The New Sudoku Players' Forum

[ronk]

Bud,

Could you please indicate the ordering of the ALS chain? It looks like the chain needs to be c - a - b - d to account for the eliminations. If so, then r9c8 doesn't actually see all the 19s in ALS c and this violates current ALS chaining elimination rules.
[...]
I think you have a forcing ALS chain or grouped-NL -- something other than a "pure" ALS chain.

Yes, set 'c' needs to be r48c8; IOW there is "endpoint overlap" at r8c8.

Code: Select all

   5     9     8  |  237    37     6   |  127      4    127
   7     3     2  |    1     4    58   |  689     69     59
   1     6     4  |  278     9   258   |  278      3    257
------------------+--------------------+---------------------
 248   248    79a |    5     6    48   |  179    179c     3
 489   478     1  | 3789  3478  3489   |    5      2      6
   6     5     3  |  279    17   129   |  179      8      4
------------------+--------------------+---------------------
 489  1478    79b |    6     2   189   |    3      5    179d
 239    12     6  |   39     5     7   |    4     19cd    8
 389   178     5  |    4   138  1389   | 12679 -167-9  1279

r9c8 -19- als:r48c8 -7- r4c3 =7= r7c3 -7- als:[r7c9,r8c8] -19- r9c8 ==> r9c8<>19

[edit: add "als:" labels to nice loop]

[Bud]

Paul and Ron,
Yes, c and d are the ends of the chain in the example. I'm not an expert on what is or isn't an ALS-XY-Chain, but it is a chain because 7 must be either in c exclusive or d. The cell eliminations are similar to those in a doubly-linked ALS-XZ.

[aran]

Code: Select all

   5     9     8  |  237    37     6   |  127      4    127
   7     3     2  |    1     4    58   |  689     69     59
   1     6     4  |  278     9   258   |  278      3    257
------------------+--------------------+---------------------
 248   248    79a |    5     6    48   |  179    179c     3
 489   478     1  | 3789  3478  3489   |    5      2      6
   6     5     3  |  279    17   129   |  179      8      4
------------------+--------------------+---------------------
 489  1478    79b |    6     2   189   |    3      5    179d
 239    12     6  |   39     5     7   |    4     19cd    8
 389   178     5  |    4   138  1389   | 12679 -167-9  1279


The hidden pair version while we are at it.
19(r8c8+r7c9)=7r7c9-7r7c3=7r4c3-7r4c8=19r48c8 : =><19>r9c8
i.e either there is a hidden pair 19 in b9, or there is a hidden pair 19 in c8.
Either way <19>r9c8.

[Bud]

Ron and Aran,
I don't see any reason to change the groupings of the ALS. The point I am trying to make is that the special case where the same RCD is in both ends of an ALS-XY-chain of any length, multidigit cell eliminations are possible as in a doubly-linked ALS-XZ. As far as rules are concerned the cell eliminations in a doubly-linked ALS-XZ do not follow the rules for the singly-linked ALS-XZ. With the original groupings, either r4c78 exclusive or r7c9r8c8 become a locked set 19 and the cell eliminations follow.

[PIsaacson]

Bud,

It's just the way ALS chains are currently defined. Your ALS c and d sets have no digits that are entirely seen by (peers of) each other. Without rearranging ALS c from r4c78 to r48c8, the eliminations are simply not legal (r9c8 doesn't see all the 19s). I know of no other way to explain this in terms of ALS chains. For a fact, the eliminations are legal, but not with the ALS chain as you have it defined.

I'm not skilled at group NLs, so I took at stab at proving chains (which I'm also not very good at...)

r4c8=9 => r4c3=7 => r7c3=9 => ALS d {17} => r9c8<>179
r4c8=7 => r4c3=9 => r7c3=7 => ALS d {19} => r9c8<>179
r4c8=1 => ALS d {79} => r9c8<>179

I deduce from this that r9c8=6 and I also conclude that r4c7 does not contribute to the logic.

New insight using Allan Barker's set/link-set logic: Adding r4c7 as a base cell results in r6c7<>1. The 3 possible values for r4c8 with the ALS c r4c78 all result in r6c7<>1. Without the r4c7 cell, set/link-set logic shows just the r9c8<>19.

