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Ron Moore

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

Posted: Tue Nov 21, 2006 4:31 pm    Post subject: Three Digit Deadly Pattern, 3 April 2006

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.)
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

 Posted: Thu Nov 23, 2006 5:31 am    Post subject: 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 Here is mine... http://www.dailysudoku.com/sudoku/forums/viewtopic.php?p=6359&highlight=&sid=acec6d3647185a689172a2e95944a8dc#6359
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