutators but they're not the most commonly used but they do come up in certain ZBLL's
      The algs i use for the other 2-corner cases are
      R U R' D R U' R' D'
      and
      R U' R' D R U R' D'
      which are just the first and 2nd halves of the E perm.

  • @faz555
    @faz555 3 ปีที่แล้ว

    hey eiden

  • @remigaborit2486
    @remigaborit2486 2 ปีที่แล้ว

    0:31 Hello and thank you for your work. An error in your (incomplete) algorithm is (z2) R' F R F' B U' F' U F' B' F2

    • @EidenCubing
      @EidenCubing  2 ปีที่แล้ว +2

      Thank you so much! I've cancelled the two F moves at the end of your algorithm, so I think we end up with the same thing.

  • @emacipadung8054
    @emacipadung8054 2 ปีที่แล้ว

    Com + comnju = conj+com+conj'