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Blindfold Cubing - Other

Messy.

Big Cubes (4x4x4 and up)

4x4x4

Speed memorization for big cube blindfold solving requires a good memorization method. Chris Hardwick explains his here.

1. Centers: Mike Hughey's guide, Daniel Beyer's guide. Learn how to use commutators to 3-cycle centers. Learn "interchangeable piece" and "lone piece."

2. Corners, just as in 3x3x3, except possibly parity.

3. Edges (there's only permutation): One method is to use free commutators. The concepts from centers can be used for this method. Also see Chris Hardwick's demo solve. The other method is a variation of M2: Erik's r2 guide.

Puzzles I have solved blindfolded - in chronological order

  • 3x3x3
  • 2x2x2
  • Pyramorphix
  • 2x2x3 (wide tower; simulated on a 4x4x4)
  • Homer Simpson Head (Another 2x2x2 variation)
  • 4x4x4
  • Pyraminx (worked out a method myself. inspired by Stefan.)
  • 3x3x2 (simulated on a 3x3x3)
  • 2x2x3 (silm tower i.e. fully working 2x2x3)
  • 2x2x4 (simulated on a 4x4x4)

    Outline of Blindfold Methods for Other Puzzles

    Once you know how 3x3x3 blindfold solving works, it's relatively easy to apply the same concepts to solve other puzzles blindfolded. I only provide outlines and useful algorithms.

    2x2x3

    2x2x3 is very simple because the orientation of every piece stays the same; you only need to worry about the permutation. Hold the puzzle so that the 2x2 squares are U and D layers. I did this on a 2x2x3 "Franken" tower from Time Traveler's Web Page I won at US National 2004, but you can always simulate this on a regular 3x3x3. I always get distracted by the U and D layer edges when I do this though....

    Pre-memorization - Prior to memorization, hold the cuboid in such a way that each middle edge has a correct orientation. If it does not, you can simply fix this by performing the rotation y. From here, you can perform y2 and x2 to match as many of the corners as possible. Once you have done this, memorize the permutation.

    Solving the corners - This is the algorithm I use to switch corner 2 and 3 (URF and URB) without disturbing the middle edges:

  • R2UR2U'R2-DU'-R2U'R2UR2-D'
    This algorithm can be used as a 3-corner cycle also. For example, if I wanted to perform the cycle (134) i.e. ULF->URF->URB->ULF, I would first do U and then use this algorithm, then do U2. As this algorithm fixes 2 corners, all 8 corners can be solved with at most 4 times of using it.

    Solving the middle edges - There are only a couple of cases for this step, each one of which can be solved with a short algorithm. Let us name the edges in the following manner: starting at the Front Left edge and going counter clockwise, name 1, 2, 3, 4. These are the algorithms I use:
    1.(23) [R2U2]x3
    2.(321) ER2E'R2
    3.(14)(23) [R2E2]x2
    It should be simple enough to determine which ones should be used for the permutation you get.

    3x3x2

    Introduction - 3x3x2 also does not have correct or incorrect orientations, and you are only left to worry about the permutation of 8 corners and 8 edges. The real 3x3x2 cuboid would be a Rubik's Domino, but I use a 3x3x3 to simulate this by restricting the moves within (R2L2F2B2UD) group. IF I ever get to try this on a real Domino, I will probably be really confused not knowing where each piece has to go. =D What? How did I take that pic then? Um, good question.... ;D Ok, let's jump right into algorithms.

    Solving the corners - Here's the algorithm I use to swap corners 2 and 3 (look at solution for 2x2x3):

  • R2UR2U'R2-DU'-R2U'R2UR2-D'-F2U2F2U2F2U2
    It's the same algorithm I used for 2x2x3, with an edge fix at the end. In the case of 3x3x2, this algorithm cannot be used to perform a 3-corner cycle. Too bad, I'll just have to find a better algorithm. *sigh*

    Solving the edges - Again, I only have a 2 edge-swap algorithm, but this one is well-known and very short:

  • R2U2R2U2R2U2
    Oh, that one.... That's right, I don't have a 3-edge cycle alg. *cries* Someone, PLEASE, find one.

    [Edit] Only 1 day after I wrote the above cry for help, Stefan Pochmann came to the rescue! =D He sent me a 3-cycle for both the corners and the edges on 3x3x2, which are listed below. He also recommended me using ACube, which I have downloaded and am currently struggling with. ^^;; Thank you, Stefan!

