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D

Daniel Lin

Guest
a perm

R2 D R2 D' F2 R2 U' R2 U F2

is it possible to put this in comm notation?
 

xyzzy

Member
Joined
Dec 24, 2015
Messages
2,878
a perm

R2 D R2 D' F2 R2 U' R2 U F2

is it possible to put this in comm notation?

Sort of cheating, but you could do this:

[R2 U F2 U' F2, {mirror across x=z plane}]
= R2 U F2 U' F2 R2 U' R2 U F2

R2 U F2 U' F2 causes a bowtie-shaped 4-cycle of edges between the U and D layers. This cycle shape is preserved under the mirroring, so applying the inverse after the mirroring causes the U- and D-layer edges to be solved again. It also sends the DFL, DFR, DBR corners to the U layer in a symmetrical manner (so again, the commutator doesn't affect those corners), but it sends the UFL, UFR, UBR corners to the D layer in an asymmetrical manner, which ends up being an A-perm at the end of the commutator.

R2 U # F2 U' F2 R2 U' R2 U F2

Insert [E', F2] at # to get rid of the 3-cycle of edges in the E slice and simplify to get R2 D R2 D' F2 R2 U' R2 U F2.
 
D

Daniel Lin

Guest
Sort of cheating, but you could do this:

[R2 U F2 U' F2, {mirror across x=z plane}]
= R2 U F2 U' F2 R2 U' R2 U F2

R2 U F2 U' F2 causes a bowtie-shaped 4-cycle of edges between the U and D layers. This cycle shape is preserved under the mirroring, so applying the inverse after the mirroring causes the U- and D-layer edges to be solved again. It also sends the DFL, DFR, DBR corners to the U layer in a symmetrical manner (so again, the commutator doesn't affect those corners), but it sends the UFL, UFR, UBR corners to the D layer in an asymmetrical manner, which ends up being an A-perm at the end of the commutator.

R2 U # F2 U' F2 R2 U' R2 U F2

Insert [E', F2] at # to get rid of the 3-cycle of edges in the E slice and simplify to get R2 D R2 D' F2 R2 U' R2 U F2.
wow. that's cool.
so [R2 U F2 U' F2, {mirror across x=z plane}] is just R2 U F2 U' F2 and the inverse of the mirror, right?
You put it in brackets so I thought it was A B A' B'
 

xyzzy

Member
Joined
Dec 24, 2015
Messages
2,878
You put it in brackets so I thought it was A B A' B'

Writing it out in full, it'd be R2 U F2 U' F2 {mirror} F2 U F2 U' R2 {mirror}, and if you hold up the cube to a mirror (at a 45-degree angle) and try to make those moves on the cube in the mirror, those five moves between the {mirror}s end up being simplified to R2 U' R2 U F2.

Another alg that can be written as a commutator with a reflection is this T-perm alg:
[R2 u R2 u' R2, {mirror across x=z plane}] = R2 u R2 u' R2 F2 u' F2 u F2
 

guysensei1

Member
Joined
Nov 24, 2013
Messages
5,143
Location
singapore
WCA
2014WENW01
A 1LLL I found by messing around with S moves,
S' R U R' S U R U2 R'
The inverse is good for OLLCP?

EDIT: similarly,
R' F R U R' F' r y' R U' R' S
 
Last edited:
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Daniel Lin

Guest
Wow I found a simple COLL/ZBLL alg just by tweaking a Pi alg

R' U2 R U2 R' F' R U R' U' R' F R2

This might be better than the standard alg, too lazy to switch though
 
D

Daniel Lin

Guest
Triple post

rotationless N perm
R D' F2 R2 U' L F2 L' U R2 F2 D R'

this is just a setup to a ZBLL
 
D

Daniel Lin

Guest
why am I the only one posting here
found a 1LLL alg : R L r D r' U2 r D' r' U2 L' R'
 

hagner

Member
Joined
Jul 26, 2016
Messages
68
Location
finland
I'm chocked that i couldn't find a thread like this.
so yeah... if you invent an alg you can post it here
ill start with R U R´ F´ L´ U´ L F
anti-p nb-perm white non-oriented bar in B
mirrors and reverses also suggested
not useful for me but maybe for someone

nb! i have no idea how useful or useless algs like this are.
since i´ve only been cubing for 6 months
 
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