# Smallest Identity Maneuver

#### siva.shanmukh

##### Member
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

##### Member
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)

#### Swordsman Kirby

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

#### Stefan

##### Member
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

##### Member
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

##### Member
M E M' U' M E' M' U

#### cuBerBruce

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

16 QTM XD

R4

#### siva.shanmukh

##### Member
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

##### Member
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.

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