After two interestingly matching sequences in this post, Sudokuwiki and Beeby toss in their towels. This allows an exploration of an offbeat version of the exclusively Systematic Sudoku Single Alternative Sue de Coq. The trial formulation is different, and the trial is the first in the review to capture the solution instead of a contradicion based removal.

The later right pages of ultrahardcore review have become tougher on the bypass. And line marking doesn’t usually start with 5f: lines. Fortunately, a hidden pair clears a little of the congestion

Advanced methods start with a 5-chain ANL, by both solvers, and a Sudokuwiki 597-wing.

Beeby does XYZ-wings as ALS_XZ. Here’s the ALS_57.

Beeby’s next step is an ALS full Nelson, its ALS Wing It’s not the imbedded AIC that’s hard to spot, its the 3 or ore ALS with single or aligned value groups.

Sudokuwiki does it as a Death Blossom. The victims grab 9’s from both ALS which in turn grab both values from the stem r2c6.

Beeby flashes its ALS advantage by going one ALS value group better in the same ALS for clue C3

The assault on the 9’s continues with an AIC boomer from 2r7c5 and a unque rectangle, type 4. In Beeby notes the boomer is a simple discontinuous loop, implying a unforced 1-way from 9r6c5

The next Beeby ALS wing is a ANL with ALS value group of four candidates. How do you define the problem to have the code look for such a thing?

The two solvers push their bailout buttons after one more jab at ultrahardcore 267. Beeby ends with an ordinary 1-way in which finds 4r6 is false when it sees a true 4r6c4, when a false 4r6c4 starts an AIC making it false.

The Sudokuwiki contribution is the 5-pattern orphan 5r1c1.

Here is the 5 panel showing many West to East freeforms from 5r3c1 and the failure of any from 5r1c1. Reason through the why not and you’re onto patterns by freeform.

The solvers having had their say and not much to color, I’m on the hunt for a trial. And theres a possible Single Alternate Sue de Coq at Wr5 as marked above. The contents formula is

Wr5 = 8(2+4)(7+9) + [7(89+98) +897 ]

That second term is a bit awkward to test. The progress already made on r5, and tracing technology, gives us a way out. Either Wr5 = 8(2+4)(7+9) with 2 or 4 present, or if 2 and 4 are both absent, then in r5, Wr5 is a naked triple, r5c6 is 5, r5c9 is 4 and r5c4 is 2. A trial failure doesn’t gain much, only a 4r5c9 removal, but given the investment so far, we give it a shot. And wouldn’t you know it, this time, we reach no contradiction, only the solution.

Next week, we start ultrahardcore 311.