## ALS Chains -A Tutorial ASI#3

Advanced methods and approaches for solving Sudoku puzzles
PIsaacson wrote: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  1279r9c8 -19- als:r48c8 -7- r4c3 =7= r7c3 -7- als:[r7c9,r8c8] -19- r9c8 ==> r9c8<>19`
[edit: add "als:" labels to nice loop]
Last edited by ronk on Sat Apr 04, 2009 9:47 pm, edited 1 time in total.
ronk
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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.
Bud

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Joined: 24 August 2008

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.
aran

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Joined: 02 March 2007

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.
Bud

Posts: 56
Joined: 24 August 2008

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.
Last edited by PIsaacson on Sat Apr 04, 2009 11:35 pm, edited 1 time in total.
PIsaacson

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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.
Bud

Posts: 56
Joined: 24 August 2008

PIsaacson wrote: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

PIsaacson wrote: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.

aran wrote: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.

Bud wrote: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
ronk
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ronk wrote: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
PIsaacson

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PIsaacson wrote:
ronk wrote: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.
ronk
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ronk wrote:
aran wrote: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.
aran

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aran wrote: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 .
ronk
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