# Is there a way to evenly distribute face turns for 12 flip?

Back in Jan 1995 Mike Reid found this process for the 12 flip:

R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3 24q

This process has 24 q turns, so I'm wondering could there be a 24 q turn process that evenly distributes the turns so that each side turns 4 q? The idea just seemed elegant to me, 6 faces each turning 4 q turns.

Mark

## Comment viewing options

### Superflip maneuvers

Bruce Norskog stated here in an earlier thread that there are only four sets of 24q superflip maneuvers. All members of a set may be interconverted by inversion, symmetry conjugation and cyclic shifts. These conversions would not change an even distribution of turns of the six faces. A quick look at the three 24q superflip maneuvers I found for the supergroup coset below do not fit the bill. Perhaps Bruce Norskog could provide a definitive answer to your query.

I compared Reid's maneuver above to my maneuvers. Reid's maneuver may be derived from the second maneuver of my three:

```Reid:

R' U2 B L' F U' B D F U D' L D2 F' R B' D F' U' B' U D'

MacKenzie:

F2 R F' B U F D B' U R' D B2 L' F' B D' B' U' F D' L U'

inverse:

U L' D F' U B D B' F L B2 D' R U' B D' F' U' B' F R' F2

Conjugate:

(U L' D F' U B D B' F)' * (U L' D F' U B D B' F L B2 D' R U' B D' F' U' B' F R' F2) * (U L' D F' U B D B' F) =

L B2 D' R U' B D' F' U' B' F R' F2 U L' D F' U B D B' F

Symmetry Conjugate: S4x

R' U2 B L' F U' B D F U D' L D2 F' R B' D F' U' B' U D'
```

### I found a 24q superflip maneu

I found a 24q superflip maneuver (also attributed to Reid) on Randelshofer's site which does not appear to map onto any of my three:

U R2 F' R D' L B' (R U')2 D F' U F' · U' D' · B L' · F' B' · D' L'

This would be an instance of Norskog's fourth type so I would say that the answer to your query is no. There are no 24q superflip maneuvers composed of four q-turns of all six faces.

### I agree

Reid has claimed long claimed that all minimal maneuvers (up to cyclic shifting, inversion, and conjugation by symmetries of the cube) are known (FTM & QTM). He makes that claim here. As none of the optimal QTM maneuvers have turns of the 6 faces distributed evenly, I agree with B MacKenzie's answer.

### Sym Distinct Superflip Maneuvers

Looking at Reid's maneuvers on the page you referenced, it looks like the contention that there are only four symmetry distinct maneuvers may not be correct. In particular, he list only one 24q/23f maneuver. After performing the cyclic shift of my 24q/23f maneuver to line up the one double turn, there is no 1 to 1 mapping of turns to convert his maneuver to mine or to the inverse of mine. So mine is fundamentally different from his.

```U R2 F' R D' L B' R U' R U' D  F' U F' U' D' B L' F' B' D' L' Reid
R F2 D' B U  R L  B L' U D  R  D' R D  U' F' D F' L  B' U  F' MacKenzie
D F2 R' F U' B L' F D' F U  D' R' D R' D' U' L B' L' R' U' B' MacKenzie inverse
```
I am still working on the ftm superflip coset. I'm 85% through depth 23f. That should finish up in the next week or so. Once my spare computer is free I'll have to look into this.

Whoops, never mind. (After a night's sleep) commuting two pairs of parallel turns allows Reid's maneuver to be converted to mine by conjugating with a two-fold rotation.

```U R2 F' R D' L B' R U' R U' D  F' U F' U' D' B L' F' B' D' L' Reid
D F2 R' F U' B L' F D' F D' U  R' D R' D' U' L B' R' L' U' B' MacKenzie inverse

1   R  R' U  U' F  F' L  L' D  D' B  B'   x, y, z  E
21  F  F' D  D' R  R' B  B' U  U' L  L'   z,-y, x  C2xz

```

### Sym Distinct Superflip Maneuvers

Having failed when comparing these two sequences manually, I knocked together some code this morning to sym reduce superflip turn sequences. I then went ahead and iterated through the depth 24q search tree and sym reduced all the superflip solutions. I found exactly four symmetry distinct solutions in keeping with Reid's result:

```Symmetry Reduced Optimal Superflip Maneuvers

metric: ftm
min depth: 20

1920 B2 D2 F2 R  L' F  L2 U  D  L  D2 L  F2 R D2 L  F' B' L2 U
1920 B2 D2 B' R2 U' D' R' U2 R' B2 L' U2 R' F B  R2 D' R  L' U2
time 24 min 44.07 sec