![]() If 2 ALS have a common restricted digit that means you only have 1 more digits than cells to work with. I think a more generalised proof of rule 1 would go something like this: Rule 1: and must occur in it, therefore 9 can be removed from B6 So following the consequences through shows the 8 in A5 must go. B5 would become 1 and since 6 and 8 are removed form J5 as well we are left with a 1 also in J5 - two 1s in the column. A3 would become 4, forcing J3 to be 6 and that removed 6 from J5. If 8 were the solution, we'd quickly get a contradiction in at least one of the two sets. Making an interesting observation is one thing, but what's the proof? Think of the 8 in A5 in the example above. Now, it so happens that any other 8 on the grid that can 'see' all the 8s in both ALSs can be removed. The Z part of the rule involves any other candidate found in both ALSs but not a restricted common that, a candidate that still appears in both ALSs and is not exclusive to one or the other. Let's call any restricted common candidate X. In the two sets in the example above, 6 is a restricted common because 6 in one set will remove it in the other. We call this candidate a restricted common. If there is a common candidate found in both sets and this common candidate is among those cells that can 'see' each other, this candidate can exist only in one set or the other. We also need a mixture of candidates in both sets. Their sizes don't matter, but they ought to be able to 'see' each other that is, have some cells that share a unit (row J in this example). To make use of Almost Locked Sets, we're going to need two or more of them.
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