This post gets us into coloring and and a look at the Beeby ALS-wing with an ALS_XZ map. Also the solvers illustrate a new pattern analysis resource.
Picking up from last post, Beeby now comes in with a complex 1-way. That’s how it starts, with a bv slink partner seeing other 3’s. It works out as a 1-way ANL with a branch coming off to whack 4r4c9 and create the necessary r4 slink.
Next is a Beeby ALS_wing, a 3 node ALS chain ANL. You would probably start this chain at the bv r5c5, noticing the winks to single value groups in two ALS on the map. That would lead you to discover that the shared value 8 and the slinks with the 8-groups and the entry candidates. One group is a box group, and the other is a line group.
Were there any other ALS wings?
Of course, human solvers don’t “spot” these. They spot possible beginnings and keep chains going, looking back for branches to create slinks.
The removal erases the r1 ALS 2356 on the ALS-map, but it leaves all the AS required for the ALS- wing that Beeby finds next. This gives you an opportunity to search for that ALS-wing on that map before looking. Go back and do that while I create a little diversion by discussing the trade-off between marking and not marking the multi-candidate value groups on the map. The markings would definitely aid you in building the inference chain, but would would add considerable unproductive clutter to the map.
Then Beeby uses a branch enhanced AIC for three eliminations: a complex 1-way from 8r8c1 to confirm 4r4c1, and extending it to confirm 3r2c1, and having removed two c1 8’s by 1-way logic, invoking the boxline SWc1 to remove 8r7c2.
Parts of that structure show up again in this complex 1-way from 2r2c6 which, true or false, extinguishes 2r2c1.
We hopefully start a second cluster as a simple 1-way removes 2r4c2 whether 2r1c2 is true or false.
Then comes the very complex 1-way, the AIC starting from 3r2c1 and sprouting four slink making branches.
Its removal of 3r2c9 allows this overlapped ALS_62.
Beeby now continues with a puzzling “complex” 1-way with an unnecessary forward branch. The Beeby annotation shows an ending slink to 7r3c7, which is valid, but unnecessary. The forward branch from 7r7c4 could be used to reinforce that the chain assumes 7r3c4 is false. It might be what Beeby does when the starting slink is a line slink, rather than a bv slink.
Sudokuwiki records this one as a digit forcing chain.
Now consider the labeling over abundance. This ALS_36 is also an ALS boomer from 3r7c7, and a unbranched ALS 1-way from 6r1c7. It depends on what you were looking for when you found it.
Wouldn’t you know it? Both ALS were added to the map.
If you can wade through one more ultra complex 1-way, we can pick off another 3 candidate. It starts at 3r2c1 again, and uses much or the same structure as the previous one.
Now the pattern analysis twist. The red/orange cluster expands with the 3 clue, and Sudokuwiki claims that 2r2c3 is an orphan.
Checking it out on the 2-panel, the North to South freeforms are restricted, but there are two patterns including 2r2c3. Note that an attempted freeform through r3c9 is stymied, because it leaves no entry into the East box. Now we know that patterns including 2r2c3 must also visit r3c4, r4c6 and r6c9.
If any combination of these three cells exclude all patterns of any value other than 2, 2r2c3 is an orphan!
Sudokuwiki notes don’t explain that reasoning, and enumerating all patterns of all other values is not normally feasible, but here we notice that 7-patterns are quite restricted, and that delivers the goods.
Going North, all 7 patterns visit r3c4, with one exception, the short-dashed one in the right panel, from r8c2. That pattern visits r6c9. So 2r2c3 excludes all 7 patterns, an unforgivable offense.
Beeby is silent here, but in a coming ultrahardcore in the review, it does acknowledge this kind of rationale for removing an orphan candidate. For future reference, we’ll call this removal method the toxic orphan.
Looking at the diagram above, there is a much easier way to eliminate the “orphan”, by a combination of coloring and the Sue de Coq. NEr2 contains 89(2+7) if red is true or 69(2+7) if it is orange. In either case, the matching 2 or 7 in the r2 remainder must come from r2c6.
We continue from here next week to a solution. That means you may need a preview of 355, the next review ultrahardcore.