7.8.2

Summary:

This puzzle consists of 13 triangular pieces
forming 3 superposing hexagons.
Clicking one of 3 hexagon center(A,B,C)
rotates 6 pieces around it.
Color schemes:
There are 5 color-schemes, namely, A,B,C,D and E.
ABCDE
solutions 1×31×32×12×31


Solved states:

The puzzle is conisidered as solved if colors on both sides of all 12 piece borders are same.
There are 1 or 2 visibly different solved states with or without their rotations to left or to right by 120°.
The color-scheme E has only one solved states, because all pieces are orient-sensitive and the sum of twists of pieces is always a multiple of 3. So, you have to know the correct orientation to solve.


Mathematical properties:

· The permutation group of pieces is S13.
· Any piece can be at any of 13 places, and can be twisted by n×120°.
· The sum of twists is always (n×360°) =0°.
So the total number of possible scrambles is: (13!) ×(3^12) = 3,309,294,160,972,800.


Solution:
These puzzles can be solved in 2 phases.
1. Fix 7 pieces around an edge.

This can be done fairly intuitively within 20~30 moves.
After that, there remains only (6!)×(3^5)=174,960 different states.

2. Fix remaining 6 pieces around a hexagon center.
The second phase is a computer-assisted 3-steps reduction to subgroups(174,960→3,888→144→1)
2-1. (0~8)moves.
Reduce the permutation of pieces to a sugroup of order 16
(direct product of D8 and C2), so that, on the picture at right:
- "1" and "6" can be swapped.
- "2-3" and "4-5" are always parted in left-right or right-left.
At this stage, the orientation of pieces doesn't matter.
After this, the number of states is reduced to 16×(3^5)=3,888.
1 2 3 4 5 6
gap> g:= Group((1,6),(2,3),(2,4)(3,5));
Group([ (1,6), (2,3), (2,4)(3,5) ])
gap> Size(g);
16

2-2. (0~11) moves.
Fix the orientation of pieces preserving the permutation group,
so that, on the picture at right:
- "10" and "60" are always on the vertical center line.
- "20-30" and "40-50" are symmetrically placed on both sides of vertical center line.
After this, the number of states is reduced to 16×(3^2)=144.
 10 11 12 20 21 22 30 31 32 40 41 42 50 51 52 60 61 62
2-3. (0~30) moves.
This can be done by combinations of these sequences.
SequenceCycle notation
B4,C1,B1,C5,A1,C5,A5,C1,A5,B1,A1,(40,41,42)(50,52,51)
A4,C2,A2,C3,B4,A4,C3,A2,B2,C1,A4,(20,30)(21,31)(22,32)
A1,B1,C2,A2,B3,A4,C4,B3,A2,B2,A1,(40,51)(41,52)(42,50)
A1,C1,B3,A3,C1,B5,C5,A3,B3,C5,(40,31,41,32,42,30)(50,22,52,21,51,20)
A1,B1,A5,B5,A5,C5,A1,C1,(40,22,41,20,42,21)(50,31,52,30,51,32)
B1,A1,B5,A5,C5,A5,C1,A1,(40,21,42,20,41,22)(50,32,51,30,52,31)
C1,B3,A3,C1,B1,C5,A3,B3,C5,A5,(40,30,42,32,41,31)(50,20,51,21,52,22)
A5,C2,A3,C5,A3,C3,A3,C5,A3,C2,A1,(40,32,50,21)(41,30,51,22)(42,31,52,20)
A5,C4,A3,C1,A3,C3,A3,C1,A3,C4,A1,(40,21,50,32)(41,22,51,30)(42,20,52,31)
A5,C4,A3,C1,A3,C3,A3,C1,A3,C4,A4,(10,60)(11,61)(12,62)(20,31)(21,32)(22,30)
A5,C2,A3,C5,A3,C3,A3,C5,A3,C2,A4,(40,50)(41,51)(42,52)(10,60)(11,61)(12,62)
B4,C1,B1,C5,A1,C5,A5,C1,A5,B1,A4,(40,32,41,30,42,31)(50,21,52,20,51,22)(10,60)(11,61)(12,62)
A1,B1,C2,A2,B3,A4,C4,B3,A2,B2,A4,(40,32,51,22)(41,30,52,20)(42,31,50,21)(10,60)(11,61)(12,62)
A4,C2,A2,C3,B4,A4,C3,A2,B2,C1,A1,(40,22,51,32)(41,20,52,30)(42,21,50,31)(10,60)(11,61)(12,62)
C1,B3,A3,C1,B1,C5,A3,B3,C5,A2,(40,41,42)(50,52,51)(20,22,21)(30,31,32)(12,62)(10,60)(11,61)
A1,C1,B3,A3,C1,B2,C5,A3,B3,C5,(40,42,41)(50,51,52)(20,21,22)(30,32,31)(12,62)(10,60)(11,61)
A1,B1,A5,B5,A5,C5,A1,C1,A3,(40,50)(41,51)(42,52)(10,60)(11,61)(12,62)(20,30)(21,31)(22,32)
A3,(40,32)(41,30)(42,31)(50,21)(51,22)(52,20)(10,60)(11,61)(12,62)
B1,A1,B5,A5,C5,A5,C1,A4,(40,51)(41,52)(42,50)(10,60)(11,61)(12,62)(20,31)(21,32)(22,30)
gap> g:=Group(
(40,51)(41,52)(42,50),
(40,32,51,22)(41,30,52,20)(42,31,50,21)(10,60)(11,61)(12,62),
(40,51)(41,52)(42,50)(10,60)(11,61)(12,62)(20,31)(21,32)(22,30)
);
<permutation group with 3 generators>
gap> Size(g);
144
gap> IdSmallGroup(g);
[ 144, 186 ]
Unless you are extremely unlucky, you can solve this puzzle within 50 moves by this method.