Oh, the puzzle looks very different after all the basics...
I found a different step back here:
- Code: Select all
23 23 23
,-----------------,--------------------,----------------------,
| 89 47 4789 |#23–78 5 6 |#23489 #23489 1 | 23
| 1 56 5689 | 4 239 2389 | 7 235689 #2356 |
| 89 2 3 | 78 179 189 | 46 456 456 |
:-----------------+--------------------+----------------------:
|#23 1356 12568 |#237 12367 4 | 23689 23689 #236 |
| 4 136 1268 | 9 1236 123 | 5 2368 7 |
| 7 9 26 | 5 8 23 | 2346 1 23–46 |
:-----------------+--------------------+----------------------:
|#23 1347 12479 |#238 2349 23589 | 2346 234567 #23456 |
| 6 347 2479 | 1 2349 2359 | 234 23457 8 |
| 5 8 24 | 6 234 7 | 1 234 9 |
'-----------------'--------------------'----------------------'
23c149\r47 needs (at least) one of each in r1c4, r26c9
Either r2c9 is not used and r1c4 and r6c9 are a RP, or the digit in there is forced into r1c4 and the other digit must take r6c9.
So RP 23r1c4&r6c9.
Or: Each digit can only appear 3 times in #-marked cells (r147c149b3/2).
23r1 23c149 \ 2#[3] 3#[3] + r1c4 r6c9 => RP 23r1c4&r6c9
(we get the RP by comparing the digits of the truths: 22223333 and links: 222333 + the cells)
Marek