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27 Nov 2006 Nightmare, Quick Uniqueness Solution

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

PostPosted: Sat Dec 09, 2006 10:50 pm    Post subject: 27 Nov 2006 Nightmare, Quick Uniqueness Solution Reply with quote

This is the position in the 27 Nov 2006 Nightmare, after initial eliminations from basic techniques and a "369" naked triple in r4c467.
Code:

.---------------------.---------------------.---------------------.
| 4      2589   256   | 2689   689    689   | 7      1      3     |
| 7     *89     136   | 5      134689 14689 | 2      46    *89+6  |
| 236   *89+2   1236  | 124689 134689 7     | 4689   5     *89+6  |
:---------------------+---------------------+---------------------:
| 1      27     4     | 69     5      369   | 36     8      2-7   |
| 58     3      57    | 1468   2      1468  | 46     9      67    |
| 28     6      9     | 48     7      348   | 5      234    1     |
:---------------------+---------------------+---------------------:
| 256    2457   2567  | 3      14689  145689| 1689   267    2689  |
| 36     47     8     | 14679  1469   2     | 1369   367    5     |
| 9      1      23567 | 678    68     568   | 368    2367   4     |
'---------------------'---------------------'---------------------'

To avoid a rectangular deadly pattern of "89" cells in r23c29 (marked with "*"), either r3c2 must be "2", or one of r23c9 must be "6". All escape routes quickly lead to the common conclusion that r4c9 <> 7.
    (89=2)r23c2 - (2=7)r4c2 - (7)r4c9

    (89=6)r23c9 - (6=7)r5c9 - (7)r4c9
After (7)r4c9 is eliminated, the solution is easily completed, with nothing beyond naked pairs needed.

*** Edit ***

After looking this over I think I could have used a single AIC, maybe:

(7=2)r4c2 - (2=89)r23c2 - UR - (89=6)r23c9 - (6=7)r5c9 => r4c9 <> 7
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