Last Layer with 3 Faces

[Stefan]

So is my proof good enough?

Not by a long shot.

 

[blah]

So is my proof good enough?

Not by a long shot.

How about now?

Setup moves for edges (HTM optimal):
UB: R2 U' R2
BU: U F' R' F U2 F2 U
UL:
LU: U2 R' U F' U
UF: R2 U R2
FU: R F R U2 R2
FL: U' F2 U' R U2
LF: U' F U
FD: U2 F' R F U2
DF: U' F2 U
FR: U2 R U2
RF: U' F' U
DR: U2 R2 U2
RD: U2 R U F' U
BR: U2 R' U2
RB: U2 R2 U F' U

Setup moves for corners (HTM optimal):
URB: U F U'
RBU: R2
BUR: R' F
UFR: F
FRU: U F2 U'
RUF: R'
ULF: F R'
LFU: F2
FUL: U' R' U
DFL: F'
FLD: F' R F
LDF: F2 R'
DRF: F' R'
RFD:
FDR: R F
DBR: R2 F
BRD: R F' R'
RDB: R

Parity: U R U' R F2 U R U R U' R' U' F2 R2

 

[Stefan]

Okay, that does convince me (for me with my knowledge of how old pochmann works, it's alright as proof). Although I must admit I only checked the first few setups. Checking all the remaining ones would've been tedious. Not as much as you finding them, of course. Anyway, this is exactly what makes it an unelegant proof. It's not like "three Rperms give me a U turn, so the cube can be solved with Rperms alone", you know? Well, you can shorten it by saying you can solve corners (and thereby also take care of parity) because their movement is unrestricted. But the edges part would still be tedious and unelegant.

 

[blah]

Okay, that does convince me (for me with my knowledge of how old pochmann works, it's alright as proof). Although I must admit I only checked the first few setups. Checking all the remaining ones would've been tedious. Not as much as you finding them, of course. Anyway, this is exactly what makes it an unelegant proof. It's not like "three Rperms give me a U turn, so the cube can be solved with Rperms alone", you know? Well, you can shorten it by saying you can solve corners (and thereby also take care of parity) because their movement is unrestricted. But the edges part would still be tedious and unelegant.

Yeah I didn't like my proof because of the elegance issue too

I've had another idea though, with a missing part I haven't had the time to think about.
1. Missing part: Solve FL, FD, RD, RB
2. ELS
3. U/H/Z-perms to solve U edges
4. Commutators for corners (haven't thought about proving this yet, but my gut tells me there's a trivial proof)

That said, this isn't much more elegant either, if at all

 

[mrCage]

Edge 2-flip:
F' U F2 U' R' F' U' R2 U R' F2 R2 F2 R2

Corner 2-twist: (2 mirrored sunes, sort of):
R' U' R U' R' U2 R U2 - R U R' U R U2 R' U2

Corner 3-cycle:
R' F U2 F' R F R' U2 R F'

Edge 3-cycle:
R U R U R U' R' U' R' U'

Apart from the edge 2-flip it's all basic knowledge

Per

 

[Stefan]

Per, what's that post supposed to be?

 

[mrCage]

Per, what's that post supposed to be?

All the required building blocks for a <U,R,F> solving strategy

On the other side once <U,R,F> solving strategy has been shown, one can quite trivially extend to another version of solving without D turns entirely.

Per

And while we are at it show constructively how to do 4x4x4 solving in the following subgroups:

<U,u,R,r>, <U,u, F,f,R,r> and <U,u,F,f,R,r,B,b,L,l>

 

[TMOY]

Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...

 

[mrCage]

Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...

The last i suggested was a joke (trivial). I'm still serious about <UuRr> and <UuRrFf> though

Per

 

[mrCage]

Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...

Well, only need to show that R and L are possible. How to do the rest is quite trivial Gimme some time on this one

Per

 

[d_sprink]

I did fused-cube solves for fun, and if you know all the right-handed algs that use F,R, and U only, you can mirror them with a U' , rotate the cube so that R becomes F, and mirror it with your new F U and L. Worst comes to worst, do a 4L-LL, but it's not usually necessary. It can be done very easily.

Edit: Reading some more posts, it's funny how far this has drifted from the original subject. Just thought it was worth mentioning.

 

[calekewbs]

well, you can orient 2 edges with F R and U, so edge orientation is possible. You can orient all corners with 2-gen, so OLL is possible.

You can permute edges with 2-gen algorithms so Edge perm is possible, and you can swap two corners with T perm (which is only U R and F) so LL with only U R and F is 100 percent possible.