# J-Perm Metric



## cuBerBruce (Feb 14, 2018)

This is about solving the Rubik's Cube using only J-perms. Both the Ja and Jb versions are allowed, but only the normal AUF case that swaps two adjacent corner and two adjacent edges are allowed. Of course, the cube can be rotated into any of the 24 orientations before applying each J-perm. So there are a total of 48 J-Perms that are allowed as a "move."

Pure Flips

There are four essentially different pure 2-flips. If the edges to be flipped are adjacent (as in UF and UR), 6 J-perms are required. For the other cases, an optimal maneuver can be made using conjugation to the adjacent case. So if the two edges are centrally opposite (as in UF and DB), 10 J-perms are required, while the remaining cases require 8 J-perms.

There are 18 symmetrically distinct 4-flip cases. These cases require from 8 to 12 J-Perms. The following cases require the minimum of 8 J-perms (edges to be flipped are listed):


```
UF UR UB UL
UF UR UB FR
UF UR UB FL
UF DF UR DR
UF FR BR DB
UF UR BR DB
```

In the above list, only a representative of each symmetry equivalence class is given. The other cases can be optimally solved by conjugation to one of the 8-J-Perm cases.

It is possible to use three of the minimal 4-flip maneuvers to flip all 12 edges. This puts an upper bound on superflip of 24 J-perms.

There are 30 symmetrically distinct 6-flip cases. These all require at least 12 moves. I haven't determined the worst case.

Pure Twists

Twisting two corners requires 6 J-perms, regardless of which two you want to twist. (Of course, one must twist clockwise, while the other must twist counterclockwise.) Twisting three corners also requires 6 J-perms if the three to be twisted are all in the same face (as in UFL URF UBR), or if all three are diagonally across a face from one another (as in UFL UBR DFR). The other cases require 8 J-perms. (Of course, to twist three corners, all must be twisted in the same direction, either all clockwise or all counterclockwise.)

Corner 3-cycles

Corner 3-cycles can take from 2 J-perms (for the A-Perm cases) up to 8 J-perms (for the Per Special cases). Most cases (that require 4 or more J-perms) have optimal maneuvers that are conjugations to an A-perm. There are a few cases where there is no optimal solution of this type.

Edge 3-cycles

As with the corner 3-cycles, edge 3-cycles varied from 2 J-Perms (U-Perm) to 8 J-Perms.

God's number bounds

It is easy to show that 24 is a lower bound for God's number. The position Pons Asinorum (F2 B2 R2 L2 U2 D2), has all 12 edges needing to be moved four times to bring them to the solved position (using only J-Perms). Each J-Perm only moves two edge pieces. Therefore, there is a minimum of 12*4/2 = 24 J-perms required to solve Pons Asinorum (or Pons Asinorum composed with any of the 44,089,920 positions that leave the edges fixed). Since Pons Asinorum requires at least 24 J-Perms to solve, 24 must be a lower bound on God's number.

I'll leave it as an exercise for the reader to find a maneuver (without using computer search!) that solves Pons Asinorum in 24 J-Perms, thus proving that the distance in the J-Perm metric for Pons Asinorum is 24.

My analysis indicates 29 is an upper bound for the J-Perm metric. I show this with a 2-phase analysis. First, build a 2x2x3 block in 14 moves, and then solve the rest of the cube using only the 20 J-Perms that leave the 2x2x3 block intact.

The first analysis only considers the pieces that make up the 2x2x3 block (5 edges and two corners). There are (24*22*20*18*16)*(24*21) = 1,532,805,120 positions. The table of positions at each distance in this breadth-first search are given below.


```
distance  positions    cumulative

    0             1             1
    1            28            29
    2           620           649
    3         10906         11555
    4        164722        176277
    5       2027409       2203686
    6      18556884      20760570
    7     110289279     131049849
    8     362745098     493794947
    9     567561836    1061356783
   10     373069930    1434426713
   11      91655722    1526082435
   12       6593895    1532676330
   13        125766    1532802096
   14          3024    1532805120
```

Below is the similar table that I got for the 2nd breadth-first search, solving the last two faces.


```
distance  positions    cumulative

    0             1             1
    1            20            21
    2           314           335
    3          4216          4551
    4         53360         57911
    5        638682        696593
    6       7108323       7804916
    7      70008254      77813170
    8     551668320     629481490
    9    2887505478    3516986968
   10    7902380058   11419367026
   11   10044559862   21463926888
   12    5598207377   27062134265
   13    1105775677   28167909942
   14      49356647   28217266589
   15        282211   28217548800
```

This gives an upper bound to solve any legal position of 14 + 15 = 29 J-perms.

It might be possible to show that 282,211 antipode cases can be solved in 13 or fewer J-Perms if using all 48 J-Perms rather than only 20. This could bring the upper bound down by one.

I hypothesize that God's number is equal to the lower bound of 24.

Edit:

I've checked over 12,000 of the 282,211 antipodal positions of the the last two faces breadth first search, but using all 48 J-perms instead of the 20 that preserve the 2x2x3 block. All of these were solved in 11 or 13 J-perms. It may take months finish all 282,211 positions, but so far it seems promising that all might be solved in 13 or fewer J-perms.

I avoided giving any maneuvers so far. Part of the reason is a notation is needed. I could expand each J-perm into a sequence of ordinary face turns, but this would be rather verbose. What I've come up with for a convention is to use a three-letter combination to describe each J-perm. The first letter indicates the face common to all 4 pieces that are affected. The 2nd letter is the other face that is common to both of the corner pieces affected. The 3rd letter is the other face for the affected edge piece that is not between the 2 corners affected. Each letter is based upon the color of the center piece of the corresponding face: W=white, Y=yellow, G=green, B=blue, O=orange, R=red. A standard color scheme cube is assumed.

So WGR would represent the J-perm that can be generated by L2 B' U' B L2 F' D F' D' F2, using the WCA scramble convention for the cube orientation where U=white, F=green. (You probably prefer a different maneuver, but I simply use the optimal FTM maneuver for convenience.) Similarly the J-perm generated by R2 B U B' R2 F D' F D F2 would be represented by WGO.

To let you try using the notation, I give a few example maneuvers. These corner 3-cycles do not have an optimal maneuver (in the J-perm metric) that is a conjugation to an A-perm:


```
1) RBY WBO RBY WBO
2) WRB GRY WRB GRY
3) WBR BOY WBR BOY
4) OYG YBR OYG YBR
5) GYR OGR GYR OGR
6) OWG BOY OWG BOY
```

Edit2:
Well, I found a "last 2 faces" position that requires 15 J-perms to solve, so it is not the case that all 282,211 antipodes of the above analysis can be solved in 13 moves or less. So lowering the upper bound from 29 to 28 will require additional work than what I thought might work.

The position that was found can be generated from this scramble:

U2 R' U2 F L2 B' R D' R2 B' D' B2 L2 F' R2 U'

and can be solved using this sequence of J-perms using my above notation:

WGR WRB WGR YBR RWG WBO RBW GRW OBW BWO BRY YBR RYB WRG OBW


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