Fewest Number of moves that doesn't Affect the Cube?

[turtwig]

So I was thinking about the fewest number of moves that one could apply on a cube to keep to return to the same position. So, for example, if you applied it to a solved cube, it would become solved again.
The algorithm can't just invert what it does, (ex. R R').

 

[sqAree]

M2 E2 M2 E2 would be my first attempt.

 

[TDM]

M2 E2 M2 E2 would be my first attempt.

8 moves HTM is quite short, wouldn't be surprised if this was the shortest. There's also R U R' F' U2 L' U' L F U2 which is 10 moves.

 

[irontwig]

I think zero moves is pretty short.

 

[whauk]

This is somewhat related: https://www.speedsolving.com/forum/threads/3x3x3-permutation-theory-puzzle.14107/
Also R L R' L'. You might argue, that R and L commute, hence this was an application of inverses, but then so is M2 E2 M2 E2.

 

[theradhaxor]

M2 and E2 are commutators, as they share parts of the cube while turning. R and L don't have a intersection, so they can't be counted, But M2 and E2 follow the X,Y X',Y' rule that makes it a commutator, which actually affects the cube

 

[Attila]

R F2 U D' L2 F2 R2 D U' B2 R
R2 D2 B2 U' D' B2 U2 R2 U' D'
R2 U' L2 U2 D2 R2 U L2 U2 D2

R2 B2 R2 L2 B2 F2 L2 R2 F2 R2

Most often possible to find like this, of domino solutions (almost infinite number of possibilities)

 

[Chree]

This is from Chris Hardwick's sig... thought it was fun: R L F2 B2 L' R' U R L B2 F2 L' R' D (Added the last D to finish)

 

[adimare]

If M2 E2 M2 E2 counts then so should u D'.

 

[turtwig]

If M2 E2 M2 E2 counts then so should u D'.

Yeah, but none of the moves affect each other in any way... In M2 E2 M2 E2's case, they at least affect each other.

 

[SenorJuan]

So (R U) 105 doesn't win, then....

 

[Cale S]

So (R U) 105 doesn't win, then....

For RU 2-gen I think optimal is (R2 U2)6

 

[Felipe Rigon]

L' U' L F U2 R U R' F' U2

 

[DoctorKilgrave]

M M M M?

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