9.2.1


Summary:

This puzzle consists of 56 equilateral triangles located at each vertex of the Klein quartic. In the solved state, there appears 8 sets of 3 identic color-heptagons around 24 faces of the Klein quartic. There are no two pieces having same 3 colors. 00 10 20 50 71 31 11 01 41 72 52 61 40 21 12 30 62 51 22 42 02 60 32 70 52 20 10 01 31 71 71 41 01 10 61 52 32 12 21 42 51 62 62 02 42 21 70 32 30 00 20 22 50 31 60 11 41 40 72 61 50 40 12 11 30 51 72 22 02 00 60 70 70 10 00 12 71 50 51 01 11 02 52 72 61 21 40 20 62 30 31 42 22 41 32 60 41 20 52 61 31 01 20 41 71 31 61 10 02 12 32 70 51 42 12 02 62 51 70 21 40 00 30 60 50 22 22 11 60 30 72 40 00 40 50 72 30 11 11 22 72 50 60 00 42 10 70 62 71 12 21 01 51 32 52 02 01 21 61 71 62 20 10 42 31 52 32 41

00 10 20 50 71 31 11 01 41 72 52 61 40 21 12 30 62 51 22 42 02 60 32 70 52 20 10 01 31 71 71 41 01 10 61 52 32 12 21 42 51 62 62 02 42 21 70 32 30 00 20 22 50 31 60 11 41 40 72 61 50 40 12 11 30 51 72 22 02 00 60 70 70 10 00 12 71 50 51 01 11 02 52 72 61 21 40 20 62 30 31 42 22 41 32 60 41 20 52 61 31 01 20 41 71 31 61 10 02 12 32 70 51 42 12 02 62 51 70 21 40 00 30 60 50 22 22 11 60 30 72 40 00 40 50 72 30 11 11 22 72 50 60 00 42 10 70 62 71 12 21 01 51 32 52 02 01 21 61 71 62 20 10 42 31 52 32 41
There are 21 operations, each rotates 8 adjacent pieces:
→ [A1] →

Notations:
CO'L' M'DN A'KN' G'BL D'GJ K'C'F B'MF' OAJ' I'OL I'D'N' IK'N IB'L' R'G'J' R'CF' RM'F RA'J E'LC' H'N'M P'NA Q'L'G EJ'D QF'K PFB HJO' H'C'O E'MD' Q'AK' P'GB' QDG' EKC HBM' PO'A' T'IO' S'ID TI'K SI'B TRG S'RC' T'R'M SR'A UEL' U'HN UPN' U'QL U'E'J UQ'F U'P'F' UH'J' SHC TEM' S'QA' T'PG' SQ'D' T'E'K' S'H'B' TP'O
O'CL' NDM' R'T'M QU'L D'Q'S TOP' UN'P C'RS' E'C'L E'D'M ETM' EUL' A'QS' A'O'P ANP' AR'S I'LO H'MN' FM'R GL'Q' IS'D G'PT' F'P'U' HSC H'OC' I'N'D' GRT FQ'U G'DQ IT'O' HU'N F'CR' KEC J'ED K'E'T' JE'U' K'AQ' J'AO KA'N' JA'R B'IL' BHM' B'F'M BG'L BI'S B'GP' BFP B'H'S' JHO' K'IN J'G'R' KF'Q JGD' KI'T J'H'U K'FC'

Mathematical properties:

· The permutation group of pieces is A56.
· Any piece can be at any of 56 places, and can be spinned by 120°.
· The sum of spins is always N×360°
So, the total number of possible scrambles is (56!)/2 ×(3^56)/3 =
62,016,571,335,578,542,789,402,226,039,926,947,857,904,658,299,206,
966,778,970,892,683,058,557,007,507,160,186,224,640,000,000,000,000 .

Solution: