Great solves, everyone!
It was interesting to see the differences in the solution paths, as well as the similarities.
Unless I'm mistaken, every step in every solution path depends solely on the digits 1234 (although some of them use other digits in ALSs, their equivalent AHSs only contain 1234).
My intended path was to get the pair 2r8c3 and 3r8c7, which totuan called out immeadiately.
It was interesting to see solutions that got very close to it, even if no one actually found the same.
After the eliminations in the targets of the JE (–49r2c6, –4r8c5) and basics:
- Code: Select all
.--------------------.--------------------.----------------------.
| 2 1359 5679 | 78 1389 1379 | 4 156 15689 |
| 1369 8 4 | 123 5 123 | 269 7 169 |
| 1579 159 579 | 6 12489 12479 | 2589 125 3 |
:--------------------+--------------------+----------------------:
| 35689 23459 1 | 23 7 23469 | 23568 23456 568 |
| 368 7 268 | 5 12346 12346 | 2368 9 168 |
| 3569 23459 2569 | 123 123469 8 | 23567 123456 1567 |
:--------------------+--------------------+----------------------:
| 4 159 5789 | 78 1368 1367 | 35679 356 2 |
| 179 6 279 | 4 123 5 | 379 8 79 |
| 578 25 3 | 9 268 267 | 1 56 4 |
'--------------------'--------------------'----------------------'
There is now a number of ways to get the desired pair, I will cover two of them.
As shye and jovi both pointed out, r2c46 and r8c5 form a triple.
We can combine it with a part of the SK loop and ask where the other 2 and 3 go in these rows.
2r8c3 and 3r2c1 are linked via c2: 2r8c3 – r9c2 = (24–3)r46c2 = r1c2 – 3r2c1,
3r2c1 and 2r2c7 are linked via r2c46
and 2r2c7 and 3r8c7 are linked via c8 (symmetrically to the first option)
Thus the only remaining option is 2r8c3 and 3r8c7.
There is also a POV that doesn't require the exocet anymore, but instead uses the possible URs in r46c28 (as many people did):
Let's first eliminate 1r2c9 (1c4b6\r26c9).
Now we can discuss where 123 go in r28.
There can be three of them in the middle stack and a 1 in c1.
What we're left with are the same candidates as in the other POV, also having to contain two true candidates.
We can link three pairs of them with each other, as we did before.
Now, remember that we haven't proven that they're different digits, so we have to check the options of two 2s and two 3s.
Luckily for us they would create URs in r46c28, also leaving us with the one true option.
It's also worth mentioning that these URs would break the exocet – we could have 1 as a base digit, but the other one would have nowhere to go in r6, every remaining option would collide either with the base or the cover-houses.
Finally, the elimination that inspired it. Have a look at the puzzle Kolk by eleven (11.9 SER) and see if you can eliminate 7r8c9. And yes, it is absolutely useless for the solve.
12.3.....4.....3....3.5......42..5......8...9.6...5.7...15..2......9..6......7..8
Marek
about 4 steps:
