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15 April 2007

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

PostPosted: Wed Apr 18, 2007 8:12 pm    Post subject: 15 April 2007 Reply with quote

Start position for the 15 April 2007 Nightmare:

Code:

. 1 .|6 . .|9 . .
6 . .|3 7 .|. 4 .
. . .|. . 1|. . 6
-----+-----+-----
1 8 .|. . 2|7 . .
. 4 .|. . .|. 6 .
. . 5|7 . .|. 8 4
-----+-----+-----
3 . .|8 . .|. . .
. 2 .|. 3 9|. . 8
. . 4|. . 7|. 9 .


This is the position after initial eliminations from basic techniques:

Code:

.------------------.------------------.------------------.
| 2457  1     2378 | 6     2458  458  | 9     2357  2357 |
| 6     59    289  | 3     7     58   | 1258  4     125  |
| 2457  357   2378 | 2459  24589 1    | 2358  2357  6    |
:------------------+------------------+------------------:
| 1     8     36   | 459   4569  2    | 7     35    359  |
| 279   4     27   | 159   1589  358  | 1235  6     12359|
| 29    36    5    | 7     19    36   | 12    8     4    |
:------------------+------------------+------------------:
| 3     5679  1679 | 8     2456  456  | 2456  1257  257  |
| 57    2     167  | 45    3     9    | 456   157   8    |
| 8     56    4    | 125   1256  7    | 2356  9     235  |
'------------------'------------------'------------------'


There are several patterns in this position:

A skyscraper for digit 7, represented by this AIC:
    (7): r1c9 = r7c9 - r7c2 = r3c2 => r1c13, r3c8 <> 7
This ALS XZ rule elimination. The two nodes in the chain are not disjoint, but that's OK as long as there's a suitable restricted common digit (digit 9 in this case).
    (6=579)r79c2|r8c1 - (9=56)r29c2 => r6c2, r78c3 <> 6
There are also two "2 restricted common digits" patterns. Unfortunately, neither yields anything beyond the results of basic follow up of the previous chain, but for the record:

The first such pattern consists of ALS's r4c3489 (four cells, digits 34569) and r8c1348 (four cells, digits 14567). Digits 4 and 6 are restricted common digits -- in the two sets, digit 4 appears only in column 4, and digit 6 appears only in column 3.

The second such pattern is actually a classic Sue de Coq position as well -- viewed this way, box 5 and column 4 are the box and line, set A (in the box) is r6c5, with digits "19"; set B (in the line) is r8c4, with digits "45", and set C (in the line/box intersection) is r45c4, with digits "1459".

Once again, these don't add anything beyond the results of the second chain above. After basic followup of the eliminations from the first and second chains, the grid appears as follows:

Code:
.---------------.----------------.---------------.
| 25   1    8   |  6    25   4   | 9    37   37  |
| 6    59   29  |  3    7    8   | 125  4    125 |
| 4    7    3   | *29   259  1   | 8   #25   6   |
:---------------+----------------+---------------:
| 1    8    6   |  59   4    2   | 7    35   359 |
| 279  4    27  |  159  8    3   | 125  6    1259|
| 29   3    5   |  7    19   6   | 12   8    4   |
:---------------+----------------+---------------:
| 3    69   179 |  8    6-2  5   | 4   *127  27  |
| 57   2    17  |  4    3    9   | 6    157  8   |
| 8    56   4   | #12   126  7   | 3    9    5-2 |
'---------------'----------------'---------------'

Now the puzzle is completed with this AIC for a skyscraper for digit 2:
    (2): r7c8 = r3c8 - r3c4 = r9c4 => r7c5, r9c9 <> 2
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