The New Sudoku Players' Forum

[hobiwan]

Steve K: Thanks for the link.

Besides if eliminations were present in r4 or b4 (the AC-E and DE-B variants) you would have two steps instead of one.

I don't believe that is correct. The sets are truly locked ... whether the viewpoint is two sets or three sets.

Let me try to get that straight:

We are linking an ALS to an AALS so we need two RCs. Since we have three RCs all candidates outside the ALS but seeing all RCs can be eliminated (X=Z). That means:

A=r4c13789 - {123567}, B=r5c79 - {3579}, X=357, Z=357 => r6c8<>357, r6c9<>5

The Sue de Coq could theoretically eliminated 126 from r4c2456 (if any of them were present). But since non of r4c2456 can see both ALS/AALS in the ALS-XZ, that elimination is not possible. You would have to write it as:

A=r4c13 - {126}, B=r4c789,r5c79 - {1235679}, X=126, Z=126 => r4...

That's what I meant with "2 moves". Am I missing something?

[ronk]

Let me try to get that straight:

We are linking an ALS to an AALS so we need two RCs. Since we have three RCs all candidates outside the ALS but seeing all RCs can be eliminated (X=Z). That means:

A=r4c13789 - {123567}, B=r5c79 - {3579}, X=357, Z=357 => r6c8<>357, r6c9<>5
...
Am I missing something?

When an ALS and an [edit: AALS] are triply linked (three restricted commons), all the candidate values are locked into the union of the two sets. The same is true when two ALSs are doubly linked.

The ALS has N+1 candidates in N cells and the AALS has M+2 candidates in M cells. Together we have N+M+3 candidates in N+M cells. But the two sets are triply linked -- meaning that we really only have N+M different candidate values in the N+M cells. (The subset counting people refer to this condition as multiplicities=0.)

In this case, we know that digits 3, 5 and 7 are shared by sets A and B. Since the [edit: cells of sets A and B containing those digits] all reside in box 6, all other candidates 3, 5, and 7 in box 6 may be eliminated ... as you've shown.

But all seven different values are locked into the union of sets A and B. Thus, we also know that set B must contain the digit 9 ... and set A must contain digits 1, 2 and 6.

I'm not very good at this explanation stuff, so I hope this does the trick.

[hobiwan]

I'm not very good at this explanation stuff, so I hope this does the trick.

I think it does, thanks a lot

It looks like my ALS-XZ code was by no means general enough (my todo list just doesn't stop growing...)

[ronk]

It looks like my ALS-XZ code was by no means general enough (my todo list just doesn't stop growing...)

If you've implemented the ALS xz-rule for one ALS and one AALS, AFAIK that would be more general than anyone else has done.

[hobiwan]

If you've implemented the ALS xz-rule for one ALS and one AALS, AFAIK that would be more general than anyone else has done.

I used the wrong tense: My ALS-XZ code is by no means general enough... But if the summers stays that rainy, who knows