L8E on 3x3

[mk8]

hello, i would like to ask if a situation, where the E slice and corners are solved, and all other edges are oriented, is solvable with a nice subset of moves (say like <M U D> or <R M U D>) in a tolerable amount of moves (preferrably under 10 STM)
if yes, do the algs exist or do i have to gen them myself? i use trangium's batch solver to gen all of my algs btw, so including a subset of moves to get all those cases would be helpful

 

[Deleted member 36445]

I remember that @Hazel had those algs, but I don't know how you can find them ATM because her site is down

 

[Bruce MacKenzie]

There are 8! / 2 = 20,160 position permutations of the Up and Down edge cubies. With D4 symmetry these may be boiled down to 2644 cases. That's a lot of algorithms.

 

[mk8]

There are 8! / 2 = 20,160 position permutations of the Up and Down edge cubies. With D4 symmetry these may be boiled down to 2644 cases. That's a lot of algorithms.

well, someone in the Batch Solver thread gave me a nice subset that should include everything and the program found 98 cases... i suppose that's not all of them right?

 

[Bruce MacKenzie]

well, someone in the Batch Solver thread gave me a nice subset that should include everything and the program found 98 cases... i suppose that's not all of them right?

Not by my reckoning. I sat down and wrote some code this evening and solved all 2644 of my cases. Here are the length 4 and 5 maneuvers I found (if you're not familiar with the notation it is outlined on Walter Randelshofer's site):

U2 MF2 U2 MF2​

R2 MF2 R2 MF2​

MU' MF2 MU MR2​

MF2 D' MF2 D​

MF2 D MF2 D'​

MF' MU2 MF' MU2​

L2 MF L2 MF'​

D2 MF' D2 MF​

D' MF2 D MF2​

D MF2 D' MF2​

U' MR2 MU' MF2 D • CU​

U MR2 U2 MR2 U​

U MR2 MU MF2 D' • CU'​

R2 MU' MF2 MU L2 • CR2​

F2 U' MF2 U B2 • CF2​

F2 U MF2 U' B2 • CF2​

D2 MF2 D' MF2 D'​

D2 MF2 D MF2 D​

D' MF2 D' MF2 D2​

D' MF' D2 MF D'​

D MF2 D MF2 D2​

D MF' D2 MF D​

This is from the symmetry reduced set so each of these represents as many as eight distinct arrangements depending on the orientation of the cube to which they are applied.

 

[mk8]

Not by my reckoning. I sat down and wrote some code this evening and solved all 2644 of my cases. Here are the length 4 and 5 maneuvers I found (if you're not familiar with the notation it is outlined on Walter Randelshofer's site):

U2 MF2 U2 MF2​

R2 MF2 R2 MF2​

MU' MF2 MU MR2​

MF2 D' MF2 D​

MF2 D MF2 D'​

MF' MU2 MF' MU2​

L2 MF L2 MF'​

D2 MF' D2 MF​

D' MF2 D MF2​

D MF2 D' MF2​

U' MR2 MU' MF2 D • CU​

U MR2 U2 MR2 U​

U MR2 MU MF2 D' • CU'​

R2 MU' MF2 MU L2 • CR2​

F2 U' MF2 U B2 • CF2​

F2 U MF2 U' B2 • CF2​

D2 MF2 D' MF2 D'​

D2 MF2 D MF2 D​

D' MF2 D' MF2 D2​

D' MF' D2 MF D'​

D MF2 D MF2 D2​

D MF' D2 MF D​

This is from the symmetry reduced set so each of these represents as many as eight distinct arrangements depending on the orientation of the cube to which they are applied.

but all of the cases can be achieved with this subset of moves: <M2, S2, M' U2 M, M U2 M', S' U2 S, S U2 S'>, right? for this exact generator, the batch solver gives 68 <M U D> cases with y rotations allowed
what could be wrong?

 

[Bruce MacKenzie]

but all of the cases can be achieved with this subset of moves: <M2, S2, M' U2 M, M U2 M', S' U2 S, S U2 S'>, right? for this exact generator, the batch solver gives 68 <M U D> cases with y rotations allowed
what could be wrong?

All the 20,160 position permutations of the UP+DOWN edge cubies may be found by closure of the group formed from the two generators R U L D F2 D' L' U' R' F2 and R' L' U D B2 R' B2 U' D' R L D'. If you can form these two generators using your subset of moves then your subset of moves can form all 20,160 U+D edge cubie position permutations. I have no idea what your batch solver is doing.

Addendum:

This morning I fired up GAP and found that one can close the group with three of the length 4 maneuvers:

MU' MF2 MU MR2
MF2 D' MF2 D
L2 MF L2 MF'

If you can show that your subset of moves can form these three elements then the subset can form the whole group.

 

[PiKeeper]

but all of the cases can be achieved with this subset of moves: <M2, S2, M' U2 M, M U2 M', S' U2 S, S U2 S'>, right? for this exact generator, the batch solver gives 68 <M U D> cases with y rotations allowed
what could be wrong?

I think you need to add U and D into those subsets to make it work. Otherwise there's no way to get a case where an s layer edge is in the m layer

 

[Hazel]

I generated all the algorithms for EO in this state:

https://imgur.com/a/ANYPHN9

Some of the algs might not work though, I don't think I actually tested all of them