Symmetrical Fridrich Method

Submitted by sle on Thu, 05/22/2014 - 01:54

Update: I mentioned this on www.speedsolving.com forum and the nice folks there pointed out that this is somewhat obvious. They also raised a concerns about the extra regripping that symmetrical algorithms require. You can read about it here. But it seems to work for me, so I'll probably keep doing it.

Overview

I use the Fridrich Method to solve the Rubik's Cube, but I do it a bit differently than most people. The standard Fridrich method favors the right hand (although I imagine left handed cubers may mirror all of the moves). I decided to learn a variation of the Fridrich Method that is both more symmetrical (ambidextrous) and easier to learn - whenever an OLL or PLL had a mirror image and the algorithm for that mirror image lacked symmetry with the original algorithm I decided to memorize the original algorithm and the algorithm mirrored about the YZ plane rather than memorizing the two distinct algorithms that the standard Fridrich Method has. To illustrate I'll consider Badmephisto's cubing website to be definitive for the standard OLL and PLL. In the following tables I'll include only those algorithms where I differ from the standard Fridrich Method on Badmephisto's site. Each diagram is taken from Badmephisto's site and mirrored using ImageMagick's "convert -flop":

My google searches did not find anything relevant for this concept, but maybe that's because it's obvious.

OLL

Number Diagram Algorithm Mirror Of
9 [F' (L' U' L U) F] U' [F' (L' U' L U) F ] 8
11 [f' (L' U' L U) f] U' [F' (L' U' L U) F ] 10
33 l' U2' L U L' l 41
36 (L' U' L U') L' U2' L 35
37 [(L' U' L U') L' U2' L] [F' (L' U' L U) F] 38
40 (l' U' L U') (L' U L U') L' U2 l 39
43 l U L' U L U2' l' 34
47 F' U' L' U L2' F L' (U' L' U L) 46
50 L U F' U' L' U L F L' 51
52 L [F' (L' U' L U) F] U' L' 16
53 (L F' L F) L2' U2' y' (L F' L' F) 27
55 [F' (L' U' L U) F] U2' [(L' U' L U) (L' F' L' F)] 54
56 (L' U' L U') (L F' L' F) L' U2' L 57

PLL

Number Diagram Algorithm Mirror Of
2 x [(L U' L) D2'] [(L' U L) D2'] L2' 1
3 L2' U' [L' U' L U] (L U) (L U' L) 4
10 [L U2' L' U2'] [L F'] [L' U' L U] [L F] L2' U 9
14 [L' U' L] y L2' u L' U L U' L u' L2' 13
16 L2' u L' U L' U' L u' L2' [y' L' U L] 15
18 [L' U' L F] {[L' U' L U] [L F'] [L2' U L] U} 7