New and Shorter OLL Parity (Single Edge Flip) Algorithm! (23q)

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Uploaded by on Jun 29, 2010

In early June 2010, Christopher Mowla discovered a 23 quarter turn pure edge flip algorithm. Prior to that, there was no pure edge flip algorithm (publicly known) less than 25 quarter turn rotations.

Amazingly, Chris did not use a solver of any kind to find it. In the near future, he (I) will give a tutorial video on how to derive pure edge flip algorithms (this one included) independently from a solver. How these long algorithms work will no longer be a mystery to the cube world!

This video specifically shows the version of the algorithm which involves wide turns for easier execution.

On the 4X4X4: (WCA Notation)
x' Rw2 U2 Lw' U2 r U2 Rw U2 x' U r U' F2 U r' U Rw2 x

and on the 5X5X5: (WCA Notation)
x' Rw2 U2 (Lw' M') U2 r U2 Rw U2 x' U r U' F2 U r' U Rw2 x

However, the algorithm can be executed using only single slice turns:

(WCA Notation)
x' r2 U2 l' U2 r U2 l x U2 x' U r U' F2 U r' U r2 x

For speedsolving, all wide slice turns can be used to obtain the non-pure result (and faster time).
Speed-Form for the 4X4X4: (WCA Notation)
x' Rw2 U2 Lw' U2 Rw U2 Rw U2 x' U Rw U' F2 U Rw' U Rw2 x

Speed-Form for the 5X5X5: (WCA Notation)
x' Rw2 U2 (Lw M)' U2 Rw U2 Rw U2 x' U Rw U' F2 U Rw' U Rw2 x

This video concludes with a 25q 3-edge flip algorithm.

On the 4X4X4: (WCA Notation)
x' U' Rw r U2 Lw' U2 Rw U2 Rw U2 x' U Rw U' F2 U Rw' U Rw' r' U x

Although not shown in the video, here it is on the 5X5X5:
x' U' Rw r U2 (Lw M)' U2 Rw U2 Rw U2 x' U Rw U' F2 U Rw' U Rw' r' U x


The software used to display the cubes is Cube Twister:
http://www.randelshofer.ch/cubetwister/

The software used to screen record is Total Screen Recorder:
http://www.totalscreenrecorder.com/

Lastly, the music is "In Memoriam" from Youtube's audioswap library (I wanted to avoid copyright infringement).

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Uploader Comments (4EverTrying)

  • I can do it in 17 moves :p

  • @speedcuber1995

    As I have stated in the title of the video, 23 is referring to block quarter turns, not block half turns. This algorithm is 16 block half turns.

  • @4EverTrying Rw U2 (r) Rw U2 Rw U2 R'w U2 Lw U2 R'w U2 Rw U2 R'w U2 R'w

  • @speedcuber1995

    Thanks for sharing, but:

    1) You should write the algorithm like this to match standard notation

    Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'

    2) I know this algorithm, and I have derived it in another one of my videos (Lucas Parity). It's 17 block half turns and 25 block quarter turns (25q). The algorithm in this video is briefer in both move sets. As another commenter stated, there are 15 block half turn algorithms as well, but none of them are less than 25q.

  • i know of a 15 move on the 4x4: R2 B2 U2 L U2 R' U2 R U2 F2 R F2 L B2 R2.

    it works on all of the cubes i believe if you just use multiple layers. for example on a 7x7 you would use the first and second layer of just the second layer or first second and third layers:) oh yeah..........I AM THE SECOND COMMENT!!!!!! OH YEAH!!!!!!!:)

  • @TheAwsomecubeGuy

    Well, it's r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2. That alg is 15 btm (slice half turn moves), but that's not the move metric I am claiming my algorithm is significant in. In q turns (slice quarter turns), the 15 move alg you posted is 25q, whereas mine is 23q.

    In addition, my algorithm is optimal in plain half turn moves, being 19, as opposed to 20 if you make just the first and last moves of the 15 btm alg wide. (Some people may consider r2 to be two half turns.)

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  • this is AMAZING....brilliant!!!

  • GAHHH MAKE THIS SLOWER I CANT GO THAT FAST!!!

  • lol I went WTF when the 4x4 suddenly got solved at the end.

  • Marvellous, thanks! The best is it doesn't affect ANY of the other blocks, where as the old way affected the top layer and set you back some. Transposed into what I know it's r2 B2 l' B2 r R' B2 r B2 D r R' D' B2 D r' R D r2. Now to go and learn it...

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