This post and the next continue on uhc 91, after last week’s SASdC trial eliminated red candidates of a red/orange cluster and left two clues. Phil Beeby’s solver chisels on the crystalized structure, conducting an ALS_XZ clinic, until the slink network started in the 7’s can get a grip.
Here’s the post trial grid with, in black, a 5-chain ANL with NWr1 boxline removing r1c9.
Then, in red, a 1-way from 3r2c3 or a boomer from 5 r8c3, and a SWr9 boxline.
“1-way” will be my shorthand term for Beeby’s AIC building strategy in which an AIC starting on a slink shows a candidate seen by the starting candidate to be false.
The victim is false if the starting candidate is true, and the AIC proves it false if the starting candidate is false. We’ll call it a complex 1-way when the unidirectional AIC uses forcing branches to clear its way. That’s not the case here.
Beeby’s next is this ALS_72. In an ALS_XZ, X is the group slinked value in the two ALS, and Z is the common value locked in one of them. One candidate in the two shared Z value sets is true.
Spotting ALS_XZ depends on recognizing potential X and Z value sets in two repeated values. Here, you might spot the similar values in E and r7, the single 2 and 7 in E when 6s are excluded from red, and that 7 and 2 can be singles when r7c7 is excluded. Then you’d do the calculations on numbers of cells vs. values for the ALS. You could start with r5c8 in red, and the two cells r7c38 in green, adding cells as necessary to fill out the two ALS. More often, these systematic spotting tasks are much easier.
Next is a hidden UR off my Tools chart. Sure enough, if 5r9c3 is true, 5r8c3 is not, so in r8c9, 5 is and 4 is not, so 4r9c9 is.
A flippable rectangle and two solutions.
Interpreting solver results, the term “hidden” calls for finding the slinks between UR candidates. Candidate cell positions tell where they are not. I have to fix the “Tools”page chart. Is this one “typed”? Somebody fill me in.
Does the spotting method above fit Beeby’s next, ALS_18? Kinda? It’s hard to put into words.
Taking a break from ALS, a grouped 1-ANL on the left.
Then a 1-way from 8r7c8, or a boomer from fr9c2.
Okay! We get a clarifying example of a Beeby complex 1-way!
Starting at 4r7c2, the red branch from 2r4c1 – which is false if the starting 4 is false – destroys an interfering 2r5c9 to insert the green slink, permitting the 1-way AIC to confirm 5r8c3. So if starting 4 is false, the victim is false. Of course if the starting 4 is true, . .
The beat goes on with a couple of hidden unique rectangles.
The right one is standard slinked opposite the extra free corner, but the left one is another Beeby special: no free corner, but a slink of the opposite value across from the slinked corner. To spot these, you might see the UR slinks first, then see the UR values on that cell rectangle.
It works. Trace it out: if 9r6c4 is true, you get the 5’s on adjacent corners, and the side 9-slink is necessary.
Beeby accepts this DIY ALS_51 that violates the ALS_XZ rule that the victim value Z be present in both ALS. The reason for the rule is, if X is in the ALS containing Z , that doesn’t necessarily lock the only Z value set. In this case, 5 in the green ALS removes one value and one cell. That leaves 6 cells containing 5 values. The true 1 may be 1r6c4! When X is not a single candidate, locking depends on where Z candidates are.
Here though, Beeby’s code anticipated something else entirely. Lacking the 5, the 2 and 6 sets in the blue ALS are locked, removing 6r6c3 and therefore, 1r6c4 as well.
On an faulty ALS_XZ with one Z value set missing, check what happens when the ALS without Z loses X and is locked. It can be another heads I win; tails, you lose.
It’s not enough for Beeby. The solver doubles down on the above by excluding cell r6c5 and one value set 8. Same violation and same saving grace.
Some heavy lifting here? We continue with the Beeby ALS_XZ clinic next week.