Here is the basics launch on Stefan Heine’s ultrahardcore 91, which quickly runs my two assisting solvers out of humanly practical removals. A Single Alternate Sue de Coq trial reduces two clusters two one, and produces two critical clues. In the next posts, solvers Beeby and Sudokuwiki can then carry the trial results along their distinctive solution paths. The challenging “exercise for the reader”, a logical solution path without trial, remains in effect for 47 and 91.
By the way, Sysudoku will refer to Phillip Beeby’s solver at philsfolly.net.au as “Beeby” because Phil’s work, and the intent guiding it, is far from folly. Phil joins Andrew Stuart in providing an accessible solver based on human technique. It may go beyond practical for most of us, but it’s very usable without going there.
Two clues and a cluttered grid greet us in Stefan Heine’s ultrahardcore 91.
No room for intuition here.
So imagine finding this ALS_59. It’s humanly feasible, but is the search through so many ALS a humanly practical thing to?
While Beeby finds this one,
Andrew Stuart’s Sudokuwiki finds the equivalent grouped Almost Nice Loop on the right, and the grouped ALS boomer below, before giving up,
Beeby finds the same removal by its version of the digit forcing chain, starting from 5r2c9 ! Assuming this candidate false, it’s the same southern forcing chain to 5r9c9 as above. Beeby’s northern chain, assuming 5r2c9 is true, is different, but the important point is the absence of any rationale for starting there. To find this, you must be searching for two chains from every candidate on the grid. Humanly possible, but totally impractical. But that’s OK, you don’t have to go where the code takes you, like your computer does.
With both solvers, we have definitely reached trial territory.
First though, on my 7-panel, there’s a finned swordfish in columns 238. Victims in r8 see the fin, 7r7c8. If they were true, the fin would be removed and they would be false. Potential victim 7r5c9 confirms the fin. That means, if this candidate is true, the swordfish that threatens it does not exist.
The removals in r7 are from boxline SEr7.
Now my problem is, the solvers are directed by clues, not candidates. I don’t edit their candidates. Fortunately there’s two small and disconnected slink networks, i.e. coloring clusters, and a potential Single Alternate Sue de Coq for a trials plan.
The blue/green and red/orange clusters interact in the East box. Only one of the four colors can be true, but red and blue together is impossible.
not(red and blue) => orange or green
Just keep that in mind, because right now,
there are no uncolored 7’s to removed for seeing both orange and green. The wall Cr5 sees one alternate 26 and is logically, 4(2+6)(1+9) if 2 or 6 is present and a naked triple 149 if 2 and 6 are absent. We get to start the SASdC trial with a triple implying red (r4) and green (c8). Are you coloring yet?
The trial contradiction is 7 inference steps down, so we can’t claim we spotted it without a trial. The red failure brings orange , but doesn’t touch green.
Orange leaves a naked pair 19 in r4, removing 9r4c5 and 1r4c6.
After getting a restart from the trial clues, the two solvers go in distinctly different directions, both with instructive examples, but both winding up in the jaws of the expanded blue/green cluster. Next time, we’ll lean first on Beeby, without the ALS-wings, for a clinic on ALS_XZ and AIC.