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Three Digit Deadly Pattern, 3 April 2006

 
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Ron Moore
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Joined: 13 Aug 2006
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PostPosted: Tue Nov 21, 2006 4:31 pm    Post subject: Three Digit Deadly Pattern, 3 April 2006 Reply with quote

The position below arises in the solution of the April 3, 2006 Nightmare. There are at least three independent eliminations which can be made from uniqueness arguments.
Code:

┌──────────────┬────────────────┬───────────────┐
│#356  2   56  │ 8     7    1   │ 9   #35   4   │
│#358  4   58  │ 69    69   2   │#35   1    7   │
│ 1    9   7   │*45    3   *45  │ 6    28   28  │
├──────────────┼────────────────┼───────────────┤
│&258 &58  9   │ 25    1    7   │ 4    6    3   │
│ 256  1   3   │*2569  4   *59  │ 8    7    259 │
│ 4    7   256 │ 3     69   8   │ 1    25   259 │
├──────────────┼────────────────┼───────────────┤
│&578 &58  1   │*49    2   *49  │#357 #358  6   │
│ 9    6   258 │ 7     58   3   │ 25   4    1   │
│ 27   3   4   │ 1     58   6   │ 27   9    58  │
└──────────────┴────────────────┴───────────────┘

A potential deadly pattern based on the three pairs of digits taken from {4, 5, 9} exists in cells r357c46, marked with "*" in the diagram. r5c4 is the only cell with surplus candidates. A value of 5 or 9 would create a non-unique solution, so these can be eliminated from r5c4 (same idea as the "unique corner" elimination to avoid a non-unique rectangle). I believe one more non-basic step is needed to complete the solution. One way, with r5c4 now reduced to "26", is this grouped chain:

(2=6)r5c4 - (6=9)r6c5 - (9=2&5)r6c89 => r5c9 <> 2.

There is another elimination, not quite so direct, which comes from the potential non-unique hexagon based on the digits 3, 5, in cells r12c1, r1c8, r2c7, r7c78 (marked with # in the diagram). Observe that in box 1, digit 3 is locked into one of r1c1, r2c1, so that placing 5 in either cell would force 3 into the other. A similar situation exists in row 7 -- 3 is locked into one of r7c7 and r7c8, so placing 5 in either cell would force 3 into the other. Thus, in order to avoid the non-unique hexagon, the 5 in box 1 must lie outside the hexagon, or the 5 in row 7 must lie outside the hexagon. In other words, the 5 in box 1 must lie in one of r12c3, or the 5 in row 7 must lie in one of r7c12. In all cases, r8c3 sees a 5, so r8c3 can be reduced to "28". One more non-basic step seems to be needed here also. A simple XY wing is enough:

(5=8)r2c3 - (8=2)r8c3 - (2=5)r8c7 => r2c7 <> 5

Finally, the elimination which the Sudocue solver finds is a "unique subset" elimination which comes from the potential non-unique rectangle based on digits 5, 8 in r47c12 -- marked with "&" in the diagram. The surplus digits 2 and 7 in r47c1 combine with cell r9c1 to form a naked pair, eliminating 2 from r5c1. (The solver continues with an XY wing and empty rectangle to complete the solution.)
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Myth Jellies
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Joined: 04 Apr 2006
Posts: 42

PostPosted: Thu Nov 23, 2006 5:31 am    Post subject: Reply with quote

Interesting that we came up with similar BUG-Lite deductions on different puzzles within hours of each other. Must have been something in the air Smile

Here is mine...
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?p=6359&highlight=&sid=acec6d3647185a689172a2e95944a8dc#6359
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