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Subgroups using basic moves
Submitted by mdlazreg on Wed, 03/20/2013 - 20:20.
In QTM, the whole cube is generated using U,D,R,L,F,B moves.
If we drop some moves we end up with some subgroups. The subgroups are: 1) I [the identity] 2) U 3) U,D 4) U,F 5) U,D,F 6) U,F,R 7) U,D,F,B 8) U,D,F,R 9) U,D,F,B,R I know the depth table for subgroups 1) 2) 3) and 4): The subgroup 1) generated by "no move", has the following obvious table:
Moves Deep arrangements (q only)
0 1
------
1
The subgroup 2) generated only by the move U, has the following table:
Moves Deep arrangements (q only)
0 1
1 2
2 1
------
4
The subgroup 3) generated by U,D has the following table:
Moves Deep arrangements (q only)
0 1
1 4
2 6
3 4
4 1
------
16
The subgroup 4) generated by U,F has the following table:
Moves Deep arrangements (q only)
0 1
1 4
2 10
3 24
4 58
5 140
6 338
7 816
8 1,970
9 4,756
10 11,448
11 27,448
12 65,260
13 154,192
14 360,692
15 827,540
16 1,851,345
17 3,968,840
18 7,891,990
19 13,659,821
20 18,471,682
21 16,586,822
22 8,039,455
23 1,511,110
24 47,351
25 87
----------
73,483,200
My question is, do we have similar numbers for the remaining subgroups? even partial numbers, or total arrangements for each group? |
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