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ER Multidigit Extension Example

 
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Bud



Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida

PostPosted: Fri Mar 04, 2011 7:40 pm    Post subject: ER Multidigit Extension Example Reply with quote

000002090070000504050009000004290080600030002020070100000800050501920060080400000

After basic moves the puzzle is as shown in the diagram below. The 9 ER cells are marked with * and the 156 extension cells are marked with #. For the ER. The extensions begin at each end of the ER. One or both of these (r6c9 or r7c2) must be 9. Since (9)r6c9=(6)r6c4 or (9)r7c2=(1)r5c2=(5)r5c4, this implies that r6c4<>5 and r7c9<>9.

Code:

+--------------------+---------------------+--------------------+
| 1348   1346  368   | 13567  14568  2     | 3678   9     13678 |
| 12389  7     23689 | 136    168    16    | 5      123   4     |
| 12348  5     2368  | 1367   1468   9     | 23678  1237  13678 |
+--------------------+---------------------+--------------------+
| 137    13    4     | 2      9      156   | 367    8     3567  |
| 6      19*   5789  | 15#    3      48    | 79*    47    2     |
| 389    2     3589  |-56#    7      48    | 1      34    3569* |
+--------------------+---------------------+--------------------+
| 23479  3469* 23679 | 8      16     1367  | 23479  5     137-9 |
| 5      34    1     | 9      2      37    | 3478   6     378   |
| 2379   8     23679 | 4      156    13567 | 2379   1237  1379  |
+--------------------+---------------------+--------------------+


I have found several examples of extended turbot fish patterns as well as other single digit 4 cell non-turbot fish patterns. This is one of the simpler ones.


Last edited by Bud on Sun Mar 06, 2011 10:08 pm; edited 1 time in total
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Fri Mar 04, 2011 8:16 pm    Post subject: Re: ER Multidigit Extension Example Reply with quote

Bud wrote:
(9)r7c2=(1)r5c2
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Sat Mar 05, 2011 8:56 pm    Post subject: Reply with quote

The elimination of <5> in r6c4 depends upon the <9>s in b6 being a conjugate pair. The ER based elimination in r7 is an unrelated coincidence.
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Bud



Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida

PostPosted: Sun Mar 06, 2011 10:58 pm    Post subject: Reply with quote

Hi DAJ aqnd Asellus.
Asellus, the relationship between the two eliminations is that they are both dependent on the fact that r7c2 inclusive-or r6c9 = 9. If this were not true, neither elimination can be made. On the other hand it is a coincidence that this particular extension occurred. However, I have found several other distinct useful extensions in other puzzles. I misplaced one of these that had a total of 4digit eliminations. When I locate this, I will post it.
I was getting ready to post an extended purple cow example which is related to this because both the purple cow elimination and the extended elimination are dependent on the ends of a single digit 4-cell AIC.
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Mar 07, 2011 12:36 am    Post subject: Reply with quote

Bud,

What you have are two chains with a few cells and candidates in common. It's nothing noteworthy. In addition, it's r6c9<>6 and not r6c4<>5 as you indicate.

Code:
(9)r6c9 = r5c7 - (9  )r5c2 = (9)r7c2  =>  r7c9<>9   (simple turbot pattern)

(9)r6c9 = r5c7 - (9=1)r5c2 - (1=5)r5c4 - (5=6)r6c4  =>  r6c9<>6
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Mar 07, 2011 4:54 am    Post subject: Reply with quote

daj95376 wrote:
In addition, it's r6c9<>6 and not r6c4<>5 as you indicate.

The elimination of <5> in r6c4 or of <6> in r6c9 are equivalent since each is a consequence of the other. As Bud wrote it using the "ER pincers" (edited slightly by me for clarity), it is the elimination of <5> in r6c4 that is the most direct result:

(9)r6c9 => (6)r6c4 and/or (9)r7c2 => (1)r5c2 => (5)r5c4

The <5> elimination works precisely because the <9>s in box 6 and the <6>s in r6 are conjugate pairs with one end of each pair sharing the cell r6c9 (that is, have a weak inference). The AIC:

(5=1)r5c4 - (1=9)r5c2 - (9)r5c7=(9-6)r6c9=(6)r6c4; r6c4<>5

My point was that this elimination would not work in the general case of an ER in box 6. Suppose there were a <9> candidate in r4c9. We'd still have the ER elimination of <9> in r7c9 but would no longer have the <5> elimination. That's because we have no weak inference between (6)r6c9 and the grouped <9>s in c9 of box 6. (That's why I put "ER pincers" in quotes above. The general case would have to consider any or all of (9)r456c9 as the other pincer.) As daj points out, it is the pincers of a Turbot Fish that are being extended here (except that, as the AIC shows, we don't really need to go all the way to the r7c2 pincer to eliminate that <5>).

More generally, the term in green above could be replaced with any grouping of <9>s in r5c789. So what is actually essential is that (9)r6c9 be conjugate with the <9>s of r5 of b6. This allows some of the possible ER configurations but not all of them.

[Edit to fix typo, shown in red.]


Last edited by Asellus on Mon Mar 07, 2011 9:26 pm; edited 1 time in total
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Mar 07, 2011 7:11 am    Post subject: Reply with quote

[Withdrawn: My point seemed to overlay Asellus' in too many respects.]

Last edited by daj95376 on Mon Mar 07, 2011 9:30 pm; edited 1 time in total
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