Ron Moore
Addict
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.) |
|
FAQ
Search
Usergroups
Register
Profile
Log in to check your private messages
Log in
