- Code: Select all
*--------------------------------------------------------------*
| 124579 M12489-6 124569 | 1267 1289 1269 | 45789 45789 3 |
| 4579 M4689-6 4569 | 367 389 369 | 45789 1 2 |
| 1279 T129-8 3 | 127 4 5 | 789 789 6 |
|--------------------------+----------------+------------------|
| 12459 3 T1249-5 | 125 6 7 | 1459 2459 8 |
| 6 1249 M1279-45 | 8 125 124 | 3 24579 1479 |
| 8 124 M127-45 | 9 1235 1234 | 14567 24567 147 |
|--------------------------+----------------+------------------|
|B129 5 12689 | 4 129 1269 | 16789 3 179 |
| 149 7 14689 | 1356 1359 1369 | 2 4689 149 |
| 3 12469 12469 | 126 7 8 | 1469 469 5 |
*--------------------------------------------------------------*
Exocet 1: r7c1 r8c1 r3c2 r4c3 1249 => - 8 r3c2, - 5 r4c3, - 6 r12c2, - 45 r56c3, lclste
That was my Exocet solution for this puzzle. There are two issues of controversy here.
1. Despite 4 failing the S cell cover set test it must be possible to prove that if 4 is in a base cell it must be in at least one target cell. So it is not a J Exocet, but is an Exocet in the more general sense.
I call this a full digit expansion of the relevant digit, and there is some controversy as to how far you should go in proving this.
2. The puzzle was probably included in a list of Double Exocet puzzles before the full implications of Mirror Node eliminations were known.
As I have shown, the puzzle does not actually need the second Exocet to be solved, but if you prefer to look for Double Exocets before you consider Mirror Node eliminations, then that is fine by me.
Leren