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Earl
Joined: 30 May 2007 Posts: 677 Location: Victoria, KS
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Posted: Sat Dec 19, 2009 3:24 pm Post subject: Dec 19 DB |
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The December 19 DB has no apparent silver bullet. (Norm please prove me wrong). I had to shoot twice.
A Solution: A double finned x-wing eliminates the 2 in R7C9 and R8C8, then a skyscraper eliminates the 4 in R1C3.
Earl
Code: |
+-------+-------+-------+
| . 9 . | . . 2 | . 7 . |
| . . 8 | . . 7 | 6 . 5 |
| . . 5 | . . . | 3 9 . |
+-------+-------+-------+
| . 5 . | . . 8 | . . . |
| 3 6 2 | . . . | 7 8 9 |
| . . . | 9 . . | . 5 . |
+-------+-------+-------+
| . 7 3 | . . . | 9 . . |
| 1 . 6 | 7 . . | 4 . . |
| . 4 . | 3 . . | . 1 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Dec 19, 2009 5:09 pm Post subject: |
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Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat Dec 19, 2009 6:08 pm Post subject: |
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Similar, but the details are different. Simple coloring on 2 and a Kite on 4. |
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Sun Dec 20, 2009 12:35 am Post subject: |
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Quote: | Norm please prove me wrong |
Code: | +---------------+------------+----------------+
| 6 9 14 | 5 3 2 | 8 7 1(4) |
| 2(4) 3 8 | 1 9 7 | 6 24 5 |
| 7 12 5 | 68 68 4 | 3 9 12 |
+---------------+------------+----------------+
| 9 5 147 | 26 267 8 | 12 346 36(4) |
| 3 6 2 | 4 1 5 | 7 8 9 |
| 8(4) 18 147 | 9 267 3 | 12 5 6(4) |
+---------------+------------+----------------+
| 5 7 3 | 28 4 1 | 9 26 268 |
| 1 28 6 | 7 5 9 | 4 23 238 |
| 28 4 9 | 3 28 6 | 5 1 7 |
+---------------+------------+----------------+ |
look at the arrangement of 4's in the grid. if the 4 in r4c9 is false, then the skyscraper on 4's exists in columns 1 and 9.
(4)r1c9 = (4)r6c9 - (4)r6c1 - (1)r1c1
and this would eliminate the 4 in r2c8 right?
so what happens when the skyscraper is false and the 4 is true in r4c9?
this triggers the kite on 2 via the 3's.
red is the kite on 2's
(4-3)r4c9 = (3)r4c8 - (3=2)r8c8 - (2)r8c2 = (2)r3c2 - (2)r3c9 = (2)r2c8
either way, the 4 in r2c8 can't exist.
[(4)r1c9 = (4)r6c9 - (4)r6c1 - (1)r1c1] = (4-3)r4c9 = (3)r4c8 - (3=2)r8c8 - (2)r8c2 = (2)r3c2 - (2)r3c9 = (2)r2c8; r2c8 <> 4 |
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