The puzzle can be solved without using the central automorphism (with credit to YZF_Solver):
- Code: Select all
+------------------------+------------------------------+------------------------+
| b234 8 6 |ec1237*^ 12347 347 | 9 5 f17 |
| 7 349 349 | 8 134 5 | 16 136 2 |
| 1 a2-3 5 | 9 237 6 | g78 g378 4 |
+------------------------+------------------------------+------------------------+
| c245^ 6 8 | d127*^ 12457 47 | 3 9 e157* |
| 23459 2349 2349 | 12367 123456789 34789 | 1678 1678 15678 |
| E359~ 1 7 | 36 35689 D389#~ | 2 4 C568# |
+------------------------+------------------------------+------------------------+
| 6 G237 G23 | 4 378 1 | 5 A8-7 9 |
| 8 479 49 | 5 679 2 | 167 167 3 |
| F39 5 1 | 367 36789 EC3789#~ | 4 2 B678 |
+------------------------+------------------------------+------------------------+
Look at:
- 1r4, 1c4, 1b5 : Fireworks pattern (*) => (1)r1c4,r4c9 = (1)r4c4
- 2r4, 2c1, 2b5 : Fireworks pattern (^) => (2)r1c4,r4c1 = (2)r4c4
=> (2)r3c2 = r1c1 - (r1c4,r4c1) == (2-1)r4c4 == (r1c4,r4c9) - (1=7)r1c9 - (7=83)r3c78 => -3 r3c2; 2 placements
Similarly, look at:
- 8r6, 8c6, 8b5 : Fireworks pattern (#) => (8)r6c9,r9c6 = (8)r6c6
- 9r6, 9c1, 9b5 : Fireworks pattern (~) => (9)r6c1,r9c6 = (9)r6c6
=> (8)r7c8 = r9c9 - (r6c9,r9c6) == (8-9)r6c6 == (r6c1,r9c6) - (9=3)r9c1 - (3=27)r7c23 => -7 r7c8; 2 placements
End as in my post above, with DP(167)r258c8 => +3 r2c8; ste
or with DP(349)r258c8 => +7 r8c2; ste
Of course, the second set of (8,9) FW patterns is the symmetric of the first set of (1,2) FW. No need to use it, if the automorphism has been claimed.