Improvements for M2/R2

[Freddy88]

Hi Guys, how are you?
I'm working on some improvements for M2... I've adopted an idea of Maky who suggest to use three cycle for the nasty cases where two M edges follow. In his guide he uses commutators. I do instead do that with M.

As you can see with x or x' cube rotations you just need to memo 8-10 algs... wich is not that easy 'cause they are all plenty of M moves. But is not also that hard...most of them are quite short.

Any way the problem is that I would like to find better, more finger friendly algs for UF DB, UB BD and DB UF (shooting form FD) do you know some R based commutator who works for that three cases....that would be great!

Have a nice evaning!

Federico Soldati



UF  DB (U' M2) (R U' R' U) M U') (R U R') (M U)
BU  DB x (U' M2) (R U' R' U) M U') (R U R') (M U)x '
UF ® BU x' (U' M2) (R U' R' U) M U') (R U R') (M U)x

DB  UF U'M'RU'R'UM'U'RUR'M2U
DB  BU x U'M'RU'R'UM'U'RUR'M2Ux'
BU  UF x' U'M'RU'R'UM'U'RUR'M2Ux

BD  UF (U M' U' M) U2 (M U M' U)
DB  UB x (U M' U' M) U2 (M U M' U) x'
UB  FU x' (U M' U' M) U2 (M U M' U) x

UF  BD (U' M U' M') U2 (M' U M U')
UB  DB x (U' M U' M') U2 (M' U M U') x'
FU ® UB x' (U' M U' M') U2 (M' U M U') x

UB  UF (U M U M) U2 (M' U M' U')
DB  FU x' (U M U M) U2 (M' U M' U') x

UF  UB (U M U' M) U2 (M' U' M' U')
FU  DB x (U M U' M) U2 (M' U' M' U') x'

BU  FU (M' U2 M U2)
BD  FU x (M' U2 M U2) x'

FU ® BU (U2 M' U2 M)
FU  BD x (U2 M' U2 M) x'

UB  BD (U' M U M') U2 (M' U' M U')
BD  UB (U M' U M) U2 (M U' M' U)
BD  BU (M U2 M U2) M2
BU  BD M2 (U2 M' U2 M')

 

[KJiptner]

Thanks Frederico, I will definitely memorize that stuff Nice you've joined the forum

 

[macky]

Hey Frederico, let me reply to you here. This definitely goes along with my letter-pair-directly-to-alg philosophy. (For people who haven't seen it, here.)

For the cases you mentioned, you could always do something like

UF DB: U'l'x [U'R'U, M] x'LU or r'Ur' [URU', M] rU'r

Eliminating all thinking is the way to go.

[edit] I guess the freestyle people disagree with this. =)

 

[Lucas Garron]

[edit] I guess the freestyle people disagree with this. =)

Yeah. Which only, like, includes the WR-holder.

 

[joey]

Nah, us freestylers are braindead too.

 

[deadalnix]

Why don't you use DR'UR'U'M'URU'r'R2D' ?

 

[Freddy88]

Yeah....I'm still searching good algs for UF->DB (shooting from FD),

For the moment I've decided to change BD-> UB and inverse with the following:

UB  BD U2 (D M' D M') U2 (M' D M' D)
BD  UB (D' M D' M) U2 (M D' M D') U2

Looks good eh?

Have a great day.....and come on, propose!

 

[deadalnix]

UF -> DB can be done like xDM'R'UR'U'Rr'URU'RD'x'

 

[Freddy88]

yeah....but I want a non-cube turns alg.....cause the moves are gonna be the x turns of that algs...

 

[deadalnix]

You can set up by M' if you want to turn the cube !

 

[Freddy88]

Yeah...I don't know if it is better to do a X or X' cube rotation ora a M or M' move...I think I would prefer the hole cube rotation...anyway what I meant is that If I find a simple UF -> DB alg I then can use it wor the following algs too with just a cube rotation (or M/ M' move):

BU -> DB x A x '
UF -> BU x' A x

Because of that I'm searching a good alg (short and fast) for that case. The problem is that there are few cube solver programs with works with M slice turn also. Maybe Stefan or Somebody with good knowlege of Acube could give it a try and find the best solution for that case...including M slice turns....

Regards,

Federico

 

[Freddy88]

Hi Guys, I'm back again.....
So, I've worked out the missing solution...now my algs looks like that:

UF  DB (U' M2) (R U' R' U) (M U') (R U r' U)
BU  DB x (U' M2) (R U' R' U) (M U') (R U r' U) x '
UF ® BU x' (U' M2) (R U' R' U) (M U') (R U r' U) x

DB  UF (U' r U' R' U) (M' U') (R U R') (M2 U)
DB  BU x (U' r U' R' U) (M' U') (R U R') (M2 U) x'
BU  UF x' (U' r U' R' U) (M' U') (R U R') (M2 U) x

BD  UF (U M' U' M) U2 (M U M' U)
DB  UB x (U M' U' M) U2 (M U M' U) x'
UB  FU x' (U M' U' M) U2 (M U M' U) x

UF  BD (U' M U' M') U2 (M' U M U')
UB  DB x (U' M U' M') U2 (M' U M U') x'
FU ® UB x' (U' M U' M') U2 (M' U M U') x

UB  UF (U M U M) U2 (M' U M' U')
DB  FU x' (U M U M) U2 (M' U M' U') x

UF  UB (U M U' M) U2 (M' U' M' U')
FU  DB x (U M U' M) U2 (M' U' M' U') x'

BU  FU (M' U2 M U2)
BD  FU x (M' U2 M U2) x'

FU ® BU (U2 M' U2 M)
FU  BD x (U2 M' U2 M) x'

UB  BD U2 (D M' D M') U2 (M' D M' D)
BD  UB (D' M D' M) U2 (M D' M D') U2
BD  BU (M U2 M U2) M2
BU  BD M2 (U2 M' U2 M')

And of course:

UB s U2 M2 (s) M (s)' M U2 Switch RU and LU
s UB U2 M' (s) M' (s)' M2 U2

UF s M2 (s) M (s)' M
s UF M' (s) M' (s)' M2
s FU (s) M2 (s)' M U2 M U2

DB s M2 (s) M' (s)' M'
s DB M (s) M (s)' M2
s BD (s) M2 (s)' U2 M' U2 M'

The two main algs for UF-> DB and Inverse seems quite long but are actually pretty fast...

I also had another Idea...that could be useful for somebody who whants to keep algs memorisation at the minimum:

For UF-> DB (and Inverse) you could also do "F2 then Alg of UF-> BD and thatn F2 again....

That means that the actual memorization of algs, would become of..about five? If you don'consider the really easy 4 moves one... of course.

I hope that can be useful for you...I don't know if that is gonna be faster than Maky's commutators way...I think so though.....

Have a grat Day!

Federico

 

[Freddy88]

Ah, and remember that Instead fo X rotation you can also do M or M'....
What do you think it's better?

Regards,

Federico

 

[deadalnix]

It seems great ! I will adopt these algs.

 

[Jgig1991]

Thank you for posting this Stefan.

 

[Freddy88]

Hi everybody,

so what follow is the last change I did to my special M layer algs...the algs to be learnt are now only 4, yes, just 4!

I still have difficulties during the UB or UF and DB couples with non M layer edges.....like UB RU or LU UB......that seems to be quite time expensive at the moment....some suggestions?

Regards,

Federico

Ah, that's the list:

UF  DB F2 (U M' U' M) U2 (M U M' U) F2
BU  DB M' F2 A F2 M
UF ® BU M F2 A F2 M'

DB  UF F2 (U' M U' M') U2 (M' U M U') F2
DB  BU M' F2 A F2 M
BU  UF M F2 A F2 M'

BD  UF (U M' U' M) U2 (M U M' U)
DB  UB M' A M
UB  FU M A M'

UF  BD (U' M U' M') U2 (M' U M U')
UB  DB M' A M
FU ® UB M A M'

UB  UF (U M U M) U2 (M' U M' U')
DB  FU M A M'
BD  UB M2 A M 2

UF  UB (U M U' M) U2 (M' U' M' U')
FU  DB M' A M
UB  BD M2 A M2

BU  FU (M' U2 M U2)
BD  FU M' A M

FU ® BU (U2 M' U2 M)
FU  BD M' A M

BD  BU (M U2 M U2) M2
BU  BD M2 (U2 M' U2 M')


A stands for the first alg of the triple or couple...I did it with colors.....

 

[martijn_cube]

These are the two algs i have for BU (buffer DF) when using M2.
But i still don't really like them, i'm currently trying to use the first one.
But is there someone who has a better one?

1) (UR'U') l (U'RU) M2 (U'R'U) l' (URU')

2) (F’ D) R’ (F D’) M2 (D F’) R (D’ F)

 

[Ville Seppänen]

(U'M')*3 U'M (U'M')*4 (I like MU )

 

[martijn_cube]

wow, thats a really nice one. alot faster then my other two. I like M'U' too.

 

[DonQuixote]

BU: (0292718) 1-1- (7281019)

Kind Regards
Martin