Mith, it was exactly my point that the ways of spotting multifish and applying SET are very similar (with a minor difference in the order things are done in) with MSLS being quite similar as well. I wrote it in response to your statement that:
Viewing this as MSLS/Multifish involves looking at a lot of pencilmarked cells
I also don't think that anyone is faulting you (or anyone else), for TH puzzles taking over the thread, Eleven is just pointing out the problem.
Denis, it seems that you are (deliberately) ignoring the gap between what you think is solvable manually and the hardest puzzles found. If TH-based relations reduce a puzzle into T&E(1), it is nowhere near the hardest.
As for the above puzzle posted by Mith, here is the state after basics:
- Code: Select all
,--------------------,--------------------,---------------------,
| 4679 468 46789 | 36789 23789 #2789 |#5789 5789 1 |
| 5679 568 2 | 6789 #789 1 | 3 4 #789 |
| 179 18 3 |#789 4 5 | 2 #789 6 |
:--------------------+--------------------+---------------------:
| 4579 458 45789 | 1 6 #4789 |#4789 23 23 |
| 14679 1468 46789 | 2 #789 3 | 46789 #789 5 |
| 2 3 46789 |#789 5 4789 | 146789 1789 #789 |
:--------------------+--------------------+---------------------:
| 35 2 1 | 3789 3789 789 | 5789 6 4 |
| 346 7 46 | 5 12389 2689 | 189 12389 2389 |
| 8 9 56 | 4 1237 267 | 157 12357 237 |
'--------------------'--------------------'---------------------'
Suppose 4r4c67.
The 4|6 in r8c3 is forced into r5c12 with 1, the other 4|6 is then forced into b4p369 => –46r19c3.
5r9c3, 5r7c7, 3r7c1, 13b8p58, 2r1c5
Now the 4 at the rectangle is the only guardian of the TH => 789 must each appear at R=r14c67.
789: c3R \ r14b4 => –789r4c12
Both cells are forced to be 5s, i.e. contra.
After that the puzzle solves easily with the remaining guardians.
I would be surprised if a puzzle that gets reduced to
B3B S2B4B? (or lower) by TH-relations could be considered among the hardest, although I don't know how many puzzles are above that level and how many of them can be simplified.
Of course, sometimes it's difficult to draw the line at what relations should be used – take Kolk for example. Obviously we use the almost sk-loop and the almost PLQ, but my solution from last year (which I should really get ready to post, but I'm so unhappy with my path from 10.5 skfr onward), extensively uses that r7c89 cannot be 79. Would we want a computer solver to use it when estimating the puzzle's difficulty or is that too far?
Marek
Added: With 2r1c6=5r1c7 we get the following AIC:
5r7c1 = 5r7c7 – 5r1c7 = 2r1c6 – 2r1c5 = (12–3)r89c5 = 3b8p12 => -3r7c1
After basics, we see a 7|8|9 in r1c3 and b4p12, eliminated from the rectangle, so 2 and 5 are both needed (as 789 cannot repeat there).
I also checked with YZF_Sudoku and it used a more powerful step instead of the AIC: the 7|8|9 in r4c7 is forced into r7c456 and r789c6, so into r7c6. If there also were a 7|8|9 in r1c7, it would be forced into r7c6 in the same way, i.e. contra., hence -789r1c7
If instead of 4r4c67 we opt for eliminating 46r1456c3, we can do it in B2B with TH relations (including the OR relations between 25x at the TH rectangle for each x in 789).