  • Corners (146): (U R2 U' F2)x2
  • Edges (139): (R2 F2 R2 U2)x2

    2x2x4

    Introduction - I first simulated this on a 4x4x4. Although each 1x1x2 block can now be oriented in 3 ways, the puzzle itself is not too hard. If you can solve two 2x2x2's blindfolded, this should not be a problem, as you only need to memorize 1 set of orientation. If there is anything hard about this, it will be determining the permutation of the middle layers. This is due to the fact that there is two pieces with the same color combination, just in different order. You can easily see by tracing, and noticing that the colors flip when moves to the other layer in the same vertical slot.

    Pre-memorization Stuff - Hold the cuboid in the same way you did a 2x2x3. There's no need for middle edge orientation check.

    Memorization - Simply memorize the permutation of the middle 2 layers, the permutation of the top and bottom layers, and the orientation of the 1x1x2 "corners." Each one of these will be solved independently.

    Solving the orientation - You can do this just like you would a 2x2x2 or 3x3x3.

    Solving the middle two layers - Set up the pieces within {R2,L2,F2,B2,Uu,Dd} group. This keeps the orientation correct. The following algorithm swaps the RF and RB "edges" of the middle layers, and can also be used for a 3-cycle:

  • R2uR2u'R2u'dR2u'R2uR2d'

    Solving the top and bottom layers - This is basically the same. Now you may want to set up within {R2,L2,F2,B2,U,D} group for mental comfort, (although that doesn't afffect anything). You may use this algorithm to swap the RF and RB corners:

  • R2UR2U'R2U'DR2U'R2UR2D'

    In general, when solving a 2x2x2n cuboid blindfolded, you must memorize the orientation of 8 1x1xn blocks and n sets of permutation of 8 pieces for each 2 layers, counting from the middle, one going down and one up. (I guess for the orientation, you can just feel it.) You can solve the orientation just like you would on a cube, then fix the permutation of each set of two layers. To solve the mth set of two layers counting from the middle, you can use this algorithm to swap the top RF and RB pieces:
  • R2UmR2Um'R2Um'DR2Um'R2UmR2D'

    This algorithm is really the general form of the base of all the algorithms I use to solve a 2x2x3, 3x3x2, a 2x2x4, and in the future, probably 2x2x6 and all that stuff. Hmm, I think that was pretty basic. If you get the point, larger cuboids shouldn't be much harder. (The ones with many center pieces are always hard, of course.)

    5 Cycles

    Just as 3-cycles allow us to correct the permutation of two pieces at a time, 5-edges can solve four pieces simultaneously. This would certainly make set-up moves more complicated, but would this feasible with several different 5-cycles?

    5-cycles of edges can be just as short as 3-cycles. For example,

  • [B2R2'UR2]x2: (1 3 12 2 11)

    which is very easy to execute and also has a very regular shape of the cycle. Some variations:

  • [R2'UR2B2]x2: (1 11 12 4 3)
  • R2B2R2UR2B2R2U: (1 3 2 4 11)

    To reduce the thinking for set-up moves enough to make this at all useful, we would also need a 5-cycle involvng the middle layer. If we have something like (1 3 2 4 5), we can do any 5-edge cycle by setting up 1-4 and then seeing if the last piece ends up in E or D layer. 5-corner cycles are probably too long and too hard to set-up.

    Cool Blindfold Cubing Algorithms

    Corner Permutation
    3-cycles:
    (L2 U) [B2 U']x2 (L2 U) (B2 U B2 U') (3 1 8) Brent Morgan
    UL2UR2 U'L2UR2U2 (1 6 5) Ron van Bruchem
    U'R2UR2UF2U'R2U'R2UF2 (1 3 6) Ron van Bruchem

    Edge Permutation
    5-cycles:
    R2B2R2UR2B2R2U (1 3 2 4 11) me
    (R2U)^2 (1 4 3 2 12) pathfinder_netstorm
    Double transpositions:
    (M' y M' D M (D2 y') M D) (3 4)(9 12) Stefan Pochmann

    Edge Orientations:
    z'F'L'F-(M'U)x4-F'LFz (5 6 7 8)
    (l U l' U) * 5 (1 2 3 4 5 9) Joel
    (M2U)(rR)(dD)(rR)(dD)(rR)(uM2)y2 Pedro

    Richard Carr's not-even-Intermediate Method Algorithms

    Application to Speedcubing (An Edge Orientation Fridrich Variation)

    (Suggested in this forum posting by Ron van Bruchem)

    During the 15 seconds of inspection, determine the edge orientation in edge as defined by the group using the technique in blindfold cubing. While making the cross, completely fix edge orientation, and continue the rest of the Fridrich method. With edges all correctly oriented, F2L can be solved using R, L, and U, and the last layer will always have all edges correctly oriented. This makes the F2L extremely quick by eliminating cube turns entirely and makes it possibly to go directly to ZBLL without ZBF2L, finishing the last layer in one step. Grant posted his analysis of edge orientation to follow Ron's interesting result.



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