Smallest Identity Maneuver

[siva.shanmukh]

I don't think this was ever discussed, but have you ever thought of what the smallest non-reducable maneuver that gives an Identity is!

My defn of non-reducable maneuver is that there are no X X' / X X2 combinations in the maneuver recursively.

I recently came across this:
F R U2 B U B' R' D2 F2 D2 R' L F2 B U2 F2 B2 D2 B L' R U2 B2 F' U' (I think Feliks posted it somewhere and I found this in some forum)

This is a 25 mover. And we know of (R U R' U') x 6 which is a 24 mover. I am wondering if there are shorter ones!

 

[Swordsman Kirby]

On Jaap's page, http://www.jaapsch.net/puzzles/cayley.htm, he says the following:

The following three move sequences all have the same effect:
a. FU'R'FRF'
b. U'FRU'R'U
c. UL'U'LFU'
This gives rise to three identities of 12 quarter turns, ab', bc', and ca'. There aren't any shorter identities on the cube (except for trivial ones)

 

[Kirjava]

Relevant;

http://www.twistypuzzles.com/forum/viewtopic.php?f=8&t=5073

 

[Swordsman Kirby]

Eh, yeah, I should've just replied with (R2U2)6.

 

[Stefan]

My defn of non-reducable maneuver is that there are no X X' / X X2 combinations in the maneuver recursively.

L R L' R'

 

[Renslay]

If Stefan's solution is considered as "invalid" (nice try by the way ), here is an other one with 8 moves:
(U2 R2 L2 D2) * 2

 

[VP7]

M E M' U' M E' M' U

 

[cuBerBruce]

If you're allowed to use slice turns (and count half-turns as one move), 4 is the shortest: E2 M2 E2 M2

 

[Kirjava]

16 QTM XD

 

[applemobile]

R4

 

[siva.shanmukh]

L R L' R'

Realized I didn't cover everything I intended to.

Reducing the Maneuver :
-Always write R before L, U before D and F before B when consecutive recursively till no change in the maneuver
-Replace all X X' / X X2 combinations to "nothing" / X' respectively
-Replace all X4 kind to "nothing"

and if the maneuver is finally reduced to "nothing" it is not valid.

I would like to ponder over such maneuvers.
Lets replace slices with corresponding face turns and let us look at both QTM and HTM counts.

 

[cuBerBruce]

If you eliminate all identities that are generally regarded as trivial, the answer for QTM was answered in the first reply of this thread. The shortest FTM (aka HTM) maneuver is HTM can be found from Cube Explorer. (Running its optimal solver on the solved cube results in showing the shortest non-trivial identity instead of the highly trivial 0-length maneuver.)

QTM: 12, example: F U' R' F R F' U' R U R' F' U
FTM: 8, example: U2 R2 L2 U2 D2 R2 L2 D2

BTW, there is a thread on QTM non-trivial identities at http://cubezzz.duckdns.org/drupal/?q=node/view/114, including the complete list of 1440 12q non-trivial identities.

 

[siva.shanmukh]

BTW, there is a thread on QTM non-trivial identities at http://cubezzz.dyndns.org/drupal/?q=node/view/114, including the complete list of 1440 12q non-trivial identities.

Thanks for the link. Exactly what I was looking for.