(1) doesn't occur in a double Exocet situation;
(2) exhibits all of its possible features; and
(3) completely solves the puzzle.
That sounds tough... wait a Jiffy.... aaah ... I think I've got it:
The following puzzle is also from Champagne's huge grey area database
98.7.....6.....7....7.5.....6...4.3...98..5......2...1.2..1......5..86.......9.42
It has 2 Exocets; r2c8 r2c9 r1c3 r3c4 1249 and r3c1 r3c2 r2c4 r1c7 1234
However, for the purposes of this exercise I've ignored eliminations possible via the first Exocet.
- Code: Select all
*--------------------------------------------------------------------------------*
| 9 8 1234 | 7 346 1236 |T123 A5-126 B45-6 |
| 6 5 1234 |T4-3 8 C123 | 7 129 49 |
|B1234 B134 7 | 9 5 1236 | 12348 1268 468 |
|--------------------------+--------------------------+--------------------------|
| 5 6 28 | 1 9 4 | 28 3 7 |
| 1234 134 9 | 8 367 367 | 5 26 46 |
| 3478 347 348 | 36 2 5 | 489 689 1 |
|--------------------------+--------------------------+--------------------------|
| 3478 2 3468 | 346 1 367 | 89 5789 589 |
| 147 9 5 | 2 47 8 | 6 17 3 |
| 1378 137 1368 | 5 367 9 | 18 4 2 |
*--------------------------------------------------------------------------------*
We have a normal type of secondary equivalence r1c7==r2c6, r2c4 which produces no eliminations.
We have a tertiary equivalence r2c4==r1c89(5), r1c7 which produces 5 eliminations:
r1c8 <> 12 - base digits not in r2c4
r1c89 <> 6 - non-base digits (other than 5) in r1c89
r2c4 <> 3 - base digit in r2c4 not in either of r1c89
stte
Leren