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