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Ron Moore

Joined: 13 Aug 2006
Posts: 72
Location: New Mexico

Posted: Sun Dec 03, 2006 2:28 am    Post subject: "Unique ALS XZ Reduction," 24 Nov 2006

One of the uniqueness-based solution methods discussed in the Solving Guide is the "Unique Subset" technique. This technique is based on a pattern in which only one side of a potential non-unique polygon contains surplus candidates, and these candidates combine with other cells to form a locked (or naked) subset, such as a naked pair or naked triple, in some house(s). These candidates can then be removed from other cells in that (those) house(s).

In my solution to the 24 Nov 2006 Nightmare, in two (consecutive!) steps I found patterns in which the surplus candidates did not quite form a locked set, but instead formed an Almost Locked Set (ALS) with other cells, and these could be used with other ALS's in an ALS XZ rule elimination.
 Code: .------------------.----------------------.------------------. | 148   9     478  | 15       6      1245 | 1257  3     245  | | 6    *25    47   | 1359    *25+49  12345| 1257  125   8    | | 14   *25    3    | 7       *25+4   8    | 9     1256  2456 | :------------------+----------------------+------------------: | 89    7     2    | 16       3      16   | 4     589   59   | | 5     3     48   | 2       A48     9    | 6     7     1    | | 489   6     1    | 58       7      45   | 3     289   29   | :------------------+----------------------+------------------: | 3     1     5    | 4       A29     7    | 8    B269  B269  | | 2     4     69   | 3-5689  A589    3-56 |B15   B159   7    | | 7     8     69   | 569      1      256  | 25    4     3    | '------------------'----------------------'------------------'

In this position, the potential non-unique rectangle in r23c25 (marked with "*") contains surplus candidates 4 and 9, in column 5 only. These candidates form an ALS with r578c5 (marked with "A"), containing digits {2,4,5,8,9}. The other ALS, marked with "B", is in r7c89 and r8c78, containing digits {1,2,5,6,9}. The "restricted common" digit is X=2, and the eliminated digit is Z=5, in r8c46.

After these eliminations:
 Code: .------------------.---------------------.------------------. | 148   9     478  |A15      6     -1245 | 1257  3     245  | | 6     25    47   | 1359    2459  -12345| 1257  125   8    | | 14    25    3    | 7       245    8    | 9     1256  2456 | :------------------+---------------------+------------------: | 89    7     2    | 16      3     B16   | 4     589   59   | | 5     3     48   | 2       48     9    | 6     7     1    | | 489   6     1    |A58      7      45   | 3     289   29   | :------------------+---------------------+------------------: | 3     1     5    | 4       29     7    | 8     269   269  | | 2     4    *69   |*69+38   589   B36   | 15    159   7    | | 7     8    *69   |*69+5    1      256  | 25    4     3    | '------------------'---------------------'------------------'

Here the potential non-unique rectangle, again marked with "*", lies in r89c34, with surplus candidates 3,5,8 in column 4 only. These combine with r16c4 (marked with "A") to form an ALS containing digits {1,3,5,8}. The other ALS, marked with "B", lies in r48c6 and contains digits {1,3,6}. The "restricted common" digit in this case is X=3, one of the surplus candidates. The eliminated digit is Z=1 in r12c6.

The steps preceding and following these steps were not entirely straightforward, as some grouped chains were involved, but for brevity I won't discuss these (unless someone requests).
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

Posted: Tue Dec 05, 2006 7:44 am    Post subject:

It is an interesting idea, using UR extras in ALS type deductions, but it seems you have to be a little bit careful, because the UR provides a potential extra cell that an otherwise extra digit could find a home in. So you could have an either an ALS or a locked set. I don't think most ALS deductions would care about that though. It only means that the strong link between anded candidates in the UR+ALS is a strong-only link rather than conjugate.
 Code: .------------------.----------------------.------------------. | 148   9     478  | 15       6      1245 | 1257  3     245  | | 6    *25    47   | 1359    *25+49  12345| 1257  125   8    | | 14   *25    3    | 7       *25+4   8    | 9     1256  2456 | :------------------+----------------------+------------------: | 89    7     2    | 16       3      16   | 4     589   59   | | 5     3     48   | 2        48     9    | 6     7     1    | | 489   6     1    | 58       7      45   | 3     289   29   | :------------------+----------------------+------------------: | 3     1     5    | 4       B29     7    | 8    C269  C269  | | 2     4     69   | 3-5689  A589    3-56 |C15   C159   7    | | 7     8     69   | 569      1      256  | 25    4     3    | '------------------'----------------------'------------------'

A little bit simpler here might be to note that the only way a 2 or 5 can escape in columns 2&5 is via (5)r8c5 or (2)r7c5. One of these must be true, so stick a strong link between them and you get...

(5)r8c5 = (2)r7c5 - (2=1&6&9&5)r7c89|r8c78 ... loop => r8c46 <> 5

 Code: .------------------.---------------------.------------------. | 148   9     478  |A15      6     -1245 | 1257  3     245  | | 6     25    47   | 1359    2459  -12345| 1257  125   8    | | 14    25    3    | 7       245    8    | 9     1256  2456 | :------------------+---------------------+------------------: | 89    7     2    | 16      3     B16   | 4     589   59   | | 5     3     48   | 2       48     9    | 6     7     1    | | 489   6     1    |A58      7      45   | 3     289   29   | :------------------+---------------------+------------------: | 3     1     5    | 4       29     7    | 8     269   269  | | 2     4    *69   |*69+38   589   B36   | 15    159   7    | | 7     8    *69   |*69+5    1      256  | 25    4     3    | '------------------'---------------------'------------------'

A similar AIC-type deduction...
(1=5&8)r16c4 - (5v8=UR=3)r89c34 - (3=6&1)r48c6 => r12c6 <> 1

But, here it seems your trick...

(1&5&8=UR/ALS=3)r1689c4|r89c3 - (3=6&1)r48c6 => r12c6 <> 1

...is actually more straightforward.

Good idea, it works for me You ought to present and get some props for it on some of the other boards.
Ron Moore

Joined: 13 Aug 2006
Posts: 72
Location: New Mexico

 Posted: Thu Dec 14, 2006 7:14 pm    Post subject: Sorry for the delay in responding. I got far behind due to some computer problems and the US Thanksgiving holiday period. I have come across several interesting uniqueness arguments lately and am just getting around to posting them. I don't really think I want to start running around in other boards. My study of other boards has been rather haphazard. If anything, I'll just submit something under Solving Tips on this site, not limited to the "Unique ALS" idea, but other uniqueness arguments not convered in Ruud's Solving Guide. I'm sure most of these are "old hat" on other boards. I like your notation for the Unique ALS in AIC's. As you seem to be suggesting in your opening remarks, it's hard to put your finger on where the Unique ALS is, but your notation covers it well. I'm wondering if there's a standard notation to use for links in AIC's which come from non-rectangular deadly patterns. Tentatively I'm using - DP -
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

Posted: Sat Dec 16, 2006 11:09 pm    Post subject:

 Ron Moore wrote: I'm wondering if there's a standard notation to use for links in AIC's which come from non-rectangular deadly patterns. Tentatively I'm using - DP -

Nothing is standard. I have used -UR- and -BUG- standing for BUG-lite. I usually will add a little text to help indicate what I am doing when it is something a bit out of the ordinary.
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