algs that goes back to solved position after X times

[Micael]

Lol! I did not knew any sequence eventually return to its state. I thought I was going to explose qqwref's calculator by inputting a scramble. After many try, I was very disappointed. But I used qqtimer for scrambles, there should be a ploy...

 

[Erdos]

...

I don't see what you're getting at. I was also showing that x is not the identity through a proof by contradiction. And since x is not the identity element, we can use the contraposition of the identity's definition: applying x once to a solved state cannot lead to a solved state. So to use the speedcubing definition in terms of order would not make sense because it is not well-defined.

 

[cuBerBruce]

... applying x once to a solved state cannot lead to a solved state.

No, it seems to me you are confusing the concept of a solved state with the concept of a "move" that is an identity. It is perfectly possible for a non-identity "move" of a puzzle to change the "state" of the puzzle from one of its solved states to another one of its solved states, assuming the puzzle has multiple states that are each considered solved. I believe you are equating the term "identity" with the term "solved state," which is the flaw in your logic, it seems to me. Think of group elements as representing "moves" of a puzzle (actions on the puzzle), not static states of the puzzle.

 

[qqwref]

^ That may be true according to our definition of "solved state" in cubing, but that leads to contradictions in mathematics. That is, your definition is not well-defined. You can't claim that x is not the identity [...]

Why not simply decide that "solved states" need not be identities? That is, there are 24 states of the full rotation-included cube group that count as solved.

I thought I was going to explose qqwref's calculator by inputting a scramble. After many try, I was very disappointed. But I used qqtimer for scrambles, there should be a ploy...

Haha!

 

[Erdos]

...

I'm not necessarily "confusing" the concept. I'm defining it in that sense in order to avoid having more than one unique solved state. For example, by your definition, there are 24 solved states (I think).

 

[cuBerBruce]

I'm not necessarily "confusing" the concept. I'm defining it in that sense in order to avoid having more than one unique solved state. For example, by your definition, there are 24 solved states (I think).

Yes, for the 3x3x3 anyway, I would consider there to be 24 solved states when the orientation of the puzzle is considered to matter (and we ignore the 2048 configurations for the orientation of the center pieces as part of the state).

One of the beautiful things about treating the group elements as actions rather than as states (which are generally recognized by looking at the stickers) is that I don't have pick any one of these 24 states as being the identity. The identity is simply a "null action" rather than any particular state of the stickers. Note that how an action rearranges the pieces of the cube is totally independent of how those pieces are stickered (at least when our grip on the puzzle serves to define our reference frame for the puzzle)!

So I'm not clear what exactly you're trying to claim is not well-defined. Definitely, the <U, x, y> group is a well-defined group, and the order of each of its 1038048078587756544000 elements is well-defined. qqwref's definition of order will not always coincide with the actual group theory definition for these elements, though.

 

[mrCage]

Somewhat related.

What is the shortest sequence effectively doing nothing?? No internal repetitions allowed, like U2 D2 U2 D2 for instance.
I discovered R' D' R F D2 L D L' F' D2 almost 30 yrs ago. Any better ones??

Per

 

[Kirjava]

10 is the shortest. We had a conversation about this back on twistypuzzles about 6 years ago.

 

[mrCage]

Hmm. Before i joined. I see. Did anyone come up with any kind of proof or other interesting examples??

Per

 

[Stefan]

What about this?

well, U2 D2 R2 L2 U2 D2 R2 L2 is shorter...

 

[mrCage]

Ok, let us switch to quarter turn metric then

[R',F] [U',R] [F',U] - another 12 qtm candidate ... Any proof 12 qtm is optimal when no repetitive sequences are allowed??

Per

 

[Herbert Kociemba]

Ok, let us switch to quarter turn metric then

[R',F] [U',R] [F',U] - another 12 qtm candidate ... Any proof 12 qtm is optimal when no repetitive sequences are allowed??

Per

Yes, 12q is the shortest even if repetitive sequences which cannot be simplified in a trivial way (by relations like U D = D U or U U' = Id etc.) are allowed.

U2 R U R' F' U2 L' U' L F (12q*)

was the shortest, my optimal qtm solver gave out for the identity position. Though some consider it as a bug to give out a 12q* maneuver for the identity position....

 

[reThinking the Cube]

M' E2 M2 E2 M'

 

[uberCuber]

M' E2 M2 E2 M'

That's 16qtm and 10htm. I don't see how this is at all special?