[Bud]

Paul,
I do agree that Ron's grouping is better because it gives an extra 7 cell elimination. However, in the doubly-linked ALS-XZ cell eliminations other than the RCD's are based on locked sets rather than ALS-XZ rules. I think that the logic for the original grouping is also based on locked sets and is valid. That said I appresiate everone's help on this.

[ronk]

r4c8=9 => r4c3=7 => r7c3=9 => ALS d {17} => r9c8<>179
r4c8=7 => r4c3=9 => r7c3=7 => ALS d {19} => r9c8<>179
r4c8=1 => ALS d {79} => r9c8<>179
...
Without the r4c7 cell, set/link-set logic shows just the r9c8<>19.

Which means r9c8<>7 is not a valid elimination. That's because ... r4c8=1 => (r8c8=9 & r7c9=17) => r9c8<>19

Adding r4c7 as a base cell results in r6c7<>1.

That's simply because the <179> naked triple in r4c378 has digit <1> confined to b6. It's not the result of the larger loop.

The hidden pair version while we are at it.
19(r8c8+r7c9)=7r7c9-7r7c3=7r4c3-7r4c8=19r48c8 : =><19>r9c8
i.e either there is a hidden pair 19 in b9, or there is a hidden pair 19 in c8.
Either way <19>r9c8.

That's exactly the same chain/loop that Paul and I posted. So we now have three different languages but, unfortunately, I don't think it's a Pentacostal experience. BTW IMO you've got the same (naming) problem here as on another thread. Those are naked pairs above, not hidden pairs.

However, in the doubly-linked ALS-XZ cell eliminations other than the RCD's are based on locked sets rather than ALS-XZ rules. I think that the logic for the original grouping is also based on locked sets and is valid. That said I appresiate everone's help on this.

With or without r4c7 there are no "locked sets" in this deduction, because the loop is discontinuous.

Bud, are you known as Nienhaus on sudopedia.org

[PIsaacson]

That's simply because the <179> naked triple in r4c378 has digit <1> confined to b6. It's not the result of the larger loop.

I don't understand. Granted that digit <1> is confined to b6, but that includes r6c7. So how is the r6c7<>1 the result of the naked triple <179> in sector r4? I checked this pretty carefully in Xsudo and there is definitely no r6c7<> 1 without including r4c7, but then I'm still struggling to understand Allan's set/link-set logic in full.

r9c8<>7 is true, but my bad in that it was a "lucky" supposition. But I stand by my assertion that "I'm not skilled"...

Cheers,
Paul

[ronk]

That's simply because the <179> naked triple in r4c378 has digit <1> confined to b6. It's not the result of the larger loop.

I don't understand. Granted that digit <1> is confined to b6, but that includes r6c7. So how is the r6c7<>1 the result of the naked triple <179> in sector r4?

Yikes! It's just a locked candidate: base set 1r4; cover set 1b6. As the unnecessary naked triple: base set r4c378; cover set 79r4 and 1b6.

[aran]

The hidden pair version while we are at it.
19(r8c8+r7c9)=7r7c9-7r7c3=7r4c3-7r4c8=19r48c8 : =><19>r9c8
i.e either there is a hidden pair 19 in b9, or there is a hidden pair 19 in c8.
Either way <19>r9c8.

That's exactly the same chain/loop that Paul and I posted. So we now have three different languages but, unfortunately, I don't think it's a Pentacostal experience. BTW IMO you've got the same (naming) problem here as on another thread. Those are naked pairs above, not hidden pairs.

Pentecostal : or can Melville come to your rescue...?
Hidden pair = naked pair with clutter.
Remove the clutter, reveal the naked pair.
Your argument is : ergo, it was always a naked pair. With which I for one disagree.

Applied above : if the clutter is removed from (r8c8+r7c9) there is revealed a naked pair 19. Revealed, so it was hidden.
Or if the clutter isn't removed, then the clutter is removed from r48c8,
revealing a naked pair 19. Revealed so it too was hidden.
Two hidden pairs, or more precisely, at least one hidden pair.
Quod erat demonstrandum.

[ronk]

Hidden pair = naked pair with clutter.
Remove the clutter, reveal the naked pair.

To avoid cluttering another thread with this topic, I posted my reply here .