This post does the basics on the second Stefan Heine ultrahardcore right page under review.  Then it shows eight ways to do the five advanced moves up to forcing chains, two of them duplicated by boomerangs. It closes by  identifying removals for three Single Alternate Sue de Coq trials, holding the trials in reserve.

Sysudoku Basic begins with 4-fill which resolves to a 3-fill.

Advanced begins with a busy grid.

The 4-chain ANL, has interesting consequences. The C4 and S7 clues trigger a E7t aligned triple boxline,  and a SEc7 boxline removing three more candidates.

The 4 removal also allows a naked quad in c6 to accompany the naked triple in the S box allowed by S7 removals.

Next, a grouped 6 -ANL removing 6r4c1 is extended into a second ANL removing 6r7c1.

Then a section starting at 5r1c9 is re-used for one of Phillip Beeby’s one-way chains. If the starter candidate 5r1c1 is true, 5r4c9 is false, of course. But if the starter is false, the chain forces 8r4c9, making 5r4c9 false, regardless.

The one-way is also a grouped boomerang, starting at 8r4c1. One-ways belong in the Sysudoku AIC building phase.

There are at least two more ways to remove 5r4c9.  Both Beeby and Sudokuwiki found the grouped boomer at left, starting on 8r4c9.

By applying ALS first, both see the ALS_65. The restricted common between ALS 56 and 235679 is value 6, and 5r7c9 sees all 5’s in both ALS. Since one ALS is missing 6 in the solution, a 5 in one of the value groups is true.

Another interesting cover is that of the ALS Death Blossom, at left. The victim  5r8c9  would lock 67 in ALS 679 and 9 in ALS 569, stripping the stem r7c5.

Sudowiki’s digit forcing chain is an ALS ANL. One terminal is the ALS 5 value group.

After this, Sudokuwiki manages three cell forcing chains and four unit forcing chains before it gives up.

Instead of going there, let’s look at a Sue de Coq trial variation proposed in  Sysudoku, the Single Alternative Sue de Coq, or SASdC. Here are three examples.

In Sue de Coq, the contents of a box/line intersection is dictated by two ALS in the two bent remainders. For a Sue de Coq example, suppose r7c5 became S2.

Then the bv 57 in the r9 remainder would prevent both 5 and 7 in SEr9.  ALS 589 in the SE remainder would prevent both 8 and 9 in SEr9. SEr9 would have to contain 6 and two alternatives, 5 or 7, and 8 or 9. Neither could be missing, because each ALS can supply only one value.

The contents of Sue de Coq SEr9 can be described as:  SEr9 = 6(5+7)(8+9).

It removes 57 from r9c2 without trial, because the 5 or 7 matching that alternative must come from r9c3. Similarly the 8 or 9 must come from the ALS 589, so the other SE 9’s are removed.

But S2 didn’t happen. Instead, in the grid above, SEr9 sees only one alternative ALS, the r9c3 bv 57. The logical description of SEr9 reads:  SEr9 = 6(5+7)(8+9) + 6(89+98).

But that is enough for an often effective trial. The bv does narrow the possibilities. SEr9 cannot contain 5 and 7, so either it contains 5 or 7  (the (5+7) above), or both 5 and 7 are missing, hence the second term. In a trial of SEr9 = 6(89+98)  a contradiction removes that second term, and a Sue de Coq is in effect, for the single alternative and its remainder. If no contradiction, the puzzle may be solved. The SASdC assumption is very specific.

Looking back to the full grid above, there are three SASdC’s available, and we can calculate the removals by contradiction for all three. But just keep them in mind. We don’t need to perform a trial unless it turns out that we can use one of those removals.  A red trial  failure removes 56r3c1, a blue one removes 5r7c1 and a green one,  57r9c2.  Work these out and you’re in command of SASdC.

Next time, ultrahardcore 47 is solved with a pattern trial. Want to do it again? A reader solution to ultrahardcore 47 without trials? Have your trial free solution path in for the post after next. Identify any solver used. I hope it’s short enough to publish here for you.